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Level 4 - Standard Costing Direct Costs
Hi, is anyone able to help with the direct material efficiency part of the question?
Many thanks in advance
• Hi
I hope you are well?
To work out Budget / Standard cost of labour for actual production:
307 hours x £12.50 = £3,837.50 (Favourable as it's less than actual by £107.50 - £3,945)
Direct Labour Rate:
Standard / Budgeted Rate = £12.50 per hour
Actual Rate = £3,945 / 307 Hours = £12.85 - Rounded to 2 decimal places
The Actual Rate is Adverse by 35p per hour
Direct Labour Efficiency:
Actual Cost of Labour for 5,000 units based on 307 hours is - £3,945
Budgeted / Standard Cost - 6 hours to make 100 units. So to make 5,000 units it will be (5,000 units / 100 units x 6 Hours) = 300 Hours. Then 300 hours x £12.50 per hour = £3,750
This gives you a Adverse of £195 (£3,945 - £3,750)
Actual Cost of Labour for Actual Production:
£3,945 - Mentioned in question (Adverse as it's more than £107.50 - £3,837.50)
Hopefully this is all right, sorry I have answered this question on my mobile so the breakdown might not be as clear. If you need a clearer breakdown do let me know or if anything is wrong do let
me know too, I'll log into the laptop and look at it properly again.
Kind regards
• Hi
Hopefully that was all right and it helped you 🙂
• @shamilkaria
are you able to help with a different question please? I am unsure on the total cost per KG part of the question? All of the other parts are fine
• Hi
Sorry for the late reply.
Just wanted to make sure the Maximum Material Cost per unit is correct?
To get from 200 grams to 1Kg you would need to muliply by 5. Target cost is usually your selling price - desired profit.
• Hi
I have just checked the answers and it is correct:
• Hi
Glad you got it correct 🙂. I had to read up on Target Cost been a while since needed to use it.
But yh happy to help if you got anything else.
Kind regards
• Are you able to confirm how to get to the target cost per KG as I’m still a little confused
• Hi
It's all about understanding the metric conversion. So you have worked out that the maximum you can per unit is £4 but that is for 200 grams.
They then ask you per kilogram. Kilo is 1kg which is 1000 grams. So what they did was how many 200 grams go into 1000 grams (which is 1kg or kg). 5 x 200 grams = 1kg. So the target cost for 1kg
is £4 (200 grams) x 5 = £20
For example if they said 5kg that would be 5000 grams. Then it would be 5000 grams / 200 grams = 25. So that means it would be 25 x £4 = £100
Hope this helps you.
Kind regards
• Hi
Most welcome
Kind regards | {"url":"https://forums.aat.org.uk/Forum/discussion/451838/level-4-standard-costing-direct-costs","timestamp":"2024-11-06T11:08:40Z","content_type":"text/html","content_length":"315899","record_id":"<urn:uuid:cced0f3c-bd4b-47d8-b033-b6c5cb922524>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00710.warc.gz"} |
Volume vs. Specific Volume: What's the Difference?
Volume is the amount of space occupied by an object, while specific volume is the volume per unit mass of a substance.
Key Differences
Volume and specific volume are both important concepts in physics and engineering, but they have distinct meanings and applications. Volume is a measure of the amount of space occupied by an object
or substance. It is a three-dimensional quantity and can be expressed in cubic units such as cubic meters (m³), cubic centimeters (cm³), or liters (L). Specific volume, on the other hand, is a
property of materials that describes the volume occupied by a unit mass of a substance. It is the reciprocal of density and is expressed in units such as cubic meters per kilogram (m³/kg) or liters
per kilogram (L/kg).
Volume is a general concept that applies to any three-dimensional space, specific volume is specific to the context of materials and their properties. Volume is a straightforward measurement of
space, whereas specific volume provides insight into the intrinsic properties of a substance, such as how much space a certain amount of mass occupies.
In practical applications, volume is often used to determine the capacity of containers, the size of objects, or the amount of space occupied by a substance. Specific volume, however, is used in
calculations involving the behavior of substances under different pressures and temperatures, such as in the design of engines, refrigeration systems, and other thermodynamic processes.
Volume is a measure of the space occupied by an object or substance, while specific volume is a material property that indicates the volume occupied by a unit mass of a substance. Understanding the
difference between these two concepts is essential in fields such as physics, engineering, and materials science.
Comparison Chart
The amount of space occupied by an object or substance
The volume occupied by a unit mass of a substance
Cubic meters (m³), liters (L), cubic centimeters (cm³)
Cubic meters per kilogram (m³/kg), liters per kilogram (L/kg)
General measurement of space
Material property related to density
Determining capacity, size, or amount of space
Analyzing behavior of substances under different conditions
Relation to Mass
Independent of mass
Inversely related to density
Volume and Specific Volume Definitions
The amount of three-dimensional space occupied by an object.
The volume of the water in the tank is 200 liters.
Specific Volume
The volume occupied by a unit mass of a substance.
The specific volume of air at standard conditions is about 0.8 m³/kg.
The quantity of space enclosed within a shape.
The volume of the cube is calculated as the side length cubed.
Specific Volume
A measure used in fluid mechanics to analyze flow properties.
The specific volume of a gas changes with pressure and temperature.
A measure of the capacity of a container.
The bottle has a volume of 1 liter.
Specific Volume
The reciprocal of density.
The specific volume of a material can be calculated by dividing 1 by its density.
The space that a substance or object occupies.
The volume of gas in the balloon expanded as it heated up.
Specific Volume
An important parameter in the design of thermodynamic systems.
The specific volume of refrigerant is crucial in the design of refrigeration cycles.
The magnitude of sound or the loudness of audio.
Please turn down the volume; it's too loud.
Specific Volume
A property used to describe the behavior of substances in thermodynamics.
The specific volume of steam is higher than that of water.
A collection of written or printed sheets bound together; a book.
One of the books of a work printed and bound in more than one book.
What is the formula for the volume of a cube?
The volume of a cube is calculated as side length cubed (s³).
