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https://www.allmath.com/circumcenter-of-triangle.php | 1,686,411,038,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657720.82/warc/CC-MAIN-20230610131939-20230610161939-00761.warc.gz | 714,206,752 | 29,235 | # Circumcenter calculator - Triangle
Enter the values in the circumcenter calculator below to find the circumcenter of a triangle.
Formula:
(y - y1) = m (x - x1)
X
Y
A
B
C
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The circumcenter of a triangle calculator is used to find the point which is concurrent for all three bisectors of the triangle.
## What is the circumference of a triangle?
It is the point where the three lines bisecting the sides of the triangle meet. These lines divide the sides of the triangle into two equal parts.
This point can be inside (for acute triangle) as well as outside (for obtuse triangle) of the triangle.
## How to find the circumcenter of the triangle?
There is a whole process you need to follow to find the circumcenter of the triangle.
1. First, you need to find the midpoint of the three sides of the triangle.
2. Now find the slope of the perpendicular bisector to all these lines.
3. Make the point-slope equation of the sides. y - y1 = m(x - x1
4. Solve any two of the equations to find the point of intersection. This point is the circumcenter.
The triangle circumcenter uses the same process therefore the results are accurate.
Example:
Find the circumcenter of the following triangle.
Solution:
Step 1: Write all the points.
A = (-4,2)
B = (2,4)
C = (4,-4)
Step 2: Find the midpoint using its formula.
The formula of the midpoint is:
Midpoint = (x1 + x2) / 2 , (y1 + y2)/2
AB = (-4+2)/2 , (2+4)/2
= -2/2 , 6/2
= -1 , 3
BC = (2+4)/2 , (4-4)/2
= 6/2 , 0
= 3 , 0
CA = (4-4)/2 , (-4+2)/2
= 0/2 , -2/2
= 0 , -1
Step 3: Calculate the slope of the lines.
AB = (4 - 2) / (2 + 4)
= 2 / 6
= 1/3
Slope of AB perpendicular = -1/(1/3)
= -3
BC = (4 + 4) / (2 - 4)
= 8 / -2
= -4
Slope of BC perpendicular = -1/(4)
= 1/4
CA = (2 + 4) / (-4 - 4)
= 6 / 0
Slope of CA perpendicular = -1/(0)
= 0
Step 4: Make point-slope equations for any two points.
The values of x1 and y1 are from the midpoint values.
For AB; m = -3, x1 = -1, and y1 = 3
y - y1 = m ( x - x1 )
y - 3 = (-3)(x + 1)
y = -3x - 3 + 3
y = -3x … equation 1
For BC, m = ¼, x1 = 3, and y1 = 0
y - y1 = m ( x - x1 )
y - 0 = (1/4)(x - 3)
y = 1/4x -3/4
4y = x - 3 … equation 2
Step 5: Find the value of x by putting the value of y in equation 2.
4y = x - 3
4(-3x) = x - 3
-12x = x - 3
-13x = 3
X = -3/13
Step 6: Use the value of x to find the y.
Y = -3(-3/13)
Y = 9/13
Circumcenter = (-3/13 , 9/13)
As you can see the process is quite lengthy. The best alternative to save time is to use the circumcenter solver. | 887 | 2,540 | {"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} | 4.875 | 5 | CC-MAIN-2023-23 | longest | en | 0.856558 |
https://www.projectrhea.org/rhea/index.php?title=Inverse_MA265F12Alvarado&diff=prev&oldid=52800 | 1,611,224,299,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524743.61/warc/CC-MAIN-20210121101406-20210121131406-00088.warc.gz | 966,860,821 | 8,299 | # Inverse of a Matrix
An n x n matrix A is said to have an inverse provided there exists an n x n matrix B such that AB = BA = In. We call B the inverse of A and denote it as A-1. Thus, AA-1 = A-1A = In. In this case, A is also called nonsingular.
Example.
$A = \left(\begin{array}{cccc}4&3\\3&2\end{array}\right)$
$A^{-1} = \left(\begin{array}{cccc}-2&3\\3&-4\end{array}\right)$
$AA^{-1} = \left(\begin{array}{cccc}4&3\\3&2\end{array}\right) $$\left(\begin{array}{cccc}-2&3\\3&-4\end{array}\right) = \left(\begin{array}{cccc}1&0\\0&1\end{array}\right) and A^{-1} = \left(\begin{array}{cccc}-2&3\\3&-4\end{array}\right)$$ \left(\begin{array}{cccc}4&3\\3&2\end{array}\right) =$$\left(\begin{array}{cccc}1&0\\0&1\end{array}\right)$
### Theorem 1
The inverse of a matrix, if it exists, is unique
### Theorem 2
If A and B are both nonsingular n x n matrices (i.e. invertible), then AB is nonsingular and (AB)-1 = B-1A-1.
### Corollary 1
If A1, A2, ..., Ar are n x n nonsingular matrices, then A1A2...Ar is nonsingular an (A1A2...Ar)-1 = Ar-1Ar-1-1...A1-1.
### Theorem 3
If A is a nonsingular matrix, then A-1 is nonsingular and (A-1)-1 = A.
### Theorem 4
If A is a nonsingular matrix, then AT is nonsingular and (A-1)T = (AT)-1.
## Methods for determining the inverse of a matrix
#### 1. Shortcut for determining the inverse of a 2 x 2 matrix
If $A = \left(\begin{array}{cccc}a&b\\c&d\end{array}\right)$ then the inverse of matrix A can be found using:
$A^{-1} = \frac{1}{detA}\left(\begin{array}{cccc}d&-b\\-c&a\end{array}\right) = \frac{1}{ad - bc}\left(\begin{array}{cccc}d&-b\\-c&a\end{array}\right)$
Example
$A = \left(\begin{array}{cccc}1&2\\3&4\end{array}\right)$
$detA = ad - bc = 1 \times 4 - 2 \times 3 = -2$
$A^{-1} = \frac{1}{-2}\left(\begin{array}{cccc}4&-2\\-3&1\end{array}\right)$
$A^{-1} = \left(\begin{array}{cccc}-2&1\\\frac{3}{2}&\frac{-1}{2}\end{array}\right)$
#### 2. Augmented Matrix Method (Can be used for any n x n matrix)
Use Gauss-Jordan Elimination to transform [ A | I ] into [ I | A-1 ].
$\left(\begin{array}{ccc|ccc}a&b&c&1&0&0\\d&e&f&0&1&0\\g&h&i70&0&1\end{array}\right)$
Example
## Alumni Liaison
Basic linear algebra uncovers and clarifies very important geometry and algebra.
Dr. Paul Garrett | 865 | 2,249 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-04 | latest | en | 0.60322 |
https://www.nytimes.com/2020/03/20/health/coronavirus-data-logarithm-chart.html?smtyp=cur&smid=tw-nythealth | 1,585,676,444,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00169.warc.gz | 1,085,727,835 | 82,299 | # A Different Way to Chart the Spread of Coronavirus
Those skyrocketing curves tell an alarming story. But logarithmic graphs can help reveal when the pandemic begins to slow.
The arc of coronavirus cases in Italy is frightening, continuing to jump by hundreds each day. But a public-health official looking at those numbers will see definite signs that the nationwide lockdown, imposed to keep individuals apart and the virus from spreading, is working.
The data look very different when plotted on what is called a logarithmic scale. In a typical graph, values on the (vertical) y-axis are plotted linearly: 1, 2, 3, and so on, or 10, 20, 30, or the like. By contrast, in a logarithmic plot, each tick on the y-axis represents a tenfold increase over the previous one: 1, then 10, then 100, then 1,000, then 10,000 and so on. (The interval doesn’t have to be a factor of 10, it could be a factor of 2, or 5, or 27, or any other number, but humans seem to prefer factors of 10.)
## Bending the Curve
Logarithmic scales can emphasize the rate of change in a way that linear scales do not. Italy seems to be slowing the coronavirus infection rate, while the number of cases in the United States continues to double every few days.
45,000
cases
100,000
cases
Coronavirus cases in Italy and the U.S.
Plotted on a linear scale
The same data
Plotted on a logarithmic scale
10,000
Italy
Italy
30,000
1,000
U.S.
100
15,000
10
U.S.
0
1
Feb.
18
March
1
March
19
Feb.
18
March
1
March
19
Coronavirus cases in Italy and the U.S.
Plotted on a linear scale
The same data
Plotted on a logarithmic scale
45,000
cases
100,000
cases
10,000
Italy
Italy
30,000
1,000
U.S.
100
15,000
10
U.S.
0
1
Feb.
18
March
1
March
19
Feb.
18
March
1
March
19
Coronavirus cases in Italy and the U.S.
Plotted on a linear scale
45,000
cases
Italy
30,000
15,000
U.S.
0
Feb.
18
March
1
March
19
The same data
Plotted on a logarithmic scale
100,000
cases
10,000
Italy
1,000
U.S.
100
10
1
Feb.
18
March
1
March
19
Coronavirus cases in Italy and the U.S.
Plotted on a linear scale
45,000
cases
Italy
30,000
15,000
U.S.
0
Feb.
18
March
1
March
19
The same data
Plotted on a logarithmic scale
100,000
cases
10,000
Italy
1,000
U.S.
100
10
1
Feb.
18
March
1
March
19
By The New York Times | Data from Worldometer
Unconstrained, the coronavirus spreads exponentially, the caseload doubling at a steady rate. That curve, plotted linearly, is a skyrocketing curve. Plotted logarithmically, however, it transforms into a straight line — which means that deviations from the exponential spread of the virus become much easier to discern.
Presented this way, the data for Italy clearly show that the infection rate is no longer exponential. The straight line is now a slight downward curve indicating that the rate of increase is slowing.
At a quick glance, the rate of spread in the United States looks similar to Italy’s, at least when plotted on a linear scale. But on a logarithmic scale, it is instantly apparent that the number of Americans becoming infected continues to double every three days or so. That indicates that the limited measures taken until recently did not sever social contact enough to slow the spreading. The U.S. curve has even bent upward in the last few days — an even faster exponential growth — perhaps reflecting more widespread testing.
Italy’s experience shows that more drastic containment measures work, so the U.S. curve may start bending downward in the coming days, as measures here go into effect. (John Burn-Murdoch at The Financial Times maintains a log chart for multiple countries.) The lag between the imposition of measures and their impact on the curve could take days to a week or two, because of the incubation time before symptoms arise. If the line does not begin to bend downward, more stringent actions are probably needed.
But when it finally does, it will herald a real change in the direction of the epidemic in the United States.
# The Coronavirus Outbreak
Updated March 24, 2020
• #### How does coronavirus spread?
It seems to spread very easily from person to person, especially in homes, hospitals and other confined spaces. The pathogen can be carried on tiny respiratory droplets that fall as they are coughed or sneezed out. It may also be transmitted when we touch a contaminated surface and then touch our face.
• #### What makes this outbreak so different?
Unlike the flu, there is no known treatment or vaccine, and little is known about this particular virus so far. It seems to be more lethal than the flu, but the numbers are still uncertain. And it hits the elderly and those with underlying conditions — not just those with respiratory diseases — particularly hard.
• #### What should I do if I feel sick?
If you’ve been exposed to the coronavirus or think you have, and have a fever or symptoms like a cough or difficulty breathing, call a doctor. They should give you advice on whether you should be tested, how to get tested, and how to seek medical treatment without potentially infecting or exposing others.
• #### What if somebody in my family gets sick?
If the family member doesn’t need hospitalization and can be cared for at home, you should help him or her with basic needs and monitor the symptoms, while also keeping as much distance as possible, according to guidelines issued by the C.D.C. If there’s space, the sick family member should stay in a separate room and use a separate bathroom. If masks are available, both the sick person and the caregiver should wear them when the caregiver enters the room. Make sure not to share any dishes or other household items and to regularly clean surfaces like counters, doorknobs, toilets and tables. Don’t forget to wash your hands frequently.
• #### Should I wear a mask?
Experts are divided on how much protection a regular surgical mask, or even a scarf, can provide for people who aren’t yet sick. The W.H.O. and C.D.C. say that unless you’re already sick, or caring for someone who is, wearing a face mask isn’t necessary. And stockpiling high-grade N95 masks will make it harder for nurses and other workers to access the resources they need. But researchers are also finding that there are more cases of asymptomatic transmission than were known early on in the pandemic. And a few experts say that masks could offer some protection in crowded places where it is not possible to stay 6 feet away from other people. Masks don’t replace hand-washing and social distancing.
• #### Should I stock up on groceries?
Plan two weeks of meals if possible. But people should not hoard food or supplies. Despite the empty shelves, the supply chain remains strong. And remember to wipe the handle of the grocery cart with a disinfecting wipe and wash your hands as soon as you get home.
• #### Should I pull my money from the markets?
That’s not a good idea. Even if you’re retired, having a balanced portfolio of stocks and bonds so that your money keeps up with inflation, or even grows, makes sense. But retirees may want to think about having enough cash set aside for a year’s worth of living expenses and big payments needed over the next five years. | 1,697 | 7,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-16 | latest | en | 0.907253 |
https://www.geeksforgeeks.org/iterative-approach-to-check-if-a-binary-tree-is-perfect/?ref=rp | 1,701,484,941,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100309.57/warc/CC-MAIN-20231202010506-20231202040506-00656.warc.gz | 889,792,397 | 55,671 | # Iterative approach to check if a Binary Tree is Perfect
Given a Binary Tree, the task is to check whether the given Binary Tree is a perfect Binary Tree or not.
A Binary tree is a Perfect Binary Tree in which all internal nodes have two children and all leaves are at the same level.
Examples:
```Input :
1
/ \
2 3
/ \ / \
4 5 6 7
Output : Yes
Input :
20
/ \
8 22
/ \ / \
5 3 4 25
/ \ / \ \
1 10 2 14 6
Output : No
One leaf node with value 4 is not present at the
last level and node with value 25 has
only one child.```
We have already discussed the recursive approach. In this post, the iterative approach is discussed.
Approach: The idea is to use a queue and a variable flag, initialized to zero, to check if a leaf node has been discovered. We will check:
1. If the current node has two children then we will check for the value of flag. If the value of flag is zero then push the left and right child in the queue, but if the value of flag is one then return false because then that means a leaf node has already been found and in a perfect binary tree all the leaf nodes must be present at the last level, no leaf node should be present at any other level.
2. If the current node has no child, that means it is a leaf node, then mark flag as one.
3. If the current node has just one child then return false, as in a perfect binary tree all the nodes have two children except for the leaf nodes, which must be present at the last level of the tree.
Below is the implementation of the above approach:
## C++
`// C++ program to check if the` `// given binary tree is perfect` `#include ` `using` `namespace` `std;` `// A binary tree node` `struct` `Node {` ` ``int` `data;` ` ``Node *left, *right;` `};` `// Utility function to allocate memory for a new node` `Node* newNode(``int` `data)` `{` ` ``Node* node = ``new` `(Node);` ` ``node->data = data;` ` ``node->left = node->right = NULL;` ` ``return` `(node);` `}` `// Function to check if the given tree is perfect` `bool` `CheckPerfectTree(Node* root)` `{` ` ``queue q;` ` ``// Push the root node` ` ``q.push(root);` ` ``// Flag to check if leaf nodes have been found` ` ``int` `flag = 0;` ` ``while` `(!q.empty()) {` ` ``Node* temp = q.front();` ` ``q.pop();` ` ``// If current node has both left and right child` ` ``if` `(temp->left && temp->right) {` ` ``// If a leaf node has already been found` ` ``// then return false` ` ``if` `(flag == 1)` ` ``return` `false``;` ` ``// If a leaf node has not been discovered yet` ` ``// push the left and right child in the queue` ` ``else` `{` ` ``q.push(temp->left);` ` ``q.push(temp->right);` ` ``}` ` ``}` ` ``// If a leaf node is found mark flag as one` ` ``else` `if` `(!temp->left && !temp->right) {` ` ``flag = 1;` ` ``}` ` ``// If the current node has only one child` ` ``// then return false` ` ``else` `if` `(!temp->left || !temp->right)` ` ``return` `false``;` ` ``}` ` ``// If the given tree is perfect return true` ` ``return` `true``;` `}` `// Driver code` `int` `main()` `{` ` ``Node* root = newNode(7);` ` ``root->left = newNode(5);` ` ``root->right = newNode(6);` ` ``root->left->left = newNode(8);` ` ``root->left->right = newNode(1);` ` ``root->right->left = newNode(3);` ` ``root->right->right = newNode(9);` ` ``root->right->right->right = newNode(13);` ` ``root->right->right->left = newNode(10);` ` ``if` `(CheckPerfectTree(root))` ` ``printf``(``"Yes"``);` ` ``else` ` ``printf``(``"No"``);` ` ``return` `0;` `}`
## Java
`// Java program to check if the ` `// given binary tree is perfect ` `import` `java.util.*;` `class` `GFG` `{` ` ` `// A binary tree node ` `static` `class` `Node ` `{ ` ` ``int` `data; ` ` ``Node left, right; ` `}; ` `// Utility function to allocate memory for a new node ` `static` `Node newNode(``int` `data) ` `{ ` ` ``Node node = ``new` `Node(); ` ` ``node.data = data; ` ` ``node.left = node.right = ``null``; ` ` ``return` `(node); ` `} ` `// Function to check if the given tree is perfect ` `static` `boolean` `CheckPerfectTree(Node root) ` `{ ` ` ``Queue q = ``new` `LinkedList(); ` ` ``// add the root node ` ` ``q.add(root); ` ` ``// Flag to check if leaf nodes have been found ` ` ``int` `flag = ``0``; ` ` ``while` `(q.size() > ``0``)` ` ``{ ` ` ``Node temp = q.peek(); ` ` ``q.remove(); ` ` ``// If current node has both left and right child ` ` ``if` `(temp.left != ``null` `&& temp.right != ``null``) ` ` ``{ ` ` ``// If a leaf node has already been found ` ` ``// then return false ` ` ``if` `(flag == ``1``) ` ` ``return` `false``; ` ` ``// If a leaf node has not been discovered yet ` ` ``// add the left and right child in the queue ` ` ``else` ` ``{ ` ` ``q.add(temp.left); ` ` ``q.add(temp.right); ` ` ``} ` ` ``} ` ` ``// If a leaf node is found mark flag as one ` ` ``else` `if` `(temp.left == ``null` `&& temp.right == ``null``) ` ` ``{ ` ` ``flag = ``1``; ` ` ``} ` ` ``// If the current node has only one child ` ` ``// then return false ` ` ``else` `if` `(temp.left == ``null` `|| temp.right == ``null``) ` ` ``return` `false``; ` ` ``} ` ` ``// If the given tree is perfect return true ` ` ``return` `true``; ` `} ` `// Driver code ` `public` `static` `void` `main(String args[])` `{ ` ` ``Node root = newNode(``7``); ` ` ``root.left = newNode(``5``); ` ` ``root.right = newNode(``6``); ` ` ``root.left.left = newNode(``8``); ` ` ``root.left.right = newNode(``1``); ` ` ``root.right.left = newNode(``3``); ` ` ``root.right.right = newNode(``9``); ` ` ``root.right.right.right = newNode(``13``); ` ` ``root.right.right.left = newNode(``10``); ` ` ``if` `(CheckPerfectTree(root)) ` ` ``System.out.printf(``"Yes"``); ` ` ``else` ` ``System.out.printf(``"No"``); ` `} ` `}` `// This code is contributed by Arnab Kundu`
## Python3
`# Python3 program to check if the ` `# given binary tree is perfect ` `import` `sys` `import` `math` `# A binary tree node` `class` `Node:` ` ``def` `__init__(``self``, data):` ` ``self``.data ``=` `data` ` ``self``.left ``=` `None` ` ``self``.right ``=` `None` `# Utility function to allocate ` `# memory for a new node ` `def` `newNode(data):` ` ``return` `Node(data)` `# Function to check if the ` `# given tree is perfect` `def` `CheckPerfectTree(root):` ` ``q ``=` `[]` ` ``# Push the root node` ` ``q.append(root)` ` ``# Flag to check if leaf nodes ` ` ``# have been found ` ` ``flag ``=` `0` ` ` ` ``while``(q):` ` ``temp ``=` `q[``0``]` ` ``q.pop(``0``)` ` ``# If current node has both ` ` ``# left and right child` ` ``if` `(temp.left ``and` `temp.right):` ` ` ` ``# If a leaf node has already been found ` ` ``# then return false ` ` ``if` `(flag ``=``=` `1``):` ` ``return` `False` ` ``# If a leaf node has not been discovered yet ` ` ``# push the left and right child in the queue ` ` ``else``:` ` ``q.append(temp.left)` ` ``q.append(temp.right)` ` ``# If a leaf node is found ` ` ``# mark flag as one ` ` ``elif``(``not` `temp.left ``and` ` ``not` `temp.right):` ` ``flag ``=` `1` ` ``# If the current node has only one child ` ` ``# then return false ` ` ``elif``(``not` `temp.left ``or` ` ``not` `temp.right):` ` ``return` `False` ` ` ` ``# If the given tree is perfect` ` ``# return true` ` ``return` `True` `# Driver Code` `if` `__name__``=``=``'__main__'``:` ` ``root ``=` `newNode(``7``)` ` ``root.left ``=` `newNode(``5``)` ` ``root.left.left ``=` `newNode(``8``)` ` ``root.left.right ``=` `newNode(``1``)` ` ``root.right ``=` `newNode(``6``)` ` ``root.right.left ``=` `newNode(``3``)` ` ``root.right.right ``=` `newNode(``9``)` ` ``root.right.right.left ``=` `newNode(``10``)` ` ``root.right.right.right ``=` `newNode(``13``)` ` ``if` `CheckPerfectTree(root):` ` ``print``(``"Yes"``)` ` ``else``:` ` ``print``(``"No"``)` `# This code is contributed` `# by Vikash Kumar 37`
## C#
`// C# program to check if the ` `// given binary tree is perfect` `using` `System;` `using` `System.Collections.Generic; ` ` ` `class` `GFG` `{` ` ` `// A binary tree node ` `public` `class` `Node ` `{ ` ` ``public` `int` `data; ` ` ``public` `Node left, right; ` `}; ` `// Utility function to allocate` `// memory for a new node ` `static` `Node newNode(``int` `data) ` `{ ` ` ``Node node = ``new` `Node(); ` ` ``node.data = data; ` ` ``node.left = node.right = ``null``; ` ` ``return` `(node); ` `} ` `// Function to check if the given tree is perfect ` `static` `Boolean CheckPerfectTree(Node root) ` `{ ` ` ``Queue q = ``new` `Queue(); ` ` ``// add the root node ` ` ``q.Enqueue(root); ` ` ``// Flag to check if leaf nodes` ` ``// have been found ` ` ``int` `flag = 0; ` ` ``while` `(q.Count > 0)` ` ``{ ` ` ``Node temp = q.Peek(); ` ` ``q.Dequeue(); ` ` ``// If current node has both ` ` ``// left and right child ` ` ``if` `(temp.left != ``null` `&& ` ` ``temp.right != ``null``) ` ` ``{ ` ` ``// If a leaf node has already been found ` ` ``// then return false ` ` ``if` `(flag == 1) ` ` ``return` `false``; ` ` ``// If a leaf node has not been discovered yet ` ` ``// add the left and right child in the queue ` ` ``else` ` ``{ ` ` ``q.Enqueue(temp.left); ` ` ``q.Enqueue(temp.right); ` ` ``} ` ` ``} ` ` ``// If a leaf node is found mark flag as one ` ` ``else` `if` `(temp.left == ``null` `&&` ` ``temp.right == ``null``) ` ` ``{ ` ` ``flag = 1; ` ` ``} ` ` ``// If the current node has only one child ` ` ``// then return false ` ` ``else` `if` `(temp.left == ``null` `||` ` ``temp.right == ``null``) ` ` ``return` `false``; ` ` ``} ` ` ``// If the given tree is perfect return true ` ` ``return` `true``; ` `} ` `// Driver code ` `public` `static` `void` `Main(String []args)` `{ ` ` ``Node root = newNode(7); ` ` ``root.left = newNode(5); ` ` ``root.right = newNode(6); ` ` ``root.left.left = newNode(8); ` ` ``root.left.right = newNode(1); ` ` ``root.right.left = newNode(3); ` ` ``root.right.right = newNode(9); ` ` ``root.right.right.right = newNode(13); ` ` ``root.right.right.left = newNode(10); ` ` ``if` `(CheckPerfectTree(root)) ` ` ``Console.WriteLine(``"Yes"``); ` ` ``else` ` ``Console.WriteLine(``"No"``); ` `} ` `}` `// This code is contributed by Rajput-Ji`
## Javascript
``
Output:
`No `
Time Complexity: O(N), where N is the total number of nodes in the binary tree.
Auxiliary Space: O(N)
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# R- Introduction
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### R- Introduction
1. 1. Introduction to R Venkat Reddy
2. 2. • • • • • • • • What is R R basics Working with R R packages R Functions Importing & Exporting data Creating calculated fields R Help & Tutorials Data Analysis Course Venkat Reddy Contents 2
3. 3. • Programming “environment” • Runs on a variety of platforms including Windows, Unix and MacOS. • Provides an unparalleled platform for programming new statistical methods in an easy and straightforward manner. • Object-oriented • Open source • Excellent graphics capabilities • Supported by a large user network Data Analysis Course Venkat Reddy R 3
5. 5. Data Analysis Course Venkat Reddy R 5
6. 6. Data Analysis Course Venkat Reddy R Studio 6
7. 7. • • • • • 2+2 log(10) help(log) summary(airquality) demo(graphics) # pretty pictures... Data Analysis Course Venkat Reddy R-Demo 7
8. 8. R-Basics :Naming convention • must start with a letter (A-Z or a-z) • can contain letters, digits (0-9), and/or periods “.” • mydata different from MyData Data Analysis Course Venkat Reddy • R is a case sensitive language. 8
9. 9. R-Basics :Assignment • “<-” used to indicate assignment • Assignment to an object is denoted by ”<-” or ”->” or ”=”. • If you see a notation ”= =”, you’ll looking at a comparison operator. – Many other notations can be found from the documentation for the Base package or R. Data Analysis Course Venkat Reddy • x<-c(1,2,3,4,5,6,7) • x<-c(1:7) • x<-1:4 9
10. 10. • during an R session, all objects are stored in a temporary, working memory • Commands are entered interactively at the R user prompt. Up and down arrow keys scroll through your command history. • list objects ls() • remove objects rm() • data() Data Analysis Course Venkat Reddy Workspace 10
11. 11. • • • • • • • • • • x <- round(rnorm(10,mean=20,sd=5)) # simulate data x mean(x) m <- mean(x) m x- m # notice recycling (x - m)^2 sum((x - m)^2) data() UKgas Data Analysis Course Venkat Reddy Demo: Working with R 11
12. 12. • R consists of a core and packages. Packages contain functions that are not available in the core. • Collections of R functions, data, and compiled code • Well-defined format that ensures easy installation, a basic standard of documentation, and enhances portability and reliability • When you download R, already a number (around 30) of packages are downloaded as well. • You can use the function search to see a list of packages that are currently attached to the system, this list is also called the search path. • search() Data Analysis Course Venkat Reddy R Packages 12
13. 13. • Select the `Packages' menu and select `Install package...', a list of available packages on your system will be displayed. • Select one and click `OK', the package is now attached to your current R session. Via the library function • The library can also be used to list all the available libraries on your system with a short description. Run the function without any arguments Data Analysis Course Venkat Reddy R packages 13
14. 14. Demo: R packages Data Analysis Course Venkat Reddy • Install cluster package • Install plyr package (for string operations) 14
15. 15. Importing Data • Reading CSV file • X <-read.csv(“file.csv") • reads in data from an external file • d <- read.table("myfile", header=TRUE) • R has ODBC for connecting to other programs • R gets confused if you use a path in your code like c:mydocumentsmyfile.txt • This is because R sees "" as an escape character. Instead, use c:my documentsmyfile.txt or c:/mydocuments/myfile.txt Data Analysis Course Venkat Reddy • read.table() 15
17. 17. Exporting data • To A Tab Delimited Text File • write.table(mydata, "c:/mydata.txt", sep="t") • library(xlsReadWrite) • write.xls(mydata, "c:/mydata.xls") • To SAS • library(foreign) • write.foreign(mydata, "c:/mydata.txt", "c:/mydata.sas", package="SAS") Data Analysis Course Venkat Reddy • To an Excel Spreadsheet 17
18. 18. Demo: Exporting data Data Analysis Course Venkat Reddy • write.table(sales_data, "C:UsersVENKATGoogle DriveTrainingRDatasales_export.txt", sep="t") • write.table(sales_data, "C:UsersVENKATGoogle DriveTrainingRDatasales_export.csv", sep=",") 18
19. 19. R- Functions Numeric Functions Description abs(x) absolute value sqrt(x) square root ceiling(x) ceiling(3.475) is 4 floor(x) floor(3.475) is 3 trunc(x) trunc(5.99) is 5 round(x, digits=n) round(3.475, digits=2) is 3.48 signif(x, digits=n) signif(3.475, digits=2) is 3.5 cos(x), sin(x), tan(x) also acos(x), cosh(x), acosh(x), etc. log(x) natural logarithm log10(x) common logarithm exp(x) e^x Data Analysis Course Venkat Reddy Function 19
20. 20. • • • • y<-abs(-20) x<-Sum(y+5) Z<-Log(x) round(x,1) Data Analysis Course Venkat Reddy Demo: Numeric Functions 20
21. 21. Character Functions Description substr(x, start=n1, stop=n2) Extract or replace substrings in a character vector. x <- "abcdef" substr(x, 2, 4) is "bcd" substr(x, 2, 4) <- "22222" is "a222ef" grep(pattern, x , ignore.case=FALSE, fixed=FALSE) Search for pattern in x. If fixed =FALSE then pattern is a regular expression. If fixed=TRUE then pattern is a text string. Returns matching indices. grep("A", c("b","A","c"), fixed=TRUE) returns 2 sub(pattern, replacement, x, ignore.case =FALSE, fixed=FALSE) Find pattern in x and replace with replacement text. If fixed=FALSE then pattern is a regular expression. If fixed = T then pattern is a text string. sub("s",".","Hello There") returns "Hello.There" strsplit(x, split) Split the elements of character vector x at split. strsplit("abc", "") returns 3 element vector "a","b","c" paste(..., sep="") Concatenate strings after using sep string to seperate them. paste("x",1:3,sep="") returns c("x1","x2" "x3") paste("x",1:3,sep="M") returns c("xM1","xM2" "xM3") paste("Today is", date()) toupper(x) Uppercase Data Analysis Course Venkat Reddy Function 21
22. 22. Demo :Character Functions Data Analysis Course Venkat Reddy • cust_id<-"Cust1233416" • id<-substr(cust_id, 5,10) • Up=toupper(cust_id) 22
23. 23. Calculated Fields in R • • • • • • mydata\$sum <- mydata\$x1 + mydata\$x2 mydata\$mean <- (mydata\$x1 + mydata\$x2)/2 attach(mydata) mydata\$sum <- x1 + x2 mydata\$mean <- (x1 + x2)/2 detach(mydata) Data Analysis Course Venkat Reddy • Use the assignment operator <- to create new variables. A wide array of operators and functions are available here. 23
24. 24. • • • • • • • • sales_data\$reduce<-(sales_data\$Sales)*0.2 View(sales_data) sales_data\$new_sales<-sales_data\$Sales- sales_data\$reduce attach(petrol) ratio=Income/consum_mill_gallons View(petrol) petrol\$ratio=Income/Consum_mill_gallons View(petrol) Data Analysis Course Venkat Reddy Demo Calculated Fields in R 24
25. 25. • If you encounter a new command during the exercises, and you’d like to know what it does, please consult the documentation. All R commands are listed nowhere, and the only way to get to know new commands is to read the documentation files, so we’d like you to practise this youself. • Tutorials Each of the following tutorials are in PDF format. • P. Kuhnert & B. Venables, An Introduction to R: Software for Statistical Modeling & Computing • J.H. Maindonald, Using R for Data Analysis and Graphics • B. Muenchen, R for SAS and SPSS Users • W.J. Owen, The R Guide • D. Rossiter, Introduction to the R Project for Statistical Computing for Use at the ITC • W.N. Venebles & D. M. Smith, An Introduction to R Data Analysis Course Venkat Reddy R-Help 25
26. 26. R-Tutorials Paul Geissler's excellent R tutorial Dave Robert's Excellent Labs on Ecological Analysis Excellent Tutorials by David Rossitier Excellent tutorial an nearly every aspect of R (c/o Rob Kabacoff) MOST of these notes follow this web page format • Introduction to R by Vincent Zoonekynd • R Cookbook • Data Manipulation Reference Data Analysis Course Venkat Reddy • • • • 26
27. 27. Data Analysis Course Venkat Reddy Thank you 27 | 2,232 | 8,101 | {"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-2019-18 | latest | en | 0.723266 |
https://www.johnpatel.com/tools/hex-to-decimal-converter/ | 1,726,421,225,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00403.warc.gz | 768,326,611 | 13,198 | # Hex to Decimal Converter Online
This converter will help you to convert Hex to Decimal with the accurate and fast result. Just enter hexadecimal string inside the left box and as a result, you will get the decimal number value inside the right box very quickly.
For Example: you can type any hexadecimal value like "905" into the left box below, and you will get the decimal result "2309" inside the second box.
#### Decimal Value
Vice Versa: Decimal to Hex
Popular Translators: Binary to Text Translator | Morse Translator
## How to calculate hex to decimal
Hex is a base 16 quantity and decimal is a base 10 quantity. We have to know the decimal equal of each hex quantity digit. See under of the web page to test the hexadecimal to decimal chart. Listed below are the steps to transform hex to decimal.
First of all, Let's take one example.
```7AB is a hex number
7AB = (7 * 162) + (10 * 161) + (11 * 160)
7AB = (7 * 256) + (10 * 16) + (11 * 1)
7AB = 1792 + 160 + 11
7AB = 1963 (in decimal number)```
### Conversion Process:
• First of all, Get the decimal equal of hex from the desk.
• Multiply each digit with 16 energy of digit location. (zero based mostly, 7AB: A location is 0, B location is 1 and the 7 location is 2)
• Sum all of the multipliers.
For hex number with n digits:
### Hex to Decimal Conversion Chart Table
You can find upto 400 Hex to Decimal values to below table.
Hex Base 16 Decimal Base 10 Calculation
0 0 -
1 1 -
2 2 -
3 3 -
4 4 -
5 5 -
6 6 -
7 7 -
8 8 -
9 9 -
A 10 -
B 11 -
C 12 -
D 13 -
E 14 -
F 15 -
10 16 1×161+0×160 = 16
11 17 1×161+1×160 = 17
12 18 1×161+2×160 = 18
13 19 1×161+3×160 = 19
14 20 1×161+4×160 = 20
15 21 1×161+5×160 = 21
16 22 1×161+6×160 = 22
17 23 1×161+7×160 = 23
18 24 1×161+8×160 = 24
19 25 1×161+9×160 = 25
1A 26 1×161+10×160 = 26
1B 27 1×161+11×160 = 27
1C 28 1×161+12×160 = 28
1D 29 1×161+13×160 = 29
1E 30 1×161+14×160 = 30
1F 31 1×161+15×160 = 31
20 32 2×161+0×160 = 32
30 48 3×161+0×160 = 48
40 64 4×161+0×160 = 64
50 80 5×161+0×160 = 80
60 96 6×161+0×160 = 96
70 112 7×161+0×160 = 112
80 128 8×161+0×160 = 128
90 144 9×161+0×160 = 144
A0 160 10×161+0×160 = 160
B0 176 11×161+0×160 = 176
C0 192 12×161+0×160 = 192
D0 208 13×161+0×160 = 208
E0 224 14×161+0×160 = 224
F0 240 15×161+0×160 = 240
100 256 1×162+0×161+0×160 = 256
200 512 2×162+0×161+0×160 = 512
300 768 3×162+0×161+0×160 = 768
400 1024 4×162+0×161+0×160 = 1024
Popular Converters: cm to km | KMH to MPH | Knots to Mph | Mph to Knots | 1,006 | 2,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-38 | latest | en | 0.604149 |
https://oeis.org/A067201 | 1,571,478,489,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692723.54/warc/CC-MAIN-20191019090937-20191019114437-00058.warc.gz | 635,314,628 | 4,126 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A067201 Numbers k such that k^2 + 2 is prime. 43
0, 1, 3, 9, 15, 21, 33, 39, 45, 57, 81, 99, 105, 111, 117, 123, 147, 171, 219, 225, 237, 243, 249, 255, 273, 297, 303, 309, 321, 345, 351, 363, 369, 375, 387, 417, 423, 429, 441, 447, 453, 477, 501, 513, 549, 555, 561, 573, 603, 609, 651, 675, 681, 699, 711, 753 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS All terms > 1 are divisible by 3. - Robert Israel, Sep 05 2014 LINKS T. D. Noe, Table of n, a(n) for n=1..1000 Eric Weisstein's World of Mathematics, Near-Square Prime MAPLE select(t -> isprime(t^2+2), [0, 1, seq(3*i, i=1..1000)]); # Robert Israel, Sep 05 2014 MATHEMATICA lst={}; Do[If[PrimeQ[n^2+2], AppendTo[lst, n]], {n, 3*10^2}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 21 2008 *) Join[{0, 1}, Select[Range[3, 1000, 6], PrimeQ[#^2 + 2] &]] (* Zak Seidov, Jan 30 2014 *) PROG (PARI) select(n -> isprime(n^2+2), [1..500]) \\ Edward Jiang, Sep 05 2014 CROSSREFS Equals 6*A056900(n-2) + 3, n>1. Cf. A106571, A040976. Other sequences of the type "Numbers k such that k^2 + i is prime": A005574 (i=1), this sequence (i=2), A049422 (i=3), A007591 (i=4), A078402 (i=5), A114269 (i=6), A114270 (i=7), A114271 (i=8), A114272 (i=9), A114273 (i=10), A114274 (i=11), A114275 (i=12). Sequence in context: A272793 A273567 A162486 * A071526 A276967 A114271 Adjacent sequences: A067198 A067199 A067200 * A067202 A067203 A067204 KEYWORD nonn AUTHOR Benoit Cloitre, Feb 19 2002 STATUS approved
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Last modified October 19 04:40 EDT 2019. Contains 328211 sequences. (Running on oeis4.) | 762 | 1,945 | {"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-2019-43 | latest | en | 0.53902 |
http://math.stackexchange.com/questions/204367/is-there-a-function-like-this?answertab=votes | 1,448,498,600,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398446248.56/warc/CC-MAIN-20151124205406-00084-ip-10-71-132-137.ec2.internal.warc.gz | 134,009,428 | 20,664 | # Is there a function like this?
Let $A=[0,1]$ and $C=\{0\}\cup\{\frac{1}{n},\ n\in\mathbb{N}\}$.
i) Is there a function $f:A\rightarrow\mathbb{R}$ such that $f\in C^{r}(A)$, $r\geq 2$ and the set of critical "Values" of $f$ is $C$?
ii) Is there a function $f:A\rightarrow\mathbb{R}$ such that $f\in C^{1}(A)$ and the set of critical "Values" of $f$ is $-C\cup C$?
This is a problem from a course of differential topology that i did last year. Is related with Morse theory. I couldn't figure out any good solution for it. I appreciate some help.
EDIT: A critical value is image of a critical point, i.e. if $f'(x)=0$ then $f(x)$ is a critical value.
Thanks
-
To be sure: $C$ is the set you describe at the top of your post, while $C^r$ is the set of continuously differentiable functions of order $r$? – Adam Saltz Sep 29 '12 at 14:02
you are right.. – Tomás Sep 29 '12 at 15:24
You can obtain example (ii) by gluing two copies of example (i). To be more precise, suppose $f:[0,1]\to [0,1]$ satisfies (i) and additionally $f(0)=0=f'(0)$. Then the odd extension $f(-x)=-f(x)$ is in $C^1[-1,1]$ and satisfies (ii). Of course, $[-1,1]$ can be changed to $[0,1]$ by a linear transformation. – user31373 Sep 30 '12 at 2:56
(ii) is clearly wrong because $-C\not\subset A$. – celtschk Oct 4 '12 at 21:54
The answer to question (i) is no, the answer to question (ii) is yes. (Thanks to Tomas for pointing out a mistake in my previous argument.)
(ii) Existence of the $\mathcal{C}^1$ function. Let $F:[0,1]\to[0,1]$ be a $\mathcal{C}^1$ function such that $F(0)=0$, $F(1)=1$, $F'(0)=F'(1)=0$, and such that $F'(x)>0$ for all $x \in (0,1)$. (E.g., one could use $F(x) = \sin^2 \frac{\pi x}{2}$.)
Now let $a_n$ be a strictly decreasing positive sequence with $a_1 = 1$ and $\lim\limits_{n\to\infty} a_n = 0$. Define the function $f$ on the interval $[a_{n+1},a_n]$ by $$f(x) = \frac{1}{n+1} + \left(\frac1n - \frac{1}{n+1}\right) F\left(\frac{x-a_{n+1}}{a_{n}-a_{n+1}}\right)$$ for all $n$. I.e., $f$ is an affine version of $F$ with $f(a_n) = \frac1n$ and $f(a_{n+1}) = \frac1{n+1}$. This already gives us a (strictly increasing) $\mathcal{C}^1$ function $f:(0,1] \to \mathbb{R}$ with critical values $\{ \frac1n:n \in \mathbb{N} \}$. Extending it to $[0,1]$ by $f(0)=0$ makes the function continuous. Now we just have to find a sequence $(a_n)$ for which this function satisfies $\lim\limits_{x\to 0} f'(x) = 0$. Then the odd extension of $f$ to $[-1,1]$ is $\mathcal{C}^1$ and has critical values $C \cup (-C)$.
Letting $M = \sup F'$, we get for all $x\in[a_{n+1},a_n]$ that $$0<f'(x) \le M\frac{\frac1n - \frac{1}{n+1}}{a_n - a_{n+1}}.$$ As $x\to 0$, we get $n\to\infty$, so we have to find a sequence such that $$\lim\limits_{n\to\infty} \frac{\frac1n - \frac1{n+1}}{a_n - a_{n+1}} = 0.$$ Many sequences work here, e.g., $a_n = \frac1{\sqrt{n}}$ or $a_n = \frac{1}{1+ \ln n}$. This finishes the argument.
(i) Non-existence of the $\mathcal{C}^2$ function. Assume that $f$ is $\mathcal{C}^2$ on $[0,1]$ with critical values $C$. Then there exists a sequence of disjoint intervals $I_n = [a_n,b_n]$ with critical points $a_n$, $b_n$, no critical points in $(a_n,b_n)$, and $f(a_n) \ge \frac1n$ or $f(b_n) \ge \frac1n$. (This is shown by an elementary, but not quite trivial argument.) This implies that $$|f(a_n)-f(b_n)| \ge \frac1n - \frac1{n+1}= \frac1{n(n+1)}\ge \frac{1}{2n^2}.$$ By the Mean Value Theorem there exists $c_n \in (a_n,b_n)$ with $$|f'(c_n)|\ge \frac{1}{2n^2(b_n-a_n)}.$$ Since $f'(a_n) = 0$, another application of the Mean Value Theorem to $f'$ shows that there exists $d_n \in (a_n,c_n)$ with $$|f''(d_n)| \ge \frac{1}{2n^2(c_n-a_n)(b_n-a_n)} \ge \frac{1}{2n^2(b_n-a_n)^2}.$$ By assumption $f$ is $\mathcal{C}^2$, so $|f''| \le M$ for some constant $M$, implying that $$b_n-a_n \ge \frac{1}{n\sqrt{2M}}.$$ This gives $\sum\limits_{n=1}^\infty (b_n-a_n) = +\infty$. However, since these intervals are mutually disjoint subsets of $[0,1]$, we have $\sum\limits_{n=1}^\infty (b_n-a_n) \le 1$, which is the desired contradiction.
-
You just showed that $f'(0)=0$. I cant see why $f\in C^{2}$ on zero or even why $\in C^{1}$ on zero. – Tomás Oct 4 '12 at 12:01
You are right, there is an additional argument needed. I'll see if this construction is fixable. – Lukas Geyer Oct 4 '12 at 18:36
OK, if all of the piecewise functions (from the lemma) are affine versions of each other, it is not too hard to show that this function will be $\mathcal{C}^1$ (implying a positive answer to (ii)), but it won't be $\mathcal{C}^2$. I'll update my answer later today. – Lukas Geyer Oct 4 '12 at 18:50
OK, fixed the $\mathcal{C}^1$ construction, and showed that $\mathcal{C}^2$ is in fact not possible. Not the most elegant proof in the world, but I hope it is correct this time... :) – Lukas Geyer Oct 4 '12 at 21:32
Just to make it clear. When you change the interval $[-1,1]$ to $[0,1]$ the set of critival values still the same? – Tomás Oct 5 '12 at 0:23
In each case, the answer is yes. Check out this post by Mate on the Topology Q+A board. To adapt it to your question:
You need this lemma, which isn't so hard to prove if you already know about bump functions:
For any $a,b,c,d \in \mathbb{R}$ with $a<b$, there exists a smooth function $f:[a,b] \rightarrow \mathbb{R}$ such that $f(a) = c, f(b) = d$, and all the derivatives of $f$ vanish at $a$ and $b$.
Now let $r: \mathbb{N} \rightarrow \mathbb{Q}$ be your favorite enumeration the set $C$ (or $-C \cup C$). Extend $r$ to a smooth function $f$ on the (positive) real numbers using the lemma. This function satisfies all your requirements.
-
Hi Adam Saltz. I think your argument dont prove the statement. It is a good argument when the function $f$ is defined on $\mathbb{R}$, but in our case the function is defined on $[0,1]$. I cant see a clean way to adapt the demonstration for this case. If you do, please show me. – Tomás Sep 29 '12 at 14:59
Indeed, some care needs to be taken with the smoothness of $f$ at the endpoints $0$ or/and $1$. – user31373 Sep 30 '12 at 2:52
You are both quite right. I misread your question, Kaye and I agree that it is difficult to adapt for the reason LVK states. – Adam Saltz Oct 1 '12 at 3:55 | 2,199 | 6,206 | {"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} | 3.546875 | 4 | CC-MAIN-2015-48 | longest | en | 0.782665 |
https://puzzling.stackexchange.com/questions/20903/to-dream-or-not-to-dream-a-haiku/20905 | 1,624,355,891,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517048.78/warc/CC-MAIN-20210622093910-20210622123910-00466.warc.gz | 424,969,256 | 42,682 | # To dream, or not to dream (a haiku)
A dream, not a dream.
Tailless, tail-less. Puzzled? Makes
Quite the parable.
What am I referring to?
Hint #1:
There are a few cryptic-clue style indicators that I'm sure many of you have latched onto. The question is, what set of letters are they indicating? Pretty much the entire puzzle goes in order, so if you find yourself jumping around to try to solve it, you've strayed from the path.
Hint #2:
The answer you seek is a single, eight-letter word.
Hint #3:
Synonymize twice.
Cryptic clue, cryptic clue. Then,
It's a synonym.
• hm, "tailless" seems suspiciously like a double curtailment indicator... – Deusovi Sep 3 '15 at 20:52
• Guys, I blew it - I forgot the wordplay tag. It's been added. D: – Bailey M Sep 4 '15 at 13:57
• Arrrgh @BaileyM would you stop preempting all of my puzzle types!!! First crossword, then murder mystery, now haiku! It's bad enough that all of your puzzles are so good – NeedAName Sep 4 '15 at 14:22
• Is any of the capitalization important? – Milo P Sep 4 '15 at 21:36
• @MiloPrice the capitalization isn't important at all, no. – Bailey M Sep 8 '15 at 13:59
I think @Hackiisan is right but for the wrong reason.
Allegory
A dream
A goal
Not a dream, tailless, tail-less
Not a dream = reality. Remove "tai" (a tailless tail) from reality leaves "rely" (So "Not a dream, tailless, tail-less" means "remove tai from reality")
puzzled?
anagram of "goal rely" gives you allegory
makes quite a parable
allegory
• You and @Hackiisan are right, but you need to combine your answers - specifically, your dream and his 'not a dream'. – Bailey M Sep 21 '15 at 13:42
• Okay, I changed my explanation. Think this makes more sense – Dr Xorile Sep 21 '15 at 16:01
• whew I can breathe easier now. You got it! – Bailey M Sep 21 '15 at 16:04
Is it a
tale?
Explanation
A dream, not a dream
Fables is dreamy and fictional, but not a literal dream.
Tailless, tailless. Puzzled?
"tailless" - "less"(the "tail" end of the word) = "tail" (rhymes with "tale") "Puzzled" = search for homophone of "tail" -> "tale"
Makes quite the parable
"Tale" is a synonym of "parable"
• You're on the right track, especially with how you've approached the second and third lines. The first line can actually be interpreted a bit more literally than that. – Bailey M Sep 4 '15 at 14:56
I'm not fully convinced (which probably means it's incorrect as your puzzles go)
But could it be:
Les Miserables?
A dream, not a dream.
A reference to the song "I dreamed a dream" - and perhaps how the situation in Paris itself is not a "dream"
Tailless, tailless
Taking the last letter from tailless cryptic style we get - Tailles which are "(in France) taxes levied on the common people by the king or an overlord" which fits with the June Rebellion in Paris that Les Mis is based on.
Puzzled? Makes Quite the parable.
I was hoping Les Mis would anagram to something to nicely fit in here, but alas..
• Looking for anagrams is a good way to start when I'm the puzzler in question. :) Something's tailless, that's for sure - but it's not tailless. The added hyphen should help sort that out. – Bailey M Sep 4 '15 at 14:55
Just thinking out loud in case somebody can use this
A dream, not a dream
Daydream, an idea, a thought
Tailless, tail-less
Tailless - tail = less, otherwise tailless - the literal tail of the word (less) = tail
Puzzled? Makes quite the parable
My guess is that the combination of two words deduced from the first two clues forms the name of a well known parable
• Deducing words is a good idea, and 'parable' being related to the final answer is also a good idea. My girlfriend made the same mistake when I walked her through this one - 'daydream', while it sounds like the opposite of a dream, is actually very similar. I would classify ideas and thoughts similarly. – Bailey M Sep 4 '15 at 14:54
• How about nightmare? – wizloc Sep 4 '15 at 15:21
• I would continue to describe a nightmare as similar to a dream. – Bailey M Sep 4 '15 at 15:41
could it be,
Story, seems too easy for one of your puzzles
A dream, not a dream.
A story is can be like a dream if it is fictional
Tailless, tailless. Puzzled? Makes
Tailless, tailless sounds like tell us, tell us(At least in the south it does). puzzled? tell us what? a story. Make up a story
Quite the parable.
parables are moral stories
Hmm.
Idea #1
It's the
Martin Luther King "I Have A Dream" speech
because
A dream, not a dream.
A dream that MLK hoped to see come true.
Tailless, tail-less.
MLK's work was terminated - cut short, curtailed - by his assassination.
Puzzled? Makes Quite the parable.
The speech uses a number of rhetorical devices which have their roots in religious preaching; see https://en.wikipedia.org/wiki/I_Have_a_Dream#The_speech_and_rhetoric
and
MLK was of course a minister of religion. "You know, actually all that I do in civil rights I do because I consider it a part of my ministry." https://en.wikipedia.org/wiki/Martin_Luther_King,_Jr.#Ideas.2C_influences.2C_and_political_stances
Idea #2
It's
because
A dream, not a dream.
In the story, dreams become prophecy.
Tailless, tail-less.
Joseph's coat of many colours is typically quite a long coat - a tail-coat! But it's taken away: Joseph is tail-coat-less.
Puzzled?
Pharoah was puzzled - needed Joseph to interpret the cow dreams.
Makes Quite the parable.
The story is a Judeo-Christian-Islamic-Baha'i religious text.
Based on the new hints, here is my second attempt:
allegory
A dream, not a dream
Refers to the words analogy and reality. I believe "analogy" is dream-like in the sense of being hypothetical.
Tailless, tail-less
The subtle difference in punctuation for "tailless" refers to two different ways of removing a "tail" from each of the words.
logy = removing "anal" (tail in the literal sense) from "analogy"
real = removing "-ity" (the suffix being the tail) from "reality"
Puzzled? Makes quite the parable
Scrambling "logy + real" gives us allegory, which is a synonym for "parable"
• You and Dr Xorile are right, but you need to combine your answers - specifically, his dream and your 'not a dream'. – Bailey M Sep 21 '15 at 13:42
• Perhaps "end goal" and "reality"? "End goal - end + reality - tail" = "goal + rey" = "alegory"? (Quite the parable because it's missing an L) – Hackiisan Sep 21 '15 at 15:55
• You're brutally close. "Goal" and "reality" are the right words. What do you have to remove from "reality" to make your result anagram into "allegory"? – Bailey M Sep 21 '15 at 16:00
• Oh I see what you want me to do now. It's "goal + reality - (tail-less "tail")" = "goal + reality - tai" = "allegory". – Hackiisan Sep 21 '15 at 17:58
• Oops I guess Dr Xorile beat me to it! – Hackiisan Sep 21 '15 at 18:00 | 1,909 | 6,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-25 | latest | en | 0.899489 |
https://www.physicsforums.com/threads/surface-area-of-a-dome-ish-roof.546994/ | 1,709,289,890,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475238.84/warc/CC-MAIN-20240301093751-20240301123751-00775.warc.gz | 926,445,929 | 17,020 | # Surface area of a dome-ish roof
• joebobjoe
In summary, the homework statement is to determine the surface area of the roof of a structure if it is formed by rotating the parabola about the axis. The Attempt at a Solution states that the surface area is 1072.33 m^2. However, according to the website http://www.slideshare.net/mrsbeth63/engineering-mechanics-statics-rc-hibbeler-12th-edition-complete-solutions-ch-9" , the answer is 1365 m^2.
## Homework Statement
Determine the surface area of the roof of the structure if it is formed by rotating the parabola about the axis.
## Homework Equations
$SA=\int _0^{16}{2\pi\left ( 4 \sqrt{16-y} \right ) dy}$ (?)
## The Attempt at a Solution
$SA=\left [ -\frac{16}{3}\pi\left ( 16-y \right )^{\frac{3}{2}}\right ]_{0}^{16}$
$SA=1072.33$
So, 1072.33 m^2?
That's what I get.
Well according to page 78 of http://www.slideshare.net/mrsbeth63/engineering-mechanics-statics-rc-hibbeler-12th-edition-complete-solutions-ch-9" [Broken], the answer is 1365 m^2. I kind of understand how they did it, I just want to know why my way doesn't work.
Last edited by a moderator:
joebobjoe said:
Well according to page 78 of http://www.slideshare.net/mrsbeth63/engineering-mechanics-statics-rc-hibbeler-12th-edition-complete-solutions-ch-9" [Broken], the answer is 1365 m^2. I kind of understand how they did it, I just want to know why my way doesn't work.
Because that's not a formula for surface area. You are integrating 2*pi*f(y). You want to integrate 2*pi*f(y)*sqrt(1+f'(y)^2).
Last edited by a moderator:
Dick said:
Because that's not a formula for surface area. You are integrating 2*pi*f(y). You want to integrate 2*pi*f(y)*sqrt(1+f'(y)^2).
Why not? How does adding up the circumferences not equal the surface area of the dome.
joebobjoe said:
Why not? How does adding up the circumferences not equal the surface area of the dome.
You are adding up infinitesimal surface areas. If you just use the circumference then you are assuming a cylinder is a good approximation to the cross sectional surface area. It's not. Try applying that to a cone. You'll get the wrong answer.
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joebobjoe said:
Why not? How does adding up the circumferences not equal the surface area of the dome.
surface area is only the outer shell of a solid, which is the SA. what you are doing is finding the area under the curve and multiplying it by 2pi
SA is the arc length (you can think of it as circumference) rotated around 2pi for this problem and the formula for arc length is √(1+[y']^2)
Dick said:
You are adding up infinitesimal surface areas. If you just use the circumference then you are assuming a cylinder is a good approximation to the cross sectional surface area. It's not. Try applying that to a cone. You'll get the wrong answer.
Okay thanks.
Calculus is stupid.
## What is the formula for calculating the surface area of a dome-ish roof?
The formula for calculating the surface area of a dome-ish roof is 2πr^2 + 2πrh, where r is the radius of the dome and h is the height of the dome.
## Can the surface area of a dome-ish roof be calculated using the same formula as a regular dome?
Yes, the surface area of a dome-ish roof can be calculated using the same formula as a regular dome. The only difference is that the radius and height used in the formula may be different due to the varying shape of the dome-ish roof.
## What units should be used when calculating the surface area of a dome-ish roof?
The units used for calculating the surface area of a dome-ish roof will depend on the units used for the radius and height of the dome. If the radius is given in feet and the height is given in inches, the final answer will be in square feet.
## What factors can affect the surface area of a dome-ish roof?
The surface area of a dome-ish roof can be affected by the radius and height of the dome, as well as any additional features such as skylights or windows. Changes in the shape or design of the roof can also impact the surface area.
## How is the surface area of a dome-ish roof used in construction?
The surface area of a dome-ish roof is an important factor in construction as it helps determine the amount of materials needed for the roof. It is also used in estimating the cost of construction and can impact the structural integrity of the roof. | 1,085 | 4,336 | {"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} | 4.03125 | 4 | CC-MAIN-2024-10 | latest | en | 0.864957 |
https://ww2.mathworks.cn/help/matlab/ref/plot.html?lang=en | 1,585,706,370,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505359.23/warc/CC-MAIN-20200401003422-20200401033422-00189.warc.gz | 791,601,667 | 29,162 | plot
Syntax
``plot(X,Y)``
``plot(X,Y,LineSpec)``
``plot(X1,Y1,...,Xn,Yn)``
``plot(X1,Y1,LineSpec1,...,Xn,Yn,LineSpecn)``
``plot(Y)``
``plot(Y,LineSpec)``
``plot(___,Name,Value)``
``plot(ax,___)``
``h = plot(___)``
Description
example
````plot(X,Y)` creates a 2-D line plot of the data in `Y` versus the corresponding values in `X`. If `X` and `Y` are both vectors, then they must have equal length. The `plot` function plots `Y` versus `X`.If `X` and `Y` are both matrices, then they must have equal size. The `plot` function plots columns of `Y` versus columns of `X`.If one of `X` or `Y` is a vector and the other is a matrix, then the matrix must have dimensions such that one of its dimensions equals the vector length. If the number of matrix rows equals the vector length, then the `plot` function plots each matrix column versus the vector. If the number of matrix columns equals the vector length, then the function plots each matrix row versus the vector. If the matrix is square, then the function plots each column versus the vector.If one of `X` or `Y` is a scalar and the other is either a scalar or a vector, then the `plot` function plots discrete points. However, to see the points you must specify a marker symbol, for example, `plot(X,Y,'o')`. ```
````plot(X,Y,LineSpec)` sets the line style, marker symbol, and color. ```
example
````plot(X1,Y1,...,Xn,Yn)` plots multiple `X`, `Y` pairs using the same axes for all lines. ```
example
````plot(X1,Y1,LineSpec1,...,Xn,Yn,LineSpecn)` sets the line style, marker type, and color for each line. You can mix `X`, `Y`, `LineSpec` triplets with `X`, `Y` pairs. For example, `plot(X1,Y1,X2,Y2,LineSpec2,X3,Y3)`.```
example
````plot(Y)` creates a 2-D line plot of the data in `Y` versus the index of each value. If `Y` is a vector, then the x-axis scale ranges from 1 to `length(Y)`.If `Y` is a matrix, then the `plot` function plots the columns of `Y` versus their row number. The x-axis scale ranges from 1 to the number of rows in `Y`.If `Y` is complex, then the `plot` function plots the imaginary part of `Y` versus the real part of `Y`, such that `plot(Y)` is equivalent to `plot(real(Y),imag(Y))`. ```
````plot(Y,LineSpec)` sets the line style, marker symbol, and color.```
example
````plot(___,Name,Value)` specifies line properties using one or more `Name,Value` pair arguments. For a list of properties, see Line Properties. Use this option with any of the input argument combinations in the previous syntaxes. Name-value pair settings apply to all the lines plotted. ```
example
````plot(ax,___)` creates the line in the axes specified by `ax` instead of in the current axes (`gca`). The option `ax` can precede any of the input argument combinations in the previous syntaxes. ```
example
````h = plot(___)` returns a column vector of chart line objects. Use `h` to modify properties of a specific chart line after it is created. For a list of properties, see Line Properties.```
Examples
collapse all
Create `x` as a vector of linearly spaced values between 0 and $2\pi$. Use an increment of $\pi /100$ between the values. Create `y` as sine values of `x`. Create a line plot of the data.
```x = 0:pi/100:2*pi; y = sin(x); plot(x,y)```
Define `x` as 100 linearly spaced values between $-2\pi$ and $2\pi$. Define `y1` and `y2` as sine and cosine values of `x`. Create a line plot of both sets of data.
```x = linspace(-2*pi,2*pi); y1 = sin(x); y2 = cos(x); figure plot(x,y1,x,y2)```
Define `Y` as the 4-by-4 matrix returned by the `magic` function.
`Y = magic(4)`
```Y = 4×4 16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1 ```
Create a 2-D line plot of `Y`. MATLAB® plots each matrix column as a separate line.
```figure plot(Y)```
Plot three sine curves with a small phase shift between each line. Use the default line style for the first line. Specify a dashed line style for the second line and a dotted line style for the third line.
```x = 0:pi/100:2*pi; y1 = sin(x); y2 = sin(x-0.25); y3 = sin(x-0.5); figure plot(x,y1,x,y2,'--',x,y3,':')```
MATLAB® cycles the line color through the default color order.
Plot three sine curves with a small phase shift between each line. Use a green line with no markers for the first sine curve. Use a blue dashed line with circle markers for the second sine curve. Use only cyan star markers for the third sine curve.
```x = 0:pi/10:2*pi; y1 = sin(x); y2 = sin(x-0.25); y3 = sin(x-0.5); figure plot(x,y1,'g',x,y2,'b--o',x,y3,'c*')```
Create a line plot and display markers at every fifth data point by specifying a marker symbol and setting the `MarkerIndices` property as a name-value pair.
```x = linspace(0,10); y = sin(x); plot(x,y,'-o','MarkerIndices',1:5:length(y))```
Create a line plot and use the `LineSpec` option to specify a dashed green line with square markers. Use `Name,Value` pairs to specify the line width, marker size, and marker colors. Set the marker edge color to blue and set the marker face color using an RGB color value.
```x = -pi:pi/10:pi; y = tan(sin(x)) - sin(tan(x)); figure plot(x,y,'--gs',... 'LineWidth',2,... 'MarkerSize',10,... 'MarkerEdgeColor','b',... 'MarkerFaceColor',[0.5,0.5,0.5])```
Use the `linspace` function to define `x` as a vector of 150 values between 0 and 10. Define `y` as cosine values of `x`.
```x = linspace(0,10,150); y = cos(5*x);```
Create a 2-D line plot of the cosine curve. Change the line color to a shade of blue-green using an RGB color value. Add a title and axis labels to the graph using the `title`, `xlabel`, and `ylabel` functions.
```figure plot(x,y,'Color',[0,0.7,0.9]) title('2-D Line Plot') xlabel('x') ylabel('cos(5x)')```
Define `t` as seven linearly spaced `duration` values between 0 and 3 minutes. Plot random data and specify the format of the `duration` tick marks using the `'DurationTickFormat'` name-value pair argument.
```t = 0:seconds(30):minutes(3); y = rand(1,7); plot(t,y,'DurationTickFormat','mm:ss')```
Starting in R2019b, you can display a tiling of plots using the `tiledlayout` and `nexttile` functions. Call the `tiledlayout` function to create a 2-by-1 tiled chart layout. Call the `nexttile` function to create an axes object and return the object as `ax1`. Create the top plot by passing `ax1` to the `plot` function. Add a title and y-axis label to the plot by passing the axes to the `title` and `ylabel` functions. Repeat the process to create the bottom plot.
```% Create data and 2-by-1 tiled chart layout x = linspace(0,3); y1 = sin(5*x); y2 = sin(15*x); tiledlayout(2,1) % Top plot ax1 = nexttile; plot(ax1,x,y1) title(ax1,'Top Plot') ylabel(ax1,'sin(5x)') % Bottom plot ax2 = nexttile; plot(ax2,x,y2) title(ax2,'Bottom Plot') ylabel(ax2,'sin(15x)')```
Define `x` as 100 linearly spaced values between $-2\pi$ and $2\pi$. Define `y1` and `y2` as sine and cosine values of `x`. Create a line plot of both sets of data and return the two chart lines in `p`.
```x = linspace(-2*pi,2*pi); y1 = sin(x); y2 = cos(x); p = plot(x,y1,x,y2);```
Change the line width of the first line to 2. Add star markers to the second line. Use dot notation to set properties.
```p(1).LineWidth = 2; p(2).Marker = '*';```
Plot a circle centered at the point (4,3) with a radius equal to 2. Use `axis equal` to use equal data units along each coordinate direction.
```r = 2; xc = 4; yc = 3; theta = linspace(0,2*pi); x = r*cos(theta) + xc; y = r*sin(theta) + yc; plot(x,y) axis equal```
Input Arguments
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y values, specified as a scalar, a vector, or a matrix. To plot against specific x values you must also specify `X`.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `categorical` | `datetime` | `duration`
x values, specified as a scalar, a vector, or a matrix.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `categorical` | `datetime` | `duration`
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.
Example: `'--or'` is a red dashed line with circle markers
Line StyleDescription
`-`Solid line (default)
`--`Dashed line
`:`Dotted line
`-.`Dash-dot line
MarkerDescription
`o`Circle
`+`Plus sign
`*`Asterisk
`.`Point
`x`Cross
`s`Square
`d`Diamond
`^`Upward-pointing triangle
`v`Downward-pointing triangle
`>`Right-pointing triangle
`<`Left-pointing triangle
`p`Pentagram
`h`Hexagram
ColorDescription
`y`
yellow
`m`
magenta
`c`
cyan
`r`
red
`g`
green
`b`
blue
`w`
white
`k`
black
Target axes, specified as an `Axes` object, a `PolarAxes` object, or a `GeographicAxes` object. If you do not specify the axes and if the current axes are Cartesian axes, then the `plot` function uses the current axes. To plot into polar axes, specify the `PolarAxes` object as the first input argument or use the `polarplot` function. To plot into a geographic axes, specify the `GeographicAxes` object as the first input argument or use the `geoplot` function.
Name-Value Pair Arguments
Specify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside quotes. You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.
Example: `'Marker','o','MarkerFaceColor','red'`
The chart line properties listed here are only a subset. For a complete list, see Line Properties.
Line color, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range `[0,1]`; for example, ```[0.4 0.6 0.7]```.
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (`#`) followed by three or six hexadecimal digits, which can range from `0` to `F`. The values are not case sensitive. Thus, the color codes `'#FF8800'`, `'#ff8800'`, `'#F80'`, and `'#f80'` are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
`'red'``'r'``[1 0 0]``'#FF0000'`
`'green'``'g'``[0 1 0]``'#00FF00'`
`'blue'``'b'``[0 0 1]``'#0000FF'`
`'cyan'` `'c'``[0 1 1]``'#00FFFF'`
`'magenta'``'m'``[1 0 1]``'#FF00FF'`
`'yellow'``'y'``[1 1 0]``'#FFFF00'`
`'black'``'k'``[0 0 0]``'#000000'`
`'white'``'w'``[1 1 1]``'#FFFFFF'`
`'none'`Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
`[0 0.4470 0.7410]``'#0072BD'`
`[0.8500 0.3250 0.0980]``'#D95319'`
`[0.9290 0.6940 0.1250]``'#EDB120'`
`[0.4940 0.1840 0.5560]``'#7E2F8E'`
`[0.4660 0.6740 0.1880]``'#77AC30'`
`[0.3010 0.7450 0.9330]``'#4DBEEE'`
`[0.6350 0.0780 0.1840]``'#A2142F'`
Example: `'blue'`
Example: ```[0 0 1]```
Example: `'#0000FF'`
Line style, specified as one of the options listed in this table.
Line StyleDescriptionResulting Line
`'-'`Solid line
`'--'`Dashed line
`':'`Dotted line
`'-.'`Dash-dotted line
`'none'`No lineNo line
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
Marker symbol, specified as one of the markers in this table. By default, a chart line does not have markers. Add markers at each data point along the line by specifying a marker symbol.
ValueDescription
`'o'`Circle
`'+'`Plus sign
`'*'`Asterisk
`'.'`Point
`'x'`Cross
`'square'` or `'s'`Square
`'diamond'` or `'d'`Diamond
`'^'`Upward-pointing triangle
`'v'`Downward-pointing triangle
`'>'`Right-pointing triangle
`'<'`Left-pointing triangle
`'pentagram'` or `'p'`Five-pointed star (pentagram)
`'hexagram'` or `'h'`Six-pointed star (hexagram)
`'none'`No markers
Example: `'Marker','+'`
Example: `'Marker','diamond'`
Indices of data points at which to display markers, specified as a vector of positive integers. If you do not specify the indices, then MATLAB displays a marker at every data point.
Note
To see the markers, you must also specify a marker symbol.
Example: `plot(x,y,'-o','MarkerIndices',[1 5 10])` displays a circle marker at the first, fifth, and tenth data points.
Example: `plot(x,y,'-x','MarkerIndices',1:3:length(y))` displays a cross marker every three data points.
Example: `plot(x,y,'Marker','square','MarkerIndices',5)` displays one square marker at the fifth data point.
Marker outline color, specified as `'auto'`, an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of `'auto'` uses the same color as the `Color` property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range `[0,1]`; for example, ```[0.4 0.6 0.7]```.
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (`#`) followed by three or six hexadecimal digits, which can range from `0` to `F`. The values are not case sensitive. Thus, the color codes `'#FF8800'`, `'#ff8800'`, `'#F80'`, and `'#f80'` are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
`'red'``'r'``[1 0 0]``'#FF0000'`
`'green'``'g'``[0 1 0]``'#00FF00'`
`'blue'``'b'``[0 0 1]``'#0000FF'`
`'cyan'` `'c'``[0 1 1]``'#00FFFF'`
`'magenta'``'m'``[1 0 1]``'#FF00FF'`
`'yellow'``'y'``[1 1 0]``'#FFFF00'`
`'black'``'k'``[0 0 0]``'#000000'`
`'white'``'w'``[1 1 1]``'#FFFFFF'`
`'none'`Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
`[0 0.4470 0.7410]``'#0072BD'`
`[0.8500 0.3250 0.0980]``'#D95319'`
`[0.9290 0.6940 0.1250]``'#EDB120'`
`[0.4940 0.1840 0.5560]``'#7E2F8E'`
`[0.4660 0.6740 0.1880]``'#77AC30'`
`[0.3010 0.7450 0.9330]``'#4DBEEE'`
`[0.6350 0.0780 0.1840]``'#A2142F'`
Marker fill color, specified as `'auto'`, an RGB triplet, a hexadecimal color code, a color name, or a short name. The `'auto'` option uses the same color as the `Color` property of the parent axes. If you specify `'auto'` and the axes plot box is invisible, the marker fill color is the color of the figure.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range `[0,1]`; for example, ```[0.4 0.6 0.7]```.
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (`#`) followed by three or six hexadecimal digits, which can range from `0` to `F`. The values are not case sensitive. Thus, the color codes `'#FF8800'`, `'#ff8800'`, `'#F80'`, and `'#f80'` are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
`'red'``'r'``[1 0 0]``'#FF0000'`
`'green'``'g'``[0 1 0]``'#00FF00'`
`'blue'``'b'``[0 0 1]``'#0000FF'`
`'cyan'` `'c'``[0 1 1]``'#00FFFF'`
`'magenta'``'m'``[1 0 1]``'#FF00FF'`
`'yellow'``'y'``[1 1 0]``'#FFFF00'`
`'black'``'k'``[0 0 0]``'#000000'`
`'white'``'w'``[1 1 1]``'#FFFFFF'`
`'none'`Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
`[0 0.4470 0.7410]``'#0072BD'`
`[0.8500 0.3250 0.0980]``'#D95319'`
`[0.9290 0.6940 0.1250]``'#EDB120'`
`[0.4940 0.1840 0.5560]``'#7E2F8E'`
`[0.4660 0.6740 0.1880]``'#77AC30'`
`[0.3010 0.7450 0.9330]``'#4DBEEE'`
`[0.6350 0.0780 0.1840]``'#A2142F'`
Marker size, specified as a positive value in points, where 1 point = 1/72 of an inch.
Format for `datetime` tick labels, specified as the comma-separated pair consisting of `'DatetimeTickFormat'` and a character vector or string containing a date format. Use the letters `A-Z` and `a-z` to construct a custom format. These letters correspond to the Unicode® Locale Data Markup Language (LDML) standard for dates. You can include non-ASCII letter characters such as a hyphen, space, or colon to separate the fields.
If you do not specify a value for `'DatetimeTickFormat'`, then `plot` automatically optimizes and updates the tick labels based on the axis limits.
Example: `'DatetimeTickFormat','eeee, MMMM d, yyyy HH:mm:ss'` displays a date and time such as ```Saturday, April 19, 2014 21:41:06```.
The following table shows several common display formats and examples of the formatted output for the date, Saturday, April 19, 2014 at 9:41:06 PM in New York City.
Value of `DatetimeTickFormat`Example
`'yyyy-MM-dd'``2014-04-19`
`'dd/MM/yyyy'``19/04/2014`
`'dd.MM.yyyy'``19.04.2014`
`'yyyy年 MM月 dd日'``2014年 04月 19日`
`'MMMM d, yyyy'``April 19, 2014`
`'eeee, MMMM d, yyyy HH:mm:ss'``Saturday, April 19, 2014 21:41:06`
`'MMMM d, yyyy HH:mm:ss Z'``April 19, 2014 21:41:06 -0400`
For a complete list of valid letter identifiers, see the `Format` property for datetime arrays.
`DatetimeTickFormat` is not a chart line property. You must set the tick format using the name-value pair argument when creating a plot. Alternatively, set the format using the `xtickformat` and `ytickformat` functions.
The `TickLabelFormat` property of the datetime ruler stores the format.
Format for `duration` tick labels, specified as the comma-separated pair consisting of `'DurationTickFormat'` and a character vector or string containing a duration format.
If you do not specify a value for `'DurationTickFormat'`, then `plot` automatically optimizes and updates the tick labels based on the axis limits.
To display a duration as a single number that includes a fractional part, for example, 1.234 hours, specify one of the values in this table.
Value of `DurationTickFormat` Description
`'y'`Number of exact fixed-length years. A fixed-length year is equal to 365.2425 days.
`'d'`Number of exact fixed-length days. A fixed-length day is equal to 24 hours.
`'h'`Number of hours
`'m'`Number of minutes
`'s'`Number of seconds
Example: `'DurationTickFormat','d'` displays duration values in terms of fixed-length days.
To display a duration in the form of a digital timer, specify one of these values.
• `'dd:hh:mm:ss'`
• `'hh:mm:ss'`
• `'mm:ss'`
• `'hh:mm'`
In addition, you can display up to nine fractional second digits by appending up to nine `S` characters.
Example: `'DurationTickFormat','hh:mm:ss.SSS'` displays the milliseconds of a duration value to three digits.
`DurationTickFormat` is not a chart line property. You must set the tick format using the name-value pair argument when creating a plot. Alternatively, set the format using the `xtickformat` and `ytickformat` functions.
The `TickLabelFormat` property of the duration ruler stores the format.
Output Arguments
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One or more chart line objects, returned as a scalar or a vector. These are unique identifiers, which you can use to query and modify properties of a specific chart line. For a list of properties, see Line Properties.
Tips
• Use `NaN` and `Inf` values to create breaks in the lines. For example, this code plots the first two elements, skips the third element, and draws another line using the last two elements:
`plot([1,2,NaN,4,5])`
• `plot` uses colors and line styles based on the `ColorOrder` and `LineStyleOrder` properties of the axes. `plot` cycles through the colors with the first line style. Then, it cycles through the colors again with each additional line style.
Starting in R2019b, you can change the colors and the line styles after plotting by setting the `ColorOrder` or `LineStyleOrder` properties on the axes. You can also call the `colororder` function to change the color order for all the axes in the figure. | 6,311 | 20,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-16 | latest | en | 0.716369 |
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# Electromagnetism
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### Electromagnetism
1. 1. 1. Magnetic Field Around A Wire <ul><li>A wire contains current going into the page as shown. What will the magnetic field look like? </li></ul><ul><li>A wire contains current going right as shown. What will the magnetic field look like? </li></ul>
2. 2. 1. Magnetic Field Around A Wire <ul><li>A wire contains current going into the page as shown. </li></ul><ul><li>A wire contains current going right as shown. </li></ul>
3. 3. 2. Magnetic Field Inside a Solenoid <ul><li>This (“rule 2”) is just an application of Rule 1. </li></ul><ul><li>The fingers follow the current (positive to negative) and the thumb represents the field INSIDE the solenoid. </li></ul>Recall x = field ‘into page’ and a dot = ‘out of page’
4. 4. “ The Trick” <ul><li>Start with your hand ‘clenched’. </li></ul><ul><li>Run your fingers around in the direction of the current. </li></ul><ul><li>Think “ Does it go right or left ” </li></ul><ul><li>Does it go into or out of the page first? </li></ul>http://www.waowen.screaming.net/Maghandrules.htm
5. 5. Solenoid / Permanent Magnet comparison <ul><li>The magnetic field comes OUT of the North pole and goes to the south pole OUTSIDE. </li></ul>http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/imgmag/barsol.gif
6. 6. Therefore <ul><li>Find the direction of the magnetic field. </li></ul><ul><li>Will it attract or repel the permanent magnet? </li></ul>N S + -
7. 7. Therefore <ul><li>Find the direction of the magnetic field. </li></ul><ul><li>Will it attract or repel the permanent magnet? </li></ul>N S + -
8. 8. S N - +
9. 9. S N - +
10. 10. N S - +
11. 11. - + N S - http://www.pschweigerphysics.com/images/rhr2.jpg
12. 12. Magnetic Force on a Current <ul><li>If a wire carries a current through a magnetic field, it creates a force on the wire. The force depends on: </li></ul><ul><li>The current </li></ul><ul><li>The length of the wire </li></ul><ul><li>The strength of the magnetic field. </li></ul><ul><li>The most common application is a motor. It uses lots of wire loops to create more force. Only one wire is shown here. </li></ul>http://www.bbc.co.uk/scotland/education/bitesize/standard/img/physics/electricity/movement/motor.gif
13. 13. Right Hand Slap Rule <ul><li>Three dimensional </li></ul><ul><li>Fingers – magnetic field </li></ul><ul><li>Thumb – current </li></ul><ul><li>Slap Force on the electric current </li></ul>Force Current (+-> -) Current http://commons.wikimedia.org/wiki/File:Right-hand-rule.jpg
14. 14. N S Current into page
15. 15. N S Current into page
16. 16. Another One
17. 17. Another One
18. 18. Another One The current is parallel to the magnetic field, so no force results.
19. 21. Questions <ul><li>Pages 163 (all). </li></ul>
20. 22. New Term Review (not covered with all classes, but good practice if you haven’t seen it) <ul><li>What is a solenoid? What is it used to make? </li></ul><ul><li>Explain the right hand rule (magnetic field around a wire carrying an electric current) and the RHR to determine the magnetic field inside a solenoid. </li></ul><ul><li>How can the strength of an electromagnet be increased? </li></ul><ul><li>Explain the “motor effect” (magnetic force on a current). You should explain how to determine the direction of the force in relation to the magnetic field and current direction. </li></ul><ul><li>Explain how to increase the strength on a current-carrying wire in a magnetic field. How do motors achieve strong forces? </li></ul>
21. 23. Magnetic Forces Magnetic Fields between Permanent Magnets: http://www.acecrc.sipex.aq/access/page/?page=75ee44de-b881-102a-8ea7-0019b9ea7c60
22. 24. What causes the motor effect? As explained in class, the field lines interact with each to produce a pattern like that shown below right. Reminder: field lines can never cross. The stronger magnetic field above pushes the wire downwards. http://en.wikibooks.org/wiki/GCSE_Science/The_motor_effect
23. 25. Another One As explained in class, note that the field lines interact with each other but do not look like this, as field lines cannot cross. The stronger magnetic field on the left pushes the wire to the right.
24. 26. The DC Motor <ul><li>A DC motor applies the motor effect (strangely) to make wire spin inside a magnetic field. </li></ul><ul><li>A commutator is necessary to keep it spinning in one direction. </li></ul><ul><li>The force on the wire (and therefore the torque) changes as the position changes. </li></ul><ul><li>Usually many loops are used (coils) to increase the force; this is necessary for a real-world motor to work. </li></ul><ul><li>A motor can (usually) be used as a generator if forced around - more on this soon. </li></ul><ul><li>An AC motor is quite different (and not to be covered now!). </li></ul>
25. 27. The DC Motor http://www.walter-fendt.de/ph11e/electricmotor.htm http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=912.0
26. 28. Problems <ul><li>Electric motor page 165 </li></ul>
27. 29. Electromagnetic Induction <ul><li>If a wire is moved through a magnetic field, the magnetic field ‘pushes’ the electrons around the circuit. This creates a voltage and (if a circuit is connected) a current. </li></ul><ul><li>The voltage (and current) can be increased by increasing the length of the wire, the speed of the movement or the strength of the magnetic field. </li></ul><ul><li>The wire must ‘cut’ magnetic field lines, or else no voltage will be produced. </li></ul>
28. 30. Right Hand Rule <ul><li>The diagram showed that: </li></ul><ul><li>FINGERS = magnetic field </li></ul><ul><li>THUMB = movement </li></ul><ul><li>Slap = induced voltage, or the direction conventional current (positive to negative) will flow if a circuit is connected. </li></ul>
29. 31. Electromagnetic Induction Example Conductor (normally wire) v The “positive charges” move up and the negative charges move down. This generates a voltage difference across the wire. + -
30. 32. Electromagnetic Induction Example Calculate the current in the circuit. The “positive charges” moved anticlockwise in the animation. 10 Ω + + + + +
31. 33. Motor Effect and Electromagnetic Induction <ul><li>MOTOR EFFECT </li></ul><ul><li>Fingers = Field </li></ul><ul><li>Thumb = current </li></ul><ul><li>Slap = Force </li></ul><ul><li>EM INDUCTION </li></ul><ul><li>Fingers = Field </li></ul><ul><li>Thumb = movement / force </li></ul><ul><li>Slap = induced Voltage (EMF) </li></ul>Cause Effect Always Magnetic Field http://commons.wikimedia.org/wiki/File:Right-hand-rule.jpg
32. 34. Another One
33. 35. Another One The current is parallel to the magnetic field, so no force results.
34. 37. Generators and Alternators <ul><li>A DC motor produces direct current (electricity which flows in one direction only) when it is turned, however the flow is uneven. </li></ul><ul><li>Most generators produce alternating current, which flows backwards and forwards. The frequency is the number of times the current flows each way per second, which is determined by the number of times the generator turns per second. These generators are called alternators. </li></ul>
35. 38. Japan’s Electricity <ul><li>Mains electricity is A.C. </li></ul><ul><li>Some countries use 50HZ (eg ________). </li></ul><ul><li>Other countries use 60Hz (eg _________). </li></ul><ul><li>Japan is unusual because it uses two frequencies: 50Hz in Tokyo (and north) and 60Hz in Osaka (and south). </li></ul><ul><li>The reason is historical: generators were purchased from the US (60Hz) and Europe (50HZ) in the nineteenth century before the entire country was connected to the mains electricity (known as the ‘national grid’). </li></ul>
36. 39. Questions <ul><li>Page 169 & 171 </li></ul>
37. 40. Transformers <ul><li>An electromagnet produces a m_______ f______. </li></ul><ul><li>If the voltage is increased, the current increases, which in turn increases the magnetic field. This has the same effect as moving a magnet into the coil. </li></ul><ul><li>If the voltage – and therefore the current – is changed, the magnetic field changes. This changing magnetic field can induce a voltage – and therefore a current – in another (“secondary”) coil of wire if the magnetic field flows through it as well. </li></ul><ul><li>If the number of turns in the primary and secondary coils are different, the voltages in each coil will be different. Transformers use two coils with different numbers of turns to transform voltage. Transformers only work with _______. </li></ul>
38. 41. Transformer Diagram <ul><li>Ignore the writing – unless you can read it! Please note that a transformer is made of two coils with different numbers of turns, connected by an iron (why?) core. </li></ul>http://commons.wikimedia.org/wiki/File:750px-Transformator.png
39. 42. Transformer Equation <ul><li>Example: A transformer consisting of a primary coil of wire with 100 turns and a secondary coil of 25 turns is connected to the Japanese mains voltage (100V). Calculate the voltage of the secondary coil. </li></ul>
40. 43. <ul><li>A transformer with 50 turns in the primary (input) coil and 200 turns in the secondary (output) coil is connected to a 1.5V (AA) battery (like the one in a remote control). Calculate the output voltage. </li></ul><ul><li>This was a trick question (and practice at the equation). The answer is 0V, since a battery produces DC and transformers need AC. </li></ul><ul><li>A laptop charger has an output of 16.5V. If the output coil has 2000 turns (probably unrealistic), how many turns will the input coil need in: a) Japan (100V) b) Europe (230V) </li></ul>
41. 44. Free Power? <ul><li>Most transformers lose some power, however if the loss is n egligible , then the input power must be the same as the output power. </li></ul><ul><li>Input * input = output * output voltage current voltage current </li></ul><ul><li>Determine the ratio of the currents in the input and output coils in relation to the number of turns in each coil. </li></ul>
42. 45. Bookwork <ul><li>Answer problems page 173 & 175 (read 174-5). </li></ul><ul><li>Chapter Summary Pages 178-179 (everything). </li></ul>
43. 46. Thermal Power Stations <ul><li>Photo of a thermal power station in AiChi (not necessary): </li></ul><ul><li>http://www.thedailystar.net/newDesign/photo_gallery.php?pid=127029 </li></ul>
44. 47. Thermal Power Stations <ul><li>A thermal power station uses steam to turn a turbine, which then turns a generator to produce electricity. Most produce the steam from burning coal, oil and natural gas, but some also waste incineration or bio-fuel. Nuclear power stations use heat from nuclear reactions and geothermal power stations use heat from the Earth. </li></ul><ul><li>Most thermal power stations produce over 1000MW. The Hekinan Power Station (Aichi) produces 4100MW of power. </li></ul>
45. 48. Diagram <ul><li>A simplified diagram of a thermal power station: (note that this one burns gas, however it could burn coal, oil or biomass) http://www.bbc.co.uk/scotland/learning/bitesize/standard/physics/energy_matters/generation_of_electricity_rev1.shtml </li></ul>
46. 49. Efficiency and Waste Heat <ul><li>Laws of Physics (IB) limit the efficiency of thermal power stations; most are between 33% and 48% efficient. The rest of the energy is converted to heat. </li></ul><ul><li>This heat can be used for industrial purposes (eg??), to heat homes, to desalinate seawater, or it can be ‘dumped’ into a source of water or through a cooling tower. </li></ul>
47. 50. This diagram from Tepco is a little more complicated than is needed at IGCSE, but still worth viewing. Note the transformer to increase the voltage. http ://www.tepco.co.jp/en/challenge/energy/thermal/power-g- e.html
48. 51. An example of a coal-fired power station with cooling towers, since there isn’t a water supply to absorb the waste heat: http ://forums.xkcd.com/viewtopic.php?f=8&t=32541&start= 280
49. 52. Pros & Cons of Fossil Fuels <ul><li>Advantages of Fossil Fuels Disadvantages of Fossil Fuels </li></ul>
50. 53. Pros & Cons of Nuclear Power <ul><li>Advantages of Nuclear Power Disadvantages of Nuclear Power </li></ul>
51. 54. The Global Energy Crisis <ul><li>The developed world – and our lifestyle – is built on an abundant supply of cheap energy. </li></ul><ul><li>Our energy started with coal during the industrial revolution, and then last century oil and gas became more popular. </li></ul><ul><li>Pollution and climate change, along with the developing third world, are leading people to question how we live and use energy. </li></ul>
52. 55. Solution to the Energy Crisis <ul><li>The problem will be discussed in the coming unit, but includes pollution (including heavy metals such as mercury from coal), political issues, climate change, risk of nuclear accidents and disposal of nuclear waste. The solution/s is/are A) Stop using nuclear and burn more fossil fuels. B) Stop using fossil fuels and use more modern, cleaner, safer nuclear. C) A combination of A and B D) Reduce our energy consumption so we can use only renewable energy. We will discuss this option more in the next unit. </li></ul>
53. 56. Questions <ul><li>Page 63 (all) </li></ul><ul><li>Finish EM Revision worksheet from PHETs and save them to your shared Google folder (go to “collections shared with me” then find yoursurname yourfirstname Physics ) </li></ul><ul><li>If time permits, read 64-65 and answer questions on page 65. </li></ul> | 3,668 | 13,589 | {"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-2017-47 | latest | en | 0.660084 |
https://www.physicsforums.com/tags/flat/ | 1,712,999,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00670.warc.gz | 907,908,858 | 28,153 | What is Flat: Definition and 495 Discussions
A Flat is a Hindi thriller film, directed by Hemant Madhukar and produced by Anjum Rizvi.The film was released on 12 November 2010 under the Anjum Rizvi Film Company and Y.T Entertainment Ltd. banners.