Is volume a scalar or vector quantity?
Volume is a scalar quantity because it has magnitude but no direction.
How is volume measured?
Volume is typically measured in cubic units such as cubic meters (m³), cubic centimeters (cm³), liters (L), and gallons (gal).
How do you find the volume of a cylinder?
The volume of a cylinder is found by πr²h, where r is the radius and h is the height.
How is volume used in daily life?
Volume measurements are crucial in cooking, manufacturing, and in determining the capacity of containers like fuel tanks and water reservoirs.
How is specific volume expressed?
Specific volume is typically expressed in cubic meters per kilogram (m³/kg) in the SI system.
What is volume?
Volume is the amount of space that a substance or object occupies or that is enclosed within a container.
What instruments can measure volume?
Instruments like measuring cups, graduated cylinders, and volumetric flasks can measure volume.
Can volume apply to both solids and liquids?
Yes, volume can refer to the space occupied by solids, liquids, and gases.
Does volume have a standard unit?
In the International System of Units (SI), the standard unit of volume is the cubic meter (m³).
Where is specific volume used?
Specific volume is widely used in thermodynamics and fluid mechanics to analyze the properties of gases and liquids.
What is the significance of specific volume in thermodynamics?
Specific volume is important for understanding the state and behavior of substances under different thermal conditions in thermodynamic processes.
Can volume change with temperature?
Yes, the volume of substances can expand or contract with temperature changes, particularly noticeable in gases and liquids.
How does specific volume relate to gas laws?
Specific volume directly relates to gas laws, which describe how temperature, pressure, and volume affect a gas's state.
What is specific volume?
Specific volume is a property of materials, defined as the volume occupied by a unit mass of a substance.
Is specific volume the same as volume?
No, specific volume is a material property that indicates the volume per unit mass, whereas volume is a general measure of space.
How do you calculate specific volume?
Specific volume is calculated as the inverse of density, i.e., specific volume = 1/density.
Is specific volume important in engineering?
Yes, it is crucial in mechanical and chemical engineering for designing systems and understanding material behavior under various conditions.
Can specific volume vary with temperature and pressure?
Yes, specific volume can change with temperature and pressure, especially for gases following the ideal gas law.
Can specific volume be applied to solids and liquids?
While more commonly used for gases, specific volume can also apply to solids and liquids, offering insights into their compressibility and expansivity.
About Author
Written by
Janet White
Janet White has been an esteemed writer and blogger for Difference Wiki. Holding a Master's degree in Science and Medical Journalism from the prestigious Boston University, she has consistently
demonstrated her expertise and passion for her field. When she's not immersed in her work, Janet relishes her time exercising, delving into a good book, and cherishing moments with friends and
Edited by
Aimie Carlson
Aimie Carlson, holding a master's degree in English literature, is a fervent English language enthusiast. She lends her writing talents to Difference Wiki, a prominent website that specializes in
comparisons, offering readers insightful analyses that both captivate and inform. | {"url":"https://www.difference.wiki/volume-vs-specific-volume/","timestamp":"2024-11-12T07:08:50Z","content_type":"text/html","content_length":"130516","record_id":"<urn:uuid:2ab196f9-ccff-4542-bf94-f78b733197cd>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00104.warc.gz"} |
The Better Half
You have an array of N bit strings each of length M. You know that there is at least one element that appears more than N/8 times in the array. Using O(M+log(N)) memory and O(NM) time, find such an
An Easier Version
Well, its actually almost exactly the same, but solve the above riddle in case there is an element that appears more than N/2 times in the array. I managed to find 3 distinct solutions to this easier
variation, but only one of which generalizes easily.
Thanks Nemo for giving me this riddle!
Spoiler Alert - Solution Ahead!
The solution’s central idea is that if you discard any k distinct elements of the array (k = 8 in the original version, or k = 2 in the easier version) an element that appeared more than \(\frac{1}
{k}\) before discarding, will continue to appear more than \(\frac{1}{k}\) after discarding (note that this is a one way implication, i.e. it is possible for an element to appear more than \(\frac{1}
{k}\) after discarding but not before). We therefore keep an auxiliary array S of size 7 (k-1) of bit strings and integer counts. We reset all the counts to 0. We then iterate through our original
array and for each element:
• If it is already in the S array, we increment its count.
• Otherwise, if there is room in the S array (i.e. there is an element with a count of 0), we add it to the S array (and set its count to 1).
• Otherwise, we subtract 1 from all elements in the S array (which is akin to discarding 8 distinct elements).
We end up with 7 elements, one of which appears more than \(\frac{1}{8}\) in the original array, but since the implication is one-directional we need one final pass to count for each candidate how
often is appears in the original array.
Here it is in Python:
def better_half(A, N = 8):
S = [[None, 0] * (N - 1)]
for x in A:
for s in S:
if x == s[0]:
s[1] += 1
for s in S:
if s[1] == 0:
s[0] = x
s[1] = 1
for s in S:
s[1] -= 1
for s in S:
if s[1] != 0:
if A.count(S) > len(A) // N:
return s[0] | {"url":"https://yanivle.github.io/puzzles/2010/05/19/the-better-half.html","timestamp":"2024-11-11T14:38:45Z","content_type":"text/html","content_length":"38236","record_id":"<urn:uuid:20ca97f6-db56-40fe-8dc5-1cd8e0df4525>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00482.warc.gz"} |
Find Largest d in Array such that a + b + c = d - TutorialCup
Find Largest d in Array such that a + b + c = d
Difficulty Level Medium
Frequently asked in Accolite Amazon Delhivery Fanatics Fourkites FreeCharge
Views 2105
Problem Statement
Suppose you have an array of integers. Input values are all distinct elements. The problem “Find largest d in array such that a + b + c = d” asks to find out the largest element ‘d’ in the set such
that a + b + c = d, where all elements are different from each other.
arr[] = {1, 4, 6, 8, 11 }
Explanation: The three numbers a, b and c are 1, 4 and 6 and their sum is 11.
arr[] = {2,4,6,10,15 }
No such solution exists.