View More On Wikipedia.org
1. B What do people mean when they say “space is flat?”
Please excuse my ignorance I hear some people say that space is flat. What exactly do they mean by this?
2. Freon filled balloons go flat QUICKLY! Why?
Years ago, I filled some latex party balloons with Freon 12. I haven't repeated the experiment with the Freon 134 that is available nowadays. It was fun to throw them around because Freon is so much heavier than air--you can throw them across the room and they land on the floor and stay there...
3. B How can the Universe be flat? Ridiculous
The earth is a sphere. I can stand anywhere on its surface and look up directly above my head at the night sky and see lots of stars many light years away. I see this view wherever I stand on the earth's surface although the stars will be different ones, depending on where I'm standing...
4. POTW Flat Modules and Intersection
Let ##M## be a flat module over a commutative ring ##A##. Suppose ##X_1## and ##X_2## are submodules of an ##A##-module ##X##. Prove that ##(X_1 \cap X_2) \otimes_A M = (X_1 \otimes_A M) \cap (X_2 \otimes_A M)## as submodules of ##X\otimes_A M##. | 374 | 1,405 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-18 | latest | en | 0.917396 |
https://socratic.org/questions/how-do-you-solve-14-7y-3y-14 | 1,582,201,151,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144722.77/warc/CC-MAIN-20200220100914-20200220130914-00155.warc.gz | 577,798,072 | 6,144 | # How do you solve 14+7y=3y+14?
Dec 14, 2016
$y = 0$
#### Explanation:
Step 1) Isolate the $x$ terms on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:
$14 - 14 + 7 y - 3 y = 3 y - 3 y + 14 - 14$
$0 + 7 y - 3 y = 0 + 14 - 14$
$7 y - 3 y = 14 - 14$
Step 2) Combine like terms:
$\left(7 - 3\right) y = 0$
$4 y = 0$
Step 3) Solve for $y$ while keeping the equation balanced:
$\frac{4 y}{4} = \frac{0}{4}$
$\frac{\cancel{4} y}{\cancel{4}} = 0$
$y = 0$ | 215 | 527 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "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} | 4.5 | 4 | CC-MAIN-2020-10 | longest | en | 0.62291 |
https://passemall.com/question/accuplacer/3x--2y--15-x--3the-two-lines-given-by-5310700457033728/ | 1,701,919,991,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100632.0/warc/CC-MAIN-20231207022257-20231207052257-00868.warc.gz | 515,863,616 | 21,249 | Scan QR code or get instant email to install app
Question:
# 3x – 2y = 15 x = 3The two lines given by the equations above intersect in the xy-plane. What is the value of the y-coordinate of the point of intersection?
A -3
explanation
Substituting 3 for x in the first equation gives 3(3) − 2y = 15. This simplifies to 9 − 2y = 15. Subtracting 9 from both sides of 9 − 2y = 15 gives −2y = 6. Finally, dividing both sides of −2y = 6 by −2 gives y = −3. | 149 | 454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-50 | latest | en | 0.862428 |
https://www.enotes.com/homework-help/how-many-grams-copper-ii-fluoride-needed-make-6-7-252143 | 1,643,412,224,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00505.warc.gz | 787,775,726 | 18,109 | # How many grams of copper (II) fluoride are needed to make 6.7 liters of a 1.2 M solution?
The concentration of a solution can be expressed in many ways, including grams per liter, etc. When the concentration is expressed in terms of moles per liter and the solution has one mole of the solute in one liter of the solution its concentration is 1M.
A solution with a...
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The concentration of a solution can be expressed in many ways, including grams per liter, etc. When the concentration is expressed in terms of moles per liter and the solution has one mole of the solute in one liter of the solution its concentration is 1M.
A solution with a concentration in terms of molarity of 1.2 M can be made by dissolving 1.2 moles of the solute in every liter of the solution.
To make 6.7 liters of solution of copper (II) fluoride with molarity 1.2 we need 1.2*6.7 = 8.04 moles of copper (II) fluoride.
The molecular mass of copper (II) fluoride is 101.543 g/mole.
The mass of 8.04 moles of copper (II) fluoride is 101.543*8.04 = 816.4 g.
To make 6.7 liters of a 1.2 M solution of copper (II) fluoride, 816.4 g of copper (II) fluoride is required.
Approved by eNotes Editorial Team | 342 | 1,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-05 | latest | en | 0.93876 |
http://www.chegg.com/homework-help/questions-and-answers/question-determine-themagnetic-flux-square-loop-sideis-parallel-distance-straight-wirethat-q116381 | 1,386,382,386,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163052995/warc/CC-MAIN-20131204131732-00036-ip-10-33-133-15.ec2.internal.warc.gz | 274,684,899 | 7,207 | Find Magnetic Flux
0 pts ended
Question: Determine themagnetic flux through a square loop of side a if one sideis parallel to, and a distance a from, a straight wirethat carries a current I.
I am assuming that B = (
μ0Ienc)/(2πr)....What I don'tunderstand is why this isn't simply Φ = B*Asince A would be a2 ... And amI supposed to use the details dr and r? Or are they just thrown into confuse me? Do I need to integrate even if the B produced by Iis perpendicular to the loop ( and therefore cos(θ) =1? | 133 | 507 | {"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-2013-48 | latest | en | 0.92914 |
https://www.physicsforums.com/threads/pacemaker-and-wire-with-current.245526/ | 1,477,533,130,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721027.15/warc/CC-MAIN-20161020183841-00481-ip-10-171-6-4.ec2.internal.warc.gz | 966,936,724 | 16,900 | # Pacemaker and Wire with Current
1. Jul 17, 2008
### myersb05
A cardiac pacemaker can be affected by a static magnetic field as small as 1.7 mT. How close can a pacemaker wearer come to a long, straight wire carrying 27 A?
I know that F=ILBsin(theta). I attempted to solve using I=27, L as unknown, F=1.7, and Bwire=mupi*I/2pir with an unknown r that would be equal to the L.
I don't think there is a need to convert the 1.7mT to Teslas because my answer should be in mm.
2. Jul 17, 2008
### alphysicist
Hi myersb05,
There's no need to deal with the force equation; the question just asks at what distance is the field of a long straight wire equal to 1.7mT.
I think you will need to convert to T, since B is in the numerator and r is in the denominator.
Also, it might just be a typo, but your B equation does not look right. It should be:
$$B = \frac{\mu_0 I}{2\pi r}$$
3. Jul 17, 2008
### myersb05
Worked perfectly. Thanks again Al. That was the same problem I had last night. My equation was written improperly in my notes. I am stuck on another one if you have time.
Two long parallel conductors separated by 14.0 cm carry currents in the same direction. The first wire carries a current I1 = 3.00 A, and the second carries I2 = 8.00 A. (Assume the conductors lie in the plane of the page.)
(a) What is the magnitude of the magnetic field created by I1 and at the location of I2?
(b) What is the force per unit length exerted by I1 on I2?
C and D are the opposites of those.
I found the F/L by using uI1I2/2pid. So that came out as .000034 which is correct and applies to both wires.
I attempted to use Bwire=u8.0/2pi.14=.000011 and that was incorrect. I tried the same for the 3 amp wire and got .000004 which is also incorrect.
4. Jul 17, 2008
### alphysicist
Are you saying that 0.000004T is not the correct answer for part a? That's what I am getting; maybe you need to keep more digits (that results is roughly 6% to 7% or so off from the non-rounded answer), or maybe they want different units?
5. Jul 17, 2008
### myersb05
WebAssign is telling me that .000004 is close but that I should keep more digits. I can't come up with an answer that is any less rounded than .000004.
6. Jul 17, 2008
### alphysicist
You found that result by calculating,
$$\frac{4\pi \times 10^{-7} (3)}{2 \pi (0.14)}$$
right? The factors of $\pi$ will cancel; do you mean your calculator only gives you one digit?
I was thinking you rounded some answers to get your result, and to get the right answer you should not round at all.
7. Jul 17, 2008
### myersb05
Yea, that is exactly what I typed into my TI-89 and I got exactly .000004
8. Jul 17, 2008
### alphysicist
Well let's try to factor it out some:
$$\frac{4\pi \times 10^{-7} (3)}{2 \pi (0.14)}\to \frac{4 \times 10^{-7} (3)}{2 (0.14)} \to \frac{6}{.14} \times 10^{-7}$$
So do 6/.14, and then add on the exponential part after you have more than one digit. Does that give the right answer?
9. Jul 17, 2008
### alphysicist
I didn't even notice that you had a TI-89. Here's what you can do:
As soon as you get the answer of 0.000004, hit the up arrow, and then hit enter. That will bring the value down to the input line, with all the digits you need.
(You can also go into the mode menu and set it to show differing numbers of digits, but the above is what I usually do.)
10. Jul 17, 2008
### myersb05
Got it, thanks again Al
11. Jul 17, 2008 | 1,033 | 3,432 | {"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} | 3.5625 | 4 | CC-MAIN-2016-44 | longest | en | 0.966057 |
https://en.m.wikipedia.org/wiki/Scrub_radius | 1,606,512,935,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194171.48/warc/CC-MAIN-20201127191451-20201127221451-00089.warc.gz | 286,320,270 | 10,818 | The scrub radius is the distance in front view between the king pin axis and the center of the contact patch of the wheel, where both would theoretically touch the road. It could be positive, negative or zero.
The kingpin axis is the line between the upper and lower ball joints of the hub. On a MacPherson strut, the top pivot point is the strut bearing, and the bottom point is the lower ball joint. The inclination of the steering axis is measured as the angle between the steering axis and the centerline of the wheel. This means that if the camber angle is adjustable within the pivot points the scrub radius can be changed, this alters the width and offset of the tires on a vehicle.
If the kingpin axis intersection point is outboard of the center of the contact patch, it is negative; if inside the contact patch, it is positive. The term scrub radius derives from the fact that either in the positive or negative mode, the tire does not turn on its centerline (it scrubs the road in a turn) and due to the increased friction, more effort is needed to turn the wheel.
Large positive values of scrub radius, 4 inches/100 mm or so, were used in cars for many years. The advantage of this is that the tire rolls as the wheel is steered, which reduces the effort when parked, provided you're not on the brake.
The advantage of a small scrub radius is that the steering becomes less sensitive to braking inputs. More scrub radius adds to road feel by pushing the inside wheel into the ground.
An advantage of a negative scrub radius is that the geometry naturally compensates for split μ (mu) braking, or failure in one of the brake circuits. It also provides center point steering in the event of a tire deflation, which provides greater stability and steering control in this emergency.
## Steering axis inclination
The steering axis inclination (SAI) is the angle between the centerline of the steering axis and vertical line from center contact area of the tire (as viewed from the front).
### Effects of SAI
SAI urges the wheels to a straight ahead position after a turn. By inclining the steering axis inward (away from the wheel), it causes the spindle to rise and fall as the wheels are turned in one direction or the other. Because the tire cannot be forced into the ground as the spindle travels in an arc, the tire/wheel assembly raises the suspension and thus causes the tire/wheel assembly to seek the low (center) return point when it is allowed to return. Thus, since it has a tendency to maintain or seek a straight ahead position, less positive caster is needed to maintain directional stability. SAI adds positive camber while turning for both steering tires. A vehicle provides stable handling without any of the drawbacks of high positive caster because of SAI.
The scrub radius is the distance at the road surface between the tire center line and the SAI line extended downward through the steering axis.
## Squirm
Squirm occurs when the scrub radius is at zero. When the pivot point is in the exact center of the tire footprint, this causes scrubbing action in opposite directions when the wheels are turned. Tire wear and some instability in corners is the result.
### Applications in suspension
MacPherson strut equipped vehicles usually have a negative scrub radius. Even though scrub radius in itself is not directly adjustable, it will be changed if the upper steering axis point or spindle angle is changed when adjusting camber. This is the case on a MacPherson strut which has the camber adjustment at the steering knuckle. Because camber is usually kept within 1/4° side to side, the resulting scrub radius difference is negligible.
Negative scrub radius decreases torque steer and improves stability in the event of brake failure. SLA suspensions usually have a positive scrub radius. With this suspension, the scrub radius is not adjustable. The greater the scrub radius (positive or negative), the less the steering effort is. When the vehicle has been modified with offset wheels, larger tires, height adjustments and side to side camber differences, the scrub radius will be changed and the handling and stability of the vehicle will be affected.
## References
1. ^ Chassis handbook : fundamentals, driving dynamics, components, mechatronics, perspectives / Bernd HeiÇing, Metin Ersoy (ed.). page 25
• Reimpell, Jornsen; Helmut Stoll; Jurgen W. Betzler. The Automotive Chassis Engineering Principles. SAE International. ISBN 978-0-7680-0657-5. | 940 | 4,499 | {"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-2020-50 | latest | en | 0.936956 |
https://normgoldblatt.com/solving-system-of-equations-using-matrix-23 | 1,675,686,668,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500339.37/warc/CC-MAIN-20230206113934-20230206143934-00804.warc.gz | 446,341,262 | 6,427 | # Solving system of equations using matrix
Matrix Method is used to find the solution of the system of the equations. In the equations, all of the variables should be written in the proper order. On the appropriate sides
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## Algebra 2 solving 2x3 matrices
Solving Systems of Equations by Matrix Method Matrix Method for solving systems of equations is also known as Row Echelon Method. The matrix method is similar to the method of
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Mathematical equations can be quite difficult to solve, but with a little perseverance and patience, anyone can do it! | 506 | 2,331 | {"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-2023-06 | latest | en | 0.923 |
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## Comparing Decimals Lesson Plan: Numbers and Operations- Fractions
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## Mood and Tone Lesson Plan: Craft and Structure
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*Click to open and customize your own copy of the Mood and Tone Lesson Plan. This lesson accompanies the BrainPOP topic Mood and Tone, and supports the standard of analyzing the impact of specific...
## I vs. Me Lesson Plan: Conventions of Standard English
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*Click to open and customize your own copy of the I vs. Me Lesson Plan. This lesson accompanies the BrainPOP topic I vs. Me and supports the standard of ensuring subject-verb and pronoun-anteceden...
## Rounding Lesson Plan: Numbers and Operations in Base Ten
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## Multi-Digit Multiplication Lesson Plan: Numbers and Operations in Base Ten
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## Computer Programming Lesson Plan: Computational Thinking
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*Click to open and customize your own copy of the Computer Programming Lesson Plan. This lesson accompanies the BrainPOP Jr. topic Computer Programming, and supports the standard of breaking probl... | 649 | 3,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.804453 |
https://www.mlodysamorzad.pl/complete/8846/how-to-determine-the-capacity-of-an-impact-crushers.html | 1,624,327,173,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00142.warc.gz | 812,131,492 | 5,989 | ## how to determine the capacity of an impact crushers
Tips to maximize crushing efficiency, May 13, 2019· Feeding impact crushers Size reduction in an impact crusher relies on energy being conveyed into the rock from the rotor, and it begins with your feed The initial impact is responsible for more than 60 percent of the crushing action, with the remainder made up of impact against an adjustable breaker bar and a small amount of inter-particle ,Impact Crusher Working Principle, Starting from the base working principle that compression is the forcing of two surfaces towards one another to crush the material caught between them Impact crushing can be of two variations: gravity and dynamic An example of gravity impact would be dropping a rock onto a steel plate (similar to what goes on into an Autogenous Mill) Dynamic impact could be described as material dropping ,Mathematical modeling of a vertical shaft impact crusher ,, However, the simulation behaviour of the crushing process is possible to predict the stress applied, know the operation performance of the machine and determine the wear of the crusher [20,21,22 ,Estimate Jaw Crusher Capacity, Metallurgical ContentCapacities and Horsepower of Jaw Crusher (tons/hr)Capacities of a Gyratory Crushers (tons/hr)Typical Capacities of Twin-Roll Crushers (tons/hr)Typical Capacities of Cone CrushersTypical Capacities of Hammermills Example capacity ,How to Calculate Production Capacity of a Factory?, Sep 10, 2018· Calculate factory production capacity separately as above You can increase /change the number of machines for the products as per requirement If you get an order of multiple items at the same time, you calculate product wise capacity in pieces according to ,.
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DESIGN AND ANALYSIS OF A HORIZONTAL SHAFT IMPACT ,, An impact crusher can be further classified as Horizontal impact crusher (HSI) and vertical shaft impact crusher (VSI) based on the type of arrangement of the impact rotor and shaft Horizontal shaft impact crusher These break rock by impacting the rock with hammers/blow bars that are fixed upon the outer edge of a spinning rotor(PDF) Modelling of output and power consumption in ,, The vertical shaft impact (VSI) crusher is a commonly-used machine in aggregate production A comprehensive understanding of the physical phenomena that ,Impact Crusher & Cement Crusher Manufacturer | Stedman ,, Granite, for example, is friable but too hard, making impact crushing more expensive than other types of crushing Stedman Machine has extensive experience crushing a variety of materials, a testing facility with full-sized equipment, and more than 10,000 test reports to help you determine what type of equipment is best for your needswith the high capacity of impact crushers!, Crushers terrasource terrasource FT/FTE Flextooth ® Crushers plates are designed to provide the greatest amount of free open area, allowing the machine to obtain the highest capacity while accurately sizing the material to the specified size Our screen grates are made of abrasion-resistant materials that canhow to calculate impact crusher capacity, NP Series impact crushers - capacity NP series impact crushers achieve a higher reduction with fewer crushing stages, lowering , and find ways to enhance the quality, ease of use, and,.
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How to Determine the Aggregate Crushing Strength Value in ,, Apr 26, 2018· Apparatus needed for finding the crushing strength of aggregate Steel cylinder consisting of open ends at both sides having an internal diameter of 152mm Cylindrical measure to measure the quantity of aggregates needed for the test Tampering rod; A balance of capacity 3kgJaw Crusher, To determine the capacity of jaw and gyratory crushers, Broman [10] divided the crusher chamber into different sections and determined the volume of each section in terms of the angle that the moving jaw subtended with the vertical Broman suggested that the capacity per stroke crushed in each section would be a function of the top surface and ,how to determine the capacity of a crusher, 1 Introduction Jaw crushers are widely used crushing machines , One of the trends in design solutions of crushers ensuring the crushing ratio of about 30 involves the , crushing capacity can be determined accordinglyHow to determine Aggregate Impact Value || Aggregate ,, Sep 20, 2018· This video (Animation, Animated Video) explains how to determine aggregate impact value (Aggregate Impact Test) Aggregate test || Coarse aggregate Test (Civ,TECHNICAL NOTES 5 CRUSHERS, The capacity of the crusher is determined by its size The gape determines the maximum size of material that can be accepted Maximum size that can be accepted into the crusher is approximately 80% of the gape Jaw crushers are operated to produce a size reduction ratio between 4:1 and 9:1 Gyratory crushers.
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how do i calculate blow bar wear life in impact crushers, Crusher Wikipedia Impact crusher Impact crushers involve the use of impact rather than pressure to crush material The material is contained within a cage, with openings on the bottom, end, or side of the desired size to allow pulverized material to escape There are two types of impact crushers horizontal shaft impactor and vertical shaft ,Impact Crusher, The three types of crushers most commonly used for crushing CDW materials are the jaw crusher, the impact crusher and the gyratory crusher (Figure 44)A jaw crusher consists of two plates, with one oscillating back and forth against the other at a fixed angle (Figure 44(a)) and it is the most widely used in primary crushing stages (Behera et al, 2014)Choosing the right crusher, impact crusher Impact crushers can be used as primary crushers, secondary crushers or tertiary crushers depending on the size and technology They are equipped with beaters, also called hammers, and impact plat They operate in the following way: they are fed from the upper part, then the stones are hit by the hammers and projected towards the platGyratory and Cone Crusher, Jan 01, 2016· The open-side setting of the discharge opening was 102 cm The rate of gyration was 175 per minute Calculate 1 the capacity of crusher for an eccentric throw of 16 cm, 2 what would the eccentric throw be if the crusher capacity had to be increased to 600 t/h at the same setting, 3how to determine the capacity of an impact thickeners, how to determine the capacity of an impact crushers 2020331how to determine the capacity of an impact crushers Capacity building or capacity development is the process by which individuals and organizations obtain improve and retain the skills knowledge tools equipment and other resources needed to do their jobs allows individuals and organizations to perform at a greater capacity larger ,.
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Ow To Calculate Impact Crusher Capacity, How To Determine Impact Crusher Capacity How to calculate impact crusher capacity crushers all crusher types for your reduction needs the primary impact crusher offers high capacity and is designed to accept large feed sizes the primary impact crushers are used to process from 200 th up to 1900 th and feed sizes of up to 1830 mm 71 in the largest model primary impact crushers are ,
## Leave Your Needs
Dear friend, please fill in your message if you like to be contacted. Please note that you do not need to have a mail programme to use this function. ( The option to mark ' * ' is required ) | 1,490 | 7,455 | {"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-2021-25 | longest | en | 0.902282 |
https://www.manchester.ac.uk/study/undergraduate/courses/2021/00499/bsc-mathematics-with-financial-mathematics/course-details/MATH37002 | 1,621,370,621,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991514.63/warc/CC-MAIN-20210518191530-20210518221530-00277.warc.gz | 847,448,309 | 11,601 | Coronavirus information for applicants and offer-holders
We understand that prospective students and offer-holders may have concerns about the ongoing coronavirus outbreak. The University is following the advice from Universities UK, Public Health England and the Foreign and Commonwealth Office.
# BSc Mathematics with Financial Mathematics / Course details
Year of entry: 2021
## Course unit details:Martingales with Applications to Finance
Unit code MATH37002 10 Level 3 Semester 2 Department of Mathematics No
### Overview
An introduction to a circle of ideas and fundamental results of the theory of martingales, which play a vital role in stochastic calculus and in the modern theory of finance.
### Pre/co-requisites
Unit title Unit code Requirement type Description
Probability 2 MATH20701 Pre-Requisite Compulsory
Students are not permitted to take more than one of MATH37001 or MATH47201 for credit in the same or different undergraduate year.
Students are not permitted to take MATH47201 and MATH67201 for credit in an undergraduate programme and then a postgraduate programme.
Note that MATH67201 is an example of an enhanced level 3 module as it includes all the material from MATH37001
When a student has taken level 3 modules which are enhanced to produce level 6 modules on an MSc programme taken within the School of Mathematics, then they are limited to a maximum of two such modules (with no alternative arrangements available otherwise)
### Aims
To provide a firm grasp of a range of basic concepts and fundamental results in the theory of martingales and to give some simple applications in the rapid developing area of financial mathematics.
### Learning outcomes
On successful completion of this course unit students will be able to:
• compute integrals of simple random variables with respect to a probability measure and apply the dominated, monotone convergence theorems;
• use the computational rules of conditional expectation and prove that a given stochastic process is a martingale;
• apply Doob Optional Sampling Theorem to calculate interesting probabilities concerning some practical problems, e.g. the gambler ruin problem;
• determine when a financial market is arbitrage-free;
• find out a replicating strategy for a financial claim;
• calculate the fair price of a financial claim in a financial market.
### Syllabus
• Probability spaces, events, Ï'-fields, probability measures and random variables. Integration with respect to a probability measure. Convergence theorems (dominated, monotone and Fatou). [5]
• Conditional expectations. Fair games and martingales, submartingales and supermartingales. Doob decomposition theorem. Stopping times and the optional sampling theorem. The upcrossing inequality and the martingale convergence theorem. The Doob maximal inequality and the martingale modification theorem. [13]
• Applications. Discrete time random models in financial markets. Price processes, self-financing portfolio and value processes. Arbitrage opportunities and equivalent martingale measures. Completeness of the markets. Options and option pricing. [6]
### Assessment methods
Examination: weighting 100%
### Feedback methods
Feedback tutorials will provide an opportunity for students' work to be discussed and provide feedback on their understanding. Coursework or in-class tests (where applicable) also provide an opportunity for students to receive feedback. Students can also get feedback on their understanding directly from the lecturer, for example during the lecturer's office hour.
Further Reading: O. Kallenberg, Foundations of Modern Probability, Springer-Verlag, 2001.
Essential: N. H. Bingham and R. Kiesel, Risk-Neutral Valuation, Springer-Verlag, 1998.
Recommended: D. Williams, Probability with Martingales, Cambridge Univ. Press, 1991.
Essential: A. N. Shiryaev, Probability, Springer-Verlag, 1996.
### Study hours
Scheduled activity hours
Lectures 22
Tutorials 11
Independent study hours
Independent study 67
### Teaching staff
Staff member Role
Tusheng Zhang Unit coordinator | 847 | 4,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-21 | latest | en | 0.878183 |
https://socialcar-project.eu/ro/va-rog-ex-7-puctul-c-si-de-ear-10-tot-in-nafara-de-a-va-rog-urgent-dau-coroana-plus-20puc.14879902.html | 1,638,595,410,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00365.warc.gz | 547,900,244 | 11,665 | PaulSirbu
1
# Va rog ex 7 puctul c si de ear 10 tot in nafara de a va rog urgent dau Coroana plus +20puc
(2) Răspunsuri
miualin85
7 c) x²+10x+25=(x+5)² x²+6x+9=(x+3)² x²-x-12=x²-4x+3x-12=x(x-4)+3(x-4)=(x+3)(x-4) x²+2x-15=x²+5x-3x-15=x(x+5)-3(x+5)=(x-3)(x-5) (x+5)²/(x+3)²•(x+3)(x-4)/(x-3)(x-5)=(x+5)²/(x+3)•(x-4)/(x-3)(x-5)
avatar245743
gata cred ca asa se face | 219 | 366 | {"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-2021-49 | latest | en | 0.167409 |
https://dsp.stackexchange.com/questions/21765/manual-extraction-of-filter-phase-response-with-matlab-using-sweep-sine/21767 | 1,620,513,714,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00046.warc.gz | 222,481,748 | 39,086 | Manual Extraction of Filter Phase Response with MATLAB using sweep sine
I'm currently at a loss for why my code seems to not be working. I'm doing a sine sweep on a filter to calculate its freq and phase response. The frequency response checks out, however the phase response doesn't look right when compared to MATLABs freqz function. Plus since I know the filter is a comb, I can tell its wrong as well.
Here's what I've got so far.
% Filter Coefficients
b = [1];
a = [1, 0, 0, 0, 0, 0, 0, 0, 0, -0.96];
% sweep is my sine wave that goes from 0->fs/2
filtered_sweep = filter(b, a, sweep);
% This section correctly Grabs and plots the Freq Response
fft_sweep = fft(filtered_sweep);
freq_response = abs(fft_sweep);
freq_response = 20*log10(freq_response/1); % Convert to dB scale
freq_response = freq_response(1:end/2); % Cut to pi fs/2
figure;
plot(freq_response);
And Now here is where I try and get the Phase two different ways with the same wrong result. From what I understand to get the phase you just need the angle of the complex values. In MATLAB I thought angle() should work, but just to be safe I tried to manually compute the angle myself using atan with the imag and real components.
% Pre Allocate Vector
% Manual Angle Extraction without using Angle()
phase = zeros(1, length(fft_sweep));
for i=1:length(fft_sweep)
phase(i) = atan2(imag(fft_sweep(i)), real(fft_sweep(i)));
end
% Using angle()
phase_response = angle(fft_sweep);
phase_response = phase_response(1:end/2);
figure;
plot(unwrap(phase_response));
grid on;
Both methods give the same incorrect phase response. Any help would be appreciated.
• You're determining the phase of the system plus the phase of the sweep. – Matt L. Feb 26 '15 at 17:44
You missing one, very important step. In order to get the impulse response of your filter while using the sweep sine, you must get the output of your system $h[n]$ to excitation with sweep-sine signal $x[n]$:
$y[n]=h[n]\star x[n]$
In your case that's the filtering with filter function. Now in order to get the impulse response you must convolve your signal with the inverse filter $f[n]$. Generally that's the time-inverted sweep signal (although special kind of pre-conditioning, amplitude scaling, etc. might be done as well):
$h[n]=y[n]\star f[n]$
Now you have your impulse response that you can use for further analysis. Please take a look at this answer tackling very similar problem. | 616 | 2,444 | {"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} | 3.15625 | 3 | CC-MAIN-2021-21 | latest | en | 0.854644 |
https://www.mathworks.com/matlabcentral/cody/problems/43739-repeat-vector-values-an-arbitrary-number-of-times/solutions/1076163 | 1,590,873,222,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410352.47/warc/CC-MAIN-20200530200643-20200530230643-00220.warc.gz | 821,133,879 | 15,726 | Cody
# Problem 43739. Repeat Vector Values an Arbitrary Number of Times
Solution 1076163
Submitted on 7 Dec 2016 by Peng Liu
• Size: 12
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x1=[1 2 3 4 5 6]; x2=[2 2 1 1 2 2]; out=[1 1 2 2 3 4 5 5 6 6]; assert(isequal(your_fcn_name(x1,x2),out))
2 Pass
x1=1; x2=7; out=[1 1 1 1 1 1 1]; assert(isequal(your_fcn_name(x1,x2),out))
3 Pass
x1=1:650; x2=ones(1,650); out=1:650; assert(isequal(your_fcn_name(x1,x2),out))
4 Pass
x1=5.*ones(250,1); x2=3.*ones(250,1); out=5.*ones(750,1); assert(isequal(your_fcn_name(x1,x2),out))
5 Pass
x1=[1 500 32 780 5 8]; x2=[7 6 5 4 3 2]; out=[1,1,1,1,1,1,1,500,500,500,500,500,500,32,32,32,32,32,780,780,780,780,5,5,5,8,8]; assert(isequal(your_fcn_name(x1,x2),out)) | 402 | 906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-24 | latest | en | 0.522107 |
http://www.kwiznet.com/p/takeQuiz.php?ChapterID=10318&CurriculumID=25&Num=3.50 | 1,544,773,733,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825495.60/warc/CC-MAIN-20181214070839-20181214092339-00607.warc.gz | 428,977,978 | 3,856 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Grade 8 - Mathematics3.50 Circles Review - 2
Directions: Check the given statements are true or false.
Q 1: If a square is inscribed in a circle then the diameter of that square divides the circle into two equal semicircles.TrueFalse Q 2: The circumference of a circle is also called the length of the circle.FalseTrue Q 3: The perimeter of a semicircle is p d.TrueFalse Q 4: The diameters of a circle are perpendicular to one other.FalseTrue Q 5: The longest chord of a circle is its diameter.TrueFalse Q 6: Angle in a semicircle is 180°.FalseTrue Q 7: The circumference of a circle is equal to the product of p and the diameter of that circle.TrueFalse Q 8: If the diameter of a circle is "d" then its area is 4pd2TrueFalse Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 261 | 1,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-51 | latest | en | 0.867707 |
https://www.slideserve.com/jaunie/section-1-4 | 1,542,811,901,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039748901.87/warc/CC-MAIN-20181121133036-20181121155036-00356.warc.gz | 1,007,901,152 | 14,667 | Download Presentation
Section 1.4
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# Section 1.4 - PowerPoint PPT Presentation
Section 1.4. 2. use window [0,2] by [0,5]. Note you cannot plug 1 into this equation. However, your calculator will help you. Do you see the “hole” at (1, 1)? Go to table set, start at 0.9 and tbl = .05 and you will see the limit of 1 under table. 3. Plug the 3 into the function. 4.
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2.
use window [0,2] by [0,5]
Note you cannot plug 1 into this equation. However, your calculator will help you. Do you see the “hole” at (1, 1)? Go to table set, start at 0.9 and tbl = .05 and you will see the limit of 1 under table.
3.
Plug the 3 into the function.
4.
Plug the 5 into the function.
6.
Plug the 0 into the function for h.
7.
Plug the - 3 into the function.
8.
You need to get rid of the h in the bottom. Factor the h out of the top and cancel it with the h in the bottom. Then plug the 0 into the function for h.
11.
Calculate the left, right and limit at 0.
Or as x becomes very large the y value approaches 0. The x-axis, y = 0 is a horizontal asymptote.
Or as x approaches 3 the y value approaches . The line x = 3 is a vertical asymptote.
Or as x becomes very large the y value approaches 0. The line y = 2 is a horizontal asymptote.
Or as x approaches 3 the y value approaches . The line x = 1 is a vertical asymptote.
14. Determine if the following function is continuous or discontinuous. If it is not state why it is not.
It is continuous.
15. Determine if the following function is continuous or discontinuous. If it is not state why it is not.
c
16 . a. Draw the graph of the following function.
b. Find the limits as x approaches 3 from the left and the right.
c. Is it continuous at 3? If not state why.
17. Determine if the function is continuous or discontinuous. If it is discontinuous, state where that occurs.
18. Determine if the function is continuous or discontinuous. If it is discontinuous, state where that occurs.
OR use your calculator and table to examine what occurs at x = 0, -1 and 1.
19. Determine if the function is continuous or discontinuous. If it is discontinuous, state where that occurs.
From problem 10 we know that there is no limit. Therefore, the function is not continuous.
20. Business: Interest Compounded Continuously If you deposit \$1 into
a bank account paying 10% interest compounded continuously, a year later
its value will be
Find the limit by making a TABLE of values correct to two decimal places, thereby
finding the value of the deposit in dollars and cents.
• near absolute zero is approximated by the function :
• which expresses the conductivity as a percent.
• Find the limit of this conductivity percent as the temperature
• x approaches 0 (absolute zero) from the right . | 905 | 3,515 | {"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} | 4.46875 | 4 | CC-MAIN-2018-47 | latest | en | 0.900545 |
https://philoid.com/question/80614-using-definite-integrals-find-the-area-of-circle-x-2-y-2-a-2 | 1,685,831,258,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00366.warc.gz | 495,919,670 | 13,256 | ## Book: RD Sharma - Mathematics (Volume 2)
### Chapter: 21. Areas of Bounded Regions
#### Subject: Maths - Class 7th
##### Q. No. 15 of Exercise 21.1
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
15
##### Using definite integrals, find the area of circle x2 + y2 = a2
Given equations are :
x2 + y2 = a2 ...... (1)
Equation (1) represents a circle with centre (0,0) and radius a, so it meets the axes at (±a,0), (0,±a). A rough sketch of the curve is given below: -
We have to find the area of shaded region.
Required area
= 4(shaded region OBCO) (as the circle is symmetrical about the x - axis as well as the y - axis)
(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)
(As x is between (0,a) and the value of y varies)
(as )
Substitute
So the above equation becomes,
We know,
So the above equation becomes,
Apply reduction formula:
On integrating we get,
Undo the substituting, we get
On applying the limits we get,
Hence the area of circle x2 + y2 = a2 is equal to square units.
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https://www.scribd.com/document/177736505/Midterm2mgt-201 | 1,569,021,490,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574084.88/warc/CC-MAIN-20190920221241-20190921003241-00252.warc.gz | 1,014,402,572 | 69,657 | You are on page 1of 8
UNIVERSITY OF CALIFORNIA, DAVIS DEPARTMENT OF ECONOMICS FINANCIAL ECONOMICS SPRING 2011 PROFESSOR LARSEN MIDTERM EXAM 2
Sign the following statement. Grades will not be given to students who do not do so.
I have not cheated or helped anyone else cheat on this exam. ________________________________ Signature
Name:_______________________________
Important Notes
1. There are 7 pages to this exam (including this cover sheet). Please make sure that you have them all!! 2. Show ALL of your work. Random guesses, without accompanying work, will not receive credit. 3. Please circle your final answers. 4. If you get stuck with the math and cant figure out why, explain what you would do to answer the question. If correct, you will receive partial credit. 5. Budget your valuable time. If you are bogged down on something, it may be best to move on and return to it if you have time at the end of the exam. 6. Finally, when in doubt, use the force, it will guide you
There are 90 points possible
Multiple Choice Section (3 points each) (No Partial Credit in this section)
1. You wish to earn a return of 13% on each of two stocks, X and Y. Stock X is expected to pay a dividend of \$3 in the upcoming year while Stock Y is expected to pay a dividend of \$4 in the upcoming year. The expected growth rate of dividends for both stocks is 7%. The price of stock X _____. A. will be greater than the price of stock Y B. will be the same as the price of stock Y C. will be less than the price of stock Y D. Not enough information E. None of these is correct. 2. The present value of growth opportunities (PVGO) is equal to I) the difference between a stock's price and its no-growth value per share. II) the stock's price. III) zero if its return on equity equals the required return rate. IV) the plowback ratio times the return on equity. A. I and IV B. II and IV C. I, III, and IV D. II, III, and IV E. I and III 3. You are considering acquiring a common stock that you would like to hold for one year. You expect to receive both \$2.50 in dividends and \$28 from the sale of the stock at the end of the year. The maximum price you would pay for the stock today is _____ if you wanted to earn a 15% return. A. \$23.91 B. \$24.11 C. \$26.52 D. \$27.50 E. None of these is correct 4. Investors want high plowback ratios A. for all firms. B. whenever ROE > r. C. whenever r > ROE. D. only when they are in low tax brackets. E. whenever bank interest rates are high.
5. Low P/E ratios tend to indicate that a company will ______, ceteris paribus. A. grow quickly B. grow at the same speed as the average company C. grow slowly D. P/E ratios are unrelated to growth E. None of these is correct 6. Consider a risky portfolio, A, with an expected rate of return of 0.15 and a standard deviation of 0.15, that lies on a given indifference curve of a risk averse individual. Which one of the following portfolios might lie on the same indifference curve? A. E(r) = 0.15; Standard deviation = 0.20 B. E(r) = 0.15; Standard deviation = 0.10 C. E(r) = 0.10; Standard deviation = 0.10 D. E(r) = 0.20; Standard deviation = 0.15 E. E(r) = 0.10; Standard deviation = 0.20 7. The certainty equivalent rate of a portfolio is A. the rate that a risk-free investment would need to offer with certainty to be considered equally attractive as the risky portfolio. B. the rate that the investor must earn for certain to give up the use of his money. C. the minimum rate guaranteed by institutions such as banks. D. the rate that equates "A" in the utility function with the average risk aversion coefficient for all risk-averse investors. E. represented by the scaling factor "-.005" in the utility function. 8. You invest \$100 in a risky asset with an expected rate of return of 0.12 and a standard deviation of 0.15 and a T-bill with a rate of return of 0.05. What percentages of your money must be invested in the risky asset and the risk-free asset, respectively, to form a portfolio with an expected return of 0.09? A. 85% and 15% B. 75% and 25% C. 67% and 33% D. 57% and 43% E. cannot be determined 9. The Capital Market Line I) is a special case of the Capital Allocation Line. II) represents the opportunity set of a passive investment strategy. III) has the one-month T-Bill rate as its intercept. IV) uses a broad index of common stocks as its risky portfolio. A. I, III, and IV B. II, III, and IV C. III and IV D. I, II, and III E. I, II, III, and IV
10. Other things equal, diversification is most effective when A. securities' returns are uncorrelated. B. securities' returns are positively correlated. C. securities' returns are high. D. securities' returns are negatively correlated. E. both securities' returns are positively correlated and securities' returns are high.
Derivation Section 1. [35 Points]
Two firms, Eva Industries and Porter Inc, are both worth \$100 million dollars and have 1 million shares outstanding. Both firms have investors that expect a required return of 8%. Eva Industries return on equity is 10% while Porter Incs is 6%. Answer the following questions about them: a. (10 pts) What are next periods earnings for each of the firms? Assume they each pay out all of the earnings they make each period. What will each firms growth be? What is each of their prices per share? For Eva Industries. E(rE)=100*.10=\$10 Million or \$10 per share gE=0 (bE=0 and g=b*ROE) Since they payout everything the dividends will be the earnings and prices will be found using the no growth dividend formula (perpetuity) PE=DE/r=10/.08=\$125 Similarly for Porter Inc. EP=100*.06=\$6M gP=0 PP=6/.08=\$75
b. (10 pts) Now instead of paying out all of their dividends Eva Industries keeps an earnings retention ratio of 50% and Porter Inc keeps an earnings retention ratio of 75%. What is the new growth rate of each company? What are the dividends per share for each? What are prices per share for each company? Given the info above bE=.5 Growth is found by multiplying the return on equity by the plowback or earnings retention ratio. gE=ROEE*bE =.1*.5 =.05 or 5% Dividends paid out are the earnings that arent kept by the firm. DE=EE(1-bE) =10*.5 =\$5 Price is found using the Gordon Growth Model Formula PE=DE/(r-gE) =5/(.08-.05)=\$166.67 Similarly for Porter Inc. bP=.75 gP=.06*.75=.045 or 4.5% DP=6*(1-.75)=\$1.5 PP=1.5/(.08-.045)=\$42.86
c. (10 pts) What plowback ratio should each firm keep in order to get the highest price possible? What Price would each firm get at that ratio? Eva should have a plowback ratio of 1 or 100% because their ROE is greater than the required return ratio. When this happens it is better for the company to reinvest as much as possible. There were two possible answer for the price with this ratio. A price of infinity would suffice. Or stating that we cannot tell because the Gordon growth model doesnt apply when g>r would also work. For Porter, the plowback ratio should be 0% because ROE<r. They cant make as much as people require so they should just pay it all out to investors. Part a of this problem had a plowback of 0 so the price would be the same as it was there, \$75.
2. [30 points]
Assume there are only two portfolios in the world. A risk free portfolio with a return of 6% and a risky portfolio with a return of 14% and a standard deviation of 22%. Assume you have a risk aversion factor of 3. Throughout this problem you may assume the usual utility curve U=E(r)-1/2A2 a. (10 pts) Assume you can only invest all of your money in one portfolio. Which portfolio would you choose? What would the risk-free rate have to be to make you indifferent between it and the risky portfolio? Just compare the utilities of each asset separately. U(rf)=.06-1/2*3*02=.06-0=.06 U(rp)=;14-1/2*3*.222=.0674 U(rP)>U(rf) so you prefer the risky asset. The risk-free rate that would make you indifferent is just the certainty equivalence or the utility given by the asset you chose (the risky asset). So the return should be 6.74%.
b. (10 pts) Now assume that you can split your money between the two portfolios in any way you like. However, if you want to invest more than 100% in the risky asset you must borrow at a rate of 8%. Draw the Capital Asset Line associated with this below. Label all axis and the points and values that correspond to the risky and risk-free portfolio. What is(are) the slope(s) of this line? Recall the Slope of the CAL is (E(rP)-rf)/P=(.14-.06)/.22=.36 But here you cannot borrow at the risk free rate you can only borrow at rb=.08. So there is a kink point at the part where you need to borrow money (to the right of 100% of money in the risky asset) There the slope is (E(rP)-rb)/P=(.14-.08)/.22=.27 See the figure below.
c. (10 pts) What is your optimal fraction of your investments that you would hold in each portfolio? What is the expected return, standard deviation and 5% VaR of the optimal portfolio? (You may find the following useful: -1.65, and -2.33 are the 5th and 1st percentile of a normal distribution). The formula for optimal risky share is: E ( rp ) rf .14 .06 y* = = = .55 2 A P 3* .22 2 Thus you should hold 55% in risky and 45% in risk-free.
c = y * P = .55 * .22 = .121 or 12.1%
The VaR for 5% level is E(r)-1.65* So .104-1.65*.121=-.09565 = -9.6% So the best worst case scenario is a return of -9.6%. | 2,477 | 9,322 | {"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} | 3.140625 | 3 | CC-MAIN-2019-39 | latest | en | 0.920634 |
https://groupprops.subwiki.org/w/index.php?title=Groups_of_order_40&action=info | 1,571,199,531,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00456.warc.gz | 528,412,557 | 8,549 | # Groups of order 40
## Contents
See pages on algebraic structures of order 40| See pages on groups of a particular order
This article gives basic information comparing and contrasting groups of order 40. The prime factorization of 40 is $40 = 2^3 \cdot 5$.
## Statistics at a glance
Quantity Value
Total number of groups 14
Number of abelian groups 3
Number of nilpotent groups 5
Number of solvable groups 14
Number of simple groups 0
## The list
There are 14 groups of order 40:
Group Second part of GAP ID (GAP ID is (40,second part)) Abelian? Nilpotent?
semidirect product of Z5 and Z8 via inverse map 1 No No
cyclic group:Z40 2 Yes Yes
semidirect product of Z5 and Z8 via square map 3 No No
nontrivial semidirect product of Z5 and Q8 4 No No
direct product of D10 and Z4 5 No No
dihedral group:D40 6 No No
? 7 No No
? 8 No No
direct product of Z20 and Z2 (also direct product of Z10 and Z4) 9 Yes Yes
direct product of D8 and Z5 10 No Yes
direct product of Q8 and Z5 11 No Yes
direct product of GA(1,5) and Z2 12 No No
direct product of D10 and V4 13 No No
direct product of E8 and Z5 14 Yes Yes
## Sylow subgroups
### 5-Sylow subgroups
Combining the congruence condition on Sylow numbers and the divisibility condition on Sylow numbers, we see that the number of 5-Sylow subgroups must be congruent to 1 modulo 5 and also must be a divisor of 8. The only possibility for both of these to hold simultaneously is that there is exactly one 5-Sylow subgroup, and hence it is a normal Sylow subgroup and the 2-Sylow subgroups are its permutable complements. In particular, this means that the whole group is a semidirect product with normal subgroup equal to the 5-Sylow subgroup and quotient/complement of order 8. | 503 | 1,727 | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-43 | latest | en | 0.864314 |
https://feng.li/files/statcomp/RootFinding.R | 1,621,166,747,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00291.warc.gz | 273,370,905 | 1,349 | rm(list=ls()) #First define function g<-function(x){(3*(x^5))-(4*(x^4))+(6*(x^3))+4*x-4} #g<-function(x){abs(x)^0.5} #g<-function(x){x*exp(-x^2)-0.4*(1/(exp(x)+1))+0.2} #g<-function(x){2*(x^3)-3*(x^2)+5} #g<-function(x){(x^3)-2*(x^2)-11*x+12} #Then define first derivative dg<-function(x){(15*(x^4))-(16*(x^3))+(18*(x^2))+4} #dg<-function(x){ifelse((x>0),0.5*(x^(-0.5)),-0.5*((-x)^(-0.5)))} #dg<-function(x){(1-2*(x^2))*exp(-x^2)+0.4*(exp(x)/(exp(x)+1)^2)} #dg<-function(x){6*x*(x-1)} #dg<-function(x){3*(x^2)-4*x-11} #Code 1 Newton-Raphson using "for" loop nralgo1<-function(x0,func,deri){ epsilon<-1E-10 #Set Tolerance Level MaxIter<-500 #Maximum Number of Iterations x<-rep(NA,MaxIter+1) #A vector to store sequence of x x[1]<-x0 # Initialise for(n in 1:MaxIter){ x[n+1]<-x[n]-(func(x[n])/deri(x[n])) #Update Step if(abs(func(x[n+1]))<=epsilon){break} #Stopping Rule } if(n==MaxIter){warning('Maximum number of iterations reached without convergence')} return(x[n+1]) #Return value of x } #CODE 2: Newton-Raphson using "while" nralgo2<-function(x0,func,deri){ epsilon<-1E-10 #Tolerance Level x<-x0 #Initialise while(abs(func(x))>epsilon){ x<-x-(func(x)/deri(x)) #Update step print(x) } return(x) } #With all functions defined we can now do optimisation: x0<-1.4 #Use Code 1 root1=nralgo1(x0,g,dg) print(paste('Root using Code 1:',round(root1,3))) print(paste('Function Evaluated at this Root',round(g(root1),3))) #Using Code 2 root2<-nralgo2(x0,g,dg) print(paste('Root using Code 2:',round(root2,3))) print(paste('Function Evaluated at this Root',round(g(root2),3))) | 608 | 1,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-21 | latest | en | 0.346651 |
https://www.aqua-calc.com/convert/density/gram-per-cubic-millimeter-to-milligram-per-cubic-inch | 1,579,289,567,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250590107.3/warc/CC-MAIN-20200117180950-20200117204950-00400.warc.gz | 760,951,997 | 9,110 | # Convert grams per (cubic millimeter) to milligrams per (cubic inch)
## g/mm³ to mg/in³ (g:gram, mm:millimeter, mg:milligram, in:inch)
### Convert g/mm³ to mg/in³
#### a density conversion table
to 100 with
gram per cubic milli-
meter
×108,
milli-
gram per cubic inch
gram per cubic milli-
meter
×108,
milli-
gram per cubic inch
gram per cubic milli-
meter
×108,
milli-
gram per cubic inch
gram per cubic milli-
meter
×108,
milli-
gram per cubic inch
gram per cubic milli-
meter
×108,
milli-
gram per cubic inch
10.2213.4416.76110.08113.3
20.3223.6426.96210.28213.4
30.5233.8437.06310.38313.6
40.7243.9447.26410.58413.8
50.8254.1457.46510.78513.9
61.0264.3467.56610.88614.1
71.1274.4477.76711.08714.3
81.3284.6487.96811.18814.4
91.5294.8498.06911.38914.6
101.6304.9508.27011.59014.7
111.8315.1518.47111.69114.9
122.0325.2528.57211.89215.1
132.1335.4538.77312.09315.2
142.3345.6548.87412.19415.4
152.5355.7559.07512.39515.6
162.6365.9569.27612.59615.7
172.8376.1579.37712.69715.9
182.9386.2589.57812.89816.1
193.1396.4599.77912.99916.2
203.3406.6609.88013.110016.4
### grams per cubic millimeter to milligrams per cubic inch conversion cards
• 1
through
20
grams per cubic millimeter
• 1 g/mm³ to mg/in³ = 20 000 000 mg/in³
• 2 g/mm³ to mg/in³ = 30 000 000 mg/in³
• 3 g/mm³ to mg/in³ = 50 000 000 mg/in³
• 4 g/mm³ to mg/in³ = 70 000 000 mg/in³
• 5 g/mm³ to mg/in³ = 80 000 000 mg/in³
• 6 g/mm³ to mg/in³ = 100 000 000 mg/in³
• 7 g/mm³ to mg/in³ = 110 000 000 mg/in³
• 8 g/mm³ to mg/in³ = 130 000 000 mg/in³
• 9 g/mm³ to mg/in³ = 150 000 000 mg/in³
• 10 g/mm³ to mg/in³ = 160 000 000 mg/in³
• 11 g/mm³ to mg/in³ = 180 000 000 mg/in³
• 12 g/mm³ to mg/in³ = 200 000 000 mg/in³
• 13 g/mm³ to mg/in³ = 210 000 000 mg/in³
• 14 g/mm³ to mg/in³ = 230 000 000 mg/in³
• 15 g/mm³ to mg/in³ = 250 000 000 mg/in³
• 16 g/mm³ to mg/in³ = 260 000 000 mg/in³
• 17 g/mm³ to mg/in³ = 280 000 000 mg/in³
• 18 g/mm³ to mg/in³ = 290 000 000 mg/in³
• 19 g/mm³ to mg/in³ = 310 000 000 mg/in³
• 20 g/mm³ to mg/in³ = 330 000 000 mg/in³
• 21
through
40
grams per cubic millimeter
• 21 g/mm³ to mg/in³ = 340 000 000 mg/in³
• 22 g/mm³ to mg/in³ = 360 000 000 mg/in³
• 23 g/mm³ to mg/in³ = 380 000 000 mg/in³
• 24 g/mm³ to mg/in³ = 390 000 000 mg/in³
• 25 g/mm³ to mg/in³ = 410 000 000 mg/in³
• 26 g/mm³ to mg/in³ = 430 000 000 mg/in³
• 27 g/mm³ to mg/in³ = 440 000 000 mg/in³
• 28 g/mm³ to mg/in³ = 460 000 000 mg/in³
• 29 g/mm³ to mg/in³ = 480 000 000 mg/in³
• 30 g/mm³ to mg/in³ = 490 000 000 mg/in³
• 31 g/mm³ to mg/in³ = 510 000 000 mg/in³
• 32 g/mm³ to mg/in³ = 520 000 000 mg/in³
• 33 g/mm³ to mg/in³ = 540 000 000 mg/in³
• 34 g/mm³ to mg/in³ = 560 000 000 mg/in³
• 35 g/mm³ to mg/in³ = 570 000 000 mg/in³
• 36 g/mm³ to mg/in³ = 590 000 000 mg/in³
• 37 g/mm³ to mg/in³ = 610 000 000 mg/in³
• 38 g/mm³ to mg/in³ = 620 000 000 mg/in³
• 39 g/mm³ to mg/in³ = 640 000 000 mg/in³
• 40 g/mm³ to mg/in³ = 660 000 000 mg/in³
• 41
through
60
grams per cubic millimeter
• 41 g/mm³ to mg/in³ = 670 000 000 mg/in³
• 42 g/mm³ to mg/in³ = 690 000 000 mg/in³
• 43 g/mm³ to mg/in³ = 700 000 000 mg/in³
• 44 g/mm³ to mg/in³ = 720 000 000 mg/in³
• 45 g/mm³ to mg/in³ = 740 000 000 mg/in³
• 46 g/mm³ to mg/in³ = 750 000 000 mg/in³
• 47 g/mm³ to mg/in³ = 770 000 000 mg/in³
• 48 g/mm³ to mg/in³ = 790 000 000 mg/in³
• 49 g/mm³ to mg/in³ = 800 000 000 mg/in³
• 50 g/mm³ to mg/in³ = 820 000 000 mg/in³
• 51 g/mm³ to mg/in³ = 840 000 000 mg/in³
• 52 g/mm³ to mg/in³ = 850 000 000 mg/in³
• 53 g/mm³ to mg/in³ = 870 000 000 mg/in³
• 54 g/mm³ to mg/in³ = 880 000 000 mg/in³
• 55 g/mm³ to mg/in³ = 900 000 000 mg/in³
• 56 g/mm³ to mg/in³ = 920 000 000 mg/in³
• 57 g/mm³ to mg/in³ = 930 000 000 mg/in³
• 58 g/mm³ to mg/in³ = 950 000 000 mg/in³
• 59 g/mm³ to mg/in³ = 970 000 000 mg/in³
• 60 g/mm³ to mg/in³ = 980 000 000 mg/in³
• 61
through
80
grams per cubic millimeter
• 61 g/mm³ to mg/in³ = 1 000 000 000 mg/in³
• 62 g/mm³ to mg/in³ = 1 020 000 000 mg/in³
• 63 g/mm³ to mg/in³ = 1 030 000 000 mg/in³
• 64 g/mm³ to mg/in³ = 1 050 000 000 mg/in³
• 65 g/mm³ to mg/in³ = 1 070 000 000 mg/in³
• 66 g/mm³ to mg/in³ = 1 080 000 000 mg/in³
• 67 g/mm³ to mg/in³ = 1 100 000 000 mg/in³
• 68 g/mm³ to mg/in³ = 1 110 000 000 mg/in³
• 69 g/mm³ to mg/in³ = 1 130 000 000 mg/in³
• 70 g/mm³ to mg/in³ = 1 150 000 000 mg/in³
• 71 g/mm³ to mg/in³ = 1 160 000 000 mg/in³
• 72 g/mm³ to mg/in³ = 1 180 000 000 mg/in³
• 73 g/mm³ to mg/in³ = 1 200 000 000 mg/in³
• 74 g/mm³ to mg/in³ = 1 210 000 000 mg/in³
• 75 g/mm³ to mg/in³ = 1 230 000 000 mg/in³
• 76 g/mm³ to mg/in³ = 1 250 000 000 mg/in³
• 77 g/mm³ to mg/in³ = 1 260 000 000 mg/in³
• 78 g/mm³ to mg/in³ = 1 280 000 000 mg/in³
• 79 g/mm³ to mg/in³ = 1 290 000 000 mg/in³
• 80 g/mm³ to mg/in³ = 1 310 000 000 mg/in³
• 81
through
100
grams per cubic millimeter
• 81 g/mm³ to mg/in³ = 1 330 000 000 mg/in³
• 82 g/mm³ to mg/in³ = 1 340 000 000 mg/in³
• 83 g/mm³ to mg/in³ = 1 360 000 000 mg/in³
• 84 g/mm³ to mg/in³ = 1 380 000 000 mg/in³
• 85 g/mm³ to mg/in³ = 1 390 000 000 mg/in³
• 86 g/mm³ to mg/in³ = 1 410 000 000 mg/in³
• 87 g/mm³ to mg/in³ = 1 430 000 000 mg/in³
• 88 g/mm³ to mg/in³ = 1 440 000 000 mg/in³
• 89 g/mm³ to mg/in³ = 1 460 000 000 mg/in³
• 90 g/mm³ to mg/in³ = 1 470 000 000 mg/in³
• 91 g/mm³ to mg/in³ = 1 490 000 000 mg/in³
• 92 g/mm³ to mg/in³ = 1 510 000 000 mg/in³
• 93 g/mm³ to mg/in³ = 1 520 000 000 mg/in³
• 94 g/mm³ to mg/in³ = 1 540 000 000 mg/in³
• 95 g/mm³ to mg/in³ = 1 560 000 000 mg/in³
• 96 g/mm³ to mg/in³ = 1 570 000 000 mg/in³
• 97 g/mm³ to mg/in³ = 1 590 000 000 mg/in³
• 98 g/mm³ to mg/in³ = 1 610 000 000 mg/in³
• 99 g/mm³ to mg/in³ = 1 620 000 000 mg/in³
• 100 g/mm³ to mg/in³ = 1 640 000 000 mg/in³
#### Foods, Nutrients and Calories
ASADERO CHEESE, UPC: 216114804766 contain(s) 321 calories per 100 grams or ≈3.527 ounces [ price ]
SKIPPY, PEANUT BUTTER SPREAD, CREAMY, UPC: 048001212107 weigh(s) 270.51 gram per (metric cup) or 9.03 ounce per (US cup), and contain(s) 656 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Kon Tiki weighs 1 601.85 kg/m³ (100.00023 lb/ft³) with specific gravity of 1.60185 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Arsine, liquid [AsH3] weighs 1 689 kg/m³ (105.44083 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Rapeseed oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
The grain per square micrometer surface density measurement unit is used to measure area in square micrometers in order to estimate weight or mass in grains
Acceleration (a) of an object measures the object's change in velocity (v) per unit of time (t): a = v / t.
troy/metric tbsp to µg/pt conversion table, troy/metric tbsp to µg/pt unit converter or convert between all units of density measurement.
#### Calculators
Capsule capacity calculator: weight, volume and number of capsules | 3,362 | 7,398 | {"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-2020-05 | latest | en | 0.20184 |
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(L05)BondsSens_ns
# (L05)BondsSens_ns - Topic 5 Bond Price Sensitivity Reading...
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Primbs/Investment Science 1 Topic 5: Bond Price Sensitivity Reading: Luenberger Chapter 3, Section 5 Chapter 4, Section 8
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Primbs/Investment Science 2 Macaulay Duration Connections Bond Price Bond Price Sensitivity Sensitivity Modified Duration (yield) Fisher-Weil Duration Quasi-Modified Duration Yield Sensitivity Spot Rate Curve Sensitivity
Primbs/Investment Science 3 Question : I would like to invest my money in a bond, but I am worried that bond yields may change in the future. Should I invest in a short or long maturity bond to reduce the effects of changes in yield on the value of my bond? What about coupons? Should I choose a bond with large coupons or small coupons? Answer : We need a way to compute the sensitivity of bond prices to yield.
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Primbs/Investment Science 4 Duration Maturity time is related to the sensitivity of the bond price to its yield. When there are coupons, maturity time does not exactly correspond to sensitivity. We will define something called duration to try to “generalize” the idea that for a zero coupon bond the time to maturity captures its sensitivity to yield.
Primbs/Investment Science 5 Macaulay Duration Let’s compute the weighted average of the times to the cash flows, where we weight each cash flow by its present value (computed with the yield) divided by the total present value. t n 0 C C C C C F t 1 t 2 t n 0 t 1 t 2 PV(t 1 ) PV(t 2 ) PV(t 2 ) PV(t 2 ) Replace by present values computed using the yield. The average time is called Macaulay Duration
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Primbs/Investment Science 6 Macaulay Duration Defn : Macaulay Duration: total n n PV t t PV t t PV t t PV D ) ( ... ) ( ) ( 2 2 1 1 + + + = PV(t i ) is the present value of the cash flow at time t k computed using the yield as the interest rate!. is the present value of the entire cash flow. n t t t t w t w t w n + + + = ... 2 1 2 1 where total i t PV t PV w i ) ( = = = n i i total t PV PV 0 ) (
Primbs/Investment Science 7 Macaulay Duration Macaulay duration is a weighted average of times . It is quoted in years . In practice, when people refer to duration, they often mean Macaulay duration.
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Primbs/Investment Science 8 Macaulay Duration and Maturity Zero-Coupon bond: Macaulay Duration = Maturity Date Coupon bond: Macaulay Duration < Maturity Date You should think : A bond with Macaulay Duration D has the same yield sensitivity as a zero coupon bond with maturity D .
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Jill Tulane University ‘16, Course Hero Intern | 963 | 4,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-22 | latest | en | 0.851765 |
https://www.reference.com/math/line-fraction-called-df9884ccab3ee630 | 1,571,105,541,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00029.warc.gz | 998,684,897 | 18,354 | # What Is the Line in a Fraction Called?
Credit: Jeffrey Coolidge/The Image Bank/Getty Images
The line that separates the top and bottom numbers of a fraction is called a vinculum. Fractions are also commonly written out using a forward slash in place of a vinculum for convenience on computers.
A fraction is a number that is not whole. In other words, in decimal form, there will be some number after the decimal place. The top number of a fraction is called the numerator, while the bottom is called the denominator. Fractions can be multiplied by multiplying numerators with each other and denominators with each other, but can only be added if they have the same denominator.
Similar Articles | 142 | 701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-43 | latest | en | 0.925432 |
https://math.stackexchange.com/questions/3580993/asymptotic-of-given-functional-as-x-rightarrow-infty | 1,606,151,923,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141163411.0/warc/CC-MAIN-20201123153826-20201123183826-00685.warc.gz | 398,456,265 | 32,833 | Asymptotic of given functional as $x\rightarrow\infty$
Consider the following functional :
$$I(x,s) =\int_0^\infty\mathrm dy \frac{F(x + \mathrm iy, s) − F(x −\mathrm iy, s)}{\mathrm e^{2πy}-1},$$ where $$F(z, s) = \dfrac{\sinh(\sin^2[π\Gamma(z)/(2z)])}{z^s}$$.
Let us restrict $$s\in[0,1]$$
Can we get sharp numerical asymptotic of $$I(x)$$ as as $$x\rightarrow \infty$$?
Also, can we get quantitative upper and lower bound estimations on the functional ?
Moreover , if someone can construct a generalized function such that: $$F_*(z, s) = \dfrac{\phi(\sin^2[π\Gamma(z)/(2z)])}{z^s}$$
(1) $$\phi(x) =0$$ if $$x$$ is zero ; and finite otherwise
(2)$$I_*(x,s)\rightarrow c$$ as $$x\rightarrow\infty$$
Here $$c$$ is a constant .
• Any updates on the question? – Bambi Mar 17 at 11:18 | 276 | 791 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-50 | latest | en | 0.617906 |
http://www.heidimeijer.nl/api650-coil/ghana_semi_undergrou_27.html | 1,606,710,097,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141205147.57/warc/CC-MAIN-20201130035203-20201130065203-00037.warc.gz | 123,164,410 | 13,291 | • # ghana semi underground tank chemical volume
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### Underground storage tank - Wikipedia
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https://www.onlineschoolbase.com/2022/12/physics-form-three-topic-5-thermal.html | 1,726,418,594,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00672.warc.gz | 858,425,209 | 38,460 | # PHYSICS FORM THREE TOPIC 5: THERMAL EXPANSION
THERMAL EXPANSION
The Concept of Heat
Explain the concept of heat
Heat – Is the transfer of energy due to temperature differences.
Temperature – Is the degree of hotness or coldness of a body.
Or heat is a form of energy which raises the temperature of the substance.
SI Unit of Temperature is Kelvin (K).Conversion of centigrade into fahrerinheit given by:f = 9/5ÂșC + 32
Conversion of Fahrenheit into centigrade given by C = 5/9/ ( F – 32). Thermometer used for measurement of temperature. Maximum thermometer is the one which is used to measure the highest temperature obviously filled with mercury.
A Minimum thermometer is used to measure the lowest temperature and it is filled with alcohol. Combined maximum and minimum thermometer (Six’s thermometer) is used to measure highest and lowest temperature at the same time.
SI Unit of Heat is Joule (J) mathematically heat energy given by:
• H = MC (Q2 – Q1)
• M = Mass of substance
• C = Specific heat capacity
• Q= Final Temperature
• Q1 = Initial Temperature
Source of Thermal Energy in Everyday Life
State the source of thermal energy in everyday life
There are numerous sources of energy such as the sun, fuels, nuclear sources, geothermal, electricity among others. The most important source of thermal energy is the sun. The sun generates its energy by a process called thermonuclear fusion. Most sources of thermal energy derive their energy from the sun. Thermal energy from the sun makes life on earth possible.
Difference between Heat and Temperature
Distinguish between heat and temperature
Heatandtemperatureare related and often confused. More heat usually means a higher temperature.
Heat(symbol:Q) is energy. It is the total amount of energy (both kinetic and potential) possessed by the molecules in a piece of matter. Heat is measured in Joules.
Temperature(symbol:T) is not energy. It relates to the average (kinetic) energy of microscopic motions of a single particle in the system per degree of freedom. It is measured inKelvin (K), Celsius (C) or Fahrenheit (F).
When you heat a substance, either of two things can happen: the temperature of the substance can rise or thestateof substance can change.
HeatTemperature
DefinitionHeat is energy that is transferred from one body to another as the result of a difference in temperature.Temperature is a measure of hotness or coldness expressed in terms of any of several arbitrary scales like Celsius and Fahrenheit.
SymbolQT
UnitJoulesKelvin, Celsius or Fahrenheit
SI unitJouleKelvin
ParticlesHeat is a measure of how many atoms there are in a substance multiplied by how much energy each atom possesses.Temperature is related to how fast the atoms within a substance are moving. The ‘temperature’ of an object is like the water level – it determines the direction in which ‘heat’ will flow.
Ability to do workHeat has the ability to do work.Temperature can only be used to measure the degree of hea
Expansion of Solids
Demonstrate expansion of solids
When a solid is heated one or more of the following may occur:
• Its temperature may rise
• Its state may change
• It may Expand
Expansion of solids is the increase in dimensions of a solid when heated.
Contraction of solids is the decrease in dimensions of a solid when is cooled.
DEMONSTRATION OF EXPANSION OF SOLIDS
In order to demonstrate the expansion of solid we can use following method
1. The ball and Ring Experiment
2. The Bar and Gap Experiment
The ball and Ring Experiment
1st Case: When the ball is not heated then it will pass through the ring very easily because it has a small volume.
2nd Case: When the ball is heated, it will expand and increase in volume, so itwill not pass through the ring.
The Bar and Gap experiment
1stCase:When the Bar not heated (Not raised with temperature) then the dimension will remain constant.
2nd Case:When the Bar is heated (raised with temperature) then the dimension of the Bar will increase and eventually will not pass thought the Gap.
Expansion of Solids in terms of Kinetic Theory of Matter
Explain expansion of solids in terms of kinetic theory of matter
When asolidis heated, its atoms vibrate faster about their fixed points. The relative increase in the size of solids when heated is therefore small. Metal railway tracks have small gaps so that when the sun heats them, the tracks expand into these gaps and don’t buckle.
Forces due toexpansion
Normally Expansion and contraction is accompanied by tremendous forces; The presence of force indicates the expansion and contraction is Resisted.
Bar breaker
• Consists a strong metal blocker with a pair of vertical jaws J and a strong metal Bar R. The metal bar has a wing nut N at one end and an eye at the other end.
• The bar is placed between the two pair of Jaws
• A short cast iron C is inserted in the eye of the bar.
• The bar is then heated it expands and the wing not screwed to tighten the bar R against the jaws.
• The bar is then allowed to cool as it cools, it contracts the short cast Iron rod C which presses against the jaws of the bar breaker, Resists the contraction of the Metal bar R.
• The resistance to the contraction of the bar sets up very large forces which breaks the short cast Iron rod C in the eye of metal bar, Because the contracting bar is trying to pull itself through the small gap in the frame.
Expansivity of Different Solids
Identify expansivity of different solids
The coefficient of linear expansion or linear expansivity (x)
The amount by which the linear dimension of a given solid expands depends on:
• The length of the solid
• The temperature to which the solid is heated
• The nature of solid
Coefficient of linear Expansivity (x) is the fractional increase in length of the solids per original length per degree rise in temperature.
Mathematical formula
Thus
SI Unit of coefficient of linear expansivity (X) is per degree centigrade ÂșC -1
Example 1
A copper rod has a length of 40cm on a day when the temperature of the room is 22.3ÂșC. What will be its length become on a day when the temperature of the room is 30ÂșC ? The linear expansivity of copper is 0. 000017/ÂșC
Solution
Table which shows the coefficient of Expansivity Constant.
Substance Linear Expasivity (PerÂșC) Aluminum (Al) 0.000026 Brass 0.000019 Copper 0.000017 Iron 0.000012 Steel 0.000012 Concrete 0.000011 Glass 0.0000085 Invar ( Alloy of Iron and Nickel) 0.000001
The Application of Expansion of Solids in Daily Life
Explain the applications of expansion of solids in daily life
There is a large number of important practical applications of thermal expansion of solids; While laying the railway tracks, a small gap is left between the successive lengths of the rails. This will allow expansion during a hot day. The iron tyre is to be put on a wheel, the tyre is first heated until its diameter becomes more than that of the wheel and is then slipped over the wheel.
The Apparent Expansion of Liquids
Explain the apparent expansion of liquids
Liquids expands when heated and contracts when cooled. It is easier to observe expansion in liquids than in solids. Different liquids expand at different rates in response to the same temperature change. Liquids expand much more than solids for equal changes of temperature. Apparent expansion of liquids is always less than the true expansion of the liquid.
Demonstrate the Effects of Heat on Liquids
Demonstrate the effect of heat on liquids
Liquids unlike solids can be poured. If a liquid is poured into a vessel, it takes the shape of the vessel. For this reason a liquid can't have linear and aerial expansivity, thus liquids have only volume expansivity. Liquids molecules have kinetic energy. This energy increases if the temperature of the liquid is raised by heating. Heating causes the molecules of liquid to move faster.
Activity 1
Items Needed
• Large heat safe glass bowl
• Cooking Oil
• Food Coloring
• Two 2×4 blocks
• Candle
• Match or Lighter
Instructions
1. Begin by filling a large glass bowl with cooking oil.
2. Next, add between 5-10 drops of food coloring into the oil. Helpful Tip: Place the drops near the center of the bowl.
3. Prop the bowl up off the table using two 2×4 blocks.
4. Light a candle and carefully place it under the bowl. The flame of the candle should touch the bottom of the glass bowl.
5. Look through the side of the glass bowl and watch carefully to observe what happens. Helpful Tip: It will likely take 5 minutes before you see anything happen to the liquid/food coloring.
You can alternatively follow the experiment through the following vedio
Verification of Anomalous of Water
Verify the anomalous expansion of water
The instrument used to demonstrate anomalous expansion of water is called hope's apparatus. it consists of a brass cylinder, jacket J with a mixture of ice and salt and two thermometers A and B at regular intervals upward and at the bottom.The hope's experiment shows that water contracts as it cools down to 4 degree centigrade and then expands as it cooled further below 4 degree centigrade.
In the year 1805, the scientist T. C. Hope devised a simple arrangement, known as Hope’s apparatus, to demonstrate the anomalous behaviour of water.
Hope’s apparatus consists of a long cylindrical jar with two openings on the side, one near the top and the other near the bottom to fit thermometers in each of these openings. A metallic cylindrical air-tight trough with an outlet is also fitted onto the jar, on its central portion. Two thermometers are fitted air-tight in the two openings of the cylindrical jar. The thermometer near the bottom of the jar is T1, and the one near the top of the jar is T2. Now the cylindrical jar is filled with water. The cylindrical trough at the central portion of the jar is filled with a freezing mixture of ice and common salt.
The Application of Expansion of Liquids in Everyday Life
Explain the applications of expansion of liquids in everyday life
Water in lakes and ponds usually freezes in winter. Ice, being less dense floats on the water. This insulates the water below against heat loss to the cold air above. Water at 4 degree centigrade being most dense, remains at the bottom of the lake, while ice, being less dense than water floats on the layers of water. This enables aquatic animals to survive in the water below the ice.
The Concept of Thermal Expansion of Gases
Explain the concept of thermal expansion of gases
Gases expand when heated just like solids and liquids. This is because the average kinetic energy of the molecules in a gas is directly proportional to the absolute temperature of the gas. Heating the gas increases the kinetic energy of its molecules, making them vibrate more vigorously and occupy more space.
The Relationship between Volume and Temperature of Fixed Mass of Air at Constant Pressure
Investigate the relationship between volume and temperature of fixed mass of air at constant pressure
Three properties are important when studying the expansion of gases. These are; pressure, volume and temperature. Charles law states that the volume of a fixed mass of gas is directly proportional to the absolute (Kelvin) temperature provided the pressure remains constant. Mathematically V1T2= V2T1.
Example 2
the volume of gas at the start is recorded as 30 cm3with a temperature of 30°C. The cylinder is heated further till the thermometer records 60°C. What is the volume of gas?
Solution:
We know,V/T = constant
therefore,
V1/T1=V2/T2
V1 =30 cm3
T1 =30°C = 30+273 = 303K(remember to convert from Celsius to Kelvin)
T2 =60°C = 60+273 = 333K
V2 =?
V1/T1=V2/T2
V2=V1xT2/T1
V2=30x333/303
= 32.97 cm3
The Relationship between Pressure and Volume of a Fixed Mass of Air at Constant Temperature
Investigate the relationship between pressure and volume of a fixed mass of air at constant temperature
The relationship obtained when the temperature of a gas is held constant while the volume and pressure are varied is known as Boyle’s law. Mathematically, P1V1 = P2V2. Boyle's law states that the volume of a fixed mass of gas is inversely proportional to its pressure if the temperature is kept constant.
PressurexVolume = constant
pxV = constant
Example 3
The volume of gas at the start is 50 cm3with a pressure of 1.2 x 105Pascals. The piston is pushed slowly into the syringe until the pressure on the gauge reads 2.0 x 105Pascals. What is the volume of gas?
Solution:
We know
p x V = constant
therefore,
p1xV1= p2xV2
p1=1.2 x 105Pascals
V1=50 cm3
p2=2.0 x 105Pascals
V2=?
p1xV1= p2xV2
V2=p1xV1/p2
V2=1.2x105x50/2.0 x 105
V2= 30 cm3
The Relationship between Pressure and Temperature of a Fixed Mass of Air at Constant Volume
Investigate the relationship between pressure and temperature of a fixed mass of air at constant volume
To investigate the relationship between the pressure and the temperature of a fixed mass, the volume of the gas is kept constant. The pressure is then measured as the temperature is varied. P1/T1 = P2/T2 ,this is called pressure law. The pressure law states that the pressure of a fixed mass of a gas is directly proportional to the absolute temperature if the volume is kept constant
Example 4
Pressure of gas is recorded as 1.0 x 105N/m2at a temperature of 0°C. The cylinder is heated further till the thermometer records 150°C. What is the pressure of the gas?
Solution:
We know,p/T = constant
therefore,
p1/T1= p2/T2
p1=1.0 x 105N/m2
T1=0°C = 0+273 = 273K(remember to convert from Celsius to Kelvin)
T2=150°C = 150+273 = 423K
p2 =?
p1/T1= p2/T2
p2=p1xT2/T1
p2=1.0x105x423/273
= 1.54 x 105N/m2
The General Gas Equation from the Gas Laws
Identify the general gas equation from the gas laws
The three gas laws give the following equations:
1. pV = constant(when T is kept constant)
2. V/T = constant(when p is kept constant)
3. P/T= constant(when V is kept constant)
These 3 equations are combined to give the ideal gas equation:
Where,
• p = the pressure of the gas
• V = the volume the gas occupies
• T = the gas temperature on the Kelvin scale
From this equation we know that if a fix mass of gas has starting values of p1, V1 and T1, and then some time later has value p2, V2 and T2, the equation can be written as:
Exercise 1
Sabah pumps up her front bicycle tyre to 1.7 x 105Pa. The volume of air in the tyre at this pressure is 300 cm3. She takes her bike for a long ride during which the temperature of the air in the tyre increases from 20°C to 30°C. Calculate the new front tyre pressure assuming the tyre had no leaks and so the volume remained constant?
Absolute Scale of Temperature
Explain absolute scale of temperature
Absolute zero is the lowest temperature that can be attained theoretically. It is not possible to attain this temperature because all gases liquefy before attaining it. The kelvin scale of temperature is obtained by shifting the vertical axis to -273 degrees Celsius and renaming it 0 K. On the scale 0 degrees Celsius becomes 273 K and 100 degrees Celsius corresponds with 373 K.
Convertion of Temperature in Degrees Centigrade (Celsius) to Kelvin
Convert temperature in degrees centigrade (celsius) to kelvin
The Kelvin temperature scale takes its name after Lord Kelvin who developed it in the mid 1800s. It takes absolute zero as the starting point and temperature measurements are given the symbol K (which stands for "Kelvin"). Temperature differences on the Kelvin scale are no different to those on the Celsius (°C) scale. The two scales differ in their starting points. Thus, 0°C is 273K.
Converting from Celsius to Kelvin
• Temperature in °C + 273 = Temperature in K
Converting from Kelvin to Celsius
• Temperature in K – 273 = Temperature in °C
Example 5
The temperature of a gas is 65 degrees Celsius. Change it to the kelvin scale.
Solution
T(K) = degrees Celsius + 273, T(K) = 65+273
therefore T(K) = 338 K.
Standard Temperature and Pressure (S.T.P)
Explain standard temperature and pressure (S.T.P)
The standard temperature and pressure (S.T.P) is a set of conditions for experimental measurements to enable comparisons to be made between sets of data. The standard temperature is 0 degrees Celsius (273 K) while the standard pressure is 1 atmosphere (101300 Pa or 760 mm of mercury).
Expansion of Gas in Daily Life
Apply expansion of gas in daily life
Land and sea breezes are the result of expansion of air caused by unequal heating and cooling of adjacent land and sea surfaces. The piston engine and firing bullets from guns work under principles of expansion of gases.
...
Thanks for reading PHYSICS FORM THREE TOPIC 5: THERMAL EXPANSION
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With JEE Main & JEE Advanced is just around the corner, it is time engineering aspirants pop-up their preparations for the entrance exam. JEE Main & JEE Advanced question paper consists of questions from Mathematics, Chemistry and Physics. The folklore among students is always considering Mathematics to be a tricky subject and turn a blind eye towards it. Many Students find Mathematics a little overwhelming while preparing for the exam. Is this myth true about Mathematics? The answer is No. In one line: "Maths is tricky but not difficult". You just have to know the right way to approach it.
Physics, Chemistry & Maths have equal weightage in the JEE Main 2019, but Maths has a real edge over the other two as, in this case of an equal overall score of two or more students, the score in mathematics is a considerably a deciding factor. The syllabus of JEE Advanced Mathematics is a perfect blend of easy and challenging topics take together to make it a competitive exam. The core method of solving mathematics is simple: First Imagine, then start. My sir used to say me “Boy, Every question in mathematics is solved two times, First in Imagination then on Paper”. It’s not the time anymore when one used to spend hours in attempting one problem. Now it is an era of behaving smartly and solving MCQs rapidly.
Mathematics Syllabus
1) Algebra: Complex Numbers, Theory of Equations & Expressions, Probability, Matrix
2) Vectors and 3D Geometry.
3) Coordinate Geometry: Circle, Conic section consisting Ellipse, Parabola & Hyperbola.
4) Calculus: Functions, Limits, Continuity & Differentiability, Application of Derivatives & Integration, Indefinite & Definite Integral.
## Strategy to Clear Mathematics: How one should work
• If we go through the previous year JEE questions, they suggest that the one should pay more attention to Vectors and 3-D than giving their one or two months to difficult and vast Indefinite integration as vectors and 3-D offers very less space to the examiner to ask questions, as variety in the problem is concerned. So there is a pattern of questions which is going to be asked about this topic.
• Each year two to three questions are asked from Complex Number. Hence, Mastering complex numbers, vectors & 3-D can’t be skipped or taken lightly at any cost.
• One can make algebra easier if they can harness the ability to imagine functions as graphs and learn to apply vertical & horizontal origin shifts carefully.
• Differential calculus has a relation with roots of equations if you can use the Rolle's & Lagrange's theorems.
• Trigonometry questions require a pure practice and dedication towards questions. Do more and more questions, and suddenly one day you will realise that you are now a Master of Trigonometry. Trust me this hard work will pay when in 12th you will do Integration, and everyone else is trying to recognise the approach of trigno's question & there you will be already with your answer.
• P&C and Probability are another important topics in algebra. One has to be thorough with the basics of Bayes concept, derangements & various ways of distribution, including cases where objects are identical and all unique.
• Matrices can be related to algebraic equations. Hence a 3x3 matrix can actually be imagined as being 3-plane in 3D geometry.
• Determinants have some very awesome properties, Learn Them.
• The basic rule of learning base of Integral calculus is to learn Trigonometry with all your heart. You can’t solve an Integral calculus problem without learning Trigonometry. Integral calculus can be simplified by learning and keeping in mind all the time some questions and way to approach them. You are also supposed to keep in mind some basic varieties of integrable functions.
• One of my friends always uses his open notebook to solve the Co-ordinate questions and could do it much faster than anyone in the class. He always uses the parametric forms of the equation for every curve. And I every time ask him - Are you going to take this to JEE exam? How will you do it there? He couldn’t figure it out. But I do. He taught me one thing- If I know can remember this parametric equation during the time of paper then I can do most of the questions fairly fast. Then I started remembering them. YOU JUST CAN’T REMEMBER THESE EQUATIONS. What you have to do is derive them while doing every question. You will be amazed to see the results. It’s worthy: Weightage of co-ordinate is 1/3rd of total mathematics.
• One can learn Chemistry from even the simplistic approach one can think. You will be in any case at least average in Physics. So what remains is Mathematics. In the beginning, my teacher said to me: “You desire to get a good rank in JEE, keep mathematics your top priority and practice it regularly.”
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posted by on .
4. The scores on a university examination are normally distributed with a mean of 62 and a standard deviation of 11. If the bottom 5% of students will fail the examination, what is the lowest score, to the nearest whole number, that a student can have and pass?
• statistics - ,
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion. Use the corresponding Z score to find your answer.
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https://www.geeksforgeeks.org/detecting-negative-cycle-using-floyd-warshall/?ref=lbp | 1,701,670,980,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100525.55/warc/CC-MAIN-20231204052342-20231204082342-00236.warc.gz | 887,596,024 | 58,509 | # Detecting negative cycle using Floyd Warshall
We are given a directed graph. We need compute whether the graph has negative cycle or not. A negative cycle is one in which the overall sum of the cycle comes negative.
Negative weights are found in various applications of graphs. For example, instead of paying cost for a path, we may get some advantage if we follow the path.
Examples:
`Input : 4 4 0 1 1 1 2 -1 2 3 -1 3 0 -1Output : YesThe graph contains a negative cycle.`
We have discussed Bellman Ford Algorithm based solution for this problem.
In this post, Floyd Warshall Algorithm based solution is discussed that works for both connected and disconnected graphs.
Distance of any node from itself is always zero. But in some cases, as in this example, when we traverse further from 4 to 1, the distance comes out to be -2, i.e. distance of 1 from 1 will become -2. This is our catch, we just have to check the nodes distance from itself and if it comes out to be negative, we will detect the required negative cycle.
Implementation:
## C++
`// C++ Program to check if there is a negative weight` `// cycle using Floyd Warshall Algorithm` `#include` `using` `namespace` `std;` `// Number of vertices in the graph` `#define V 4` ` ` `/* Define Infinite as a large enough value. This ` ` ``value will be used for vertices not connected ` ` ``to each other */` `#define INF 99999` ` ` `// A function to print the solution matrix` `void` `printSolution(``int` `dist[][V]);` ` ` `// Returns true if graph has negative weight cycle` `// else false.` `bool` `negCyclefloydWarshall(``int` `graph[][V])` `{` ` ``/* dist[][] will be the output matrix that will ` ` ``finally have the shortest ` ` ``distances between every pair of vertices */` ` ``int` `dist[V][V], i, j, k;` ` ` ` ``/* Initialize the solution matrix same as input` ` ``graph matrix. Or we can say the initial values ` ` ``of shortest distances are based on shortest ` ` ``paths considering no intermediate vertex. */` ` ``for` `(i = 0; i < V; i++)` ` ``for` `(j = 0; j < V; j++)` ` ``dist[i][j] = graph[i][j];` ` ` ` ``/* Add all vertices one by one to the set of ` ` ``intermediate vertices.` ` ``---> Before start of a iteration, we have shortest` ` ``distances between all pairs of vertices such ` ` ``that the shortest distances consider only the` ` ``vertices in set {0, 1, 2, .. k-1} as intermediate ` ` ``vertices.` ` ``----> After the end of a iteration, vertex no. k is ` ` ``added to the set of intermediate vertices and ` ` ``the set becomes {0, 1, 2, .. k} */` ` ``for` `(k = 0; k < V; k++)` ` ``{` ` ``// Pick all vertices as source one by one` ` ``for` `(i = 0; i < V; i++)` ` ``{` ` ``// Pick all vertices as destination for the` ` ``// above picked source` ` ``for` `(j = 0; j < V; j++)` ` ``{` ` ``// If vertex k is on the shortest path from` ` ``// i to j, then update the value of dist[i][j]` ` ``if` `(dist[i][k] + dist[k][j] < dist[i][j])` ` ``dist[i][j] = dist[i][k] + dist[k][j];` ` ``}` ` ``}` ` ``}` ` ``// If distance of any vertex from itself` ` ``// becomes negative, then there is a negative` ` ``// weight cycle.` ` ``for` `(``int` `i = 0; i < V; i++)` ` ``if` `(dist[i][i] < 0)` ` ``return` `true``;` ` ``return` `false``; ` `}` ` ` ` ``// driver program` `int` `main()` `{` ` ``/* Let us create the following weighted graph` ` ``1` ` ``(0)----------->(1)` ` ``/|\ |` ` ``| |` ` ``-1 | | -1` ` ``| \|/` ` ``(3)<-----------(2)` ` ``-1 */` ` ` ` ``int` `graph[V][V] = { {0 , 1 , INF , INF},` ` ``{INF , 0 , -1 , INF},` ` ``{INF , INF , 0 , -1},` ` ``{-1 , INF , INF , 0}};` ` ` ` ``if` `(negCyclefloydWarshall(graph))` ` ``cout << ``"Yes"``;` ` ``else` ` ``cout << ``"No"``; ` ` ``return` `0;` `}`
## Java
`// Java Program to check if there is a negative weight` `// cycle using Floyd Warshall Algorithm` `class` `GFG` `{` ` ``// Number of vertices in the graph` ` ``static` `final` `int` `V = ``4``;` ` ` ` ``/* Define Infinite as a large enough value. This ` ` ``value will be used for vertices not connected ` ` ``to each other */` ` ``static` `final` `int` `INF = ``99999``;` ` ` ` ``// Returns true if graph has negative weight cycle` ` ``// else false.` ` ``static` `boolean` `negCyclefloydWarshall(``int` `graph[][])` ` ``{` ` ` ` ``/* dist[][] will be the output matrix that will ` ` ``finally have the shortest ` ` ``distances between every pair of vertices */` ` ``int` `dist[][] = ``new` `int``[V][V], i, j, k;` ` ` ` ``/* Initialize the solution matrix same as input` ` ``graph matrix. Or we can say the initial values ` ` ``of shortest distances are based on shortest ` ` ``paths considering no intermediate vertex. */` ` ``for` `(i = ``0``; i < V; i++)` ` ``for` `(j = ``0``; j < V; j++)` ` ``dist[i][j] = graph[i][j];` ` ` ` ``/* Add all vertices one by one to the set of ` ` ``intermediate vertices.` ` ``---> Before start of a iteration, we have shortest` ` ``distances between all pairs of vertices such ` ` ``that the shortest distances consider only the` ` ``vertices in set {0, 1, 2, .. k-1} as intermediate ` ` ``vertices.` ` ``----> After the end of a iteration, vertex no. k is ` ` ``added to the set of intermediate vertices and ` ` ``the set becomes {0, 1, 2, .. k} */` ` ``for` `(k = ``0``; k < V; k++)` ` ``{` ` ` ` ``// Pick all vertices as source one by one` ` ``for` `(i = ``0``; i < V; i++)` ` ``{` ` ` ` ``// Pick all vertices as destination for the` ` ``// above picked source` ` ``for` `(j = ``0``; j < V; j++)` ` ``{` ` ` ` ``// If vertex k is on the shortest path from` ` ``// i to j, then update the value of dist[i][j]` ` ``if` `(dist[i][k] + dist[k][j] < dist[i][j])` ` ``dist[i][j] = dist[i][k] + dist[k][j];` ` ``}` ` ``}` ` ``}` ` ` ` ``// If distance of any vertex from itself` ` ``// becomes negative, then there is a negative` ` ``// weight cycle.` ` ``for` `(i = ``0``; i < V; i++)` ` ``if` `(dist[i][i] < ``0``)` ` ``return` `true``;` ` ``return` `false``; ` ` ``}` ` ` ` ``// Driver code` ` ``public` `static` `void` `main (String[] args)` ` ``{` ` ` ` ``/* Let us create the following weighted graph` ` ``1` ` ``(0)----------->(1)` ` ``/|\ |` ` ``| |` ` ``-1 | | -1` ` ``| \|/` ` ``(3)<-----------(2)` ` ``-1 */` ` ` ` ``int` `graph[][] = { {``0``, ``1``, INF, INF},` ` ``{INF, ``0``, -``1``, INF},` ` ``{INF, INF, ``0``, -``1``},` ` ``{-``1``, INF, INF, ``0``}};` ` ` ` ``if` `(negCyclefloydWarshall(graph))` ` ``System.out.print(``"Yes"``);` ` ``else` ` ``System.out.print(``"No"``); ` ` ``}` `}` `// This code is contributed by Anant Agarwal.`
## Python3
`# Python Program to check` `# if there is a` `# negative weight` `# cycle using Floyd` `# Warshall Algorithm` ` ` `# Number of vertices` `# in the graph` `V ``=` `4` ` ` `# Define Infinite as a` `# large enough value. This ` `# value will be used ` `#for vertices not connected ` `# to each other ` `INF ``=` `99999` ` ` `# Returns true if graph has` `# negative weight cycle` `# else false.` `def` `negCyclefloydWarshall(graph):` ` ` ` ``# dist[][] will be the` ` ``# output matrix that will ` ` ``# finally have the shortest ` ` ``# distances between every` ` ``# pair of vertices ` ` ``dist``=``[[``0` `for` `i ``in` `range``(V``+``1``)]``for` `j ``in` `range``(V``+``1``)]` ` ` ` ``# Initialize the solution` ` ``# matrix same as input` ` ``# graph matrix. Or we can` ` ``# say the initial values ` ` ``# of shortest distances` ` ``# are based on shortest ` ` ``# paths considering no` ` ``# intermediate vertex. ` ` ``for` `i ``in` `range``(V):` ` ``for` `j ``in` `range``(V):` ` ``dist[i][j] ``=` `graph[i][j]` ` ` ` ``''' Add all vertices one` ` ``by one to the set of ` ` ``intermediate vertices.` ` ``---> Before start of a iteration,` ` ``we have shortest` ` ``distances between all pairs` ` ``of vertices such ` ` ``that the shortest distances` ` ``consider only the` ` ``vertices in set {0, 1, 2, .. k-1}` ` ``as intermediate vertices.` ` ``----> After the end of a iteration,` ` ``vertex no. k is ` ` ``added to the set of` ` ``intermediate vertices and ` ` ``the set becomes {0, 1, 2, .. k} '''` ` ``for` `k ``in` `range``(V):` ` ` ` ``# Pick all vertices ` ` ``# as source one by one` ` ``for` `i ``in` `range``(V):` ` ` ` ``# Pick all vertices as` ` ``# destination for the` ` ``# above picked source` ` ``for` `j ``in` `range``(V):` ` ` ` ``# If vertex k is on` ` ``# the shortest path from` ` ``# i to j, then update` ` ``# the value of dist[i][j]` ` ``if` `(dist[i][k] ``+` `dist[k][j] < dist[i][j]):` ` ``dist[i][j] ``=` `dist[i][k] ``+` `dist[k][j]` ` ` ` ``# If distance of any` ` ``# vertex from itself` ` ``# becomes negative, then` ` ``# there is a negative` ` ``# weight cycle.` ` ``for` `i ``in` `range``(V):` ` ``if` `(dist[i][i] < ``0``):` ` ``return` `True` ` ` ` ``return` `False` ` ` `# Driver code` ` ` `''' Let us create the` ` ``following weighted graph` ` ``1` ` ``(0)----------->(1)` ` ``/|\ |` ` ``| |` ` ``-1 | | -1` ` ``| \|/` ` ``(3)<-----------(2)` ` ``-1 '''` ` ` `graph ``=` `[ [``0``, ``1``, INF, INF],` ` ``[INF, ``0``, ``-``1``, INF],` ` ``[INF, INF, ``0``, ``-``1``],` ` ``[``-``1``, INF, INF, ``0``]]` ` ` `if` `(negCyclefloydWarshall(graph)):` ` ``print``(``"Yes"``)` `else``:` ` ``print``(``"No"``) ` `# This code is contributed` `# by Anant Agarwal.`
## C#
`// C# Program to check if there` `// is a negative weight cycle` `// using Floyd Warshall Algorithm` `using` `System;` `namespace` `Cycle` `{` `public` `class` `GFG` `{ ` ` ` ` ``// Number of vertices in the graph` ` ``static` `int` `V = 4;` ` ` ` ``/* Define Infinite as a large enough value. This ` ` ``value will be used for vertices not connected ` ` ``to each other */` ` ``static` `int` `INF = 99999;` ` ` ` ``// Returns true if graph has negative weight cycle` ` ``// else false.` ` ``static` `bool` `negCyclefloydWarshall(``int` `[,]graph)` ` ``{` ` ` ` ``/* dist[][] will be the output matrix that will ` ` ``finally have the shortest ` ` ``distances between every pair of vertices */` ` ``int` `[,]dist = ``new` `int``[V,V];` ` ``int` `i, j, k;` ` ` ` ``/* Initialize the solution matrix same as input` ` ``graph matrix. Or we can say the initial values ` ` ``of shortest distances are based on shortest ` ` ``paths considering no intermediate vertex. */` ` ``for` `(i = 0; i < V; i++)` ` ``for` `(j = 0; j < V; j++)` ` ``dist[i,j] = graph[i,j];` ` ` ` ``/* Add all vertices one by one to the set of ` ` ``intermediate vertices.` ` ``---> Before start of a iteration, we have shortest` ` ``distances between all pairs of vertices such ` ` ``that the shortest distances consider only the` ` ``vertices in set {0, 1, 2, .. k-1} as intermediate ` ` ``vertices.` ` ``----> After the end of a iteration, vertex no. k is ` ` ``added to the set of intermediate vertices and ` ` ``the set becomes {0, 1, 2, .. k} */` ` ``for` `(k = 0; k < V; k++)` ` ``{` ` ` ` ``// Pick all vertices as source one by one` ` ``for` `(i = 0; i < V; i++)` ` ``{` ` ` ` ``// Pick all vertices as destination for the` ` ``// above picked source` ` ``for` `(j = 0; j < V; j++)` ` ``{` ` ` ` ``// If vertex k is on the shortest path from` ` ``// i to j, then update the value of dist[i][j]` ` ``if` `(dist[i,k] + dist[k,j] < dist[i,j])` ` ``dist[i,j] = dist[i,k] + dist[k,j];` ` ``}` ` ``}` ` ``}` ` ` ` ``// If distance of any vertex from itself` ` ``// becomes negative, then there is a negative` ` ``// weight cycle.` ` ``for` `(i = 0; i < V; i++)` ` ``if` `(dist[i,i] < 0)` ` ``return` `true``;` ` ``return` `false``; ` ` ``}` ` ` ` ``// Driver code` ` ``public` `static` `void` `Main()` ` ``{` ` ` ` ``/* Let us create the following weighted graph` ` ``1` ` ``(0)----------->(1)` ` ``/|\ |` ` ``| |` ` ``-1 | | -1` ` ``| \|/` ` ``(3)<-----------(2)` ` ``-1 */` ` ` ` ``int` `[,]graph = { {0, 1, INF, INF},` ` ``{INF, 0, -1, INF},` ` ``{INF, INF, 0, -1},` ` ``{-1, INF, INF, 0}};` ` ` ` ``if` `(negCyclefloydWarshall(graph))` ` ``Console.Write(``"Yes"``);` ` ``else` ` ``Console.Write(``"No"``); ` ` ``}` `} ` `}` `// This code is contributed by Sam007.`
## Javascript
``
Output
```Yes
```
The time complexity of the Floyd Warshall algorithm is O(V^3) where V is the number of vertices in the graph. This is because the algorithm uses a nested loop structure, where the outermost loop runs V times, the middle loop runs V times and the innermost loop also runs V times. Therefore, the total number of iterations is V * V * V which results in O(V^3) time complexity.
The space complexity of the Floyd Warshall algorithm is O(V^2) where V is the number of vertices in the graph. This is because the algorithm uses a 2D array of size V x V to store the shortest distances between every pair of vertices. Therefore, the total space required is V * V which results in O(V^2) space complexity.
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Functionalists note that education has replaced some traditional family functions. (Points : 2) true or false
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,184 | 3,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-30 | longest | en | 0.953563 |
https://republicofsouthossetia.org/question/which-equation-can-be-used-to-find-the-volume-of-the-cylinder-a-cylinder-with-a-radius-of-4-inch-16112526-94/ | 1,638,480,709,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362297.22/warc/CC-MAIN-20211202205828-20211202235828-00472.warc.gz | 520,187,784 | 13,280 | ## Which equation can be used to find the volume of the cylinder? A cylinder with a radius of 4 inches and height of 16 inches. V = (4) squar
Question
Which equation can be used to find the volume of the cylinder? A cylinder with a radius of 4 inches and height of 16 inches. V = (4) squared (16) V = pi (4) squared (16) V = pi (8) squared (16) V = pi (16) squared (4) PLZ HELP I NEED THIS TEST DONE IN LIKE 5MINS
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## Answers ( )
The answer is option 2, V = π × 4 squared × 16.
Step-by-step explanation:
It is given that the formula of volume of cyclinder is V = πr²h, where r is the radius and h is the height :
Let r = 4 inches,
Let h = 16 inches,
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share
## Examplesoperations's examples
• They both explain fraction operations with a visual model of a pie, and let you practice sufficiently with picture exercises so you will teaches fractions and their operations with visual models, the sequel. — “Free Online Fraction Calculator”,
• Dark Ops ON SALE Dark Ops Batteries, Dark Ops Flashlight Accessories, Dark Ops Flashlights, Dark Ops Riflescope Mounts, Rings & Bases, Dark Ops Riflescopes. — “Dark Ops Products SALE Dark_Ops BRAND MERCHANDISE Darkops”,
• Learn what it takes to hire and train the right people to run your day-to-day operations. Your managers, business technology, staff training, customer relations strategy, and business processes all come together to make a company operate smoothly. — “Running a Business : Advice on Running a Small Business”,
• Explains the order of operations ('PEMDAS') in plain terms, points out common mistakes, and presents worked examples. — “The Order of Operations: PEMDAS”,
• Ops Ops , Roman goddess of abundance. Her festivals, Opalia on 19 August and Opiconsivia on 25 August, were close to festivals of Consus , the god of. — “Ops: Information from ”,
• Brief and Straightforward Guide: In Baseball, What Is OPS? On base percentage plus slugging percentage (OPS) is a relatively new measurement of a hitter's performance in baseball. — “In Baseball, What Is OPS?”,
• Philippine reports updated weekly. Profiles of Cabinet Secretaries. Government Directory. Media Directory. OTHER LINKS. Designed and Managed by the EDP / M.I.S. Division. Presidential Communications Operations Office (PCOO)) Malacanang, Philippines. — “Office of the Press Secretary”, ops.gov.ph
• Simple problems that introduce order of operations for basic arithmetic, including parentheses and exponents. Hints. Basic Order of Operations. Order of Operations With Parentheses. Order of Operations With Exponents. — “Dad's Worksheets - Order of Operations”,
• Definition of operations in the Online Dictionary. Meaning of operations. Pronunciation of operations. Translations of operations. operations synonyms, operations antonyms. Information about operations in the free online English dictionary and. — “operations - definition of operations by the Free Online”,
• Information and resources for the Omaha public schools, including calendar of events and links to individual schools. Congratulations to OPS students and staff for these recent accomplishments! Privacy Statement | Terms Of Use | Comments to Web Team | Register. — “Omaha Public Schools”,
• Multiplying two fractions is the easiest of any of the operations. Try the practice below to be sure you understand how to perform these operations on fractions. — “Operations on Fractions”, cstl.syr.edu
• This page lists our free online video tutorials on operations, operations word problems, and printable operations worksheets. — “Operations Help : Videos | Worksheets | Word Problems”,
• operations - definition of operations from : Jobs or tasks comprising of one or more elements or subtasks, and which are performed typically in one location. Operations transform resource or data inputs into desired goods,. — “operations definition”,
• Order of Operations. When a numerical expression involves two or more The proper application of "order of operations" is needed when working with such. — “Order of Operations”,
• One fault of OPS is that it weighs on-base average and slugging percentage equally, such as wOBA build on this distinction using linear weights, avoiding OPS' flaws. — “On-base plus slugging - Wikipedia, the free encyclopedia”,
• Partial Operations. Why Work With Abstract Entities and Binary Operations? 1. Working with Partial Operations. Why Work With Abstract Entities and Binary Operations? 1. Working with. — “Binary Operations”, www2.latech.edu
• offensive, defensive, and stability or civil support operations simultaneously as part of persistent conflict only by conducting military operations in concert with diplomatic,. — “OPERATIONS”, army.mil
• What does OPS stand for? Definition of OPS in the list of acronyms and abbreviations provided by the Free Online Dictionary and Thesaurus. — “OPS - What does OPS stand for? Acronyms and abbreviations by”,
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• bestjobsonline: bestjobsonline: Operations Support Manager - http://t.co/pjhDtG8 #jobs #FrontlineSourceGroup #Shreveport
• easterncloud: easterncloud: RT @ArabicWords Shaykh Al Albani: Suicide Operations Today Are All Haram http://t.co/KN76JNJ #islam
• CannibalCorona: CannibalCorona: RT @Anonymously37: I feel we could do more if we all quit spaming Operations for a minute. And talked to each other. I assume we don't all live in one city.
• HRNewsJobs: HRNewsJobs: Executive Coord Sr: thorough knowledge of company operations. Basic Requirements The Executive Coordin... http://t.co/XvisV4R #hr #jobs
• memeflag: memeflag: ‘Top Secret America’ A look at the military’s Joint Special Operations Command - The Washington Post http://t.co/z942lbP via @washingtonpost
• Alexweir1949: Alexweir1949: RT @cdashiell: JSOC operates above the law, has authority to assassinate http://t.co/sWetuyD
• Noalimperialism: Noalimperialism: RT @MitraCNN: Special forces from Britain, France, Jordan, Qatar on ground in Libya stepped up operations in Tripoli n other cities - NATO official
• microitcoach: microitcoach: Enterprise Small Business Sales and Operations Specialist EU ...: Google is hiring! View and apply for the Enter... http://t.co/44byAuZ
• ProjectMgmtGuy: ProjectMgmtGuy: ProjectMgmtGuy: Boston Blazers Suspend Operations, Leave a Hole in the Boston Lacrosse Scene http://t.co/VG5nVRa
• Callum_Sonic_: Callum_Sonic_: @fourzerotwo operations on COD ELITE ar set @ completely unfair times, ther set for USA aftanoon wat bout english midnite wen we need sleep?
• lexxioca: lexxioca: #Jobs #Cambridge IAN MARTIN LIMITED / THE 500 STAFFING INC In Need Of Operations, Service Supervisor With A Tech... http://t.co/UaSXYny
• ArabicWords: ArabicWords: Shaykh Al Albani: Suicide Operations Today Are All Haram http://t.co/Vd6R2DD #islam
• LashawniquaG: LashawniquaG: RT @CiFWatch: Wikileaks: Gaza City's Shifa Hospital described as "an operations center for Hamas" during Op. Cast Lead http://t.co/V45bTbP #wlfind
• srubenfeld: srubenfeld: The Washington Post profiles JSOC http://t.co/vnl2tBa
• WebDyrWS: WebDyrWS: Management Consulting Firm in USA: DistribuTek offer for comprehensive expertise in facilities and operations p... *http://t.co/z3lb7UX
• WebDyrWS: WebDyrWS: Management Consulting Firm in USA: DistribuTek offer for comprehensive expertise in facilities and operations pl... http://t.co/k58vA9c
• CargoNews1: CargoNews1: Which is the best 3PL/4PL in China? http://t.co/M5AmsBn
• Rickyrocket1: Rickyrocket1: A Leadership Magic formula: Substitute Desired goals Along with Operations Using The Discussed Wish http://t.co/VV0sX1k
• Vivienfueyd: Vivienfueyd: Windows Reseller web hosting operations and steps might be more favorite among Countless The black htc desire's ... http://t.co/kyDWOtm
• Reannaruoyr: Reannaruoyr: The Key Benefits of Windows Reseller Web Hosting Over Linux ...: Windows Reseller web hosting operations and s... http://t.co/rbwG3FE
• MissTrust90: MissTrust90: Sometimes, I dress as a mime. Everybody loves a mime. This would obviously come in quite useful during police stealth operations.
• Anonymously37: Anonymously37: I feel we could do more if we all quit spaming Operations for a minute. And talked to each other. I assume we don't all live in one city.
• RecruitingGuru: RecruitingGuru: Looking for a Director of IT Operations in Kansas City, MO http://t.co/yQ58Rx2 #job
• willrelo: willrelo: Operations Manager- Retail - IKEA - Brooklyn, NY: Will and ability for further development with... http://t.co/5XE9Vx4 #jobs #willrelo
• blazeryu: blazeryu: RT @UNPeacekeeping: What is the role of the UN General Assembly in UN Peacekeeping? Find out here http://t.co/kUUuIgr
• HourlyJobs: HourlyJobs: District Manager North Eastern Pennsylvania job at Family Dollar Operations, I... in Allentown, PA #jobs http://t.co/4h94Bzc
• thejobdaddy: thejobdaddy: Director Sales Operations - J. Patrick & Associates - Atlanta, GA - ResponsibilitiesThe Director of Sales... http://t.co/W37ddmn
• dj_mousfand: dj_mousfand: I liked a @YouTube video http://t.co/8GH2a7F Jon O'Bir feat. Fisher - Found A Way (Joint Operations Cen
• anthonydanger11: anthonydanger11: Hiring a Sr. Unix Operations Manager in Irving, TX http://t.co/EdeszFd #job
• LD10: LD10: can you name how many countries the US currently has combat operations in?
• TradeForex2FXco: TradeForex2FXco: http://t.co/BxNUCmb #Ecommerce #Operations, John Thornhill Digital Mentorship Monthly: Follow… http://t.co/ozysDak
• FSGITCONSULTING: FSGITCONSULTING: FSO 360T; a web-based technology platform proven to deliver cost efficiencies to clients and 360° views of their clients' operations.
• Romanuva: Romanuva: ‘Top Secret America’: A look at the military’s Joint Special Operations Command http://fun.ly/vstk
• FourPawAction: FourPawAction: RT @winstontryball Treibball Chief Dog of Operations ~Treibball Dog Sport http://t.co/fVvFde8 ~High Paw ♥♥missPhoebe♥♥ http://t.co/nS3Igpx
• sfmjobs: sfmjobs: #sf Director of Online Operations - 2K Games http://t.co/XoK0zsA #marketing #job
• sfmjobs: sfmjobs: #sf Director of Operations http://t.co/OguS3Sj #marketing #job
• ShamaSukul: ShamaSukul: A Continuously Innovative and Customer Centric Operations Strategy is Vital for Survival in today's Digital World http://t.co/ln1pMsi
• ExquisiteIdea: ExquisiteIdea: Which is the best 3PL/4PL in China?: http://t.co/By0MD7K
• ExquisiteIdea: ExquisiteIdea: Dealing with Human behavior within the work place and the best way to challenge the extremes. I would like to ... http://t.co/aX5UihY
• zachinglis: zachinglis: @rayhanrafiq They definitely do. Some of their stores are awful but their online and general operations are top notch.
• John_Corey: John_Corey: From hostage team to secret army http://t.co/75mnLF9
• HennersOnAMoped: HennersOnAMoped: Also, there should be minor toe operations floating around somewhere, but they will not be allowed to conflict with the driving test
• mk_allen: mk_allen: #NSHMBA Quality Operations Nurse Job (Georgia, US): Quality Operations NurseJob ID: 2011-1607... http://t.co/4zU1ESV #jobs #Healthcare
• JobsInSydney: JobsInSydney: Manufacturing Operations Support - Sydney Australia #job http://t.co/TCH1VCh
• ClarityBell: ClarityBell: Technically, all operations are on a knife edge.
• joakinen: joakinen: RT @nytjim: Amazing quote in WaPo piece on US secret army: “We’re the force that orders the universe but can’t be seen.” http://t.co/dZ5dx3G
• GraysonCounty: GraysonCounty: Curious George types hanging out around firefighting operations North of US82 and West of SH289 - PLEASE GO HOME!... http://t.co/ayzJleP
• cdashiell: cdashiell: JSOC operates above the law, has authority to assassinate http://t.co/sWetuyD
• LyonsHaydn: LyonsHaydn: @Charlie26Adam would you please send me a signed pic or a match shirt u have worn horrible year with 2major operations :/ would mean so mch
• AshleyHoover1: AshleyHoover1: Looking for an Escrow/Operations Manager in Los Angeles, CA http://t.co/iFpTHyN #job
• Laibaah: Laibaah: According to Sherry Rehman, U.S. military operations in Afghanistan is deepening the state-society rift within Pakistan. (Viva ISI proxies)
• Kristavvs: Kristavvs: #article Status of 2010 Census Operations (Part 1) http://t.co/Ezsx2v2 #marketing
• Kazukoitt: Kazukoitt: #article Status of 2010 Census Operations (Part 1) http://t.co/Qik61YP #marketing
• Genevarym: Genevarym: #article Status of 2010 Census Operations (Part 1) http://t.co/umlbA1z #marketing
• Voncilerjc: Voncilerjc: #article Status of 2010 Census Operations (Part 1) http://t.co/Q5nmw9V #marketing
• Lilliekcp: Lilliekcp: #internetmarketing Status of 2010 Census Operations (Part 1) http://t.co/xEcuQGJ #articlemarketing
• Maryalicedhi: Maryalicedhi: #article Status of 2010 Census Operations (Part 1) http://t.co/FzL3e7O #marketing
• bestjobsonline: bestjobsonline: QUALITY ASSURANCE- CONTRACT ADMINISRATOR-Operations - http://t.co/zEKwzNL #internships #CoreMedicalGroup #Salem
• Alltop_PR: Alltop_PR: Time, Tide and the Net Wait for No One by Chief of Naval Operations Adm. Gary Roughead http://t.co/Ma0R8Fc
• PublicityNews: PublicityNews: Time, Tide and the Net Wait for No One by Chief of Naval Operations Adm. Gary Roughead: In June 2011, I was ... http://t.co/FNNWBOJ #pr
• NGStevie: NGStevie: Today last day in kitchen. Tomorrow first day in CIM Operations wheeeeeeeeeeeeeeeeeee !!! FHM in 2 more weeks.........
• Morico91: Morico91: Job: AdWords Online Community Project Manager, Global Advertising Operations (Mountain View, CA) http://t.co/S39OZE8 #Jobs
• mahmoudkablan: mahmoudkablan: RT @NewsInLibya: #Libya's Ras Lanuf refinery to resume operations: (MENAFN) #Libya's Ras Lanuf Oil and Gas Processing Company's... http://t.co/O3kHItT
• actionnews17: actionnews17: Update from Emergency Ops Center 6PM Saturday St. Tammany Parish Emergency Operations Center Statement by Parish... http://t.co/qLvHaOm
• JossanSE: JossanSE: RT @NewsInLibya: #Libya's Ras Lanuf refinery to resume operations: (MENAFN) #Libya's Ras Lanuf Oil and Gas Processing Company's... http://t.co/O3kHItT
• NewsInLibya: NewsInLibya: #Libya's Ras Lanuf refinery to resume operations: (MENAFN) #Libya's Ras Lanuf Oil and Gas Processing Company's... http://t.co/O3kHItT
• tony_topping: tony_topping: Jason Bourne had sub multiple personalities one is the Bourne Identity the other David Webb and such black mind operations are very real.
• NYSalesJob: NYSalesJob: #nyc Middle/Back Office Operations Senior Manager http://t.co/kCJNh7x #sales #jobs
• Joga51: Joga51: @Energy_UKsModel Hun, what happened? Why you need that 2 operations? If you dont want to answer i'l understand.
• shagers: shagers: @nbehrens83 Real Job: IT or Business Process improvement or Operations. Usually in Management.
• SK_Jobs: SK_Jobs: Operations Technician: RSC Equipment Rental, Inc. (Saskatoon SK): "...In this position you w... http://t.co/TbJAWNj #Saskatchewan #jobs
• D_Cortorreal: D_Cortorreal: She grew up on to The Operations pricelist hell, with no mother to protect or tell. That independence is much closer than you think
• JessicaWhiote: JessicaWhiote: Search for missing fishing boat | ieyenews: The RCIPS Marine and Air Operations Units, along with the US Coast G... http://t.co/urhSF50
• tcothub: tcothub: Colombia drug-traffickers seized - Thirty-six suspects have been arrested in two operations against... http://t.co/0PkoaWG
• bestjobsonline: bestjobsonline: Marketing Operations Coordinator - http://t.co/42T4mWn #internships #JacksonNationalLifeInsuranceCompany #Denver
• ionstrategic: ionstrategic: We build business together. Learn about our Executive Vice President of Business Operations http://t.co/HnUJZo6
• Hound_JobSearch: Hound_JobSearch: #jobs #careers Specialist DNS Network Operations: This position will provide critical DNS/DHCP/IPAM support for ... http://t.co/HannHVw
• DrSamRizk: DrSamRizk: In men, nasal operations are often combined with a chin implant to create a more balanced and masculine appearance
• tony_topping: tony_topping: The Bourne Ultimatum Treadstone etc how close to real life such mind control operations are...
• TennieMielke: TennieMielke: Operations in Low Intensity Conflict: PREFACE This manual provides tactical-level guidance to brigade and batt... http://t.co/u5j4uyW
• SchwartzNow: SchwartzNow: ‘Top Secret America’: Military’s Joint Special Operations Command in NC - http://t.co/PjdtQF3
• vatyma: vatyma: “We’re the force that orders the universe but can’t be seen.” http://t.co/4hXvqgv #epicquote #USSecretArmy
## Blogs & Forumblogs and forums about operations
• “I recently got accepted to present at a nonprofit research conference about the Nonprofit Management and Operations Blog, Wiki and Brown Bag Lunch discussions as being collaborative learning and professional development tools for nonprofit managers”
— Nonprofit Management and Operations Blog | Aspiration,
• “The purpose of the Forum area is spelled out in its mission statement: The OR Forum area invites work that Papers that address prospects in areas not traditionally covered by Operations Research”
Operations Research Forum,
• “Behind the scenes at Tiger Technologies. Announcements, technical details, useful tips, thoughts on the Web hosting industry, and more”
— Tiger Technologies Blog,
• “So ends the ARRC's annual exercise - ARRCADE FUSION 09 (link to web page): ambitious, ground breaking and designed to test HQ ARRC's ability to integrate Allied Command Operations Blog”
— Ex Arrcade Fusion 09 – Putting the theory into practice,
• “Operations Manager's Blog - Windows Live Nothing else is changing at this blog; other technical articles will continue to be posted here at ops-”
Operations Manager's Blog - Windows Live, ops-
• “SWIFT Operations Forum Europe. Dates. 13 - 15 December 2010. Venue. Dolce your forum. As well as keeping you informed about our plans, the forum encourages”
— SWIFT - SWIFT Operations Forum Europe - Welcome,
• “operations course (1) operations courses (1) operations internship (1) operations (1) women computer science (1) Blog Archive. 2010 (25) November (3)”
— Leaders for Global Operations Blog, lgo-blog.mit.edu
• “Welcome to the S&OP Experts Blog Series. This series features a weekly Q&A with an industry thought leader on sales and operations planning trends and strategies. Follow-up question and answer' sessions are hosted in the S&OP section of the Supply Chain Expert Community”
— The 21st Century Supply Chain " Sales & operations planning (S&OP), | 4,973 | 19,075 | {"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-2017-47 | longest | en | 0.873118 |
http://slidegur.com/doc/144451/b.6-nonlinear-programming | 1,521,617,666,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647584.56/warc/CC-MAIN-20180321063114-20180321083114-00378.warc.gz | 255,067,985 | 12,971 | ### B.6 Nonlinear Programming
```Readings
Chapter 8
Nonlinear Optimization Models
BA 452 Lesson B.6 Nonlinear Programming
1
Overview
Overview
BA 452 Lesson B.6 Nonlinear Programming
2
Overview
Local and Global Optima of nonlinear programming problems are the optimal
solutions either over a neighborhood of the feasible set (local), or over the
entire feasible set (global).
Approximation Errors are typically measured as the sum of squares of the
differences between desired values and actual values, such as the difference
between an Index Fund and the S&P 500.
Nonlinear Assignment Problems are assignment problems where the objective
is nonlinear. For example, minimizing the distance times weight of cargo
Variance of an investment portfolio is measured as the expected value of the
sum of squares of the differences between expected rates of return and actual
rates of return.
Semi Variance of an investment portfolio modifies Variance by measuring only
the times when actual rates of return fall short of expected rates of return.
Those are the bad unexpected differences.
BA 452 Lesson B.6 Nonlinear Programming
3
Overview
Tool Summary
Distinguish between a local optimum and a global
optimum. The former is an optimum only in the
immediate neighborhood, and the latter is an optimum in
the entire feasible region.
Know that the product of two binary variables is, itself, a
binary variable. And the product is 1 if, and only if, each
separate binary variable is 1. Thus if x is a binary
variable that is 1 when you mow the yard and y is a
binary variable that is 1 when you wash the car, the
product of x and y is 1 when you mow the yard and wash
the car.
BA 452 Lesson B.6 Nonlinear Programming
4
Local and Global Optima
Local and Global Optima
BA 452 Lesson B.6 Nonlinear Programming
5
Local and Global Optima
Overview
Local and Global Optima of nonlinear programming
problems are the optimal solutions either over a
neighborhood of the feasible set (local), or over the
entire feasible set (global).
Because it is so hard for a computer program to solve
some nonlinear programs, you need not compute any
solutions.
BA 452 Lesson B.6 Nonlinear Programming
6
Local and Global Optima
Nonlinear optimization problems are those optimization
problems in which at least one term in the objective
function or a constraint is nonlinear. Nonlinear terms
include x 3 , 1/ x , 5x x , and log x .
1
2
1 2
3
BA 452 Lesson B.6 Nonlinear Programming
7
Local and Global Optima
A feasible solution is a local optimum if there are no
other feasible solutions with a better objective
function value in the immediate neighborhood.
• For a maximization problem the local optimum
corresponds to a local maximum.
• For a minimization problem the local optimum
corresponds to a local minimum.
A feasible solution is a global optimum if there are no
other feasible points with a better objective function
value in the feasible region.
Obviously, a global optimum is also a local optimum.
BA 452 Lesson B.6 Nonlinear Programming
8
Local and Global Optima
Nonlinear optimization problems can have multiple
local optimal solutions, in which case we want to find
the best local optimum.
Nonlinear problems with multiple local optima are
difficult to solve with optimization software.
In these cases, the software can get “stuck” and
terminate at a local optimum.
There can be a severe penalty (loss of objective) for
confusing a local optimum for a global optimum.
Developing software to find global optima is an
active research area.
BA 452 Lesson B.6 Nonlinear Programming
9
Local and Global Optima
2 ( X 2 ( Y 1)2 )
Consider the function f ( X , Y ) 3(1 X ) e
10( X /5 X Y )e
3
e( ( X 1)
2
Y 2 )
5
( X 2 Y 2 )
/3.
The shape of this function is shown on the next slide.
The hills and valleys in the graph show that this
function has several local maximums and local
minimums.
There are three local maximums, one of which is the
global maximum.
BA 452 Lesson B.6 Nonlinear Programming
10
Local and Global Optima
2 ( X 2 ( Y 1)2 )
f ( X , Y ) 3(1 X ) e
10( X /5 X Y )e
3
5
( X 2 Y 2 )
BA 452 Lesson B.6 Nonlinear Programming
e
( ( X 1)2 Y 2 )
/3
11
Local and Global Optima
Consider the function f ( X , Y ) X 2 Y 2 .
The shape of this function is shown on the next slide.
A function that is bowl-shaped down is called a
concave function.
The maximum value for this particular function is 0
and the point (0, 0) gives the optimal value of 0.
Functions such as this one have a single local
maximum that is also a global maximum.
A nonlinear problem with that type of objective
function is relatively easy to maximize.
BA 452 Lesson B.6 Nonlinear Programming
12
Local and Global Optima
Concave Function f ( X , Y ) X 2 Y 2
BA 452 Lesson B.6 Nonlinear Programming
13
Local and Global Optima
Consider the function f ( X , Y ) X 2 Y 2 .
The shape of this function is shown on the next slide.
A function that is bowl-shaped up is called a convex
function.
The minimum value for this particular function is 0
and the point (0, 0) gives the optimal value of 0.
Functions such as this one have a single local
minimum that is also a global minimum.
This type of nonlinear problem is relatively easy to
minimize.
BA 452 Lesson B.6 Nonlinear Programming
14
Local and Global Optima
Convex Function f ( X , Y ) X 2 Y 2
40
20
Z
4
2
0
X
-2
-4
-4
0
-2
2
4
Y
BA 452 Lesson B.6 Nonlinear Programming
15
Approximation Errors
Approximation Errors
BA 452 Lesson B.6 Nonlinear Programming
16
Approximation Errors
Overview
Approximation Errors are typically measured as the sum
of squares of the differences between predicted or
desired values and actual values, such as the difference
between the returns on an Index Fund and the past
performance of a broad market index, such as the S&P
500.
BA 452 Lesson B.6 Nonlinear Programming
17
Approximation Errors
Index funds are a very popular investment vehicle in
the mutual fund industry.
Vanguard 500 Index Fund is the largest mutual fund
in the U.S. with over \$70 billion in net assets in 2005.
An index fund is an example of passive asset
management, which emphasizes past performance.
The key idea behind an index fund is to construct a
portfolio of stocks, mutual funds, or other securities
that closely matches the past performance of a broad
market index such as the S&P 500.
Behind the popularity of index funds is research that
basically says “you can’t beat the market”.
BA 452 Lesson B.6 Nonlinear Programming
18
Approximation Errors
Question: Lymann Brothers has a substantial number of
clients who wish to own a mutual fund portfolio that closely
matches the performance of the S&P 500 stock index.
A manager at Lymann Brothers has selected five mutual
funds (shown on the next slide) that will be considered for
inclusion in the portfolio. The manager must decide what
percentage of the portfolio should be invested in each
mutual fund.
BA 452 Lesson B.6 Nonlinear Programming
19
Approximation Errors
Mutual Fund Performance in 4 Selected Years
Mutual Fund
Annual Returns (Planning Scenarios)
Year 1
Year 2
Year 3
Year 4
International Stock
Large-Cap Blend
Mid-Cap Blend
Small-Cap Blend
Intermediate Bond
25.64
15.31
18.74
14.19
7.88
27.62
18.77
18.43
12.37
9.45
5.80
11.06
6.28
-1.92
10.56
-3.13
4.75
-1.04
7.32
3.31
S&P 500
13.00
12.00
7.00
2.00
BA 452 Lesson B.6 Nonlinear Programming
20
Approximation Errors
Formulate and solve a non-linear programming problem to
determine what percentage of the portfolio should be
invested in each mutual fund so that the portfolio closely
matches the performance of the S&P 500 stock index. Use
the sum of squared deviations to measure the closeness of
the portfolio to the index.
BA 452 Lesson B.6 Nonlinear Programming
21
Approximation Errors
Answer: Define the 9 Decision Variables
IS = proportion of portfolio invested in international stock
LC = proportion of portfolio invested in large-cap blend
MC = proportion of portfolio invested in mid-cap blend
SC = proportion of portfolio invested in small-cap blend
IB = proportion of portfolio invested in intermediate bond
R1 = portfolio return for scenario 1 (year 1)
R2 = portfolio return for scenario 2 (year 2)
R3 = portfolio return for scenario 3 (year 3)
R4 = portfolio return for scenario 4 (year 4)
BA 452 Lesson B.6 Nonlinear Programming
22
Approximation Errors
Fund Year 1 Year 2 Year 3 Year 4
IS
25.64 27.62 5.80 -3.13
LC
15.31 18.77 11.06 4.75
MC 18.74 18.43 6.28 -1.04
SC
14.19 12.37 -1.92 7.32
IB
7.88 9.45 10.56 3.31
SP
13.00 12.00
7.00 2.00
Define the objective function
Min (R1 – 13)2 + (R2 – 12)2 + (R3 – 7)2 + (R4 – 2)2
Define the constraints (including non-negativity)
25.64IS + 15.31LC + 18.74MC + 14.19SC + 7.88IB = R1
27.62IS + 18.77LC + 18.43MC + 12.37SC + 9.45IB = R2
5.80IS + 11.06LC + 6.28MC - 1.92SC + 10.56IB = R3
- 3.13IS + 4.75LC - 1.04MC + 7.32SC + 3.31IB = R4
IS + LC + MC + SC + IB = 1
IS, LC, MC, SC, IB > 0
BA 452 Lesson B.6 Nonlinear Programming
23
Approximation Errors
Optimal Solution for Lymann Brothers Example
• R1 = 12.51
(12.51% portfolio return for scenario 1)
• R2 = 12.90
(12.90% portfolio return for scenario 2)
• R3 = 7.13
( 7.13% portfolio return for scenario 3)
• R4 = 2.51
( 2.51% portfolio return for scenario 4)
• IS = 0
( 0.0% of portfolio in international stock)
• LC = 0
( 0.0% of portfolio in large-cap blend)
• MC = .332
(33.2% of portfolio in mid-cap blend)
• SC = .161 (16.1% of portfolio in small-cap blend)
• IB = .507
(50.7% of portfolio in intermediate bond)
Lymann Brothers Portfolio Return vs. S&P 500 Return
Scenario
Portfolio Return
S&P 500 Return
1
2
3
4
12.51
12.90
7.13
2.51
13.00
12.00
7.00
2.00
BA 452 Lesson B.6 Nonlinear Programming
24
Nonlinear Assignment
Nonlinear Assignment
BA 452 Lesson B.6 Nonlinear Programming
25
Nonlinear Assignment
Overview
Nonlinear Assignment Problems are assignment problems
where the objective is nonlinear. For example, minimizing
docks.
BA 452 Lesson B.6 Nonlinear Programming
26
Nonlinear Assignment
docks is giving in the table below:
1
2
3
1 |
0
100
150
2 | 100
0
50
3 | 150
50
0
Three tankers currently at sea are coming into Long
Beach. Each tanker needs its own dock. Currently
tankers 2 and 3 are empty. Tanker 1 has cargo that must
be loaded onto the other tankers: 60 tons to Tanker 2
and 80 tons to Tanker 3.
Formulate then solve a non-linear programming problem
to minimize the product of tonnage times distance,
BA 452 Lesson B.6 Nonlinear Programming
27
Nonlinear Assignment
This is an extension of the assignment problems
considered earlier. Like assigning “agents” to “tasks”,
now assign tankers to docks.
Using standard notation for an assignment problem, let
XIJ = 1 if Tanker I is assigned to Dock J, and 0 if not.
The constraints are just like the standard assignment
problems.
! Each tanker must be assigned to a dock;
X11 + X12 + X13 = 1; !Tanker 1;
X21 + X22 + X23 = 1; !Tanker 2;
X31 + X32 + X33 = 1; !Tanker 3;
! Each dock can be assigned to at most one tanker;
X11 + X21 + X31 < 1; !Dock 1;
X12 + X22 + X32 < 1; !Dock 2;
X13 + X23 + X33 < 1; !Dock 3;
BA 452 Lesson B.6 Nonlinear Programming
28
Nonlinear Assignment
The difference from the standard assignment problem is
the objective function is now nonlinear. The reason is
the cost of assigning Tanker I to Dock J is no longer
constant. For example, the cost of assigning Tanker 3 to
Dock J depends on where Tanker 1 is assigned,
because that assignment affects how far the 80 tons of
cargo must move from Tanker 1 to Tanker 3.
For example, consider the result of assigning Tanker 1 to
Dock 2 and Tanker 3 to Dock 1.
meters. And Tanker 1 must transfer 80 tons of goods to
Tanker 3. This means that 80 tons must be moved 100
meters. To capture that in the objective function, write
100*80*X12*X31. Since both X12 = 1 and X31 = 1 their
product is 1, and the product of distance and tonnage
moved (100*80) is the term 100*80*X12*X31.
BA 452 Lesson B.6 Nonlinear Programming
29
Nonlinear Assignment
The complete objective function is
MIN = 100*60*X11*X22 + 150*60*X11*X23
+ 100*60*X12*X21 + 50*60*X12*X23
+ 150*60*X13*X21 + 50*60*X13*X22
! Those are the costs of moving 60 tons from Tanker 1 to 2;
+ 100*80*X11*X32 + 150*80*X11*X33
+ 100*80*X12*X31 + 50*80*X12*X33
+ 150*80*X13*X31 + 50*80*X13*X32;
! Those are the costs of moving 80 tons from Tanker 1 to 3;
For example, compute the cost of assigning Tanker 1 to
Dock 2, Tanker 2 to Dock 3, and Tanker 3 to Dock 1.
Variables are X12 = X23 = X31 = 1, with all other XIJ = 0.
The objective function is then
100*60*0*0 + 150*60*0*1
+ 100*60*1*0 + 50*60*1*1 + … = 50*60 + 100*80 = 11,000
BA 452 Lesson B.6 Nonlinear Programming
30
Nonlinear Assignment
Optimal assignment:
Tanker 1 to Dock 2
Tanker 2 to Dock 1
Tanker 3 to Dock 3
@BIN(X33); specifies X33 as a binary variable
BA 452 Lesson B.6 Nonlinear Programming
31
Variance
Variance
BA 452 Lesson B.6 Nonlinear Programming
32
Variance
Overview
Variance of an investment portfolio is measured as the
expected value of the sum of squares of the differences
between expected rates of return and actual rates of
return.
BA 452 Lesson B.6 Nonlinear Programming
33
Variance
The key tradeoff in most portfolio optimization models
is between risk and return.
The index fund model (Lymann Brothers example)
presented earlier managed the tradeoff passively (ex
post, or after the fact).
The Markowitz mean-variance portfolio model
provides a very convenient way for an investor to
trade off risk versus return actively (ex ante, or before
the fact).
We demonstrate the Markowitz portfolio model by
extending the Lymann Brothers example.
BA 452 Lesson B.6 Nonlinear Programming
34
Variance
In the Lymann Brothers example there were four
scenarios and the return under each scenario was
defined by the variables R1, R2, R3, and R4.
If ps is the probability of scenario s, and there are n
scenarios, then the expected return for the portfolio R
is
n
R ps Rs
s 1
If we assume that the four scenarios in the Lymann
Brothers model are equally likely, then
1
R Rs or
s 1 4
4
1 4
R Rs
4 s 1
BA 452 Lesson B.6 Nonlinear Programming
35
Variance
The measure of risk most often associated with the
Markowitz model is the variance of the portfolio.
For our example, the portfolio variance is
n
Var ps ( Rs R )2
s 1
For our example, the four planning scenarios are
equally likely. Thus,
1
Var ( Rs R )2 or
s 1 4
4
1 4
Var ( Rs R )2
4 s 1
BA 452 Lesson B.6 Nonlinear Programming
36
Variance
The portfolio variance is the average of the sum of
the squares of the deviations from the mean value
under each scenario.
The larger the variance value, the more widely
dispersed the scenario returns are around the
average return value.
If the portfolio variance equaled zero, then every
scenario return Ri would have to equal.
BA 452 Lesson B.6 Nonlinear Programming
37
Variance
There are two alternative ways to formulate the
Markowitz model:
• (1) Minimize the variance of the portfolio subject to
constraints on the expected return, and
• (2) Maximize the expected return of the portfolio
subject to a constraint on risk (we did that in a
previous lesson).
We demonstrate the first (1) formulation, assuming
that Lymann Brothers’ client requires the expected
BA 452 Lesson B.6 Nonlinear Programming
38
Approximation Errors
Question: Lymann Brothers has a substantial number of
clients who wish to own a mutual fund portfolio with
minimum variance and an expected return on 9 percent.
A manager at Lymann Brothers has selected five mutual
funds (shown on the next slide) that will be considered for
inclusion in the portfolio. The manager must decide what
percentage of the portfolio should be invested in each
mutual fund.
BA 452 Lesson B.6 Nonlinear Programming
39
Approximation Errors
Mutual Fund Performance in 4 Selected Years
Mutual Fund
Annual Returns (Planning Scenarios)
Year 1
Year 2
Year 3
Year 4
International Stock
Large-Cap Blend
Mid-Cap Blend
Small-Cap Blend
Intermediate Bond
25.64
15.31
18.74
14.19
7.88
27.62
18.77
18.43
12.37
9.45
5.80
11.06
6.28
-1.92
10.56
BA 452 Lesson B.6 Nonlinear Programming
-3.13
4.75
-1.04
7.32
3.31
40
Approximation Errors
Formulate and solve a non-linear programming problem to
determine what percentage of the portfolio should be
invested in each mutual fund so that the portfolio has
minimum variance and an expected return of 9 percent.
Assume all scenarios are equally likely.
BA 452 Lesson B.6 Nonlinear Programming
41
Approximation Errors
Answer: Define the 9 Decision Variables
IS = proportion of portfolio invested in international stock
LC = proportion of portfolio invested in large-cap blend
MC = proportion of portfolio invested in mid-cap blend
SC = proportion of portfolio invested in small-cap blend
IB = proportion of portfolio invested in intermediate bond
R1 = portfolio return for scenario 1 (year 1)
R2 = portfolio return for scenario 2 (year 2)
R3 = portfolio return for scenario 3 (year 3)
R4 = portfolio return for scenario 4 (year 4)
BA 452 Lesson B.6 Nonlinear Programming
42
Variance
Define the Objective Function
Minimize the portfolio variance:
1 4
Min ( Rs R )2
4 s 1
Fund Year 1 Year 2 Year 3 Year 4
IS
25.64 27.62 5.80 -3.13
LC
15.31 18.77 11.06 4.75
MC 18.74 18.43 6.28 -1.04
SC
14.19 12.37 -1.92 7.32
IB
7.88 9.45 10.56 3.31
Constrain the return for
each scenario:
25.64IS + 15.31LC + 18.74MC + 14.19SC + 7.88IB = R1
27.62IS + 18.77LC + 18.43MC + 12.37SC + 9.45IB = R2
5.80IS + 11.06LC + 6.28MC 1.92SC + 10.56IB = R3
3.13IS + 4.75LC 1.04MC + 7.32SC + 3.31IB = R4
BA 452 Lesson B.6 Nonlinear Programming
43
Variance
Constrain that all money be invested:
IS + LC + MC + SC + IB = 1
Define the expected return for the portfolio:
1 4
Rs = R
4 s 1
percent:
R 9
Constrain non-negativity:
IS, LC, MC, SC, IB > 0
BA 452 Lesson B.6 Nonlinear Programming
44
Variance
Optimal Solution
• R1 = 10.63% portfolio return for scenario 1
• R2 = 12.20% portfolio return for scenario 2
• R3 = 8.93% portfolio return for scenario 3
• R4 = 4.24% portfolio return for scenario 4
• Rbar = 9.00% expected portfolio return
• IS =
0.00% of portfolio in international stock
• LC = 25.10% of portfolio in large-cap blend
• MC = 0.00%of portfolio in mid-cap blend
• SC = 14.1% of portfolio in small-cap blend
• IB =
60.8% of portfolio in intermediate bond
100.0% of portfolio
BA 452 Lesson B.6 Nonlinear Programming
45
Semi Variance
Semi Variance
BA 452 Lesson B.6 Nonlinear Programming
46
Semi Variance
Overview
Semi Variance of an investment portfolio modifies
Variance by measuring only the times when actual rates
of return fall short of expected rates of return. Those
are the times when unexpected returns are bad.
.
BA 452 Lesson B.6 Nonlinear Programming
47
Semi Variance
Most investors are happy when their returns are
above average, but want to avoid returns below
average.
In the previous Markowitz porfolio optimization
problem, we minimized variance, which is the sum of
squares of the deviations from the mean.
Now change the objective to minimize the semivariance, which is the sum of squares of those
deviations from the mean that are below the mean.
BA 452 Lesson B.6 Nonlinear Programming
48
Semi Variance
As before, constrain the return
for each scenario:
Fund Year 1 Year 2 Year 3 Year 4
IS
25.64 27.62 5.80 -3.13
LC
15.31 18.77 11.06 4.75
MC 18.74 18.43 6.28 -1.04
SC
14.19 12.37 -1.92 7.32
IB
7.88 9.45 10.56 3.31
SP
13.00 12.00
7.00 2.00
25.64IS + 15.31LC + 18.74MC + 14.19SC + 7.88IB = R1
27.62IS + 18.77LC + 18.43MC + 12.37SC + 9.45IB = R2
5.80IS + 11.06LC + 6.28MC 1.92SC + 10.56IB = R3
3.13IS + 4.75LC 1.04MC + 7.32SC + 3.31IB = R4
Then, define and constrain the expected return for the
portfolio:
1 4
Rs = R
R 9
4 s 1
BA 452 Lesson B.6 Nonlinear Programming
49
Semi Variance
But instead of minimizing the variance
1 4
Min ( Rs R)2
4 s 1
Decompose deviations into positive or negative terms
R1 - Rbar = DP1 – DN1,
R2 - Rbar = DP2 – DN2,
R3 - Rbar = DP3 – DN3,
R4 - Rbar = DP4 – DN4.
Then minimize the sum of the squares of the negative
deviations, which is the semi-variance.
BA 452 Lesson B.6 Nonlinear Programming
50
Approximation Errors
Question: Lymann Brothers has a substantial number of
clients who wish to own a mutual fund portfolio with
minimum semivariance and an expected return on 9
percent.
A manager at Lymann Brothers has selected five mutual
funds (shown on the next slide) that will be considered for
inclusion in the portfolio. The manager must decide what
percentage of the portfolio should be invested in each
mutual fund.
BA 452 Lesson B.6 Nonlinear Programming
51
Approximation Errors
Mutual Fund Performance in 4 Selected Years
Mutual Fund
Annual Returns (Planning Scenarios)
Year 1
Year 2
Year 3
Year 4
International Stock
Large-Cap Blend
Mid-Cap Blend
Small-Cap Blend
Intermediate Bond
25.64
15.31
18.74
14.19
7.88
27.62
18.77
18.43
12.37
9.45
5.80
11.06
6.28
-1.92
10.56
BA 452 Lesson B.6 Nonlinear Programming
-3.13
4.75
-1.04
7.32
3.31
52
Approximation Errors
Formulate and solve a non-linear programming problem to
determine what percentage of the portfolio should be
invested in each mutual fund so that the portfolio has
minimum semivariance and an expected return of 9
percent. Assume all scenarios are equally likely.
BA 452 Lesson B.6 Nonlinear Programming
53
Semi Variance
BA 452 Lesson B.6 Nonlinear Programming
54
Semi Variance
As an error check, make
sure that, for each variable,
either DPi = 0 or DNi = 0.
DN1 = DN2 = 0.
DP3 = DP4 = 0.
BA 452 Lesson B.6 Nonlinear Programming
55
BA 452
Quantitative Analysis
End of Lesson B.6
BA 452 Lesson B.6 Nonlinear Programming
56
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https://strategywiki.org/wiki/Professor_Layton_and_the_Last_Specter/Puzzles_076-100 | 1,670,520,926,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00756.warc.gz | 532,720,198 | 24,978 | Caution! This page contains all the hints and solutions for every puzzle from 76 to 100. Scroll carefully or you might spoil the answer of a puzzle for yourself. The solutions are hidden behind spoiler tags, so only take a peek if you're desperate.
## Puzzle 076
Name: Swap Meet
Trigger: Talk to Finch
Location: Staircase Bridge
Chapter: 7
Picarats: 50
Description: Starting with A and moving clockwise,this circle of people is arranged in the following order: female, female, male, female, female, male, male. Swap just two people to create a new arrangement where, starting with A, you will create the same male-female pattern whether you count every person or every other person. Can you make the swap?
• Hint 1: Try imagining the group as a row instead of a circle. That might make things a little clearer in your mind.
• Hint 2: You can make a lot of swaps that result in similar patterns. But don't forget that to solve this particular puzzle, you must start counting form the person labeled A.
• Hint 3: Swap out the person labeled C...but with whom?
• Super Hint: The new pattern you must create is female, female, female, male, female, male, male The order will also be the same if you only count every fourth person. Once you make your swap, try counting it off.
Answer: Swap C with D.
## Puzzle 077
Name: How the Bells Toll
Trigger: Talk to Sebastian
Location: Closed Factory
Chapter: 5
Picarats: 35
Description: Below are three bells, each of which has one red and one blue hammer attached. Beneath each bell is a counter, and each one is rigged such that its count will increase according to certain rules whenever the bell above it is struck. The counters have already reached the numbers shown below. Can you figure out how to use the red and blue hammers to ring the bells and make all three counters display the same number?
• Hint 1: Each time you ring a bell with a red or blue hammer, the counter goes up by a fixed amount. Try to work out these amounts by using each hammer and noting the outcome.
• Hint 2: Regardless of the bell, the same-colored hammer will always increase the counter by the same amount. If you ring the bells and watch the counters carefully, you should be able to work out the function of each hammer.
• Hint 3: There is a way to make all three numbers match in just six moves. Once you work out the pattern, look for the simplest possible method reaching the same number on each counter.
• Super Hint: The lowest common value for all three bells is 16. Now that you have a goal set for you, all that's left to do is work out which hammers to hit in order to teach it.
First, sound the upper bell with the red hammer. Second, sound the lower-left bell with the red hammer. Third, sound the lower-left bell again with the blue hammer. Fourth, sound the upper bell with the blue hammer. And fifth, sound the lower-right bell with the red hammer 2 times.
## Puzzle 078
Name: Lucky Table Cloth
Trigger: Talk to
Location: Foyer
Chapter: 4
Picarats: 60
Description: Your lucky tablecloth is damages and you want to salvage it. You must start by cutting off the shaded area. Next you must make two straight cuts along the dotted lines and up with three pieces. Sewing these pieces together will give you a smaller square tablecloth with five complete four-leaf clovers! Draw two lines on the cloth to show where you should make your cuts.
• Hint 1: Try thinking about where the cloth needs to be joined to create five complete four-leaf clovers. That should give you some ideas for how to form the square.
• Hint 2: One of the pieces will contain exactly four squares of material.
• Hint 3: One of the lines you need to cut starts from top-right side of the shaded square area that you cut off first and travels down diagonally to the right. The piece created by this cut is an isosceles triangle.
• Super Hint: The second cut starts from the lower-left corner and heads up diagonally to the right. If you combine this cut with the one described in Hint 3, the three pieces created can be combined to create a rectangular tablecloth with five four-leaf clovers on it!
Cut the tablecloth according to the picture.
## Puzzle 079
Name: Sweet Sums
Trigger: Talk to Aunt Taffy
Location: Market Entrance
Chapter: 8
Picarats: 50
Description: "My children have been very kind to each other lately, so I decided to give them some candy as a reward. I have four jars of candy, A, B, C, and D. The combined number of candies in jars A and B is equal to twice the number in jar C. The combined number of candies in jars B and D equals twice the number in jar A. If you take three candies from jar D and put them in jar A, jar A will contain twice the number in jar B. Which jar contains six pieces of candy?"
• Hint 1: All you need to do is figure out which jar contains six pieces of candy. Even if you can't work it out using complicated mathematical calculations, you should still be able to solve it by figuring out an alternate method for determining the number of candies in each jar.
• Hint 2: Here's the puzzle in equations:
1. A + B = 2C
2. B + D = 2A
3. A + 3 = 2B
4. D 3
This might help you spot a detail that you may have missed.
• Hint 3: A + 3 = 2B
2B must be an even number because 2 is one of its factors. That means A + 3 is also an even number. Subtracting 3 from an even number will always result in an odd number, which means A must be odd. So if A (odd) + B = 2C (even), then what is B, an odd or even number?
• Super Hint: Odd + Odd = Even | Odd + Even = Odd
From this simple rile, you can figure out that the number of candies in jar B must be odd and that the number in jar D must also be odd. This means that all of the jars except A contain an odd number of candies. Therefore, the number of candies in jar C must be...
Jar C contains the six pieces of candy.
## Puzzle 080
Name: Paint the Plinth
Trigger: Talk to Shackwell
Location: Black Market
Chapter: 7
Picarats: 55
Description: You must paint several cubes that will form a base platform when put together. All cubes are the same size, you have to paint the visible sides of the base. For example, if you painted a base made up of two cubes, you would only paint four sides of each cube. If you wanted to make a base made up of cubes painted on one side, two sides, three sides, four sides, what is the lowest number of cubes you would need? You can place cubes on top of each other.
• Hint 1: If you place just one cube on the ground, then five sides will be showing. If you place a cube next to it, both cubes will have four sides showing. Try out some other configurations, and think about which sides of each cube would be showing.
• Hint 2: What configuration would you need to paint a cube on only one side? You would need to surround the cube so that only the top surface is showing.
• Hint 3: Once you've worked out how to paint a cube on only one side, the next task is the cube with two sides painted. You will need to cover three sides of that cube in addition to the side facing the ground.
• Super Hint: You need four cubes to cover the sides of the cube you paint on one side. You can use one of these four cubes as the cube you paint on two sides-- simply place two more cubes on either side of it. The basic configuration is actually all you need to get cubes painted on one side, two sides, three sides, and four sides.
The lowest number of cubes you'd need is 7 cubes.
## Puzzle 081
Name: Stick & Move
Trigger: Talk to Dean Delmona
Location: Lobby
Chapter: 7
Picarats: 45
Description: A man is moving a water jug using a round stick and a long plank. If the circumference of the stick is 20 cm, how far ahead, in centimeters, would the water jug be after one complete turn of the stick? This is assuming that the plank, stick, and jug do not slip out of place or get stuck.
• Hint 1: Remember that the stick isn't staying in one spot--it moves forward 20 cm in one turn. Think about what else is moving.
• Hint 2: The position of the plank changes with the position of the stick and vice versa. What does the man have to do to the plank to make the stick move?
• Hint 3: Even as the stick moves forward, it's still rolling toward the back of the plank. Think about what this means for the final position of the jug. If it's too hard to imagine, try it out using a round object nearby.
• Super Hint: The plank rests on the stick. When the stick moves 20 cm along the ground, it slides the plank 20 cm forward at the same time. Taken together, this is the total distance moved by the jug.
The water jug would be 40 cm ahead.
## Puzzle 082
Name: Block Ostrich
Trigger: Talk to Rosa
Location: Layton's Office
Chapter: 7
Picarats: 30
Description: Rearrange the pieces making up the shape of Layton's hat into the shape of an ostrich. You can move, flip, or rotate the pieces in any direction.
• Hint 1: Let's start with the ostrich's head. Grab the piece that makes up the left portion of the brim of Professor Layton's hat. This piece will fit perfectly in the space formed by the ostrich's head.
• Hint 2: Use the L-shaped piece to make up the ostrich's legs. Be careful how you handle them--a kick from an ostrich is powerful enough to take out a lion!
• Hint 3: Now it's time to put those flightless wings in place. Flip the piece shaped like a backward "J" and rotate it so it can fit into the body section. Then take the smaller of the two remaining pieces and place it in the space you just created.
• Super Hint: Only one piece left! Simply flip that piece and rotate it until it fits into the remaining space for the neck and shoulders. Now there's an ostrich that can flaunt its feathers proudly!
Place the blocks according to the picture here.
## Puzzle 083
Name: Silver Marbles
Trigger: Talk to Monica
Location: Second-Floor Desk
Chapter: 7
Picarats: 35
Description: To solve this puzzle, you need to remove all but one of the silver marbles below. You can jump a marble horizontally or vertically over an adjacent marble and into an empty space to remove the jumped-over marble from the board. Also, you can move or jump over any marble except those marked with an X.
• Hint 1: To begin with, the leftmost marble in the third row from the bottom needs to be moved to the right.
• Hint 2: After following Hint 1, the next marbles to move are the ones second and third from the right in the third row from the bottom. These marbles will now be on the third row from the top. Jump the one in the second column from the right over to the central column.
• Hint 3: Continuing from Hint 2, the next step is to remove the marble in the very center by jumping over it form the left. You can then move the top marble of the central column down to take its place.
• Super Hint: Did you follow all of the other hints? If so, you should be almost there. First of all, jump the marble in the very center to the right. Then take the marble on the far left of the central row and jump it to the right. Next, jump the marble in the second bottom row of the central column upward. Only a couple of moves left now. Try working those out on your own.
First, move the second marble from the 3rd row (3rd row from bottom) to the 5th row. Second, the left-most marble to the right. Third, the same marble from the first to the 3rd row. Fourth, the current left-most marble to the right. Fifth, the same marble from the fourth to the 5th row. Sixth, the marble from the 4th row next to the marble marked with an X, move to the left. Seventh, the marble from the 5th row move below. Eighth, the marble from the 2nd row move upward. Ninth, the right-most marble move to the left. Tenth, the same marble from ninth move above. Eleventh, move the top-most marble downwards. I hope that wasn't too confusing... ^_^
## Puzzle 084
Name: Sixteen Tulips
Trigger: Talk to Augustus
Location: Florist
Chapter: 7
Picarats: 40
Description: You've planted 16 magic tulip bulbs in this 4x4 flower bed. Watering a space will cause all the magic bulbs in the same row and column to sprout if they are belowground, or recede if they're aboveground. Can you figure out how to fill the entire flower bed with tulips?
• Hint 1: You could just keep trying different spots until you happened upon the answer, but that would take a while. This puzzle can actually be solved in as few as as five waterings, if you work out the most efficient method.
• Hint 2: To get going, here's the first step: begin by watering the top-left corner. That'll make a lot of tulips bloom!
• Hint 3: The second spot to water is in the third column from the left, third row from the top. After that you should water the spot in the second column from the left, second row from the top. You only need to water two more spots after that to fill your flower bed with beautiful tulips!
• Super Hint: Continuing from Hint 3, water the spot in the top-right corner. You only need to water one more spot to finish! You probably already know which one it is.
Water the flower bed according to this picture.
## Puzzle 085
Name: Bus Stop
Trigger: Talk to Mick
Location: Museum Entrance
Chapter: 7
Picarats: 55
Description: The buses at this stop leave at the times indicated by the schedule below. The first column marks the hour, and the next three mark the minutes. All the buses start at this bus stop and take exactly one hour to return. What is the minimum number of buses necessary to service this bus stop?
• Hint 1: The key to this puzzle is that it takes a bus one hour to return to this stop. Look at the timetable, and figure out which times a single bus could cover.
• Hint 2: If the first bus leaves at 7:05, it will return at 8:05. So different buses will be required for 7:15 and 7:40. Follow this logic to figure out how many buses are needed in total.
• Hint 3: Three buses are needed from 7:00. Two of the buses return in time for the 08:10 and 08:20 departures, but an extra bus will be required to cover the 08:30 departure. How many buses are needed in all?
• Super Hint: Four buses are required for the timetable to run smoothly until 8:30, but one extra bus is needed to depart at 11:10. You should be able to figure out if any more buses are needed for the rest of the timetable.
The minimum number of buses necessary is 5 buses.
## Puzzle 086
Name: Window Boarder
Trigger: Talk to Chappy
Location: Museum, Second Floor
Chapter: 7
Picarats: 60
Description: You want to board up this broken window, but much to your dismay the only pieces of wood you can find have holes in them. Place the boards over the window so that the broken square part is completely covered and no holes show through. No boards may extend beyond the square window frame, but you can flip and rotate each piece of wood if necessary.
• Hint 1: The three T-shaped wooden planks go in the top-left, top-right, and bottom-right corners of the window. As to which T-shaped wooden plank belong where...you'll have to work that out on your own.
• Hint 2: Place the J-shaped wooden plank with a hole in its bend in the top-left corner. Don't flip it, just rotate it 180 degrees before placing it down. Position it so hat the hole in the board lies over the middle pane in the top row. Next, take the remaining J-shaped wooden plank, rotate it 90 degrees clockwise, and place it so that the hole covers the pane in the bottom-right corner of the window.
• Hint 3: Rotate the glove-looking wooden plank (sorry) with one hole 180 degrees and place it in the bottom-left corner. When oriented properly, the hole will be on the top right of the board. Next, take the glove-looking wooden plank (again sorry) with two holes and place it in the top-right corner. There's no need to flip or rotate it, just put it down as it is. The two holes in the board should lie against the right edge of the window frame.
• Super Hint: Place the T-shaped wooden plank with no holes on the bottom right, the T-shaped wooden plank with hole on the top right, and the T-shaped wooden plank with two holes on the top left. Once you start trying to position these last few boards so that no holes show through, the rest should fit into place naturally. Now finish the job!
Place the wooden planks according to this picture.
## Puzzle 087
Name: Flip Maze 2
Trigger: Talk to Hans
Location: Twisted Light
Chapter: 7
Picarats: 50
Description: Here's another double-sided maze! The goal is to move the ball to the end of the maze, which is marked by a star in the lower-right corner. When the ball is dropped into one of the holes, it will appear on the other side of the maze. This maze is a bit bigger than the last one. Can you get the ball to the star?
• Hint 1: You've probably heard this hint before, but it's a good idea to use the Memo function to keep track of your progress for this type of puzzle. By doing so, you'll be able to keep track of where you've been and avoid going around and around in circles.
• Hint 2: This isn't so much a hint as it is some goo, old-fashioned encouragements! If you stay focused and keep at it, you should be able to solve this puzzle. If you can't figure it out, then you can always use the next hint.
• Hint 3: From the start, drop the ball through the hole down and to the right or the one straight down from the start. Whichever one you choose, you'll end up at the same junction.
• Super Hint: Have you found the crossroads yet? It's on the path that runs slightly to the bottom right of the center. Drop the ball through the hole four squares from the bottom and three squares from the right. From here, the rest should be a breeze.
Get the ball to the star by following the picture.
## Puzzle 088
Name: Fuel for the Fire
Trigger: Tap the fireplace
Location: Chief's Office
Chapter: 7
Picarats: 45
Description: In an average 365-day year, the family in this house burns one bundle of wood per day in the fireplace. In the three-month period starting this month, the family will burn three more bundles of wood than in the three-month period after it. From this information alone, can you guess what the current month is? Express your answer as a number.
• Hint 1: In order for one three-month period to burn more wood than another, it must also have more days. Three more bundles of wood means three more days in the period.
• Hint 2: What month has the fewest days in any given year? If you're having trouble, check a calendar.
• Hint 3: On non-leap years, February has only 28 days,making it the shortest month. Which three-month period containing February has the fewest number of days?
• Super Hint: The three-month period starting in February has only 89 days, fewer than any other consecutive three-month. That should be enough to work out the answer right?
The current month is 11/November.
## Puzzle 089
Name: Proper Storage
Trigger: Tap the crate of bottles
Location: Cellar
Chapter: 6
Picarats: 50
Description: In order to store this fine wine in an appropriate manner, the following rules must be observed:
1. White wine should always be stored that there is only one bottle on any horizontal, vertical, or diagonal line.
2. Red wine should also be stored so that there is only one bottle on the same horizontal, vertical, or diagonal line.
3. The two bottles already in the crate cannot be moved.
Can you figure out a way to store it?
• Hint 1: You can solve this puzzle in two ways, but whichever way you do it, it's easier to figure out using the Memo function. Start from either of the two immovable bottles, and draw lines to show where the other bottles cannot be placed.
• Hint 2: If you try to put both of the red- and white-wine bottles at the same time, you will just confuse yourself. Figure out all of one type before going on to the next.
• Hint 3: To solve this puzzle, you need to make sure there is a red- and white-wine bottle on each horizontal and vertical line.
• Super Hint: Lets's concentrate on each color one at a time, imagining that the rows are numbered left to right and the columns top to bottom. Place a white-wine bottle in row 3, column 2. Now, place another in row 2, column 4. Place a red-wine bottle in row 4, column 2. Now place another in row 3, column 4.
Place the bottles according to this picture.
## Puzzle 090
Name: Puppet Master
Trigger: Talk to Rosa
Location: Layton's Office
Chapter: 7
Picarats: 45
Description: Can you fit the pieces of Professor Layton's hat together to form the puppet's head without any of the pieces overlapping or leaving any empty spaces? The pieces can be moved, flipped, and rotated in any direction.
• Hint 1: First, rotate the L-shaped piece and make it fit into the puppet's nose and mouth area. That should help you out quite a bit!
• Hint 2: Take the left-hand side of the hat's brim, and move it to the four squares on the bottom right of the puppet's head so that the piece sticking out touches that eye
• Hint 3: Flip the backward-J-shaped piece and rotate it so that the long part fits into the top-right corner of the head.
• Super Hint: After following the previous hints, it should be obvious where to place the small two-blocks piece. Putting in the final piece should also be child's play!
Place the pieces according to this picture.
## Puzzle 091
Name: Evacuate!
Trigger: Talk to Greppe
Chapter: 8
Picarats: 50
Description: You need to evacuate the neighborhood, and fast! Luckily, you were able to recruit nine friends to help you out:
• Two whose voices carry for 3 squares
• Three whose voices carry for 2 squares
• Four whose voices carry for 1 square
Where should you place each person so that the evacuation message reaches every square of the neighborhood?
• Hint 1: Try to think about how to place your friends so that there is as little overlap as possible (but a little overlap is still OK). The easiest way to avoid overlap is to place the people first and work from there.
• Hint 2: Think of the neighborhood as a grid of 10 columns and 7 rows. Grab the two people whose voices carry the farthest and place them as follows:
• In the middle square of the column farthest to the right.
• Three squares up from the bottom in the column farthest to the left.
• Hint 3: Place the three people whose voices carry two squares as follows:
• In the very top square of the third column from the left.
• On the second row from the top, fourth column from the right.
• On the bottom row of the fourth column from the right.
Use the remaining volunteers to fill in the gaps!
• Super Hint: If you followed the previous hints, you should be able to complete the puzzle by placing the rest of the volunteers in the remaining spaces. The last few people may overlap, but don't worry about that. As long as the message gets out to everyone in the neighborhood, the evacuation will be a success!
Place the friends according to this picture.
## Puzzle 092
Name: Key Quandary
Trigger: Tap the door
Location: Interrogation Cell
Chapter: 7
Picarats: 30
Description: Layton, Luke, and Emmy need to escape the police station! Luckily, Toppy has brought them a key ring, but only one of the keys fits the lock. Which of these keys opens the door? The door won't open unless the key can be fully inserted and turned.
• Hint 1: Think about how you will turn the key once you insert it. There might be a trick to turning the key
• Hint 2: The shape of the end of key B will keep it from fitting into the keyhole.
• Hint 3: The bit on the end key C is too big to fit into the keyhole.
• Super Hint: It's OK if you can't insert the key all the way until you've turned it. You may be able to insert a key and turn it partway before you insert it the rest of the way.
The key that opens the door is A.
## Puzzle 093
Name: Escape Route
Trigger: Tap the arrow to try to go to Police Reception
Location: Police HQ Hall
Chapter: 7
Picarats: 40
Description: Layton, Luke, and Emmy need to get the key and reach the exit without alerting the guards. there is a path that will allow the group to pick up the key on their way to the exit without passing in front of any guarded doorway. Unfortunately, the corridors are all one-way, so they must walk in the direction of the arrows. Draw a path that shows how the group can make their escape!
• Hint 1: Start the big escape by following these directions: right, down, down. That path avoids detection by the guards. Now how to get that key...
• Hint 2: If you followed the previous hint, get the key by going in these directions: right, down, left. All that's left to do is to escape without being seen.
• Hint 3:After you've followed the hints to reach the key, go in these directions: left, down, right, right, right. Just a little more winding around the rooms to avoid the guards, and then you'll reach the exit.
• Super Hint: After you've followed the previous hints, go in these directions: up, up, up! From there, you should be able to figure out how to avoid the last few guards on your way to the exit!
Draw your escape route according to this picture.
## Puzzle 094
Name: Policing the streets
Trigger: Talk to Hans
Location: Twisted Light
Chapter: 8
Picarats: 40
Description: Four police officers are assigned to guard duty. In order to maintain peace, the officers must keep a collective eye on every inch of pavement in town. They can see down every road as far as it extends, as long as it remains a straight line. Place each officer on a junction marked with a red X so that they can protect the entire town.
• Hint 1: Place an officer on the top road at the second junction from the right. This will allow him to keep an eye on several roads, including diagonal ones.
• Hint 2: Place another officer on the second road from the bottom, at the second junction from the left.
• Hint 3: Place the third officer on the second road from top, at the first junction from the right.
• Super Hint: The junction where you place the final officer doesn't have a diagonal road. The Memo function may help you determine which roads are already being covered by the three other officers.
Place the police officers according to this picture.
## Puzzle 095
Name: House Hunting
Trigger: Talk to Maggie
Location: Grand Plaza
Chapter: 8
Picarats: 25
Description: "The houses in this town are numbered from 1 to 88. I asked a boy three questions to find out where he lives:
• Is your house number higher than 44?
•Can your house number be divided by four?
• Can I multiply a number by itself to get your house number?
The boy said yes to all three questions. What is the number of the boy's house?"
• Hint 1: If you think about it, this puzzle isn't all that difficult. the key is to work out which question gives you the fewest possible answers.
• Hint 2: Try to limit the number of possible answers by thinking about the third question.
• Hint 3: Which numbers between 44 and 88 can be found by multiplying a number by itself? There aren't many to choose from.
• Super Hint: You should have worked out the choices 49, 64 and 81 from Hint 3. If you know which number can be divided by four, you have your answer.
The boy's house number is 64.
## Puzzle 096
Name: Seven Towers
Trigger: Talk to Otaki-San
Location: Grand Plaza
Chapter: 8
Picarats: 35
Description: Seven towers stand on a waterway, each connected by a bridge. One day the towers were tested to see which ones could be reached after crossing exactly six bridges, starting from shore A or shore B. It was discovered that all of the towers could be reached in this way except one. Keeping in mind that the same bridge cannot be crossed twice in one trip, which tower cannot be reached after crossing over exactly six bridges?
• Hint 1: Use the Memo function to mark off any towers you can reach from either shore in exactly six moves without crossing the same bridge twice.
• Hint 2: You can reach the bottom-left and bottom-right towers in exactly six moves, so neither of those is the answer.
• Hint 3:You can also reach the towers on the far left and far right in exactly six moves. That leaves only three towers: the top left, the top right, and the one in the middle.
• Super Hint: The tower in the middle can be reached in exactly six moves. That leaves only the towers on top right and the top left.
The top left tower cannot be reached after crossing over exactly six bridges.
## Puzzle 097
Name: How Many Friends?
Trigger: Talk to Sean
Location: Quiet Townscape
Chapter: 8
Picarats: 30
Description: A girl is having some friends from school over for a party and wants to share her special chocolates with them. If she gives them out so that everyone has three chocolates each, there will be six left over. However, if she tries to give everyone five chocolates each, she'll be short y six chocolates. How many friends are coming to her party?
• Hint 1: For these kinds of problems, you need to make some equations. Including the girl, the number of people at the part will be A. Then, if each person gets three chocolates and six are left over, the total number of chocolates can be expressed as 3A + 6.
• Hint 2: It's not possible for all guests to be given five chocolates , because the girl is short by six. This means that the total number of chocolates can be expressed as 5A - 6. Since both this expression and the one from Hint 1 represent the number of chocolates, we can say: 3A + 6 = 5A - 6 Solve this equation to find A.
• Hint 3: 5A - 3A = 6 + 6
2A = 12
A = 6
The total number of people at the party, including the little girl, will be six.
• Super Hint: If you thought the answer was six, you'll have to think again. Reread the question. It asks for how many friends will come to the party. That means you have to take the little girl away from the total. The answer is six people minus one.
5 friends will be coming to her party.
## Puzzle 098
Name: Symbol Code
Trigger: Talk to Tweeds
Location: Descending Path
Chapter: 8
Picarats: 50
Description: All that is known about this secret code is what's written on the back of the piece of paper: "These symbols follow a rule and are listed in order from top to bottom." As shown by the question mark, the sixth symbol of the code is missing. Can you figure out which symbol, A-E, is part of the code?
• Hint 1: Once you figure out the rule, it would be theoretically possible to create the 100th or 200th symbol in the sequence. Doing so, however, would be nearly impossible!
• Hint 2: Don't get distracted by the shapes of the symbols. There's only one feature of each symbol that you need to worry about: Take C, for example. It's possible to swap C with the fourth symbol in the sequence without breaking the code's rule. See if you can figure out what features these two symbols have in common.
• Hint 3: The symbols of this code represent numbers. In order from top to bottom, they represent 0, 1, 2, 3, 4... See if you can figure out how each symbol represents each number.
• Super Hint: Try tracing each symbol. For most of them, there will be points where you can't go any farther forward. The first symbol in the sequence has none of these end points and represents 0. The next symbol has one end point, so it represents 1. How many end points would you expect the sixth symbol have?
The sixth symbol is D.
## Puzzle 099
Name: A Watched Pot...
Trigger: Talk to Dugen
Location: Excavation Base
Chapter: 8
Picarats: 40
Description: As an appraiser of antiques browsed an antique shop, he was approached by the shopkeeper. "One of these four pots is known as the 9:15 Pot. If you can workout which one it is, the I'll believe that you're a top pot appraiser." Which pot is the 9:15 Pot.
• Hint 1: These pots do all look rather suspicious, don't they? Look closely at the patterns on the outside of each before you make your decision.
• Hint 2: There are digital and analog clock, and of course both kinds are capable of showing the time 9:15. How would that time look on either kind of clock?
• Hint 3: So 9:15 could be a time in the morning or in the evening. Would it change things if this 9:15 were a time in the evening?
• Super Hint: How would 9:15 p.m. look on a 24-hour clock?
The 9:15 clock is pot D.
## Puzzle 100
Name: Cat Catcher
Trigger: Talk to Hugo
Location: Apartment Row
Chapter: 8
Picarats: 35
Description: A naughty tomcat has been causing trouble all over town. Now he's in spot B, hiding from his owner, who'es trying her best to track him down. The owner starts at spot A, but after she moves, the cat jumps to an adjacent spot. If the owner plans her route well,she'l be able to catch the cat on her fourth move. What path should she follow for her first three moves?
• Hint 1: The cat's owner has four possible moves at the start. Two of these moves will guide her to the right spot. The key to think puzzle is that the owner will move before the cat. Try to figure out which is the best place to be if you want to catch that cat.
• Hint 2: The best place to be if you want to catch the cat is in the spot that has the most connections. the cat has a number of positions it could land on by the third move, so the owner should aim at the one with the most options.
• Hint 3: Where could the cat be after his third move? There are four possible places it could be hiding. once you know that, it should be obvious where the owner has to go on her third move.
• Super Hint: By the third move, the cat will be at either C, E, F, or H. Regardless of where the cat is on his third move, there is only one place that guarantees the cat's owner can catch him on her fourth move. | 8,052 | 33,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-49 | latest | en | 0.913906 |
http://www.cse.iitd.ernet.in/~suban/vision/tutorial/node30.html | 1,508,660,951,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825154.68/warc/CC-MAIN-20171022075310-20171022095310-00511.warc.gz | 425,282,571 | 4,333 | Next: Projective Bases for the Up: Some Standard Cross Ratios Previous: Cross Ratios and Projective
## Cross Ratios of Pencils of Lines
Let U and V be two lines in the projective plane, defined by their dual coordinate vectors. Consider a set of lines defined by:
with the defined up to scale as usual. belongs to the pencil of lines through the intersection of U and V (c.f. section 2.2.1), so each Wi passes through this point. As in section 3.1.1, the cross ratio is defined to be the cross ratio of the four homogeneous coordinate pairs . Dually to points, we have:
Theorem: The cross ratio of any four lines of a pencil is invariant under collineations.
Cross ratios of collinear points and coincident lines are linked as follows:
Theorem: The cross ratio of four lines of a pencil equals the cross ratio of their points of intersection with an arbitrary fifth line transversal to the pencil (i.e. not through the pencil's centre) -- see fig. 3.1.
In fact, we already know that the cross ratios of the intersection points must be the same for any two transversal lines, since the lines correspond bijectively to one another under a central projection, which is a collineation.
The simplest way to establish the result is to recall the line intersection formulae of section 2.2.1. If U and Vare the basis 3-vectors for the line pencil, and L is the transversal line, the intersection points of L with U, V and are respectively , and . In other words, the coordinates of a line in the basis are the same as the coordinates of its intersection with L in the basis . Hence, the two cross ratios are the same.