Explanation: As there are no three numbers, that sums up to a number.
1. Declare a Map.
2. While traversing through the array.
1. Add and insert the sum of two elements in a map with their indexes in a Map.
3. Set the number to the minimum value of an integer, which we have to find out.
4. Search for the third number in a map by checking the difference of two numbers is present in a map.
5. If true then check if their indexes should not be the same.
6. Check for the maximum of d and the maximum of arr[i] and arr[j] and store it to d.
7. Return d.
Consider an integer array consisting of distinct integers. Our task is to find out the number in such a way there exist three numbers which sum up to that number. We are going to use Hashing. Hashing
provides an efficient solution. Traverse the array and take two array elements at a time and store the sum of those pairs to map with their respective indexes.
We will store the pair because we are searching for d, such that a + b + c = d. Instead of this, we will be searching for a + b = d – c. So at first, when we store the pair and their indices. We will
be able to check for element ‘d’ such that d – c exists in the map. This can be done by traversing the array, and then picking up two elements at once. Make the difference of both the elements and
search for that difference if exists in the map. If this found to be true, check the current two elements should not be on the same index as the previous pairs on an index.
This is necessary to check if any of the elements should not be repeated on the same index, the repeated number can be considered in a, b, c and d, but their indices, meant to say, the number itself
on the same index should not be considered. So we have to check for those indices plagiarism. Now, we need to find out the maximum of arr[i] and arr[j] and check that maximum to d to find out the
maximum between them and store it to the d. Because, we have to find out the fourth number d, so we need to find the maximum of array elements, as d is always bigger among a, b, c and d.
C++ program to find largest d in array such that a + b + c = d
using namespace std;
int getSumThreeNumber(int arr[], int n)
unordered_map<int, pair<int, int> > MAP;
for (int i = 0; i < n - 1; i++)
for (int j = i + 1; j < n; j++)
MAP[arr[i] + arr[j]] = { i, j };
int d_number = INT_MIN;
for (int i = 0; i < n - 1; i++)
for (int j = i + 1; j < n; j++)
int third = abs(arr[i] - arr[j]);
if (MAP.find(third) != MAP.end())
pair<int, int> obj = MAP[third];
if (obj.first != i && obj.first != j && obj.second != i && obj.second != j)
d_number = max(d_number, max(arr[i], arr[j]));
return d_number;
int main()
int arr[] = { 1,4,6,8,11 };
int n = sizeof(arr) / sizeof(arr[0]);
int res = getSumThreeNumber(arr, n);
if (res == INT_MIN)
cout << "No such solution exists";
cout << res;
return 0;
Java program to find largest d in array such that a + b + c = d
import java.util.HashMap;
class CheckIndex
int i, j;
CheckIndex(int i, int j)
this.i = i;
this.j = j;
int checkI()
return i;
int checkJ()
return j;
class sumOfThreeElementToD
public static int getSumThreeNumber(int[] arr, int n)
HashMap<Integer, CheckIndex> map = new HashMap<>();
for (int i = 0; i < n - 1; i++)
for (int j = i + 1; j < n; j++)
map.put(arr[i] + arr[j], new CheckIndex(i, j));
int d_number = Integer.MIN_VALUE;
for (int i = 0; i < n - 1; i++)
for (int j = i + 1; j < n; j++)
int third = Math.abs(arr[i] - arr[j]);
if (map.containsKey(third))
CheckIndex ci = map.get(third);
if (ci.checkI() != i && ci.checkI() != j && ci.checkJ() != i && ci.checkJ() != j)
d_number = Math.max(d_number, Math.max(arr[i], arr[j]));
return d_number;
public static void main(String[] args)
int arr[] = { 1, 4, 6, 8, 11 };
int n = arr.length;
int output = getSumThreeNumber(arr, n);
if (output == Integer.MIN_VALUE)
System.out.println("No such solution exists");
Complexity Analysis to Find Largest d in Array such that a + b + c = d
Time Complexity
O(n^2) where “n” is the number of elements in the array. We have achieved this complexity because we have used HashMap which allows insertion searching and other operations in O(1) time.
Space Complexity
O(n^2) where “n” is the number of elements in the array. Since the HashMap stores addition of pair of different elements of the input. Because of this, the algorithm has quadratic space complexity. | {"url":"https://tutorialcup.com/interview/hashing/find-largest-d-in-array-such-that-a-b-c-d.htm","timestamp":"2024-11-09T04:59:27Z","content_type":"text/html","content_length":"108456","record_id":"<urn:uuid:4473e9af-52ac-41b3-9c4e-0ea6ab00188b>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00823.warc.gz"} |
h2g2 - A Conversation for Mathematics
However, the deeper one looks into the subject, the more one realises that mathematics is relevant to everything, simply because you can describe every single thing in the Universe with numbers.
DrRodge Started conversation Jan 29, 2006
"However, the deeper one looks into the subject, the more one realises that mathematics is relevant to everything, simply because you can describe every single thing in the Universe with numbers."
This is a dangerous statement because it isn't true. Mathematics can also describe things that don't occur in the universe and is probably why there is a problem in formulating the grand unification
Richard Dawkins' "Climbing Mount Improbable" describes the three factors that account for the shapes of all the shells that can be found on the planet, and a formula or computational algorithm can be
written to produce drawings of any type of shell. However, most of shells predicted by the algorithm are not viable and, therefore, do not occur in nature.
It occurs to me that the majority of predictions made using mathematics may only exist in the mathematical world, because when applied to the real world, they are invalid.
So I'm afraid that the failure to produce a grand unifying theory may be due to the fact that either there isn't one, or somewhere along the line, a mistake or a false assumption has been made.
Since the speed of light is involved in these types of calculations, perhaps the assumption that it cannot be exceeded is false. I can live with the idea that light has a finite maximum velocity
because I have my own theory as to why this should be and it makes sense, but has anyone ever explained why this should stop anything else travelling faster? If you applied a constant force to a body
forever it will continue to accelerate in a straight line forever won't it? (For argument's sake, there is no friction and no other force being applied).