To establish the result, we will consider a particular transversal line: the line at infinity. Let ax0 + bx1 + cx2 = 0 be the equation defining U and dx0 + ex1 + fx2 = 0 the one corresponding to V. Therefore the one associated to is:
This line intersect the line at infinity at a point with coordinates:
If we consider two reference points M and N being the intersection with line U and V, i.e. the points with coordinates
the generic point can therefore be described as . Therefore the intersection of line Wi with the line at infinity has the coordinates on the line [M, N], and therefore from the definition their cross ratios coincide.
Another way to prove the result is to show that cross ratios of lengths along the transversal line can be replaced by cross ratios of angle sines, and hence are independent of the transversal line chosen (c.f fig. 3.1):
(3.3)
However, this is quite a painful way to compute a cross ratio. A more elegant method uses determinants:
Theorem (Möbius): Let be any four lines intersecting in O, and be any four points respectively on these lines, then
(3.4)
where denotes the determinant of the matrix whose columns are the homogeneous coordinate vectors of points O, Ai and Aj.
This 19th century result extends gracefully to higher dimensions. To prove it, let (a,b, 1), (x,y, 1) and (u,v, 1) be the normalized affine coordinate vectors of O, Ai and Aj. Then
The vector's lengths cancel out of the cross ratio of these terms, and if the coordinate vectors do not have affine normalization, the scale differences cancel out too.
Next: Projective Bases for the Up: Some Standard Cross Ratios Previous: Cross Ratios and Projective
Bill Triggs
1998-11-13 | 775 | 3,318 | {"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} | 4.40625 | 4 | CC-MAIN-2017-43 | latest | en | 0.914636 |
https://mathematica.stackexchange.com/questions/206618/is-it-possible-to-make-decompose-work-with-coefficients-containing-radicals | 1,597,099,013,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738699.68/warc/CC-MAIN-20200810205824-20200810235824-00501.warc.gz | 403,216,541 | 34,259 | # Is it possible to make Decompose work with coefficients containing radicals?
It appears that Decompose works reliably only on polynomials with integer coefficients, although it seems to be not mentioned in the docs. Can I make it work (or implement an alternative that would work) on polynomials with coefficients containing radicals? For example,
Decompose[3 + 3 √2 + (14 + 4 √2) x + (12 + 26 √2) x^2 +
(56 + 8 √2) x^3 + (8 + 48 √2) x^4 + 48 x^5 + 16 √2 x^6, x]
(* {3 + 3 √2 + (14 + 4 √2) x + (4 + 12 √2) x^2 + 8 x^3, x + √2 x^2} *)
Is there a known algorithmic solution to this problem at all?
One way to decompose a polynomial $$p(x)$$ is to factor $$p'(x)$$ over a field and exploit $$\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$$.
The following code does just that to find all candidate $$g'(x)$$:
extendedDecompose[poly_, x_, extension_:Automatic] /; PolynomialQ[poly, x] :=
Module[{ext, res},
ext = If[extension === Automatic, CoefficientList[poly, x], extension];
res = FixedPoint[
Flatten[Prepend[Rest[#], iExtendedDecompose[First[#], x, ext]]]&,
{poly}
];
res = res //. {a___, x^n_., x^m_., b___} :> {a, x^(n + m), b};
simplifyCoefficients[res, x]
]
extendedDecompose[expr_, ___] := {expr}
iExtendedDecompose[poly_, x_, extension_] :=
Module[{deg, facs, cands, dg, d, a, gcoeffs, f, g, sys, sol, res},
deg = Exponent[poly, x];
If[deg <= 1 || PrimeQ[deg], Return[{poly}]];
facs = Select[FactorList[D[poly, x], Extension -> extension], Exponent[#[[1]], x] > 0&];
If[Length[facs] < 2, Return[{poly}]];
cands = Select[Times @@@ Subsets[Join @@ ConstantArray @@@ facs], 1 <= Exponent[#, x] <= 0.5deg&];
SetAttributes[a, Listable];
Catch[
Do[
dg = Exponent[gprime, x]+1;
d = deg/dg;
f = fromCoefficients[a[Range[0, d]], x];
gcoeffs = Prepend[CoefficientList[gprime, x]/Range[dg], 0];
gcoeffs /= SelectFirst[gcoeffs, Not @* PossibleZeroQ];
g = fromCoefficients[gcoeffs, x];
sys = Thread[CoefficientList[Expand[f /. x -> g] - poly, x] == 0];
sol = Solve[sys, a[Range[0, d]]];
If[ListQ[sol] && Length[sol] > 0,
Throw[{f /. First[sol], g}]
],
{gprime, cands}
];
{poly}
]
]
SetAttributes[simplifyCoefficients, Listable];
simplifyCoefficients[poly_, x_] :=
fromCoefficients[Simplify[CoefficientList[poly, x]], x]
fromCoefficients[coeffs_, x_] := coeffs . PowerRange[1, x^(Length[coeffs] - 1), x]
poly = 3 + 3 √2 + (14 + 4 √2) x + (12 + 26 √2) x^2 +
(56 + 8 √2) x^3 + (8 + 48 √2) x^4 + 48 x^5 + 16 √2 x^6;
decomp = extendedDecompose[poly, x]
{3 (1 + √2) + (14 + 4 √2) x + (4 + 12 √2) x^2 + 8 x^3, x + √2 x^2}
PossibleZeroQ[poly - (First[decomp] /. x -> Last[decomp])]
True
A nested example:
poly2 = Collect[poly /. x -> x^4 - x + 1, x];
decomp2 = extendedDecompose[poly2, x]
{3 (47 + 35 √2) + (-478 - 368 √2) x + (540 + 436 √2) x^2 - 8 (25 + 22 √2) x^3,
x - (2 x^2)/(4 + √2), x - x^4}
Fold[#1 /. x -> #2 &, decomp2] - poly2 // PossibleZeroQ
True
• This is also the method used by Decompose (due to Alagar and Tranh if I remember correctly). It was implemented well before the Extension option in Factor and we never really revisited it. – Daniel Lichtblau Sep 22 '19 at 14:02 | 1,124 | 3,102 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-34 | latest | en | 0.688349 |
http://mathoverflow.net/questions/85311/optimization-of-a-specific-polynomial | 1,469,808,130,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831769.86/warc/CC-MAIN-20160723071031-00164-ip-10-185-27-174.ec2.internal.warc.gz | 157,760,480 | 15,903 | Optimization of a Specific Polynomial
I have a polynomial:
$$f(x_1 \dots x_n) = \prod_{i=1}^n (c_ix_i + 1) - \frac{1}{2}c_0\sum_{i=1}^nx_i^2$$
Given some values for $c_0 \dots c_n$, I'd like to choose the maximizing values for $x_1 \dots x_n$. I'm not concerned with the actual maximum value of $f(x_1 \dots x_n)$, only the values of the inputs $x_1 \dots x_n$, so if there is an equivalent function maximized at the same values I value the solution just as much. The values for any $c$ and any $x$ can be any real value.
I'm not familiar with any easy ways to do this with a polynomial of this type. My best guess is gradient ascent, but I'm not sure the function is convex.
Can anyone tell me if the function is convex, or failing that, if there's another way to find an (approximate) global argmax?
-
What values are these variables allowed to take? – Gjergji Zaimi Jan 10 '12 at 6:43
Note that if there are at least 3 nonzero coefficients among $c_1\dots c_n$ then $f$ is unbounded, both from above and from below. – Pietro Majer Jan 10 '12 at 9:56
@Pietro Majer - I'm missing something; Can you explain why this is the case? – Charles Parker Jan 11 '12 at 1:22
@Gjergji Zaimi - The variables must take real values. I've edited the question. – Charles Parker Jan 11 '12 at 1:25
Presumably you want the subscripts in your product and sum to range over $i=1,\ldots,n$, rather than have each term pinned to its value at sub$_n$. – Joseph O'Rourke Jan 11 '12 at 12:03
As noted by Pietro in the comments only a few combinations of $n$ and coefficients will even have a finite maximum, unless you bound your domain. For example, consider $c_0=-2$ and all other $c_i=0$. Even this is unbounded above.
Now, let's assume you either bound your domain, or have ensured that the polynomial is bounded above. The problem of finding the argmax is computationally "difficult" for large $n$ (as in: the cryptography guys worry about this stuff a lot at least in finite fields), so numerical optimization methods do probably make sense. Assuming your coefficients are sufficiently generic, a sequential quadratic programming technique will definitely work much better than gradient-climbing. | 596 | 2,185 | {"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} | 3.28125 | 3 | CC-MAIN-2016-30 | latest | en | 0.86287 |
http://umapiano.com/gpm-to-hxhydf/74jaix.php?id=leverage-score-hat-matrix-d8f61e | 1,623,971,096,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00605.warc.gz | 44,827,848 | 12,226 | Because it contains the "leverages" that help us identify extreme x values! H = X ( XTX) –1XT. So, where is the connection between these two concepts: The leverage score of a particular row or observation in the dataset will be found in the corresponding entry in the diagonal of the hat matrix. For reporting purposes, it would therefore be advisable to analyze the data twice — once with and once without the red data point — and to report the results of both analyses. Similarly, the (i,j)-cross-leverage scores are equal to the off-diagonal elements of this projection matrix, i.e., cij = (PA)ij = U(i),U(j) . Posted by oolongteafan1 on January 15, 2018 January 31, 2018. Not used, if method=highest.ranks. Clearly, O(nd2) time suffices to compute all the statis- The statistical leverage scores are widely used for detecting outliers and influential data [ 27], [28], [13]. As we know from our investigation of this data set in the previous section, the red data point does not affect the estimated regression function all that much. If we actually perform the matrix multiplication on the right side of this equation: we can see that the predicted response for observation i can be written as a linear combination of the n observed responses y1, y2, ..., yn: $\hat{y}_i=h_{i1}y_1+h_{i2}y_2+...+h_{ii}y_i+ ... + h_{in}y_n \;\;\;\;\; \text{ for } i=1, ..., n$. The matrix displayed on the right shows the resulting change in the fitted ... important to recognize that the sum of leverages for a set of observations equals the number of variables in the design matrix. The proportionality constant used is called Leverage which is denoted by h i.Hence each data point has a leverage value. hii of H may be interpreted as the amount of leverage excreted by the ith observation yi on the ith fitted value ˆ yi. Here are some important properties of the leverages: The first bullet indicates that the leverage hii quantifies how far away the ith x value is from the rest of the x values. %�쏢 Let's try our leverage rule out an example or two, starting with this data set (influence3.txt): Of course, our intution tells us that the red data point (x = 14, y = 68) is extreme with respect to the other x values. You might recall from our brief study of the matrix formulation of regression that the regression model can be written succinctly as: Therefore, the predicted responses can be represented in matrix notation as: And, if you recall that the estimated coefficients are represented in matrix notation as: then you can see that the predicted responses can be alternatively written as: That is, the predicted responses can be obtained by pre-multiplying the n × 1 column vector, y, containing the observed responses by the n × n matrix H: Do you see why statisticians call the n × n matrix H "the hat matrix?" ��?�����ӏk�I��5au�D��i��������]�{rIi08|#l��2�yN��n��2Ⱦ����(��v傌��{ƂK>߹OB�j\�j:���n�Z3�~�m���Zҗ5�=u���'-��Qt��C��"��9Й�цI��d2���x��� \AL� ���L;�QiPoj?�xL8���� [^���2�]#� �m��SGN��em��,τ�g�e��II)�p����(����rE�~Y-�N����xo�#Lt��9:Y��k2��7��+KE������gx�Q���& ab�;� 9[i��l��Xe���:H�rX��xM/�_�(,��ӫ��&�qz���>C"'endstream Default: 1. That is, if hii is small, then the observed response yi plays only a small role in the value of the predicted response $$\hat{y}_i$$. Z(L*��°��uT�c��1�ʊ�; *�J�bX�"��Fw�7P9�F1Q��ǖ�$����Z���*����AF��\:�7Z��?-�k,�T^�4�~�vX��P��ol��UB=t81?��i;� ... Then and where the hat matrix is the projection matrix onto the column space of ,, <> A vector with the diagonal Hat matrix values, the leverage of each observation. The sum of the h ii equals k+1, the number of parameters (regression coefficients including the intercept). And, as we move from the x values near the mean to the large x values the leverages increase again. When n is large, Hat matrix is a huge (n * n). In some applications, it is expensive to sample the entire response vector. If the ith x value is far away, the leverage hii will be large; and otherwise not. x��WM�7˄fW���H��H�&i���H q �p%�&��H���U�SͰZ%���.�U��+W��ж��7�_��������_�Ok+��>�t�����[��:TJWݟ�EU���H)U>E!C����������)CT����]�����[[g����� Let's see how this the leverage rule works on this data set (influence4.txt): Of course, our intution tells us that the red data point (x = 13, y = 15) is extreme with respect to the other x values. I don't know of a specific function or package off the top of my head that provides this info in a nice data … If a data point i, is moved up or moved down, the corresponding fitted value y i ’moves proportionally to the change in y i. tistical leverage scores of a matrix A are equal to the diagonal elements of the projection matrix onto the span of its columns. 16 0 obj alpha=0 is equivalent to method="top.scores". This entry in the hat matrix will have a direct influence on the way entry$y_i$will result in$\hat y_i$( high-leverage of the$i\text{-th}$… i��lx�w#��I[ӴR�����i��!�� Npx�mS�N��NS�-��Q��j�,9��Q"B���ͮ��ĵS2^B��z���ԠL_�E~ݴ�w��P�C�y��W-�t�vw�QB#eE��L�0���x/�H�7�^!�tp�&{���@�(c�9(�+ -I)S�&���X��I�. In the linear regression model, the leverage score for the i t h data unit is defined as: h i i = (H) i i, the i t h diagonal element of the hat matrix H = X (X ⊤ X) − 1 X ⊤, where ⊤ denotes the matrix transpose. For matrix with rows denote the leverage score of row by. Computing an explicit leave-one-observation-out (LOOO) loop is included but no influence measures are currently computed from it. """ Let's take another look at the following data set (influence2.txt): this time focusing only on whether any of the data points have high leverage on their predicted response. Leverages only take into account the extremeness of the x values, but a high leverage observation may or may not actually be influential. stream The diagonal elements of H are the leverage scores, that is, Hi,i is the leverage of the ith sample. See x2fx for a description of this matrix and for a description of the order in which terms appear. x�}T�n�0��N� v��iy$b��~-P譆nMO)R�@ 5 0 obj Let's see if our intuition agrees with the leverages. What does your intuition tell you? The American Statistician , 32(1):17-22, 1978. vector is then by= Hy, where H = XX† is the hat matrix. <> Remember, a data point has large influence only if it affects the estimated regression function. I can't find a proof anywhere. tells a different story this time. To identify a leverage point, a hat matrix: H= X(X’X)-1 X’ is used. Rather than looking at a scatter plot of the data, let's look at a dotplot containing just the x values: Three of the data points — the smallest x value, an x value near the mean, and the largest x value — are labeled with their corresponding leverages. The hat matrix projects the outcome variable(s) ... was increased by one unit and PCs and scores recomputed. 6 0 obj 3 are, up to scaling, equal to the diagonal elements of the so-called “hat matrix,” i.e., the projection matrix onto the span of the top k right singular vectors of A (19, 20). These quantities are of interest in recently-popular problems such as matrix completion and Nystrom-based low-rank¨ 0 ≤ h i i ≤ 1 ∑ i = 1 n h i i = p, where p is the number of coefficients in the regression model, and n is the number of observations. Leverage Values • Outliers in X can be identified because they will have large leverage values. The leverage score for subject i can be expressed as the ith diagonal of the following hat matrix: (6.26) H = X X ′ V Θ ˆ − 1 X − X ′ V Θ ˆ − 1 . Let's see! Therefore: Now, the leverage of the data point, 0.311, is greater than 0.286. In the case study, we manually inspect the most influential samples, and find that influence sketching pointed us to new, previously unidentified pieces of malware.1 I. weighted if true, leverage scores are computed with weighting by the singular values. In this section, we learn more about "leverages" and how they can help us identify extreme x values. Best used whith method=top.scores. So for observation $i$ the leverage score will be found in $\bf H_{ii}$. The leverage score is also known as the observation self-sensitivity or self-influence, because of the equation Privacy and Legal Statements 8 2.1 Leverage Average leverages We showed in the homework that the trace of the hat matrix equals the number of coe cients we estimate: trH = p+ 1 (17) But the trace of any matrix is the sum of its diagonal entries, trH = Xn i=1 H ii (18) so the trace of the hat matrix is the sum of each point’s leverage. • Leverage considered large if it is bigger than 576 @cache_readonly def hat_matrix_diag (self): """ Diagonal of the hat_matrix for GLM Notes-----This returns the diagonal of the hat matrix that was provided as argument to GLMInfluence or computes it using the results method get_hat_matrix. """ In this case, there are n = 21 data points and k+1 = 2 parameters (the intercept β0 and slope β1). Let the data matrix be X (n * p), Hat matrix is: Hat = X(X'X)^{-1}X' where X' is the transpose of X. stream We did not call it "hatvalues" as R contains a built-in function with such a name. As you can see, the two x values furthest away from the mean have the largest leverages (0.176 and 0.163), while the x value closest to the mean has a smaller leverage (0.048). endobj The hat matrix in regression and ANOVA. So computing it is time consuming. endobj As such, they have a natural statistical interpretation as a “leverage score” or “influence score” associated with each of the data points ( … Do any of the x values appear to be unusually far away from the bulk of the rest of the x values? Alternatively, model can be a matrix of model terms accepted by the x2fx function. Let's use the above properties — in particular, the first one — to investigate a few examples. stream where the weights hi1, hi2, ..., hii, ..., hin depend only on the predictor values. A refined rule of thumb that uses both cut-offs is to identify any observations with a leverage greater than $$3 (k+1)/n$$ or, failing this, any observations with a leverage that is greater than $$2 (k+1)/n$$ and very isolated. The hat matrix H is defined in terms of the data matrix X: H = X ( XTX) –1XT. %PDF-1.2 Do any of the x values appear to be unusually far away from the bulk of the rest of the x values? The leverage of observation i is the value of the i th diagonal term, hii , of the hat matrix, H, where. It's for this reason that the hii are called the "leverages.". On the other hand, if hii is large, then the observed response yi plays a large role in the value of the predicted response $$\hat{y}_i$$. In this case k should be set to its default value. Therefore, the data point should be flagged as having high leverage, as it is: In this case, we know from our previous investigation that the red data point does indeed highly influence the estimated regression function. Source code for regressors.stats. Should be positive. And, as we move from the x values near the mean to the large x values the leverages increase again (the last leverage in the list corresponds to the red point). As with many statistical "rules of thumb," not everyone agrees about this $$3 (k+1)/n$$ cut-off and you may see $$2 (k+1)/n$$ used as a cut-off instead. The i th diagonal of the above matrix is the leverage score for subject i displaying the degree of the case’s difference from others in one or more independent variables. Moreover, we find that influential samples are especially likely to be mislabeled. You can use this matrix to specify other models including ones without a constant term. The leverage score is also known as the observation self-sensitivity or self-influence, because of the equation $h_{ii} = \frac{\partial\widehat{y\,}_i}{\partial y_i},$ which states that the leverage of the i -th observation equals the partial derivative of the fitted i -th dependent value $\widehat{y\,}_i$ with respect to the measured i -th dependent value $y_i$ . x��UKkA&��1���n\5ڞ�}��ߏ� ��b��z�(+$��uϣk�� 2�������j�����]����������6�K��l��Ȼ�y{�T��)���s\�H�]���0ͅ�A���������k�w�x��!�7H�0�����Y+� ��@ϑ}�w!Jo�Ar�(�4�aq�U� There is such an important distinction between a data point that has high leverage and one that has high influence that it is worth saying it one more time: Copyright © 2018 The Pennsylvania State University 639 Leverage scores and matrix sketches for machine learning. 1 Leverage.This is a measure of how unusual the X value of a point is, relative to the X observations as a whole. and determines the fitted or predicted values since. The hat matrix is also known as the projection matrix because it projects the vector of observations, y, onto the vector of predictions, , thus putting the "hat" on y. For robust fitting problem, I want to find outliers by leverage value, which is the diagonal elements of the 'Hat' matrix. INTRODUCTION 23 0 obj I think you're looking for the hat values. In this talk we will discuss the notion of leverage scores: a simple statistic that reveals columns (or rows) of a matrix that lie well within the subspace spanned by the top prin-cipal components. Again, of the three labeled data points, the two x values furthest away from the mean have the largest leverages (0.153 and 0.358), while the x value closest to the mean has a smaller leverage (0.048). matrixchernoffbound Morespecifically,togetasubspaceembedding,wesample eachcolumnaiwithprobability˝(ai) logn ϵ2. The statistical leverage scores of a matrix A are the squared row-norms of the matrix containing its (top) left singular vectors and the coherence is the largest leverage score. Use hatvalues(fit).The rule of thumb is to examine any observations 2-3 times greater than the average hat value. The function returns the diagonal values of the Hat matrix used in linear regression. The diagonal terms satisfy. Well, all we need to do is determine when a leverage value should be considered large. �G�!� Let's take another look at the following data set (influence3.txt): What does your intuition tell you here? H = A(ATA)-1AT is the “hat” matrix, i.e.$�萒�Q�:�yp�Д�l�e O����J��%@����57��4��K4k5�༗)%�S�*$�=4��lo.�T*D�g��G�K����*gfVX����U�� �SRN[>'x_�ZB����Bl�����t���t8ZF�d0!sj�R� kd[ A common rule is to flag any observation whose leverage value, hii, is more than 3 times larger than the mean leverage value: $\bar{h}=\frac{\sum_{i=1}^{n}h_{ii}}{n}=\frac{k+1}{n}$. Hat matrix H = A(ATA)−1AT Leverage scores ℓ j(A) = H jj 1 ≤ j ≤ m Singular Value Decomposition A = U ΣVT UT U =I n Hat matrix H = UUT ℓ j(A) = keT j Uk 2 1 ≤ j ≤ m QR decomposition A = Q R QTQ =In Hat matrix H = QQT ℓ j(A) = keT Qk2 1 ≤ j ≤ m Definition. endobj The statistical leverage scores of a matrix A are the squared row-norms of the matrix containing its (top) left singular vectors and the coherence is the largest leverage score. ����i\�>���-=O��-� W��Nq�A��~B�DQ��D�UC��e:��L�D�ȩ{}*�T�Tf�0�j��=^����q1�@���V���8�;�"�|��̇v��A���K����85�s�t��&kjF��>�ne��(�)������n;�.���9]����WmJ��8/��x!FPhڹ�� How? And, that's exactly what happens in this statistical software output: A word of caution! 15 0 obj That is, are any of the leverages hii unusually high? Leverage of a point has an absolute minimum of 1=n, and we can see that the red point is right in the middle of the points on the X axis, and has a residual of 0.05. Let's see! 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The rest of the leverage score will be large ; and otherwise not unusual. Included but no influence measures are currently computed from it. '' in which terms appear large influence only it... 0.311, is the leverage score will be large ; and otherwise.! Posted by oolongteafan1 on January 15, 2018 January 31, 2018 January 31,.! Denote the leverage of the x observations as a whole vector with the leverages hii high... I$ the leverage hii will be large ; and otherwise not it seems as if the ith.!, why do we care about the hat matrix is a number between 0 and 1 inclusive. As having high leverage — in particular, the first one leverage score hat matrix to investigate a few examples how the. Your intuition tell you here as R contains a built-in function with a. S )... was increased by one unit and PCs and scores.... Be large ; and otherwise not will be large ; and otherwise not i.Hence each data,..., hi2,..., hii,..., hin depend only on the third property mentioned above Outliers x... Slope β1 ) reduces predictive accuracy all the way down to 90.24.... Such a name highly influence the estimated regression function, leverage scores computed. Expensive to sample the entire response vector 28 ], [ 13 ] learn more . '' as R contains a built-in function with such a name value is far away the! Number between 0 and 1, inclusive ], [ 13 ] leverages. increased by unit. | 6,416 | 24,349 | {"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": 3, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-25 | latest | en | 0.883248 |
https://www.tes.com/teaching-resources/blog/tes-maths-top-resources-october | 1,500,656,188,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423787.24/warc/CC-MAIN-20170721162430-20170721182430-00175.warc.gz | 865,148,483 | 15,864 | # TES Maths: Top resources for October
Craig Barton
03rd October 2016
## Engage secondary students with this selection of the best lesson and activity ideas, as chosen by TES Maths
With a month’s worth of teaching under your belt, it’s time to make sure that your pupils have really grasped everything they’ve learnt so far. These stimulating resources are ideal for cementing understanding in the areas of proportion, quadrilaterals, graphs and much more.
Ensure your learners are on the right track with these hand-picked lesson ideas, created and shared by the generous and talented members of the TES Maths community.
Craig Barton, TES Maths adviser
### Codebreakers
• Proportion codebreaker
Get practising a range of GCSE-level questions on direct and indirect proportion, including those that require them to generate formulae, with this humorous activity.
• Simultaneous equations coded message
Consolidate work on simultaneous equations using these differentiated worksheets, requiring pupils to crack the code by matching equations with solutions.
### Shapes and angles
• Properties of quadrilaterals
Secure pupils’ understanding of the properties of quadrilaterals, as well as the methods used to find missing angles, with this well-structured lesson and activities.
• Parallel line maze
Using their knowledge of key angle facts, challenge learners to plot the path from one side of the maze to the other. Answers provided.
### Graphs
• Cubic graph practice
Help GCSE and A-level students get to grips with the properties of cubic graphs by setting this unique matching task.
• Finding the equation of a line
Also relevant to the new GCSE specification, pupils must calculate the equation of a line between two given points in order to decode the hidden message.
### Homework
• KS3 and 4 homework booklets
Revise a wide range of topics every week with this collection of workbooks for Years 7 to 11, including a table to help pupils to assess their learning.
• Differentiated homework pack
With over 600 different activity sheets, these booklets not only cover all aspects of the new GCSE specification, but also encourage learners to record their own targets and progress.
This blog post is featured in the October maths newsletter from TES Resources.
If you'd like to receive any of our resources newsletters, all you have to do is edit your email preferences when logged into TES.com. Simply go to your preference centre and tick the boxes that are relevant to you. Let us know if you have any problems. | 521 | 2,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-30 | longest | en | 0.929987 |
https://businesscasestudies.co.uk/analysis-of-profitability-liquidity-and-performance/ | 1,653,737,034,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00378.warc.gz | 196,719,870 | 92,215 | The profit of a business is the difference between its revenues and its costs. It is important to consider two main types of profit:
1.Gross profit – this is calculated by deducting the cost of sales of a business from its sales revenue (turnover).
2.Operating profit – is calculated by then taking away overhead expenses from gross profit.
Given the above figures it is possible to analyse the profitability of Better Hotels Plc in the two years. To do this we need to calculate how much of every pound spent by customers in the hotels is profit. This is calculated in the following way:
1. Gross profit % (i.e. how many pence in each £1 of customer spending is profit). This is calculated by:
For Better Hotels in 2004 this is:
For Better Hotels in 2005 this is:
The profit margin i.e. operating profit %
is calculated by:
For Better Hotels in 2004 this is:
For Better Hotels in 2005 this is:
By examining the profit figures you can see that Better Hotels is more profitable in 2005 than it was in 2004.
Gross profit % has gone up from 60% to 75%, and Operating profit % has increased from 30% to 40%.
Profitability
Using these profitability calculations you are able to compare business profits in one year compared with others, and also compare the profitability of different businesses.
Another important measure of how well a business is being run is how liquid it is. To do this you need to look at the current assets and current liabilities in the balance sheet.
The following shows part of the balance sheet for Better Hotels in 2004 and 2005:
Extract from Balance Sheet 31st Dec 2004 by examining the two balance sheets it is possible to see that in 2005, Better Hotels has a more liquid assets relative to current liabilities.
In 2004 the ratio of current assets to current liabilities was:
80:40 (i.e. £2 for every £1)
In 2005 the ratio was:
90:40 (i.e. £2.50 for every £1)
It is important for businesses to have a good liquidity position because, should people that the business owes money to (current liabilities) press for payment it is essential to have the liquidity to pay up. A liquid asset is one that can quickly be turned into cash.
Working capital
We use the term working capital to describe the difference between current assets and current liabilities. A business has working capital if its current assets are greater than its current liabilities. Working capital is required for the day-to-day running of a business – paying bills, wages etc.
A business performs well when it has:
*high and rising sales
*high and rising profits
*good control over its costs
*a good liquidity/working capital position. | 579 | 2,656 | {"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-2022-21 | longest | en | 0.964611 |
https://ccp4wiki.org/te6t62a/ | 1,628,172,118,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155925.8/warc/CC-MAIN-20210805130514-20210805160514-00366.warc.gz | 175,887,222 | 16,526 | Multiplication Worksheets: 2nd Math. Pythagorean Theorem Word Problems Grade 8 Connect The Dots Kindergarten We Bring The Good Multiplication Worksheets To Life | CCP4 Wiki
2nd Math
Published at Wednesday, December 16th 2020. by in Multiplication Worksheets.
The alphabet song: This remains a lovely way to practice the alphabet. Sing it slowly and sing it often. If you have a large alphabet chart and point out to each letter while you sing, it will be of great value. You can give all children letter cards in order (alphabet flashcards); they can hold up each letter as it is sung. Show a magazine or picture book to children. Ask them to identify all instances of the given letter in any page. Hand out letter cards to all children. Call out a letter. The child with that card has to come in front of the class and display the letter. Divide the class into two groups. Give one group letter cards. Give other group various objects. The first group will hold up a letter. The second group should hold up an object that starts with that letter.
Remember that with the so many worksheets available, choose one that is best suited for a certain lesson. Plan ahead what type of worksheet to use for a given day, depending on what you plan to teach. There are many free worksheets available, especially online, but still the best worksheet is one that you personally draft. This way, you are able to match the level of difficulty of the activity in accordance to the performance level of your own students. It is not bad to reuse worksheets for another batch of students, but once in a while it is also better to vary the activities you give to kids. Worksheets can be made for fun if it is attuned to the current interests of kids. The kids will respond better to activities close to their own interests.
Minute Math Multiplication
Care should be taken to give children worksheets that they are capable of doing. This involves understanding and monitoring the child continually, since the level of attainment of different children would often be quite different. The worksheet should challenge the child but not overwhelm her. If the worksheet is too easy or too repetitive, it may bore the child and she would not be happy. If the activity is too difficult it would frustrate her and she would not like to take up more sheets. The children should love to do the worksheets; they should not be thrust upon them. Also doing only worksheets alone repeatedly would not be very productive. You should have a range of physical games and activities as well that would reinforce the concepts learned.
Birthday celebrations. On the child has birthday, the class has a party. The children gather around as the teacher tells the story of the child has birth and life up till now. Invariably the story begins with the child has spirit looking down on the earth, and deciding to go down to join the people there. Candles may be lit as the story is told, another candle for each year of life. Then the children have a special cake that has been baked by the teacher. Food: All the food is completely natural, with no strange chemicals or additives. The menu emphasizes whole grains & organic foods. Usually the children help with the preparation. Snacks are frequently oatmeal or some other porridge, fruit, tea, and water.
Well made kindergarten worksheets can be interesting for children to do and can be of great use in reinforcing basic concepts. Completing the activity in a worksheet can give children an immense sense of accomplishment. Kindergarten ABC worksheets should have different activities to help children identify the various letters of the alphabet. The activities may involve very simple things like coloring, ticking, drawing a line to match items etc. Using attractive illustrations and cartoon characters would make it more fun for children. The activities should be graded, i.e initial activities should be very simple and easy (but should be fun with good pictures etc, so as to interest the child); later worksheets may involve a little bit more work.
If you are looking for printable worksheets for your preschool child, the array of choices can be a little intimidating. You may just be looking for a few pages to keep your child occupied with something more constructive than yet another half hour in front of the TV, or you may feel it is time you started helping your child learn the basic skills she or he will need for school. Whatever your motivation for looking for worksheets for preschool, there are a few points to consider before you decide which ones you want.
No plastic. Children this age are still integrating their senses (as you know if you know a child with sensory integration difficulties, as we did), and it is very helpful for them to be able to match up texture, weight, color and pattern consistently. Natural materials such as wood always look and feel the same; the child gets a consistent message. Plastics have dazzling colors and any number of strange textures and weights. Furthermore, natural materials are simply more comfortable for a young child. Dress-up area. Over here is a rack of costumes hanging, and a bin stuffed with crowns, boas, sashes, and capes. All costumes are made of cotton, wool, silk, and other natural fibers. Nature table. Children are forever finding treasures in nature: pine cones, rocks, feathers, flowers, shells of cicadas, autumn leaves. On this table they are arranged lovingly -- a sort of altar to nature. | 1,097 | 5,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-31 | longest | en | 0.939571 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1344/2/a/h/1/1/ | 1,713,157,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00686.warc.gz | 777,787,449 | 77,680 | # Properties
Label 1344.2.a.h.1.1 Level $1344$ Weight $2$ Character 1344.1 Self dual yes Analytic conductor $10.732$ Analytic rank $0$ Dimension $1$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [1344,2,Mod(1,1344)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(1344, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0, 0, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("1344.1");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$1344 = 2^{6} \cdot 3 \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1344.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$10.7318940317$$ Analytic rank: $$0$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 672) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## Embedding invariants
Embedding label 1.1 Character $$\chi$$ $$=$$ 1344.1
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q-1.00000 q^{3} +2.00000 q^{5} -1.00000 q^{7} +1.00000 q^{9} +O(q^{10})$$ $$q-1.00000 q^{3} +2.00000 q^{5} -1.00000 q^{7} +1.00000 q^{9} +4.00000 q^{11} +6.00000 q^{13} -2.00000 q^{15} -2.00000 q^{17} -4.00000 q^{19} +1.00000 q^{21} -4.00000 q^{23} -1.00000 q^{25} -1.00000 q^{27} +2.00000 q^{29} +8.00000 q^{31} -4.00000 q^{33} -2.00000 q^{35} +10.0000 q^{37} -6.00000 q^{39} -2.00000 q^{41} -8.00000 q^{43} +2.00000 q^{45} +1.00000 q^{49} +2.00000 q^{51} +10.0000 q^{53} +8.00000 q^{55} +4.00000 q^{57} +12.0000 q^{59} -10.0000 q^{61} -1.00000 q^{63} +12.0000 q^{65} +8.00000 q^{67} +4.00000 q^{69} +12.0000 q^{71} +2.00000 q^{73} +1.00000 q^{75} -4.00000 q^{77} +1.00000 q^{81} -12.0000 q^{83} -4.00000 q^{85} -2.00000 q^{87} +6.00000 q^{89} -6.00000 q^{91} -8.00000 q^{93} -8.00000 q^{95} +2.00000 q^{97} +4.00000 q^{99} +O(q^{100})$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 0 0
$$3$$ −1.00000 −0.577350
$$4$$ 0 0
$$5$$ 2.00000 0.894427 0.447214 0.894427i $$-0.352416\pi$$
0.447214 + 0.894427i $$0.352416\pi$$
$$6$$ 0 0
$$7$$ −1.00000 −0.377964
$$8$$ 0 0
$$9$$ 1.00000 0.333333
$$10$$ 0 0
$$11$$ 4.00000 1.20605 0.603023 0.797724i $$-0.293963\pi$$
0.603023 + 0.797724i $$0.293963\pi$$
$$12$$ 0 0
$$13$$ 6.00000 1.66410 0.832050 0.554700i $$-0.187167\pi$$
0.832050 + 0.554700i $$0.187167\pi$$
$$14$$ 0 0
$$15$$ −2.00000 −0.516398
$$16$$ 0 0
$$17$$ −2.00000 −0.485071 −0.242536 0.970143i $$-0.577979\pi$$
−0.242536 + 0.970143i $$0.577979\pi$$
$$18$$ 0 0
$$19$$ −4.00000 −0.917663 −0.458831 0.888523i $$-0.651732\pi$$
−0.458831 + 0.888523i $$0.651732\pi$$
$$20$$ 0 0
$$21$$ 1.00000 0.218218
$$22$$ 0 0
$$23$$ −4.00000 −0.834058 −0.417029 0.908893i $$-0.636929\pi$$
−0.417029 + 0.908893i $$0.636929\pi$$
$$24$$ 0 0
$$25$$ −1.00000 −0.200000
$$26$$ 0 0
$$27$$ −1.00000 −0.192450
$$28$$ 0 0
$$29$$ 2.00000 0.371391 0.185695 0.982607i $$-0.440546\pi$$
0.185695 + 0.982607i $$0.440546\pi$$
$$30$$ 0 0
$$31$$ 8.00000 1.43684 0.718421 0.695608i $$-0.244865\pi$$
0.718421 + 0.695608i $$0.244865\pi$$
$$32$$ 0 0
$$33$$ −4.00000 −0.696311
$$34$$ 0 0
$$35$$ −2.00000 −0.338062
$$36$$ 0 0
$$37$$ 10.0000 1.64399 0.821995 0.569495i $$-0.192861\pi$$
0.821995 + 0.569495i $$0.192861\pi$$
$$38$$ 0 0
$$39$$ −6.00000 −0.960769
$$40$$ 0 0
$$41$$ −2.00000 −0.312348 −0.156174 0.987730i $$-0.549916\pi$$
−0.156174 + 0.987730i $$0.549916\pi$$
$$42$$ 0 0
$$43$$ −8.00000 −1.21999 −0.609994 0.792406i $$-0.708828\pi$$
−0.609994 + 0.792406i $$0.708828\pi$$
$$44$$ 0 0
$$45$$ 2.00000 0.298142
$$46$$ 0 0
$$47$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$48$$ 0 0
$$49$$ 1.00000 0.142857
$$50$$ 0 0
$$51$$ 2.00000 0.280056
$$52$$ 0 0
$$53$$ 10.0000 1.37361 0.686803 0.726844i $$-0.259014\pi$$
0.686803 + 0.726844i $$0.259014\pi$$
$$54$$ 0 0
$$55$$ 8.00000 1.07872
$$56$$ 0 0
$$57$$ 4.00000 0.529813
$$58$$ 0 0
$$59$$ 12.0000 1.56227 0.781133 0.624364i $$-0.214642\pi$$
0.781133 + 0.624364i $$0.214642\pi$$
$$60$$ 0 0
$$61$$ −10.0000 −1.28037 −0.640184 0.768221i $$-0.721142\pi$$
−0.640184 + 0.768221i $$0.721142\pi$$
$$62$$ 0 0
$$63$$ −1.00000 −0.125988
$$64$$ 0 0
$$65$$ 12.0000 1.48842
$$66$$ 0 0
$$67$$ 8.00000 0.977356 0.488678 0.872464i $$-0.337479\pi$$
0.488678 + 0.872464i $$0.337479\pi$$
$$68$$ 0 0
$$69$$ 4.00000 0.481543
$$70$$ 0 0
$$71$$ 12.0000 1.42414 0.712069 0.702109i $$-0.247758\pi$$
0.712069 + 0.702109i $$0.247758\pi$$
$$72$$ 0 0
$$73$$ 2.00000 0.234082 0.117041 0.993127i $$-0.462659\pi$$
0.117041 + 0.993127i $$0.462659\pi$$
$$74$$ 0 0
$$75$$ 1.00000 0.115470
$$76$$ 0 0
$$77$$ −4.00000 −0.455842
$$78$$ 0 0
$$79$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$80$$ 0 0
$$81$$ 1.00000 0.111111
$$82$$ 0 0
$$83$$ −12.0000 −1.31717 −0.658586 0.752506i $$-0.728845\pi$$
−0.658586 + 0.752506i $$0.728845\pi$$
$$84$$ 0 0
$$85$$ −4.00000 −0.433861
$$86$$ 0 0
$$87$$ −2.00000 −0.214423
$$88$$ 0 0
$$89$$ 6.00000 0.635999 0.317999 0.948091i $$-0.396989\pi$$
0.317999 + 0.948091i $$0.396989\pi$$
$$90$$ 0 0
$$91$$ −6.00000 −0.628971
$$92$$ 0 0
$$93$$ −8.00000 −0.829561
$$94$$ 0 0
$$95$$ −8.00000 −0.820783
$$96$$ 0 0
$$97$$ 2.00000 0.203069 0.101535 0.994832i $$-0.467625\pi$$
0.101535 + 0.994832i $$0.467625\pi$$
$$98$$ 0 0
$$99$$ 4.00000 0.402015
$$100$$ 0 0
$$101$$ 10.0000 0.995037 0.497519 0.867453i $$-0.334245\pi$$
0.497519 + 0.867453i $$0.334245\pi$$
$$102$$ 0 0
$$103$$ −8.00000 −0.788263 −0.394132 0.919054i $$-0.628955\pi$$
−0.394132 + 0.919054i $$0.628955\pi$$
$$104$$ 0 0
$$105$$ 2.00000 0.195180
$$106$$ 0 0
$$107$$ 12.0000 1.16008 0.580042 0.814587i $$-0.303036\pi$$
0.580042 + 0.814587i $$0.303036\pi$$
$$108$$ 0 0
$$109$$ 2.00000 0.191565 0.0957826 0.995402i $$-0.469465\pi$$
0.0957826 + 0.995402i $$0.469465\pi$$
$$110$$ 0 0
$$111$$ −10.0000 −0.949158
$$112$$ 0 0
$$113$$ −14.0000 −1.31701 −0.658505 0.752577i $$-0.728811\pi$$
−0.658505 + 0.752577i $$0.728811\pi$$
$$114$$ 0 0
$$115$$ −8.00000 −0.746004
$$116$$ 0 0
$$117$$ 6.00000 0.554700
$$118$$ 0 0
$$119$$ 2.00000 0.183340
$$120$$ 0 0
$$121$$ 5.00000 0.454545
$$122$$ 0 0
$$123$$ 2.00000 0.180334
$$124$$ 0 0
$$125$$ −12.0000 −1.07331
$$126$$ 0 0
$$127$$ 8.00000 0.709885 0.354943 0.934888i $$-0.384500\pi$$
0.354943 + 0.934888i $$0.384500\pi$$
$$128$$ 0 0
$$129$$ 8.00000 0.704361
$$130$$ 0 0
$$131$$ 20.0000 1.74741 0.873704 0.486458i $$-0.161711\pi$$
0.873704 + 0.486458i $$0.161711\pi$$
$$132$$ 0 0
$$133$$ 4.00000 0.346844
$$134$$ 0 0
$$135$$ −2.00000 −0.172133
$$136$$ 0 0
$$137$$ 18.0000 1.53784 0.768922 0.639343i $$-0.220793\pi$$
0.768922 + 0.639343i $$0.220793\pi$$
$$138$$ 0 0
$$139$$ −20.0000 −1.69638 −0.848189 0.529694i $$-0.822307\pi$$
−0.848189 + 0.529694i $$0.822307\pi$$
$$140$$ 0 0
$$141$$ 0 0
$$142$$ 0 0
$$143$$ 24.0000 2.00698
$$144$$ 0 0
$$145$$ 4.00000 0.332182
$$146$$ 0 0
$$147$$ −1.00000 −0.0824786
$$148$$ 0 0
$$149$$ −6.00000 −0.491539 −0.245770 0.969328i $$-0.579041\pi$$
−0.245770 + 0.969328i $$0.579041\pi$$
$$150$$ 0 0
$$151$$ −16.0000 −1.30206 −0.651031 0.759051i $$-0.725663\pi$$
−0.651031 + 0.759051i $$0.725663\pi$$
$$152$$ 0 0
$$153$$ −2.00000 −0.161690
$$154$$ 0 0
$$155$$ 16.0000 1.28515
$$156$$ 0 0
$$157$$ −2.00000 −0.159617 −0.0798087 0.996810i $$-0.525431\pi$$
−0.0798087 + 0.996810i $$0.525431\pi$$
$$158$$ 0 0
$$159$$ −10.0000 −0.793052
$$160$$ 0 0
$$161$$ 4.00000 0.315244
$$162$$ 0 0
$$163$$ −16.0000 −1.25322 −0.626608 0.779334i $$-0.715557\pi$$
−0.626608 + 0.779334i $$0.715557\pi$$
$$164$$ 0 0
$$165$$ −8.00000 −0.622799
$$166$$ 0 0
$$167$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$168$$ 0 0
$$169$$ 23.0000 1.76923
$$170$$ 0 0
$$171$$ −4.00000 −0.305888
$$172$$ 0 0
$$173$$ −6.00000 −0.456172 −0.228086 0.973641i $$-0.573247\pi$$
−0.228086 + 0.973641i $$0.573247\pi$$
$$174$$ 0 0
$$175$$ 1.00000 0.0755929
$$176$$ 0 0
$$177$$ −12.0000 −0.901975
$$178$$ 0 0
$$179$$ 20.0000 1.49487 0.747435 0.664335i $$-0.231285\pi$$
0.747435 + 0.664335i $$0.231285\pi$$
$$180$$ 0 0
$$181$$ −2.00000 −0.148659 −0.0743294 0.997234i $$-0.523682\pi$$
−0.0743294 + 0.997234i $$0.523682\pi$$
$$182$$ 0 0
$$183$$ 10.0000 0.739221
$$184$$ 0 0
$$185$$ 20.0000 1.47043
$$186$$ 0 0
$$187$$ −8.00000 −0.585018
$$188$$ 0 0
$$189$$ 1.00000 0.0727393
$$190$$ 0 0
$$191$$ 12.0000 0.868290 0.434145 0.900843i $$-0.357051\pi$$
0.434145 + 0.900843i $$0.357051\pi$$
$$192$$ 0 0
$$193$$ −14.0000 −1.00774 −0.503871 0.863779i $$-0.668091\pi$$
−0.503871 + 0.863779i $$0.668091\pi$$
$$194$$ 0 0
$$195$$ −12.0000 −0.859338
$$196$$ 0 0
$$197$$ −6.00000 −0.427482 −0.213741 0.976890i $$-0.568565\pi$$
−0.213741 + 0.976890i $$0.568565\pi$$
$$198$$ 0 0
$$199$$ 16.0000 1.13421 0.567105 0.823646i $$-0.308063\pi$$
0.567105 + 0.823646i $$0.308063\pi$$
$$200$$ 0 0
$$201$$ −8.00000 −0.564276
$$202$$ 0 0
$$203$$ −2.00000 −0.140372
$$204$$ 0 0
$$205$$ −4.00000 −0.279372
$$206$$ 0 0
$$207$$ −4.00000 −0.278019
$$208$$ 0 0
$$209$$ −16.0000 −1.10674
$$210$$ 0 0
$$211$$ 16.0000 1.10149 0.550743 0.834675i $$-0.314345\pi$$
0.550743 + 0.834675i $$0.314345\pi$$
$$212$$ 0 0
$$213$$ −12.0000 −0.822226
$$214$$ 0 0
$$215$$ −16.0000 −1.09119
$$216$$ 0 0
$$217$$ −8.00000 −0.543075
$$218$$ 0 0
$$219$$ −2.00000 −0.135147
$$220$$ 0 0
$$221$$ −12.0000 −0.807207
$$222$$ 0 0
$$223$$ −8.00000 −0.535720 −0.267860 0.963458i $$-0.586316\pi$$
−0.267860 + 0.963458i $$0.586316\pi$$
$$224$$ 0 0
$$225$$ −1.00000 −0.0666667
$$226$$ 0 0
$$227$$ −20.0000 −1.32745 −0.663723 0.747978i $$-0.731025\pi$$
−0.663723 + 0.747978i $$0.731025\pi$$
$$228$$ 0 0
$$229$$ −18.0000 −1.18947 −0.594737 0.803921i $$-0.702744\pi$$
−0.594737 + 0.803921i $$0.702744\pi$$
$$230$$ 0 0
$$231$$ 4.00000 0.263181
$$232$$ 0 0
$$233$$ −14.0000 −0.917170 −0.458585 0.888650i $$-0.651644\pi$$
−0.458585 + 0.888650i $$0.651644\pi$$
$$234$$ 0 0
$$235$$ 0 0
$$236$$ 0 0
$$237$$ 0 0
$$238$$ 0 0
$$239$$ 12.0000 0.776215 0.388108 0.921614i $$-0.373129\pi$$
0.388108 + 0.921614i $$0.373129\pi$$
$$240$$ 0 0
$$241$$ −30.0000 −1.93247 −0.966235 0.257663i $$-0.917048\pi$$
−0.966235 + 0.257663i $$0.917048\pi$$
$$242$$ 0 0
$$243$$ −1.00000 −0.0641500
$$244$$ 0 0
$$245$$ 2.00000 0.127775
$$246$$ 0 0
$$247$$ −24.0000 −1.52708
$$248$$ 0 0
$$249$$ 12.0000 0.760469
$$250$$ 0 0
$$251$$ 4.00000 0.252478 0.126239 0.992000i $$-0.459709\pi$$
0.126239 + 0.992000i $$0.459709\pi$$
$$252$$ 0 0
$$253$$ −16.0000 −1.00591
$$254$$ 0 0
$$255$$ 4.00000 0.250490
$$256$$ 0 0
$$257$$ −18.0000 −1.12281 −0.561405 0.827541i $$-0.689739\pi$$
−0.561405 + 0.827541i $$0.689739\pi$$
$$258$$ 0 0
$$259$$ −10.0000 −0.621370
$$260$$ 0 0
$$261$$ 2.00000 0.123797
$$262$$ 0 0
$$263$$ −12.0000 −0.739952 −0.369976 0.929041i $$-0.620634\pi$$
−0.369976 + 0.929041i $$0.620634\pi$$
$$264$$ 0 0
$$265$$ 20.0000 1.22859
$$266$$ 0 0
$$267$$ −6.00000 −0.367194
$$268$$ 0 0
$$269$$ 2.00000 0.121942 0.0609711 0.998140i $$-0.480580\pi$$
0.0609711 + 0.998140i $$0.480580\pi$$
$$270$$ 0 0
$$271$$ −32.0000 −1.94386 −0.971931 0.235267i $$-0.924404\pi$$
−0.971931 + 0.235267i $$0.924404\pi$$
$$272$$ 0 0
$$273$$ 6.00000 0.363137
$$274$$ 0 0
$$275$$ −4.00000 −0.241209
$$276$$ 0 0
$$277$$ 18.0000 1.08152 0.540758 0.841178i $$-0.318138\pi$$
0.540758 + 0.841178i $$0.318138\pi$$
$$278$$ 0 0
$$279$$ 8.00000 0.478947
$$280$$ 0 0
$$281$$ −30.0000 −1.78965 −0.894825 0.446417i $$-0.852700\pi$$
−0.894825 + 0.446417i $$0.852700\pi$$
$$282$$ 0 0
$$283$$ 12.0000 0.713326 0.356663 0.934233i $$-0.383914\pi$$
0.356663 + 0.934233i $$0.383914\pi$$
$$284$$ 0 0
$$285$$ 8.00000 0.473879
$$286$$ 0 0
$$287$$ 2.00000 0.118056
$$288$$ 0 0
$$289$$ −13.0000 −0.764706
$$290$$ 0 0
$$291$$ −2.00000 −0.117242
$$292$$ 0 0
$$293$$ −6.00000 −0.350524 −0.175262 0.984522i $$-0.556077\pi$$
−0.175262 + 0.984522i $$0.556077\pi$$
$$294$$ 0 0
$$295$$ 24.0000 1.39733
$$296$$ 0 0
$$297$$ −4.00000 −0.232104
$$298$$ 0 0
$$299$$ −24.0000 −1.38796
$$300$$ 0 0
$$301$$ 8.00000 0.461112
$$302$$ 0 0
$$303$$ −10.0000 −0.574485
$$304$$ 0 0
$$305$$ −20.0000 −1.14520
$$306$$ 0 0
$$307$$ −4.00000 −0.228292 −0.114146 0.993464i $$-0.536413\pi$$
−0.114146 + 0.993464i $$0.536413\pi$$
$$308$$ 0 0
$$309$$ 8.00000 0.455104
$$310$$ 0 0
$$311$$ 8.00000 0.453638 0.226819 0.973937i $$-0.427167\pi$$
0.226819 + 0.973937i $$0.427167\pi$$
$$312$$ 0 0
$$313$$ −14.0000 −0.791327 −0.395663 0.918396i $$-0.629485\pi$$
−0.395663 + 0.918396i $$0.629485\pi$$
$$314$$ 0 0
$$315$$ −2.00000 −0.112687
$$316$$ 0 0
$$317$$ 2.00000 0.112331 0.0561656 0.998421i $$-0.482113\pi$$
0.0561656 + 0.998421i $$0.482113\pi$$
$$318$$ 0 0
$$319$$ 8.00000 0.447914
$$320$$ 0 0
$$321$$ −12.0000 −0.669775
$$322$$ 0 0
$$323$$ 8.00000 0.445132
$$324$$ 0 0
$$325$$ −6.00000 −0.332820
$$326$$ 0 0
$$327$$ −2.00000 −0.110600
$$328$$ 0 0
$$329$$ 0 0
$$330$$ 0 0
$$331$$ 8.00000 0.439720 0.219860 0.975531i $$-0.429440\pi$$
0.219860 + 0.975531i $$0.429440\pi$$
$$332$$ 0 0
$$333$$ 10.0000 0.547997
$$334$$ 0 0
$$335$$ 16.0000 0.874173
$$336$$ 0 0
$$337$$ −14.0000 −0.762629 −0.381314 0.924445i $$-0.624528\pi$$
−0.381314 + 0.924445i $$0.624528\pi$$
$$338$$ 0 0
$$339$$ 14.0000 0.760376
$$340$$ 0 0
$$341$$ 32.0000 1.73290
$$342$$ 0 0
$$343$$ −1.00000 −0.0539949
$$344$$ 0 0
$$345$$ 8.00000 0.430706
$$346$$ 0 0
$$347$$ 28.0000 1.50312 0.751559 0.659665i $$-0.229302\pi$$
0.751559 + 0.659665i $$0.229302\pi$$
$$348$$ 0 0
$$349$$ −18.0000 −0.963518 −0.481759 0.876304i $$-0.660002\pi$$
−0.481759 + 0.876304i $$0.660002\pi$$
$$350$$ 0 0
$$351$$ −6.00000 −0.320256
$$352$$ 0 0
$$353$$ 6.00000 0.319348 0.159674 0.987170i $$-0.448956\pi$$
0.159674 + 0.987170i $$0.448956\pi$$
$$354$$ 0 0
$$355$$ 24.0000 1.27379
$$356$$ 0 0
$$357$$ −2.00000 −0.105851
$$358$$ 0 0
$$359$$ −12.0000 −0.633336 −0.316668 0.948536i $$-0.602564\pi$$
−0.316668 + 0.948536i $$0.602564\pi$$
$$360$$ 0 0
$$361$$ −3.00000 −0.157895
$$362$$ 0 0
$$363$$ −5.00000 −0.262432
$$364$$ 0 0
$$365$$ 4.00000 0.209370
$$366$$ 0 0
$$367$$ −24.0000 −1.25279 −0.626395 0.779506i $$-0.715470\pi$$
−0.626395 + 0.779506i $$0.715470\pi$$
$$368$$ 0 0
$$369$$ −2.00000 −0.104116
$$370$$ 0 0
$$371$$ −10.0000 −0.519174
$$372$$ 0 0
$$373$$ 34.0000 1.76045 0.880227 0.474554i $$-0.157390\pi$$
0.880227 + 0.474554i $$0.157390\pi$$
$$374$$ 0 0
$$375$$ 12.0000 0.619677
$$376$$ 0 0
$$377$$ 12.0000 0.618031
$$378$$ 0 0
$$379$$ 8.00000 0.410932 0.205466 0.978664i $$-0.434129\pi$$
0.205466 + 0.978664i $$0.434129\pi$$
$$380$$ 0 0
$$381$$ −8.00000 −0.409852
$$382$$ 0 0
$$383$$ −8.00000 −0.408781 −0.204390 0.978889i $$-0.565521\pi$$
−0.204390 + 0.978889i $$0.565521\pi$$
$$384$$ 0 0
$$385$$ −8.00000 −0.407718
$$386$$ 0 0
$$387$$ −8.00000 −0.406663
$$388$$ 0 0
$$389$$ −38.0000 −1.92668 −0.963338 0.268290i $$-0.913542\pi$$
−0.963338 + 0.268290i $$0.913542\pi$$
$$390$$ 0 0
$$391$$ 8.00000 0.404577
$$392$$ 0 0
$$393$$ −20.0000 −1.00887
$$394$$ 0 0
$$395$$ 0 0
$$396$$ 0 0
$$397$$ −34.0000 −1.70641 −0.853206 0.521575i $$-0.825345\pi$$
−0.853206 + 0.521575i $$0.825345\pi$$
$$398$$ 0 0
$$399$$ −4.00000 −0.200250
$$400$$ 0 0
$$401$$ 10.0000 0.499376 0.249688 0.968326i $$-0.419672\pi$$
0.249688 + 0.968326i $$0.419672\pi$$
$$402$$ 0 0
$$403$$ 48.0000 2.39105
$$404$$ 0 0
$$405$$ 2.00000 0.0993808
$$406$$ 0 0
$$407$$ 40.0000 1.98273
$$408$$ 0 0
$$409$$ −22.0000 −1.08783 −0.543915 0.839140i $$-0.683059\pi$$
−0.543915 + 0.839140i $$0.683059\pi$$
$$410$$ 0 0
$$411$$ −18.0000 −0.887875
$$412$$ 0 0
$$413$$ −12.0000 −0.590481
$$414$$ 0 0
$$415$$ −24.0000 −1.17811
$$416$$ 0 0
$$417$$ 20.0000 0.979404
$$418$$ 0 0
$$419$$ −12.0000 −0.586238 −0.293119 0.956076i $$-0.594693\pi$$
−0.293119 + 0.956076i $$0.594693\pi$$
$$420$$ 0 0
$$421$$ 34.0000 1.65706 0.828529 0.559946i $$-0.189178\pi$$
0.828529 + 0.559946i $$0.189178\pi$$
$$422$$ 0 0
$$423$$ 0 0
$$424$$ 0 0
$$425$$ 2.00000 0.0970143
$$426$$ 0 0
$$427$$ 10.0000 0.483934
$$428$$ 0 0
$$429$$ −24.0000 −1.15873
$$430$$ 0 0
$$431$$ 36.0000 1.73406 0.867029 0.498257i $$-0.166026\pi$$
0.867029 + 0.498257i $$0.166026\pi$$
$$432$$ 0 0
$$433$$ −6.00000 −0.288342 −0.144171 0.989553i $$-0.546051\pi$$
−0.144171 + 0.989553i $$0.546051\pi$$
$$434$$ 0 0
$$435$$ −4.00000 −0.191785
$$436$$ 0 0
$$437$$ 16.0000 0.765384
$$438$$ 0 0
$$439$$ 8.00000 0.381819 0.190910 0.981608i $$-0.438856\pi$$
0.190910 + 0.981608i $$0.438856\pi$$
$$440$$ 0 0
$$441$$ 1.00000 0.0476190
$$442$$ 0 0
$$443$$ −28.0000 −1.33032 −0.665160 0.746701i $$-0.731637\pi$$
−0.665160 + 0.746701i $$0.731637\pi$$
$$444$$ 0 0
$$445$$ 12.0000 0.568855
$$446$$ 0 0
$$447$$ 6.00000 0.283790
$$448$$ 0 0
$$449$$ 18.0000 0.849473 0.424736 0.905317i $$-0.360367\pi$$
0.424736 + 0.905317i $$0.360367\pi$$
$$450$$ 0 0
$$451$$ −8.00000 −0.376705
$$452$$ 0 0
$$453$$ 16.0000 0.751746
$$454$$ 0 0
$$455$$ −12.0000 −0.562569
$$456$$ 0 0
$$457$$ −6.00000 −0.280668 −0.140334 0.990104i $$-0.544818\pi$$
−0.140334 + 0.990104i $$0.544818\pi$$
$$458$$ 0 0
$$459$$ 2.00000 0.0933520
$$460$$ 0 0
$$461$$ 26.0000 1.21094 0.605470 0.795868i $$-0.292985\pi$$
0.605470 + 0.795868i $$0.292985\pi$$
$$462$$ 0 0
$$463$$ 16.0000 0.743583 0.371792 0.928316i $$-0.378744\pi$$
0.371792 + 0.928316i $$0.378744\pi$$
$$464$$ 0 0
$$465$$ −16.0000 −0.741982
$$466$$ 0 0
$$467$$ −36.0000 −1.66588 −0.832941 0.553362i $$-0.813345\pi$$
−0.832941 + 0.553362i $$0.813345\pi$$
$$468$$ 0 0
$$469$$ −8.00000 −0.369406
$$470$$ 0 0
$$471$$ 2.00000 0.0921551
$$472$$ 0 0
$$473$$ −32.0000 −1.47136
$$474$$ 0 0
$$475$$ 4.00000 0.183533
$$476$$ 0 0
$$477$$ 10.0000 0.457869
$$478$$ 0 0
$$479$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$480$$ 0 0
$$481$$ 60.0000 2.73576
$$482$$ 0 0
$$483$$ −4.00000 −0.182006
$$484$$ 0 0
$$485$$ 4.00000 0.181631
$$486$$ 0 0
$$487$$ −32.0000 −1.45006 −0.725029 0.688718i $$-0.758174\pi$$
−0.725029 + 0.688718i $$0.758174\pi$$
$$488$$ 0 0
$$489$$ 16.0000 0.723545
$$490$$ 0 0
$$491$$ 36.0000 1.62466 0.812329 0.583200i $$-0.198200\pi$$
0.812329 + 0.583200i $$0.198200\pi$$
$$492$$ 0 0
$$493$$ −4.00000 −0.180151
$$494$$ 0 0
$$495$$ 8.00000 0.359573
$$496$$ 0 0
$$497$$ −12.0000 −0.538274
$$498$$ 0 0
$$499$$ 16.0000 0.716258 0.358129 0.933672i $$-0.383415\pi$$
0.358129 + 0.933672i $$0.383415\pi$$
$$500$$ 0 0
$$501$$ 0 0
$$502$$ 0 0
$$503$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$504$$ 0 0
$$505$$ 20.0000 0.889988
$$506$$ 0 0
$$507$$ −23.0000 −1.02147
$$508$$ 0 0
$$509$$ 18.0000 0.797836 0.398918 0.916987i $$-0.369386\pi$$
0.398918 + 0.916987i $$0.369386\pi$$
$$510$$ 0 0
$$511$$ −2.00000 −0.0884748
$$512$$ 0 0
$$513$$ 4.00000 0.176604
$$514$$ 0 0
$$515$$ −16.0000 −0.705044
$$516$$ 0 0
$$517$$ 0 0
$$518$$ 0 0
$$519$$ 6.00000 0.263371
$$520$$ 0 0
$$521$$ 38.0000 1.66481 0.832405 0.554168i $$-0.186963\pi$$
0.832405 + 0.554168i $$0.186963\pi$$
$$522$$ 0 0
$$523$$ −28.0000 −1.22435 −0.612177 0.790721i $$-0.709706\pi$$
−0.612177 + 0.790721i $$0.709706\pi$$
$$524$$ 0 0
$$525$$ −1.00000 −0.0436436
$$526$$ 0 0
$$527$$ −16.0000 −0.696971
$$528$$ 0 0
$$529$$ −7.00000 −0.304348
$$530$$ 0 0
$$531$$ 12.0000 0.520756
$$532$$ 0 0
$$533$$ −12.0000 −0.519778
$$534$$ 0 0
$$535$$ 24.0000 1.03761
$$536$$ 0 0
$$537$$ −20.0000 −0.863064
$$538$$ 0 0
$$539$$ 4.00000 0.172292
$$540$$ 0 0
$$541$$ 26.0000 1.11783 0.558914 0.829226i $$-0.311218\pi$$
0.558914 + 0.829226i $$0.311218\pi$$
$$542$$ 0 0
$$543$$ 2.00000 0.0858282
$$544$$ 0 0
$$545$$ 4.00000 0.171341
$$546$$ 0 0
$$547$$ 32.0000 1.36822 0.684111 0.729378i $$-0.260191\pi$$
0.684111 + 0.729378i $$0.260191\pi$$
$$548$$ 0 0
$$549$$ −10.0000 −0.426790
$$550$$ 0 0
$$551$$ −8.00000 −0.340811
$$552$$ 0 0
$$553$$ 0 0
$$554$$ 0 0
$$555$$ −20.0000 −0.848953
$$556$$ 0 0
$$557$$ −30.0000 −1.27114 −0.635570 0.772043i $$-0.719235\pi$$
−0.635570 + 0.772043i $$0.719235\pi$$
$$558$$ 0 0
$$559$$ −48.0000 −2.03018
$$560$$ 0 0
$$561$$ 8.00000 0.337760
$$562$$ 0 0
$$563$$ 20.0000 0.842900 0.421450 0.906852i $$-0.361521\pi$$
0.421450 + 0.906852i $$0.361521\pi$$
$$564$$ 0 0
$$565$$ −28.0000 −1.17797
$$566$$ 0 0
$$567$$ −1.00000 −0.0419961
$$568$$ 0 0
$$569$$ 18.0000 0.754599 0.377300 0.926091i $$-0.376853\pi$$
0.377300 + 0.926091i $$0.376853\pi$$
$$570$$ 0 0
$$571$$ −16.0000 −0.669579 −0.334790 0.942293i $$-0.608665\pi$$
−0.334790 + 0.942293i $$0.608665\pi$$
$$572$$ 0 0
$$573$$ −12.0000 −0.501307
$$574$$ 0 0
$$575$$ 4.00000 0.166812
$$576$$ 0 0
$$577$$ 2.00000 0.0832611 0.0416305 0.999133i $$-0.486745\pi$$
0.0416305 + 0.999133i $$0.486745\pi$$
$$578$$ 0 0
$$579$$ 14.0000 0.581820
$$580$$ 0 0
$$581$$ 12.0000 0.497844
$$582$$ 0 0
$$583$$ 40.0000 1.65663
$$584$$ 0 0
$$585$$ 12.0000 0.496139
$$586$$ 0 0
$$587$$ −12.0000 −0.495293 −0.247647 0.968850i $$-0.579657\pi$$
−0.247647 + 0.968850i $$0.579657\pi$$
$$588$$ 0 0
$$589$$ −32.0000 −1.31854
$$590$$ 0 0
$$591$$ 6.00000 0.246807
$$592$$ 0 0
$$593$$ −42.0000 −1.72473 −0.862367 0.506284i $$-0.831019\pi$$
−0.862367 + 0.506284i $$0.831019\pi$$
$$594$$ 0 0
$$595$$ 4.00000 0.163984
$$596$$ 0 0
$$597$$ −16.0000 −0.654836
$$598$$ 0 0
$$599$$ −36.0000 −1.47092 −0.735460 0.677568i $$-0.763034\pi$$
−0.735460 + 0.677568i $$0.763034\pi$$
$$600$$ 0 0
$$601$$ 10.0000 0.407909 0.203954 0.978980i $$-0.434621\pi$$
0.203954 + 0.978980i $$0.434621\pi$$
$$602$$ 0 0
$$603$$ 8.00000 0.325785
$$604$$ 0 0
$$605$$ 10.0000 0.406558
$$606$$ 0 0
$$607$$ 8.00000 0.324710 0.162355 0.986732i $$-0.448091\pi$$
0.162355 + 0.986732i $$0.448091\pi$$
$$608$$ 0 0
$$609$$ 2.00000 0.0810441
$$610$$ 0 0
$$611$$ 0 0
$$612$$ 0 0
$$613$$ 10.0000 0.403896 0.201948 0.979396i $$-0.435273\pi$$
0.201948 + 0.979396i $$0.435273\pi$$
$$614$$ 0 0
$$615$$ 4.00000 0.161296
$$616$$ 0 0
$$617$$ 18.0000 0.724653 0.362326 0.932051i $$-0.381983\pi$$
0.362326 + 0.932051i $$0.381983\pi$$
$$618$$ 0 0
$$619$$ 20.0000 0.803868 0.401934 0.915669i $$-0.368338\pi$$
0.401934 + 0.915669i $$0.368338\pi$$
$$620$$ 0 0
$$621$$ 4.00000 0.160514
$$622$$ 0 0
$$623$$ −6.00000 −0.240385
$$624$$ 0 0
$$625$$ −19.0000 −0.760000
$$626$$ 0 0
$$627$$ 16.0000 0.638978
$$628$$ 0 0
$$629$$ −20.0000 −0.797452
$$630$$ 0 0
$$631$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$632$$ 0 0
$$633$$ −16.0000 −0.635943
$$634$$ 0 0
$$635$$ 16.0000 0.634941
$$636$$ 0 0
$$637$$ 6.00000 0.237729
$$638$$ 0 0
$$639$$ 12.0000 0.474713
$$640$$ 0 0
$$641$$ 26.0000 1.02694 0.513469 0.858108i $$-0.328360\pi$$
0.513469 + 0.858108i $$0.328360\pi$$
$$642$$ 0 0
$$643$$ −28.0000 −1.10421 −0.552106 0.833774i $$-0.686176\pi$$
−0.552106 + 0.833774i $$0.686176\pi$$
$$644$$ 0 0
$$645$$ 16.0000 0.629999
$$646$$ 0 0
$$647$$ −16.0000 −0.629025 −0.314512 0.949253i $$-0.601841\pi$$
−0.314512 + 0.949253i $$0.601841\pi$$
$$648$$ 0 0
$$649$$ 48.0000 1.88416
$$650$$ 0 0
$$651$$ 8.00000 0.313545
$$652$$ 0 0
$$653$$ −14.0000 −0.547862 −0.273931 0.961749i $$-0.588324\pi$$
−0.273931 + 0.961749i $$0.588324\pi$$
$$654$$ 0 0
$$655$$ 40.0000 1.56293
$$656$$ 0 0
$$657$$ 2.00000 0.0780274
$$658$$ 0 0
$$659$$ −44.0000 −1.71400 −0.856998 0.515319i $$-0.827673\pi$$
−0.856998 + 0.515319i $$0.827673\pi$$
$$660$$ 0 0
$$661$$ −18.0000 −0.700119 −0.350059 0.936727i $$-0.613839\pi$$
−0.350059 + 0.936727i $$0.613839\pi$$
$$662$$ 0 0
$$663$$ 12.0000 0.466041
$$664$$ 0 0
$$665$$ 8.00000 0.310227
$$666$$ 0 0
$$667$$ −8.00000 −0.309761
$$668$$ 0 0
$$669$$ 8.00000 0.309298
$$670$$ 0 0
$$671$$ −40.0000 −1.54418
$$672$$ 0 0
$$673$$ 50.0000 1.92736 0.963679 0.267063i $$-0.0860531\pi$$
0.963679 + 0.267063i $$0.0860531\pi$$
$$674$$ 0 0
$$675$$ 1.00000 0.0384900
$$676$$ 0 0
$$677$$ 18.0000 0.691796 0.345898 0.938272i $$-0.387574\pi$$
0.345898 + 0.938272i $$0.387574\pi$$
$$678$$ 0 0
$$679$$ −2.00000 −0.0767530
$$680$$ 0 0
$$681$$ 20.0000 0.766402
$$682$$ 0 0
$$683$$ −28.0000 −1.07139 −0.535695 0.844411i $$-0.679950\pi$$
−0.535695 + 0.844411i $$0.679950\pi$$
$$684$$ 0 0
$$685$$ 36.0000 1.37549
$$686$$ 0 0
$$687$$ 18.0000 0.686743
$$688$$ 0 0
$$689$$ 60.0000 2.28582
$$690$$ 0 0
$$691$$ −12.0000 −0.456502 −0.228251 0.973602i $$-0.573301\pi$$
−0.228251 + 0.973602i $$0.573301\pi$$
$$692$$ 0 0
$$693$$ −4.00000 −0.151947
$$694$$ 0 0
$$695$$ −40.0000 −1.51729
$$696$$ 0 0
$$697$$ 4.00000 0.151511
$$698$$ 0 0
$$699$$ 14.0000 0.529529
$$700$$ 0 0
$$701$$ −30.0000 −1.13308 −0.566542 0.824033i $$-0.691719\pi$$
−0.566542 + 0.824033i $$0.691719\pi$$
$$702$$ 0 0
$$703$$ −40.0000 −1.50863
$$704$$ 0 0
$$705$$ 0 0
$$706$$ 0 0
$$707$$ −10.0000 −0.376089
$$708$$ 0 0
$$709$$ −38.0000 −1.42712 −0.713560 0.700594i $$-0.752918\pi$$
−0.713560 + 0.700594i $$0.752918\pi$$
$$710$$ 0 0
$$711$$ 0 0
$$712$$ 0 0
$$713$$ −32.0000 −1.19841
$$714$$ 0 0
$$715$$ 48.0000 1.79510
$$716$$ 0 0
$$717$$ −12.0000 −0.448148
$$718$$ 0 0
$$719$$ 8.00000 0.298350 0.149175 0.988811i $$-0.452338\pi$$
0.149175 + 0.988811i $$0.452338\pi$$
$$720$$ 0 0
$$721$$ 8.00000 0.297936
$$722$$ 0 0
$$723$$ 30.0000 1.11571
$$724$$ 0 0
$$725$$ −2.00000 −0.0742781
$$726$$ 0 0
$$727$$ −40.0000 −1.48352 −0.741759 0.670667i $$-0.766008\pi$$
−0.741759 + 0.670667i $$0.766008\pi$$
$$728$$ 0 0
$$729$$ 1.00000 0.0370370
$$730$$ 0 0
$$731$$ 16.0000 0.591781
$$732$$ 0 0
$$733$$ 14.0000 0.517102 0.258551 0.965998i $$-0.416755\pi$$
0.258551 + 0.965998i $$0.416755\pi$$
$$734$$ 0 0
$$735$$ −2.00000 −0.0737711
$$736$$ 0 0
$$737$$ 32.0000 1.17874
$$738$$ 0 0
$$739$$ −32.0000 −1.17714 −0.588570 0.808447i $$-0.700309\pi$$
−0.588570 + 0.808447i $$0.700309\pi$$
$$740$$ 0 0
$$741$$ 24.0000 0.881662
$$742$$ 0 0
$$743$$ −28.0000 −1.02722 −0.513610 0.858024i $$-0.671692\pi$$
−0.513610 + 0.858024i $$0.671692\pi$$
$$744$$ 0 0
$$745$$ −12.0000 −0.439646
$$746$$ 0 0
$$747$$ −12.0000 −0.439057
$$748$$ 0 0
$$749$$ −12.0000 −0.438470
$$750$$ 0 0
$$751$$ 16.0000 0.583848 0.291924 0.956441i $$-0.405705\pi$$
0.291924 + 0.956441i $$0.405705\pi$$
$$752$$ 0 0
$$753$$ −4.00000 −0.145768
$$754$$ 0 0
$$755$$ −32.0000 −1.16460
$$756$$ 0 0
$$757$$ 26.0000 0.944986 0.472493 0.881334i $$-0.343354\pi$$
0.472493 + 0.881334i $$0.343354\pi$$
$$758$$ 0 0
$$759$$ 16.0000 0.580763
$$760$$ 0 0
$$761$$ 30.0000 1.08750 0.543750 0.839248i $$-0.317004\pi$$
0.543750 + 0.839248i $$0.317004\pi$$
$$762$$ 0 0
$$763$$ −2.00000 −0.0724049
$$764$$ 0 0
$$765$$ −4.00000 −0.144620
$$766$$ 0 0
$$767$$ 72.0000 2.59977
$$768$$ 0 0
$$769$$ 26.0000 0.937584 0.468792 0.883309i $$-0.344689\pi$$
0.468792 + 0.883309i $$0.344689\pi$$
$$770$$ 0 0
$$771$$ 18.0000 0.648254
$$772$$ 0 0
$$773$$ 42.0000 1.51064 0.755318 0.655359i $$-0.227483\pi$$
0.755318 + 0.655359i $$0.227483\pi$$
$$774$$ 0 0
$$775$$ −8.00000 −0.287368
$$776$$ 0 0
$$777$$ 10.0000 0.358748
$$778$$ 0 0
$$779$$ 8.00000 0.286630
$$780$$ 0 0
$$781$$ 48.0000 1.71758
$$782$$ 0 0
$$783$$ −2.00000 −0.0714742
$$784$$ 0 0
$$785$$ −4.00000 −0.142766
$$786$$ 0 0
$$787$$ 4.00000 0.142585 0.0712923 0.997455i $$-0.477288\pi$$
0.0712923 + 0.997455i $$0.477288\pi$$
$$788$$ 0 0
$$789$$ 12.0000 0.427211
$$790$$ 0 0
$$791$$ 14.0000 0.497783
$$792$$ 0 0
$$793$$ −60.0000 −2.13066
$$794$$ 0 0
$$795$$ −20.0000 −0.709327
$$796$$ 0 0
$$797$$ 42.0000 1.48772 0.743858 0.668338i $$-0.232994\pi$$
0.743858 + 0.668338i $$0.232994\pi$$
$$798$$ 0 0
$$799$$ 0 0
$$800$$ 0 0
$$801$$ 6.00000 0.212000
$$802$$ 0 0
$$803$$ 8.00000 0.282314
$$804$$ 0 0
$$805$$ 8.00000 0.281963
$$806$$ 0 0
$$807$$ −2.00000 −0.0704033
$$808$$ 0 0
$$809$$ 10.0000 0.351581 0.175791 0.984428i $$-0.443752\pi$$
0.175791 + 0.984428i $$0.443752\pi$$
$$810$$ 0 0
$$811$$ 4.00000 0.140459 0.0702295 0.997531i $$-0.477627\pi$$
0.0702295 + 0.997531i $$0.477627\pi$$
$$812$$ 0 0
$$813$$ 32.0000 1.12229
$$814$$ 0 0
$$815$$ −32.0000 −1.12091
$$816$$ 0 0
$$817$$ 32.0000 1.11954
$$818$$ 0 0
$$819$$ −6.00000 −0.209657
$$820$$ 0 0
$$821$$ 10.0000 0.349002 0.174501 0.984657i $$-0.444169\pi$$
0.174501 + 0.984657i $$0.444169\pi$$
$$822$$ 0 0
$$823$$ 32.0000 1.11545 0.557725 0.830026i $$-0.311674\pi$$
0.557725 + 0.830026i $$0.311674\pi$$
$$824$$ 0 0
$$825$$ 4.00000 0.139262
$$826$$ 0 0
$$827$$ −4.00000 −0.139094 −0.0695468 0.997579i $$-0.522155\pi$$
−0.0695468 + 0.997579i $$0.522155\pi$$
$$828$$ 0 0
$$829$$ 14.0000 0.486240 0.243120 0.969996i $$-0.421829\pi$$
0.243120 + 0.969996i $$0.421829\pi$$
$$830$$ 0 0
$$831$$ −18.0000 −0.624413
$$832$$ 0 0
$$833$$ −2.00000 −0.0692959
$$834$$ 0 0
$$835$$ 0 0
$$836$$ 0 0
$$837$$ −8.00000 −0.276520
$$838$$ 0 0
$$839$$ −16.0000 −0.552381 −0.276191 0.961103i $$-0.589072\pi$$
−0.276191 + 0.961103i $$0.589072\pi$$
$$840$$ 0 0
$$841$$ −25.0000 −0.862069
$$842$$ 0 0
$$843$$ 30.0000 1.03325
$$844$$ 0 0
$$845$$ 46.0000 1.58245
$$846$$ 0 0
$$847$$ −5.00000 −0.171802
$$848$$ 0 0
$$849$$ −12.0000 −0.411839
$$850$$ 0 0
$$851$$ −40.0000 −1.37118
$$852$$ 0 0
$$853$$ −42.0000 −1.43805 −0.719026 0.694983i $$-0.755412\pi$$
−0.719026 + 0.694983i $$0.755412\pi$$
$$854$$ 0 0
$$855$$ −8.00000 −0.273594
$$856$$ 0 0
$$857$$ −18.0000 −0.614868 −0.307434 0.951569i $$-0.599470\pi$$
−0.307434 + 0.951569i $$0.599470\pi$$
$$858$$ 0 0
$$859$$ −12.0000 −0.409435 −0.204717 0.978821i $$-0.565628\pi$$
−0.204717 + 0.978821i $$0.565628\pi$$
$$860$$ 0 0
$$861$$ −2.00000 −0.0681598
$$862$$ 0 0
$$863$$ 52.0000 1.77010 0.885050 0.465495i $$-0.154124\pi$$
0.885050 + 0.465495i $$0.154124\pi$$
$$864$$ 0 0
$$865$$ −12.0000 −0.408012
$$866$$ 0 0
$$867$$ 13.0000 0.441503
$$868$$ 0 0
$$869$$ 0 0
$$870$$ 0 0
$$871$$ 48.0000 1.62642
$$872$$ 0 0
$$873$$ 2.00000 0.0676897
$$874$$ 0 0
$$875$$ 12.0000 0.405674
$$876$$ 0 0
$$877$$ −14.0000 −0.472746 −0.236373 0.971662i $$-0.575959\pi$$
−0.236373 + 0.971662i $$0.575959\pi$$
$$878$$ 0 0
$$879$$ 6.00000 0.202375
$$880$$ 0 0
$$881$$ −42.0000 −1.41502 −0.707508 0.706705i $$-0.750181\pi$$
−0.707508 + 0.706705i $$0.750181\pi$$
$$882$$ 0 0
$$883$$ −16.0000 −0.538443 −0.269221 0.963078i $$-0.586766\pi$$
−0.269221 + 0.963078i $$0.586766\pi$$
$$884$$ 0 0
$$885$$ −24.0000 −0.806751
$$886$$ 0 0
$$887$$ −16.0000 −0.537227 −0.268614 0.963248i $$-0.586566\pi$$
−0.268614 + 0.963248i $$0.586566\pi$$
$$888$$ 0 0
$$889$$ −8.00000 −0.268311
$$890$$ 0 0
$$891$$ 4.00000 0.134005
$$892$$ 0 0
$$893$$ 0 0
$$894$$ 0 0
$$895$$ 40.0000 1.33705
$$896$$ 0 0
$$897$$ 24.0000 0.801337
$$898$$ 0 0
$$899$$ 16.0000 0.533630
$$900$$ 0 0
$$901$$ −20.0000 −0.666297
$$902$$ 0 0
$$903$$ −8.00000 −0.266223
$$904$$ 0 0
$$905$$ −4.00000 −0.132964
$$906$$ 0 0
$$907$$ −48.0000 −1.59381 −0.796907 0.604102i $$-0.793532\pi$$
−0.796907 + 0.604102i $$0.793532\pi$$
$$908$$ 0 0
$$909$$ 10.0000 0.331679
$$910$$ 0 0
$$911$$ −12.0000 −0.397578 −0.198789 0.980042i $$-0.563701\pi$$
−0.198789 + 0.980042i $$0.563701\pi$$
$$912$$ 0 0
$$913$$ −48.0000 −1.58857
$$914$$ 0 0
$$915$$ 20.0000 0.661180
$$916$$ 0 0
$$917$$ −20.0000 −0.660458
$$918$$ 0 0
$$919$$ 8.00000 0.263896 0.131948 0.991257i $$-0.457877\pi$$
0.131948 + 0.991257i $$0.457877\pi$$
$$920$$ 0 0
$$921$$ 4.00000 0.131804
$$922$$ 0 0
$$923$$ 72.0000 2.36991
$$924$$ 0 0
$$925$$ −10.0000 −0.328798
$$926$$ 0 0
$$927$$ −8.00000 −0.262754
$$928$$ 0 0
$$929$$ 22.0000 0.721797 0.360898 0.932605i $$-0.382470\pi$$
0.360898 + 0.932605i $$0.382470\pi$$
$$930$$ 0 0
$$931$$ −4.00000 −0.131095
$$932$$ 0 0
$$933$$ −8.00000 −0.261908
$$934$$ 0 0
$$935$$ −16.0000 −0.523256
$$936$$ 0 0
$$937$$ 10.0000 0.326686 0.163343 0.986569i $$-0.447772\pi$$
0.163343 + 0.986569i $$0.447772\pi$$
$$938$$ 0 0
$$939$$ 14.0000 0.456873
$$940$$ 0 0
$$941$$ −6.00000 −0.195594 −0.0977972 0.995206i $$-0.531180\pi$$
−0.0977972 + 0.995206i $$0.531180\pi$$
$$942$$ 0 0
$$943$$ 8.00000 0.260516
$$944$$ 0 0
$$945$$ 2.00000 0.0650600
$$946$$ 0 0
$$947$$ −44.0000 −1.42981 −0.714904 0.699223i $$-0.753530\pi$$
−0.714904 + 0.699223i $$0.753530\pi$$
$$948$$ 0 0
$$949$$ 12.0000 0.389536
$$950$$ 0 0
$$951$$ −2.00000 −0.0648544
$$952$$ 0 0
$$953$$ −54.0000 −1.74923 −0.874616 0.484817i $$-0.838886\pi$$
−0.874616 + 0.484817i $$0.838886\pi$$
$$954$$ 0 0
$$955$$ 24.0000 0.776622
$$956$$ 0 0
$$957$$ −8.00000 −0.258603
$$958$$ 0 0
$$959$$ −18.0000 −0.581250
$$960$$ 0 0
$$961$$ 33.0000 1.06452
$$962$$ 0 0
$$963$$ 12.0000 0.386695
$$964$$ 0 0
$$965$$ −28.0000 −0.901352
$$966$$ 0 0
$$967$$ 8.00000 0.257263 0.128631 0.991692i $$-0.458942\pi$$
0.128631 + 0.991692i $$0.458942\pi$$
$$968$$ 0 0
$$969$$ −8.00000 −0.256997
$$970$$ 0 0
$$971$$ −28.0000 −0.898563 −0.449281 0.893390i $$-0.648320\pi$$
−0.449281 + 0.893390i $$0.648320\pi$$
$$972$$ 0 0
$$973$$ 20.0000 0.641171
$$974$$ 0 0
$$975$$ 6.00000 0.192154
$$976$$ 0 0
$$977$$ −22.0000 −0.703842 −0.351921 0.936030i $$-0.614471\pi$$
−0.351921 + 0.936030i $$0.614471\pi$$
$$978$$ 0 0
$$979$$ 24.0000 0.767043
$$980$$ 0 0
$$981$$ 2.00000 0.0638551
$$982$$ 0 0
$$983$$ 24.0000 0.765481 0.382741 0.923856i $$-0.374980\pi$$
0.382741 + 0.923856i $$0.374980\pi$$
$$984$$ 0 0
$$985$$ −12.0000 −0.382352
$$986$$ 0 0
$$987$$ 0 0
$$988$$ 0 0
$$989$$ 32.0000 1.01754
$$990$$ 0 0
$$991$$ 8.00000 0.254128 0.127064 0.991894i $$-0.459445\pi$$
0.127064 + 0.991894i $$0.459445\pi$$
$$992$$ 0 0
$$993$$ −8.00000 −0.253872
$$994$$ 0 0
$$995$$ 32.0000 1.01447
$$996$$ 0 0
$$997$$ 62.0000 1.96356 0.981780 0.190022i $$-0.0608559\pi$$
0.981780 + 0.190022i $$0.0608559\pi$$
$$998$$ 0 0
$$999$$ −10.0000 −0.316386
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 1344.2.a.h.1.1 1
3.2 odd 2 4032.2.a.f.1.1 1
4.3 odd 2 1344.2.a.r.1.1 1
7.6 odd 2 9408.2.a.cb.1.1 1
8.3 odd 2 672.2.a.b.1.1 1
8.5 even 2 672.2.a.f.1.1 yes 1
12.11 even 2 4032.2.a.n.1.1 1
16.3 odd 4 5376.2.c.m.2689.2 2
16.5 even 4 5376.2.c.s.2689.2 2
16.11 odd 4 5376.2.c.m.2689.1 2
16.13 even 4 5376.2.c.s.2689.1 2
24.5 odd 2 2016.2.a.k.1.1 1
24.11 even 2 2016.2.a.l.1.1 1
28.27 even 2 9408.2.a.g.1.1 1
56.13 odd 2 4704.2.a.m.1.1 1
56.27 even 2 4704.2.a.be.1.1 1
By twisted newform
Twist Min Dim Char Parity Ord Type
672.2.a.b.1.1 1 8.3 odd 2
672.2.a.f.1.1 yes 1 8.5 even 2
1344.2.a.h.1.1 1 1.1 even 1 trivial
1344.2.a.r.1.1 1 4.3 odd 2
2016.2.a.k.1.1 1 24.5 odd 2
2016.2.a.l.1.1 1 24.11 even 2
4032.2.a.f.1.1 1 3.2 odd 2
4032.2.a.n.1.1 1 12.11 even 2
4704.2.a.m.1.1 1 56.13 odd 2
4704.2.a.be.1.1 1 56.27 even 2
5376.2.c.m.2689.1 2 16.11 odd 4
5376.2.c.m.2689.2 2 16.3 odd 4
5376.2.c.s.2689.1 2 16.13 even 4
5376.2.c.s.2689.2 2 16.5 even 4
9408.2.a.g.1.1 1 28.27 even 2
9408.2.a.cb.1.1 1 7.6 odd 2 | 19,395 | 34,460 | {"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} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.404218 |
http://psychology.wikia.com/wiki/Light_refraction | 1,501,268,986,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550977093.96/warc/CC-MAIN-20170728183650-20170728203650-00222.warc.gz | 263,641,505 | 80,471 | ## FANDOM
34,203 Pages
Refraction is the change in direction of a wave due to a change in its speed. This is most commonly seen when a wave passes from one medium to another. Refraction of light is the most commonly seen example, but any type of wave can refract when it interacts with a medium, for example when sound waves pass from one medium into another or when water waves move into water of a different depth.
## ExplanationEdit
In optics, refraction occurs when light waves travel from a medium with a given refractive index to a medium with another. At the boundary between the media, the wave's phase velocity is altered, it changes direction, and its wavelength increases or decreases but its frequency remains constant. For example, a light ray will refract as it enters and leaves glass; understanding of this concept led to the invention of lenses and the refracting telescope.
Refraction can be seen when looking into a bowl of water. Air has a refractive index of about 1.0003, and water has a refractive index of about 1.33. If a person looks at a straight object, such as a pencil or straw, which is placed at a slant, partially in the water, the object appears to bend at the water's surface. This is due to the bending of light rays as they move from the water to the air. Once the rays reach the eye, the eye traces them back as straight lines (lines of sight). The lines of sight (shown as dashed lines) intersect at a higher position than where the actual rays originated. This causes the pencil to appear higher and the water to appear shallower than it really is. The depth that the water appears to be when viewed from above is known as the apparent depth. This is an important consideration for spearfishing from the surface because it will make the target fish appear to be in a different place, and the fisher must aim lower to catch the fish.
The diagram on the right shows an example of refraction in water waves. Ripples travel from the left and pass over a shallower region inclined at an angle to the wavefront. The waves travel more slowly in the shallower water, so the wavelength decreases and the wave bends at the boundary. The dotted line represents the normal to the boundary. The dashed line represents the original direction of the waves. The phenomenon explains why waves on a shoreline never hit the shoreline at an angle. Whichever direction the waves travel in deep water, they always refract towards the normal as they enter the shallower water near the beach.
Refraction is also responsible for rainbows and for the splitting of white light into a rainbow-spectrum as it passes through a glass prism. Glass has a higher refractive index than air and the different frequencies of light travel at different speeds (dispersion), causing them to be refracted at different angles, so that you can see them. The different frequencies correspond to different colors observed.
While refraction allows for beautiful phenomena such as rainbows it may also produce peculiar optical phenomena, such as mirages and Fata Morgana. These are caused by the change of the refractive index of air with temperature.
Snell's law is used to calculate the degree to which light is refracted when traveling from one medium to another.
Recently some metamaterials have been created which have a negative refractive index. With metamaterials, we can also obtain the total refraction phenomena when the wave impedances of the two media are matched. There is no reflected wave.
Also, since refraction can make objects appear closer than they are, it is responsible for allowing water to magnify objects. First, as light is entering a drop of water, it slows down. If the water's surface is not flat, then the light will be bent into a new path. This round shape will bend the light outwards and as it spreads out, the image you see gets larger.
A useful analogy in explaining the refraction of light would be to imagine a marching band as they march from pavement (a fast medium) into mud (a slower medium) The marchers on the side that runs into the mud first will slow down first. This causes the whole band to pivot slightly toward the normal (make a smaller angle from the normal).
## OphthalmologyEdit
In medicine, particularly ophthalmology and optometry, refraction (also known as refractometry) is a clinical test in which a phoropter is used to determine the eye's refractive error and the best corrective lenses to be prescribed. A series of test lenses in graded optical powers or focal lengths are presented to determine which provide the sharpest, clearest vision.[1]
## AcousticsEdit
In underwater acoustics, refraction is the bending or curving of a sound ray that results when the ray passes through a sound speed gradient from a region of one sound speed to a region of a different speed. The amount of ray bending is dependent upon the amount of difference between sound speeds, that is, the variation in temperature, salinity, and pressure of the water.[2] Similar acoustics effects are also found in the Earth's atmosphere. | 1,059 | 5,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-30 | latest | en | 0.948071 |
https://support.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.ode.bvp_ps_lin_solve.html | 1,723,035,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694449.36/warc/CC-MAIN-20240807111957-20240807141957-00438.warc.gz | 442,890,309 | 11,130 | # naginterfaces.library.ode.bvp_ps_lin_solve¶
naginterfaces.library.ode.bvp_ps_lin_solve(a, b, c, bmat, y, bvec, f)[source]
bvp_ps_lin_solve finds the solution of a linear constant coefficient boundary value problem by using the Chebyshev integration formulation on a Chebyshev Gauss–Lobatto grid.
For full information please refer to the NAG Library document for d02ue
https://support.nag.com/numeric/nl/nagdoc_30.1/flhtml/d02/d02uef.html
Parameters
afloat
, the lower bound of domain .
bfloat
, the upper bound of domain .
cfloat, array-like, shape
The Chebyshev coefficients , , for the right-hand side of the boundary value problem. Usually these are obtained by a previous call of bvp_ps_lin_coeffs().
bmatfloat, array-like, shape
must contain the coefficients , for , for , in the problem formulation of Notes.
yfloat, array-like, shape
The points, , for , where the boundary conditions are discretized.
bvecfloat, array-like, shape
The values, , for , in the formulation of the boundary conditions given in Notes.
ffloat, array-like, shape
The coefficients, , for , in the formulation of the linear boundary value problem given in Notes. The highest order term, , needs to be nonzero to have a well posed problem.
Returns
bmatfloat, ndarray, shape
The coefficients have been scaled to form an equivalent problem defined on the domain .
ffloat, ndarray, shape
The coefficients have been scaled to form an equivalent problem defined on the domain .
ucfloat, ndarray, shape
The Chebyshev coefficients in the Chebyshev series representations of the solution and derivatives of the solution to the boundary value problem. The elements contain the coefficients representing the solution , for . contains the coefficients representing the th derivative of , for .
residfloat
The maximum residual resulting from substituting the solution vectors returned in into the equations representing the linear boundary value problem and associated boundary conditions.
Raises
NagValueError
(errno )
On entry, .
Constraint: .
(errno )
On entry, .
Constraint: is even.
(errno )
On entry, and .
Constraint: .
(errno )
On entry, .
(errno )
On entry, .
Constraint: .
(errno )
Internal error while unpacking matrix during iterative refinement.
(errno )
Singular matrix encountered during iterative refinement.
Warns
NagAlgorithmicWarning
(errno )
During iterative refinement, the maximum number of iterations was reached.
and .
(errno )
During iterative refinement, convergence was achieved, but the residual is more than . .
Notes
bvp_ps_lin_solve solves the constant linear coefficient ordinary differential problem
subject to a set of linear constraints at points , for :
where , is an matrix of constant coefficients and are constants. The points are usually either or .
The function is supplied as an array of Chebyshev coefficients , for the function discretized on Chebyshev Gauss–Lobatto points (as returned by bvp_ps_lin_cgl_grid()); the coefficients are normally obtained by a previous call to bvp_ps_lin_coeffs(). The solution and its derivatives (up to order ) are returned, in the form of their Chebyshev series representation, as arrays of Chebyshev coefficients; subsequent calls to bvp_ps_lin_cgl_vals() will return the corresponding function and derivative values at the Chebyshev Gauss–Lobatto discretization points on . Function and derivative values can be obtained on any uniform grid over the same range by calling the interpolation function bvp_ps_lin_grid_vals().
References
Clenshaw, C W, 1957, The numerical solution of linear differential equations in Chebyshev series, Proc. Camb. Phil. Soc. (53), 134–149
Coutsias, E A, Hagstrom, T and Torres, D, 1996, An efficient spectral method for ordinary differential equations with rational function coefficients, Mathematics of Computation (65(214)), 611–635
Greengard, L, 1991, Spectral integration and two-point boundary value problems, SIAM J. Numer. Anal. (28(4)), 1071–80
Lundbladh, A, Hennigson, D S and Johannson, A V, 1992, An efficient spectral integration method for the solution of the Navier–Stokes equations, Technical report FFA–TN, 1992–28, Aeronautical Research Institute of Sweden
Muite, B K, 2010, A numerical comparison of Chebyshev methods for solving fourth-order semilinear initial boundary value problems, Journal of Computational and Applied Mathematics (234(2)), 317–342 | 1,003 | 4,413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-33 | latest | en | 0.827049 |
http://oeis.org/A240752/internal | 1,552,968,138,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201885.28/warc/CC-MAIN-20190319032352-20190319054352-00477.warc.gz | 154,174,989 | 2,681 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A240752 First differences of digit sums of squares, cf. A004159. 5
%I
%S 1,3,5,-2,0,2,4,-3,-1,-8,3,5,7,0,-7,4,6,-10,1,-6,5,7,0,2,-5,6,-1,1,-6,
%T -4,7,-9,11,-5,-3,8,1,-6,-4,-2,9,2,4,-3,-10,1,3,-4,-2,0,2,4,6,-1,-8,3,
%U 5,-2,0,-7,4,6,8,-8,-6,5,7,-9,2,-5,-3,8,1,3,-4
%N First differences of digit sums of squares, cf. A004159.
%C a(n) = A007953(A000290(n+1)) - A007953(A000290(n)).
%C a(A202089(n)) = 0; a(A239878(n)) = 1; a(A240754(n)) = -1.
%H Reinhard Zumkeller, <a href="/A240752/b240752.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A004159(n+1) - A004159(n).
%t Differences[Total[IntegerDigits[#]]&/@(Range[0,80]^2)] (* _Harvey P. Dale_, Mar 10 2019 *)
%o a240752 n = a240752_list !! (n-1)
%o a240752_list = zipWith (-) (tail a004159_list) a004159_list
%K sign,base,changed
%O 1,2
%A _Reinhard Zumkeller_, Apr 12 2014
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Last modified March 18 23:08 EDT 2019. Contains 321305 sequences. (Running on oeis4.) | 532 | 1,307 | {"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-2019-13 | latest | en | 0.604454 |
https://www.kfs.oeaw.ac.at/mediawiki/index.php/Programmer_Guide/Command_Reference/NUMCHECK | 1,571,789,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987826436.88/warc/CC-MAIN-20191022232751-20191023020251-00531.warc.gz | 960,127,369 | 7,209 | # NUM, NUMCHECK
Warning: Display title "NUM, NUMCHECK" overrides earlier display title "NUMCHECK".
command return value value of RC
`NUM expression` value of expression or
empty string if the evaluation fails
0
error code
`NUMCHECK expression` value of expression or
empty string if the evaluation fails
0
warning code
The expression will be evaluated numerically, and the result (textual representation) will be returned. The expression may consist of the following parts:
decimal numbers
`123.456`, `17.5e3`, `100`, `-312.123`, `1e-3`,…
hexa-decimal numbers
`0x1234`, `0xabc`, `0XabC`, …
special constants
`pi` (=3.1415...), `e` (=2.71828...),
`rand` (a linear distributed pseudo random number r; -1≤r<1)
`lran` (a linear distributed pseudo random number r; 0≤r<1)
numerical operators
`-a` (negate), `a+b` (add), `a-b` (subtract), `a*b` (multiply), `a/b` (divide), `a%b` (modulo), `a^b` (power)
logical operators (bitwise)
`!a` (not), `a&b` (and), `a|b` (or), `a^b` (exclusive or)
brackets
`(expression)`
functions
`setlran(seed)` linear distributed pseudo random number r (0≤r<1)seed is the start value (0≤seed<<1) `sin(x)`, `cos(x)`, `tan(x)` sine, cosine or tangent of x `asin(x)`, `acos(x)`, `atan(x)` inverse sine, cosine or tangent of x `exp(x)` computes `ex` `log(x)` common logarithm of x (base `10`) `ln(x)` natural logarithm of x (base `e`) `sqrt(x)` square root of x (base 10) `abs(x)` absolute value of x (base 10) `int(x)` integer part of x, the fractional part is truncated(range of x: 32bit signed integer) `round(x)` round to the nearest integer of x(range of x: 32bit signed integer) `db(x)` convert level to factor (`10x/20`) `sinc(x)`, `sinx(x)` sinc function: `sin(x) / x` `sign(x)` sign of x; returns -1 if `x<0`, otherwise 1 `hz2bark(x)` convert x from Hertz to Bark `bark2hz(x)` convert x from Bark to Hertz `bit(n)` returns the integer with only bit n (0≤n<32) is set and all other are cleared; can be used to generate bit masks `floor(x)` returns a floating-point value representing the largest integer that is less than or equal to x(range of x: 64bit float) `iserr(rc)`, `iserror(rc)` returns 1 if rc is a STx error code, otherwise 0this function can be used to test the completion code `RC` of a command `iswarn(rc)`, `iswarning(rc)` returns 1 if rc is a STx warning code, otherwise 0this function can be used to test the completion code `RC` of a command `npow2(n)` next power of 2; returns the smallest possible value `2m ≥ n`
Notes
• In case of the expression being syntactically ill-formed, an error (`NUM`) or warning (`NUMCHECK`) is reported.
• The special numerical objects (vectors, matrices) available in EVAL-expressions, can not be used in the `NUM` expressions. Only the use of numerical constants and variables is possible.
• The same expression syntax is used for the INT and INTCHECK, for the numerical expressions in conditions (e.g. IF or WHILE) and for the evaluation of numerical arguments of commands.
• Numerical expressions are evaluated in 64bit floating point precission.
• For logical (operators `!, &, |, ^`) and integer expressions (functions `round, int, bit`) 32bit signed integers are used.
INT, INTCHECK, EVAL, EVALCHECK, SEGMENT
Examples
```// #wave = wave item
#t := num \$#wave[!length] / \$#wave[!srate] // duration in seconds
// #t = frame length in seconds, #o = overlap in percent
#lfrm := int \$#t * \$#wave[!srate] // frame length in samples
#lhop := int \$#t * (1 - \$#o / 100) * \$#wave[!srate] // hopsize in samples
#lfft := int npow2(\$#lfrm) // fft length
#df := num \$#wave[!srate] / \$#lfft // frequency resolution
```
```// using expressions in a IF statement
if 'db(\$#level)' > 1 then
conlog 'level must be lower or equal 0dB'
#level := 0
end
``` | 1,125 | 3,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-43 | latest | en | 0.592287 |
https://www.riansclub.com/open-system-closed-system-isolated-system/ | 1,621,134,544,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991659.54/warc/CC-MAIN-20210516013713-20210516043713-00610.warc.gz | 997,274,184 | 51,833 | # Open System, Closed System, And Isolated Systems [Thermodynamic Systems]
A system is a collection of matter, atoms, molecules, particles, or anything that can be quantified and measurable. For example, water in a bottle, or fuel in the engine. And a thermodynamic system is a collection of anything whose thermodynamic properties are of interest. There are different thermodynamic systems like open system, closed system, and isolated system. So in this article, we will learn about all those different types of thermodynamic systems.
But before we learn about the types of thermodynamic systems, we should have a broad idea about what is a thermodynamic system and what are the different elements that contribute to the thermodynamic system.
## What Is a Thermodynamic System?
A thermodynamic system is a collection of matter and/or radiation confined in space by a wall with defined permeability that separates it from the surroundings. The surroundings may include other thermodynamic systems or non-thermodynamic systems.
A thermodynamic system consists of the following elements.
• System
• Boundary
• Surroundings
• Universe
We already learned about systems. Everything outside of the system is called surroundings and the wall that separates the system from the surroundings is called the boundary.
### System
A system is where all thermodynamic process takes place and where you can analyze the matter or particle. The system is the region where you can measure the matter or properties.
For example, If you keep water in a bottle then this is called a system
### Boundary
A boundary is a real or imaginary surface that separates the system from the surroundings. A boundary can be fixed or movable and the shape and size of the boundary depend on the size of the system.
For Example, if you keep water in a bottle, then the bottle is the boundary.
### Sorroundings
Everything external to the system is called surrounding. For example, if you keep a water bottle inside a car then the interior of the car can be termed as surroundings.
### Universe
Everything external to the surroundings is called the universe. For example, if you keep a water bottle in a car, then the outer space of the car is the universe.
## Types Of Thermodynamic System
Following are the types of thermodynamic systems.
• Open system
• Closed system
• Isolated system
### Open System
An open system is something where the transfer of mass and energy can take place. When we talk about energy, in thermodynamics it is the heat energy.
#### Properties Of Open System
• Transfer of mass is possible
• Transfer of heat energy is possible
• An open system is open to its surroundings
#### Example of Open System
• Boiling water in a pan
• Water in a bottle with its cap open
• Internal combustion engine
• Compressors
• Turbines
• Water pumps
• Boilers
### Closed System
A closed system is a thermodynamic system where the energy transfer can take place but the mass can not be transferred. The energy can be transferred in the form of heat or work.
#### Properties Of Closed System
• Transfer of energy in the form of heat or work is possible
• Absolutely no transfer of mass
• The physical boundary is required for a closed system
#### Example Of Closed System
• Piston in a cylinder
• Pressure cooker
• Oxygen cylinders
• Gas cylinders
• Sealed water bottles
### Isolated System
In isolated system neither the mass or the energy transfer can take place. So it is totally separated from the surroundings.
#### Properties Of Isolated System
• No mass transfer
• No energy transfer
• No interaction with the surroundings
#### Example Of Isolated System
• Thermos
• Insulated water heaters
• Our universe
## Conclusion: Open System, Closed System And Isolated System
I hope you got a fair idea about different thermodynamic systems and the properties of it. If you still have any questions or queries, please don’t hesitate to write in the comment system and I will be happy to assist.
You may also like to read: Intensive Properties and Extensive Properties
## Frequently Asked Questions (FAQ) On Thermodynamic Systems
### What is An Open system?
A thermodynamic closed system is a collection of matter, molecules, or particles from where both mass and energy transfer can take place.
### What Is A Closed System?
A thermodynamic open system is a collection of matter, molecules or particles from where no mass transfer can take place, but energy transfer can happen in the form of heat or work
### What Is An Isolated System?
A thermodynamic isolated system is a collection of matter, molecules, or particles from where neither mass nor energy transfer can take place.
Scroll to Top | 964 | 4,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-21 | longest | en | 0.92817 |
https://www.teacherspayteachers.com/Product/Solving-Inequalities-Notes-2276212 | 1,495,991,492,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463610342.84/warc/CC-MAIN-20170528162023-20170528182023-00388.warc.gz | 1,202,895,389 | 22,899 | # Main Categories
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# Solving Inequalities Notes
Product Rating
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PDF (Acrobat) Document File
709 KB|2 pages
Product Description
This is a one-sided notes page on solving one-step, two-step & multi-step inequalities in one variable. A total of 14 examples are included and start off easy with one-step inequalities and move on to two-step, variables on both sides and distributive property examples. The point is not so much to teach the student how to solve (since they should already know how to do this), but to reinforce the new concept of having to flip the sign or not. The examples can be used for in-class guided practice.
Check out my Why do we need to flip the inequality sign? Discovery Activity (free) written to be used in conjunction with these notes. Key included. 2 pages.
This resource is also available as a part of my Inequalities Unit Bundle.
All of my products are color-coded and sorted by topic. To find more products on Inequalities, please return to my store and look for the lime green background or choose the Inequalities category on the left-hand side.
My other Algebra 1 Inequalities products include:
Intro to Inequalities Notes
Solving Inequalities Notes
Systems of Inequalities Notes & Practice
One-Variable Inequalities Foldable
Graphing Linear Inequalities Foldable
Intro to Inequalities Practice
Solving Inequalities Practice
Inequality Mini Word Problems Practice
Graphing Inequalities in Slope-Intercept Form
Graphing Inequalities in Standard Form
Graphing Inequalities Mixed Practice
Checking Solutions of Inequalities in Standard Form Practice (free)
Where Should We Shade? Group Activity
Why do we need to flip the inequality sign? Discovery Activity (free)
One-Variable Inequalities Matching Activity
Graphing Linear Inequalities Group Activity
Inequalities Review Activity
Inequalities Quiz
Graphing Inequalities Quiz
Inequalities Test
Inequalities Exit Tickets (or Warm-ups)
Inequalities Project
This purchase is for one teacher and for classroom use only. All rights reserved. No part of this product may be reproduced, transmitted or distributed without express permission from the author. This product may not be posted or uploaded to the Internet in any form including classroom websites or network drives. If other teachers, colleagues, teams, schools or districts wish to use this resource, additional licenses must be purchased and can be done so for half price.
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\$1.25 | 564 | 2,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-22 | latest | en | 0.882257 |
https://brilliant.org/problems/sheepish-camels-can-bark/ | 1,524,421,407,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945637.51/warc/CC-MAIN-20180422174026-20180422194026-00011.warc.gz | 575,496,798 | 17,691 | # Sheepish Camels Can Bark!
Algebra Level 3
Ahmed and Ali were camel-drivers who wanted to become shepherds. So they went to the market and sold all their camels. The amount of money (in dinars) that they received for each camel was the same as the total number of camels they owned. With that money, they bought as many sheep as possible at 10 dinars a sheep. With the money that was left they bought a goat.
On their way home they got in a fight and decided to split up. When they divided the sheep there was one sheep left. So Ali said to Ahmed, "I take the last sheep and you can get the goat".
"That's not fair," said Ahmed, "a goat costs less than a sheep".
"OK," Ali said, "then I will give you one of my dogs and then we are even".
Ahmed agreed that the trade was fair. (since 1 sheep - 1 dog = 1 goat + 1 dog)
What's the cost (in dinars) of a dog?
× | 227 | 866 | {"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-2018-17 | longest | en | 0.993812 |
https://www.colorschemer.com/birthday-calculator/ | 1,627,126,772,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150264.90/warc/CC-MAIN-20210724094631-20210724124631-00520.warc.gz | 716,791,207 | 7,463 | Birthday Calculator
Birthday Calculator helps you to calculate the number of days until your next birthday. Check how many days are remaining based upon your birth date.
What is Birthday Calculator?
“When You Are Old.”
Birthdays are the anniversary of the particular day on which a person was born. It is often treated as an event and celebrated with our loved ones. In other words, it is a tradition of marking an anniversary of a living being, organizations, and even of the fictional characters.
Birthdays are indeed memorable, and people wait for them. Some start the countdown while others count the months left for their special occasion. However, in both cases, one is required to count it on their own, and if it is about counting each day, it could be a mess.
And to clear out this mess, we have our handy yet innovative tool called Birthday Calculator.
Birthday Calculator is a tool via which one can count the months and days with 100% accuracy. The tool is programmed in such a manner that it always comes with an accurate answer without any error.
How to Use Birthday Calculator?
Are you exhausted of counting your special day on your own? It could be a frantic task due to the odd and even months. However, with this astounding tool engineered by our team, one can find the answers in an instant. For this, below are the mentioned steps to be followed.
• First of all, enter the date of birth in the required field.
• You have to enter your birth year, date, and month in their respective fields.
• Click on calculate
• And you're done.
Yes, it was this simple. By following the steps mentioned above, you can calculate your birthday without indulging yourself in any calculations.
Example of Birthday Calculator:
To understand how this unusual calculator works, let us take an example. However, note that this example does not refer to any event or day. It is entirely a random example based on a hypothetical situation.
Ashley is a seven years girl who is always excited about her birthday. However, she is a special child who may find difficulty calculating numbers on her own at this age. Thus, she keeps annoying her Mother by asking how many days and months are left for her birthday.
Being a working woman, her mother doesn't have enough time to calculate it every time she asks. And hence, she uses this smart Birthday Calculator for telling her the accurate month and days left.
Ashley was born on 2008, 1st June.
Her Mother, checking her days left for her birthday on 23rd March 2021 gets an answer which is Two months and eight days.
Conclusion:
Gone are the days when you have to count your birthday on your fingertips. after actualizing this tool's existence, you can anytime know the months and days left for your birthday in a click. So if you are also excited to know the time left for your birthday, use this tool now. | 593 | 2,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-31 | latest | en | 0.968908 |
https://community.fabric.microsoft.com/t5/DAX-Commands-and-Tips/DAX-measure-with-calculated-column-and-measure-as-a-single/td-p/755922 | 1,716,171,802,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00693.warc.gz | 173,870,891 | 114,681 | cancel
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Frequent Visitor
## DAX measure with calculated column and measure as a single measure
Hi All,
I am trying to create a single measure which combines the result of a calculated column with a measure. I have done it with a calculated column and measure but it is required that both logic be in a measure.
Calculated Coumn = IF(Tbl[Time]) <= 11, "AM", IF(Tbl[Time]) >= 12 && Tbl[Time] < 20 , "PM", FM))
Measure := CALCULATE(COUNTROWS(Tbl))
The table is as follow:
Time
4/3/2019 1:45
4/4/2019 1:45
4/5/2019 1:45
4/6/2019 1:45
4/3/2019 16:45
4/3/2019 16:45
4/3/2019 16:45
4/3/2019 23:45
4/3/2019 23:45
4/3/2019 23:46
4/3/2019 23:50
The result below:
Label Time Count
AM 1-11 4
PM 12-19 3
FM 20-23 4
Many Thanks
5 REPLIES 5
Super User
Hi @power_firm ,
Create 3 measures and use SUMX to total each category. I then added the three measures to a multi row card and turned off the Category Labels to get this result.
AM - 4
PM - 3
FM - 4
AM = CONCATENATE("AM - ", SUMX(Table1, IF(Hour(Table1[Time]) <=11, 1)))
PM = CONCATENATE("PM - ", SUMX(Table1, IF(Hour(Table1[Time])>=12 && Hour(Table1[Time])<20, 1)))
FM = CONCATENATE("FM - ", SUMX(Table1, IF(Hour(Table1[Time])>=20, 1)))
No Calculated Columns needed!
If this post helps, consider marking it as the solution.
Proud to be a Super User!
Super User
@power_firm
Or if you want one measure, one string: AM - 4 PM - 3 FM - 4 with a space in between.
All Time Periods = CONCATENATE( [AM] & " " & [PM] & " ", [FM]) Works in a Card with the labels turned off.
Proud to be a Super User!
Frequent Visitor
I really appreciate everyone input on the issue, i found an alternative solution coding both the measure and the calculated column in one DAX measure. The measure is a concatenated string that conmbines the measure and calculated column together.
TransCountByPeriod = VAR TotalTransaction = CALCULATE(COUNTROWS(Sheet1))
VAR UT = ADDCOLUMNS(VALUES(Sheet1[Systime]), "UTP", HOUR(Sheet1[Systime]))
IF([UTP] >= 0 && [UTP] < 11, "Early AM",
IF([UTP] >= 11 && [UTP] < 17, "Afternoon PM",
IF([UTP] >= 17 && [UTP] < 20, "Early Evening PM",
"Evening"))))
VAR Summary = SUMMARIZE(TransPeriod, [TimePeriod], "UniquePeriod", COUNTROWS(Sheet1))
VAR Results = ADDCOLUMNS(Summary, "Result", [TimePeriod] & " " & [UniquePeriod])
RETURN
IF(TotalTransaction > 0,
CONCATENATEX(
SELECTCOLUMNS(Results,
"Result", [Result]
),
[Result],
CONCATENATE(" ", UNICHAR(10))
)
)
Community Champion
Are you trying to return a table?
Best Regards,
Mariusz
If this post helps, then please consider Accepting it as the solution.
Please feel free to connect with me.
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Hi @Mariusz
Thanks for the response.
I am trying to return a string as a measure that concatenate the label and count into a single measure. It should be similar to the result below:
LabelwithCount
AM - 4
PM - 3
FM - 4
Thanks
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# when the formula to find the wind chill temperature is given by
w = 33 - (10.45 + 10√V - V) (33 - T over 22 and W is wind chill temperature (temperature with no wind) and T is actual temperature in Celcius, V is wind speeds in m/sec
a) T = 10 degrees Celcius, v = 9m/sec
b) T = 0 degrees celcius, v - 15m/sec
c) T = - 10 degrees celcius, v = 20m/sec
how do you figure this out step by step | 146 | 423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2014-41 | latest | en | 0.83643 |
https://www.styleforum.net/threads/cars-we-drive.17839/page-275 | 1,516,106,621,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886416.17/warc/CC-MAIN-20180116105522-20180116125522-00653.warc.gz | 1,025,808,786 | 25,134 | # Cars We Drive!
Discussion in 'Fine Living, Home, Design & Auto' started by Bert1568, Jul 18, 2006.
1. ### HRoiStylish Dinosaur
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yes sir - e90 M3, not M5
2. ### KurtS90Senior Member
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I think thats an M3 sedan?
Edit: Failed to check next page...
Last edited: Apr 15, 2012
3. ### GreenFrogStylish Dinosaur
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guys that's a 328i
Last edited: Apr 15, 2012
4. ### Sir FDistinguished Member
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Nice, so what did you change the rims?
5. ### KookzSenior Member
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Can't tell if trolling or serious...
All the torque in the world means nothing about how fast or quick a car is.
6. ### Arthur PESenior Member
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it's an e92 (sorry e90, e92 is the coupe) M3 sedan
and very nice
~420 HP (DIN) and 295 lb-ft torque
revs to 8400 or so
50/50 wt distribution
Last edited: Apr 15, 2012
7. ### Arthur PESenior Member
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torque is all that matters as far as acceleration, HP is associated with constant top speed, but really torque is doing the work
torque is the force which drives/propels the car, an instantaneous primary quantity
power is the rate of application of said torque, a derived/secondary quantity, P = T x w (w =2 x Pi x rev/sec) in our system P = T x rpm/5252
5252 = 60 sec/min x 550 lb ft/sec (HP conversion) / 2 Pi
F = ma
F in this case is thrust, which is axle torque divided by wheel radius (torque = force x radius, it's a vector, so the cross product, some call torque moment)
so if gearing is say 6 (tranny x diff) and engine torque is 300 axle torque is 1800 lb ft, if we say 15% losses ~1530 net
if the tire radius is 12" or 1 ft thrust = 1530 lb-ft/1ft = 1530 lb (force)
so a = F/m, so the more torque or the lower the weight the faster it accelerates
as far as top speed, work = f x distance, so it is still torque, which not only moves the car against gravity/inertia, but against wind resistance and mechanical resistance (gear/engine friction, tire slip/friction, etc.)
Last edited: Apr 15, 2012
8. ### A YDistinguished Member
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I'm not sure how to respond to this. The entire post you quoted is serious and correct. Your use of torque without reference to RPMs (or gearing) is meaningless.
To move a car requires force. That force is torque, so torque is what matters when you're trying to figure out how fast is a car. Torque is affected by RPMs because for equal amounts of torque generated at two different RPMs, torque at the higher RPM is more useful because it can be geared down more, multiplying its effect. Gears do the same thing, too. To accelerate a car quickly requires high torque in the RPM range in which you're accelerating.
Power only comes into it as an expression of the torque at a particular RPM (horsepower = torque*RPM/5252). In cars, it is an abstract quantity that is indirectly related to the driving experience. It's a good shorthand for expressing a certain quality of a car, but unless you drive around near redline, with the throttle on the floor, it isn't easily relatable to the everyday driving experience of a car.
9. ### KookzSenior Member
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This is my whole point in quoting your post. The use of torque without reference to RPM and gearing is meaningless, but torque WITH reference to those IS the very definition of (horse)power.
Everything you mention in your second paragraph is horsepower, you just seem to keep wanting to call it "torque at higher RPM".
Power is work over time, torque has no reference to time and thus has no bearing on the quickness of a car.
Code:
``` Volkswagen GTI Volkswagen Golf TDI
0-30 (sec): 2.8 3.0
0-45 (sec): 4.7 5.3
0-60 (sec): 7.0 8.7
0-75 (sec): 9.7 13.2
1/4 Mile (sec @ mph): 15.0 @ 95.2 16.4 @ 83.7
0-60 With 1-ft Rollout (sec): 6.6 8.3
30-0 (ft): 34 31
60-0 (ft): 130 121
Skid Pad Lateral Acceleration (g): 0.87 0.86
Slalom 65.5 69.2
```
Last edited: Apr 15, 2012
10. ### Arthur PESenior Member
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force (or torque) has a time relative component
F = mass x accel
mass = kg (or slug)
accel = m/sec^2 (or ft/sec^2)
it is the second derivative of position/displacement relative to time (d2s/dt2) times mass
if a car weighing 2000 lbs and with 2000 lbs of thrust (net axle torque / tire radius) it will accelerate at 1 g (ignoring tire slip/friction, drag, etc.)
32 ft/sec-sec or 32 ft/sec^2
if linear and 1 gear gets us to 60 mph (88 ft/sec, btw velocity is the first derivative of displacement) it will take ~ 88 ft/sec / 32 ft/sec-sec ~ 2.6 sec (units check, ft cancels, 1 set of sec cancel, leaving sec)
I size and select engines for various applications and always deal in torque
btw : engine torque = V x mep / 4Pi
V = displacement
mep, mean effective pressure ~ compression ratio x atm pressure x vol eff (x boost ratio (net) if forced induction)
4 = after 3 and before 5, base 10
Pi = 3.141596....not as in apple
Last edited: Apr 15, 2012
11. ### KookzSenior Member
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Arthur, you mention axle torque which is absolutely correct, but that is a product of the power, not torque.
In a fixed gear, a car will accelerate fastest at peak torque. However, a car will always accelerate fastest at peak HP given the ability to change gearing (which is why we have transmissions) because peak HP means maximum wheel torque, and has no relation to engine torque. An F1 engine makes less torque than a VW Golf TDI.
12. ### KookzSenior Member
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I take it that the PE means what I think it means (I'm a CivE) but you are telling me that torque is force, and that is incorrect. Pounds are a unit of force, and newtons are a unit of force, but ft-lbs and N/m are not units of force.
13. ### Arthur PESenior Member
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axle torque = crank torque x gearing (torque multiplication)
power = torque x rotational speed = T x w (w = 2 Pi (a radian/rev) x n = rev/sec)
power is (kind of) torque per unit time ~ torque x 1/sec
or the rate of torque application
if an F1 engine makes 800 HP at 18000 then T = 5252/18000 x 800 ~ 250 lb ft? (don't feel like getting a slide rule)
Last edited: Apr 15, 2012
14. ### Arthur PESenior Member
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torque is a force, a rotational force (T = F x r) a moment
lb-ft is torque (or force about a moment)
ft-lbs is energy or work as in distance x force or feet x pounds
wiki
Torque, moment or moment of force, is the tendency of a force to rotate an object about an axis, fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist to an object.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
it's not a linear force, like thrust, but thrust x radius = torque
Last edited: Apr 15, 2012
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https://www.coursehero.com/file/7113509/We-know-that-the-liquid-level-is-above-87-but-less-than-88/ | 1,493,391,155,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122992.88/warc/CC-MAIN-20170423031202-00173-ip-10-145-167-34.ec2.internal.warc.gz | 885,414,503 | 24,453 | Lab manual
# We know that the liquid level is above 87 but less
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Unformatted text preview: uch I have. Mathematical Functions and Significant Figures Often, ma th is required using numbers of various significant figures. We’ll cover only two basic classes of mathematical functions: (1) Addition and Subtraction: When adding or subtracting two or more numbers, do not worry so much about the number of significant figures. Instead, keep only the smallest number of digits after the decimal point that are significant in any of the numbers. For instance, when adding 101.2336 and 207.44, the answer will have only two places past the decimal point (308.67). (2) Multiplication and Division: When multiplying or dividing, keep only the maximum number of significant figures as is in any of the digits involved. For example, 101.2336 has 7 significant figures, but 207.44 has only 5 significant figures, so our product can have only 5 significant figures (21,000.). Be careful not to confuse exact numbers for inexact. For example, we know there are 12 inches in 1 foot, so if we are using this conversion factor to convert 3.0252 ft to inches, we might think we can only have two significant figures because of the factor “12.” However, this is an exact definition, and as such, it can have as many significant figures as we want. In other words, even though we don’t bother writing the zeros, there are really “12.000000000000000…” inches in 1 foot because this is an exact definition. Thus, our answer will be 36.302 inches (5 significant figures). Reading Instruments There are two types of instruments; analog and digital. Digital instruments are easy, just write down every number they give you, including zeros. This automatically Dakota State University page 230 of 232 Significant Figures General Chemistry I and II Lab Manual gives you the number of significant figures. However, analog devices are a little more tricky. Analogue devices have some form of scale, with an indicator. In an old-fashioned thermometer, the scale is on the side, with the indicator being the level of the liquid. Other instruments, like voltmeters, for example, had a scale (usually with a portion mirrored so you always look at it from the same angle by lining up the pointer so you cannot see the image) with a pointer. Whenever you have an instrument like this, you can always estimate one significant figure more than the scale on the instrument. Take the following example; suppose you are measuring the liquid in a graduated cylinder, with markings every 0.1 mL, as shown in the figure to the left. We know that the liquid level is above 8.7, but less than 8.8; so what is it? (Forgive the squiggly line; it was drawn by hand.) Well, how far up does it look to you? Maybe 70% of the way? OK, so y record 8.77 in your records. Don’t ou worry that the las...
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## This note was uploaded on 09/18/2012 for the course CHEMISTRY 1010 taught by Professor Kumar during the Fall '11 term at WPI.
Ask a homework question - tutors are online | 719 | 3,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-17 | longest | en | 0.899522 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=82635 | 1,369,166,649,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700497024/warc/CC-MAIN-20130516103457-00096-ip-10-60-113-184.ec2.internal.warc.gz | 313,799,910 | 1,001 | ```Question 113455
There are 3 equations with 3 unknowns, so it should be possible
to find x, y, and z.
{{{ x + 5y - z = 4}}}
{{{3x + 4y + z = 4}}}
{{{ x - y + z = 0}}}
----------------
{{{ x + 5y - z =4}}}
{{{ x- y + z =0}}}
{{{2x + 4y = 4}}}
{{{4y = 4 - 2x}}}
{{{y = 1 - (1/2)x}}}
subtract this equation from the 2nd one
{{{3x + 4y + z =4}}}
{{{2x + 4y = 4}}}
{{{x + z = 0}}}
{{{z = -x}}}
Now plug values for y and z into the 3rd equation
{{{ x - y + z = 0}}}
{{{x - (1 - (1/2)x) + (-x) = 0}}}
{{{x - 1 + (1/2)x - x = 0}}}
{{{(1/2)x = 1}}}
{{{x = 2}}}
{{{z = -2}}}
{{{y = 1 - (1/2)*2}}}
{{{y = 0}}}
x = 2
y = 0
z = -2
----------
{{{ x + 5y - z = 4}}}
{{{3x + 4y + z = 4}}}
{{{ x - y + z = 0}}}
-------------------
{{{2 + 5*0 - (-2) = 4}}}
{{{2 + 2 = 4}}}
-------------------
{{{3*2 + 4*0 + (-2) = 4}}}
{{{6 - 2 = 4}}}
------------------
{{{2 - 0 + (-2) = 0}}}
{{{2 - 2 = 0}}}
-------------------
OK ``` | 436 | 905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2013-20 | latest | en | 0.44521 |
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#### Resources tagged with Estimating and approximating similar to Power Countdown:
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### Power Countdown
##### Stage: 4 Challenge Level:
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How many generations would link an evolutionist to a very distant ancestor?
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Estimate these curious quantities sufficiently accurately that you can rank them in order of size
### More or Less?
##### Stage: 4 Challenge Level:
Are these estimates of physical quantities accurate?
### Big and Small Numbers in Biology
##### Stage: 4 Challenge Level:
Work with numbers big and small to estimate and calulate various quantities in biological contexts.
### Big and Small Numbers in Physics
##### Stage: 4 Challenge Level:
Work out the numerical values for these physical quantities.
### Does This Sound about Right?
##### Stage: 3 Challenge Level:
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10 graphs of experimental data are given. Can you use a spreadsheet to find algebraic graphs which match them closely, and thus discover the formulae most likely to govern the underlying processes?
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How might you use mathematics to improve your chances of guessing the number of sweets in a jar?
### Using the Haemocytometer
##### Stage: 4 Challenge Level:
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##### Stage: 3 Challenge Level:
Can you work out how many of each kind of pencil this student bought?
### Some Games That May Be Nice or Nasty for Two
##### Stage: 2 and 3 Challenge Level:
Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your oponent.
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Squaring the Circle
##### Stage: 3 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
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##### Stage: 2 and 3 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
### Dicey Operations
##### Stage: 2 and 3 Challenge Level:
Who said that adding, subtracting, multiplying and dividing couldn't be fun? | 756 | 3,515 | {"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} | 3.609375 | 4 | CC-MAIN-2015-40 | longest | en | 0.824029 |
https://coversuper.net/bike/what-size-bike-fits-a-5-10-man.html | 1,679,675,540,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00101.warc.gz | 240,839,659 | 10,298 | What Size Bike Fits A 5 10 Man?
Is 26 inch bike good for what height?
Is 26 inch bike good for what height? Bikes that are 26 inches in size work well for children and adults who range from approximately five feet to six-feet tall. They can also be appropriate for some taller people, but the best way to find out if it’s a good fit is by testing ride one!
Is 26 inch bike good for adults?
“ 26” wheel size is perfect for most of adults with a height above 6 feet. Most of the touring bikes and hybrid bikes come in 700C metric wheels, which are also known as 29-inch wheels.
How do you know what size bike to buy for a man?
Method Two: Calculating the bike size
1. Take off your shoes and stand with your legs about 15-20 cm(6” – 8”) apart. Measure the height from the ground to your crotch.
2. Be sure of the type of bicycle you want: Mountain bike, city bike or road bike.
3. Now you can take your calculator and calculate the right size:
What size bike should a 6 foot man ride?
People between 6 feet tall and 6 feet 2 inches tall need a bike between 17 and 19 inches in height. Those between 6 feet 2 inches and 6 feet 4 inches should look for a bike between 19 and 21 inches in height. Riders over 6 feet 4 inches tall should purchase a bike that is 21 inches or higher.
You might be interested: Quick Answer: What Size Bike Is For A 3 Year Old?
How do you determine bike size?
How to calculate a bike size? Measure your inseam!
1. Stand close to a wall, your feet should be 6-8″ (15-20 cm) apart.
2. Place a large, hardcover book between your legs – it will simulate the saddle.
3. Mark where the book’s spine touches the wall.
4. Measure the distance from that point to the floor – that’s your inseam.
How do I determine bike frame size?
Adult bikes are measured by their frame size. Most manufacturers now measure the frame from the centre of the crank axle, to the top of the seat tube. Most road bikes are measured in centimetres (cm), whereas mountain bikes are generally measured using inches (in). | 501 | 2,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-14 | latest | en | 0.945434 |
https://discuss.leetcode.com/topic/36289/python-o-1-solution-96-6 | 1,516,560,920,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890823.81/warc/CC-MAIN-20180121175418-20180121195418-00311.warc.gz | 664,247,392 | 15,794 | # Python O(1) Solution 96.6%
• ``````class Solution(object):
def isPowerOfThree(self, n):
return n > 0 and 1162261467 % n == 0``````
• hi,
nice solution! Do you get 1162261467 programatically, or how do you get it?
I'm guessing that's the greatest integer power of 3, Am I right?
• You are right, this is the maximum integer of pow 3 in python 2.7.
I tested out the value knowing the maximum int is 2*31 - 1
Next val 3
1162261467 is a long int.
• I tried to get the value by using the following code, keep multiply by 3 until it overflows.
``````int i = 1;
while(i * 3 > i){
i *= 3;
}
System.out.println(i);``````
• In python, this way doesn't work. overflow only happens when your full memory is full
• Also you could use the following math approach to get the max 32 bit integer power of 'x' number:
pow( x,int( log(0x7FFFFFFF)/log(x) ) ).
In this case x= 3, 0x7FFFFFFF isthe maximum 32 bit integer and the change of base with log(0x7FFFFFFF)/log(x) = logx(0x7FFFFFFF) which yields the number to which you have to power 'x' to get the biggest 32 bit integer.
Note that int() returns the floor of a number passed as parameter.
• nice job! but it's not portable i think.
• Easy Python Solution
``````class Solution(object):
def isPowerOfThree(self, n):
if n <= 0:
return False
while n % 3 == 0:
n = n / 3
return True if n == 1 else False``````
• Yes, it is an easy solution. But your complexity is O(logn). Besides, the return statement can be written as return n == 1. You don't have to make if test.
• You are correct on both observations.
I am learning Python and found solving problems is the best way to do it.
• Yes, indeed, I learned a lot here as well.
• Why not use pow(x,int(log(0x7FFFFFFF,x)))
• It's exactly the same, just a different way to do it so that if you didn't had a built in like log(x, base) you would still be able to get it mathematically
• How do you know 3* 1162261467 is a long int? But on my system, it is a int.
• We're dealing with the assumption that we're working with 32-bit integers
• I find the max int in python is sys.maxint,which is 9223372036854775807
• That's not true when you're using languages like java, c, c++ and so many others. I'm not completely sure, but in my experience working with this OJ (leetcode), this web page uses int as the standard 32bit word
• It's not going to be portable and i haven't checked it , but i think you could use something like pow(x, (log(sys.maxint, x) )) if you feel more comfortable, it would be able to deal with the biggest int allowed in python, wich is bigger than the one yielded by a 32bit word
• @am78 no loop as the demand
• @am78
The question mentioned that loop and recursion are not allowed.
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 788 | 2,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-05 | latest | en | 0.905113 |
https://www.savidude.com/post/to-noobs-with-love-machine-learning-simple-linear-regression | 1,638,351,364,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00208.warc.gz | 1,046,306,246 | 161,357 | Search
• Savidu Dias
# To Noobs with Love: Machine Learning: Simple Linear Regression
Welcome to my blog series on Machine Learning, where I plan on writing about the basic concepts of machine learning all the way up to some really advanced stuff (hopefully). As a machine learning noob myself, I will write these posts as I get to study more about the concepts of machine learning. Consider these to be my own notes as I study machine learning. Let’s start off with Simple Linear Regression.
## What is Simple Linear Regression?
You might be familiar with plotting line graphs with one X axis and one Y axis. The values in the X axis are sometimes called “independent variables”, while the values in the Y axis are called “dependent variables”. Simple Linear Regression plots one independent variable X against one dependent variable Y in a line graph.
To explain things in a more formal way, Simple Linear Regression is a statistical method that allows us to summarize the relationship between two variables.
## Simple Linear Regression Formula
Regression Analysis is a major part of data science, which is the process used to find equations that match a particular data set. Consider the following chart showing the relationship between the years of experience of a bunch of employees in a company, and their salary.
This type of representation is called a “scatter plot”. Each cross (x) in this diagram represents a single employee, where the X axis represents their years of experience, and the Y axis represents their salary. If we study the graph closely, we can see that the data appears to form a straight line.
When such data appears to form a straight line, we can use Simple Linear Regression to predict the salary of a future employee based on their experience. If we recall the algebra you learned when you were 5 years old, you’ll remember that the equation for a straight line is y = mx + c. However, statistics generally prefer to use the following equation.
## y = b0 + b1x
y represents the dependent variable
x represents the independent variable
b0 and b1 are constants and are parameters (or coefficients) that need to be estimated from the data.
b0 is known as the intercept. This is the point in which the straight line touches the Y axis. In our example, this would be the predicted salary of a fresh graduate joining with no experience.
b1 indicates the slope of the line. This shows the increase in salary per year.
## Best Fitting Line
The red line from the graphs above is known as the “Best Fitting Line”. This line represents the model for Simple Linear Regression.
The task of developing a Simple Linear Regression model is to come up with a best fitting line that represents a collection of data.
Once we draw this line, the model can easily predict the salary of a new employee based on how much experience they have.
The best fitting line above shows that a new employee with x1 years of experience should get a salary of y1.
How do we draw this best fitting line?
Let’s take a look at one employee. yi represents the salary of the employee, and yi^ represents what their salary should be according to the model. In technical terms:
• yi is the actual observation
• yi^ is the modeled observation
To figure out how good this line is, we take the sum of (yi - yi^)2 for all plotted values in the graph. This can be represented by this equation.
Linear regression draws all possible lines, gets the sum of all of them and uses this information to find the line having the minimum value for the sum of squares. This is called the ordinary least squares method.
And that is all you have to know about Simple Linear Regression! Here’s a quick recap of all you need to know about predicting values with Simple Linear Regression Model:
2. Plot them on a graph
3. Figure out the best fitting line
4. Predict stuff | 792 | 3,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-49 | longest | en | 0.933687 |
https://ww2.mathworks.cn/matlabcentral/cody/problems/11-back-and-forth-rows/solutions/558324 | 1,597,019,022,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738595.30/warc/CC-MAIN-20200809222112-20200810012112-00057.warc.gz | 554,686,612 | 17,891 | Cody
# Problem 11. Back and Forth Rows
Solution 558324
Submitted on 16 Jan 2015 by Yannick
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 4; a = [ 1 2 3 4; 8 7 6 5; 9 10 11 12; 16 15 14 13]; assert(isequal(a,back_and_forth(n)));
b = 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 0 0 0 5 0 0 0 0 0 0 0 0 b = 1 2 3 4 0 0 6 5 0 0 0 0 0 0 0 0 b = 1 2 3 4 0 7 6 5 0 0 0 0 0 0 0 0 b = 1 2 3 4 8 7 6 5 0 0 0 0 0 0 0 0 b = 1 2 3 4 8 7 6 5 9 0 0 0 0 0 0 0 b = 1 2 3 4 8 7 6 5 9 10 0 0 0 0 0 0 b = 1 2 3 4 8 7 6 5 9 10 11 0 0 0 0 0 b = 1 2 3 4 8 7 6 5 9 10 11 12 0 0 0 0 b = 1 2 3 4 8 7 6 5 9 10 11 12 0 0 0 13 b = 1 2 3 4 8 7 6 5 9 10 11 12 0 0 14 13 b = 1 2 3 4 8 7 6 5 9 10 11 12 0 15 14 13 b = 1 2 3 4 8 7 6 5 9 10 11 12 16 15 14 13
2 Pass
%% n = 5; a = [ 1 2 3 4 5; 10 9 8 7 6; 11 12 13 14 15; 20 19 18 17 16; 21 22 23 24 25]; assert(isequal(a,back_and_forth(n)));
b = 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 0 0 0 7 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 0 0 8 7 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 0 9 8 7 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 0 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 0 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 0 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 0 0 0 0 0 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 0 0 0 0 16 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 0 0 0 17 16 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 0 0 18 17 16 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 0 19 18 17 16 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 0 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 0 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 22 0 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 22 23 0 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 22 23 24 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 22 23 24 25 | 2,052 | 2,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-34 | latest | en | 0.594273 |
https://www.interstellarmedium.org/extinction/extinction_computations/ | 1,624,013,875,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487636559.57/warc/CC-MAIN-20210618104405-20210618134405-00207.warc.gz | 758,271,314 | 6,779 | Extinction computations
Extinction computations
NICER
NICER is a method used to estimate extinction based on the reddening of the light of background stars, that is, how much redder stars appear to be compared to their intrinsic colours. The reddening is caused by dust clouds between the stars and the observer and thus gives a measure for the amount of interstellar matter along that line of sight. NICER is a way to optimally combine measurements of several clouds. The method is described in an article by Lombardi and Alves (2001) .
The calculation of a map of extinction consists of two step. First, one estimates the extinction for every star in the field. Second, one calculates extinction values for each pixel in the map. Unlike stars, pixels form a regular grid on the sky and each pixel value is obtained as a weighted average over the extinction values of stars near that pixel. Both steps should be easy to parallelise: extinction estimates of individual stars can be calculated in parallel and similarly the averages for each pixel can be calculated in parallel.
We wrote a Python program that used two OpenCL kernels that correspond to the two steps mentioned above. The task of the main program is only to read in the input data, call the kernels, and write the results to files. Here is the listing of the first kernel, where the work items together loop over all the stars. The kernel is listed (with minimal comments) just to show what kind of calculations it involves. The kernel calls only one external routine, solve_cramer2, which calculates the inverse of a matrix (not shown).
``````
__kernel void nicer(
__global float *K, // [NB], extinction relative to Av
__global float *RCOL, // average colours in reference area
__global float *RCOV, // covariance of colour in ref. area
__global float *MAG, // magnitudes [NS*NB]
__global float *dMAG, // magnitude error estimates
__global float *AV, // output: estimate Av [NS]
__global float *dAV // output: estimated dAv [NS]
)
{
int i, j, ss = get_global_id(0) ; // one pixel
int gs = get_global_size(0) ;
if (ss>=NS) return ; // id > NS, the number of ON field stars
float C[NB*NB] ; // NB = number of bands
float av, b[10] ;
for(int s=ss; s<NS; s+=gs) { // work items loop together over all NS stars
for(i=0; i<NC; i++) { // NC = number of colours
C[NC*NB+i] = -K[i] ;
C[i*NB+NC] = -K[i] ;
}
for(i=0; i<NB; i++) b[i] = 0.0f ;
b[NC] = -1.0f ;
for(i=0; i<NC; i++) {
for(j=0; j<NC; j++) {
C[i*NB+j] = 0.0f ;
}
}
for(i=0; i<NC; i++) {
C[i*NB+i] = dMAG[s*NB+i]*dMAG[s*NB+i] + dMAG[s*NB+i+1]*dMAG[s*NB+i+1] ;
}
C[NC+NB*NC] = 0.0f ;
for(i=0; i<NC-1; i++) {
C[i*NB + i+1] = -dMAG[s*NB+i+1]*dMAG[s*NB+i+1] ;
C[(i+1)*NB+i] = -dMAG[s*NB+i+1]*dMAG[s*NB+i+1] ;
}
for(i=0; i<NC; i++) {
for(j=0;j<NC; j++) {
C[i*NB+j] += RCOV[i*NC+j*j] ;
}
}
float C0=C[0], C1=C[1], C4=C[4] ;
float det ;
solve_cramer3(C, b, &det) ;
av = 0.0 ;
for(i=0; i<NC; i++) av += ((MAG[s*NB+i]-MAG[s*NB+(i+1)])-RCOL[i]) * b[i] ;
AV[s] = av ;
dAV[s] = sqrt(b[0]*b[0]*C0 + b[1]*b[1]*C4 + 2.0f*b[0]*b[1]*C1) ;
}
}
``````
The second kernel calculates for each pixel a weighted average of the Av values of individual stars. This time work items (threads) loop over the map pixels. The calculation is done actually two times, the second loop dropping outliers based on user-defined thresholds (CLIP_UP and CLIP_DOWN).
``````__kernel void smooth(
__global float *RA, // coordinates of the stars [NS]
__global float *DE, // -"-
__global float *A, // Av of individual stars
__global float *dA, // dAv of individual stars
__global float *SRA, // coordinates of the pixels [NPIX]
__global float *SDE, // -"-
__global float *SA, // Average Av for a pixel
__global float *dSA // error estimate
)
{
int j, idd = get_global_id(0) ; // index of smoothed value, single pixel
int gs = get_global_size(0) ;
if (idd>=NPIX) return ;
// calculate weighted average with sigma-clipping
float ra, de ; // centre of the beam
float cosy = cos(de) ; // we can use the plane approximation for distances
float w, dx, dy, weight, sum, s2, uw, d2, ave, std, count ;
float K = 4.0f*log(2.0f)/(FWHM*FWHM) ; // radian^-2
const float LIMIT2 = 9.0f*FWHM*FWHM ; // ignore stars further than sqrt(LIMIT2)
for(int id=idd; id<NPIX; id+=gs) {
ra = SRA[id] ;
de = SDE[id] ;
weight = sum = s2 = uw = count = 0.0f ;
for(j=0; j<NS; j++) {
dx = cosy* (RA[j]-ra) ;
dy = (DE[j]-de) ;
d2 = dx*dx + dy*dy ;
if (d2<LIMIT2) {
w = exp(-K*d2) / (dA[j]*dA[j]) ;
weight += w ;
sum += w*A[j] ; // weighted sum
uw += A[j] ; // unweighted sum
s2 += A[j]*A[j] ; // for unweighted standard deviation
count += 1.0f ;
}
}
ave = sum/weight ; // weighted average
std = sqrt(s2/count - (uw/count)*(uw/count)) ; // unweighted std
// Repeat, this time with sigma clipping
weight = sum = s2 = uw = count = 0.0f ;
for(j=0; j<NS; j++) {
dx = cosy* (RA[j]-ra) ;
dy = DE[j]-de ;
d2 = dx*dx+dy*dy ;
if ((d2<LIMIT2) && (A[j]>(ave-CLIP_DOWN*std)) && (A[j]<(ave+CLIP_UP*std)) ) {
w = exp(-K*d2) / (dA[j]*dA[j]) ;
weight += w ;
sum += w*A[j] ;
uw += A[j] ;
count += 1.0f ;
s2 += w*w*dA[j]*dA[j] ; // weighted
}
}
if (count>1.0) {
SA[id] = sum/weight ; // weighted average
dSA[id] = sqrt(s2)/weight ; // weighted
} else { // count <= 1
if (weight>0.0) {
SA[id] = sum/weight ; dSA[id] = 999.0 ;
} else {
SA[id] = 0.0 ; dSA[id] = 999.0 ;
}
}
}
}
``````
Interestingly the two kernels scale differently on CPU and GPU. We computed a series of test cases, starting with a map that was 0.5×0.5 degrees in size and had a pixel size of 1.0 arcmin. The number of stars over the area was 75 000. We scaled the problem by increasing either the number of stars (which affects both kernels) or, alternatively, only the number of pixels (affecting the second kernel). The resulting run times are shown in the plot below.
On the x-axis value of 1 corresponds to the original problem size as described above. Here CPU refers to a run with a CPU with 6 hyperthreaded cores. When the number of stars is increased, the scaling is similar for both CPU and GPU, the latter being faster by a factor of 3-4. On the other hand, when the number of pixel is increased and a larger fraction of time is spend on the second kernel, the GPU run time barely increases at all. However, here the number of pixels is increased only up to ~60 000 pixels and by that time GPU is again almost a factor of 4 faster than the 6-core CPU.
The tests were run on a laptop with a six core i7-8700K CPU running at 3.70 GHz and with an Nvidia GTX-1080 GPU. | 2,147 | 6,801 | {"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-2021-25 | latest | en | 0.865881 |
https://fenicsproject.discourse.group/t/imposing-dirichlet-bc-on-the-boundary-of-a-hole-embedded-in-a-continuum-body/6215/2 | 1,708,813,505,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00835.warc.gz | 248,489,548 | 6,291 | Imposing Dirichlet bc on the boundary of a hole embedded in a continuum body
Hello,
I am solving an elasticity problem in which the domain is a rectangle (150x120) with two circular voids (radius = 4) (shaped like a videotape). I would like to apply Dirichlet BC on the boundary of the voids and to apply other Dirichlet BC on the right boundary of the rectangle. When I run my code I get the error:
*** Warning: Found no facets matching domain for boundary condition.
I checked the similar question asked here:https://fenicsproject.discourse.group/t/dirichlet-bcs-on-an-annulus/1711
but I was hoping for a new answer since 2 years have passed. I understand the problem is probably with the BC on the voids.
So, the way I wrote it is:
``````# Boundary conditions
right= CompiledSubDomain("near(x[0], 150.0) && x[1] < 120.0 && x[1] >= 0.0 && on_boundary")
tondo_top = CompiledSubDomain("(x[0]-25.00)*(x[0]-25.00)+(x[1]-87.50)*(x[1]-87.50)-16.00 < 0.001 && (x[0]-25.00)*(x[0]-25.00)+(x[1]-87.50)*(x[1]-87.50)-16.00 > -0.001 && x[1]-87.50 >= 0.0 && on_boundary")
tondo_bot = CompiledSubDomain("(x[0]-25.00)*(x[0]-25.00)+(x[1]-32.50)*(x[1]-32.50)-16.00 < 0.001 && (x[0]-25.00)*(x[0]-25.00)+(x[1]-32.50)*(x[1]-32.50)-16.00 > -0.001 && x[1]-32.50 >= 0.0 && on_boundary")
load_top = Expression("t", t = 0.0, degree=1)
load_bot = Expression("t", t = 0.0, degree=1)
bcright= DirichletBC(W, Constant((0.0,0.0)), right)
bc_u = [bc_tondo_top, bc_tondo_bot, bcright]
boundaries = MeshFunction("size_t", mesh, mesh.topology().dim() - 1)
boundaries.set_all(0)
tondo_top.mark(boundaries,1)
ds = Measure("ds")(subdomain_data=boundaries)
n = FacetNormal(mesh)
``````
The reason I am defining “n” is because at the end of the code I ask for the dot product of a tensor and “n” (this is not necessary).
Can anyone help?
As you have not supplied your mesh file, it is not very easy to help you. Please not that if you create your mesh with an external mesh creator (say GMSH) you can mark your boundaries in the creation step and import them, as covered in:
2 Likes
I changed the lines:
``````tondo_top = CompiledSubDomain("(x[0]-25.00)*(x[0]-25.00)+(x[1]-87.50)*(x[1]-87.50)-16.00 < 0.001 && (x[0]-25.00)*(x[0]-25.00)+(x[1]-87.50)*(x[1]-87.50)-16.00 > -0.001 && x[1]-87.50 >= 0.0 && on_boundary")
tondo_bot = CompiledSubDomain("(x[0]-25.00)*(x[0]-25.00)+(x[1]-32.50)*(x[1]-32.50)-16.00 < 0.001 && (x[0]-25.00)*(x[0]-25.00)+(x[1]-32.50)*(x[1]-32.50)-16.00 > -0.001 && x[1]-32.50 >= 0.0 && on_boundary")
``````
with the following:
``````def tondo_top(x, on_boundary):
return abs((x[0]-25.00)**2+(x[1]-87.50)**2 - 16.00) >= 1E-6 and on_boundary
def tondo_bot(x, on_boundary):
return abs((x[0]-25.00)**2+(x[1]-87.50)**2 - 16.00) >= 1E-6 and on_boundary
``````
and the error disappeared. Unfortunately, now I cannot assign any label to boundaries and get an error for “n”. I tried to fix it as dokken suggested by generating the mesh on freefem (.mesh file) and labeling there the various boundaries.
As you are not staying what error you are obtaining, and you do not provide a minimal code example reproducing the error, I cannot provide you much more help. | 1,097 | 3,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-10 | latest | en | 0.800381 |
https://thisisbeep.com/how-long-is-sats-arithmetic-test/ | 1,695,907,616,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00421.warc.gz | 608,869,895 | 12,180 | # How long is SATs arithmetic test?
## How long is SATs arithmetic test?
Paper 1 is an arithmetic test which should take around 15 minutes. It will consist of 25 marks. Paper 2 involves reasoning, problem solving and mathematical fluency. This paper has 35 marks available and will last for approximately 35 minutes.
## How long is KS2 arithmetic test?
30 minutes
You have 30 minutes to complete this test. Work as quickly and as carefully as you can. Put your answer in the box for each question.
What level is Key Stage 2?
Key Stage 2 – ages 7-11 (Years 3-6)
What age group is Key Stage 2?
Key Stage 2 is designed for students aged 8 to 10 years. The core subjects remain English, Mathematics and Science. Literacy skills now move towards reading for meaning and teaching students how to produce reports, recounts, instructions and stories in writing.
### What months are Year 2 SATs?
The key stage 2 tests are timetabled from Monday 9 May to Thursday 12 May 2022….Key stage 2.
Date Activity
Monday 9 May 2022 English grammar, punctuation and spelling papers 1 and 2
Tuesday 10 May 2022 English reading
Wednesday 11 May 2022 Mathematics papers 1 and 2
Thursday 12 May 2022 Mathematics paper 3
### What year group is Key Stage 2?
Key Stage 2 – ages 7-11 (Years 3-6) Key Stage 3 – ages 11-14 (Years 7-9)
Are Key Stage 2 SATs Cancelled?
Updated 20 March 2021: In recognition of the challenges posed by the pandemic, the Department of Education has now cancelled both KS1 and KS2 SATs tests for 2021.
What is the difference between Key Stage 1 and Key Stage 2?
As aforementioned, primary school encompasses Key Stages 1 and 2; KS1 students are mostly aged between 5 and 7 (Years 1 and 2), and those aged 7 to 11 fall into the KS2 category (Years 3, 4, 5 and 6).
## Is there a KS2 SAT arithmetic paper for teachers?
This has been devised to support teachers in the run up to SATs. There is also one for KS1 SATs Generate as many different KS2 SAT Arithmetic papers as you need – all questions directly based on the 2017 SAT (difficulty and characteristics of each questions are preserved as much as possible).
## How to prepare for Year 6 Maths SATs?
Achieve good grades by practising with the latest maths Sats papers. The year 6 maths SATs papers are an excellent resource to help children to prepare for their SATs exams at the end of year 6.
How many questions are on the KS2 test?
There are 25 questions in each test. Each test ranges in difficulty from easy to challenging. Each test covers the spectrum of KS2 content that is able to be assessed in an arithmetic test. Individual questions can be selected for use in other ability or year groups as you can select the strand you want to assess.
What is a revision mat for Year 6?
A revision mat made for Year 6 to help them remember key facts and rules. Two pages that can be printed back to back and laminated so children can regularly refer back to it. Included on the mat are: Fraction, decimal and percentage equivalents Measurement facts Angle types Averages Or… | 739 | 3,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-40 | latest | en | 0.957148 |
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Topic: ZFC and God
Replies: 390 Last Post: Jan 31, 2013 4:22 PM
Messages: [ Previous | Next ]
Virgil Posts: 8,833 Registered: 1/6/11
Re: ZFC and God
Posted: Jan 26, 2013 4:50 PM
In article <ke0gdo\$5s2\$1@Kil-nws-1.UCIS.Dal.Ca>,
gus gassmann <gus@nospam.com> wrote:
> On 25/01/2013 10:58 PM, Virgil wrote:
> > In article <87ip6kvhk1.fsf@phiwumbda.org>,
> > "Jesse F. Hughes" <jesse@phiwumbda.org> wrote:
> >
> >> WM <mueckenh@rz.fh-augsburg.de> writes:
> >>
> >>>> I'm not going to bother working through your addled analogy.
> >>>
> >>> You need not. Just ask yourself whether or not it is possible to
> >>> define in ZFC the set of all terminating decimal representations of
> >>> the real numbers of the unit interval. If you think that it is not
> >>> possible, then you should try to learn it. If you know it already,
> >>> then we can formally restrict ourselves to working in this set until
> >>> we discover a digit that is not defined in an element of this set.
> >>>
> >>> Your further questions then turn out meaningless.
> >>
> >> I asked how you define terminating decimal representation. How is
> >> that meaningless?
> >>
> >> Here's the definition I suggested again. Please tell me if you agree
> >> with it, and if not, what definition you have in mind.
> >>
> >> Let x be a real number in [0,1]. We say that x has a terminating
> >> decimal representation iff there is an f:N -> {0,...,9} such
> >> that
> >>
> >> x = sum_i f(i) * 10^-i,
> >>
> >> and
> >>
> >> (En)(Am > n)(f(m) = 0) or (En)(Am > n)(f(m) = 9)
> >>
> >> If x has no terminating decimal representation, then we say that x is
> >> non-terminating.
> >>
> >> We cannot continue unless I know what you mean by terminating decimal
> >> representation.
> >
> > WM finds precise careful definitions far too restricting for his
> > maunderings in Wolkenmuekenheim, so will resist either providing one
> > himself or accepting anyone else's.
>
> I suspect he is threatened by them. He cannot work through the (En) and
> (Am) notation and all that stuff, so he denies, denies, denies. He is
> all bluster, with absolutely zero understanding or hope of understanding
> even the simplest mathematical concepts.
>
> That's why he cannot let himself be pinned down by Jesse's definitions.
> He cannot understand them, so he cannot control them. He is, above all,
> a control freak.
It appears hat Jesse has finally managed to get WM to produce a concrete
definition and has used it to good advantage.
--
Date Subject Author
1/23/13 Aatu Koskensilta
1/23/13 Jesse F. Hughes
1/23/13 mueckenh@rz.fh-augsburg.de
1/23/13 Jesse F. Hughes
1/23/13 mueckenh@rz.fh-augsburg.de
1/23/13 Jesse F. Hughes
1/23/13 Jesse F. Hughes
1/23/13 mueckenh@rz.fh-augsburg.de
1/23/13 Jesse F. Hughes
1/24/13 mueckenh@rz.fh-augsburg.de
1/24/13 Virgil
1/24/13 Jesse F. Hughes
1/24/13 mueckenh@rz.fh-augsburg.de
1/24/13 Jesse F. Hughes
1/24/13 Jesse F. Hughes
1/24/13 mueckenh@rz.fh-augsburg.de
1/24/13 Jesse F. Hughes
1/24/13 mueckenh@rz.fh-augsburg.de
1/24/13 Virgil
1/24/13 Jesse F. Hughes
1/24/13 Jesse F. Hughes
1/24/13 Virgil
1/24/13 mueckenh@rz.fh-augsburg.de
1/24/13 Jesse F. Hughes
1/24/13 mueckenh@rz.fh-augsburg.de
1/24/13 Virgil
1/24/13 Jesse F. Hughes
1/25/13 mueckenh@rz.fh-augsburg.de
1/25/13 Virgil
1/25/13 mueckenh@rz.fh-augsburg.de
1/25/13 Virgil
1/25/13 mueckenh@rz.fh-augsburg.de
1/25/13 Virgil
1/25/13 Jesse F. Hughes
1/25/13 mueckenh@rz.fh-augsburg.de
1/25/13 Jesse F. Hughes
1/25/13 mueckenh@rz.fh-augsburg.de
1/25/13 Jesse F. Hughes
1/25/13 mueckenh@rz.fh-augsburg.de
1/25/13 Virgil
1/25/13 Jesse F. Hughes
1/25/13 Virgil
1/26/13 gus gassmann
1/26/13 Virgil
1/27/13 gus gassmann
1/26/13 mueckenh@rz.fh-augsburg.de
1/26/13 Jesse F. Hughes
1/26/13 mueckenh@rz.fh-augsburg.de
1/26/13 Jesse F. Hughes
1/27/13 mueckenh@rz.fh-augsburg.de
1/27/13 Virgil
1/27/13 mueckenh@rz.fh-augsburg.de
1/27/13 Virgil
1/27/13 Jesse F. Hughes
1/27/13 mueckenh@rz.fh-augsburg.de
1/27/13 Jesse F. Hughes
1/27/13 mueckenh@rz.fh-augsburg.de
1/27/13 Jesse F. Hughes
1/27/13 Jesse F. Hughes
1/27/13 mueckenh@rz.fh-augsburg.de
1/27/13 Jesse F. Hughes
1/27/13 mueckenh@rz.fh-augsburg.de
1/27/13 Jesse F. Hughes
1/27/13 mueckenh@rz.fh-augsburg.de
1/27/13 Jesse F. Hughes
1/27/13 mueckenh@rz.fh-augsburg.de
1/27/13 Jesse F. Hughes
1/28/13 mueckenh@rz.fh-augsburg.de
1/28/13 mueckenh@rz.fh-augsburg.de
1/28/13 Virgil
1/28/13 mueckenh@rz.fh-augsburg.de
1/28/13 mueckenh@rz.fh-augsburg.de
1/28/13 Jesse F. Hughes
1/29/13 mueckenh@rz.fh-augsburg.de
1/29/13 Jesse F. Hughes
1/29/13 mueckenh@rz.fh-augsburg.de
1/29/13 Virgil
1/29/13 Virgil
1/30/13 mueckenh@rz.fh-augsburg.de
1/30/13 Virgil
1/30/13 mueckenh@rz.fh-augsburg.de
1/30/13 Virgil
1/31/13 mueckenh@rz.fh-augsburg.de
1/31/13 Virgil
1/31/13 mueckenh@rz.fh-augsburg.de
1/31/13 Virgil
1/28/13 mueckenh@rz.fh-augsburg.de
1/28/13 Virgil
1/29/13 mueckenh@rz.fh-augsburg.de
1/29/13 Virgil
1/29/13 mueckenh@rz.fh-augsburg.de
1/29/13 Virgil
1/30/13 mueckenh@rz.fh-augsburg.de
1/30/13 Virgil
1/27/13 Virgil
1/27/13 Virgil
1/27/13 Virgil
1/27/13 Virgil
1/27/13 Virgil
1/27/13 Virgil
1/26/13 Virgil
1/26/13
1/25/13 Virgil
1/25/13 Virgil
1/25/13 mueckenh@rz.fh-augsburg.de
1/25/13 Virgil
1/24/13 Virgil
1/24/13 Virgil
1/24/13 mueckenh@rz.fh-augsburg.de
1/24/13 Virgil
1/25/13 mueckenh@rz.fh-augsburg.de
1/25/13 Virgil
1/25/13 mueckenh@rz.fh-augsburg.de
1/25/13 Virgil
1/23/13 Virgil
1/23/13 Virgil
1/23/13 David Bernier
1/23/13 Jesse F. Hughes
1/23/13 mueckenh@rz.fh-augsburg.de
1/23/13 Jesse F. Hughes
1/23/13 Virgil
1/23/13 Virgil
1/23/13 ross.finlayson@gmail.com
1/23/13 Aatu Koskensilta
1/23/13 mueckenh@rz.fh-augsburg.de
1/23/13 Virgil
1/23/13 Aatu Koskensilta
1/23/13 mueckenh@rz.fh-augsburg.de
1/23/13 Virgil
1/23/13 Aatu Koskensilta
1/23/13 mueckenh@rz.fh-augsburg.de
1/23/13 Virgil | 2,336 | 5,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-17 | longest | en | 0.88087 |
https://blogs.fangraphs.com/the-mathematical-improbability-of-parity/ | 1,621,219,821,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991557.62/warc/CC-MAIN-20210517023244-20210517053244-00096.warc.gz | 163,195,237 | 29,570 | # The Mathematical Improbability of Parity
Here’s something you hear a lot that also has the benefit of being true: baseball is a sport of haves and have nots. There are super teams scattered around the league, the Dodgers and Yankees and Astros of the world. There are plenty of teams that aren’t trying to compete this year; the Tigers and Royals spring to mind, but it’s not like there aren’t others.
In a technical sense, however, the league achieved a rare degree of parity over the weekend — and as we all know, being technically correct is the best kind of correct. After each team played three games, the entire league stood at either 1-2 or 2-1, with fifteen teams apiece in each camp. In that odd, specific sense, this is one of the best years ever for parity in baseball.
Come again? Per no less an authority than MLB.com, this is the first time in the last 66 years that no team started 3-0 in their first three games. In that contrived sense, then, this is the most parity since 1954. Given that it was far easier to have no team start 3-0 then (there were only 16 teams), you would even be justified in saying that this was the most balanced start of all time.
That sounds, without putting too much thought into it, very impressive. 1954! Man hadn’t landed on the moon. The LOOGY hadn’t been invented, or the personal computer. It was a very different time.
As fun as it would be to leave it at “Wow, that was crazy,” I thought I’d spoil the fun with a little math. First things first — what if we think every team is evenly matched? Let’s leave home field advantage out for now — we’re just approximating anyway, and that makes the math cleaner. The math for a single series is easy; if each game is a coin flip, all we need to do is find the odds of getting either three heads or three tails in a row.
That works out to 0.5*0.5*0.5, or 0.125, for each outcome. In other words, any given three-game series between evenly matched teams ends in a sweep 25% of the time; with each team sweeping the other 12.5% of the time. From there it’s a snap to work out the odds of it happening to the whole league.
The odds of neither team sweeping are 75%. We need that to happen in 15 straight series. Easy, peasy — 0.75 raised to the 15th power will do it. That’s 1.3%, which is indeed unlikely. To get to a 50% chance of that happening at least once, you’d need to play 54 seasons. To get to a 90% chance, you’d need to play 176 seasons. It’s unlikely, is my point.
“It’s unlikely” is always going to be our answer. But let’s get a little more precise in our estimation. Every team isn’t a coin flip against every other team. The Dodgers, for example — the juggernaut Dodgers, who added no less a presence than Mookie Betts to an already loaded squad — faced off against the rebuilding Giants. Even ignoring home field advantage, that’s hardly an even matchup.
To estimate each team’s chances in each series, we can use a neat little mathematical trick called the odds ratio. Odds ratio is a mathematical way of answering, essentially, this question: what happens when a .600 team plays a .400 team? The answer clearly isn’t a 60% chance at winning; a .600 team is, in theory, 60% to beat an average team, and their opponent is worse than average. The formula looks like this:
Team A Win Percentage = (A – A * B) / (A + B – 2 * A * B)
A is the first team’s winning percentage, B is the second team’s winning percentage, and the output will give you Team A’s chances of beating Team B in a single game. We’re ignoring starting pitching, naturally, which is just the cost of doing business at such a high level; if you want to zoom in and take each matchup into account, you no doubt can, but that’s more detail than we’re getting into today.
Okay — let’s use the real Dodgers and Giants win percentages to work out the odds. Not their actual records this year, of course — that’s wrong on, well, tons of levels. First of all, they’re both .500. Second of all, those games were against each other. Third of all, that doesn’t reflect their true talent. I don’t have time to explain why I don’t have time to explain the rest of the reasons it won’t work, but just trust me on this one; we’re going to need a projection.
Luckily, our very own website has the goods. On our projected standings page, you can see a “rest of season win percentage” column driven by our Depth Charts playing time and skill projections. Handily enough for us, those win percentages are against neutral opposition, which means we can plug them right into our formula.
The Dodgers are a .606 true talent team. The Giants are a .447 true talent team. Plug them into the formula, and the Dodgers are 65.5% likely to win a random game against the Giants. Armed with that, we can work out the probabilities of a sweep. The Dodgers have a 28.1% chance of sweeping, while the Giants check in at a mere 4.1%. Add them up, and that’s a 32.1% chance of a sweep occurring.
You know what’s coming next. We’re going to do the same thing for each of the 15 matchups that took place over the weekend. Away winning percentage, home winning percentage, and Shazam! We have our odds of a sweep:
Sweep Odds By Series
Away Home Home WP Away Sweep % Home Sweep % Sweep %
Giants Dodgers 65.6% 4.1% 28.2% 32.3%
Yankees Nationals 49.0% 13.3% 11.8% 25.0%
Pirates Cardinals 56.4% 8.3% 17.9% 26.2%
Braves Mets 48.3% 13.8% 11.3% 25.1%
Tigers Reds 61.3% 5.8% 23.0% 28.8%
Blue Jays Rays 58.4% 7.2% 20.0% 27.1%
Marlins Phillies 56.7% 8.1% 18.2% 26.3%
Brewers Cubs 51.7% 11.3% 13.8% 25.1%
Royals Indians 58.9% 6.9% 20.5% 27.4%
Orioles Red Sox 62.9% 5.1% 24.8% 30.0%
Rockies Rangers 49.1% 13.2% 11.8% 25.0%
Twins White Sox 46.3% 15.5% 9.9% 25.4%
Diamondbacks Padres 52.3% 10.9% 14.3% 25.2%
Mariners Astros 65.9% 4.0% 28.6% 32.6%
Angels Athletics 52.2% 10.9% 14.2% 25.1%
From there, it’s easy. The odds of not having a sweep in a given series are just one minus the odds of a sweep. We multiply the odds of each series ending without a sweep together to figure out the chances of no one being 3-0 after three games. It works out to a 0.863% chance. To put it in the same terms as earlier, you’d need to play 80 seasons to have a 50% chance of seeing this at least once. To get to a 90% chance of seeing it at least once, you’d need to play 266 seasons. It’s not very likely, is the point.
You might notice something interesting about the odds of each series ending in a sweep. The lowest odds anywhere were 25%. That also happens to be the chances of an evenly matched series ending in a sweep, and that’s no coincidence. Take a look at the range of sweep probabilities for a given home team winning percentage:
The graph is symmetrical, of course, because we don’t care which team is doing the sweeping. The key is that closer matchups lead to lower chances of a sweep. That’s intuitive, but it’s still neat to me. It might seem like the added prospects of a sweep by a better underdog would offset the lost sweep percentage from a less dominant favorite, but it simply doesn’t work that way.
Think of it like this: imagine a 10-by-10 grid, with each square representing a possible outcome of a two-game series. The first game will run along the x axis at the bottom, and the second along the y axis along the left side. The left and lower sides represent home team wins, the upper and right sides away team wins. It looks like so:
Odds of a Two-Game Sweep
Odds 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%
100% 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2 0-2
90% 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2 0-2
80% 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2 0-2
70% 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2 0-2
60% 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2 0-2
50% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
40% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
30% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
20% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
10% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
I’ve shaded in the squares that represent sweeps by either team. Next, let’s make the home team 60% to win the second game:
Odds of a Two-Game Sweep
Odds 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%
100% 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2 0-2
90% 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2 0-2
80% 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2 0-2
70% 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2 0-2
60% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
50% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
40% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
30% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
20% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
10% 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1 1-1
As you can see, the same number of cells remain shaded. You could move this to 80%, or 40%, and get the same result. Here, we added five cells to the home team’s sweeps while subtracting five from the away team’s. Perfectly balanced, you might say.
But watch what happens when we match the first game’s winning percentage with the second game:
Odds of a Two-Game Sweep
Odds 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%
100% 1-1 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2
90% 1-1 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2
80% 1-1 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2
70% 1-1 1-1 1-1 1-1 1-1 1-1 0-2 0-2 0-2 0-2
60% 2-0 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1
50% 2-0 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1
40% 2-0 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1
30% 2-0 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1
20% 2-0 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1
10% 2-0 2-0 2-0 2-0 2-0 2-0 1-1 1-1 1-1 1-1
We just got rid of four sweep cells for the visiting team while adding six for the home team. That’s more total sweeps, and since the move is symmetrical, any move away from a 50% winning percentage does some version of this, even if it’s to a lower number. Move from a 50% to 51% winner percentage? It’s this same thing, only with a 100-by-100 grid. A three-game series? That’s a three-dimensional grid, but it’s a grid all the same, and a move off of 50% will always increase the proportion of shaded parts.
What does any of this matter in real life? Oh, it doesn’t. It doesn’t at all. You don’t get to know teams’ winning percentages before each series, you don’t get to draw these perfect boxes that show win probabilities. Different pitchers muddle with the math. There’s nothing really actionable about this discovery. But it’s neat, and I spent a bit thinking about it, and I like neat math tricks. So here we are! With a neat math trick about the shocking parity in this weird season.
Ben is a writer at FanGraphs. He can be found on Twitter @_Ben_Clemens.
Member
I was hoping this was about some of the parity account’s that post on this web sight. Could of used a laugh with every thing going on this world right now.
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24
25
26/**
27\page weighted_statistics Weighted Statistics
28
29\section Introduction
30There are several different reasons why a statistical analysis needs
31to adjust for weighting. In literature reasons are mainly diveded in
32to groups.
33
34The first group is when some of the measurements are known to be more
35precise than others. The more precise a measurement is, the larger
36weight it is given. The simplest case is when the weight are given
37before the measurements and they can be treated as deterministic. It
38becomes more complicated when the weight can be determined not until
39afterwards, and even more complicated if the weight depends on the
40value of the observable.
41
42The second group of situations is when calculating averages over one
43distribution and sampling from another distribution. Compensating for
44this discrepency weights are introduced to the analysis. A simple
45example may be that we are interviewing people but for economical
46reasons we choose to interview more people from the city than from the
47countryside. When summarizing the statistics the answers from the city
48are given a smaller weight. In this example we are choosing the
49proportions of people from countryside and people from city being
50intervied. Hence, we can determine the weights before and consider
51them to be deterministic. In other situations the proportions are not
52deterministic, but rather a result from the sampling and the weights
53must be treated as stochastic and only in rare situations the weights
54can be treated as independent of the observable.