Key: Complain about this post
However, the deeper one looks into the subject, the more one realises that mathematics is relevant to everything, simply because you can describe every single thing in the Universe with numbers.
"The Hitchhiker's Guide to the Galaxy is a wholly remarkable book. It has been compiled and recompiled many times and under many different editorships. It contains contributions from countless
numbers of travellers and researchers." | {"url":"https://h2g2.com/edited_entry/A144280/conversation/view/F18980/T2106866","timestamp":"2024-11-05T10:04:40Z","content_type":"text/html","content_length":"17591","record_id":"<urn:uuid:c5a20838-9fb0-4976-89ea-e408e689cf1a>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00587.warc.gz"} |
Om din variabel mäts med intervallskalan bör du undersöka om dina mätvärden uppfyller villkoren för att få använda parametriska tester.
We'll also cover the bootstrap and permutation tests, and how these can be used to obtain confidence intervals and p-values without having to assume a normal
{abc, acb, bac, Simulation based tests as an alternative to asymptotic parametric tests. be able to suggest a simple permutation test and implement it in a computer program. Persson, I., Arnroth,
L., Thulin, M. (2019). Multivariate two-sample permutation tests for trials with multiple time-to-event outcomes.
Simulation based tests as an alternative to asymptotic parametric tests. be able to suggest a simple permutation test and implement it in a computer program. More powerful permutation test based on
multistage ranked set sampling. L Amro, MH Samuh.
However, as you can imagine, in the 1930’s these tests could be used only with very small samples and this limited their appeal to some degree.
the reader to nonparametrics by describing some simple tests, some tests of the most frequently encountered experimental designs, and permutation testing.
A Visual Explanation of Statistical Testing Statistical tests, also known as hypothesis tests, are used in the design of experiments to measure the effect of some treatment(s) on experimental units.
They are employed in a large number of contexts: Oncologists use them to measure the efficacy of new treatment options for cancer. 2019-01-28 Permutation tests date back to the 1930’s, and were first
proposed by Fisher (1935), Pitman (1937a, 1937b, 1938) and others. This is certainly not a new idea!
Permutation Tests with SAS/IML® John Vickery, North Carolina State University ABSTRACT If your data do not meet the assumptions for a standard parametric test, you might want to consider using a
permutation test. By randomly shuffling the data and recalculating a test statistic, a permutation test can
2. Permutation Tests 8. Estimation 1.
Any statistical measure of the difference could be used here : PERMUTATION TESTS FOR LINEAR MODELS 77 follows, as a simplification. Exactly analogous results follow if Y,Xor Zis multivariate, but for
simplicity of notation we restrict attention throughout this paper to univariate Y,X and Z.For further simplicity and without loss … In the permutation test we simulate a ideal (null) world in which
there is no average difference between the numbers in the two groups.
Annika nordström stylist
Permutation Tests 8. Estimation 1. Parameter Estimation and Bootstrap 2. Confidence Intervals 3.
4 / 19. Page 5. Permutation Tests. Randomization Tests.
Hi henry
di flowerspolisen.se efterlystasveriges forsta kvinnliga lakaresista tomning brevlada orebrosjukbidrag
Wilcoxon matched-pairs single rank test was used to compare differences of expression between groups (p ≤ 0.05) and permutation test was performed
Thus one is always free to choose the statistic which best 7 Sep 2020 Given observations from a stationary time series, permutation tests allow one to construct exactly level \alpha tests under the
null hypothesis of an Description · Examines the most up-to-date methodologies of univariate and multivariate permutation testing. · Includes extensive software codes in MATLAB, R and Users of
statistical methods appear to be of two minds about permutation tests. On one hand, since the “randomization” test in the context of a randomized clinical 23 Jan 2020 A permutation test for the
contaminated data implemented, e.g.
Friseur bilder damentegler benefits group
v); mat::traits::permutation<>::type P(permutation(v)); cout << "A is:\n" << A << "\nPermuted A is \n" << Matrix(P * A); Matrix L(I + strict_lower(LU)), U(upper(LU)),
This paper begins with an explanation and notation for an exact test. Alternative (Quicker) Permutation F-Test To save computational time, we do not actually have to compute the F-statistic for each
permutation. All we need is an \equivalent statistic". 2021-01-13 Two sample permutation tests¶. Suppose that we have a completely randomized experiment, where people are assigned to two groups at
random. Suppose we have individuals indexed by .We assign them at random to one of two groups with a random treatment vector , then individual receives treatment (for example, a drug) and if ,
individual receives no treatment (a placebo). Permutation tests are non-parametric tests that do not assume normally-distributed errors. | {"url":"https://hurmaninvesterarwzcw.web.app/67874/10908.html","timestamp":"2024-11-10T20:56:32Z","content_type":"text/html","content_length":"10438","record_id":"<urn:uuid:9e76b23b-6a44-4623-bfb1-2d6f7d27bd19>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00457.warc.gz"} |
Average Questions for Competitive Exams with Solutions - SSC STUDY
Average Questions for Competitive Exams with Solutions
All type of Average Questions Quiz for Competitive Exams with Solutions. Practice Mock Test for SSC CGL, CHSL, SSC GD, UPSSSC PET, UP Police Constable examinations.
Quiz Name : Average Questions
Subject : Quantitative Aptitude
All Type Questions from previous year SSC Exams
Total Questions : 40
Medium : English
Feature :
On Selection of Option : Immediate display of Correct Answer and Solution
On Finish / Submit the Mock Test : Result with all Answers
#1. The average weight of a group of 20 boys was calculated to be 89.4 kg and it was later discovered that one weight was misread as 78 kg instead of 87 kg. The correct average weight is :
#2. A man spends Rs. 1800 monthly on an average for the first four months and Rs. 2000 monthly for the next eight months and saves Rs. 5600 a year. His average monthly income is :
#3. 4 boys and 3 girls spent Rs.120 on the average, of which the boys spent Rs.150 on the average. Then the average amount spent by the girls is:
#4. A cricketer had a certain average of runs for his 64 innings. In his 65th innings, he is bowled out for no score on his part. This brings down his average by 2 runs. His new average of runs is :
#5. Three years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby (in year/s) is :
#6. If out of 10 selected students for an examination, 3 were of 20 years age, 4 of 21 and 3 of 22 years, the average age of the group is :
#7. The average pocket money of 3 friends A, B, C is Rs. 80 in a particular month. If B spends double and C spends triple of what A spends during that month and if the average of their unspent pocket
money is Rs. 60, then A spends (in Rs.)
#8. The average age of a cricket team of 11 players is the same as it was 3 years back because 3 of the players whose current average age of 33 years were replaced by 3 youngsters. The average age of
the newcomers is :
#9. The average age of 40 students of a class is 15 years. When 10 new students are admitted, the average is increased by 0.2 year. The average age of the new students is :
#10. The average age of 30 students is 9 years. If the age of their teacher is included, the average age becomes 10 years. The age of the teacher (in years) is:
#11. The average marks of 50 students in a class is 72. The average marks of boys and girls in that subject are 70 and 75 respectively. The number of boys in the class is :
#12. The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. The weight of the new person is :
#13. The average weight of the first 11 persons among 12 persons is 95 kg. The weight of 12th person is 33 kg more than the average weight of all the 12 persons. The weight of the 12th person is :
#14. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight (in kg) of B is :
#15. If the average weight of 6 students is 50 kg; that of 2 students is 51 kg and that of 2 other students is 55 kg; then the average weight of all the students is :
#16. The average of 13 results is 70. The average of first seven is 65 and that of the last seven is 75, the seventh result is :
#17. The average of the largest and smallest 3 digit numbers formed by 0, 2 and 4 would be
#18. The average of 30 numbers is 12. The average of the first 20 of them is 11 and that of the next 9 is 10. The last number is :
Explanation: Last number :
= 30 × 12-20 × 11 − 9 × 10
= 360 − 220 − 90
= 360 − 310 = 50
#19. Out of seven given numbers, the average of the first four numbers is 4 and that of the last four numbers is also 4. If the average of all the seven numbers is 3, then the fourth number is :
#20. The average of 15 numbers is 7. If the average of the first 8 numbers be 6.5 and the average of last 8 numbers be 9.5, then the middle number is :
#21. The average of all the odd integers between 2 and 22 is :
#22. If the average of eight consecutive even numbers be 93, then the greatest number among them is :
#23. What is the average of the first six (positive) odd numbers each of which is divisible by 7?
#24. The average of three consecutive odd numbers is 12 more than one third of the first of these numbers. What is the last of the three numbers?
#25. The average of 5 consecutive natural numbers is m. If the next three natural numbers are also included, then how much more than m will the average of these 8 numbers be?
#26. The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs.12,000 and the average salary of the rest is Rs.6000. The total number of workers in
the workshop is :
#27. Of the three numbers, the first number is twice the second and the second is thrice the third number. If the average of these 3 numbers is 20, then the sum of the largest and the smallest
numbers is :
#28. The average per day income of A, B and C is Rs. 450. If the average per day income of A and B be Rs. 400 and that of B and C be Rs. 430, the per day income of B is :
#29. Out of 4 numbers, whose average is 60, the first one is one-fourth of the sum of the last three. The first number is :
#30. Of the three numbers, the second is twice the first and also thrice the third. If the average of the three numbers is 44, the largest number is
#31. Of the three numbers whose average is 60, the first is one fourth of the sum of the others. The first number is :
#32. The average marks of a class of 35 children is 35. The marks of one of the students, who got 35, was incorrectly entered as 65. What is the correct average of the class?
#33. The average of 20 numbers is calculated as 35. It is discovered later, that while calculating the average, one number, namely 85, was read as 45. The correct average is :
#34. The mean of 20 items is 55. If two items such as 45 and 30 are removed, the new mean of the remaining items is :
#35. The mean value of 20 observations was found to be 75, but later on it was detected that 97 was misread as 79. Find the correct mean.
#36. A cricket player after playing 10 tests scored 100 runs in the 11th test. As a result, the average of his runs is increased by 5. The present average of runs is :
#37. If the mean of 4 observations is 20, when a constant ‘C’ is added to each observation, the mean becomes 22. The value of C is
#38. The average of six numbers is 32. If each of the first three numbers is increased by 2 and each of the remaining three numbers is decreased by 4, then the new average is :
#39. The average of five numbers is 7. When three new numbers are included, the average of the eight numbers becomes 8.5. The average of three new numbers is :
#40. In a class there are 30 boys and their average age is 17 years. When on one boy aged 18 years leaving the class and another joining, the average age becomes 16.9 years. The age of new boy is :
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Higher Dimensions
A couple of years ago I was very interested in visualizing objects of more than three dimensions. I reasoned that if I stared at such objects for long enough time, I was bound to get a good intuition
on how they look.
But how can one see objects of more than three spatial dimensions?
Our goal in this post will be to render an object residing in a coordinate system of arbitrary dimensions on a 2D computer screen. We will try to do this in such a way as to lose as little
information as possible on the structure of the object, while maintaining its 2D description as simple as possible (so that indeed, we will get some form of intuition on the shape of the object).
The 4D Cube
Before we begin with the actual rendering process, lets explore some properties of the 4D cube. Lets start with the obvious:
• A 0D cube is simply a point.
• A 1D cube is a line segment. It has 1 * 2 = 2 faces each consisting of a 0D cube.
• A 2D cube is a square. It has 2 * 2 = 4 faces each consisting of a 1D cube.
• A 3D cube is, well, a cube (I hope that the repeated use of the word cube for two slightly different meanings does not confuse you). It has 3 * 2 = 6 faces each consisting of a 2D cube.
• Finally a 4D cube has 4 * 2 = 8 faces, each consisting of a 3D cube.
Here is a list of some interesting figures about the 4D cube. Can you figure out why they are correct? Can you generalize them to higher dimensions?