55
56Since there are various origins for a weight occuring in a statistical
57analysis, there are various ways to treat the weights and in general
58the analysis should be tailored to treat the weights correctly. We
59have not chosen one situation for our implementations, so see specific
60function documentation for what assumtions are made. Though, common
61for implementations are the following:
62
63 - Setting all weights to unity yields the same result as the
64non-weighted version.
65 - Rescaling the weights does not change any function.
66 - Setting a weight to zero is equivalent to removing the data point.
67
68An important case is when weights are binary (either 1 or 0). Then we
69get the same result using the weighted version as using the data with
70weight not equal to zero and the non-weighted version. Hence, using
71binary weights and the weighted version missing values can be treated
72in a proper way.
73
74\section AveragerWeighted
75
76
77
78\subsection Mean
79
80For any situation the weight is always designed so the weighted mean
81is calculated as \f$m=\frac{\sum w_ix_i}{\sum w_i} \f$, which obviously
82fulfills the conditions above.
83
84
85
86In the case of varying measurement error, it could be motivated that
87the weight shall be \f$w_i = 1/\sigma_i^2 \f$. We assume measurement error
88to be Gaussian and the likelihood to get our measurements is
89\f$L(m)=\prod 90(2\pi\sigma_i^2)^{-1/2}e^{-\frac{(x_i-m)^2}{2\sigma_i^2}} \f$. We
91maximize the likelihood by taking the derivity with respect to \f$m \f$ on
92the logarithm of the likelihood \f$\frac{d\ln L(m)}{dm}=\sum 93\frac{x_i-m}{\sigma_i^2} \f$. Hence, the Maximum Likelihood method yields
94the estimator \f$m=\frac{\sum w_i/\sigma_i^2}{\sum 1/\sigma_i^2} \f$.
95
96
97\subsection Variance
98In case of varying variance, there is no point estimating a variance
99since it is different for each data point.
100
101Instead we look at the case when we want to estimate the variance over
102\f$f\f$ but are sampling from \f$f' \f$. For the mean of an observable \f$O \f$ we
103have \f$\widehat O=\sum\frac{f}{f'}O_i=\frac{\sum w_iO_i}{\sum 104w_i} \f$. Hence, an estimator of the variance of \f$X \f$ is
105
106\f$107s^2 = <X^2>-<X>^2= 108\f$
109
110\f$111 = \frac{\sum w_ix_i^2}{\sum w_i}-\frac{(\sum w_ix_i)^2}{(\sum w_i)^2}= 112\f$
113
114\f$115 = \frac{\sum w_i(x_i^2-m^2)}{\sum w_i}= 116\f$
117
118\f$119 = \frac{\sum w_i(x_i^2-2mx_i+m^2)}{\sum w_i}= 120\f$
121
122\f$123 = \frac{\sum w_i(x_i-m)^2}{\sum w_i} 124\f$
125
126This estimator fulfills that it is invariant under a rescaling and
127having a weight equal to zero is equivalent to removing the data
128point. Having all weights equal to unity we get \f$\sigma=\frac{\sum 129(x_i-m)^2}{N} \f$, which is the same as returned from Averager. Hence,
130this estimator is slightly biased, but still very efficient.
131
132\subsection standard_error Standard Error
133The standard error squared is equal to the expexted squared error of
134the estimation of \f$m\f$. The squared error consists of two parts, the
135variance of the estimator and the squared bias:
136
137\f$138<m-\mu>^2=<m-<m>+<m>-\mu>^2= 139\f$
140\f$141<m-<m>>^2+(<m>-\mu)^2 142\f$.
143
144In the case when weights are included in analysis due to varying
145measurement errors and the weights can be treated as deterministic, we
146have
147
148\f$149Var(m)=\frac{\sum w_i^2\sigma_i^2}{\left(\sum w_i\right)^2}= 150\f$
151\f$152\frac{\sum w_i^2\frac{\sigma_0^2}{w_i}}{\left(\sum w_i\right)^2}= 153\f$
154\f$155\frac{\sigma_0^2}{\sum w_i}, 156\f$
157
158where we need to estimate \f$\sigma_0^2 \f$. Again we have the likelihood
159
160\f$161L(\sigma_0^2)=\prod\frac{1}{\sqrt{2\pi\sigma_0^2/w_i}}\exp{(-\frac{w_i(x-m)^2}{2\sigma_0^2})} 162\f$
163and taking the derivity with respect to
164\f$\sigma_o^2\f$,
165
166\f$167\frac{d\ln L}{d\sigma_i^2}= 168\f$
169\f$170\sum -\frac{1}{2\sigma_0^2}+\frac{w_i(x-m)^2}{2\sigma_0^2\sigma_o^2} 171\f$
172
173which
174yields an estimator \f$\sigma_0^2=\frac{1}{N}\sum w_i(x-m)^2 \f$. This
175estimator is not ignoring weights equal to zero, because deviation is
176most often smaller than the expected infinity. Therefore, we modify
177the expression as follows \f$\sigma_0^2=\frac{\sum w_i^2}{\left(\sum 178w_i\right)^2}\sum w_i(x-m)^2\f$ and we get the following estimator of
179the variance of the mean \f$\sigma_0^2=\frac{\sum w_i^2}{\left(\sum 180w_i\right)^3}\sum w_i(x-m)^2\f$. This estimator fulfills the conditions
181above: adding a weight zero does not change it: rescaling the weights
182does not change it, and setting all weights to unity yields the same
183expression as in the non-weighted case.
184
185In a case when it is not a good approximation to treat the weights as
186deterministic, there are two ways to get a better estimation. The
187first one is to linearize the expression \f$\left<\frac{\sum 188w_ix_i}{\sum w_i}\right>\f$. The second method when the situation is
189more complicated is to estimate the standard error using a
190bootstrapping method.
191
192\section AveragerPairWeighted
193Here data points come in pairs (x,y). We are sampling from \f$f'_{XY}\f$
194but want to measure from \f$f_{XY}\f$. To compensate for this decrepency,
195averages of \f$g(x,y)\f$ are taken as \f$\sum \frac{f}{f'}g(x,y)\f$. Even
196though, \f$X\f$ and \f$Y\f$ are not independent \f$(f_{XY}\neq f_Xf_Y)\f$ we
197assume that we can factorize the ratio and get \f$\frac{\sum 198w_xw_yg(x,y)}{\sum w_xw_y}\f$
199\subsection Covariance
200Following the variance calculations for AveragerWeighted we have
201\f$Cov=\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sum w_xw_y}\f$ where
202\f$m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}\f$
203
204\subsection Correlation
205
206As the mean is estimated as
207\f$208m_x=\frac{\sum w_xw_yx}{\sum w_xw_y} 209\f$,
210the variance is estimated as
211\f$212\sigma_x^2=\frac{\sum w_xw_y(x-m_x)^2}{\sum w_xw_y} 213\f$.
214As in the non-weighted case we define the correlation to be the ratio
215between the covariance and geometrical average of the variances
216
217\f$218\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sqrt{\sum w_xw_y(x-m_x)^2\sum 219w_xw_y(y-m_y)^2}} 220\f$.
221
222
223This expression fulfills the following
224 - Having N equal weights the expression reduces to the non-weighted expression.
225 - Adding a pair of data, in which one weight is zero is equivalent
226to ignoring the data pair.
227 - Correlation is equal to unity if and only if \f$x\f$ is equal to
228\f$y\f$. Otherwise the correlation is between -1 and 1.
229
230\section Score
231
232\subsection Pearson
233
234\f$\frac{\sum w(x-m_x)(y-m_y)}{\sqrt{\sum w(x-m_x)^2\sum w(y-m_y)^2}}\f$.
235
236See AveragerPairWeighted correlation.
237
238\subsection ROC
239
240An interpretation of the ROC curve area is the probability that if we
241take one sample from class \f$+\f$ and one sample from class \f$-\f$, what is
242the probability that the sample from class \f$+\f$ has greater value. The
243ROC curve area calculates the ratio of pairs fulfilling this
244
245\f$246\frac{\sum_{\{i,j\}:x^-_i<x^+_j}1}{\sum_{i,j}1}. 247\f$
248
249An geometrical interpretation is to have a number of squares where
250each square correspond to a pair of samples. The ROC curve follows the
251border between pairs in which the samples from class \f$+\f$ has a greater
252value and pairs in which this is not fulfilled. The ROC curve area is
253the area of those latter squares and a natural extension is to weight
254each pair with its two weights and consequently the weighted ROC curve
255area becomes
256
257\f$258\frac{\sum_{\{i,j\}:x^-_i<x^+_j}w^-_iw^+_j}{\sum_{i,j}w^-_iw^+_j} 259\f$
260
261This expression is invariant under a rescaling of weight. Adding a
262data value with weight zero adds nothing to the exprssion, and having
263all weight equal to unity yields the non-weighted ROC curve area.
264
265\subsection tScore
266
267Assume that \f$x\f$ and \f$y\f$ originate from the same distribution
268\f$N(\mu,\sigma_i^2)\f$ where \f$\sigma_i^2=\frac{\sigma_0^2}{w_i}\f$. We then
269estimate \f$\sigma_0^2\f$ as
270\f$271\frac{\sum w(x-m_x)^2+\sum w(y-m_y)^2} 272{\frac{\left(\sum w_x\right)^2}{\sum w_x^2}+ 273\frac{\left(\sum w_y\right)^2}{\sum w_y^2}-2} 274\f$
275The variance of difference of the means becomes
276\f$277Var(m_x)+Var(m_y)=\\\frac{\sum w_i^2Var(x_i)}{\left(\sum 278w_i\right)^2}+\frac{\sum w_i^2Var(y_i)}{\left(\sum w_i\right)^2}= 279\frac{\sigma_0^2}{\sum w_i}+\frac{\sigma_0^2}{\sum w_i}, 280\f$
281and consequently the t-score becomes
282\f$283\frac{\sum w(x-m_x)^2+\sum w(y-m_y)^2} 284{\frac{\left(\sum w_x\right)^2}{\sum w_x^2}+ 285\frac{\left(\sum w_y\right)^2}{\sum w_y^2}-2} 286\left(\frac{1}{\sum w_i}+\frac{1}{\sum w_i}\right), 287\f$
288
289For a \f$w_i=w\f$ we this expression get condensed down to
290\f$291\frac{w\sum (x-m_x)^2+w\sum (y-m_y)^2} 292{n_x+n_y-2} 293\left(\frac{1}{wn_x}+\frac{1}{wn_y}\right), 294\f$
295in other words the good old expression as for non-weighted.
296
297\subsection FoldChange
298Fold-Change is simply the difference between the weighted mean of the
299two groups \f$\frac{\sum w_xx}{\sum w_x}-\frac{\sum w_yy}{\sum w_y}\f$
300
301\subsection WilcoxonFoldChange
302Taking all pair samples (one from class \f$+\f$ and one from class \f$-\f$)
303and calculating the weighted median of the distances.
304
305\section Distance
306
307A \ref concept_distance measures how far apart two ranges are. A Distance should
308preferably meet some criteria:
309
310 - It is symmetric, \f$d(x,y) = d(y,x) \f$, that is distance from \f$311 x \f$ to \f$y \f$ equals the distance from \f$y \f$ to \f$x \f$.
312 - Zero self-distance: \f$d(x,x) = 0 \f$
313 - Triangle inequality: \f$d(x,z) \le d(x,y) + d(y,z) \f$
314
315\subsection weighted_distance Weighted Distance
316
317Weighted Distance is an extension of usual unweighted distances, in
318which each data point is accompanied with a weight. A weighted
319distance should meet some criteria:
320
321 - Having all unity weights should yield the unweighted case.
322 - Rescaling the weights, \f$w_i = Cw_i \f$, does not change the distance.
323 - Having a \f$w_x = 0 \f$ the distance should ignore corresponding
324 \f$x \f$, \f$y \f$, and \f$w_y \f$.
325 - A zero weight should not result in a very different distance than a
326 small weight, in other words, modifying a weight should change the
327 distance in a continuous manner.
328 - The duplicate property. If data is coming in duplicate such that
329 \f$x_{2i}=x_{2i+1} \f$, then the case when \f$w_{2i}=w_{2i+1} \f$
330 should equal to if you set \f$w_{2i}=0 \f$.
331
332The last condition, duplicate property, implies that setting a weight
333to zero is not equivalent to removing the data point. This behavior is
334sensible because otherwise we would have a bias towards having ranges
335with small weights being close to other ranges. For a weighted
336distance, meeting these criteria, it might be difficult to show that
337the triangle inequality is fulfilled. For most algorithms the triangle
338inequality is not essential for the distance to work properly, so if
339you need to choose between fulfilling triangle inequality and these
340latter criteria it is preferable to meet the latter criteria.
341
342In test/distance_test.cc there are tests for testing these properties.
343
344\section Kernel
345\subsection polynomial_kernel Polynomial Kernel
346The polynomial kernel of degree \f$N\f$ is defined as \f$(1+<x,y>)^N\f$, where
347\f$<x,y>\f$ is the linear kernel (usual scalar product). For the weighted
348case we define the linear kernel to be
349\f$<x,y>=\frac{\sum {w_xw_yxy}}{\sum{w_xw_y}}\f$ and the
350polynomial kernel can be calculated as before
351\f$(1+<x,y>)^N\f$.
352
353\subsection gaussian_kernel Gaussian Kernel
354We define the weighted Gaussian kernel as \f$\exp\left(-N\frac{\sum 355w_xw_y(x-y)^2}{\sum w_xw_y}\right)\f$.
356
357\section Regression
358\subsection Naive
359\subsection Linear
360We have the model
361
362\f$363y_i=\alpha+\beta (x-m_x)+\epsilon_i, 364\f$
365
366where \f$\epsilon_i\f$ is the noise. The variance of the noise is
367inversely proportional to the weight,
368\f$Var(\epsilon_i)=\frac{\sigma^2}{w_i}\f$. In order to determine the
369model parameters, we minimimize the sum of quadratic errors.
370
371\f$372Q_0 = \sum \epsilon_i^2 373\f$
374
375Taking the derivity with respect to \f$\alpha\f$ and \f$\beta\f$ yields two conditions
376
377\f$378\frac{\partial Q_0}{\partial \alpha} = -2 \sum w_i(y_i - \alpha - 379\beta (x_i-m_x)=0 380\f$
381
382and
383
384\f$\frac{\partial Q_0}{\partial \beta} = -2 \sum 385w_i(x_i-m_x)(y_i-\alpha-\beta(x_i-m_x)=0 386\f$
387
388or equivalently
389
390\f$391\alpha = \frac{\sum w_iy_i}{\sum w_i}=m_y 392\f$
393
394and
395
396\f$\beta=\frac{\sum w_i(x_i-m_x)(y-m_y)}{\sum 397w_i(x_i-m_x)^2}=\frac{Cov(x,y)}{Var(x)} 398\f$
399
400Note, by having all weights equal we get back the unweighted
401case. Furthermore, we calculate the variance of the estimators of
402\f$\alpha\f$ and \f$\beta\f$.
403
404\f$405\textrm{Var}(\alpha )=\frac{w_i^2\frac{\sigma^2}{w_i}}{(\sum w_i)^2}= 406\frac{\sigma^2}{\sum w_i} 407\f$
408
409and
410\f$411\textrm{Var}(\beta )= \frac{w_i^2(x_i-m_x)^2\frac{\sigma^2}{w_i}} 412{(\sum w_i(x_i-m_x)^2)^2}= 413\frac{\sigma^2}{\sum w_i(x_i-m_x)^2} 414\f$
415
416Finally, we estimate the level of noise, \f$\sigma^2\f$. Inspired by the
417unweighted estimation
418
419\f$420s^2=\frac{\sum (y_i-\alpha-\beta (x_i-m_x))^2}{n-2} 421\f$
422
423we suggest the following estimator
424
425\f$s^2=\frac{\sum w_i(y_i-\alpha-\beta (x_i-m_x))^2}{\sum 426w_i-2\frac{\sum w_i^2}{\sum w_i}} \f$
427
428*/
429
430
431
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# January 10, 2001 13:09
g65-ch2
L IMITS AND
C ONTINUITY
## The problem of defining and calculating instantaneous rates
such as speed and acceleration attracted almost all the
mathematicians of the seventeenth century.
Morris Kline
he development of calculus in the seventeenth century by Newton and Leibniz provided scientists with their
first real understanding of what is meant by an instantaneous rate of change such as velocity and acceleration.
Once the idea was understood conceptually, efficient computational methods followed, and science took a quantum
leap forward. The fundamental building block on which
rates of change rest is the concept of a limit, an idea that
is so important that all other calculus concepts are now
based on it.
In this chapter we will develop the concept of a limit in
stages, proceeding from an informal, intuitive notion to a
precise mathematical definition. We will also develop theorems and procedures for calculating limits, and we will
conclude the chapter by using the limits to study continuous curves.
108
g65-ch2
## 2.1 LIMITS (AN INTUITIVE APPROACH)
The concept of a limit is the fundamental building block on which all other calculus
concepts are based. In this section we will study limits informally, with the goal of
developing an intuitive feel for the basic ideas. In the following three sections we
will focus on the computational methods and precise definitions.
## INSTANTANEOUS VELOCITY AND
THE SLOPE OF A CURVE
Recall from Formula (11) of Section 1.5 that if a particle moves along an s-axis, then the
average velocity vave over the time interval from t0 to t1 is defined as
vave =
=
pe
(t1, s1)
Slo
s = f (t)
s1 s0
(t0, s0)
t1 t0
t
t0
t1
v av
s1 s0
s
=
t
t1 t0
where s0 and s1 are the s-coordinates of the particle at times t0 and t1 , respectively. Geometrically, vave is the slope of the line joining the points (t0 , s0 ) and (t1 , s1 ) on the position
versus time curve for the particle (Figure 2.1.1).
Suppose, however, that we are not interested in average velocity over a time interval,
but rather the velocity vinst at a specific instant in time. It is not a simple matter of applying
Formula (1), since the displacement and the elapsed time in an instant are both zero. However, intuition suggests that the velocity at an instant t = t0 can be approximated by finding
the position of the particle at a time t1 just before, or just after, time t0 and computing the
average velocity over the brief time interval between the two moments. That is,
Figure 2.1.1
vinst vave =
(1)
s1 s0
t1 t 0
(2)
## provided t = t1 t0 is small. Moreover, if we are able to make very precise measurements,
the closer t1 is to t0 , the better vave approximates vinst . That is, as we sample at times t1 ,
closer and closer to t0 , vave approaches a limiting value that we understand to be vinst .
Example 1 Suppose that a ball is thrown vertically upward and the height in feet of the
ball t seconds after its release is modeled by the function
s(t) = 16t 2 + 29t + 6,
0t 2
What is a reasonable estimate for the instantaneous velocity of the ball at time t = 0.5 s?
Solution. At any time 0 t 2 we may envision the height s(t) of the ball as a position
on a (vertical) s-axis, where s = 0 corresponds to ground level (Figure 2.1.2). The height
of the ball at time t = 0.5 s is s(0.5) = 16.5 ft, and the height of the ball 0.01 s later is
s(0.51) = 16.6284 ft. Therefore, the average velocity of the ball over the time interval from
t = 0.5 to t = 0.51 is
0.1284
16.6284 16.5
vave =
=
= 12.84 ft/s
0.51 0.5
0.01
Figure 2.1.2
Similarly, the height of the ball 0.49 s after its release is s(0.49) = 16.3684 ft, and the
average velocity of the ball over the time interval from t = 0.49 to t = 0.5 is
vave =
0.1316
16.3684 16.5
=
= 13.16 ft/s
0.49 0.5
0.01
Consequently, we would expect the instantaneous velocity of the ball at time t = 0.5 to be
between 12.84 ft/s and 13.16 ft/s. To improve our estimate of this instantaneous velocity,
we can compute the average velocity
vave (t1 ) =
s(t1 ) 16.5
16t12 + 29t1 + 6 16.5
16t12 + 29t1 10.5
=
=
t1 0.5
t1 0.5
t1 0.5
for values of t1 even closer to 0.5. Table 2.1.1 displays the results of several such computa-
g65-ch2
2.1
109
Table 2.1.1
time t1 (s)
vave(t1) =
## 16t12 + 29t1 10.5
(ft/s)
t1 0.5
0.5010
0.5005
0.5001
0.4999
0.4995
0.4990
12.9840
12.9920
12.9984
13.0016
13.0080
13.0160
tions. It appears from these computations that a reasonable estimate for the instantaneous
velocity of the ball at time t = 0.5 s is 13 ft/s.
s
=v
av
Slo
pe
=v
ins
e
lop
t0
t1
Figure 2.1.3
LIMITS
## The domain of the height function s(t) = 16t 2 + 29t + 6 in Example
1 is the closed interval [0, 2]. Why do we not consider values of t less than 0 or greater than
2 for this function? In Table 2.1.1, why is there not a value of vave (t1 ) for t1 = 0.5?
We can interpret vinst geometrically from the interpretation of vave as the slope of the
line joining the points (t0 , s0 ) and (t1 , s1 ) on the position versus time curve for the particle.
When t = t1 t0 is small, the points (t0 , s0 ) and (t1 , s1 ) are very close to each other on
the curve. As the sampling point (t1 , s1 ) is selected closer to our anchoring point (t0 , s0 ),
the slope vave more nearly approximates what we might reasonably call the slope of the
position curve at time t = t0 . Thus, vinst can be viewed as the slope of the position curve at
time t = t0 (Figure 2.1.3). We will explore this connection more fully in Section 3.1.
In Example 1 it appeared that choosing values of t1 close to (but not equal to) 0.5 resulted
in values of vave (t1 ) that were close to 13. One way of describing this behavior is to say that
the limiting value of vave (t1 ) as t1 approaches 0.5 is 13 or, equivalently, that 13 is the limit
of vave (t1 ) as t1 approaches 0.5. More generally, we will see that the concept of the limit of
a function provides a foundation for the tools of calculus. Thus, it is appropriate to start a
study of calculus by focusing on the limit concept itself.
The most basic use of limits is to describe how a function behaves as the independent
variable approaches a given value. For example, let us examine the behavior of the function
f(x) = x 2 x + 1
for x-values closer and closer to 2. It is evident from the graph and table in Figure 2.1.4 that
the values of f(x) get closer and closer to 3 as values of x are selected closer and closer
to 2 on either the left or the right side of 2. We describe this by saying that the limit of
x 2 x + 1 is 3 as x approaches 2 from either side, and we write
lim (x 2 x + 1) = 3
x 2
(3)
Observe that in our investigation of limx 2 (x 2 x + 1) we are only concerned with the
values of f(x) near x = 2 and not the value of f(x) at x = 2.
This leads us to the following general idea.
2.1.1 LIMITS (AN INFORMAL VIEW). If the values of f(x) can be made as close as
we like to L by taking values of x sufficiently close to a (but not equal to a), then we
write
lim f(x) = L
(4)
x a
110
g65-ch2
## Limits and Continuity
y
f (x)
y = f (x) = x 2 x + 1
3
f (x)
1.0
1.5
1.9
1.95
1.99
1.995
1.999
2.001
2.005
2.01
2.05
2.1
2.5
3.0
Left side
Right side
Figure 2.1.4
## Equation (4) is also commonly written as
f(x) L as x a
With this notation we can express (3) as
x 2 x + 1 3 as x 2
In order to investigate limx a f(x), we ask ourselves the question, If x is close to,
but different from, a, is there a particular number to which f(x) is close? This question
presumes that the function f is defined everywhere near a, in other words, that f is
defined at all points x in some open interval containing a, except possibly at x = a. The
value of f at a, if it exists at all, is not relevant to the determination of limx a f(x). Many
important applications of the limit concept involve contexts in which the domain of the
function excludes a. Indeed, our discussion of instantaneous velocity concluded that vinst
could be interpreted as a limit of the average velocities, even though the average velocity
at an instant is not defined.
The process of determining a limit generally involves a discovery phase, followed by
a verification phase. The discovery phase begins with sampled x-values, and ends with
a conjecture for the limit. Figure 2.1.4 illustrates the discovery phase for the problem of
finding the value of limx 2 (x 2 x + 1). We sampled values for x near 2 and found that
the corresponding values of f(x) were close to 3. Indeed, values of x nearer to 2 produced
values of f(x) closer to 3. Our conjecture that limx 2 (x 2 x + 1) = 3 concluded the
discovery phase for this limit. However, a complete treatment of any limit also involves a
verification phase in which it is shown that the conjectured limit is actually correct. For
example, consider our conjecture that limx 2 (x 2 x + 1) = 3. We can only sample a
relatively few values of x near 2, even by using a graphing utility. We cannot sample all
values of x near 2, for no matter how close to 2 we take an x-value, there are infinitely
many values of x nearer yet to 2. To verify that limx 2 (x 2 x + 1) is indeed 3, we need
to resort to an analysis that can overcome this dilemma. This analysis will require a more
mathematically precise definition of limit and is the focus of Section 2.4. In this section,
we concentrate on the discovery phase for limit problems.
Example 2 Make a conjecture about the value of the limit
x
lim
x 0
x+11
(5)
g65-ch2
2.1
111
## Solution. Observe that the function
f(x) =
x+11
is not defined at x = 0. However, f is defined for x > 1, x = 0, so the domain of f contains values of x everywhere near 0. Table 2.1.2 shows samples of x-values approaching
0 from the left side and from the right side. In both cases the values of f(x), calculated to
six decimal places, appear to get closer and closer to 2, and hence we conjecture that
x
lim
=2
(6)
x 0
x+11
A graphing utility could be used to produce Figure 2.1.5, providing more evidence in support
of our conjecture.
In the next section we will see that the graph of f(x) is identical to that
1
x
-1
Figure 2.1.5
Table 2.1.2
0.01
0.001
0.0001
0.00001
f(x)
1.994987
1.999500
1.999950
1.999995
0.00001
0.0001
0.001
0.01
2.000005
2.000050
2.000500
2.004988
Left side
Right side
## Using a graphing utility, find a window about x = 0 in which all values
of f(x) are within 0.5 of y = 2. Find a window in which all values of f(x) are within 0.1
of y = 2.
Example 3 Make a conjecture about the value of the limit
lim
x 0
sin x
x
(7)
Solution. The function f(x) = (sin x)/x is not defined at x = 0, but, as discussed previously, this has no bearing on the limit. With the help of a calculating utility set in radian
mode, we obtain the table in Figure 2.1.6.
sin x
=1
(8)
x
The result is consistent with the graph of f(x) = (sin x)/x shown in the figure. Later in this
chapter we will give a geometric argument to prove that our conjecture is correct.
lim
x 0
x
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.01
Figure 2.1.6
y=
sin x
x
0.84147
0.87036
0.89670
0.92031
0.94107
0.95885
0.97355
0.98507
0.99335
0.99833
0.99998
y
1
f(x)
y = f (x) =
sin x
x
x
## As x approaches 0 from the left
or right, f(x) approaches 1.
112
g65-ch2
## Limits and Continuity
SAMPLING PITFALLS
x
x
x
x
x
x
=
=
=
=
=
1
0.1
0.01
0.001
0.0001
.
.
.
## Use a calculating utility to sample x-values closer to 0 than in Table ??.
Does the limit change if x is in degrees?
Although numerical and graphical evidence is helpful for guessing at limits, we can be
misled by an insufficient or poorly selected sample. For example, the table in Figure 2.1.7
shows values of f(x) = sin(/x) at selected values of x on both sides of 0. The data
incorrectly suggest that
=0
lim sin
x 0
x
The fact that this is incorrect is evidenced by the graph of f shown in the figure. This graph
indicates that as x 0, the values of f oscillate between 1 and 1 with increasing rapidity,
and hence do not approach a limit. The data are deceiving because the table consists only
of sample values of x that are x-intercepts for f(x).
p
x
p
10p
100p
1000p
10,000p
.
.
.
p
f (x) = sin x
( )
sin(p) = 0
sin(10p) = 0
sin(100p) = 0
sin(1000p) = 0
sin(10,000p) = 0
.
.
.
y
1
y = sin p
x
( )
x
-1
-1
Figure 2.1.7
Numerical evidence can lead to incorrect conclusions about limits because of roundoff
error or because the sample of values used is not extensive enough to give a good indication
of the behavior of the function. Thus, when a limit is conjectured from a table of values, it
is important to look for corroborating evidence to support the conjecture.
ONE-SIDED LIMITS
y
1
x
-1
|x|
y= x
Figure 2.1.8
The limit in (4) is commonly called a two-sided limit because it requires the values of f(x)
to get closer and closer to L as values of x are taken from either side of x = a. However,
some functions exhibit different behaviors on the two sides of an x-value a, in which case
it is necessary to distinguish whether values of x near a are on the left side or on the right
side of a for purposes of investigating limiting behavior. For example, consider the function
1, x > 0
|x|
f(x) =
=
x
1, x < 0
(Figure 2.1.8). Note that x-values approaching 0 and to the right of 0 produce f(x) values
that approach 1 (in fact, they are exactly 1 for all such values of x). On the other hand, xvalues approaching 0 and to the left of 0 produce f(x) values that approach 1. We describe
these two statements by saying that the limit of f(x) = |x|/x is 1 as x approaches 0 from
the right and that the limit of f(x) = |x|/x is 1 as x approaches 0 from the left. We
denote these limits by writing
|x|
|x|
lim
= 1 and
lim
= 1
(910)
x 0+ x
x 0 x
With this notation, the superscript + indicates a limit from the right and the superscript
indicates a limit from the left.
This leads to the following general idea.
g65-ch2
2.1
## Limits (An Intuitive Approach)
113
2.1.2 ONE-SIDED LIMITS (AN INFORMAL VIEW). If the values of f(x) can be made
as close as we like to L by taking values of x sufficiently close to a (but greater than a),
then we write
lim+ f(x) = L
(11)
x a
which is read the limit of f(x) as x approaches a from the right is L. Similarly, if the
values of f(x) can be made as close as we like to L by taking values of x sufficiently
close to a (but less than a), then we write
lim f(x) = L
(12)
x a
## which is read the limit of f(x) as x approaches a from the left is L.
Expressions (11) and (12), which are called one-sided limits, are also commonly written as
f(x) L as x a +
and
f(x) L as x a
## respectively. With this notation (9) and (10) can be expressed as
|x|
|x|
1 as x 0
1 as x 0+ and
x
x
In general, there is no guarantee that a function will have a limit at a specified location. If
the values of f(x) do not get closer and closer to some single number L as x a, then
we say that the limit of f(x) as x approaches a does not exist (and similarly for one-sided
limits). For example, the two-sided limit limx 0 |x|/x does not exist because the values of
f(x) do not approach a single number as x 0; the values approach 1 from the left and
1 from the right.
In general, the following condition must be satisfied for the two-sided limit of a function
to exist.
## THE RELATIONSHIP BETWEEN
ONE-SIDED LIMITS AND
TWO-SIDED LIMITS
2.1.3 THE RELATIONSHIP BETWEEN ONE-SIDED AND TWO-SIDED LIMITS. The twosided limit of a function f(x) exists at a if and only if both of the one-sided limits exist
at a and have the same value; that is,
lim f(x) = L if and only if
lim f(x) = L = lim+ f(x)
x a
x a
x a
REMARK.
Sometimes, one or both of the one-sided limits may fail to exist (which, in
turn, implies that the two-sided limit does not exist). For example, we saw earlier that the
one-sided limits of f(x) = sin(/x) do not exist as x approaches 0 because the function
keeps oscillating between 1 and 1, failing to settle on a single value. This implies that the
two-sided limit does not exist as x approaches 0.
Example 4 For the functions in Figure 2.1.9, find the one-sided and two-sided limits at
x = a if they exist.
y
y = f (x)
y = f (x)
2
1
Figure 2.1.9
y = f (x)
114
g65-ch2
## Limits and Continuity
Solution. The functions in all three figures have the same one-sided limits as x a, since
the functions are identical, except at x = a. These limits are
lim f(x) = 3
x a+
lim f(x) = 1
and
x a
In all three cases the two-sided limit does not exist as x a because the one-sided limits
are not equal.
Example 5 For the functions in Figure 2.1.10, find the one-sided and two-sided limits
at x = a if they exist.
y
y = f (x)
y = f (x)
y = f (x)
Figure 2.1.10
## Solution. As in the preceding example, the value of f at x = a has no bearing on the
limits as x a, so that in all three cases we have
lim f(x) = 2
x a+
lim f(x) = 2
and
x a
Since the one-sided limits are equal, the two-sided limit exists and
lim f(x) = 2
x a
## INFINITE LIMITS AND VERTICAL
ASYMPTOTES
Sometimes one-sided or two-sided limits will fail to exist because the values of the function
increase or decrease indefinitely. For example, consider the behavior of the function f(x) =
1/x for values of x near 0. It is evident from the table and graph in Figure 2.1.11 that as
x-values are taken closer and closer to 0 from the right, the values of f(x) = 1/x are
positive and increase indefinitely; and as x-values are taken closer and closer to 0 from the
left, the values of f(x) = 1/x are negative and decrease indefinitely. We describe these
y
1
y= x
1
y= x
1
x
1
x
lim x =
lim x = +
x 0
x
1
x
x 0 +
0.1
0.01
0.001
0.0001 0
0.0001
0.001
0.01
0.1
10
100
1000
10,000
10,000
1000
100
10
Left side
Figure 2.1.11
Right side
g65-ch2
2.1
115
## limiting behaviors by writing
1
1
lim+ = + and
lim =
x 0 x
x 0 x
More generally:
2.1.4 INFINITE LIMITS (AN INFORMAL VIEW). If the values of f(x) increase indefinitely as x approaches a from the right or left, then we write
lim f(x) = +
or
x a+
lim f(x) = +
x a
as appropriate, and we say that f(x) increases without bound, or f(x) approaches
+, as x a + or as x a . Similarly, if the values of f(x) decrease indefinitely as x
approaches a from the right or left, then we write
lim f(x) =
or
x a+
lim f(x) =
x a
as appropriate, and say that f(x) decreases without bound, or f(x) approaches , as
x a + or as x a . Moreover, if both one-sided limits are +, then we write
lim f(x) = +
x a
## and if both one-sided limits are , then we write
lim f(x) =
x a
REMARK.
It should be emphasized that the symbols + and are not real numbers. The
phrase f(x) approaches + is akin to saying that f(x) approaches the unapproachable;
it is a colloquialism for f(x) increases without bound. The symbols + and are used
here to encapsulate a particular way in which limits fail to exist. To say, for example, that
f(x) + as x a + is to indicate that limx a + f(x) does not exist, and to say further
that this limit fails to exist because values of f(x) increase without bound as x approaches
a from the right. Furthermore, since + and are not numbers, it is inappropriate to
manipulate these symbols using rules of algebra. For example, it is not correct to write
(+) (+) = 0.
Example 6 For the functions in Figure 2.1.12, describe the limits at x = a in appropriate
limit notation.
1
f (x) = x a
x
(a)
1
f (x) = x a
f (x) =
(b)
f (x) =
1
(x a)2
x
1
(x a)2
(c)
(d)
Figure 2.1.12
Solution (a). In Figure 2.1.12a, the function increases indefinitely as x approaches a from
the right and decreases indefinitely as x approaches a from the left. Thus,
1
1
lim
= + and
lim
=
x a+ x a
x a x a
116
g65-ch2
## Limits and Continuity
Solution (b). In Figure 2.1.12b, the function increases indefinitely as x approaches a from
both the left and right. Thus,
1
1
1
lim
= lim+
= lim
= +
2
2
x a (x a)
x a (x a)
x a (x a)2
Solution (c). In Figure 2.1.12c, the function decreases indefinitely as x approaches a from
the right and increases indefinitely as x approaches a from the left. Thus,
1
1
lim+
= and
lim
= +
x a x a
x a x a
## Solution (d ). In Figure 2.1.12d, the function decreases indefinitely as x approaches a
from both the left and right. Thus,
1
1
1
= lim+
= lim
=
lim
x a (x a)2
x a (x a)2
x a (x a)2
Geometrically, if f(x) + as x a or x a + , then the graph of y = f(x) rises
without bound and squeezes closer to the vertical line x = a on the indicated side of x = a.
If f(x) as x a or x a + , then the graph of y = f(x) falls without bound and
squeezes closer to the vertical line x = a on the indicated side of x = a. In these cases, we
call the line x = a a vertical asymptote. (Asymptote comes from the Greek asymptotos,
meaning nonintersecting. We will soon see that taking asymptote to be synonymous
with nonintersecting is a bit misleading.)
2.1.5 DEFINITION. A line x = a is called a vertical asymptote of the graph of a
function f if f(x) + or f(x) as x approaches a from the left or right.
Example 7 The four functions graphed in Figure 2.1.12 all have a vertical asymptote at
x = a, which is indicated by the dashed vertical lines in the figure.
## LIMITS AT INFINITY AND
HORIZONTAL ASYMPTOTES
Thus far, we have used limits to describe the behavior of f(x) as x approaches a. However,
sometimes we will not be concerned with the behavior of f(x) near a specific x-value, but
rather with how the values of f(x) behave as x increases without bound or decreases without
bound. This is sometimes called the end behavior of the function because it describes how
the function behaves for values of x that are far from the origin. For example, it is evident
from the table and graph in Figure 2.1.13 that as x increases without bound, the values of
y
1
y= x
=0
lim
x x
x
1
y= x
=0
lim
x+ x
1
x
1
x
1000
100
10
10
100
1000
10,000
...
0.01
0.1
f(x)
0.1
0.01
0.001
0.0001
...
x . . . 10,000
Figure 2.1.13
g65-ch2
2.1
## Limits (An Intuitive Approach)
117
f(x) = 1/x are positive, but get closer and closer to 0; and as x decreases without bound,
the values of f(x) = 1/x are negative, and also get closer and closer to 0. We indicate these
limiting behaviors by writing
1
1
lim
= 0 and
lim
=0
x + x
x x
More generally:
2.1.6 LIMITS AT INFINITY (AN INFORMAL VIEW). If the values of f(x) eventually get
closer and closer to a number L as x increases without bound, then we write
lim f(x) = L
x +
or
f(x) L as x +
(13)
Similarly, if the values of f(x) eventually get closer and closer to a number L as x
decreases without bound, then we write
lim f(x) = L
or
f(x) L as x
(14)
## Geometrically, if f(x) L as x +, then the graph of y = f(x) eventually gets
closer and closer to the line y = L as the graph is traversed in the positive direction (Figure 2.1.14a); and if f(x) L as x , then the graph of y = f(x) eventually gets
closer and closer to the line y = L as the graph is traversed in the negative x-direction
(Figure 2.1.14b). In either case we call the line y = L a horizontal asymptote of the graph
of f . For example, the function in Figure 2.1.13 all have y = 0 as a horizontal asymptote.
y
y
Horizontal asymptote
y=L
Horizontal asymptote
(a)
y=L
(b)
Figure 2.1.14
function f if
y=
3x + 1
x
lim f(x) = L
x +
y=3
Figure 2.1.15
TO EXIST
or
lim f(x) = L
## Sometimes the existence of a horizontal asymptote of a function f will be readily apparent
from the formula for f . For example, it is evident that the function
3x + 1
1
f(x) =
=3+
x
x
has a horizontal asymptote at y = 3 (Figure 2.1.15), since the value of 1/x approaches 0 as
x + or x . For more complicated functions, algebraic manipulations or special
techniques that we will study in the next section may have to be applied to confirm the
existence of horizontal asymptotes.
Limits at infinity can fail to exist for various reasons. One possibility is that the values of
f(x) may increase or decrease without bound as x + or as x . For example, the
values of f(x) = x 3 increase without bound as x + and decrease without bound as
118
g65-ch2
## Limits and Continuity
y
x ; and for f(x) = x 3 the values decrease without bound as x + and increase
without bound as x (Figure 2.1.16). We denote this by writing
Increases
without
bound
x
y = x3
lim x 3 = +,
x +
Decreases
without
bound
lim x 3 = ,
lim (x 3 ) = ,
lim (x 3 ) = +
x +
More generally:
2.1.8 INFINITE LIMITS AT INFINITY (AN INFORMAL VIEW). If the values of f(x) increase without bound as x + or as x , then we write
y
Increases
without
bound
y = x 3
lim f(x) = +
x +
or
lim f(x) = +
## as appropriate; and if the values of f(x) decrease without bound as x + or as
x , then we write
Decreases
without
bound
lim f(x) =
Figure 2.1.16
x +
or
lim f(x) =
as appropriate.
y
Limits at infinity can also fail to exist because the graph of the function oscillates indefinitely in such a way that the values of the function do not approach a fixed number and do
not increase or decrease without bound; the trigonometric functions sin x and cos x have
this property, for example (Figure 2.1.17). In such cases we say that the limit fails to exist
because of oscillation.
y = sin x
x
There is no limit as
x + or x .
Figure 2.1.17
## EXERCISE SET 2.1
Graphing Calculator
CAS
(a) lim f(x)
(b) lim+ f(x)
(c) lim f(x)
x 3
x 3
x 3
(d) f(3)
(f ) lim f(x).
x
x +
(a) lim g(x)
(b) lim+ g(x)
(c) lim g(x)
x 4
x
y
y
x 4
x 4
(d) g(4)
(f ) lim g(x).
x +
y = g(x)
y = f (x)
3
1
x
4
10
Figure Ex-1
(a) lim f(x)
(b) lim+ f(x)
(c) lim f(x)
x 2
x 2
x 2
(d) f(2)
x
(f ) lim f(x).
x +
Figure Ex-3
(a) lim g(x)
(b) lim+ g(x)
(c) lim g(x)
x 0
x 0
x 0
(d) g(0)
x
y = f (x)
(f ) lim g(x).
x +
y = g(x)
4
2
x
x
Figure Ex-2
Figure Ex-4
g65-ch2
2.1
## 5. For the function F graphed in the accompanying figure, find
(a) lim F (x) (b) lim + F (x) (c) lim F (x)
x 2
x 2
x 2
(d) F (2)
x
(f ) lim F (x).
x +
119
(a) lim f(x)
(c) lim f(x)
(b) lim+ f(x)
x 3
x 3
x 3
(d) f(3)
x
y = F(x)
(f ) lim f(x).
x +
y = f (x)
4
3
x
x
-2
Figure Ex-5
(a) lim F (x)
(c) lim F (x)
(b) lim+ F (x)
x 3
x 3
x 3
(d) F (3)
x
(f ) lim F (x).
x +
Figure Ex-9
(a) lim f(x)
(b) lim+ f(x)
(c) lim f(x)
x 0
x 0
x 0
(d) f(0)
y = F(x)
x
(f ) lim f(x).
x +
y = f (x)
x
3
x
3
-2
Figure Ex-6
(a) lim (x)
(b) lim + (x)
(c) lim (x)
x 2
x 2
x 2
(d) (2)
x
(f ) lim (x).
x +
y = f(x)
Figure Ex-10
(a) lim G(x)
(b) lim+ G(x)
(c) lim G(x)
x 0
x 0
x 0
(d) G(0)
x
(f ) lim G(x).
x +
y = G(x)
2
x
1
Figure Ex-7
Figure Ex-11
(a) lim (x)
(b) lim+ (x)
(c) lim (x)
x 4
x 4
x 4
(d) (4)
x
(f ) lim (x).
x +
(a) lim G(x)
(b) lim+ G(x)
(c) lim G(x)
x 0
x
(f ) lim G(x).
x +
y = f(x)
y
y = G(x)
x
4
Figure Ex-8
x 0
x 0
(d) G(0)
Figure Ex-12
120
g65-ch2
## Limits and Continuity
13. Consider the function g graphed in the accompanying figure. For what values of x0 does lim g(x) exist?
In Exercises 1922: (i) Make a guess at the limit (if it exists) by evaluating the function at the specified x-values.
the function over an appropriate interval. (iii) If you have a
CAS, then use it to find the limit. [Note: For the trigonometric functions, be sure to set your calculating and graphing
x x0
y = g(x)
2
x
4
## 19. (a) lim
20.
21.
22.
x 1
14. Consider the function f graphed in the accompanying figure. For what values of x0 does lim f(x) exist?
x x0
y = f (x)
x
3
Figure Ex-14
## In Exercises 1518, sketch a possible graph for a function f
with the specified properties. (Many different solutions are
possible.)
23.
(i) f(0) = 2 and f(2) = 1
(ii) lim f(x) = + and lim+ f(x) =
x 1
x 1
x +
16.
## 24. Consider the motion of the ball described in Example 1. By
interpreting instantaneous velocity as a limit of average velocity, make a conjecture for the value of the instantaneous
velocity of the ball 0.75 s after its release.
## (i) f(0) = f(2) = 1
(ii) lim f(x) = + and lim+ f(x) = 0
x 2
x 2
x 1
x 1
## In Exercises 25 and 26: (i) Approximate the y-coordinates
of all horizontal asymptotes of y = f(x) by evaluating f at the x-values 10, 100, 1000, 100,000, and
100,000,000. (ii) Confirm your conclusions by graphing
y = f(x) over an appropriate interval. (iii) If you have a
CAS, then use it to find the horizontal asymptotes.
x +
## 17. (i) f(x) = 0 if x is an integer and f(x) = 0 if x is not an
integer
(ii) lim f(x) = 0 and lim f(x) = 0
x +
18.
## (i) f(x) = 1 if x is a positive integer and f(x) = 1 if
x > 0 is not a positive integer
(ii) f(x) = 1 if x is a negative integer and f(x) = 1
if x < 0 is not a negative integer
(iii) lim f(x) = 1 and lim f(x) = 1
x +
x1
; x = 2, 1.5, 1.1, 1.01, 1.001, 0, 0.5, 0.9,
x3 1
0.99, 0.999
x+1
; x = 2, 1.5, 1.1, 1.01, 1.001, 1.0001
(b) lim+ 3
x 1 x 1
x+1
(c) lim 3
; x = 0, 0.5, 0.9, 0.99, 0.999, 0.9999
x 1 x 1
x+11
(a) lim
; x = 0.25, 0.1, 0.001,
x 0
x
0.0001
x+1+1
; x = 0.25, 0.1, 0.001, 0.0001
(b) lim+
x 0
x
x+1+1
(c) lim
; x = 0.25, 0.1, 0.001,
x 0
x
0.0001
sin 3x
(a) lim
; x = 0.25, 0.1, 0.001, 0.0001
x 0
x
cos x
(b) lim
; x = 0, 0.5, 0.9, 0.99, 0.999,
x 1 x + 1
1.5, 1.1, 1.01, 1.001
tan(x + 1)
; x = 0, 0.5, 0.9, 0.99, 0.999,
(a) lim
x 1
x+1
1.5, 1.1, 1.01, 1.001
sin(5x)
(b) lim
; x = 0.25, 0.1, 0.001, 0.0001
x 0 sin(2x)
Consider the motion of the ball described in Example 1. By
interpreting instantaneous velocity as a limit of average velocity, make a conjecture for the value of the instantaneous
velocity of the ball 0.25 s after its release.
Figure Ex-13
15.
2x + 3
x+4
x2 + 1
(c) f(x) =
x+1
3 x
(b) f(x) = 1 +
x
g65-ch2
2.1
Velocity
lim f(1/t) = L
t 0
lim f(x) and
x +
1.1/x 2
f(x) = 1 + x 2
## Graph f in the window [1, 1] [2.5, 3.5] and use the
calculators trace feature to make a conjecture about the
limit of f as x 0.
(b) Graph f in the window [0.001, 0.001][2.5, 3.5] and
use the calculators trace feature to make a conjecture
about the limit of f as x 0.
(c) Graph f in the window [0.000001, 0.000001]
[2.5, 3.5] and use the calculators trace feature to make
a conjecture about the limit of f as x 0.
(d) Later we will be able to show that
1.1/x 2
lim 1 + x 2
3.00416602
v = n(t)
v = e(t)
(Relativity)
t
Time
Figure Ex-27
## 28. Let T = f(t) denote the temperature of a baked potato t
minutes after it has been removed from a hot oven. The accompanying figure shows the temperature versus time curve
for the potato, where r is the temperature of the room.
(a) What is the physical significance of lim+ f(t)?
x 0
evidence (as revealed by the graphs you obtained) to
t 0
t +
## Roundoff error is one source of inaccuracy in calculator
and computer computations. Another source of error, called
catastrophic subtraction, occurs when two nearly equal numbers are subtracted, and the result is used as part of another
calculation. For example, by hand calculation we have
T
400
T = f (t )
## However, a calculator that can only store 14 decimal digits
produces a value of 0 for this computation, since the numbers being subtracted are identical in the first 14 digits. Catastrophic subtraction can sometimes be avoided by rearranging
formulas algebraically, but your best defense is to be aware
that it can occur. Watch out for it in the next exercise.
t
Time (min)
Figure Ex-28
## In Exercises 29 and 30: (i) Conjecture a limit from numerical
evidence. (ii) Use the substitution t = 1/x to express the
limit as an equivalent limit in which t 0+ or t 0 , as
appropriate. (iii) Use a graphing utility to make a conjecture
1
29. (a) lim x sin
x +
x
2 x
(c) lim
1+
x
x
cos(/x)
/x
(c) lim (1 2x)1/x
x +
x
lim f(x)?
## 32. (a) Do any of the trigonometric functions, sin x, cos x,
tan x, cot x, sec x, csc x, have horizontal asymptotes?
(b) Do any of them have vertical asymptotes? Where?
(Classical)
121
## 26. (a) f(x) =
Temperature (F)
1 x
x2 1
(b)
f(x)
=
2
+
5x 2 + 1
x
sin x
(c) f(x) =
x
27. Assume that a particle is accelerated by a constant force.
The two curves v = n(t) and v = e(t) in the accompanying
figure provide velocity versus time curves for the particle
as predicted by classical physics and by the special theory
of relativity, respectively. The parameter c designates the
speed of light. Using the language of limits, describe the
differences in the long-term predictions of the two theories.
1x
(b) lim
x + 1 + x
(b) lim
x +
x
1+x
f(x) =
x sin x
x3
## Make a conjecture about the limit of f as x 0+ by
evaluating f(x) at x = 0.1, 0.01, 0.001, 0.0001.
(b) Evaluate f(x) at x = 0.000001, 0.0000001,
0.00000001, 0.000000001, 0.0000000001, and make
another conjecture.
(c) What flaw does this reveal about using numerical evidence to make conjectures about limits?
(d) If you have a CAS, use it to show that the exact value
of the limit is 16 .
122
g65-ch2
## 35. (a) The accompanying figure shows two different views of
the graph of the function in Exercise 34, as generated
by Mathematica. What is happening?
0.166667
0.166666
## (b) Use your graphing utility to generate the graphs, and
see whether the same problem occurs.
0.166666
0.166666
## (c) Would you expect a similar problem to occur in the
vicinity of x = 0 for the function
f(x) =
-0.01
-0.005
1 cos x
?
x
0.005
0.01
0.166667
0.166667
See if it does.
0.166667
0.166667
0.166667
-0.001 -0.0005
0.0005 0.001
Erratic graph generated by Mathematica
Figure Ex-35
## 2.2 COMPUTING LIMITS
In this section we will discuss algebraic techniques for computing limits of many functions. We base these results on the informal development of the limit concept discussed
in the preceding section. A more formal derivation of these results is possible after
Section 2.4.
## First we will obtain the limits of some simple functions.
Then we will develop a repertoire of theorems that will enable us to use the limits
of those simple functions as building blocks for finding limits of more complicated
functions.
We start with the cases of a constant function f(x) = k, the identity function f(x) = x,
and the reciprocal function f(x) = 1/x.
2.2.1
THEOREM.
lim k = k
x a
lim
x 0
1
=
x
## Let a and k be real numbers.
lim x = a
x a
lim+
x 0
1
= +
x
The four limits in Theorem 2.2.1 should be evident from inspection of the function graphs
shown in Figure 2.2.1.
In the case of the constant function f(x) = k, the values of f(x) do not change as x
varies, so the limit of f(x) is k, regardless of at which number a the limit is taken. For
example,
lim 3 = 3,
x 25
lim 3 = 3,
x 0
lim 3 = 3
g65-ch2
## cyan magenta yellow black
2.2
y
y=x
y = f (x) = k
lim k = k
x
1
lim x =
lim x = a
x a
1
x
1
x
1
y= x
f (x) = x
123
1
y= x
f (x) = x
Computing Limits
lim x = +
x 0
x a
x 0 +
Figure 2.2.1
Since the identity function f(x) = x just echoes its input, it is clear that f(x) = x a
as x a. In terms of our informal definition of limits (2.1.1), if we decide just how close
to a we would like the value of f(x) = x to be, we need only restrict its input x to be just
as close to a.
The one-sided limits of the reciprocal function f(x) = 1/x about 0 should conform
with your experience with fractions: making the denominator closer to zero increases the
magnitude of the fraction (i.e., increases its absolute value). This is illustrated in Table 2.2.1.
Table 2.2.1
values
x
1/x
1
1
x
1/x
1
1
conclusion
## 0.1 0.01 0.001 0.0001 . . .
10 100 1000 10,000 . . .
0.1
0.01
0.001
0.0001 . . .
10
100
1000
10,000 . . .
## As x 0 the value of 1/x
decreases without bound.
As x 0 + the value of 1/x
increases without bound.
The following theorem, parts of which are proved in Appendix G, will be our basic tool
for finding limits algebraically.
2.2.2
THEOREM.
## Let a be a real number, and suppose that
lim f(x) = L1
lim g(x) = L2
and
x a
x a
That is, the limits exist and have values L1 and L2 , respectively. Then,
(a) lim [f(x) + g(x)] = lim f(x) + lim g(x) = L1 + L2
x a
(b)
(c)
(d )
(e)
x a
x a
x a
lim [f(x)g(x)] =
x a
lim
x a
lim
x a
x a
x a
lim f(x)
lim g(x) = L1 L2
x a
x a
lim f(x)
L1
f(x)
x a
=
=
,
g(x)
lim g(x)
L2
n
x a
f(x) =
lim f(x) =
x a
n
provided L2 = 0
L1 ,
124
g65-ch2
## A casual restatement of this theorem is as follows:
(a) The limit of a sum is the sum of the limits.
(b) The limit of a difference is the difference of the limits.
(c)
## The limit of a product is the product of the limits.
(d ) The limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero.
(e)
REMARK.
## LIMITS OF POLYNOMIALS AND
RATIONAL FUNCTIONS AS x a
Although results (a) and (c) in Theorem 2.2.2 are stated for two functions, they
hold for any finite number of functions. For example, if the limits of f(x), g(x), and h(x)
exist as x a, then the limit of their sum and the limit of their product also exist as x a
and are given by the formulas
lim [f(x) + g(x) + h(x)] = lim f(x) + lim g(x) + lim h(x)
x a
x a
lim [f(x)g(x)h(x)] =
x a
x a
x a
lim f(x)
x a
lim g(x)
x a
lim h(x)
x a
## In particular, if f(x) = g(x) = h(x), then this yields
3
lim [f(x)]3 = lim f(x)
x a
x a
More generally, if n is a positive integer, then the limit of the nth power of a function is the
nth power of the functions limit. Thus,
lim x n =
x a
n
lim x
x a
= an
(1)
For example,
lim x 4 = 34 = 81
x 3
Another useful result follows from part (c) of Theorem 2.2.2 in the special case when
one of the factors is a constant k:
lim (k f(x)) = lim k lim f(x) = k lim f(x)
(2)
x a
x a
x a
x a
and similarly for limx a replaced by a one-sided limit, limx a + or limx a . Rephrased,
this last statement says:
A constant factor can be moved through a limit symbol.
Example 1 Find lim (x 2 4x + 3) and justify each step.
x 5
Solution. First note that limx 5 x 2 = 52 = 25 by Equation (1). Also, from Equation (2),
## limx 5 4x = 4(limx 5 x) = 4(5) = 20. Since limx 5 3 = 3 by Theorem 2.2.1, we may
appeal to Theorem 2.2.2(a) and (b) to write
lim (x 2 4x + 3) = lim x 2 lim 4x + lim 3 = 25 20 + 3 = 8
x 5
x 5
x 5
x 5
However, for conciseness, it is common to reverse the order of this argument and simply
g65-ch2
## cyan magenta yellow black
2.2
write
lim (x 2 4x + 3) = lim x 2 lim 4x + lim 3
x 5
x 5
x 5
2
lim x
x 5
x 5
4 lim x + lim 3
x 5
x 5
= 52 4(5) + 3
Computing Limits
125
Theorem 2.2.1
=8
REMARK.
## In our presentation of limit arguments, we will adopt the convention of providing
just a concise, reverse argument, bearing in mind that the validity of each equality may be
conditional upon the successful resolution of the remaining limits.
Our next result will show that the limit of a polynomial p(x) at x = a is the same as
the value of the polynomial at x = a. This greatly simplifies the computation of limits of
polynomials by allowing us to simply evaluate the polynomial.
2.2.3
THEOREM.
## For any polynomial
p(x) = c0 + c1 x + + cn x n
and any real number a,
lim p(x) = c0 + c1 a + + cn a n = p(a)
x a
Proof.
lim p(x) = lim c0 + c1 x + + cn x n
x a
x a
x a
x a
x a
## = lim c0 + c1 lim x + + cn lim x n
x a
x a
x a
= c0 + c1 a + + cn a = p(a)
n
Recall that a rational function is a ratio of two polynomials. Theorem 2.2.3 and Theorem
2.2.2(d) can often be used in combination to compute limits of rational functions.
5x 3 + 4
.
x 2 x 3
## Example 2 Find lim
Solution.
lim (5x 3 + 4)
5x 3 + 4
x 2
lim
=
x 2 x 3
lim (x 3)
Theorem 2.2.2(d )
x 2
5 23 + 4
= 44
=
23
2.2.4
Theorem 2.2.3
THEOREM.
## Consider the rational function
n(x)
f(x) =
d(x)
where n(x) and d(x) are polynomials. For any real number a,
(a) if d(a) = 0, then lim f(x) = f(a).
x a
(b) if d(a) = 0 but n(a) = 0, then lim f(x) does not exist.
x a
126
g65-ch2
n(x)
d(x)
lim n(x)
## lim f(x) = lim
x a
x a
x a
Theorem 2.2.2(d )
lim d(x)
x a
n(a)
= f(a)
d(a)
Theorem 2.2.3
If d(a) = 0 and n(a) = 0, then we again appeal to your experience with fractions. For
values of x sufficiently near a, the value of n(x) will be near n(a) and not zero. Thus, since
0 = d(a) = limx a d(x), as values of x approach a, the magnitude (absolute value) of the
fraction n(x)/d(x) will increase without bound, so limx a f(x) does not exist.
As an illustration of part (b) of Theorem 2.2.4, consider
5x 3 + 4
x 3 x 3
Note that limx 3 (5x 3 + 4) = 5 33 + 4 = 139 and limx 3 (x 3) = 3 3 = 0. It is
evident from Table 2.2.2 that
5x 3 + 4
lim
x 3 x 3
does not exist.
lim
Table 2.2.2
values
x
5x3 + 4
x3
x
+4
x3
5x3
2.99
2.999
conclusion
2.9999
## 13,765.45 138,865.04 1,389,865.00
...
5x3 + 4
decreases
x3
without bound as x 3.
The value of
...
3.01
3.001
3.0001
...
14,035.45
139,135.05
1,390,135.00
...
The value of
5x3 + 4
increases
x3
without bound as x 3+.
In Theorem 2.2.4(b), where the limit of the denominator is zero but the limit of the
numerator is not zero, the response does not exist can be elaborated upon in one of the
following three ways.
## The limit may be +.
The limit may be from one side and + from the other.
Figure 2.2.2 illustrates these three possibilities graphically for rational functions of the form
1/(x a), 1/(x a)2 , and 1/(x a)2 .
Example 3 Find
2x
(a) lim
x 4 (x 4)(x + 2)
(b) lim+
x 4
2x
(x 4)(x + 2)
(c) lim
x 4
2x
(x 4)(x + 2)
Solution. With n(x) = 2 x and d(x) = (x 4)(x + 2), we see that n(4) = 2 and
d(4) = 0. By Theorem 2.2.4(b), each of the limits does not exist. To be more specific, we
g65-ch2
## cyan magenta yellow black
2.2
1
y= xa
y=
1
(x a)2
Computing Limits
y=
1
(x a)2
127
lim x a = +
x a+
lim
x a
lim
=
x a x a
1
= +
(x a)2
lim
x a
1
=
(x a)2
Figure 2.2.2
+ + +
analyze the sign of the ratio n(x)/d(x) near x = 4. The sign of the ratio, which is given
in Figure 2.2.3, is determined by the signs of 2 x, x 4, and x + 2. (The method of
test values, discussed in Appendix A, provides a simple way of finding the sign of the ratio
here.) It follows from this figure that as x approaches 4 from the left, the ratio is always
positive; and as x approaches 4 from the right, the ratio is always negative. Thus,
2x
2x
lim
= + and
lim
=
x 4 (x 4)(x + 2)
x 4+ (x 4)(x + 2)
Because the one-sided limits have opposite signs, all we can say about the two-sided limit
is that it does not exist.
0+ + x
2
Sign of
2x
(x 4)(x + 2)
Figure 2.2.3
The missing case in Theorem 2.2.4 is when both the numerator and the denominator of a
rational function f(x) = n(x)/d(x) have a zero at x = a. In this case, n(x) and d(x) will
each have a factor of x a, and canceling this factor may result in a rational function to
which Theorem 2.2.4 applies.
0/0
x2 4
.
x 2 x 2
## Example 4 Find lim
Solution. Since 2 is a zero of both the numerator and denominator, they share a common
factor of x 2. The limit can be obtained as follows:
x2 4
(x 2)(x + 2)
lim
= lim
= lim (x + 2) = 4
x 2 x 2
x 2
x 2
x2
REMARK.
Although correct, the second equality in the preceding computation needs some
justification, since canceling the factor x 2 alters the function by expanding its domain.
However, as discussed in Example 5 of Section 1.2, the two functions are identical, except at
x = 2 (Figure 1.2.9). From our discussions in the last section, we know that this difference
has no effect on the limit as x approaches 2.
Example 5 Find
x 2 6x + 9
(a) lim
x 3
x3
2x + 8
x 4 x 2 + x 12
(b) lim
x 2 3x 10
x 5 x 2 10x + 25
(c) lim
Solution (a). The numerator and the denominator both have a zero at x = 3, so there is a
common factor of x 3. Then,
x 2 6x + 9
(x 3)2
= lim
= lim (x 3) = 0
lim
x 3
x 3 x 3
x 3
x3
128
g65-ch2
## Limits and Continuity
Solution (b). The numerator and the denominator both have a zero at x = 4, so there is
a common factor of x (4) = x + 4. Then,
2
2x + 8
2(x + 4)
2
lim
= lim
= lim
=
x 4 x 2 + x 12
x 4 (x + 4)(x 3)
x 4 x 3
7
Solution (c). The numerator and the denominator both have a zero at x = 5, so there is a
common factor of x 5. Then,
(x 5)(x + 2)
x+2
x 2 3x 10
lim 2
= lim
= lim
x 5 x 5
x 5 x 10x + 25
x 5 (x 5)(x 5)
However,
lim (x + 2) = 7 = 0
x 5
lim (x 5) = 0
and
x 5
By Theorem 2.2.4(b),
x 2 3x 10
x+2
= lim
2
x 5 x 10x + 25
x 5 x 5
does not exist.
lim
## The case of a limit of a quotient,
f(x)
lim
x a g(x)
where limx a f(x) = 0 and limx a g(x) = 0, is called an indeterminate form of type
0/0. Note that the limits in Examples 4 and 5 produced a variety of answers. The word
indeterminate here refers to the fact that the limiting behavior of the quotient cannot
be determined without further study. The expression 0/0 is just a mnemonic device
to describe the circumstance of a limit of a quotient in which both the numerator and
denominator approach 0.
## Example 6 Find lim
x 0
x
x+11
Solution. Recall that in Example 2 of Section 2.1 we conjectured this limit to be 2. Note
that this limit expression is an indeterminate form of type 0/0, so Theorem 2.2.2(d) does
not apply. One strategy for resolving this limit is to first rationalize the denominator of the
function. This yields
x
x( x + 1 + 1)
= x + 1 + 1, x = 0
=
(x + 1) 1
x+11
Therefore,
x
lim
= lim ( x + 1 + 1) = 2
x 0
x + 1 1 x 0
LIMITS OF PIECEWISE-DEFINED
FUNCTIONS
For functions that are defined piecewise, a two-sided limit at an x-value where the formula
changes is best obtained by first finding the one-sided limits at that number.
Example 7 Let
1 (x + 2),
f(x) = x 2 5,
x + 13,
x < 2
2 < x 3
x>3
Find
(a) lim f(x)
x 2
x 0
x 3
g65-ch2
2.2
Computing Limits
129
## Solution (a). As x approaches 2 from the left, the formula for f is
f(x) =
1
x+2
so that
1
=
x 2
x 2 x + 2
As x approaches 2 from the right, the formula for f is
lim f(x) = lim
f(x) = x 2 5
so that
lim f(x) = lim+ (x 2 5) = (2)2 5 = 1
x 2+
x 2
## Thus, limx 2 f(x) does not exist.
Solution (b). As x approaches 0 from either the left or the right, the formula for f is
f(x) = x 2 5
Thus,
lim f(x) = lim (x 2 5) = 02 5 = 5
x 0
x 0
## Solution (c). As x approaches 3 from the left, the formula for f is
f(x) = x 2 5
so that
lim f(x) = lim (x 2 5) = 32 5 = 4
x 3
x 3
f(x) = x + 13
so that
lim+ f(x) = lim+
x 3
x 3
x + 13 =
lim+ (x + 13) =
x 3
3 + 13 = 4
lim f(x) = 4
x 3
(b) lim+
(a) lim 7
x 8
x 0
(c) lim 3x
x 2
y 3
(a) lim f(x)
(b) lim f(x)
x 5
x 5
## (c) lim+ f(x)
x 0
3. Given that
lim f(x) = 2,
x a
x 0
lim g(x) = 4,
x a
lim h(x) = 0
x a
find the limits that exist. If the limit does not exist, explain
why.
(a) lim [f(x) + 2g(x)]
(b) lim [h(x) 3g(x) + 1]
x a
x a
x a
(e) lim
x a
(g) lim
x a
3
6 + f(x)
3f(x) 8g(x)
h(x)
x a
2
g(x)
7g(x)
(h) lim
x a 2f(x) + g(x)
(f ) lim
x a
## 4. Use the graphs of f and g in the accompanying figure to
find the limits that exist. If the limit does not exist, explain
why.
130
g65-ch2
## (b) lim [f(x) + g(x)]
31. Verify the limit in Example 1 of Section 2.1. That is, find
## 16t12 + 29t1 10.5
t1 0.5
t1 0.5
32. Let s(t) = 16t 2 + 29t + 6. Find
x 0
x 2
1 + g(x)
f(x)
(h) lim f(x)
f(x)
1 + g(x)
(g) lim+ f(x)
(f ) lim
(e) lim
x 2
x 2
x 0
lim
x 0
x 0
lim
x 0
s(t) s(1.5)
t 1.5
33. Let
y = f (x)
t 1.5
y = g(x)
f(x) =
1
(y 1)(y 2)
y+1
x 2 16
x 4 x 4
7. lim
x 1
x1
6. lim
x 3
x 2x
x+1
8. lim
x 0
6x 9
x 3 12x + 3
9. lim+
x 1
t +8
t +2
3
10. lim
t 2
x 2 + 6x + 5
11. lim 2
x 1 x 3x 4
x 2 4x + 4
12. lim 2
x 2 x + x 6
t + 3t 12t + 4
t 3 4t
x
15. lim+
x 3 x 3
x
17. lim
x 3 x 3
x
19. lim 2
x 2 x 4
t + t 5t + 3
t 3 3t + 2
x
16. lim
x 3 x 3
x
18. lim+ 2
x 2 x 4
x
20. lim 2
x 2 x 4
13. lim
t 2
21. lim+
y 6
23. lim
y 6
y+6
y 2 36
y+6
y 2 36
3x
2x 8
25. lim
x2
27. lim+
1
|2 x|
x 4
x 2
x9
29. lim
x 9
x3
14. lim
t 1
22. lim
y+6
y 2 36
24. lim+
3x
x 2 2x 8
26. lim
3x
2x 8
y 6
x 4
x 4 x2
28. lim
x 3
30. lim
y 4
1
|x 3|
4y
2 y
x 3
x 3
t 0
t <0
t 2,
t 2,
t 0
Find
(a) lim g(t)
y 2
x>3
x 3
g(t) =
5. lim
x3
Find
(a) lim f(x)
34. Let
Figure Ex-4
x 1,
3x 7,
t 0
## (c) lim g(t).
t 0
x 1
.
x1
(a) Find lim f(x).
x 1
36. Let
x 9,
f(x) = x + 3
k,
x = 3
x = 3
x 3
## (b) With k assigned the value limx 3 f (x), show that
f (x) can be expressed as a polynomial.
37. (a) Explain why the following calculation is incorrect.
1
1
1
1
lim+
2 = lim+ lim+ 2
x 0
x 0 x
x 0 x
x
x
= + (+) = 0
1
1
2 = .
x
x
x 0
1
1
+ 2 .
38. Find lim
x 0
x
x
## In Exercises 39 and 40, first rationalize the numerator, then
find the limit.
x+42
x2 + 4 2
39. lim
40. lim
x 0
x 0
x
x
41. Let p(x) and q(x) be polynomials, and suppose q(x0 ) = 0.
Discuss the behavior of the graph of y = p(x)/q(x) in the
vicinity of x = x0 . Give examples to support your conclusions.
g65-ch2
2.3
131
## 2.3 COMPUTING LIMITS: END BEHAVIOR
In this section we will discuss algebraic techniques for computing limits at for
many functions. We base these results on the informal development of the limit concept
discussed in Section 2.1. A more formal development of these results is possible after
Section 2.4.
## SOME BASIC LIMITS
The behavior of a function toward the extremes of its domain is sometimes called its end
behavior. Here we will use limits to investigate the end behavior of a function as x or
as x +. As in the last section, we will begin by obtaining limits of some simple functions
and then use these as building blocks for finding limits of more complicated functions.
2.3.1
THEOREM.
## Let k be a real number.
lim k = k
lim k = k
x +
lim x =
lim x = +
lim
x +
1
=0
x
lim
x +
1
=0
x
The six limits in Theorem 2.3.1 should be evident from inspection of the function graphs
in Figure 2.3.1.
x
y
y =x
y
x
f (x) = x
y = f (x) = k
f (x) = x
y=x
lim x =
lim k = k, lim k = k
x +
lim x = +
x +
1
y= x
1
y= x
1
x
x
1
x
lim x = 0
Figure 2.3.1
lim x = 0
x +
132
g65-ch2
## Limits and Continuity
The limits of the reciprocal function f (x) = 1/x should make sense to you intuitively,
based on your experience with fractions: increasing the magnitude of x makes its reciprocal
closer to zero. This is illustrated in Table 2.3.1.
Table 2.3.1
values
x
1/x
1
1
x
1/x
1
1
conclusion
## 10 100 1000 10,000 . . .
0.1 0.01 0.001 0.0001 . . .
10
100
1000
10,000 . . .
0.1
0.01
0.001
0.0001 . . .
## As x the value of 1/x
increases toward zero.
As x + the value of 1/x
decreases toward zero.
The following theorem mirrors Theorem 2.2.2 as our tool for finding limits at algebraically. (The proof is similar to that of the portions of Theorem 2.2.2 that are proved in
Appendix G.)
2.3.2
THEOREM.
lim f(x) = L1
x +
Suppose that
and
lim g(x) = L2
x +
That is, the limits exist and have values L1 and L2 , respectively. Then,
(a) lim [f(x) + g(x)] = lim f(x) + lim g(x) = L1 + L2
x +
(c)
x +
(d )
(e)
x +
x +
## lim [f(x) g(x)] = lim f(x) lim g(x) = L1 L2
x +
x +
x +
lim [f(x)g(x)] =
lim f(x)
lim g(x) = L1 L2
(b)
x +
x +
lim f(x)
f(x)
L1
x +
, provided L2 = 0
=
=
x + g(x)
L2
lim g(x)
x +
lim n f(x) = n lim f(x) = n L1 , provided L1 > 0 if n is even.
lim
x +
x +
## Moreover, these statements are also true if x .
REMARK.
As in the remark following Theorem 2.2.2, results (a) and (c) can be extended to
sums or products of any finite number of functions. In particular, for any positive integer n,
n
n
lim (f(x))n =
lim f(x)
lim (f(x))n =
lim f(x)
x +
x +
## Also, since limx + (1/x) = 0, if n is a positive integer, then
1
1 n
1
1 n
lim
= lim
=0
lim
= lim
=0
x + x n
x + x
x x n
x x
(1)
For example,
1
1
lim
= 0 and
lim
=0
x + x 4
x x 4
Another useful result follows from part (c) of Theorem 2.3.2 in the special case where
one of the factors is a constant k:
lim f(x) = k
lim f(x)
(2)
lim (k f(x)) =
lim k
x +
x +
x +
x +
g65-ch2
2.3
## Computing Limits: End Behavior
133
and similarly, for limx + replaced by limx . Rephrased, this last statement says:
A constant factor can be moved through a limit symbol.
In Figure 2.3.2 we have graphed the polynomials of the form x n for n = 1, 2, 3, and 4.
Below each figure we have indicated the limits as x + and as x . The results in
the figure are special cases of the following general results:
LIMITS OF x n AS x
lim x n = +,
n = 1, 2, 3, . . .
x +
lim x =
n
n = 1, 3, 5, . . .
+,
n = 2, 4, 6, . . .
(4)
y=
x3
y = x4
y = x2
y=x
x
-4
(3)
x
-4
-8
-8
lim x = +
x
-4
-8
lim x 2 = +
lim x 3 = +
-4
-8
lim x 4 = +
x +
x +
x +
x +
lim x =
lim x 2 = +
lim x 3 =
lim x 4 = +
Figure 2.3.2
Multiplying x n by a positive real number does not affect limits (3) and (4), but multiplying
by a negative real number reverses the sign.
Example 1
lim 2x 5 = +,
x +
lim 7x 6 = ,
x +
LIMITS OF POLYNOMIALS AS
x
lim 2x 5 =
lim 7x 6 =
There is a useful principle about polynomials which, expressed informally, states that:
The end behavior of a polynomial matches the end behavior of its highest degree term.
More precisely, if cn = 0 then
lim c0 + c1 x + + cn x n = lim cn x n
x
lim
x +
c0 + c1 x + + cn x n = lim cn x n
x +
(5)
(6)
We can motivate these results by factoring out the highest power of x from the polynomial
134
g65-ch2
## and examining the limit of the factored expression. Thus,
c0
c1
+
+
+
c
c0 + c 1 x + + c n x n = x n
n
xn
x n1
As x or x +, it follows from (1) that all of the terms with positive powers of x
in the denominator approach 0, so (5) and (6) are certainly plausible.
Example 2
lim (7x 5 4x 3 + 2x 9) = lim 7x 5 =
x
8
## LIMITS OF RATIONAL FUNCTIONS
AS x
A useful technique for determining the end behavior of a rational function f(x) = n(x)/d(x)
is to factor and cancel the highest power of x that occurs in the denominator d(x) from
both n(x) and d(x). The denominator of the resulting fraction then has a (nonzero) limit
equal to the leading coefficient of d(x), so the limit of the resulting fraction can be quickly
determined using (1), (5), and (6). The following examples illustrate this technique.
Example 3 Find lim
x +
3x + 5
.
6x 8
Solution. Divide the numerator and denominator by the highest power of x that occurs
in the denominator; that is, x 1 = x. We obtain
lim (3 + 5/x)
x(3 + 5/x)
3 + 5/x
3x + 5
x +
= lim
= lim
=
lim
x + 6x 8
x + x(6 8/x)
x + 6 8/x
lim (6 8/x)
x +
x +
x +
## lim 6 lim 8/x
x +
x +
3 + 5 lim 1/x
x +
6 8 lim 1/x
x +
3 + (5 0)
1
=
6 (8 0)
2
Example 4 Find
4x 2 x
x 2x 3 5
(a) lim
5x 3 2x 2 + 1
x
3x + 5
(b) lim
Solution (a). Divide the numerator and denominator by the highest power of x that occurs
in the denominator, namely x 3 . We obtain
4x 2 x
x 3 (4/x 1/x 2 )
4/x 1/x 2
=
lim
=
lim
lim
x 2x 3 5
x x 3 (2 5/x 3 )
x 2 5/x 3
=
lim (2 5/x 3 )
(4 0) 0
0
= =0
2 (5 0)
2
## Solution (b). Divide the numerator and denominator by x to obtain
5x 3 2x 2 + 1
5x 2 2x + 1/x
= lim
= +
x
x
3x + 5
3 + 5/x
where the final step is justified by the fact that
5
1
5x 2 2x +,
0, and 3 + 3
x
x
as x .
lim
g65-ch2
x +
Solution.
lim
x +
3x + 5
=
6x 8
=
2.3
lim
3
135
3x + 5
.
6x 8
3
## cyan magenta yellow black
x +
3x + 5
6x 8
Theorem 2.3.2(e)
1
2
Example 3
Example 6 Find
x2 + 2
x2 + 2
(a) lim
(b) lim
x + 3x 6
x 3x 6
In both parts it would be helpful to manipulate the function so that the powers of x are
transformed to powers of 1/x. This can be achieved
in both cases by dividing the numerator
and denominator by |x| and using the fact that x 2 = |x|.
## Solution (a). As x +, the values of x under consideration are positive, so we can
replace |x| by x where helpful. We obtain
x2 + 2
x 2 + 2/|x|
x 2 + 2/ x 2
lim
= lim
= lim
x + 3x 6
x + (3x 6)/|x|
x + (3x 6)/x
lim
1 + 2/x 2
2
1 + 2/x
x +
=
= lim
x + 3 6/x
lim (3 6/x)
x +
lim (1 +
x +
lim (3 6/x)
x +
1 + (2 0)
1
=
=
3
3 (6 0)
1
x
-2 -1
2/x 2 )
lim 1 + 2 lim 1/x 2
x +
x +
=
lim 3 6 lim 1/x
x +
x +
-1
x 2 + 2/|x|
x 2 + 2/ x 2
x2 + 2
lim
= lim
= lim
x (3x 6)/|x|
x (3x 6)/(x)
x 3x 6
1 + 2/x 2
1
= lim
=
x 3 + 6/x
3
y = x 6 + 5 x 3
(a)
y
4
y = 2.5
2
1
x
-1
-1
## Use a graphing utility to explore the end behavior of
2
x +2
f(x) =
3x 6
Your investigation should support the results of Example 6.
Example 7 Find
(a) lim ( x 6 + 5 x 3 )
x +
y=
x6
5x 3
(b)
Figure 2.3.3
x 3,
x0
(b) lim ( x 6 + 5x 3 x 3 )
x +
## x 6 + 5 x 3 and g(x) = x 6 + 5x 3 x 3 for
x 0 are shown in Figure 2.3.3. From the graphs we might conjecture that the limits are 0
and 2.5, respectively. To confirm this, we treat each function as a fraction with denominator
136
g65-ch2
## 1 and rationalize the numerator.
x6 + 5 + x3
3
3
6
6
lim ( x + 5 x ) = lim ( x + 5 x )
x +
x +
x6 + 5 + x3
(x 6 + 5) x 6
5
= lim
= lim
6
3
6
x +
x
x +5+x
x + 5 + x3
5/x 3
= lim
x +
1 + 5/x 6 + 1
=
lim (
x +
x6
5x 3
0
x ) = lim (
x +
x6
x 6 = x 3 for x > 0
=0
1+0+1
5x 3
x 6 + 5x 3 + x 3
x )
x 6 + 5x 3 + x 3
3
(x 6 + 5x 3 ) x 6
5x 3
= lim
= lim
x +
x +
x 6 + 5x 3 + x 3
x 6 + 5x 3 + x 3
5
= lim
x +
1 + 5/x 3 + 1
=
1+0+1
x 6 = x 3 for x > 0
5
2
REMARK.
## Example 7 illustrates an indeterminate form of type . Exercises 3134
explore more examples of this type.
Graphing Calculator
## 1. In each part, find the limit by inspection.
(a) lim (3)
(b) lim (2h)
x
(g) lim
lim f(x) = 3,
x +
lim g(x) = 5,
x +
lim h(x) = 0
x +
find the limits that exist. If the limit does not exist, explain
why.
(a) lim [f(x) + 3g(x)]
(b) lim [h(x) 4g(x) + 1]
x +
x +
x +
(g) lim
x +
3
## (d) lim [g(x)]
x +
3
(f ) lim
x + g(x)
6f(x)
(h) lim
x + 5f(x) + 3g(x)
5 + f(x)
3h(x) + 4
x2
lim f(x) = 7,
lim g(x) = 6
find the limits that exist. If the limit does not exist, explain
why.
3
f(x) +
g(x)
x
g(x)
f(x)
xf(x)
(h) lim
x (2x + 3)g(x)
(f ) lim
f(x)g(x)
5.
7.
4. Given that
x
(e) lim
3. Given that
(e) lim
x
(a) lim f(x)
(b) lim f(x)
x +
h +
x +
9.
lim (3 x)
6. lim
lim (1 + 2x 3x )
x
lim
5
x +
x +
3x + 1
11. lim
x + 2x 5
3
13. lim
y y + 4
15.
lim
x x 2
17.
lim
x +
x2
+ 2x + 1
2 + 3x 5x 2
1 + 8x 2
1
5
x
x +
10. lim
5x
x
5x 2 4x
x + 2x 2 + 3
1
14. lim
x + x 12
12. lim
5x 2 + 7
x + 3x 2 x
7
5
3 3s 4s
18. lim
s +
2s 7 + 1
16. lim
g65-ch2
2.3
19.
21.
lim
lim
5x 2 2
x+3
2y
7 + 6y 2
20. lim
x +
22. lim
y +
5x 2 2
x+3
2y
7 + 6y 2
3x 4 + x
3x 4 + x
24. lim
2
x x 8
x + x 2 8
5
7 6x
5 2t 3
26. lim 2
25. lim
x + x + 3
t t + 1
3
6t
x + 4x 3
28. lim
27. lim
3
t + 7t + 3
x 1 x 2 + 7x 3
29. Let
2
x<0
2x + 5,
3
f(x) =
3 5x , x 0
1 + 4x + x 3
Find
(a) lim f(x)
(b) lim f(x).
23.
## Computing Limits: End Behavior
137
40. Find
c0 + c 1 x + + c n x n
x + d0 + d1 x + + dm x m
where cn = 0 and dm = 0. [Hint: Your answer will depend
on whether m < n, m = n, or m > n.]
lim
lim
x +
30. Let
2 + 3t
t < 1,000,000
5t 2 + 6 ,
g(t) =
2
5t
Find
(a) lim g(t)
(b) lim g(t).
t
t +
## In Exercises 3134, find the limits.
32. lim ( x 2 3x x)
lim ( x 2 + 3 x)
x +
x +
2
33. lim ( x + ax x)
x +
34. lim ( x 2 + ax x 2 + bx)
31.
## The notion of an asymptote can be extended to include curves
as well as lines. Specifically, we say that f(x) is asymptotic
to g(x) as x + if
lim [f(x) g(x)] = 0
x +
## and that f(x) is asymptotic to g(x) as x if
lim [f(x) g(x)] = 0
## Informally stated, if f(x) is asymptotic to g(x) as x +,
then the graph of y = f(x) gets closer and closer to the graph
of y = g(x) as x +, and if f(x) is asymptotic to g(x) as
x , then the graph of y = f(x) gets closer and closer
to the graph of y = g(x) as x . For example, if
2
f(x) = x 2 +
and g(x) = x 2
x1
then f(x) is asymptotic to g(x) as x + and as x
since
1
lim [f(x) g(x)] = lim
=0
x +
x + x 1
1
=0
lim [f(x) g(x)] = lim
x
x x 1
This asymptotic behavior is illustrated in the following figure,
which also shows the vertical asymptote of f(x) at x = 1.
y
20
15
x +
## 35. Discuss the limits of p(x) = (1 x) as x + and
x for positive integer values of n.
36. Let p(x) = (1 x)n and q(x) = (1 x)m . Discuss the
limits of p(x)/q(x) as x + and x for positive
integer values of m and n.
n
## 37. Let p(x) be a polynomial of degree n. Discuss the limits
of p(x)/x m as x + and x for positive integer
values of m.
38. In each part, find examples of polynomials p(x) and q(x)
that satisfy the stated condition and such that p(x) +
and q(x) + as x +.
p(x)
p(x)
(a) lim
=1
(b) lim
=0
x + q(x)
x + q(x)
p(x)
= +
(d) lim [p(x) q(x)] = 3
(c) lim
x + q(x)
x +
39. Assuming that m and n are positive integers, find
2 + 3x n
x 1 x m
lim
or m > n.]
y = f (x)
10
y = g(x)
x
-4
-3
-2
-1
-5
-10
## In Exercises 4146, determine a function g(x) to which f(x)
is asymptotic as x + or x . Use a graphing utility
to generate the graphs of y = f(x) and y = g(x) and identify
all vertical asymptotes.
x2 2
x3 x + 3
42. f(x) =
x2
x
x 3 + 3x 2 + x 1
43. f(x) =
x3
x5 x3 + 3
44. f(x) =
x2 1
1
x3 x2 + 2
45. f(x) = sin x +
46. f(x) =
x1
x1
41. f(x) =
138
g65-ch2
## 2.4 LIMITS (DISCUSSED MORE RIGOROUSLY)
Thus far, our discussion of limits has been based on our intuitive feeling of what it
means for the values of a function to get closer and closer to a limiting value. However, this level of informality can only take us so far, so our goal in this section is to
define limits precisely. From a purely mathematical point of view these definitions are
needed to establish limits with certainty and to prove theorems about them. However,
they will also provide us with a deeper understanding of the limit concept, making it
possible for us to visualize some of the more subtle properties of functions.
In Sections 2.1 to 2.3 our emphasis was on the discovery of values of limits, either through the
sampling of selected x-values or through the application of limit theorems. In the preceding
sections we interpreted limx a f(x) = L to mean that the values of f(x) can be made
as close as we like to L by selecting x-values sufficiently close to a (but not equal to a).
Although this informal definition is sufficient for many purposes, we need a more precise
definition to verify that a conjectured limit is actually correct, or to prove the limit theorems
in Sections 2.2 and 2.3. One of our goals in this section is to give the informal phrases as
close as we like to L and sufficiently close to a a precise mathematical interpretation.