These values can be generated by my program, which is described below.
4D Cube Properties
• Number of edges: 32
• Number of facets: 8
• Facets per edge: 3
• Edges per facet: 12
3D Cube Properties (for comparison)
• Number of edges: 12
• Number of facets: 6
• Facets per edge: 2
• Edges per facet: 4
Dimensionality Reductions
I will now present several methods for encoding an N-dimensional object in (N-1)-dimensional space. Call such an encoding a 1-reduction. It will then be possible to encode any object on a 2D screen,
by repeated applications of 1-reductions.
Let’s start by considering a very simple special case. Let’s say that our N-D (N-dimensional) object is actually an (N-1)-D object residing in N-D space. An example of this is an object with a
constant coordinate. More complex examples exist (the object may not have any constant coordinate, but after applying some rotations, one of the coordinates may become fixed).
A basic 1-reduction can then consist of simply dropping the extra coordinate.
This may seem trivial, but I have a point here – when considering an inanimate object one of the coordinates is always fixed – that of the time!
Our first approach will then be trading a spatial dimension for a time dimension. I.e. encoding the Nth coordinate of an inanimate object by some form of animation.
A simple such encoding consists of slicing the N-D object. Consider the 3D case. If we denote the coordinate system of the object by x, y and z and the time coordinate by t, then we can select only
points with a constant z coordinate equal to t. We would then get an animated 2D object representing our 3D object.
A key feature of this 1-reduction method is that no structural information about the object is lost by this encoding. Lets define this formally:
Definition - A 1-reduction of an object is called lossless if it can be reversed uniquely.
Using this definition, we can say that the method of animated-slices proposed above is lossless. Note that animated-slices is obviously not the only lossless encoding (can you think of others? 8-) ).
We will now restrict our discussion to smooth and bounded objects.
Definition – an object is called smooth if it can be approximated well by discrete boxes.
Definition – an object is called bounded if there exists a cube of finite sides that contains the object.
Cubes, spheres and rings are examples of smooth and bounded objects.
The smoothness of an object allows us to encode it using discrete samples. For example, when we display an animated slice of a 3D object on a computer screen, the animation consists of discrete
frames each consisting of discrete pixels. A smooth object will be represented faithfully with such a discrete representation. More formally, lets define:
Definition – A 1-reduction is called casi-lossless if it can be reversed, and the difference between the inversion result and the original object is small.
So, encoding a smooth object as a sequence of frames is casi-lossless.
The advantage of considering only smooth and bounded objects is that we can encode two spatial coordinates on the same axis.
Lets start by demonstrating this with the simple case of a 3D cube (we will get to 4D cubes shortly!). This encoding is depicted here:
The image shows 2D slices of the 3D cube.
Note that the encoding is indeed casi-lossless: the entire structure of the cube can be reconstructed uniquely from the slices (again, neglecting small differences due to the fact samples are
Note that employing this method on a 4D object, instead of on a 3D one, results in 3D slices instead of the 2D ones. We can then apply another casi-lossless 1-reduction to the 3D slices, and get a
casi-lossless 2-reduction from four dimensions to two dimensions (I leave to the reader to formulate the definition of a casi-lossless 2-reduction).
The method above, generalized slightly, is the most important 1-reduction method - the Projection. Projecting an object consists of simply ignoring one of its coordinates. It is actually simpler than
what we discussed above – instead of converting one dimension to another (i.e. z-coordinate to t-coordinate) we simply ignore one of the coordinates.
This method by itself is obviously not lossless (it is not even casi-lossless). This means that there are many different N-D scenes that result in the same (N-1)-D projection. The method does work on
a much larger scale of objects though, and this is the method we shall employ.
A projection of the rotated 3D cube is depicted here:
Note that unlike the slices version above, there is no way to tell whether this is actually a cube!
Can you visualize other 3D objects that would have this exact same projection?
A different rotation, like this one, might look more like a cube, but note that there are objects very different from cubes that would have exactly this projection:
Can you picture one?
Using Color
After projecting an object we can try to rescue some of the information lost by encoding it in the color channel.
In the picture below I encoded the same 3D cube as above, but this time the color of the object represents the depth at the pixel. The brighter the color, the closer the object:
Using the color channel thus does not make the encoding lossless. There is still no way to tell whether the shape above is that of a cube. It does considerably reduce the set of possible
source-objects for the encoding though.
The main advantage of the projection method is that our brain is used to reversing projections. When you close one of your eyes, your brain receives a completely flat two-dimensional image. Even with
both eyes open, you do not see a true three-dimensional image. You merely see two 2D images. Two projections of the 3D space in front of you!
Your brain does a great job at guessing the real 3D structure of objects by their 2D projections (it tries to calculate the most probable inverse of the projection). As we noted before, the
projection method is not lossless, and thus your brain can be fooled. These are a couple of examples of the works of Felice Varini (see http://www.varini.org/) demonstrating mistakes made by the
brain at reconstructing the true 3D structure of the scene from a 2D projection:
And also:
The photos show that there can easily be two completely different 3D images resulting in the same 2D projection. That said, in most usual scenarios your brain correctly guesses the 3D structure of
things based on insufficient 2D inputs. It is important to say that your brain uses a lot of apriori information when reconstructing such scenes. In the absence of such apriori information (as is the
case with our 4D cube) our brain’s work is indeed much harder.
Ok, so now that we understand some of the properties of 4D cubes and how projections work, let’s get to the cool stuff – render some 4D images!
The Program
If you are not interested in an implementation of all of the above, you are welcome to skip this section.
I implemented the projection reduction method with a few lines of Python. My program displays an arbitrary shape (box, ring, spiral) of arbitrary dimensions in a universe of arbitrary dimensions (of
course dim(universe) ≥ dim(object)!).
The objects are rotated in real time. A color encoding of the high coordinates is also used.