This will enable us to replace the informal definition of limit given in Definition 2.1.1 with
a more fully developed version that may be used in proofs.
To start, consider the function f graphed in Figure 2.4.1a for which f(x) L as x a.
We have intentionally placed a hole in the graph at x = a to emphasize that the function
f need not be defined at x = a to have a limit there. Also, to simplify the discussion, we
have chosen a function that is increasing on an open interval containing a.
y
y = f (x)
y = f (x)
L+
L
L
L
(a)
y = f (x)
L+
x0
(b)
x1
x0
x1
(c)
Figure 2.4.1
To motivate an appropriate definition for a two-sided limit, suppose that we choose any
positive number, say ", and draw horizontal lines from L + " and L " on the y-axis to the
curve y = f(x) and then draw vertical lines from those points on the curve to the x-axis. As
shown in Figure 2.4.1b, let x0 and x1 be points where the vertical lines intersect the x-axis.
Next, imagine that x gets closer and closer to a (from either side). Eventually, x will
lie inside the interval (x0 , x1 ), which is marked in green in Figure 2.4.1c; and when this
happens, the value of f(x) will fall between L " and L + ", marked in red in the figure.
Thus, we conclude:
If f(x) L as x a, then for any positive number ", we can find an open interval
(x0 , x1 ) on the x-axis that contains a and has the property that for each x in that
interval (except possibly for x = a), the value of f(x) is between L " and L + ".
## Consider the limit, limx 0 (sin x)/x, conjectured to be 1 in Example
3 of Section 2.1. Draw a figure similar to Figure 2.4.1 that illustrates the preceding analysis
for this limit.
g65-ch2
2.4
## Limits (Discussed More Rigorously)
139
What is important about this result is that it holds no matter how small we make ".
However, making " smaller and smaller forces f(x) closer and closer to Lwhich is
precisely the concept we were trying to capture mathematically.
Observe that in Figure 2.4.1c the interval (x0 , x1 ) extends farther on the right side of a
than on the left side. However, for many purposes it is preferable to have an interval that
extends the same distance on both sides of a. For this purpose, let us choose any positive
number that is smaller than both x1 a and a x0 , and consider the interval (a , a + ).
This interval extends the same distance on both sides of a and lies inside of the interval
(x0 , x1 ) (Figure 2.4.2). Moreover, the condition L " < f(x) < L + " holds for every
x in this interval (except possibly x = a), since this condition holds on the larger interval
(x0 , x1 ). This is illustrated by graphing f in the window (a , a + ) (L ", L + ") and
observing that the graph exits the window at the sides, not at the top or bottom (except
possibly at x = a).
Example 1 Let f(x) = 12 x + 14 sin(x /2). It can be shown that lim f(x) = L = 0.75.
x 1
Let " = 0.05.
y = f (x)
L+
(a)
Use a graphing utility to find an open interval (x0 , x1 ) containing a = 1 such that for
each x in this interval, f(x) is between L " = 0.75 " = 0.75 0.05 = 0.70 and
L + " = 0.75 + " = 0.75 + 0.05 = 0.80.
(b)
Find a value of such that f(x) is between 0.70 and 0.80 for every x in the interval
(1 , 1 + ).
L
L
x
x0
x1
d
d
(
(
(
x0
Solution (a). Figure 2.4.3 displays the graph of f . With a graphing utility, we discover
that (to five decimal places) the points (0.90769, 0.70122) and (1.09231, 0.79353) are
on the graph of f . Suppose that we take x0 = 0.908 and x1 = 1.09. Since the graph
of f rises from left to right, we see that for x0 = 0.908 < x < 1.090 = x1 , we have
0.90769 < x < 1.09231 and therefore 0.7 < 0.70122 < f(x) < 0.79353 < 0.8.
a+d
x
(
x1
Figure 2.4.2
Solution (b). Since x1 a = 1.09 1 = 0.09 and a x0 = 1 0.908 = 0.902, any value
or that is less than 0.09 will be acceptable. For example, for = 0.08, if x belongs to the
interval (1 , 1 + ) = (0.92, 1.08), then f(x) will lie between 0.70 and 0.80.
Note that the condition L " < f(x) < L + " can be expressed as
|f(x) L| < "
and the condition that x lies in the interval (a , a + ), but x = a, can be expressed as
y
0 < |x a| <
y=
1
x
2
1
4
px
sin
2
( )
## Thus, we can summarize this discussion in the following definition.
0.5
2.4.1 LIMIT DEFINITION. Let f(x) be defined for all x in some open interval containing the number a, with the possible exception that f(x) need not be defined at a. We
will write
x
0
0.5
lim f(x) = L
x a
Figure 2.4.3
if given any number " > 0 we can find a number > 0 such that
|f(x) L| < "
REMARK.
if
0 < |x a| <
With this definition we have made the transition from informal to formal in
the definition of a two-sided limit. The phrase as close as we like to L has been given
quantitative meaning by the number " > 0, and the phrase sufficiently close to a has been
140
g65-ch2
## Limits and Continuity
made precise by the number > 0. Commonly known as the "- definition of a limit,
Definition 2.4.1 was developed primarily by the German mathematician Karl Weierstrass
in the nineteenth century.
The definitions for one-sided limits are similar to Definition 2.4.1. For example, in the
definition of limx a + f(x) we assume that f(x) is defined for all x in an interval of the
form (a, b) and replace the condition 0 < |x a| < by the condition a < x < a + .
Comparable changes are made in the definition of limx a f(x).
In the preceding sections we illustrated various numerical and graphical methods for
guessing at limits. Now that we have a precise definition to work with, we can actually
confirm the validity of those guesses with mathematical proof. Here is a typical example of
such a proof.
Example 2 Use Definition 2.4.1 to prove that lim (3x 5) = 1.
x 2
Solution. We must show that given any positive number ", we can find a positive number
such that
| (3x 5) 1 | < "
f(x)
if
0 < |x 2 | <
(1)
There are two things to do. First, we must discover a value of for which this statement
holds, and then we must prove that the statement holds for that . For the discovery part we
begin by simplifying (1) and writing it as
|3x 6| < "
if
0 < |x 2| <
Next, we will rewrite this statement in a form that will facilitate the discovery of an appropriate :
3|x 2| < "
if
0 < |x 2| <
|x 2| < " /3
if
0 < |x 2| <
(2)
It should be self-evident that this last statement holds if = " /3, which completes the
discovery portion of our work. Now we need to prove that (1) holds for this choice of .
However, statement (1) is equivalent to (2), and (2) holds with = " /3, so (1) also holds
with = " /3. This proves that limx 2 (3x 5) = 1.
KARL WEIERSTRASS (18151897). Weierstrass, the son of a customs officer, was born in Ostenfelde, Germany.
As a youth Weierstrass showed outstanding skills in languages and mathematics. However, at the urging of his
dominant father, Weierstrass entered the law and commerce program at the University of Bonn. To the chagrin of
his family, the rugged and congenial young man concentrated instead on fencing and beer drinking. Four years
later he returned home without a degree. In 1839 Weierstrass entered the Academy of Munster to study for a career
in secondary education, and he met and studied under an excellent mathematician named Christof Gudermann.
Gudermanns ideas greatly influenced the work of Weierstrass. After receiving his teaching certificate, Weierstrass
spent the next 15 years in secondary education teaching German, geography, and mathematics. In addition, he
taught handwriting to small children. During this period much of Weierstrasss mathematical work was ignored
because he was a secondary schoolteacher and not a college professor. Then, in 1854, he published a paper of
major importance that created a sensation in the mathematics world and catapulted him to international fame
overnight. He was immediately given an honorary Doctorate at the University of Konigsberg and began a new
career in college teaching at the University of Berlin in 1856. In 1859 the strain of his mathematical research
caused a temporary nervous breakdown and led to spells of dizziness that plagued him for the rest of his life.
Weierstrass was a brilliant teacher and his classes overflowed with multitudes of auditors. In spite of his fame,
he never lost his early beer-drinking congeniality and was always in the company of students, both ordinary and
brilliant. Weierstrass was acknowledged as the leading mathematical analyst in the world. He and his students
opened the door to the modern school of mathematical analysis.
g65-ch2
2.4
## Limits (Discussed More Rigorously)
141
REMARK.
This example illustrates the general form of a limit proof: We assume that we
are given a positive number ", and we try to prove that we can find a positive number such
that
|f(x) L| < "
if
0 < |x a| <
(3)
This is done by first discovering , and then proving that the discovered works. Since
the argument has to be general enough to work for all positive values of ", the quantity
has to be expressed as a function of ". In Example 2 we found the function = " /3
by some simple algebra; however, most limit proofs require a little more algebraic and
logical ingenuity. Thus, if you find our ensuing discussion of "- proofs challenging, do
not become discouraged; the concepts and techniques are intrinsically difficult. In fact, a
precise understanding of limits evaded the finest mathematical minds for more than 150
years after the basic concepts of calculus were discovered.
x 0
## Solution. Note that the domain of x is 0 x, so it is valid to discuss the limit as x 0+ .
We must show that given " > 0, there exists a > 0 such that
or more simply,
## x < " if 0 < x <
But, by squaring both sides of the inequality
x < "2
if
(4)
x < ", we can rewrite (4) as
0<x<
(5)
It should be self-evident that (5) is true if = " ; and since (5) is areformulation of (4), we
have shown that (4) holds with = " 2 . This proves that limx 0+ x = 0.
2
REMARK.
THE VALUE OF
IS NOT UNIQUE
## In this example the
limit from the left and the two-sided limit do not exist at
x = 0 because the domain of x includes no numbers to the left of 0.
In preparation for our next example, we note that the value of in Definition 2.4.1 is not
unique; once we have found a value of that fulfills the requirements of the definition, then
any smaller positive number 1 will also fulfill those requirements. That is, if it is true that
|f(x) L| < "
if
0 < |x a| <
|f(x) L| < "
if
0 < |x a| < 1
## This is because {x : 0 < |x a| < 1 } is a subset of {x : 0 < |x a| < } (Figure 2.4.4),
and hence if |f(x) L| < " is satisfied for all x in the larger set, then it will automatically
be satisfied for all x in the subset. Thus, in Example 2, where we used = " /3, we could
have used any smaller value of such as = " /4, = " /5, or = " /6.
(
(
(
(
a d1 a a + d1 a + d
Figure 2.4.4
## Example 4 Prove that lim x 2 = 9.
x 3
Solution. We must show that given any positive number ", we can find a positive number
such that
|x 2 9| < "
if
0 < |x 3| <
(6)
Because |x 3| occurs on the right side of this if statement, it will be helpful to factor the
left side to introduce a factor of |x 3|. This yields the following alternative form of (6)
|x + 3||x 3| < "
if
0 < |x 3| <
(7)
142
g65-ch2
## Using the triangle inequality, we see that
|x + 3| = |(x 3) + 6| |x 3| + 6
Therefore, if 0 < |x 3| < then
|x + 3||x 3| (|x 3| + 6)|x 3| < ( + 6)
It follows that (7) will be satisfied for any positive value of such that ( + 6) ". Let
us agree to restrict our attention to positive values of such that 1. (This is justified
because of our earlier observation that once a value of is found, then any smaller positive
value of can be used.) With this restriction, ( + 6) 7, so that (7) will be satisfied as
long as it is also the case that 7 ". We can achieve this by taking to be the minimum
of the numbers " /7 and 1, which is sometimes written as = min(" /7, 1). This proves that
limx 3 x 2 = 9.
REMARK.
LIMITS AS x
You may have wondered how we knew to make the restriction 1 (as opposed
to 12 or 5, for example). Actually, it does not matter; any restriction of the form
c would work equally well.
## In Section 2.1 we discussed the limits
lim f(x) = L
x +
and
lim f(x) = L
from an intuitive viewpoint. We interpreted the first statement to mean that the values of
f(x) eventually get closer and closer to L as x increases indefinitely, and we interpreted the
second statement to mean that the values of f(x) eventually get closer and closer to L as x
decreases indefinitely. These ideas are captured more precisely in the following definitions
and are illustrated in Figure 2.4.5.
2.4.2 DEFINITION. Let f(x) be defined for all x in some infinite open interval extending in the positive x-direction. We will write
lim f(x) = L
x +
if given any number " > 0, there corresponds a positive number N such that
|f(x) L| < "
if
x>N
2.4.3 DEFINITION. Let f(x) be defined for all x in some infinite open interval extending in the negative x-direction. We will write
lim f(x) = L
if given any number " > 0, there corresponds a negative number N such that
|f(x) L| < "
if
x<N
To see how these definitions relate to our informal concepts of these limits, suppose
that f(x) L as x +, and for a given " let N be the positive number described in
Definition 2.4.2. If x is allowed to increase indefinitely, then eventually x will lie in the
interval (N, +), which is marked in green in Figure 2.4.5a; when this happens, the value
of f(x) will fall between L " and L + ", marked in red in the figure. Since this is true for
all positive values of " (no matter how small), we can force the values of f(x) as close as
we like to L by making N sufficiently large. This agrees with our informal concept of this
limit. Similarly, Figure 2.4.5b illustrates Definition 2.4.3.
g65-ch2
2.4
143
L+e
f (x)
L
L+e
f (x)
L
Le
Le
(a)
(b)
Figure 2.4.5
## Example 5 Prove that lim
x +
1
= 0.
x
Solution. Applying Definition 2.4.2 with f(x) = 1/x and L = 0, we must show that
given " > 0, we can find a number N > 0 such that
1
0 < " if x > N
x
(8)
Because x + we can assume that x > 0. Thus, we can eliminate the absolute values in
this statement and rewrite it as
1
< " if x > N
x
or, on taking reciprocals,
1
x>
if x > N
(9)
"
It is self-evident that N = 1/" satisfies this requirement, and since (9) is equivalent to (8)
for x > 0, the proof is complete.
INFINITE LIMITS
In Section 2.1 we discussed limits of the following type from an intuitive viewpoint:
lim f(x) = +,
x a
lim f(x) = +,
x a+
lim f(x) = +,
x a
lim f(x) =
(10)
lim f(x) =
(11)
lim f(x) =
(12)
x a
x a+
x a
Recall that each of these expressions describes a particular way in which the limit fails to
exist. The + indicates that the limit fails to exist because f(x) increases without bound,
and the indicates that the limit fails to exist because f(x) decreases without bound.
These ideas are captured more precisely in the following definitions and are illustrated in
Figure 2.4.6.
2.4.4 DEFINITION. Let f(x) be defined for all x in some open interval containing a,
except that f(x) need not be defined at a. We will write
lim f(x) = +
x a
if given any positive number M, we can find a number > 0 such that f(x) satisfies
f(x) > M
if
0 < |x a| <
144
g65-ch2
y
a+d
M
x
a+d
## f (x) < M if 0 < | x a | < d
(a)
(b)
Figure 2.4.6
2.4.5 DEFINITION. Let f(x) be defined for all x in some open interval containing a,
except that f(x) need not be defined at a. We will write
lim f(x) =
x a
if given any negative number M, we can find a number > 0 such that f(x) satisfies
f(x) < M
if
0 < |x a| <
To see how these definitions relate to our informal concepts of these limits, suppose
that f(x) + as x a, and for a given M let be the corresponding positive number
described in Definition 2.4.4. Next, imagine that x gets closer and closer to a (from either
side). Eventually, x will lie in the interval (a , a + ), which is marked in green in
Figure 2.4.6a; when this happens the value of f(x) will be greater than M, marked in red in
the figure. Since this is true for any positive value of M (no matter how large), we can force
the values of f(x) to be as large as we like by making x sufficiently close to a. This agrees
with our informal concept of this limit. Similarly, Figure 2.4.6b illustrates Definition 2.4.5.
REMARK.
The definitions for the one-sided limits are similar. For example, in the definition
of limx a f(x) = + we assume that f(x) is defined for all x in some interval of the
form (c, a) and replace the condition 0 < |x a| < by the condition a < x < a.
1
= +.
x 0 x2
## Example 6 Prove that lim
Solution. Applying Definition 2.4.4 with f(x) = 1/x 2 and a = 0, we must show that
given a number M > 0, we can find a number > 0 such that
1
> M if 0 < |x 0| <
(13)
x2
or, on taking reciprocals and simplifying,
1
x2 <
if 0 < |x| <
(14)
M
But x 2 < 1/M if |x| < 1/ M, so that = 1/ M satisfies (14). Since (13) is equivalent
to (14), the proof is complete.
lim f(x) = +,
x +
lim f(x) = +,
lim f(x) =
x +
lim f(x) = ?
(15)
g65-ch2
2.4
## Limits (Discussed More Rigorously)
145
Graphing Calculator
## 1. (a) Find the largest open interval, centered at the origin on
the x-axis, such that for each x in the interval the value
of the function f(x) = x + 2 is within 0.1 unit of the
number f(0) = 2.
(b) Find the largest open interval, centered at x = 3, such
that for each x in the interval the value of the function f(x) = 4x 5 is within 0.01 unit of the number
f(3) = 7.
(c) Find the largest open interval, centered at x = 4, such
that for each x in the interval the value of the function f(x) = x 2 is within 0.001 unit of the number
f(4) = 16.
2. In each part, find the largest open interval, centered at
x = 0, such that for each x in the interval the value of
f(x) = 2x + 3 is within " units of the number f(0) = 3.
(a) " = 0.1
## (c) " = 0.0012
3. (a) Find the values of x1 and x2 in the accompanying figure.
0 < |x 4| < .
## that the inequality |f(x) 2| < 0.05 can be rewritten as
1.95 < x 3 4x + 5 < 2.05, and estimate the values of x
for which x 3 4x + 5 = 1.95 and x 3 4x + 5 = 2.05.]
6. Use
the method of Exercise 5 to find a number such that
| 5x + 1 4| < 0.5 if 0 < |x 3| < .
## 7. Let f(x) = x + x with L = limx 1 f(x) and let " = 0.2.
Use a graphing utility and its trace feature to find a positive
number such that |f(x) L| < " if 0 < |x 1| < .
8. Let f(x) = (sin 2x)/x and use a graphing utility to conjecture the value of L = limx 0 f(x). Then let " = 0.1 and
use the graphing utility and its trace feature to find a positive
number such that |f(x) L| < " if 0 < |x| < .
In Exercises 918, a positive number " and the limit L of
a function f at a are given. Find a number such that
|f(x) L| < " if 0 < |x a| < .
9. lim 2x = 8; " = 0.1
x 4
1
x = 1; " = 0.1
x 2 2
10. lim
x 1
x 3
x2 4
= 4; " = 0.05
x 2 x 2
x2 1
14. lim
= 2; " = 0.05
x 1 x + 1
## 16. lim x = 3; " = 0.001
15. lim x 2 = 16; " = 0.001
13. lim
y = x
2 + 0.05
2
2 0.05
x 4
x1
x2
x 9
1
1
17. lim = ; " = 0.05
x 5 x
5
x 0
## In Exercises 1932, use Definition 2.4.1 to prove that the
stated limit is correct.
Figure Ex-3
## 4. (a) Find the values of x1 and x2 in the accompanying figure.
(b) Find a positive number such that |(1/x) 1| < 0.1 if
0 < |x 1| < .
19. lim 3x = 15
## 22. lim (2 3x) = 5
x +x
=1
x 0
x
25. lim 2x 2 = 2
x2 9
= 6
x 3 x + 3
26. lim (x 2 5) = 4
x 5
x 2
23. lim
1
y= x
x 1
1
=3
27. lim
x 1/3 x
29. lim x = 2
1 + 0.1
1
1 0.1
x
x1 1
x2
## Not drawn to scale
Figure Ex-4
5. Generate the graph of f(x) = x 3 4x + 5 with a graphing utility, and use the graph to find a number such that
|f(x) 2| < 0.05 if 0 < |x 1| < . [Hint: Show
x 3
x 1
24. lim
x 3
1
= 1
x+1
30. lim x + 3 = 3
x 4
x 6
x + 2, x = 1
31. lim f(x) = 3, where f(x) =
x 1
10,
x=1
28. lim
x 2
32. lim (x 2 + 3x 1) = 9
x 2
33. (a) Find the smallest positive number N such that for each
x in the interval (N, +), the value of the function
f(x) = 1/x 2 is within 0.1 unit of L = 0.
146
g65-ch2
## Limits and Continuity
y
(b) Find the smallest positive number N such that for each
x in the interval (N, +), the value of f(x) = x /(x+1)
is within 0.01 unit of L = 1.
1
y=3
x
(c) Find the largest negative number N such that for each
x in the interval (, N), the value of the function
f(x) = 1/x 3 is within 0.001 unit of L = 0.
x1
(d) Find the largest negative number N such that for each
x in the interval (, N), the value of the function
f(x) = x /(x + 1) is within 0.01 unit of L = 1.
34. In each part, find the smallest positive value of N such that
for each x in the interval (N, +), the function f(x) = 1/x 3
is within " units of the number L = 0.
(a) " = 0.1
(b) " = 0.01
(c) " = 0.001
35. (a) Find the values of x1 and x2 in the accompanying figure.
(b) Find a positive number N such that
x2
1 + x 2 1 < "
for x > N.
(c) Find a negative number N such that
x2
<"
1
1 + x2
x2
Figure Ex-36
## In Exercises 3740, a positive number " and the limit L of a
function f at + are given. Find a positive number N such
that |f(x) L| < " if x > N .
1
= 0; " = 0.01
x2
1
= 0; " = 0.005
38. lim
x + x + 2
x
39. lim
= 1; " = 0.001
x + x + 1
4x 1
40. lim
= 2; " = 0.1
x + 2x + 5
37.
lim
x +
## In Exercises 4144, a positive number " and the limit L of a
function f at are given. Find a negative number N such
that |f(x) L| < " if x < N.
for x < N.
y =
1
x2
1 + x2
x
x1
1
= 0; " = 0.005
x+2
1
42. lim 2 = 0; " = 0.01
x x
4x 1
= 2; " = 0.1
43. lim
x 2x + 5
x
44. lim
= 1; " = 0.001
x x + 1
41.
x2
Figure Ex-35
## 36. (a) Find the values of x1 and x2 in the accompanying figure.
(b) Find a positive number N such that
1
0 = 1 < "
3 x
3x
## In Exercises 4552, use Definition 2.4.2 or 2.4.3 to prove that
the stated limit is correct.
45.
47.
for x > N.
(c) Find a negative number N such that
1
0 = 1 < "
3x
3 x
49.
51.
53.
for x < N.
lim
1
1
=0
46. lim
=0
x x
x2
1
1
=0
48. lim
=0
lim
x x + 2
x + x + 2
x
x
lim
=1
50. lim
=1
x + x + 1
x x + 1
4x 1
4x 1
lim
=2
52. lim
=2
x 2x + 5
x + 2x + 5
(a) Find the largest open interval, centered at the origin on
the x-axis, such that for each x in the interval, other
lim
x +
g65-ch2
2.5
## than the center, the values of f(x) = 1/x 2 are greater
than 100.
(b) Find the largest open interval, centered at x = 1, such
that for each x in the interval, other than the center, the
values of the function
f(x) = 1/|x 1|
are greater than 1000.
(c) Find the largest open interval, centered at x = 3, such
that for each x in the interval, other than the center, the
values of the function
f(x) = 1/(x 3)2
are less than 1000.
(d) Find the largest open interval, centered at the origin on
the x-axis, such that for each x in the interval, other
than the center, the values of f(x) = 1/x 4 are less
than 10,000.
54. In each part, find the largest open interval, centered at
x = 1, such that for each x in the interval the value of
f(x) = 1/(x 1)2 is greater than M.
(a) M = 10
(b) M = 1000
(c) M = 100,000
In Exercises 5560, use Definition 2.4.4 or 2.4.5 to prove that
the stated limit is correct.
1
55. lim
= +
x 3 (x 3)2
1
57. lim
= +
x 0 |x|
1
59. lim 4 =
x 0
x
1
56. lim
=
x 3 (x 3)2
1
58. lim
= +
x 1 |x 1|
1
60. lim 4 = +
x 0 x
## In Exercises 6166, use the remark following Definition 2.4.1
to prove that the stated limit is correct.
61. lim+ (x + 1) = 3
x 2
63. lim+ x 4 = 0
Continuity
147
## 62. lim (3x + 2) = 5
x 1
64. lim
x 4
x 0
x,
65. lim+ f(x) = 2, where f(x) =
x 2
3x,
x,
66. lim f(x) = 6, where f(x) =
x 2
3x,
x = 0
x>2
x2
x>2
x2
## In Exercises 67 and 68, use the remark following Definitions
2.4.4 and 2.4.5 to prove that the stated limit is correct.
67. (a) lim+
1
=
1x
(b) lim
1
= +
1x
## 68. (a) lim+
1
= +
x
(b) lim
1
=
x
x 1
x 0
x 1
x 0
For Exercises 69 and 70, write out definitions of the four limits in (18), and use your definitions to prove that the stated
limits are correct.
69. (a) lim (x + 1) = +
(b) lim (x + 1) =
## 70. (a) lim (x 2 3) = +
(b) lim (x 3 + 5) =
x +
x +
x
x
## 71. Prove the result in Example 4 under the assumption that
2 rather than 1.
72. (a) In Definition 2.4.1 there is a condition requiring that
f(x) be defined for all x in some open interval containing a, except possibly at a itself. What is the purpose
of this requirement?
x 0
## x = 0.1 a correct statement?
(c) Is lim
x 0.01
2.5 CONTINUITY
A moving object cannot vanish at some point and reappear someplace else to continue its motion. Thus, we perceive the path of a moving object as an unbroken curve,
without gaps, breaks, or holes. In this section, we translate unbroken curve into a
precise mathematical formulation called continuity, and develop some fundamental
properties of continuous curves.
DEFINITION OF CONTINUITY
Recall from Theorem 2.2.3 that if p(x) is a polynomial and c is a real number, then
limx c p(x) = p(c) (see Figure 2.5.1). Together with Theorem 2.2.2, we are able to
calculate limits of a variety of combinations of functions by evaluating the combination.
That is, we saw many examples of functions f (x) such that limx c f(x) = f (c) if f(x)
is defined on an interval containing a number c. In this case, function values f(x) can be
guaranteed to be near f(c) for any x-value selected close enough to c. (See Exercise 53 for
a precise formulation of this statement.)
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## lim p(x) = p(c)
p(c)
xc
c
Figure 2.5.1
On the other hand, we have also seen functions for which this nice property is not true.
For example,
sin(/x), x = 0
f(x) =
0,
x=0
does not satisfy limx 0 f(x) = f (0), since limx 0 f(x) fails to exist.
y
1
x
-1
-1
Figure 2.5.2
The term continuous is used to describe the useful circumstance where the calculation
of a limit can be accomplished by mere evaluation of the function.
2.5.1 DEFINITION. A function f is said to be continuous at x = c provided the
following conditions are satisfied:
1. f(c) is defined.
2. lim f(x) exists.
x c
3.
## lim f(x) = f(c).
x c
If one or more of the conditions of this definition fails to hold, then we will say that f
has a discontinuity at x = c. Each function drawn in Figure 2.5.3 illustrates a discontinuity
at x = c. In Figure 2.5.3a, the function is not defined at c, violating the first condition
of Definition 2.5.1. In Figures 2.5.3b and 2.5.3c, limx c f(x) does not exist, violating
the second condition of Definition 2.5.1. In Figure 2.5.3d, the function is defined at c and
limx c f(x) exists, but these two values are not equal, violating the third condition of
Definition 2.5.1.
From such graphs we can develop an intuitive, geometric feel for where a function is
continuous and where it is discontinuous. Observe that continuity at c may fail due to a
break in the graph of the function, either due to a hole or to a jump as in Figure 2.5.3, or
perhaps due to a wild oscillation as in Figure 2.5.2. Although the intuitive interpretation of
f is continuous at c as the graph of f is unbroken at c lacks precision, it is a useful
guide in most circumstances.
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## cyan magenta yellow black
2.5
y
Continuity
149
y = f (x)
y = f (x)
y = f (x)
y = f (x)
(a)
(b)
(c)
(d)
Figure 2.5.3
REMARK.
Note that the third condition of Definition 2.5.1 really implies the first two
conditions, since it is understood in the statement limx c f(x) = f (c) that the limit on
the left exists, the expression f(c) on the right is defined and has a finite value, and that
quantitites on the two sides are equal. Thus, when we want to establish continuity of a
function at a point our usual procedure will be to establish the validity of the third condition
only.
Example 1 Determine whether the following functions are continuous at x = 2.
2
x2 4
x 4, x =
2
, x = 2
2
x 4
f(x) =
h(x) = x 2
,
g(x) = x 2
x2
4,
3,
x = 2,
x=2
Solution. In each case we must determine whether the limit of the function as x 2 is
the same as the value of the function at x = 2. In all three cases the functions are identical,
except at x = 2, and hence all three have the same limit at x = 2, namely
x2 4
= lim (x + 2) = 4
x 2
x 2
x 2
x 2 x 2
x 2
The function f is undefined at x = 2, and hence is not continuous at x = 2 (Figure 2.5.4a).
The function g is defined at x = 2, but its value there is g(2) = 3, which is not the same as
the limit as x approaches z; hence, g is also not continuous at x = 2 (Figure 2.5.4b). The
value of the function h at x = 2 is h(2) = 4, which is the same as the limit as x approaches
z; hence, h is continuous at x = 2 (Figure 2.5.4c). (Note that the function h could have
been written more simply as h(x) = x + 2, but we wrote it in piecewise form to emphasize
its relationship to f and g.)
lim f(x) = lim g(x) = lim h(x) = lim
y = f (x)
y = g(x)
y = h (x)
4
3
x
2
x
2
(a)
x
2
(b)
(c)
Figure 2.5.4
CONTINUITY IN APPLICATIONS
In applications, discontinuities often signal the occurrence of important physical phenomena. For example, Figure 2.5.5a is a graph of voltage versus time for an underground cable
that is accidentally cut by a work crew at time t = t0 (the voltage drops to zero when the line
150
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## Limits and Continuity
y (Units of inventory)
V (Voltage)
y1
y0
t
Line
cut
t0
Restocking occurs
(a)
(b)
Figure 2.5.5
is cut). Figure 2.5.5b shows the graph of inventory versus time for a company that restocks
its warehouse to y1 units when the inventory falls to y0 units. The discontinuities occur at
those times when restocking occurs.
Given the possible physical significance of discontinuities, it is important to be able to
identify discontinuities for specific functions, and to be able to make general statements
about the continuity properties of entire families of functions. This is our next goal.
## CONTINUITY ON AN INTERVAL AND
CONTINUITY OF POLYNOMIALS
If a function f is continuous at each number in an open interval (a, b), then we say that f is
continuous on (a, b). This definition applies to infinite open intervals of the form (a, +),
(, b), and (, +). In the case where f is continuous on (, +), we will say that
f is continuous everywhere.
The general procedure for showing that a function is continuous everywhere is to show
that it is continuous at an arbitrary real number. For example, we showed in Theorem 2.2.3
that if p(x) is a polynomial and a is any real number, then
lim p(x) = p(a)
x a
2.5.2
THEOREM.
## Solution. We can write |x| as
0
|x| =
if
x>0
if
x=0
if
x<0
so |x| is the same as the polynomial x on the interval (0, +) and is the same as the
polynomial x on the interval (, 0). But polynomials are continuous everywhere, so
x = 0 is the only possible discontinuity for |x|. Since |0| = 0, to prove the continuity at
x = 0 we must show that
lim |x| = 0
(1)
x 0
Because the formula for |x| changes at 0, it will be helpful to consider the one-sided limits
at 0 rather than the two-sided limit. We obtain
lim |x| = lim+ x = 0
x 0+
x 0
and
x 0
x 0
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## cyan magenta yellow black
2.5
Continuity
151
The following theorem, which is a consequence of Theorem 2.2.2, will enable us to reach
conclusions about the continuity of functions that are obtained by adding, subtracting,
multiplying, and dividing continuous functions.
SOME PROPERTIES OF
CONTINUOUS FUNCTIONS
## 2.5.3 THEOREM. If the functions f and g are continuous at c, then
(a) f + g is continuous at c.
(b) f g is continuous at c.
(c) f g is continuous at c.
(d ) f /g is continuous at c if g(c) = 0 and has a discontinuity at c if g(c) = 0.
We will prove part (d ). The remaining proofs are similar and will be omitted.
Proof. First, consider the case where g(c) = 0. In this case f(c)/g(c) is undefined, so
the function f /g has a discontinuity at c.
Next, consider the case where g(c) = 0. To prove that f /g is continuous at c, we must
show that
f(x)
f(c)
lim
=
(2)
x c g(x)
g(c)
Since f and g are continuous at c,
lim f(x) = f(c)
x c
and
x c
lim
x c
lim f(x)
f(c)
f(x)
x c
=
=
g(x)
lim g(x)
g(c)
x c
## which proves (2).
Since polynomials are continuous everywhere, and since rational functions are ratios of
polynomials, part (d ) of Theorem 2.5.3 yields the following result.
CONTINUITY OF RATIONAL
FUNCTIONS
2.5.4 THEOREM. A rational function is continuous at every number where the denominator is nonzero.
Example 3 For what values of x is there a hole or a gap in the graph of
y=
x2 9
?
x 2 5x + 6
Solution. The function being graphed is a rational function, and hence is continuous at
every number where the denominator is nonzero. Solving the equation
x 2 5x + 6 = 0
yields discontinuities at x = 2 and at x = 3.
CONTINUITY OF COMPOSITIONS
## If you use a graphing utility to generate the graph of the equation in
this example, then there is a good chance that you will see the discontinuity at x = 2 but
not at x = 3. Try it, and explain what you think is happening.
The following theorem, whose proof is given in Appendix G, will be useful for calculating
limits of compositions of functions.
152
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## 2.5.5 THEOREM. If limx c g(x) = L and if the function f is continuous at L, then
limx c f(g(x)) = f(L). That is,
lim f(g(x)) = f lim g(x)
x c
x c
## This equality remains valid if limx c is replaced everywhere by one of limx c+ ,
limx c , limx + , or limx .
In words, this theorem states:
A limit symbol can be moved through a function sign provided the limit of the expression inside the function sign exists and the function is continuous at this limit.
Example 4 We know from Example 2 that the function |x| is continuous everywhere;
thus, it follows that if limx a g(x) exists, then
lim |g(x)| = lim g(x)
(3)
x a
x a
That is, a limit symbol can be moved through an absolute value sign, provided the limit of
the expression inside the absolute value signs exists. For example,
lim |5 x 2 | = lim (5 x 2 ) = | 4| = 4
x 3
x 3
## The following theorem is concerned with the continuity of compositions of functions;
the first part deals with continuity at a specific number, and the second part with continuity
everywhere.
2.5.6 THEOREM.
(a) If the function g is continuous at c, and the function f is continuous at g(c), then
the composition f g is continuous at c.
(b) If the function g is continuous everywhere and the function f is continuous everywhere, then the composition f g is continuous everywhere.
Proof. We will prove part (a) only; the proof of part (b) can be obtained by applying part
(a) at an arbitrary number c. To prove that f g is continuous at c, we must show that the
value of f g and the value of its limit are the same at x = c. But this is so, since we can
write
lim (f g)(x) = lim f(g(x)) = f( lim g(x)) = f(g(c)) = (f g)(c)
x c
x c
x c
Theorem 2.5.5
We know from Example 2 that the function |x| is continuous everywhere. Thus, if g(x)
is continuous at c, then by part (a) of Theorem 2.5.6, the function |g(x)| must also be
continuous at c; and, more generally, if g(x) is continuous everywhere, then so is |g(x)|.
Stated informally:
y = | 4 x 2|
g is continuous at c.
5
4
3
2
1
-4 -3 -2 -1
x
1
## For example, the polynomial g(x) = 4 x 2 is continuous everywhere, so we can conclude
that the function |4 x 2 | is also continuous everywhere (Figure 2.5.6).
Figure 2.5.6
Can the absolute value of a function that is not continuous be continuous? Justify your answer.
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## cyan magenta yellow black
2.5
Continuity
153
Because Definition 2.5.1 involves a two-sided limit, that definition does not generally apply
at the endpoints of a closed interval [a, b] or at the endpoint of an interval of the form
[a, b), (a, b], (, b], or [a, +). To remedy this problem, we will agree that a function
is continuous at an endpoint of an interval if its value at the endpoint is equal to the appropriate one-sided limit at that endpoint. For example, the function graphed in Figure 2.5.7 is
continuous at the right endpoint of the interval [a, b] because
AND RIGHT
y
y = f (x)
x b
x a+
Figure 2.5.7
lim f(x) = f(c)
x c
lim f(x) = f(c)
x c+
## Using this terminology we define continuity on a closed interval as follows.
2.5.7 DEFINITION. A function f is said to be continuous on a closed interval [a, b]
if the following conditions are satisfied:
1. f is continuous on (a, b).
2. f is continuous from the right at a.
3. f is continuous from the left at b.
## We leave it for you to modify this definition appropriately so that it
applies to intervals of the form [a, +), (, b], (a, b], and [a, b).
Example 5 What can you say about the continuity of the function f(x) =
9 x2?
Solution. Because the natural domain of this function is the closed interval [3, 3], we
will need to investigate the continuity of f on the open interval (3, 3) and at the two
endpoints. If c is any number in the interval (3, 3), then it follows from Theorem 2.2.2(e)
that
lim f(x) = lim 9 x 2 = lim (9 x 2 ) = 9 c2 = f(c)
x c
x c
x c
which proves f is continuous at each number in the interval (3, 3). The function f is also
continuous at the endpoints since
lim f(x) = lim 9 x 2 = lim (9 x 2 ) = 0 = f(3)
x 3
x 3
x 3
lim + (9 x 2 ) = 0 = f(3)
lim + f(x) = lim + 9 x 2 =
x 3
x 3
x 3
## Thus, f is continuous on the closed interval [3, 3].
THE INTERMEDIATE-VALUE
THEOREM
Figure 2.5.8 shows the graph of a function that is continuous on the closed interval [a, b].
The figure suggests that if we draw any horizontal line y = k, where k is between f(a)
and f(b), then that line will cross the curve y = f(x) at least once over the interval [a, b].
Stated in numerical terms, if f is continuous on [a, b], then the function f must take on
every value k between f(a) and f(b) at least once as x varies from a to b. For example,
the polynomial p(x) = x 5 x + 3 has a value of 3 at x = 1 and a value of 33 at x = 2.
Thus, it follows from the continuity of p that the equation x 5 x + 3 = k has at least one
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## Limits and Continuity
y
solution in the interval [1, 2] for every value of k between 3 and 33. This idea is stated more
precisely in the following theorem.
f (b)
k
## 2.5.8 THEOREM (Intermediate-Value Theorem). If f is continuous on a closed interval
[a, b] and k is any number between f(a) and f(b), inclusive, then there is at least one
number x in the interval [a, b] such that f(x) = k.
f (a)
x
x b
Figure 2.5.8
## APPROXIMATING ROOTS USING
THE INTERMEDIATE-VALUE
THEOREM
f (a) > 0
b
a
f (x) = 0
Figure 2.5.9
A variety of problems can be reduced to solving an equation f(x) = 0 for its roots. Sometimes it is possible to solve for the roots exactly using algebra, but often this is not possible
and one must settle for decimal approximations of the roots. One procedure for approximating roots is based on the following consequence of the Intermediate-Value Theorem.
2.5.9 THEOREM. If f is continuous on [a, b], and if f(a) and f(b) are nonzero and
have opposite signs, then there is at least one solution of the equation f(x) = 0 in the
interval (a, b).
f (b) < 0
Although this theorem is intuitively obvious, its proof depends on a mathematically precise
development of the real number system, which is beyond the scope of this text.
## This result, which is illustrated in Figure 2.5.9, can be proved as follows.
Proof. Since f(a) and f(b) have opposite signs, 0 is between f(a) and f(b). Thus, by
the Intermediate-Value Theorem there is at least one number x in the interval [a, b] such
that f(x) = 0. However, f(a) and f(b) are nonzero, so x must lie in the interval (a, b),
which completes the proof.
Before we illustrate how this theorem can be used to approximate roots, it will be helpful
to discuss some standard terminology for describing errors in approximations. If x is an
approximation to a quantity x0 , then we call
" = |x x0 |
the absolute error or (less precisely) the error in the approximation. The terminology in
Table 2.5.1 is used to describe the size of such errors:
Table 2.5.1
error
description
|x x0 |
| x x0 |
| x x0 |
| x x0 |
0.1
0.01
0.001
0.0001
## x approximates x0 with an error of at most 0.1.
x approximates x0 with an error of at most 0.01.
x approximates x0 with an error of at most 0.001.
x approximates x0 with an error of at most 0.0001.
| x x0 |
| x x0 |
| x x0 |
| x x0 |
0.5
0.05
0.005
0.0005
## x approximates x0 to the nearest integer.
x approximates x0 to 1 decimal place (i.e., to the nearest tenth).
x approximates x0 to 2 decimal places (i.e., to the nearest hundredth).
x approximates x0 to 3 decimal places (i.e., to the nearest thousandth).
## Example 6 The equation
x3 x 1 = 0
cannot be solved algebraically very easily because the left side has no simple factors.
However, if we graph p(x) = x 3 x 1 with a graphing utility (Figure 2.5.10), then we
are led to conjecture that there is one real root and that this root lies inside the interval [1, 2].
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## cyan magenta yellow black
2.5
Continuity
155
The existence of a root in this interval is also confirmed by Theorem 2.5.9, since p(1) = 1
and p(2) = 5 have opposite signs. Approximate this root to two decimal-place accuracy.
y
2
Solution. Our objective is to approximate the unknown root x0 with an error of at most
x
2
y = x3 x 1
Figure 2.5.10
0.005. It follows that if we can find an interval of length 0.01 that contains the root, then the
midpoint of that interval will approximate the root with an error of at most 0.01/2 = 0.005,
which will achieve the desired accuracy.
We know that the root x0 lies in the interval [1, 2]. However, this interval has length
1, which is too large. We can pinpoint the location of the root more precisely by dividing
the interval [1, 2] into 10 equal parts and evaluating p at the points of subdivision using
a calculating utility (Table 2.5.2). In this table p(1.3) and p(1.4) have opposite signs, so
we know that the root lies in the interval [1.3, 1.4]. This interval has length 0.1, which is
still too large, so we repeat the process by dividing the interval [1.3, 1.4] into 10 parts and
evaluating p at the points of subdivision; this yields Table 2.5.3, which tells us that the root
is inside the interval [1.32, 1.33] (Figure 2.5.11). Since this interval has length 0.01, its
midpoint 1.325 will approximate the root with an error of at most 0.005. Thus, x0 1.325
to two decimal-place accuracy.
Table 2.5.2
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
f (x)
0.77
0.47
0.10
0.34
0.88
1.50
2.21
3.03
3.96
1.4
Table 2.5.3
1.3
1.31
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
## f (x) 0.103 0.062 0.020 0.023
0.066
0.110
0.155
0.201
0.248
0.296 0.344
y = p(x) = x 3 x 1
0.02
0.01
x
1.322 1.324 1.326 1.328 1.330
-0.01
-0.02
Figure 2.5.11
APPROXIMATING ROOTS BY
ZOOMING WITH A GRAPHING
UTILITY
The method illustrated in Example 6 can also be implemented with a graphing utility as
follows.
Step 1.
## Figure 2.5.12a shows the graph of f in the window [5, 5] [5, 5]
with xScl = 1 and yScl = 1. That graph places the root between
x = 1 and x = 2.
Step 2.
## Since we know that the root lies between x = 1 and x = 2, we will
zoom in by regraphing f over an x-interval that extends between
these values and in which xScl = 0.1. The y-interval and yScl are not
critical, as long as the y-interval extends above and below the x-axis.
Figure 2.5.12b shows the graph of f in the window [1, 2] [1, 1]
with xScl = 0.1 and yScl = 0.1. That graph places the root between
x = 1.3 and x = 1.4.
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## Limits and Continuity
Step 3.
Since we know that the root lies between x = 1.3 and x = 1.4, we
will zoom in again by regraphing f over an x-interval that extends
between these values and in which xScl = 0.01. Figure 2.5.12c shows
the graph of f in the window [1.3, 1.4] [0.1, 0.1] with xScl =
0.01 and yScl = 0.01. That graph places the root between x = 1.32
and x = 1.33.
Step 4.
Since the interval in Step 3 has length 0.01, its midpoint 1.325 approximates the root with an error of at most 0.005, so x0 1.325 to
two decimal-place accuracy.
[5, 5] [5, 5]
xScl = 1, yScl = 1
[1, 2] [1, 1]
xScl = 0.1, yScl = 0.1
## [1.3, 1.4] [0.1, 0.1]
xScl = 0.01, yScl = 0.01
(a)
(b)
(c)
Figure 2.5.12
REMARK.
To say that x approximates x0 to n decimal places does not mean that the first
n decimal places of x and x0 will be the same when the numbers are rounded to n decimal
places. For example, x = 1.084 approximates x0 = 1.087 to two decimal places because
|x x0 | = 0.003(<0.005). However, if we round these values to two decimal places, then
we obtain x 1.08 and x0 1.09. Thus, if you approximate a number to n decimal places,
then you should display that approximation to at least n + 1 decimal places to preserve the
accuracy.
## Use a graphing or calculating utility to show that the root x0 in Example
6 can be approximated as x0 1.3245 to three decimal-place accuracy.
Graphing Calculator
## In Exercises 14, let f be the function whose graph is shown.
On which of the following intervals, if any, is f continuous?
(a) [1, 3]
(b) (1, 3)
(c) [1, 2]
(d) (1, 2)
(e) [2, 3]
(f) (2, 3)
For each interval on which f is not continuous, indicate which
conditions for the continuity of f do not hold.
1.
2.
4.
x
1
x
1
3.
x
1
## In Exercises 5 and 6, find all values of c such that the specified
function has a discontinuity at x = c. For each such value of
c, determine which conditions of Definition 2.5.1 fail to be
satisfied.
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2.5
## 5. (a) The function f in Exercise 1 of Section 2.1.
(b) The function F in Exercise 5 of Section 2.1.
(c) The function f in Exercise 9 of Section 2.1.
6. (a) The function f in Exercise 2 of Section 2.1.
(b) The function F in Exercise 6 of Section 2.1.
(c) The function f in Exercise 10 of Section 2.1.
7. Suppose that f and g are continuous functions such that
f(2) = 1 and lim [f(x) + 4g(x)] = 13. Find
(a) g(2)
x 2
x 2
## 8. Suppose that f and g are continuous functions such that
lim g(x) = 5 and f(3) = 2. Find lim [f(x)/g(x)].
x 3
x 3
## 9. In each part sketch the graph of a function f that satisfies
the stated conditions.
(a) f is continuous everywhere except at x = 3, at which
point it is continuous from the right.
(b) f has a two-sided limit at x = 3, but it is not continuous
at x = 3.
(c) f is not continuous at x = 3, but if its value at x = 3
is changed from f(3) = 1 to f(3) = 0, it becomes
continuous at x = 3.
(d) f is continuous on the interval [0, 3) and is defined on
the closed interval [0, 3]; but f is not continuous on the
interval [0, 3].
10. Find formulas for some functions that are continuous on the
intervals (, 0) and (0, +), but are not continuous on the
interval (, +).
11. A student parking lot at a university charges \$2.00 for the
first half hour (or any part) and \$1.00 for each subsequent
half hour (or any part) up to a daily maximum of \$10.00.
(a) Sketch a graph of cost as a function of the time parked.
(b) Discuss the significance of the discontinuities in the
graph to a student who parks there.
12. In each part determine whether the function is continuous
or not, and explain your reasoning.
(a) The Earths population as a function of time
(b) Your exact height as a function of time
(c) The cost of a taxi ride in your city as a function of the
distance traveled
(d) The volume of a melting ice cube as a function of time
In Exercises 1324, find the values of x (if any) at which f
is not continuous.
13. f(x) = x 3 2x + 3
## 14. f(x) = (x 5)17
x
15. f(x) = 2
x +1
x
16. f(x) = 2
x 1
17. f(x) =
x4
x 2 16
18. f(x) =
3x + 1
x 2 + 7x 2
19. f(x) =
x
|x| 3
20. f(x) =
5
2x
+
x
x+4
Continuity
157
x+3
22. f(x) = 2
21. f(x) = |x 3 2x 2 |
|x + 3x|
2x + 3, x 4
23. f(x) =
16
7 + , x > 4
x
3 , x = 1
24. f(x) = x 1
3,
x=1
25. Find a value for the constant k, if possible, that will make
the function continuous everywhere.
7x 2, x 1
(a) f(x) =
kx 2 ,
x>1
2
kx ,
x2
(b) f(x) =
2x + k, x > 2
26. On which of the following intervals is
1
f(x) =
x2
continuous?
(a) [2, +) (b) (, +) (c) (2, +)
(d) [1, 2)
## A function f is said to have a removable discontinuity at
x = c if limx c f(x) exists but f is not continuous at x = c,
either because f is not defined at c or because the definition
for f(c) differs from the value of the limit. This terminology
will be needed in Exercises 2730.
27. (a) Sketch the graph of a function with a removable discontinuity at x = c for which f(c) is undefined.
(b) Sketch the graph of a function with a removable discontinuity at x = c for which f(c) is defined.
28. (a) The terminology removable discontinuity is appropriate because a removable discontinuity of a function f
at x = c can be removed by redefining the value of
f appropriately at x = c. What value for f(c) removes
the discontinuity?
(b) Show that the following functions have removable discontinuities at x = 1, and sketch their graphs.
1, x > 1
2
x 1
f(x) =
and g(x) = 0, x = 1
x1
1, x < 1
(c) What values should be assigned to f(1) and g(1) to
remove the discontinuities?
In Exercises 29 and 30, find the values of x (if any) at which
f is not continuous, and determine whether each such value
is a removable discontinuity.
|x|
x
x2
(c) f(x) =
|x| 2
(b) f(x) =
x 2 + 3x
x+3
158
g65-ch2
(b) f(x) =
(c) f(x) =
x2 4
x3 8
2x 3,
x2
## 45. The accompanying figure shows the graph of y = x 4 +x1.
Use the method of Example 6 to approximate the xintercepts with an error of at most 0.05.
x2,
x>2
3x 2 + 5, x =
1
6,
x=1
31. (a) Use a graphing utility to generate the graph of the function f(x) = (x + 3)/(2x 2 + 5x 3), and then use
the graph to make a conjecture about the number and
locations of all discontinuities.
(b) Check your conjecture by factoring the denominator.
32. (a) Use a graphing utility to generate the graph of the function f(x) = x /(x 3 x + 2), and then use the graph to
make a conjecture about the number and locations of
all discontinuities.
(b) Use the Intermediate-Value Theorem to approximate
the location of all discontinuities to two decimal places.
33. Prove that f(x) = x 3/5 is continuous everywhere, carefully
justifying each step.
34. Prove that f(x) = 1/ x 4 + 7x 2 + 1 is continuous everywhere, carefully justifying each step.
[5, 4] [3, 6]
xScl = 1, yScl = 1
Figure Ex-45
## 46. Use a graphing utility to solve the problem in Exercise 45
by zooming.
47. The accompanying figure shows the graph of y = 5xx 4 .
Use the method of Example 6 to approximate the roots of
the equation 5 x x 4 = 0 to two decimal-place accuracy.
## 35. Let f and g be discontinuous at c. Give examples to show
that
(a) f + g can be continuous or discontinuous at c
(b) fg can be continuous or discontinuous at c.
36. Prove Theorem 2.5.4.
37. Prove:
(a) part (a) of Theorem 2.5.3
(b) part (b) of Theorem 2.5.3
(c) part (c) of Theorem 2.5.3.
[5, 4] [3, 6]
xScl = 1, yScl = 1
Figure Ex-47
38. Prove: If f and g are continuous on [a, b], and f(a) > g(a),
f(b) < g(b), then there is at least one solution of the equation f(x) = g(x) in (a, b). [Hint: Consider f(x) g(x).]
## 48. Use a graphing utility to solve the problem in Exercise 47
by zooming.
2
49. Use the
fact that 5 is a solution of x 5 = 0 to approximate 5 with an error of at most 0.005.
## 39. Give an example of a function f that is defined on a closed
interval, and whose values at the endpoints have opposite
signs, but for which the equation f(x) = 0 has no solution
in the interval.
a
b
+
=0
x1 x3
## 40. Use the Intermediate-Value Theorem to show that there is a
square with a diagonal length that is between r and 2r and
an area that is half the area of a circle of radius r.
41. Use the Intermediate-Value Theorem to show that there is
a right circular cylinder of height h and radius less than r
whose volume is equal to that of a right circular cone of
In Exercises 42 and 43, show that the equation has at least
one solution in the given interval.
42. x 3 4x + 1 = 0; [1, 2]
## has at least one solution in the interval (1, 3).
51. A sphere of unknown radius x consists of a spherical core
and a coating that is 1 cm thick (see the accompanying figure). Given that the volume of the coating and the volume of
the core are the same, approximate the radius of the sphere
to three decimal-place accuracy.
x
1 cm
43. x 3 +x 2 2x = 1; [1, 1]
44. Prove: If p(x) is a polynomial of odd degree, then the equation p(x) = 0 has at least one real solution.
Figure Ex-51
g65-ch2
2.6
## 52. A monk begins walking up a mountain road at 12:00 noon
and reaches the top at 12:00 midnight. He meditates and
rests until 12:00 noon the next day, at which time he begins
walking down the same road, reaching the bottom at 12:00
midnight. Show that there is at least one point on the road
159
## that he reaches at the same time of day on the way up as on
the way down.
53. Let f be defined at c. Prove that f is continuous at c if, given
" > 0, there exists a > 0 such that |f(x) f(c)| < " if
|x c| < .
## 2.6 LIMITS AND CONTINUITY OF TRIGONOMETRIC FUNCTIONS
In this section we will investigate the continuity properties of the trigonometric functions, and we will discuss some important limits involving these functions.
CONTINUITY OF TRIGONOMETRIC
FUNCTIONS
Before we begin, recall that in the expressions sin x, cos x, tan x, cot x, sec x, and csc x it
is understood that x is in radian measure.
In trigonometry, the graphs of sin x and cos x are drawn as continuous curves (Figure 2.6.1). To actually prove that these functions are continuous everywhere, we must show
that the following equalities hold for every real number c:
lim sin x = sin c
x c
and
## lim cos x = cos c
(12)
x c
Although we will not formally prove these results, we can make them plausible by considering the behavior of the point P (cos x, sin x) as it moves around the unit circle. For this
purpose, view c as a fixed angle in radian measure, and let Q(cos c, sin c) be the corresponding point on the unit circle. As x c (i.e., as the angle x approaches the angle c), the point
P moves along the circle toward Q, and this implies that the coordinates of P approach the
corresponding coordinates of Q; that is, cos x cos c, and sin x sin c (Figure 2.6.2).
y
1
x
x
O
-1
y = sin x
-1
y = cos x
Figure 2.6.1
Q(cos c, sin c)
Formulas (1) and (2) can be used to find limits of the remaining trigonometric functions
by expressing them in terms of sin x and cos x; for example, if cos c = 0, then
sin c
sin x
=
= tan c
cos x
cos c
Thus, we are led to the following theorem.
lim tan x = lim
x c
P(cos x, sin x)
x c
2.6.1 THEOREM. If c is any number in the natural domain of the stated trigonometric
function, then
lim sin x = sin c
x c
Figure 2.6.2
x c
x c
x c
x c
## lim cot x = cot c
x c
It follows from this theorem, for example, that sin x and cos x are continuous everywhere
and that tan x is continuous, except at the points where it is undefined.
160
g65-ch2
## Example 1 Find the limit
2
x 1
lim cos
x 1
x1
Solution. Recall from the last section that since the cosine function is continuous everywhere,
lim cos(g(x)) = cos( lim g(x))
x 1
x 1
## provided limx 1 g(x) exists. Thus,
2
x 1
= lim cos(x + 1) = cos lim (x + 1) = cos 2
lim cos
x 1
x 1
x 1
x1
## In Section 2.1 we used the numerical evidence in Table ?? to conjecture that
sin x
=1
(3)
lim
x 0 x
However, it is not a simple matter to establish this limit with certainty. The difficulty is
that the numerator and denominator both approach zero as x 0. As discussed in Section
2.2, such limits are called indeterminate forms of type 0/0. Sometimes indeterminate forms
of this type can be established by manipulating the ratio algebraically, but in this case no
simple algebraic manipulation will work, so we must look for other methods.
The problem with indeterminate forms of type 0/0 is that there are two conflicting
influences at work: as the numerator approaches 0 it drives the magnitude of the ratio
toward 0, and as the denominator approaches 0 it drives the magnitude of the ratio toward
(depending on the sign of the expression). The limiting behavior of the ratio is determined
by the precise way in which these influences offset each other. Later in this text we will
discuss general methods for attacking indeterminate forms, but for the limit in (3) we can
use a method called squeezing.
In the method of squeezing one proves that a function f has a limit L at a number c by
trapping the function between two other functions, g and h, whose limits at c are known to be
L (Figure 2.6.3). This is the idea behind the following theorem, which we state without proof.
y = h(x)
y = f (x)
L
y = g(x)
x
c
Figure 2.6.3
2.6.2
## g(x) f(x) h(x)
for all x in some open interval containing the number c, with the possible exception that
the inequalities need not hold at c. If g and h have the same limit as x approaches c, say
lim g(x) = lim h(x) = L
x c
y=
1
sin x
x
lim f(x) = L
lim
x0
y=
1
1 cos x
x
lim
x0
Figure 2.6.4
x c
sin x
x =1
y
x c
## then f also has this limit as x approaches c, that is,
1 cos x
=0
x
The Squeezing Theorem also holds for one-sided limits and limits at
+ and . How do you think the hypotheses of the theorem would change in those cases?
The usefulness of the Squeezing Theorem will be evident in our proof of the following
theorem (Figure 2.6.4).
2.6.3
(a)
THEOREM.
lim
x 0
sin x
=1
x
(b)
lim
x 0
1 cos x
=0
x
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2.6
## Limits and Continuity of Trigonometric Functions
161
However, before giving the proof, it will be helpful to review the formula for the area A of
a sector with radius r and a central angle of radians (Figure 2.6.5). The area of the sector
can be derived by setting up the following proportion to the area of the entire circle:
A
area of the sector
central angle of the sector
=
=
area of the circle
central angle of the circle
r 2
2
Area = A
u
r
A = 12 r 2
(4)
Figure 2.6.5
## Now we are ready for the proof of Theorem 2.6.3.
Proof (a). In this proof we will interpret x as an angle in radian measure, and we will
assume to start that 0 < x < /2. It follows from Formula (4) that the area of a sector of
radius 1 and central angle x is x /2. Moreover, it is suggested by Figure 2.6.6 that the area
of this sector lies between the areas of two triangles, one with area (tan x)/2 and one with
area (sin x)/2. Thus,
tan x
x
sin x
2
2
2
Multiplying through by 2/(sin x) yields
1
x
1
sin x
cos x
and then taking reciprocals and reversing the inequalities yields
sin x
1
(5)
x
Moreover, these inequalities also hold for /2 < x < 0, since replacing x by x in (5)
and using the identities sin(x) = sin x and cos(x) = cos x leaves the inequalities
unchanged (verify). Finally, since the functions cos x and 1 both have limits of 1 as x 0,
it follows from the Squeezing Theorem that (sin x)/x also has a limit of 1 as x 0.
cos x
(1, tan x)
(cos x, sin x)
tan x
(1, 0)
x
1
sin x
1
Area of triangle
tan x
2
1
Area of sector
x
2
1
Area of triangle
sin x
2
Figure 2.6.6
Proof (b). For this proof we will use the limit in part (a), the continuity of the sine function,
and the trigonometric identity sin2 x = 1 cos2 x. We obtain
1 cos x
sin2 x
1 cos x 1 + cos x
= lim
= lim
lim
x 0
x 0
x 0 (1 + cos x)x
x
x
1 + cos x
0
sin x
sin x
= lim
lim
= (1)
=0
x 0 x
x 0 1 + cos x
1+1
Example 2 Find
tan x
sin 2
(b) lim
(a) lim
x 0 x
0
(c) lim
x 0
sin 3x
sin 5x
162
g65-ch2
## Limits and Continuity
Solution (a).
tan x
= lim
x 0 x
x 0
lim
sin x
1
x
cos x
= (1)(1) = 1
Solution (b). The trick is to multiply and divide by 2, which will make the denominator
the same as the argument of the sine function [ just as in Theorem 2.6.3(a)]:
sin 2
sin 2
sin 2
= lim 2
= 2 lim
0
2
2
Now make the substitution x = 2 , and use the fact that x 0 as 0. This yields
lim
lim
sin 2
sin 2
sin x
= 2 lim
= 2 lim
= 2(1) = 2
0 2
x 0 x
Solution (c).
sin 3x
sin 3x
3
sin 3x
3
31
x
3x
= lim
=
= lim
=
lim
x 0 sin 5x
x 0 sin 5x
x 0
51
5
sin 5x
5
x
5x
Use a graphing utility to confirm the limits in the last example graphically, and if you have a CAS, then use it to obtain the limits.
## Example 3 Make conjectures about the limits
1
1
(a) lim sin
(b) lim x sin
x 0
x 0
x
x
x
-1
and confirm your conclusions by generating the graphs of the functions near x = 0 using a
graphing utility.
Solution (a). Since 1/x + as x 0+ , we can view sin(1/x) as the sine of an angle
-1
that increases indefinitely as x 0+ . As this angle increases, the function sin(1/x) keeps
oscillating between 1 and 1 without approaching a limit. Similarly, there is no limit from
the left since 1/x as x 0 . These conclusions are consistent with the graph of
y = sin(1/x) shown in Figure 2.6.7a. Observe that the oscillations become more and more
rapid as x approaches 0 because 1/x increases (or decreases) more and more rapidly as x
approaches 0.
y = sin 1
x
()
(a)
y
y = |x|
## Solution (b). If x > 0, x x sin(1/x) x, and if x < 0, x x sin(1/x) x.
Thus, for x = 0, |x| x sin(1/x) |x|. Since both |x| 0 and |x| 0 as x 0,
the Squeezing Theorem applies and we can conclude that x sin(1/x) 0 as x 0. This is
illustrated in Figure 2.6.7b.
REMARK.
y = |x|
y = x sin
(b)
Figure 2.6.7
1
x
()
## It follows from part (b) of this example that the function
x sin(1/x), x = 0
f(x) =
0,
x=0
is continuous at x = 0, since the value of the function and the value of the limit are the
same at 0. This shows that the behavior of a function can be very complex in the vicinity
of an x-value c, even though the function is continuous at c.
g65-ch2
2.6
## EXERCISE SET 2.6
Graphing Calculator
163
CAS
35. lim
## In Exercises 110, find the discontinuities, if any.
1. f(x) = sin(x 2 2)
2. f(x) = cos
3. f(x) = cot x
4. f(x) = sec x
5. f(x) = csc x
6. f(x) =
7. f(x) = | cos x|
1
9. f(x) =
1 2 sin x
x 0
x
x
1 + sin2 x
8. f(x) = 2 + tan2 x
3
10. f(x) =
5 + 2 cos x
## 11. Use Theorem 2.5.6 to show that the following functions
are continuous everywhere by expressing them as compositions of simpler functions that are known to be continuous.
(a) sin(x 3 + 7x + 1)
(b) |sin x|
(d) 3 + sin 2x
(c) cos3 (x + 1)
(e) sin(sin x)
(f ) cos5 x 2 cos3 x + 1
12. (a) Prove that if g(x) is continuous everywhere, then so are
sin(g(x)), cos(g(x)), g(sin(x)), and g(cos(x)).
(b) Illustrate the result in part (a) with some of your own
choices for g.
1
13. lim cos
x +
x
x
15. lim sin
x +
2 3x
sin 3
17. lim
0
2
14. lim sin
x +
x
16. lim
h0
sin h
2h
sin
18. lim+ 2
0
sin2 x
x 0 3x 2
20. lim
sin x
21. lim+
x 0 5 x
22. lim
tan 7x
x 0 sin 3x
h
25. lim
h 0 tan h
sin2
0
sin h
26. lim
h 0 1 cos h
x 0
23. lim
2
0 1 cos
x 0
sin 6x
sin 8x
24. lim
x
1
cos 2 x
27. lim
28. lim
0 cos
1 cos 5h
31. lim
h 0 cos 7h 1
1
33. lim+ cos
x 0
x
t2
t 0 1 cos2 t
1
32. lim+ sin
x 0
x
29. lim
36. lim
x 5
x 0
30. lim
x 2 3 sin x
34. lim
x 0
x
sin(x 5)
x 2 25
37. lim
x 2
sin(2x 4)
x2 4
sin(x 2 + 3x + 2)
sin(x 2 + 3x + 2)
39. lim
x 1
x 2
x3 + 1
x+2
40. Find a value for the constant k that makes
sin 3x , x = 0
x
f(x) =
k,
x=0
38. lim
continuous at x = 0.
41. Find a nonzero value for the constant k that makes
tan kx ,
x<0
x
f(x) =
3x + 2k 2 , x 0
continuous at x = 0.
42. Is
sin x
|x|
19. lim
## In Exercises 3639: (i) Construct a table to estimate the limit
by evaluating the function near the limiting value. (ii) Find
the exact value of the limit.
## Find the limits in Exercises 1335.
2x + sin x
x
sin x ,
|x|
f(x) =
1,
x = 0
x=0
continuous at x = 0?
43. In each part, find the limit by making the indicated substitution.
1
1
(a) lim x sin ; t =
x +
x
x
1
1
; t=
(b) lim x 1 cos
x
x
x
x
(c) lim
. [Hint: Let t = x.]
x sin x
cos(/x)
; t= .
44. Find lim
x 2 x 2
2
x
sin(x)
tan x1
.
46. Find lim
.
45. Find lim
x 1 x 1
x /4 x/4
47. Use the Squeezing Theorem to show that
50
=0
lim x cos
x 0
x
and illustrate the principle involved by using a graphing utility to graph y = |x|, y = |x|, and y = x cos(50/x) on
the same screen in the window [1, 1] [1, 1].
48. Use the Squeezing Theorem to show that
50
=0
lim x 2 sin
3
x 0
x
164
g65-ch2
## and illustrate the principle involved by using a graphing
utility to graph y = x 2 , y = x 2 , and y = x 2 sin(50/ 3 x ) on
the same screen in the window [0.5, 0.5] [0.25, 0.25].
49. Sketch the graphs of y = 1 x 2 , y = cos x, and y = f(x),
where f is a function that satisfies the inequalities
1 x 2 f(x) cos x
for all x in the interval (/2, /2). What can you say about
the limit of f(x) as x 0? Explain your reasoning.
50. Sketch the graphs of y = 1/x, y = 1/x, and y = f(x),
where f is a function that satisfies the inequalities
1
1
f(x)
x
x
for all x in the interval [1, +). What can you say about the
limit of f(x) as x +? Explain your reasoning.
51. Find formulas for functions g and h such that g(x) 0 and
h(x) 0 as x + and such that
sin x
g(x)
h(x)
x
for positive values of x. What can you say about the limit
sin x
?
lim
x + x
52. Draw pictures analogous to Figure 2.6.3 that illustrate the
Squeezing Theorem for limits of the forms limx + f(x)
and limx f(x).
Recall that unless stated otherwise the variable x in trigonometric functions such as sin x and cos x is assumed to be in
radian measure. The limits in Theorem 2.6.3 are based on
that assumption. Exercises 53 and 54 explore what happens
to those limits if degree measure is used for x.
53. (a) Show that if x is in degrees, then
sin x
lim
=
x 0 x
180
(b) Confirm that the limit in part (a) is consistent with the
results produced by your calculating utility by setting
the utility to degree measure and calculating (sin x)/x
for some values of x that get closer and closer to 0.
54. What is the limit of (1cos x)/x as x 0 if x is in degrees?
55. It follows from part (a) of Theorem 2.6.3 that if is small
(near zero) and measured in radians, then one should expect
the approximation
sin
to be good.
(a) Find sin 10 using a calculating utility.
(b) Estimate sin 10 using the approximation above.
56. (a) Use the approximation of sin that is given in Exercise 55 together with the identity cos 2 = 1 2 sin2
with = /2 to show that if is small (near zero)
## and measured in radians, then one should expect the
approximation
cos 1 12 2
to be good.
(b) Find cos 10 using a calculating utility.
(c) Estimate cos 10 using the approximation above.
57. It follows from part (a) of Example 2 that if is small (near
zero) and measured in radians, then one should expect the
approximation
tan
to be good.
(a) Find tan 5 using a calculating utility.
(b) Find tan 5 using the approximation above.
58. Referring to the accompanying figure, suppose that the angle of elevation of the top of a building, as measured from
a point L feet from its base, is found to be degrees.
(a) Use the relationship h = L tan to calculate the height
of a building for which L = 500 ft and = 6 .
(b) Show that if L is large compared to the building height
h, then one should expect good results in approximating
h by h L /180.
(c) Use the result in part (b) to approximate the building
height h in part (a).
a
L
h
Figure Ex-58
## 59. (a) Use the Intermediate-Value Theorem to show that the
equation x = cos x has at least one solution in the interval [0, /2].
(b) Show graphically that there is exactly one solution in
the interval.
(c) Approximate the solution to three decimal places.
60. (a) Use the Intermediate-Value Theorem to show that the
equation x + sin x = 1 has at least one solution in the
interval [0, /6].
(b) Show graphically that there is exactly one solution in
the interval.
(c) Approximate the solution to three decimal places.
61. In the study of falling objects near the surface of the Earth,
the acceleration g due to gravity is commonly taken to be
9.8 m/s2 or 32 ft/s2 . However, the elliptical shape of the
Earth and other factors cause variations in this constant that
are latitude dependent. The following formula, known as the
Geodetic Reference Formula of 1967, is commonly used to
predict the value of g at a latitude of degrees (either north
or south of the equator):
g = 9.7803185(1.0 + 0.005278895 sin2
0.000023462 sin4 ) m/s2
(a) Observe that g is an even function of . What does this
suggest about the shape of the Earth, as modeled by the
Geodetic Reference Formula?
g65-ch2
## cyan magenta yellow black
Supplementary Exercises
of 38 and 39 .
62. Let
f(x) =
1
0
165
## (a) Make a conjecture about the limit of f(x) as x 0.
(b) Make a conjecture about the limit of xf(x) as x 0.
if x is a rational number
if x is an irrational number
SUPPLEMENTARY EXERCISES
1. For the function f graphed in the accompanying figure, find
the limit if it exists.
(a) lim f(x)
(b) lim f(x)
(c) lim f(x)
x 1
x 2
x 3
(f ) lim f(x)
## (i) lim f(x)
x +
x 4
x 3
x 0
x 3
y
3
2
1
-1
x
1
Figure Ex-1
2. (a) Find a formula for a rational function that has a vertical asymptote at x = 1 and a horizontal asymptote at
y = 2.
(b) Check your work by using a graphing utility to graph
the function.
3. (a) Write a paragraph or two that describes how the limit of
a function can fail to exist at x = a. Accompany your
description with some specific examples.
(b) Write a paragraph or two that describes how the limit
of a function can fail to exist as x + or x .
Also, accompany your description with some specific
examples.
(c) Write a paragraph or two that describes how a function
can fail to be continuous at x = a. Accompany your
description with some specific examples.
4. Show that the conclusion of the Intermediate-Value Theorem may be false if f is not continuous on the interval
[a, b].
5. In each part, evaluate the function for the stated values of x,
and make a conjecture about the value of the limit. Confirm
your conjecture by finding the limit algebraically.
x2
(a) f(x) = 2
; lim f(x); x = 2.5, 2.1, 2.01,
x 4 x 2+
2.001, 2.0001, 2.00001
tan 4x
; lim f(x); x = 1.0, 0.1, 0.01,
(b) f(x) =
x 0
x
0.001, 0.0001, 0.00001
6. In each part, find the horizontal asymptotes, if any.
2x 7
x 3 x 2 + 10
(b)
y
=
x 2 4x
3x 2 4x
2x 2 6
(c) y = 2
x + 5x
7. (a) Approximate the value for the limit
3x 2 x
lim
x 0
x
to three decimal places by constructing an appropriate
table of values.
(b) Confirm your approximation using graphical evidence.
(a) y =
8. According to Ohms law, when a voltage of V volts is applied across a resistor with a resistance of R ohms, a current
of I = V /R amperes flows through the resistor.
(a) How much current flows if a voltage of 3.0 volts is applied across a resistance of 7.5 ohms?
(b) If the resistance varies by 0.1 ohm, and the voltage
remains constant at 3.0 volts, what is the resulting range
of values for the current?
(c) If temperature variations cause the resistance to vary
by from its value of 7.5 ohms, and the voltage remains constant at 3.0 volts, what is the resulting range
of values for the current?
(d) If the current is not allowed to vary by more than
" = 0.001 ampere at a voltage of 3.0 volts, what variation of from the value of 7.5 ohms is allowable?
(e) Certain alloys become superconductors as their temperature approaches absolute zero (273 C), meaning
that their resistance approaches zero. If the voltage remains constant, what happens to the current in a superconductor as R 0+ ?
9. Suppose that f is continuous on the interval [0, 1] and that
0 f(x) 1 for all x in this interval.
(a) Sketch the graph of y = x together with a possible
graph for f over the interval [0, 1].
(b) Use the Intermediate-Value Theorem to help prove that
there is at least one number c in the interval [0, 1] such
that f(c) = c.
10. Use algebraic methods to find
1 cos
t 1
(a) lim tan
(b) lim
0
t 1
t 1
(2x 1)5
+ 2x 7)(x 3 9x)
sin( + )
.
(d) lim cos
0
2
(c)
lim
x + (3x 2
166
g65-ch2
## 11. Suppose that f is continuous on the interval [0, 1], that
f(0) = 2, and that f has no zeros in the interval. Prove that
f(x) > 0 for all x in [0, 1].
12. Suppose that
x 4 + 3,
f(x) =
x 2 + 9,
sin2 (kx)
,
x 0
x2
sin x
25. lim+
x 0
x
26. lim
k = 0
3x sin(kx)
, k = 0
x
2x + x sin 3x
28. lim
x + 5x 2 2x + 1
29. One dictionary describes a continuous function as one
whose value at each point is closely approached by its values
at neighboring points.
(a) How would you explain the meaning of the terms
neighboring points and closely approached to a
nonmathematician?
(b) Write a paragraph that explains why the dictionary definition is consistent with Definition 2.5.1.
27. lim
x 0
x2
x>2
## Is f continuous everywhere? Justify your conclusion.
13. Show that the equation x 4 + 5x 3 + 5x 1 = 0 has at least
two real solutions in the interval [6, 2].
## 14. Use the Intermediate-Value Theorem to approximate 11
the root directly with a calculating utility.
15. Suppose that f is continuous at x0 and that f(x0 ) > 0. Give
either an "- proof or a convincing verbal argument to show
that there must be an open interval containing x0 on which
f(x) > 0.
16. Sketch the graph of f(x) = |x 2 4|/(x 2 4).
17. In each part, approximate the discontinuities of f to three
decimal places.
x+1
(a) f(x) = 2
x + 2x 5
x+3
(b) f(x) =
|2 sin x x|
18. In Example 3 of Section 2.6 we used the Squeezing Theorem
to prove that
1
lim x sin
=0
x 0
x
## 30. (a) Show by rationalizing the numerator that
1
x2 + 4 2
lim
=
x 0
x2
4
(b) Evaluate f(x) for
x = 1.0, 0.1, 0.01, 0.001, 0.0001, 0.00001
and explain why the values are not getting closer and
closer to the limit.
(c) The accompanying figure shows the graph of f generated with a graphing utility and zooming in on the
origin. Explain what is happening.
## Why couldnt we have obtained the same result by writing
1
1
= lim x lim sin
lim x sin
x 0
x 0
x 0
x
x
1
= 0?
= 0 lim sin
x 0
x
In Exercises 19 and 20, find lim f(x), if it exists, for
x a
## [5, 5] [ 0.1, 0.5]
xScl = 1, yScl = 0.1
## [ 0.5, 0.5] [ 0.1, 0.5]
xScl = 0.1, yScl = 0.1
a = 0, 5+ , 5 , 5, 5, , +
19. (a) f(x) =
5x
## 20. (a) f(x) = (x + 5)/(x 2 25)
(x 5)/|x 5|, x = 5
(b) f(x) =
0,
x=5
In Exercises 2128, find the indicated limit, if it exists.
tan ax
sin bx
sin 3x
22. lim
x 0 tan 3x
sin 2
23. lim
0 2
21. lim
x 0
## [5 10 6, 5 106 ] [0.1, 0.5]
xScl = 10 6, yScl = 0.1
Figure Ex-30
(a = 0, b = 0)
24. lim
x 0
x sin x
1 cos x
## In Exercises 3136, approximate the limit of the function
by looking at its graph and calculating values for some appropriate choices of x. Compare your answer with the value
produced by a CAS.
g65-ch2
## cyan magenta yellow black
Supplementary Exercises
C
33. lim
x 0
2x 8
x 3 x 3
32. lim
sin x sin 1
2
(1.001)1/x
C 34. lim x
x 1
x 0+
x1
x+ x x
35. lim
36.
x +
lim
x +
3x + 5 x
1/x
## (a) The equation x 3 x 1 = 0 has only one real solution.
Show that this equation can be written as
3
x = g(x) = x + 1
(b) Graph y = x and y = g(x) in the same coordinate
system for 1 x 3.
(c) Starting with an arbitrary estimate x1 , make a sketch
that shows the location of the successive iterates
x2 = g(x1 ), x3 = g(x2 ), . . .
lim
x 0
sin x
=1
x
sin x
< 0.001
1
x
## (d) Use x1 = 1 and calculate x2 , x3 , . . . , continuing until
you obtain two consecutive values that differ by less
than 104 . Experiment with other starting values such
as x1 = 2 or x1 = 1.5.
y
y=x
## if 0 < |x| < . Estimate the largest such .
38. If \$1000 is invested in an account that pays 7% interest
compounded n times each year, then in 10 years there will
be 1000(1 + 0.07/n)10n dollars in the account. How much
money will be in the account in 10 years if the interest is
compounded quarterly (n = 4)? Monthly (n = 12)? Daily
(n = 365)? Estimate the amount of money that will be in
the account in 10 years if the interest is compounded continuously, that is, as n +?
39. There are various numerical methods other than the method
discussed in Section 2.5 to obtain approximate solutions of
equations of the form f(x) = 0. One such method requires
that the equation be expressed in the form x = g(x), so that
a solution x = c can be interpreted as the value of x where
the line y = x intersects the curve y = g(x), as shown in the
accompanying figure. If x1 is an initial estimate of c and the
graph of y = g(x) is not too steep in the vicinity of c, then
a better approximation can be obtained from x2 = g(x1 )
(see the figure). An even better approximation is obtained
from x3 = g(x2 ), and so forth. The formula xn+1 = g(xn )
for n = 1, 2, 3, . . . generates successive approximations
x2 , x3 , x4 , . . . that get closer and closer to c.
167
y = g(x)
c x3
x2
x1
Figure Ex-39
## 40. The method described in Exercise 39 will not always work.
(a) The equation x 3 x 1 = 0 can be expressed as
x = g(x) = x 3 1. Graph y = x and y = g(x) in
the same coordinate system. Starting with an arbitrary
estimate x1 , make a sketch illustrating the locations of
the successive iterates x2 = g(x1 ), x3 = g(x2 ), . . . .
(b) Use x1 = 1 and calculate the successive iterates xn for
n = 2, 3, 4, 5, 6.
In Exercises 41 and 42, use the method of Exercise 39 to
approximate the roots of the equation.
41. x 5 x 2 = 0
42. x cos x = 0 | 48,159 | 139,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-10 | latest | en | 0.910325 |
https://www.physicsforums.com/threads/bomb-explodes-at-rest-conservation-of-motion-problem.619956/ | 1,508,602,354,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824820.28/warc/CC-MAIN-20171021152723-20171021172723-00294.warc.gz | 991,665,375 | 15,231 | # Bomb explodes at rest, conservation of motion problem
1. Jul 10, 2012
### T531
1. The problem statement, all variables and given/known data
A bomb at rest with 1.0x10^4 J explodes into three pieces.
The first piece is 1.0 kg and travels in the positive y direction with a velocity of 60 m/s.
The second piece is 4.0 kg and travels in the positive x direction at 40 m/s.
Find the third piece's mass, velocity, and angle.
Diagram attached, but link just in case:
http://www.flickr.com/photos/82417987@N08/7545812162/in/photostream
2. Relevant equations
momentum(p) = mass*velocity
p(before) = p(after)
Potential Energy = Kinetic Energy
KE =1/2mv^2
3. The attempt at a solution
KE = 1/2*m*v^2
1.0*10^4 = 1/2*(1)*(60)^2 + 1/2*(4)*(40)^2 + 1/2*(m)*(v)^2 <- where m and v are the unknown mass and velocity
Apply conservation of momentum in the x-direction:
The first piece is only in y, so it is not included in the equation. The x-mom eqn becomes:
4.0kg*40m/s + m(ofx)*v(of x) = 0
Now the y-mom equation:
1.0kg*60m/s +m(of y)*v(of y) = 0
I know: m(of x)=m(of y) because both are piece 3. And v^2 = v(of x)^2 + v(of y)^2.
Not sure how to find v(of x) and v(of y)
#### Attached Files:
• ###### Physics Problem.jpg
File size:
17.6 KB
Views:
63
2. Jul 10, 2012
### nasu
Solve the last two equations for vx and vy. Then plug into the energy conservation and solve for m.
3. Jul 10, 2012
### T531
Okay, so
v(of x) = -160/m
v(of y) = -60/m
1.0x10^4 J = (m)*[(160/m)^2+(60/m)^2]
1.0x10^4 J = m*(25600/m^2 + 3600/m^2) = 29200/m
1.0x10^4 J*m=29200
m=2.92kg
then do I use KE = mv^2 ?
if so 1.0x10^4 J = (2.92kg)v^2
therefore v = 58.5 m/s
then v(of x) = -160/m = -160/(2.92 kg) = -54.8m m/s
and v(of y) = -60/m = -60/(2.92 kg) = -20.5 m/s
so theta = inverse tan(-20.5/-54.8) = 20.5 degrees
So:
Mass = 2.92 kg
Velocity = 58.5 m/s
Theta = 20.5 degrees
Can someone verify that?
Thanks
Last edited: Jul 10, 2012 | 719 | 1,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-43 | longest | en | 0.79946 |
https://www.stem.org.uk/resources/elibrary/resource/31691/interactive-data-resources | 1,716,930,900,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059148.63/warc/CC-MAIN-20240528185253-20240528215253-00799.warc.gz | 852,111,608 | 12,105 | Tooltip
These resources have been reviewed and selected by STEM Learning’s team of education specialists for factual accuracy and relevance to teaching STEM subjects in UK schools.
# Interactive Data Resources
This resource from SMILE contains eight interactive spreadsheets as described below:
Rolling one dice: simulates the rolling of one dice and then records the number in the frequency table and simultaneously in the bar graph
The difference when rolling two dice: simulates the rolling of two dice and then recording the difference of the two numbers in a frequency table and simultaneously in the bar graph
The sum when rolling two dice: simulates the rolling of two dice and then records the sum of the two numbers in the frequency table and simultaneously in the bar graph
Using an assumed mean: demonstrates how an assumed mean can be used to calculate the mean height of five students
Calculating the mean from a frequency table: demonstrates how to calculate the mean of up to seven numbers from data in a frequency table. The values can be any set of consecutive numbers with the first number between 0 and 10
Calculating the mean: demonstrates how to calculate the mean of up to seven numbers. The value of each number can be any integer between 1 and 10. The calculation of each set of numbers is displayed. If the calculation involves decimals, these are initially displayed correct to 10 decimal places and then rounded to two decimal places
Calculating the median and range: demonstrates how to calculate the median and range of up to seven numbers. The value of each number can be an integer between 1 and 10
Calculating the mode: demonstrates how to calculate the mode of up to 10 numbers. The value of each number can be an integer between 1 and 7.
#### Show health and safety information
Please be aware that resources have been published on the website in the form that they were originally supplied. This means that procedures reflect general practice and standards applicable at the time resources were produced and cannot be assumed to be acceptable today. Website users are fully responsible for ensuring that any activity, including practical work, which they carry out is in accordance with current regulations related to health and safety and that an appropriate risk assessment has been carried out. | 446 | 2,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.913791 |
https://www.audiomasterclass.com/newsletter/reader-s-e-mail | 1,560,804,086,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998580.10/warc/CC-MAIN-20190617203228-20190617225228-00354.warc.gz | 663,671,939 | 11,321 | ## Steven Willis sheds further light on the relationship between bytes, kilobytes and megabytes...
First, I love the newsletters. I have learned a lot from these emails.
Now, your article on the bits and bytes. I may be able to clear a few things up for your readers. As a programmer and hardware designer, I am fully aware of the difference between bits and bytes and also why 1KB is not 1000 bytes.
Here's the breakdown. Since computers are binary, there are only two states a bit can be in. To make the math a little easier, the kilobyte was not defined as 1000 bytes. It was defined as a power of 2. In this case 2 to the 10th power equals 1024. Electrical engineers also like to think in decades or powers of ten.
## FREE EBOOK - Equipping Your Home Recording Studio
The next big step is 1 megabyte or 1,048,576 bytes or 2 to the 20th power. 1 gigabyte is 1,073,741,824 bytes or 2 to the 30th power. 1 terabyte is 2 to the 40th, 1 peta is 2 to the 50th, 1 exabyte is 2 to the 60th.
Electrical engineers seem to like multiples of ten, software programmers seem to like multiples of eight.
Many thanks for the advice and excellent industry information.
Steven Willis | 305 | 1,177 | {"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-2019-26 | longest | en | 0.947664 |
https://www.expertsmind.com/library/a-mixing-chamber-steadily-mixes-a-hot--water-stream-at-5630868.aspx | 1,685,888,337,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649986.95/warc/CC-MAIN-20230604125132-20230604155132-00157.warc.gz | 824,920,476 | 14,956 | ### A mixing chamber steadily mixes a hot water stream at
Assignment Help Mechanical Engineering
##### Reference no: EM13630868
A mixing chamber steadily mixes a hot water stream at 80°C and 0.5kg/s rate with a cold water stream at 20°C. If the leaving mixture is at 42°C and mixing happens at 250kPa, determine the mass flow rate of the cold water stream.
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All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd | 1,091 | 4,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-23 | latest | en | 0.911162 |
http://slideplayer.com/slide/5662550/ | 1,585,815,372,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506870.41/warc/CC-MAIN-20200402080824-20200402110824-00101.warc.gz | 167,022,440 | 17,142 | # 1.8 Properties of Real Numbers. Commutative (Addition) The “you can switch it around and it just don’t matter” property a + b = b + a.
## Presentation on theme: "1.8 Properties of Real Numbers. Commutative (Addition) The “you can switch it around and it just don’t matter” property a + b = b + a."— Presentation transcript:
1.8 Properties of Real Numbers
Commutative (Addition) The “you can switch it around and it just don’t matter” property a + b = b + a
Commutative (Multiplication) The “you can switch it around and it just don’t matter” property a b = b a
Associative (Addition) Order of numbers stay, parenthesis shift (a + b) + c = a + (b + c)
Associative (Multiplication) Order of numbers stay, parenthesis shift (a b) c = a (b c)
Identity Property (Addition) The “zero” property What can you add to a number to get the SAME EXACT NUMBER? a + 0 = a
Identity Property (Multiplication) The “one” property What can you multiply a number by to get the SAME EXACT NUMBER? a 1 = a
Inverse (Addition) For every value a, there is an equal but opposite a (additive inverse) a + (-a) = 0
Inverse (Multiplication) For every value a, there is an equal but reciprocal of a (multiplicative inverse) a ( ) = 0
BOOKS! PAGE 55 CHECK UNDERSTANDING (MIDDLE OF PAGE) A - F
Homework Page 57 #32 – 43 ALL #45, 46
Download ppt "1.8 Properties of Real Numbers. Commutative (Addition) The “you can switch it around and it just don’t matter” property a + b = b + a."
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https://community.deeplearning.ai/t/week-2-exercise-8/20670 | 1,709,455,815,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00270.warc.gz | 169,949,671 | 7,382 | # Week 2 - Exercise 8
My code passed all tests in exercises 1 -8, including “model_test(model)” but after that the running the cell:
logistic_regression_model = model(train_set_x, train_set_y, test_set_x, test_set_y, num_iterations=2000, learning_rate=0.005, print_cost=True)
shows a message:
ValueError Traceback (most recent call last)
in
----> 1 logistic_regression_model = model(train_set_x, train_set_y, test_set_x, test_set_y, num_iterations=2000, learning_rate=0.005, print_cost=True)
in model(X_train, Y_train, X_test, Y_test, num_iterations, learning_rate, print_cost)
38 w, b = initialize_with_zeros(X_train.shape[0])
39
—> 40 params, grads, costs = optimize(w, b, X_train, Y_test, num_iterations, learning_rate, print_cost=False)
41
42 w = params[“w”]
in optimize(w, b, X, Y, num_iterations, learning_rate, print_cost)
36 # YOUR CODE STARTS HERE
37
—> 38 grads, cost = propagate(w, b, X, Y)
39
40 # YOUR CODE ENDS HERE
in propagate(w, b, X, Y)
31
32 A = 1/(1+np.exp(-(np.dot(w.T,X)+b)))
—> 33 cost = - np.sum(Y * np.log(A) + (1-Y) * np.log(1-A)) / m
34
35 # YOUR CODE ENDS HERE
ValueError: operands could not be broadcast together with shapes (1,50) (1,209)
Hi @jawart I think in the cost function calculation needs to be reviewed. Particular attention to the matrix transposition in order to have a dimensionally correct matrix products. Hope this helps
Hi,
is it possible if it is the same cost function formula, which has given correct output in excercise 5 and passed tests there?
Hi again, I have found the reason.
It was “_test” instead of Y_train as function argument.
2 Likes
Right, you can see the bug in the code you showed in your original post here. Also note that there’s another bug in that call to optimize from model: you will never print the cost values, even if they ask for them.
In Exercise 8:
1. where are we reading the w, b, X_test/train and Y_test/train data from?
2. params, grads, costs = optimize(w, b, X, Y, num_iterations, learning_rate, print_cost=True) - do the input parameters (w, b, X, Y) change for train and test cases?
3. Y_prediction_train = predict(w, b, X_train) - similarly, which w and b are inputs to the function Y_prediction_train
4. Is the solution only in 4 lines of code (as shown in the comments)?
Thank you!
1. When you are implementing the model function, the X_train, Y_train, X_test and Y_test values are simply passed to you as parameters. Just use the values as passed. If you want to know where they came from, they are either synthetic values created for the purposes of testing your code or they are the real image data. See the early sections in the notebook for how they are read in and prepared for use by the model function. The w and b values are initialized to
zeros by calling the initialization routine that you wrote earlier.
2. When you call optimize from model, you just pass in the current values of all the parameters. It’s a mistake to “hard-code” the value of print_cost as you show. What if the test case passes “False”?
3. The w and b you use for the predictions are the trained values that you get back from calling optimize, right?
4. Those “number of lines” are just suggestions, so don’t stress if your code takes more lines. The thing to stress about is when the test cases fail.
Really appreciate your explanations! I initialize w and b, and then call the optimize twice (using (X_train, Y_train), and (X_test, Y_test)). I do this because I then call the predict function to get the predicted y using ‘Y_prediction_test = predict(w, b, X_test)’.
## I encounter the following error:
ValueError Traceback (most recent call last)
in
1 from public_tests import *
2
----> 3 model_test(model)
~/work/release/W2A2/public_tests.py in model_test(target)
113 y_test = np.array([0, 1, 0])
114
→ 115 d = target(X, Y, x_test, y_test, num_iterations=50, learning_rate=0.01)
116
117 assert type(d[‘costs’]) == list, f"Wrong type for d[‘costs’]. {type(d[‘costs’])} != list"
in model(X_train, Y_train, X_test, Y_test, num_iterations, learning_rate, print_cost)
35 # YOUR CODE STARTS HERE
36 w, b = initialize_with_zeros(dim)
—> 37 params, grads, costs = optimize(w, b, X_test, Y_test, num_iterations, learning_rate, print_cost)
38 w = params[“w”]
39 b = params[“b”]
in optimize(w, b, X, Y, num_iterations, learning_rate, print_cost)
35 # grads, cost = …
36 # YOUR CODE STARTS HERE
—> 37 grads, cost = propagate(w, b, X, Y)
38
39 # YOUR CODE ENDS HERE
in propagate(w, b, X, Y)
29 # cost = …
30 # YOUR CODE STARTS HERE
—> 31 A = sigmoid(np.dot(w.T,X)+b)
32
33 cost= -1*(np.dot(Y, np.log(A.T)) + np.dot((1-Y), np.log(1-A.T)))/m
<array_function internals> in dot(*args, **kwargs)
ValueError: shapes (1,2) and (4,3) not aligned: 2 (dim 1) != 4 (dim 0)
Could I get some suggestions on how to troubleshoot it? Thank you
It is a mistake to call optimize twice. You don’t need to call it with X_test. The point is that you train on the training data (X_train and Y_train). Then you compute the predictions using the trained model (w and b) on both X_train and X_test to see how the model performs. | 1,397 | 5,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-10 | latest | en | 0.80549 |
http://math.stackexchange.com/questions/133737/semisimplicity-of-a-polynomial-ring | 1,469,815,887,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831770.41/warc/CC-MAIN-20160723071031-00177-ip-10-185-27-174.ec2.internal.warc.gz | 152,460,055 | 18,693 | # Semisimplicity of a polynomial ring
Given a ring $R$ (with 1 and not necessarily commutative) when is the polynomial ring $R[x]$ semisimple?
For example if R is a Noetherian integral domain then R[x] is not semisimple. Indeed, $R[x]$ is Noetherian but then if we assume that $R[x]$ is semisimple this would imply that $R[x]$ is Artinian which is not.
Question: is there a ring $R$ (with $1$ and not necessarily commmutative) such that $R[x]$ is semisimple? or is $R[x]$ never semisimple for any ring $R$?
-
$\mathbb Z[X]$ is not a PID (for example, the ideal $(2,X)$ is not principal) so there is something wrong with your second paragraph! – Mariano Suárez-Alvarez Apr 19 '12 at 3:42
It would be useful if you make explicit what definition of semisimplicity you have in mind... – Mariano Suárez-Alvarez Apr 19 '12 at 3:42
@Mariano Suárez-Alvarez: sorry, just edited. Well I've read several equivalences: i.e every submodule is a direct summand, isomorphic to a direct sum of simple modules,etc. – user6495 Apr 19 '12 at 3:44
Please add that information to the question itself. – Mariano Suárez-Alvarez Apr 19 '12 at 3:47
If we understand semisimple to mean all modules are projective then $R[X]$ is never semisimple.
Indeed, if we let $S=R[X]/(X)$, the canonical $R[X]$-linear projection map $f:R[X]\to S$ is not split: there is no $R[X]$-linear map $s:S\to R[X]$ such that $f\circ s$ is the identity of $S$. To see this notice that the element $u=f(1_R)\in S$ is an element such that $X\cdot u=0$ and that $s$ would necessarily be injective, so that we would have an element $v=s(u)\in R[X]$ such that $X\cdot v=0$.
-
If you don't assume $R$ has a $1$, this argument still works taking $u=f(w)$ for any non-zero $w\in R$. Artinianness follows, in the usual situation, by the existence of $1$, so this is indeed a separate obstacle to semisimplicity. – Mariano Suárez-Alvarez Apr 19 '12 at 3:51
cheers, very nice! – user6495 Apr 19 '12 at 3:52
Your post already contains the answer. Polynomial rings are not Artinian, since $$(x)\supset(x^2)\supset(x^3)\supset\ldots$$ But semisimple rings are Artinian.
In fact, the Wedderburn-Artin Theorem states that all semisimple rings are direct sums of matrix rings over division algebras. Polynomial rings are certainly not of this form.
-
thanks, can't believe I forgot that fact! – user6495 Apr 19 '12 at 3:51
Yay! Let's do all the characterizations. All semisimple rings are Von Neumann regular. But, if you choose $x\in R[x]$ then you need to be able to find a $p(x)\in R[x]$ such that $x=x^2p(x)$ but this is clearly impossible.
- | 792 | 2,592 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2016-30 | latest | en | 0.871364 |
https://slideplayer.com/slide/7519223/ | 1,639,005,427,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00122.warc.gz | 578,192,600 | 25,291 | # Section 2.4 Representing Data.
## Presentation on theme: "Section 2.4 Representing Data."— Presentation transcript:
Section 2.4 Representing Data
Objectives Create graphs to reveal patterns in data. Interpret graphs.
Graphs When data is analyzed, scientists are often looking for patterns. Displaying data in a TABLE (or chart) may or may not reveal a pattern. Creating a GRAPH, however, helps to reveal patterns that may exist.
Graphs A graph is a visual display of data.
27.37% Na A graph is a visual display of data. There are different types of graphs. A circle graph or pie chart is useful to show parts of a whole. 57.14% O 14.30% C
Graphs A bar graph is used to show how a quantity varies across categories. The quantity being measured (dependent variable) appears on the y-axis and the categories (independent variable) appears on the x-axis.
Graphs A line graph represents the points of intersection of data for 2 variables. The independent variable is plotted on the x-axis and the dependent variable is on the y-axis. Line graphs are used to the exclusion of all others in chemistry.
Line Graphs Because data points can be very scattered, a line connecting all of them cannot be drawn. A line must be drawn so that the same number of points fall above & below the line. This is called a BEST-FIT line.
Creating a Line Graph Two (straight) axes are drawn on graph paper and labeled. The x-axis represents the variable that is manipulated (changed) while the y-axis represents the variable that responds to the changes in the manipulated variable. Each label must include the name of the variable AND the unit attached to that variable.
Creating a Line Graph A scale for each axis must be selected and assigned. Determine the highest & lowest values of the data. Subtract the values to find the range of the data. (NOTE: If you want the axis to start at 0, you must use 0 as the lowest value!) Count the number of boxes along the axis. Divide the range value by the number of boxes to find the value that will be represented by each box. Assign the determined scale to the axis, labeling clearly with the appropriate values. It may not be necessary to label every box but label enough boxes so that the scale is evident.
Creating a Line Graph Important Points to Note:
The scale for each axis must be determined separately. Each axis, therefore, will have its own scale and units. They can, but do not have to be, the same. A “break in the scale” symbol can NEVER be used on a line graph in Chemistry. Choose a scale that is easy to plot & easy to read. The scale and labels must be consistent along the axis – the value of each “box” must remain the same. The scale cannot be changed as the axis is labeled. Scales do not have to start at zero. It is very important, therefore, to assign a value for each axis at their start. A properly scaled axis will not have any data points that fall “off-scale”. If you find this happens, you must re-scale your axis.
Creating a Line Graph Data points are plotted on the graph.
The values of the independent & dependent variables form ordered pairs (x,y). The ordered pairs can be plotted on the graph from their corresponding x-axis or y-axis. Dependent Variable Y-axis (x, y) y x Independent variable X-axis
Creating a Line Graph A “best-fit” line is drawn.
A straight “best-fit” line is almost always drawn. Sometimes a “best-fit” curve is needed. Instructions must be read carefully to determine which to draw. “Dot-to-dot” lines are NEVER drawn. Remember, the data points do not need to be on the line or curve drawn. In fact, sometimes none of the data points will actually be on the line/curve drawn.
Interpreting a Line Graph
Calculating Slope A straight line always has a constant slope. 2 points ON THE LINE (NOT data points) must be chosen: (x1, y1) & (x2, y2) These points can be used with this formula to calculate slope: y2- y1 or Δy x2-x Δx Note: when reporting the slope, include the units (y/x)!
Interpreting a Line Graph
What does the slope tell you? If a slope can be determined (that is, the line is straight), there is a linear relationship between the variables. If the line slopes upward, the slope is positive. This means the dependent (y) variable increases as the independent (x) variable increases. This represents a DIRECT relationship. If the line slopes downward, the slope is negative. This means the dependent (y) variable decreases as the independent (x) variable increases. This represents an INVERSE relationship.
Interpreting a Line Graph
If a slope cannot be determined (that is, the “line” is curved), there is a nonlinear relationship between the variables. As can be seen in this graph, as the independent variable (x) increases, the dependent variable (y) decreases. This curve, therefore, shows an inverse relationship.
Interpreting a Line Graph
Extrapolation When a “best-fit” line is drawn, the points on the line are considered continuous data points. That means you can read data from any place along the line. If the line that was drawn is extended beyond the plotted points, it has been extrapolated. Estimate values for the variables can be read from an extrapolated line.
Practice Problems Plot the data in each table. Explain whether the graphs represent direct or inverse relationships. Table Table 2 Effect of Pressure on Gas Effect of Pressure on Gas Pressure (mm Hg) Volume (mL) 3040 5.0 1520 10.0 1013 15.0 760 20.0 Pressure (mm Hg) Temperature (K) 3040 1092 1520 546 1013 410 760 273 | 1,255 | 5,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-49 | latest | en | 0.894528 |
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Source: pinterest.com
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# Seating Arrangement case-11
Exam Name Seating Arrangement case-11 NABARD BANK 2009 PUBLIC 3 rahul(226)
.
Question: Case 11)
A, B, C, D, E, F, G and H are eight friends sitting around a circular table facing the centre. A sits second to the left of D, who is third to the left of E. C sits third to the right of G, who is not an immediate neighbour of E. H sits third to the right of B, who sits second to the right of G.
Who sits between D and C?
Answer:Only B
Question: Case 11)
A, B, C, D, E, F, G and H are eight friends sitting around a circular table facing the centre. A sits second to the left of D, who is third to the left of E. C sits third to the right of G, who is not an immediate neighbour of E. H sits third to the right of B, who sits second to the right of G.
Who sits second to the right of E?
Answer:F
Question: Case 11)
A, B, C, D, E, F, G and H are eight friends sitting around a circular table facing the centre. A sits second to the left of D, who is third to the left of E. C sits third to the right of G, who is not an immediate neighbour of E. H sits third to the right of B, who sits second to the right of G.
What is the position of A with respect to H?
Answer:Second to the right
Question: Case 11)
A, B, C, D, E, F, G and H are eight friends sitting around a circular table facing the centre. A sits second to the left of D, who is third to the left of E. C sits third to the right of G, who is not an immediate neighbour of E. H sits third to the right of B, who sits second to the right of G.
Four of the following five are alike based upon their seating arrangements and so form a group. Which is the one that does not belong to that group?
Answer:DA
Question: Case 11)
A, B, C, D, E, F, G and H are eight friends sitting around a circular table facing the centre. A sits second to the left of D, who is third to the left of E. C sits third to the right of G, who is not an immediate neighbour of E. H sits third to the right of B, who sits second to the right of G.
Which of the following pairs has the second person sitting to the immediate left of the first person?
Answer:None of these | 584 | 2,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-49 | latest | en | 0.945138 |
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Missions impossible (Posted on 2014-11-17)
A certain chain store sells chocolate bars in packets of 17 and 9 only. Clearly, you could not get 8 or 25 bars. Find all quantities of bars that you cannot buy.
Generalize for m and n, mutually prime integers, m>n.
No Solution Yet Submitted by Ady TZIDON No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
re: Start of a General Solution Comment 3 of 3 |
(In reply to Start of a General Solution by Omri)
ok, just show that above a certain number N any given quantity is representable in terms of m and n.
e.g. if there were only 5 and 3 dollar bills then 7 is the highest amount that cannot be paid: 5+3, 3+3+3, 5+5 and so on by adding 3's only.
JUST GENERALIZE FOR m AND n.
Posted by Ady TZIDON on 2014-11-25 17:50:24
Search: Search body:
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# Runge-Kutta method
Posted 9 years ago
How I can use mathematica to get the answer of this question (Using the Runge-Kutta method of order 2 solve x'(t) = -x (t) + t^0.1 (1.1+t) x (0)=0 whose solution is , x(t)=t^1.1 solve the question on [0,5] printing the solution and the errors at x= 1,3,5 use step- sizes h=0.1 ,0.05,0.025 ,0.0125,0.00625 calculate the ratios by which the error decrease when h is halved
Please show code for what you have already tried!(Or is the question a homework problem you're asking us to do for you?)Have you already tried something simpler, such as applying Euler's Method?And why do you want to use Runge-Kutta here to get a numerical solution since you already know an exact solution—which you could find in Mathematica from: DSolve[{x'[t] == -x[t] + t^(1/10) (11/10 + t), x[0] == 0}, x[t], t] | 279 | 895 | {"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} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.942066 |
https://calories-info.com/pineapple-calories-kcal/ | 1,656,419,184,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00255.warc.gz | 198,004,416 | 63,429 | # Pineapple: Calories and Nutrition Analyse
## How many calories in pineapple?
100g of pineapple has about 50 calories (kcal).
Calories per: ounce | one pineapple | slice | cup, chunks
For example, medium size pineapple (700 g) has about 350 calories.
It is about 15% of daily calories intake for adult person with medium weight and medium activity (for calculation we assumed 2400 kcal daily intake).
Scroll down for details and nutrition tables.
To visualize how much 350 kcal actually is, take in mind that calories amount from pineapple is similar to calories amount from ie.:
• 4 glasses of Coca Cola (220 ml glass)
• 3.5 slices of cheese
• 2.5 glasses of milk
• 18 cubes of sugar
Take a look at tables below to see details about pineapple nutrition.
### Pineapple: calories and nutrition per 100g (and per ounce)
per 100 gper ouncemedium size pineapple (700 g)
Calories50 14.18 350
Carbs Total13.52 g3.83 g94.6 g
Dietary fiber1.4 g0.4 g9.8 g
Fat0.12 g< 0.1 g0.8 g
Protein0.54 g0.15 g3.8 g
## pineapple - vitamins per 100g
• Vit A58 IU
• Vit B1 (Thiamine)0.079 mg
• Vit B2 (riboflavin)0.018 mg
• Vit B3 (Niacin)0.5 mg
• Vit B60.112 mg
• Vit B9 (Folic acid)18 mcg
• Vit C47.8 mg
• Vit E0.02 mg
• Vit K0.07 mg
• 94% CARBS
• 4% PROTEIN
• 2% FAT
## pineapple - minerals per 100g
• Potassium109 mg
• Phosphorus8 mg
• Magnessium12 mg
• Calcium13 mg
• Sodium1 mg
• Iron0.29 mg
• Zink0.12 mg
• Beta karoten35 mg
### How many calories in 1, 2, 3 or 5 pineapples?
As I wrote before medium size pineapple (700 g) has 350 calories. It is easy to count that two pineapples have about 700 calories and three pineapples have about 1050 calories. In table below you can also see calories amount for four and five pineapples.
• Medium size pineapple (700 g)350 kcal
• Slice of pineapple (60g)30 kcal
• Pineapple cup, chunks (165g)83 kcal
• Ounce (oz) of pineapple14 kcal
• Half of medium size pineapple175 kcal
• Small size pineapple (560g)280 kcal
• Big size pineapple (910g)455 kcal
• Two medium size pineapples700 kcal
• Three medium size pineapples1050 kcal
• Four medium size pineapples1400 kcal
• Five medium size pineapples1750 kcal
### Protein in pineapple
Pineapple has 0.54 g protein per 100g.
When you multiplay this value with weight of medium size pineapple (700 g) you can see that you will get about 3.8 g of protein.
### Carbs in pineapple
Pineapple has 13.52 g carbohydrates per 100g.
In the same way as for protein we can calculate that medium size pineapple (700 g) has about 94.6 g of carbs.
### Fat in pineapple
Pineapple has 0.12 g fat per 100g.
So it is easy to count that medium size pineapple (700 g) has about 0.8 g of fat.
## medium size pineapple (700 g) has:
350kcalFor burning these calories you have to:
Bike50 min.Bike
Horse ride64 min.Horse ride
Play tennis34 min.Tennis
Run35 min.Run
Swim41 min.Swim
## Interesting charts - compare pineapple with other fruits
When you look at charts below you will see how pineapple looks like in comparsion to other products from its category. When you click on selected product you will se detailed comparsion.
Loquat (1) more...Figs (2) more...Mangosteen (2.9) more...Green grape (3.2) more...Pear (4.2) more...Apple (4.6) more...Nectarine (5.4) more...Peach (6.6) more...Persimmon (7.5) more...Watermelon (8.1) more...Banana (8.7) more...Plum (9.5) more...Blueberries (9.7) more...Apricot (10) more...Avocado (10) more...Pomegranate (10.2) more...Grapes (10.8) more...Cherimoya (12.6) more...Cranberry (13.3) more...Jackfruit (13.7) more...Sapodilla (14.7) more...Quince fruit (15) more...Honeydew melon (18) more...Plantain (18.4) more...Durian (19.7) more...Chokeberry (20) more...Blackberries (21) more...Cherry (21) more...Rose apple (22.3) more...Raspberry (26.2) more...Mandarin (Tangerine) (26.7) more...Gooseberry (27.7) more...Mango (27.7) more...Lime (29.1) more...Passion fruit (30) more...Grapefruit (31.2) more...Fejhoa (32.9) more...Mixed berries (33.8) more...Carambola (34.4) more...Black lilac (36) more...Mulberry (36.4) more...Cantaloupe (36.7) more...Melon (36.7) more...Kumquats (43.9) more...Pineapple (47.8) more...Orange (48.5) more...Cutie clementine (48.8) more...Clementine (48.8) more...Lemon (53) more...Strawberries (58.8) more...Papaya (60.9) more...Pomelo (61) more...Jujube (69) more...Lychee (71.5) more...Kiwi (92.7) more...Blackcurrant (181) more...Guava (228) more...Acerola fruit (1677.6) more...
Vitamin C in Pineapple Based On Fruits Category
Lemon (1) more...Cherimoya (2) more...Fejhoa (2) more...Strawberries (7) more...Pear (12) more...Mulberry (12) more...Raspberry (16) more...Mango (17) more...Durian (23) more...Banana (26) more...Apple (27) more...Blueberries (32) more...Pineapple (35) more...Cranberry (36) more...Grapes (39) more...Kiwi (52) more...Jackfruit (61) more...Avocado (62) more...Figs (85) more...Mandarin (Tangerine) (101) more...Nectarine (150) more...Kumquats (155) more...Peach (162) more...Plum (190) more...Persimmon (253) more...Papaya (276) more...Watermelon (303) more...Guava (374) more...Grapefruit (686) more...Passion fruit (743) more...Apricot (1094) more...Melon (2020) more...
Beta Carotene in Pineapple Based On Fruits Category
Watermelon (0.4) more...Honeydew melon (0.8) more...Cantaloupe (0.9) more...Grapes (0.9) more...Melon (0.9) more...Green grape (0.9) more...Pomelo (1) more...Acerola fruit (1.1) more...Lychee (1.3) more...Pineapple (1.4) more...Plum (1.4) more...Peach (1.5) more...Jackfruit (1.5) more...Cutie clementine (1.7) more...Loquat (1.7) more...Clementine (1.7) more...Mulberry (1.7) more...Grapefruit (1.7) more...Plantain (1.7) more...Nectarine (1.7) more...Papaya (1.7) more...Mandarin (Tangerine) (1.8) more...Mango (1.8) more...Mangosteen (1.8) more...Quince fruit (1.9) more...Apricot (2) more...Strawberries (2) more...Cherry (2.1) more...Apple (2.4) more...Blueberries (2.4) more...Orange (2.5) more...Banana (2.6) more...Lemon (2.8) more...Lime (2.8) more...Carambola (2.8) more...Figs (2.9) more...Cherimoya (3) more...Kiwi (3) more...Pear (3.1) more...Mixed berries (3.5) more...Persimmon (3.6) more...Durian (3.8) more...Pomegranate (4) more...Gooseberry (4.3) more...Blackcurrant (4.3) more...Cranberry (4.6) more...Chokeberry (5.3) more...Blackberries (5.3) more...Sapodilla (5.3) more...Guava (5.4) more...Fejhoa (6.4) more...Kumquats (6.5) more...Raspberry (6.5) more...Avocado (6.7) more...Black lilac (7) more...Passion fruit (10.4) more...
Fiber in Pineapple Based On Fruits Category
Pomelo (0.04) more...Quince fruit (0.1) more...Pear (0.12) more...Pineapple (0.12) more...Cranberry (0.13) more...Grapefruit (0.14) more...Honeydew melon (0.14) more...Cutie clementine (0.15) more...Watermelon (0.15) more...Clementine (0.15) more...Grapes (0.16) more...Green grape (0.16) more...Apple (0.17) more...Cantaloupe (0.19) more...Melon (0.19) more...Loquat (0.2) more...Lime (0.2) more...Cherry (0.2) more...Jujube (0.2) more...Peach (0.25) more...Papaya (0.26) more...Mango (0.27) more...Plum (0.28) more...Rose apple (0.3) more...Lemon (0.3) more...Acerola fruit (0.3) more...Figs (0.3) more...Orange (0.3) more...Strawberries (0.3) more...Mandarin (Tangerine) (0.31) more...Nectarine (0.32) more...Banana (0.33) more...Blueberries (0.33) more...Carambola (0.33) more...Mixed berries (0.35) more...Plantain (0.35) more...Mulberry (0.39) more...Apricot (0.4) more...Blackcurrant (0.41) more...Lychee (0.44) more...Blackberries (0.49) more...Black lilac (0.5) more...Chokeberry (0.5) more...Kiwi (0.52) more...Gooseberry (0.58) more...Mangosteen (0.58) more...Fejhoa (0.6) more...Jackfruit (0.64) more...Raspberry (0.65) more...Cherimoya (0.68) more...Passion fruit (0.7) more...Kumquats (0.86) more...Guava (0.95) more...Sapodilla (1.1) more...Pomegranate (1.17) more...Durian (5.33) more...Avocado (14.66) more...
Fat in Pineapple Based On Fruits Category
Rose apple (0.07) more...Grapefruit (0.08) more...Carambola (0.08) more...Orange (0.09) more...Pomelo (0.11) more...Apple (0.12) more...Mango (0.13) more...Cutie clementine (0.14) more...Fejhoa (0.14) more...Clementine (0.14) more...Mandarin (Tangerine) (0.15) more...Persimmon (0.15) more...Pear (0.17) more...Plum (0.17) more...Honeydew melon (0.17) more...Acerola fruit (0.2) more...Cantaloupe (0.21) more...Melon (0.21) more...Watermelon (0.24) more...Peach (0.25) more...Papaya (0.25) more...Cranberry (0.25) more...Banana (0.26) more...Guava (0.26) more...Cherimoya (0.27) more...Loquat (0.28) more...Blueberries (0.28) more...Nectarine (0.28) more...Pineapple (0.29) more...Mangosteen (0.3) more...Pomegranate (0.3) more...Kiwi (0.31) more...Lychee (0.31) more...Gooseberry (0.31) more...Green grape (0.36) more...Grapes (0.36) more...Cherry (0.36) more...Figs (0.37) more...Apricot (0.39) more...Strawberries (0.41) more...Durian (0.43) more...Jujube (0.48) more...Mixed berries (0.51) more...Avocado (0.55) more...Plantain (0.55) more...Lemon (0.6) more...Jackfruit (0.6) more...Lime (0.6) more...Blackberries (0.62) more...Raspberry (0.69) more...Quince fruit (0.7) more...Sapodilla (0.8) more...Kumquats (0.86) more...Chokeberry (1.18) more...Blackcurrant (1.54) more...Passion fruit (1.6) more...Black lilac (1.6) more...Mulberry (1.85) more...
Iron in Pineapple Based On Fruits Category
Apple (5) more...Rose apple (5) more...Black lilac (5) more...Pomelo (6) more...Blueberries (6) more...Lime (6) more...Cranberry (6) more...Green grape (7) more...Pear (7) more...Plum (7) more...Grapes (7) more...Lemon (8) more...Quince fruit (8) more...Peach (9) more...Fejhoa (9) more...Persimmon (9) more...Mango (9) more...Grapefruit (9) more...Nectarine (9) more...Apricot (10) more...Cutie clementine (10) more...Watermelon (10) more...Lychee (10) more...Gooseberry (10) more...Clementine (10) more...Honeydew melon (10) more...Orange (10) more...Jujube (10) more...Carambola (10) more...Cherry (11) more...Mandarin (Tangerine) (12) more...Pineapple (12) more...Cantaloupe (12) more...Pomegranate (12) more...Melon (12) more...Sapodilla (12) more...Strawberries (13) more...Loquat (13) more...Mangosteen (13) more...Kiwi (17) more...Figs (17) more...Cherimoya (17) more...Acerola fruit (18) more...Mulberry (18) more...Kumquats (20) more...Blackberries (20) more...Chokeberry (20) more...Papaya (21) more...Guava (22) more...Raspberry (22) more...Blackcurrant (24) more...Banana (27) more...Passion fruit (29) more...Avocado (29) more...Durian (30) more...Plantain (36) more...Jackfruit (37) more...
Magnesium in Pineapple Based On Fruits Category
Pineapple (8) more...Mangosteen (9.21) more...Papaya (10) more...Peach (11) more...Apple (11) more...Pear (11) more...Guava (11) more...Quince fruit (11) more...Acerola fruit (11) more...Carambola (12) more...Cranberry (13) more...Plum (16) more...Persimmon (17) more...Grapefruit (18) more...Fejhoa (19) more...Jackfruit (21) more...Cherry (21) more...Banana (22) more...Apricot (23) more...Jujube (23) more...Cherimoya (26) more...Nectarine (26) more...Loquat (27) more...Gooseberry (27) more...Lychee (31) more...Pomegranate (36) more...Durian (39) more...Avocado (52) more...Chokeberry (55) more...Blackcurrant (59) more...Passion fruit (68) more...
Phosphorus in Pineapple Based On Fruits Category
Lime (102) more...Apple (107) more...Pineapple (109) more...Watermelon (112) more...Pear (119) more...Quince fruit (119) more...Rose apple (123) more...Carambola (133) more...Grapefruit (135) more...Lemon (138) more...Acerola fruit (146) more...Raspberry (151) more...Strawberries (153) more...Mango (156) more...Plum (157) more...Chokeberry (158) more...Persimmon (161) more...Blackberries (162) more...Mandarin (Tangerine) (166) more...Lychee (171) more...Fejhoa (172) more...Clementine (177) more...Cutie clementine (177) more...Orange (179) more...Papaya (182) more...Kumquats (186) more...Peach (190) more...Grapes (191) more...Green grape (191) more...Sapodilla (193) more...Mulberry (194) more...Gooseberry (198) more...Nectarine (201) more...Pomelo (216) more...Cherry (222) more...Honeydew melon (228) more...Figs (232) more...Pomegranate (236) more...Jujube (250) more...Apricot (259) more...Loquat (266) more...Melon (267) more...Cantaloupe (267) more...Black lilac (280) more...Cherimoya (287) more...Jackfruit (303) more...Kiwi (312) more...Blackcurrant (322) more...Passion fruit (348) more...Banana (358) more...Guava (417) more...Durian (436) more...Mangosteen (48) more...Avocado (485) more...Plantain (487) more...Blueberries (77) more...Cranberry (85) more...
Potassium in Pineapple Based On Fruits Category
Apple (0.26) more...Pear (0.38) more...Acerola fruit (0.4) more...Quince fruit (0.4) more...Cranberry (0.4) more...Mangosteen (0.41) more...Loquat (0.43) more...Sapodilla (0.44) more...Papaya (0.47) more...Mango (0.5) more...Pineapple (0.54) more...Honeydew melon (0.54) more...Persimmon (0.58) more...Rose apple (0.6) more...Watermelon (0.6) more...Black lilac (0.66) more...Strawberries (0.67) more...Mixed berries (0.7) more...Plum (0.7) more...Lime (0.7) more...Green grape (0.72) more...Grapes (0.72) more...Blueberries (0.74) more...Figs (0.75) more...Pomelo (0.76) more...Grapefruit (0.77) more...Mandarin (Tangerine) (0.81) more...Lychee (0.83) more...Cantaloupe (0.84) more...Melon (0.84) more...Cutie clementine (0.85) more...Clementine (0.85) more...Gooseberry (0.88) more...Peach (0.91) more...Orange (0.94) more...Fejhoa (0.98) more...Kiwi (1) more...Carambola (1.04) more...Nectarine (1.06) more...Cherry (1.06) more...Banana (1.09) more...Lemon (1.1) more...Raspberry (1.2) more...Jujube (1.2) more...Plantain (1.3) more...Blackberries (1.39) more...Apricot (1.4) more...Chokeberry (1.4) more...Blackcurrant (1.4) more...Mulberry (1.44) more...Durian (1.47) more...Cherimoya (1.57) more...Pomegranate (1.67) more...Jackfruit (1.72) more...Kumquats (1.88) more...Avocado (2) more...Passion fruit (2.2) more...Guava (2.55) more...
Protein in Pineapple Based On Fruits Category
Loquat (1) more...Apple (1) more...Quince fruit (1) more...Lychee (1) more...Apricot (1) more...Raspberry (1) more...Figs (1) more...Banana (1) more...Gooseberry (1) more...Blackberries (1) more...Strawberries (1) more...Cutie clementine (1) more...Pomelo (1) more...Pear (1) more...Blueberries (1) more...Watermelon (1) more...Persimmon (1) more...Melon (1) more...Pineapple (1) more...Plum (1) more...Clementine (1) more...Mango (2) more...Lime (2) more...Carambola (2) more...Green grape (2) more...Guava (2) more...Blackcurrant (2) more...Mandarin (Tangerine) (2) more...Cranberry (2) more...Lemon (2) more...Durian (2) more...Fejhoa (3) more...Jackfruit (3) more...Jujube (3) more...Kiwi (3) more...Pomegranate (3) more...Plantain (4) more...Chokeberry (4) more...Black lilac (6) more...Avocado (7) more...Mangosteen (7) more...Acerola fruit (7) more...Cherimoya (7) more...Papaya (8) more...Mulberry (10) more...Kumquats (10) more...Sapodilla (12) more...Cantaloupe (16) more...Honeydew melon (18) more...
Sodium in Pineapple Based On Fruits Category
Fejhoa (0.006) more...Lychee (0.011) more...Pear (0.012) more...Cranberry (0.012) more...Carambola (0.014) more...Apple (0.017) more...Melon (0.017) more...Loquat (0.019) more...Quince fruit (0.02) more...Blackberries (0.02) more...Rose apple (0.02) more...Jujube (0.02) more...Acerola fruit (0.02) more...Papaya (0.023) more...Peach (0.024) more...Cherry (0.027) more...Kiwi (0.027) more...Plum (0.028) more...Apricot (0.03) more...Lime (0.03) more...Persimmon (0.03) more...Banana (0.031) more...Watermelon (0.033) more...Nectarine (0.034) more...Pomelo (0.034) more...Kumquats (0.037) more...Honeydew melon (0.038) more...Lemon (0.04) more...Gooseberry (0.04) more...Cantaloupe (0.041) more...Grapefruit (0.043) more...Blackcurrant (0.05) more...Mangosteen (0.054) more...Mango (0.058) more...Sapodilla (0.058) more...Mandarin (Tangerine) (0.058) more...Figs (0.06) more...Plantain (0.062) more...Avocado (0.067) more...Pomegranate (0.067) more...Guava (0.067) more...Grapes (0.069) more...Green grape (0.069) more...Pineapple (0.079) more...Cutie clementine (0.086) more...Clementine (0.086) more...Orange (0.087) more...Cherimoya (0.101) more...Jackfruit (0.105) more...Durian (0.374) more...
Vitamin B1 in Pineapple Based On Fruits Category
Honeydew melon (0.012) more...Carambola (0.016) more...Fejhoa (0.018) more...Pineapple (0.018) more...Cantaloupe (0.019) more...Lemon (0.02) more...Persimmon (0.02) more...Lime (0.02) more...Sapodilla (0.02) more...Cranberry (0.02) more...Strawberries (0.022) more...Loquat (0.024) more...Pear (0.025) more...Kiwi (0.025) more...Apple (0.026) more...Plum (0.026) more...Melon (0.026) more...Pomelo (0.027) more...Nectarine (0.027) more...Papaya (0.027) more...Gooseberry (0.03) more...Clementine (0.03) more...Quince fruit (0.03) more...Rose apple (0.03) more...Cutie clementine (0.03) more...Peach (0.031) more...Grapefruit (0.031) more...Cherry (0.033) more...Mandarin (Tangerine) (0.036) more...Raspberry (0.038) more...Apricot (0.04) more...Guava (0.04) more...Orange (0.04) more...Jujube (0.04) more...Blueberries (0.041) more...Figs (0.05) more...Blackcurrant (0.05) more...Pomegranate (0.053) more...Mangosteen (0.054) more...Jackfruit (0.055) more...Mango (0.057) more...Black lilac (0.06) more...Acerola fruit (0.06) more...Lychee (0.065) more...Grapes (0.07) more...Green grape (0.07) more...Banana (0.073) more...Plantain (0.076) more...Kumquats (0.09) more...Mulberry (0.101) more...Avocado (0.13) more...Passion fruit (0.13) more...Cherimoya (0.131) more...Durian (0.2) more...
Vitamin B2 in Pineapple Based On Fruits Category
Apple (0.091) more...Lemon (0.1) more...Persimmon (0.1) more...Cranberry (0.101) more...Cherry (0.154) more...Pear (0.157) more...Watermelon (0.178) more...Loquat (0.18) more...Grapes (0.188) more...Green grape (0.188) more...Quince fruit (0.2) more...Lime (0.2) more...Sapodilla (0.2) more...Grapefruit (0.204) more...Pomelo (0.22) more...Orange (0.274) more...Mangosteen (0.286) more...Pomegranate (0.293) more...Fejhoa (0.295) more...Gooseberry (0.3) more...Blackcurrant (0.3) more...Papaya (0.338) more...Kiwi (0.341) more...Carambola (0.367) more...Mandarin (Tangerine) (0.376) more...Strawberries (0.386) more...Figs (0.4) more...Acerola fruit (0.4) more...Plum (0.417) more...Blueberries (0.418) more...Honeydew melon (0.418) more...Kumquats (0.429) more...Black lilac (0.5) more...Pineapple (0.5) more...Mango (0.584) more...Raspberry (0.598) more...Apricot (0.6) more...Lychee (0.603) more...Mulberry (0.62) more...Cutie clementine (0.636) more...Clementine (0.636) more...Cherimoya (0.644) more...Blackberries (0.646) more...Banana (0.665) more...Plantain (0.672) more...Cantaloupe (0.734) more...Melon (0.734) more...Rose apple (0.8) more...Peach (0.806) more...Jujube (0.9) more...Jackfruit (0.92) more...Durian (1.074) more...Guava (1.084) more...Nectarine (1.125) more...Passion fruit (1.5) more...Avocado (1.738) more...
Vitamin B3 (Niacin) in Pineapple Based On Fruits Category
Acerola fruit (0.009) more...Carambola (0.017) more...Peach (0.025) more...Nectarine (0.025) more...Pear (0.028) more...Plum (0.029) more...Blackberries (0.03) more...Kumquats (0.036) more...Pomelo (0.036) more...Sapodilla (0.037) more...Papaya (0.038) more...Quince fruit (0.04) more...Apple (0.041) more...Mangosteen (0.041) more...Lime (0.043) more...Watermelon (0.045) more...Strawberries (0.047) more...Cherry (0.049) more...Mulberry (0.05) more...Blueberries (0.052) more...Grapefruit (0.053) more...Apricot (0.054) more...Raspberry (0.055) more...Cranberry (0.057) more...Orange (0.063) more...Chokeberry (0.065) more...Blackcurrant (0.066) more...Fejhoa (0.067) more...Cantaloupe (0.072) more...Melon (0.072) more...Cutie clementine (0.075) more...Pomegranate (0.075) more...Clementine (0.075) more...Mandarin (Tangerine) (0.078) more...Lemon (0.08) more...Gooseberry (0.08) more...Jujube (0.081) more...Grapes (0.086) more...Green grape (0.086) more...Honeydew melon (0.088) more...Loquat (0.1) more...Persimmon (0.1) more...Lychee (0.1) more...Passion fruit (0.1) more...Guava (0.11) more...Pineapple (0.112) more...Figs (0.113) more...Mango (0.134) more...Black lilac (0.23) more...Plantain (0.242) more...Avocado (0.257) more...Cherimoya (0.257) more...Durian (0.316) more...Jackfruit (0.329) more...Banana (0.367) more...
Vitamin B6 in Pineapple Based On Fruits Category
Green grape (0.002) more...Lime (0.008) more...Mandarin (Tangerine) (0.016) more...Honeydew melon (0.019) more...Cantaloupe (0.021) more...Plantain (0.022) more...Clementine (0.024) more...Cutie clementine (0.024) more...Cranberry (1) more...Grapes (2) more...Apple (3) more...Watermelon (3) more...Quince fruit (3) more...Peach (4) more...Cherry (4) more...Plum (5) more...Nectarine (5) more...Black lilac (6) more...Figs (6) more...Gooseberry (6) more...Mulberry (6) more...Blueberries (6) more...Pear (7) more...Persimmon (8) more...Blackcurrant (8) more...Apricot (9) more...Lemon (11) more...Carambola (12) more...Grapefruit (13) more...Loquat (14) more...Lychee (14) more...Mango (14) more...Sapodilla (14) more...Passion fruit (14) more...Kumquats (17) more...Pineapple (18) more...Banana (20) more...Melon (21) more...Raspberry (21) more...Fejhoa (23) more...Cherimoya (23) more...Jackfruit (24) more...Strawberries (24) more...Blackberries (25) more...Kiwi (25) more...Mangosteen (31) more...Durian (36) more...Papaya (37) more...Pomegranate (38) more...Orange (39) more...Guava (49) more...Avocado (81) more...
Vitamin B9 (Folic Acid) in Pineapple Based On Fruits Category
Passion fruit (0.02) more...Pineapple (0.02) more...Honeydew melon (0.02) more...Watermelon (0.05) more...Cantaloupe (0.05) more...Melon (0.05) more...Lychee (0.07) more...Banana (0.1) more...Figs (0.11) more...Pear (0.12) more...Quince fruit (0.12) more...Grapefruit (0.13) more...Kumquats (0.15) more...Lemon (0.15) more...Carambola (0.15) more...Fejhoa (0.16) more...Apple (0.18) more...Green grape (0.19) more...Grapes (0.19) more...Mandarin (Tangerine) (0.2) more...Clementine (0.2) more...Cutie clementine (0.2) more...Lime (0.22) more...Plum (0.26) more...Cherimoya (0.27) more...Strawberries (0.29) more...Papaya (0.3) more...Jackfruit (0.34) more...Blueberries (0.57) more...Pomegranate (0.6) more...Guava (0.73) more...Peach (0.73) more...Persimmon (0.73) more...Nectarine (0.77) more...Mulberry (0.87) more...Mango (1.12) more...Blackberries (1.17) more...Cranberry (1.2) more...Raspberry (1.42) more...Kiwi (1.46) more...Chokeberry (1.5) more...Avocado (2.07) more...
Vitamin E in Pineapple Based On Fruits Category
Lime (0.001) more...Cantaloupe (0.003) more...Honeydew melon (0.003) more...Green grape (0.015) more...Plantain (0.029) more...Pineapple (0.07) more...Lychee (0.4) more...Banana (0.5) more...Passion fruit (0.7) more...Apple (2.2) more...Nectarine (2.2) more...Strawberries (2.2) more...Melon (2.5) more...Guava (2.6) more...Persimmon (2.6) more...Papaya (2.6) more...Peach (2.6) more...Apricot (3.3) more...Fejhoa (3.5) more...Mango (4.2) more...Pear (4.5) more...Quince fruit (4.5) more...Figs (4.7) more...Cranberry (5.1) more...Plum (6.4) more...Mulberry (7.8) more...Raspberry (7.8) more...Jackfruit (10) more...Grapes (14.6) more...Pomegranate (16.4) more...Blueberries (19.3) more...Blackberries (19.8) more...Avocado (21) more...Kiwi (40.3) more...
Vitamin K in Pineapple Based On Fruits Category
Acerola fruit (0.01) more...Apple (0.04) more...Mango (0.04) more...Quince fruit (0.04) more...Loquat (0.05) more...Jujube (0.05) more...Lemon (0.06) more...Fejhoa (0.06) more...Orange (0.06) more...Lychee (0.07) more...Grapes (0.07) more...Grapefruit (0.07) more...Cherry (0.07) more...Papaya (0.08) more...Passion fruit (0.1) more...Watermelon (0.1) more...Pear (0.1) more...Plum (0.1) more...Sapodilla (0.1) more...Cranberry (0.1) more...Black lilac (0.11) more...Persimmon (0.11) more...Pineapple (0.12) more...Gooseberry (0.12) more...Mulberry (0.12) more...Carambola (0.12) more...Kiwi (0.14) more...Strawberries (0.14) more...Banana (0.15) more...Figs (0.15) more...Blueberries (0.16) more...Cherimoya (0.16) more...Kumquats (0.17) more...Peach (0.17) more...Nectarine (0.17) more...Melon (0.18) more...Apricot (0.2) more...Mangosteen (0.21) more...Guava (0.23) more...Blackcurrant (0.27) more...Durian (0.28) more...Pomegranate (0.35) more...Raspberry (0.42) more...Jackfruit (0.42) more...Chokeberry (0.53) more...Blackberries (0.53) more...Avocado (0.64) more...
Zinc in Pineapple Based On Fruits Category
Rose apple (5.7) more...Carambola (6.73) more...Watermelon (7.6) more...Acerola fruit (7.69) more...Strawberries (7.7) more...Papaya (8) more...Cantaloupe (8.16) more...Avocado (8.53) more...Melon (8.6) more...Honeydew melon (9.09) more...Lemon (9.32) more...Peach (9.54) more...Chokeberry (9.6) more...Blackberries (9.61) more...Pomelo (9.62) more...Mulberry (9.8) more...Gooseberry (10.18) more...Lime (10.54) more...Nectarine (10.55) more...Grapefruit (10.7) more...Apricot (11.12) more...Plum (11.42) more...Orange (11.89) more...Raspberry (11.94) more...Mixed berries (11.97) more...Cutie clementine (12.02) more...Clementine (12.02) more...Loquat (12.14) more...Cranberry (12.2) more...Fejhoa (13) more...Mandarin (Tangerine) (13.34) more...Pineapple (13.52) more...Apple (13.81) more...Pear (13.81) more...Quince fruit (13.81) more...Guava (14.3) more...Blueberries (14.49) more...Kiwi (14.66) more...Blackcurrant (15.38) more...Kumquats (15.9) more...Cherry (16.1) more...Lychee (16.53) more...Mango (17) more...Cherimoya (17.71) more...Mangosteen (17.91) more...Grapes (18) more...Green grape (18.1) more...Black lilac (18.4) more...Persimmon (18.59) more...Pomegranate (18.7) more...Figs (19.18) more...Sapodilla (19.9) more...Jujube (20.53) more...Banana (22.84) more...Passion fruit (23.38) more...Jackfruit (23.5) more...Durian (27.09) more...Plantain (31.89) more...
Carbohydrates in Pineapple Based On Fruits Category
Rose apple (25) more...Lemon (29) more...Watermelon (30) more...Lime (30) more...Carambola (31) more...Strawberries (32) more...Acerola fruit (32) more...Cantaloupe (34) more...Melon (34) more...Honeydew melon (36) more...Pomelo (38) more...Peach (39) more...Grapefruit (42) more...Blackberries (43) more...Mulberry (43) more...Papaya (43) more...Gooseberry (44) more...Nectarine (44) more...Plum (46) more...Cranberry (46) more...Cutie clementine (47) more...Loquat (47) more...Chokeberry (47) more...Clementine (47) more...Apricot (48) more...Mixed berries (49) more...Orange (49) more...Pineapple (50) more...Apple (52) more...Raspberry (52) more...Mandarin (Tangerine) (53) more...Fejhoa (55) more...Blueberries (57) more...Quince fruit (57) more...Pear (58) more...Kiwi (61) more...Blackcurrant (63) more...Cherry (63) more...Lychee (66) more...Guava (68) more...Green grape (69) more...Grapes (69) more...Persimmon (70) more...Mango (70) more...Kumquats (71) more...Black lilac (73) more...Mangosteen (73) more...Figs (74) more...Cherimoya (75) more...Jujube (79) more...Pomegranate (83) more...Sapodilla (83) more...Banana (89) more...Jackfruit (95) more...Passion fruit (97) more...Plantain (122) more...Durian (147) more...Avocado (160) more...
Calories in Pineapple Based On Fruits Category
Plantain (3) more...Carambola (3) more...Pomelo (4) more...Banana (5) more...Lychee (5) more...Apple (6) more...Durian (6) more...Peach (6) more...Plum (6) more...Blueberries (6) more...Nectarine (6) more...Honeydew melon (6) more...Watermelon (7) more...Persimmon (8) more...Cranberry (8) more...Pear (9) more...Cantaloupe (9) more...Melon (9) more...Green grape (10) more...Mango (10) more...Grapes (10) more...Pomegranate (10) more...Cherimoya (10) more...Quince fruit (11) more...Passion fruit (12) more...Avocado (12) more...Acerola fruit (12) more...Mangosteen (12) more...Apricot (13) more...Pineapple (13) more...Cherry (13) more...Mixed berries (14) more...Strawberries (16) more...Loquat (16) more...Fejhoa (17) more...Guava (18) more...Papaya (20) more...Sapodilla (21) more...Jujube (21) more...Grapefruit (22) more...Gooseberry (25) more...Raspberry (25) more...Lemon (26) more...Chokeberry (28) more...Rose apple (29) more...Blackberries (29) more...Cutie clementine (30) more...Clementine (30) more...Lime (33) more...Kiwi (34) more...Jackfruit (34) more...Figs (35) more...Mandarin (Tangerine) (37) more...Black lilac (38) more...Mulberry (39) more...Orange (40) more...Blackcurrant (55) more...Kumquats (62) more...
Calcium in Pineapple Based On Fruits Category
Cherimoya (5) more...Fejhoa (6) more...Pomelo (8) more...Strawberries (12) more...Lemon (22) more...Pear (23) more...Mulberry (25) more...Raspberry (33) more...Mangosteen (35) more...Quince fruit (40) more...Jujube (40) more...Durian (44) more...Apple (54) more...Blueberries (54) more...Pineapple (58) more...Sapodilla (60) more...Cranberry (60) more...Carambola (61) more...Banana (64) more...Green grape (66) more...Grapes (66) more...Mixed berries (70) more...Persimmon (81) more...Kiwi (87) more...Jackfruit (110) more...Figs (142) more...Avocado (146) more...Chokeberry (214) more...Blackberries (214) more...Blackcurrant (230) more...Orange (230) more...Kumquats (290) more...Gooseberry (290) more...Peach (326) more...Nectarine (332) more...Rose apple (339) more...Plum (345) more...Watermelon (569) more...Black lilac (600) more...Guava (624) more...Cherry (640) more...Mandarin (Tangerine) (681) more...Mango (765) more...Papaya (950) more...Plantain (1127) more...Grapefruit (1150) more...Passion fruit (1274) more...Loquat (1528) more...Apricot (1926) more...Melon (3382) more...Acerola fruit (7671) more...
Vitamin A in Pineapple Based On Fruits Category
All information about nutrition on this website was created with help of information from the official United States Department of Agriculture database. | 11,295 | 29,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | longest | en | 0.751333 |
http://mathhelpforum.com/discrete-math/77074-mathematical-induction-help-print.html | 1,524,487,454,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945942.19/warc/CC-MAIN-20180423110009-20180423130009-00207.warc.gz | 205,907,933 | 3,648 | # Mathematical Induction help
• Mar 5th 2009, 08:41 AM
tokio
Mathematical Induction help
1^3+2^3+...+n^3 = [n(n+1)/2]^2 n>= 1
If anyone has a link to write in mathematical nottaion on this forum, i will appreciate it, thanks.
• Mar 5th 2009, 08:50 AM
running-gag
Hi
Have a look to the LATEX tutorial in the LATEX forum
Let P(n) be $\displaystyle 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$
Check that P(1) is true
Suppose that P(k) is true and show that P(k+1) is true
• Mar 5th 2009, 09:25 AM
tokio
Thanks for the help, but i need help on the induction stage, when i do n+1. Here is what i have.
$\displaystyle \frac{n(n+1)^2+2(n+1)^3}{2} = \frac{((n+1(n+2)))}{2}^2$
• Mar 5th 2009, 10:38 AM
running-gag
Quote:
Originally Posted by tokio
Thanks for the help, but i need help on the induction stage, when i do n+1. Here is what i have.
$\displaystyle \frac{n(n+1)^2+2(n+1)^3}{2} = \frac{((n+1(n+2)))}{2}^2$
You suppose that P(n) is true, which means that $\displaystyle 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$
You want to prove that P(n+1) is true : $\displaystyle 1^3+2^3+...+n^3+(n+1)^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$
$\displaystyle 1^3+2^3+...+n^3+(n+1)^3 =(1^3+2^3+...+n^3)+(n+1)^3 =\left(\frac{n(n+1)}{2}\right)^2+ (n+1)^3$
Factor (n+1)²/4
• Mar 6th 2009, 09:58 AM
tokio
Quote:
Originally Posted by running-gag
You suppose that P(n) is true, which means that $\displaystyle 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$
You want to prove that P(n+1) is true : $\displaystyle 1^3+2^3+...+n^3+(n+1)^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$
$\displaystyle 1^3+2^3+...+n^3+(n+1)^3 =(1^3+2^3+...+n^3)+(n+1)^3 =\left(\frac{n(n+1)}{2}\right)^2+ (n+1)^3$
Factor (n+1)²/4
Im still stuck on the induction stage, It seems like if i still factor that i still cant get both sides of the quation to be logically equilvalent.
• Mar 6th 2009, 11:21 AM
Proof by induction
Hello tokio
Quote:
Originally Posted by tokio
Im still stuck on the induction stage, It seems like if i still factor that i still cant get both sides of the quation to be logically equilvalent.
Here's the bit you need:
$\displaystyle \left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3$
$\displaystyle = \frac{n^2\color{red}(n+1)^2}{\color{red}4} + \frac{4(n+1)\color{red}(n+1)^2}{\color{red}4}$
$\displaystyle = \color{red}\frac{(n+1)^2}{4}\color{black}(n^2 + 4(n+1))$
$\displaystyle = \frac{(n+1)^2}{4}(n+2)^2$
Can you complete it now?
• Mar 6th 2009, 06:22 PM
tokio
Quote:
Hello tokioHere's the bit you need:
$\displaystyle \left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3$
$\displaystyle = \frac{n^2\color{red}(n+1)^2}{\color{red}4} + \frac{4(n+1)\color{red}(n+1)^2}{\color{red}4}$
$\displaystyle = \color{red}\frac{(n+1)^2}{4}\color{black}(n^2 + 4(n+1))$
$\displaystyle = \frac{(n+1)^2}{4}(n+2)^2$
Can you complete it now?
Thanks for the help but i still cant believe im confused. Im assume that your just working onthe right hand side, i dont see where you proved for (n+1).
• Mar 6th 2009, 10:08 PM
Proof by induction
Hello tokio
Quote:
Originally Posted by tokio
Thanks for the help but i still cant believe im confused. Im assume that your just working onthe right hand side, i dont see where you proved for (n+1).
That's correct. I thought it was just the manipulation of the RHS that you couldn't do. The complete proof is:
Let $\displaystyle P(n)$ be the propositional function: $\displaystyle 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$
Then $\displaystyle P(n) \Rightarrow 1^3+2^3+...+n^3+(n+1)^3 =(1^3+2^3+...+n^3)+ (n+1)^3 =\left(\frac{n(n+1)}{2}\right)^2+ (n+1)^3$
$\displaystyle = \dots$ (as in my previous posting)
$\displaystyle = \frac{(n+1)^2}{4}(n+2)^2$
$\displaystyle = \left(\frac{(n+1)([n+1]+1)}{2}\right)^2$
$\displaystyle \Rightarrow P(n+1)$
$\displaystyle P(1)$ is $\displaystyle 1^3 = \left(\frac{1 \cdot 2}{2} \right)^2$, which is true. Hence $\displaystyle P(n), \forall n \in \mathbb{N}$. | 1,548 | 3,944 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-17 | latest | en | 0.727283 |
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# Question Video: Modeling Addition on a Number Line to Find the Total Mathematics • Second Year of Primary School
Count on to add. Find the total. 12 + 6 = ＿.
00:49
### Video Transcript
Count on to add. Find the total, 12 plus six.
Number 12 has already been highlighted. That’s where we need to begin counting. And we need to add six. That means we should start at 12 and count on six more.
One, two, three, four, five, six, we landed on number 18. So we can say that 12 add six equals 18. We found our answer by counting forward six more places on the number line.
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https://byjus.com/icse-selina-solution-concise-mathematics-class-6-chapter-2-estimation-exercise-2a/ | 1,656,577,426,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00525.warc.gz | 189,835,969 | 156,589 | # Selina Solutions Concise Mathematics Class 6 Chapter 2: Estimation Exercise 2(A)
Mathematics plays a vital role in the life of students in boosting their confidence at the time of examination. Rounding off a number is the main topic explained under this exercise. Students who want to improve their skills in the subject, are advised to follow the Selina Solutions Concise Mathematics Class 6 Chapter 2 Estimation Exercise 2(A) PDF. These Solutions, which are well structured by subject experts, help students in enhancing their performance in the examination. Download the PDF of Selina Solutions Concise Mathematics Class 6 Chapter 2 Estimation Exercise 2(A) PDF, from the links provided here.
## Selina Solutions Concise Mathematics Class 6 Chapter 2 Estimation Exercise 2(A) Download PDF
### Access another exercise of Selina Solutions Concise Mathematics Class 6 Chapter 2 Estimation
Exercise 2(B) Solutions
### Access Selina Solutions Concise Mathematics Class 6 Chapter 2: Estimation Exercise 2(A)
#### Exercise 2(A) page no: 11
1. Round off each of the following to the nearest ten:
(i) 62
(ii) 265
(iii) 543
(iv) 8261
(v) 6294
Solution:
If the ones place digit is less than 5, replace ones digit by 0, and keep the other digit as the same
And if the digit at ones place is 5 or more than 5, increase tens digit by 1 and replace ones digit by 0
(i) 62 to the nearest ten is 60
(ii) 265 to the nearest ten is 270
(iii) 543 to the nearest ten is 540
(iv) 8261 to the nearest ten is 8260
(v) 6294 to the nearest ten is 6290
2. Round off each of the following to the nearest hundred:
(i) 748
(ii) 784
(iii) 2667
(iv) 5432
(v) 6388
Solution:
If the digit at tens place is less than 5, replace each one of tens and ones digits by 0 and keep the other digits as the same
If the tens digit is 5 or more than 5, increase the hundreds digit by 1 and replace each of tens and ones digit by 0
(i) 748 to the nearest hundred is 700
(ii) 784 to the nearest hundred is 800
(iii) 2667 to the nearest hundred is 2700
(iv) 5432 to the nearest hundred is 5400
(v) 6388 to the nearest hundred is 6400
3. Round off each of the following to the nearest thousand:
(i) 6475
(ii) 6732
(iii) 25352
(iv) 32568
(v) 9248
Solution:
Observe the hundreds digit of the given number
If the hundreds digit is less than 5, replace each one of hundreds, tens, and ones digits by 0 and keep the other digits as same
If hundreds digit is 5 or more than 5 in the given number, increase thousand digit by 1 and replace each other digit on its right by 0
(i) 6475 to the nearest thousand is 6000
(ii) 6732 to the nearest thousand is 7000
(iii) 25352 to the nearest thousand is 25000
(iv) 32568 to the nearest thousand is 33000
(v) 9248 to the nearest thousand is 9000
4.Round off
(i) 578 to the nearest ten.
(ii) 578 to the nearest hundred.
(iii) 4327 to the nearest thousand.
(iv) 32974 to the nearest ten-thousand.
(v) 27487 to the nearest ten-thousand.
Solution:
(i) 578 to the nearest ten is 580
(ii) 578 to the nearest hundred is 600
(iii) 4327 to the nearest thousand is 4000
(iv) 32974 to the nearest ten-thousand is 30000
(v) 27487 to the nearest ten- thousand is 30000
5. Round off each of the following to the nearest ten, nearest hundred and nearest thousand.
(i) 864
(ii) 1249
(iii) 54, 547
(iv) 68, 076
(v) 56, 293
Solution:
(i) 864 to the nearest ten is 860
864 to the nearest hundred is 900
864 to the nearest thousand is 1000
(ii) 1249 to the nearest ten is 1250
1249 to the nearest hundred is 1200
1249 to the nearest thousand is 1000
(iii) 54547 to the nearest ten is 54550
54547 to the nearest hundred is 54500
54547 to the nearest thousand is 55000
(iv) 68076 to the nearest ten is 68080
68076 to the nearest hundred is 68100
68076 to the nearest thousand is 68000
(v) 56293 to the nearest ten is 56290
56293 to the nearest hundred is 56300
56293 to the nearest thousand is 56000
6. Round off the following to the nearest tens ;
(i) ₹ 562
(ii) 837 m
(iii) 545 cm
(iv) ₹ 27
Solution:
(i) ₹ 562 to the nearest ten is ₹ 560
(ii) 837 m to the nearest ten is 840 m
(iii) 545 cm to the nearest ten is 550 cm
(iv) ₹ 27 to the nearest ten is ₹ 30
7.List all the numbers which can be round off to 30.
Solution:
26, 27, 28, 29,31, 32, 33, 34 are the numbers that can be rounded off to 30
8.List all the numbers which can be rounded off to 50.
Solution:
46, 47, 48, 49, 51, 52, 53, 54 are the numbers that can be rounded off to 50
9. Write the smallest and the largest numbers which are rounded off to 80.
Solution:
75 is the smallest number which is rounded off to 80 and 84 is the largest number which is rounded off to 80
10.Write the smallest and the largest numbers which are rounded off to 130.
Solution:
125 is the smallest number which is rounded off to 130 and 134 is the largest number which is rounded off to 130 | 1,422 | 4,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-27 | longest | en | 0.857782 |
https://theuvocorp.com/a-suppose-f-is-one-to-one-function-with-domain-and-4/ | 1,638,860,126,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00595.warc.gz | 622,819,808 | 9,246 | # A Suppose F Is One To One Function With Domain And
(a) Suppose f is a one-to-one function with domain and range B. How is the inverse function defined? What is the domain of f-1? What is the range of f-1?
(b) If you are given a formula for f, how do you find a formula for f-1?
(c) If you are given the graph of f, how do you find the graph of f-1?
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# Accommodating Struggling Learners in Mathematics - PowerPoint PPT Presentation
Accommodating Struggling Learners in Mathematics. How can we, as teachers support our struggling scholars so that they can still have success? . Challenges that Struggling Learners Face . Difficulty retrieving arithmetic facts Difficulty understanding mathematical concepts
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### Accommodating Struggling Learners in Mathematics
How can we, as teachers support our struggling scholars so that they can still have success?
Difficulty retrieving arithmetic facts
Difficulty understanding mathematical concepts
Struggle to find and appropriately use alternate strategies
Students struggle to identify language to follow directions and or solve story problems.
Math Instruction Issues that Impact students with math learning Problems
• Spiral Curriculum
• If you don’t get it now, eventually you will.
• Struggling students need consistent practice with basic skills before moving on
• Teaching to mastery
• Students may not get to mastery
• What skills can we incorporate to support them if they can’t?
• Reforms that are cyclical in nature
• Shifts to Common Core
• Student’s home life
Concrete: each math skill is first modeled with concrete materials such as chips, counters, pattern blocks, etc.
Representational: The math concept is then modeled with tallies, dots, circles, etc
Abstract: The math concept is modeled at the abstract level with numbers and symbols. These should be used with the drawings
Teaching Math to Students with Disabilities. Accessed March 25, 2014. http://view.officeapps.live.com/op/view.aspx?src=http%3A%2F%2Fwww.specialed.ccsu.edu%2Fnicoll-senft%2FTeaching%2520Math%2520to%2520Students%2520with%2520Disabilities.ppt
Implementation of C-R-A Instruction Instruction)
Describe and model the new math concept
Allow for multiple opportunities to use concrete objects
Provide many practice opportunities where students will draw their solutions or use pictures to solve
When students demonstrate mastery by drawing solutions, describe and model how to use and perform skills using only math symbols
Again, give students multiple practice opportunities by performing the sills with numbers and symbols
Give students periodic practice to support maintenance of their acquire skill
Teaching Math to Students with Disabilities. Accessed March 25, 2014. http://view.officeapps.live.com/op/view.aspx?src=http%3A%2F%2Fwww.specialed.ccsu.edu%2Fnicoll-senft%2FTeaching%2520Math%2520to%2520Students%2520with%2520Disabilities.ppt
C-R-A Instruction Instruction)
How can you use the C-R- A Instruction in both small groups and full groups while in your classroom?
Explicit Modeling Instruction)
Provides a clear and accessible format for acquiring and understanding the mathematics concept
Allows students to become independent learners and problem solvers.
Teacher both describes and models the math skill/concept.
Teacher clearly describes features of the math concept or steps in performing math skill.
Teacher breaks math concept/skill into learnable parts.
Teacher describes/models using multi-sensory techniques.
Teacher engages students in learning through demonstrating enthusiasm, through maintaining a lively pace, through periodically questioning students, and through checking for student understanding.
Provides students with a learning map
Teaching Math to Students with Disabilities. Accessed March 25, 2014. http://view.officeapps.live.com/op/view.aspx?src=http%3A%2F%2Fwww.specialed.ccsu.edu%2Fnicoll-senft%2FTeaching%2520Math%2520to%2520Students%2520with%2520Disabilities.ppt
Identify what students will learn, both visually and auditory
Discuss the importance of the concepts
Break the skill into 3-4 learnable parts
Describe each piece with visual devices and examples
Cue students to essential attributes of the skill that you are modeling
Teacher Interaction
Think Aloud
Teaching Math to Students with Disabilities. Accessed March 25, 2014. http://view.officeapps.live.com/op/view.aspx?src=http%3A%2F%2Fwww.specialed.ccsu.edu%2Fnicoll-senft%2FTeaching%2520Math%2520to%2520Students%2520with%2520Disabilities.ppt
Implementing Explicit Instruction Instruction)
Ensure that your students have the prerequisite skills to perform the skill.
Break down the skill into logical and learnable parts (Ask yourself, "what do I do and what do I think as I perform the skill?").
Provide a meaningful context for the skill (e.g. word or story problem suited to the age & interests of your students).
Provide visual, auditory, kinesthetic (movement), and tactile means for illustrating important aspects of the concept/skill (e.g. visually display word problem and equation, orally cue students by varying vocal intonations, point, circle, highlight computation signs or important information in story problems).
"Think aloud" as you perform each step of the skill (i.e. say aloud what you are thinking as you problem-solve).
Link each step of the problem solving process (e.g. restate what you did in the previous step, what you are going to do in the next step, and why the next step is important to the previous step).
Periodically check student understanding with questions, remodeling steps when there is confusion.
Maintain a lively pace while being conscious of student information processing difficulties (e.g. need additional time to process questions).
Model a concept/skill at least three times before beginning to scaffold your instruction.
Teaching Math to Students with Disabilities. Accessed March 25, 2014. http://view.officeapps.live.com/op/view.aspx?src=http%3A%2F%2Fwww.specialed.ccsu.edu%2Fnicoll-senft%2FTeaching%2520Math%2520to%2520Students%2520with%2520Disabilities.ppt
Explicit Modeling Instruction)
Take a moment to identify how you could use explicit modeling in either small or full group instruction to support scholars
Authentic Mathematics Learning Contexts Instruction)
• Explicitly connect the target skills to a relevant context, promoting a deeper understanding for students
• Teacher describes/models math concept/skill
• Embedded within contexts that are meaningful to your students.
• Consider your student's age-related interests, cultural/community interests, and common experiences your students share.
• Examples:
• Think Story Problems!
• Shopping at the mall to look at geometric shapes
• Identifying amount needed to buy something
Teaching Math to Students with Disabilities. Accessed March 25, 2014. http://view.officeapps.live.com/op/view.aspx?src=http%3A%2F%2Fwww.specialed.ccsu.edu%2Fnicoll-senft%2FTeaching%2520Math%2520to%2520Students%2520with%2520Disabilities.ppt
Critical Elements to Authentic Learning Contexts Instruction)
Contexts that are meaningful for the students you are teaching (age, interests, experiences). Contexts may be school related, family related, or community related.
Explicit teacher modeling of math skill is incorporated.
Relevance of math concept/skill to the authentic context is clearly demonstrated.
Student practice of math skill follows teacher instruction. Teacher monitors, provides specific corrective feedback, remodels as needed, and provides positive reinforcement.
Teaching Math to Students with Disabilities. Accessed March 25, 2014. http://view.officeapps.live.com/op/view.aspx?src=http%3A%2F%2Fwww.specialed.ccsu.edu%2Fnicoll-senft%2FTeaching%2520Math%2520to%2520Students%2520with%2520Disabilities.ppt
Implementation Instruction)
Teacher chooses appropriate context within which to teach target math concept/skill.
Teacher activates student prior knowledge of authentic context, identifies the math concept/skill students will learn, and explicitly relates the target math concept/skill to the meaningful context.
Teacher explicitly models math concept/skill within authentic context.
Teacher involves students by prompting student thinking about how the math concept/skill is relevant to the authentic context.
Teacher checks for student understanding.
Students receive opportunities to apply math concept or perform math skill within authentic context. Teacher monitors, provides specific corrective feedback, remodels math concept/skill as needed, and provides positive reinforcement.
Teacher provides review and closure, explicitly re-stating how the target math skill relates to the authentic context and remodeling the skill.
Students receive multiple opportunities to apply math concept or practice math skill after initial instructional activity.
Incorporating the teacher instruction strategies, Building Meaningful Student Connections, Explicit Teacher Modeling, & Scaffolding Instruction when teaching within authentic contexts can be very effective.
Teaching Math to Students with Disabilities. Accessed March 25, 2014. http://view.officeapps.live.com/op/view.aspx?src=http%3A%2F%2Fwww.specialed.ccsu.edu%2Fnicoll-senft%2FTeaching%2520Math%2520to%2520Students%2520with%2520Disabilities.ppt
Authentic Mathematics Learning Contexts Instruction)
How can you incorporate this into your instruction?
Math Intervention and Accommodation Strategies for Struggling Learners
• Intensify Instruction
• More modeling
• Use concrete learning opportunities
• Break tasks down into smaller steps
• Step-by-Step Strategies
• Give extra support for a temporary amount of time
• Provide opportunities for practice and feedback.
Center on Instruction. Intensive interventions for Students Struggling in Reading and Math. 2012. Accessed on March 23, 2014. http://www.centeroninstruction.org/files/Intensive%20Interventions%20for%20Students%20Struggling%20in%20Reading%20%26%20Math.pdf
Teach Using Multiple Instructional Examples Struggling Learners
• Model “I do, We do, You do”
• Focus on selecting and sequencing your instructional examples
• Scaffolding will be needed when acquiring new skills for students to master and be successful on.
• Teach to a wide range of knowledge so that students can carry their mastery oer to other concepts.
• Scaffold in a way that students can master simple to complex tasks
Center on Instruction. Mathematics Instruction For Students with Learning Disabilities of Difficulty learning Mathematics. Accessed on April 2, 2014. http://www.centeroninstruction.org/files/Mathematics%20Instruction%20LD%20Guide%20for%20Teachers.pdf
Have Students “Talk it Out” Struggling Learners
Encourage students to think aloud their decision making
Students can verbalize steps in a solution format using self questioning. This can be done before, during, and/or after they have solved
Verbalization may help anchor skills and strategies for those who are LD. Verbalization can also facilitate student self-regulation when solving.
Center on Instruction. Mathematics Instruction For Students with Learning Disabilities of Difficulty learning Mathematics. Accessed on April 2, 2014. http://www.centeroninstruction.org/files/Mathematics%20Instruction%20LD%20Guide%20for%20Teachers.pdf
Visual representations from both teachers and students can help explain and clarify problems.
Visuals can be effective when combined with explicit instruction.
Center on Instruction. Mathematics Instruction For Students with Learning Disabilities of Difficulty learning Mathematics. Accessed on April 2, 2014. http://www.centeroninstruction.org/files/Mathematics%20Instruction%20LD%20Guide%20for%20Teachers.pdf
Teach Students to ProblemsSolve Using Heuristic Strategies
• Can address computational skills, problem solving, and fractions
• A Heuristic Strategy approach can be in steps
• A way to organize information
• Include student discussion and reflection
• Read the problem, circle numbers, underline key words, solve the problem, check your work
• Discuss answers, why does it work?
Center on Instruction. Mathematics Instruction For Students with Learning Disabilities of Difficulty learning Mathematics. Accessed on April 2, 2014. http://www.centeroninstruction.org/files/Mathematics%20Instruction%20LD%20Guide%20for%20Teachers.pdf
Ongoing assessment and evaluation can help teachers measure the rhythm of student growth and help them fine tune instruction to meet student’s needs.
Center on Instruction. Mathematics Instruction For Students with Learning Disabilities of Difficulty learning Mathematics. Accessed on April 2, 2014. http://www.centeroninstruction.org/files/Mathematics%20Instruction%20LD%20Guide%20for%20Teachers.pdf
Visual Aids and Guides with Step by Step Instructions
Peer Buddy
Sit near teacher or other area for less distractions
• Provide student with kinesthetic approaches
• Provide multiple ways to solve
• Ex: 2+5 can be done with fingers, touch math, blocks, and tallies.
• Provide extra examples of vocabulary
• Differentiate task by giving them more practice on skills
Now it is your turn! Students
Using your student population group students into the struggling, average, and above average learners.
Think about these groups strengths and weaknesses while reviewing IEP and your curriculum
Design and model a lesson or skill most still struggle with using C-R-A, Explicit Modeling, or Authentic Concepts. Think about what other accommodations you can use to support scholars as well
Be prepared to share by stating your lesson’s objective, how you would use the curricular framework of your choice, how you would implement other accommodations, what do you think will go well upon implementation, as well as what challenges may arise? | 3,028 | 14,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-51 | latest | en | 0.892294 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/9200/2/a/cd/ | 1,675,614,832,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500273.30/warc/CC-MAIN-20230205161658-20230205191658-00674.warc.gz | 882,122,151 | 56,476 | # Properties
Label 9200.2.a.cd Level $9200$ Weight $2$ Character orbit 9200.a Self dual yes Analytic conductor $73.462$ Analytic rank $0$ Dimension $3$ CM no Inner twists $1$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$9200 = 2^{4} \cdot 5^{2} \cdot 23$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 9200.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$73.4623698596$$ Analytic rank: $$0$$ Dimension: $$3$$ Coefficient field: 3.3.621.1 Defining polynomial: $$x^{3} - 6x - 3$$ x^3 - 6*x - 3 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 920) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + \beta_1 q^{3} + (\beta_1 + 1) q^{7} + (\beta_{2} + \beta_1 + 1) q^{9}+O(q^{10})$$ q + b1 * q^3 + (b1 + 1) * q^7 + (b2 + b1 + 1) * q^9 $$q + \beta_1 q^{3} + (\beta_1 + 1) q^{7} + (\beta_{2} + \beta_1 + 1) q^{9} + ( - \beta_{2} + \beta_1 - 1) q^{11} + (\beta_{2} - \beta_1) q^{13} + (\beta_{2} - \beta_1 + 1) q^{17} + (2 \beta_{2} + \beta_1 - 1) q^{19} + (\beta_{2} + 2 \beta_1 + 4) q^{21} + q^{23} + 3 q^{27} + (\beta_{2} + 1) q^{29} + (\beta_{2} + 3 \beta_1 - 2) q^{31} + (2 \beta_{2} - \beta_1 + 5) q^{33} - 4 q^{37} + ( - 2 \beta_{2} - 5) q^{39} + ( - 2 \beta_{2} - 3 \beta_1 - 2) q^{41} + (2 \beta_{2} + 6) q^{43} + (\beta_{2} + 5) q^{47} + (\beta_{2} + 3 \beta_1 - 2) q^{49} + ( - 2 \beta_{2} + \beta_1 - 5) q^{51} + ( - 4 \beta_{2} - 2) q^{53} + ( - \beta_{2} + 2 \beta_1 + 2) q^{57} + (2 \beta_{2} + 2 \beta_1 - 2) q^{59} + (\beta_{2} - \beta_1 + 9) q^{61} + (\beta_{2} + 4 \beta_1 + 4) q^{63} + ( - 4 \beta_{2} + 4 \beta_1 + 4) q^{67} + \beta_1 q^{69} + (2 \beta_{2} + 3 \beta_1 - 2) q^{71} + ( - \beta_{2} + 4 \beta_1 + 5) q^{73} + (\beta_{2} + 4) q^{77} + ( - 3 \beta_{2} - 3) q^{81} + ( - 2 \beta_{2} - 4 \beta_1 - 4) q^{83} + ( - \beta_{2} + 2 \beta_1 - 1) q^{87} + ( - 2 \beta_{2} - 10) q^{89} + ( - \beta_{2} - \beta_1 - 5) q^{91} + (2 \beta_{2} + 2 \beta_1 + 11) q^{93} + ( - 3 \beta_{2} + \beta_1 - 3) q^{97} + (3 \beta_1 - 3) q^{99}+O(q^{100})$$ q + b1 * q^3 + (b1 + 1) * q^7 + (b2 + b1 + 1) * q^9 + (-b2 + b1 - 1) * q^11 + (b2 - b1) * q^13 + (b2 - b1 + 1) * q^17 + (2*b2 + b1 - 1) * q^19 + (b2 + 2*b1 + 4) * q^21 + q^23 + 3 * q^27 + (b2 + 1) * q^29 + (b2 + 3*b1 - 2) * q^31 + (2*b2 - b1 + 5) * q^33 - 4 * q^37 + (-2*b2 - 5) * q^39 + (-2*b2 - 3*b1 - 2) * q^41 + (2*b2 + 6) * q^43 + (b2 + 5) * q^47 + (b2 + 3*b1 - 2) * q^49 + (-2*b2 + b1 - 5) * q^51 + (-4*b2 - 2) * q^53 + (-b2 + 2*b1 + 2) * q^57 + (2*b2 + 2*b1 - 2) * q^59 + (b2 - b1 + 9) * q^61 + (b2 + 4*b1 + 4) * q^63 + (-4*b2 + 4*b1 + 4) * q^67 + b1 * q^69 + (2*b2 + 3*b1 - 2) * q^71 + (-b2 + 4*b1 + 5) * q^73 + (b2 + 4) * q^77 + (-3*b2 - 3) * q^81 + (-2*b2 - 4*b1 - 4) * q^83 + (-b2 + 2*b1 - 1) * q^87 + (-2*b2 - 10) * q^89 + (-b2 - b1 - 5) * q^91 + (2*b2 + 2*b1 + 11) * q^93 + (-3*b2 + b1 - 3) * q^97 + (3*b1 - 3) * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$3 q + 3 q^{7} + 3 q^{9}+O(q^{10})$$ 3 * q + 3 * q^7 + 3 * q^9 $$3 q + 3 q^{7} + 3 q^{9} - 3 q^{11} + 3 q^{17} - 3 q^{19} + 12 q^{21} + 3 q^{23} + 9 q^{27} + 3 q^{29} - 6 q^{31} + 15 q^{33} - 12 q^{37} - 15 q^{39} - 6 q^{41} + 18 q^{43} + 15 q^{47} - 6 q^{49} - 15 q^{51} - 6 q^{53} + 6 q^{57} - 6 q^{59} + 27 q^{61} + 12 q^{63} + 12 q^{67} - 6 q^{71} + 15 q^{73} + 12 q^{77} - 9 q^{81} - 12 q^{83} - 3 q^{87} - 30 q^{89} - 15 q^{91} + 33 q^{93} - 9 q^{97} - 9 q^{99}+O(q^{100})$$ 3 * q + 3 * q^7 + 3 * q^9 - 3 * q^11 + 3 * q^17 - 3 * q^19 + 12 * q^21 + 3 * q^23 + 9 * q^27 + 3 * q^29 - 6 * q^31 + 15 * q^33 - 12 * q^37 - 15 * q^39 - 6 * q^41 + 18 * q^43 + 15 * q^47 - 6 * q^49 - 15 * q^51 - 6 * q^53 + 6 * q^57 - 6 * q^59 + 27 * q^61 + 12 * q^63 + 12 * q^67 - 6 * q^71 + 15 * q^73 + 12 * q^77 - 9 * q^81 - 12 * q^83 - 3 * q^87 - 30 * q^89 - 15 * q^91 + 33 * q^93 - 9 * q^97 - 9 * q^99
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{3} - 6x - 3$$ :
$$\beta_{1}$$ $$=$$ $$\nu$$ v $$\beta_{2}$$ $$=$$ $$\nu^{2} - \nu - 4$$ v^2 - v - 4
$$\nu$$ $$=$$ $$\beta_1$$ b1 $$\nu^{2}$$ $$=$$ $$\beta_{2} + \beta _1 + 4$$ b2 + b1 + 4
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−2.14510 −0.523976 2.66908
0 −2.14510 0 0 0 −1.14510 0 1.60147 0
1.2 0 −0.523976 0 0 0 0.476024 0 −2.72545 0
1.3 0 2.66908 0 0 0 3.66908 0 4.12398 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$1$$
$$5$$ $$1$$
$$23$$ $$-1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 9200.2.a.cd 3
4.b odd 2 1 4600.2.a.y 3
5.b even 2 1 1840.2.a.t 3
20.d odd 2 1 920.2.a.g 3
20.e even 4 2 4600.2.e.r 6
40.e odd 2 1 7360.2.a.cb 3
40.f even 2 1 7360.2.a.ca 3
60.h even 2 1 8280.2.a.bo 3
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
920.2.a.g 3 20.d odd 2 1
1840.2.a.t 3 5.b even 2 1
4600.2.a.y 3 4.b odd 2 1
4600.2.e.r 6 20.e even 4 2
7360.2.a.ca 3 40.f even 2 1
7360.2.a.cb 3 40.e odd 2 1
8280.2.a.bo 3 60.h even 2 1
9200.2.a.cd 3 1.a even 1 1 trivial
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(9200))$$:
$$T_{3}^{3} - 6T_{3} - 3$$ T3^3 - 6*T3 - 3 $$T_{7}^{3} - 3T_{7}^{2} - 3T_{7} + 2$$ T7^3 - 3*T7^2 - 3*T7 + 2 $$T_{11}^{3} + 3T_{11}^{2} - 15T_{11} + 12$$ T11^3 + 3*T11^2 - 15*T11 + 12
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{3}$$
$3$ $$T^{3} - 6T - 3$$
$5$ $$T^{3}$$
$7$ $$T^{3} - 3 T^{2} - 3 T + 2$$
$11$ $$T^{3} + 3 T^{2} - 15 T + 12$$
$13$ $$T^{3} - 18T - 29$$
$17$ $$T^{3} - 3 T^{2} - 15 T - 12$$
$19$ $$T^{3} + 3 T^{2} - 33 T + 48$$
$23$ $$(T - 1)^{3}$$
$29$ $$T^{3} - 3 T^{2} - 6 T + 12$$
$31$ $$T^{3} + 6 T^{2} - 42 T - 249$$
$37$ $$(T + 4)^{3}$$
$41$ $$T^{3} + 6 T^{2} - 60 T - 69$$
$43$ $$T^{3} - 18 T^{2} + 72 T + 32$$
$47$ $$T^{3} - 15 T^{2} + 66 T - 76$$
$53$ $$T^{3} + 6 T^{2} - 132 T - 536$$
$59$ $$T^{3} + 6 T^{2} - 36 T - 32$$
$61$ $$T^{3} - 27 T^{2} + 225 T - 596$$
$67$ $$T^{3} - 12 T^{2} - 240 T + 2944$$
$71$ $$T^{3} + 6 T^{2} - 60 T - 203$$
$73$ $$T^{3} - 15 T^{2} - 42 T + 588$$
$79$ $$T^{3}$$
$83$ $$T^{3} + 12 T^{2} - 60 T - 64$$
$89$ $$T^{3} + 30 T^{2} + 264 T + 608$$
$97$ $$T^{3} + 9 T^{2} - 69 T - 138$$ | 3,675 | 6,897 | {"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} | 2.890625 | 3 | CC-MAIN-2023-06 | latest | en | 0.43737 |
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