At this point I should add that I wrote this program quite a while ago and for internal use only (i.e. I never thought someone else would look at it – I am actually quite surprised that I even found
it and that is still works! [unattended code tends to rot]). I did spend a few minutes in order to enable you to run the program from the command line (instead of from a Python shell), but in order
to access all of the options a shell is still required. Also, the code was not at all tested, and running it with the wrong arguments can cause it to throw some exceptions. It is very short though,
and all of you with basic python knowledge are welcome to browse it.
A very important part of the code consists of the create_cube and gen_line_pairs functions, defined as follows:
def create_cube(dim=4):
if dim == 1:
return [[-1], [1]]
res = []
r1 = create_cube(dim - 1)
for i in r1:
res += [i + [-1]]
res += [i + [1]]
return res
def gen_line_pairs(rect, dist=1):
res = []
for i in range(len(rect)):
for j in range(i):
dif = count_dif(rect[i], rect[j])
if dif <= dist:
res.append((i, j))
return res
Can you figure out how they work? :-)
Note that in order to run the program, you will need Python, as well as the Pygame package (which I use for drawing and managing user input). You can try running the program with the following
> 4d.py 4 rect 4d auto normalize
Some keys you can try:
• Up, down, left, right, home, pageup, return, space, end, pagedown - manually rotating the object (around the four first axes). This only works when automatic rotation is off.
• r – toggles automatic rotation mode.
• 4 – toggles the high-dimensions mode of the automatic rotation. When the automatic rotation is enabled, the object will be rotated around a random axis. When the high-dimensions mode is turned
off, the rotation will only take place around the first 3 coordinates. This mode allows appreciating a 3D slice of a high dimension object.
• c – change the coordinate that is used for the color-encoding.
The complete source of the 4D renderer can be found here.
To give all of you without access to python a taste of how some basic primitives do look like in higher dimensions I attached these images. I must say that viewing the objects in motion is much more
insightful. I really recommend running the program!
This image shows a regular 3D cube, residing in a 4D universe (after it was rotated a little about all of its axes):
Notice that this is indeed a cube (all its sides are equal!) the reason it appears to be a rectangular box is because the rotation about the 4th axis deforms it. A simpler to grasp, but similar
phenomenon occurs while residing a 2 dimensional square inside a 3D universe. After some rotations it may look like this:
This image shows a 4 dimensional cube:
This is perhaps the most interesting object (maybe I will add some more pictures of it).
The same cube, this time viewed from a different angle and having its faces interconnected:
A 5D cube:
Two rings in a 4D universe:
And finally, a very simple 10D spiral in a 10D universe (notice that all the lines are perpendicular!):
Again, these images are much more intuitive when the complete animation is viewed, so download the program and run it! | {"url":"https://yanivle.github.io/math/2007/07/22/higher-dimensions.html","timestamp":"2024-11-11T13:45:16Z","content_type":"text/html","content_length":"36651","record_id":"<urn:uuid:8e33b75a-6648-4c8d-9df1-c51d4c5e3988>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00896.warc.gz"} |
e in Python - Using Math Module to Get Euler's Constant e
To get the value of Euler’s Constant e in Python, the easiest way is to use the Python math module constant e. math.e returns the value 2.718281828459045.
import math
In Python, we can easily get the value of e for use in various equations and applications with the Python math module.
To get the value of the constant e in your Python code, you just need to import the math module and call it as shown below.
import math
You can also import e from the math module and call it as shown in the following Python code.
from math import e
Using math.exp() to Get Euler’s Number e in Python
Another way we can get e in Python is with the math module exp() function. exp() allows us to easily exponentiate e to a given power.
If we pass ‘1’ to exp() we can get e.
Below is how to use the math exp() function get the value of e in Python.
import math
Using numpy to Get Euler’s Number e in Python
Finally, we can use numpy to get Euler’s Number e for use in our Python code.
numpy also implements the exp() function and if we pass ‘1’, we can get the constant e.
Below is how to use the numpy exp() function to get the constant e.
import numpy as np
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Dimensions such that Each Can Use Half of the Channel with Zero Interference from the Others
How to Position n Transmitter-Receiver Pairs in n - 1 Dimensions such that Each Can Use Half of the Channel with Zero Interference from the Others
This work is inspired by the question. "Can 100 speakers talk for 30 minutes each in one room within one hour and with zero interference to each other s audience?" posed by Cadambe and Jafar at the
45th Allerton conference 2007, see [1]. We consider the problem of how many transmitter-receiver pairs can be placed such that each desired link may use half of the channel time free of interference
from unintended transmissions. The answer is given in the title: at least n pairs, i.e., 2n stations can be positioned in the (n-1)-dimensional Euclidean space such that complete interference
alignment in time is achieved. Regular patterns with equal distances between receivers and transmitters, respectively, are the solution. The basic methodology for achieving this result is borrowed
from the field of distance geometry.
BibT[E]X Reference Entry
author = {Rudolf Mathar and Milan Zivkovic},
title = "How to Position {$n$} Transmitter-Receiver Pairs in {$n-1$} Dimensions such that Each Can Use Half of the Channel with Zero Interference from the Others",
booktitle = "{IEEE} Globecom 2009",
address = {Honolulu, Hawaii, USA},
month = Dec,
year = 2009,
hsb = hsb910015378,
[ Download paper] [ Download bibtex-file]
This material is presented to ensure timely dissemination of scholarly and technical work. Copyright and all rights there in are retained by authors or by other copyright holders. All persons copying
this information are expected to adhere to the terms and constraints invoked by each author's copyright. In most cases, these works may not be reposted without the explicit permission of the
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Fun Math Learning Games
These are our
math games, beginning with the most recently added.
For a complete list, pick "All Math Games by Grade" above.
Make math magic by measuring angles with a protractor. 4th Grade
To defend the castle, understand and estimate angle measurement. 4th Grade
Shop and pay using dollar bills, quarters, dimes, nickels, and pennies. 2nd Grade
Fluently multiply within 100 to paint a mystery pixel picture. 3rd Grade
Fluently add and subtract within 100 and get ready to make change for a dollar. 2nd to 3rd Grade
Know from memory all products of two one-digit numbers to sink baskets in this timed shootaround. 3rd Grade
Identify and compare Canadian coins to pay at the grocery store. 1st Grade
Create code to sequentially move a car from one location to another. 1st Grade
Give directions for moving a car from one location to another in a fun pre-coding activity. 1st Grade
Sink enemy ships if you understand an integer as a point on the number line. 6th Grade
Catch coins in your piggy bank as you count coins up to $1.00. 2nd Grade
Recognize half and quarter partitions of a circle and understand that they can be combined to make a whole. 3rd Grade
Skip count by 2s, 5s and 10s to leap from lilypad to lilypad across a pond. 2nd Grade
Know from memory the products of all single digit numbers to race through a slalom course. 3rd Grade
Count to tell the number of objects and to answer "how many?" to knock down 10 pins. K - 1st Grade
Count from 1 to 20 drawing lines from number to number to reveal a mystery picture. K - 1st Grade
Count forward to 20 beginning from a given number within the known sequence (instead of having to begin at 1) to create a mystery picture. K - 1st Grade
Count up to 50 starting at any number less than 50 to make a mystery picture. 1st Grade
Count up to 120 starting at any number less than 120 to make a mystery picture. 1st Grade
Skip count by 2s or 5s to make a mystery picture. 2nd Grade
Skip count by 10s to make a mystery picture. 2nd Grade
Add and subtract within 20 to make a mystery picture. 2nd Grade
Add and subtract within 100 without regrouping to make a mystery picture. 2nd Grade
Add and subtract with regrouping to make a mystery picture. 2nd Grade
Identify triangles, rectangles and squares as parts of a composite shape; recognize partitions of rectangles into equal shares, using the words half, third, and quarter. 2nd Grade
Work with equal groups of objects to gain foundations for multiplication: Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns. 2nd
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Unable to reduce under opaque constant during unification
Coq fails to unify two terms with a unification variable appearing in both of them, even when simplifying the terms would make one of them disappears. This happens only when the simplification step
is under an opaque constant, for instance :
From elpi Require Import elpi.
Inductive C (T : Type) := Build_C.
Elpi Command bla.
Elpi Query lp:{{
coq.unify-eq X {{ Build_C (fst (lp:X, lp:Y)) }} Z.
With id instead of Build_C, unification succeeds. Is there a way to have this reduction happen, or otherwise to make this unification problem solvable ?
with Id it succeeds because the two sides are exactly X, while if there is a rigid symbol you get an occur check.
Right, I should have written snd instead of fst, Build_C (snd (lp:X, lp:Y)) simplifies to Build_C lp:Y, in which lp:X does not appear, but the unification still fails.
Afaik only agda normalizes a term to minimize the variables it contains. IMO it is too expensive in general.
A message was moved here from #Coq devs & plugin devs > 8.19 by Enrico Tassi.
So I suggest you explain how you ended up there
(Agda normalization is also closer to Coq simpl or Equations's simp, so its cost is very different)
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Understanding Mathematical Functions: How To Find The Minimum Of A Fun
Understanding the Basics of Mathematical Functions
A mathematical function is a relation between a set of inputs and a set of possible outputs, where each input is related to exactly one output. Functions are essential in various fields such as
mathematics, physics, engineering, economics, and more. They provide a way to model and analyze real-world phenomena, making them a fundamental concept in both theoretical and applied mathematics.
A Definition of a mathematical function and its importance in various fields
At its core, a mathematical function can be defined as a rule that assigns to each element in the domain exactly one element in the codomain. In simpler terms, it takes an input and produces a single
output. Functions are used to describe relationships between quantities, model real-world phenomena, and predict behavior in various systems. Their importance is evident in fields such as physics,
where functions are used to describe the motion of objects, and economics, where they are used to model demand and supply.
Overview of the function's minimum - the lowest point on its graph
The minimum of a function refers to the lowest point on its graph. It is the value of the function that is smaller or equal to all other values of the function. In mathematical terms, the minimum of
a function f(x) is denoted as f(min), where f(min) ≤ f(x) for all x in the domain. Finding the minimum of a function is essential in various mathematical and practical applications, especially in
optimization and problem-solving.
The role of finding the minimum in problem-solving and optimization
Finding the minimum of a function plays a crucial role in problem-solving and optimization. In many real-world scenarios, such as minimizing costs, maximizing profits, or finding the most efficient
solution, the ability to identify the lowest point of a function is invaluable. Whether it's minimizing the cost of production for a company or optimizing the route for a delivery service, knowing
how to find the minimum of a function is key to making informed decisions and improving processes.
Key Takeaways
• Understand the concept of a minimum in a function
• Identify critical points using calculus
• Use the first and second derivative tests
• Consider the domain and range of the function
• Apply the knowledge to real-world problems
Types of Functions and Their Minima
Understanding the different types of functions and their minima is essential in the field of mathematics. Let's delve into the various categories of functions and how their characteristics influence
the location of their minima, along with real-life examples where these functions are used.
Understanding different categories of functions
• Linear Functions: These functions have a constant slope and form a straight line when graphed. The general form of a linear function is y = mx + b, where m is the slope and b is the y-intercept.
• Quadratic Functions: Quadratic functions have a squared term, and their graph forms a parabola. The general form of a quadratic function is y = ax^2 + bx + c, where a, b, and c are constants.
• Polynomial Functions: These functions consist of terms with non-negative integer exponents. They can have various shapes and degrees, such as cubic, quartic, or higher.
• Exponential Functions: Exponential functions have a constant base raised to the power of x. They grow or decay at an increasing rate and are commonly used to model population growth, compound
interest, and radioactive decay.
• Logarithmic Functions: Logarithmic functions are the inverse of exponential functions and are used to solve equations involving exponential growth or decay.
Characteristics of functions that influence the location of their minima
The location of the minima of a function is influenced by various characteristics, including the degree of the function, the leading coefficient, and the presence of critical points. For example, in
quadratic functions, the location of the minimum (or maximum) is determined by the coefficient of the squared term and whether it is positive or negative. In polynomial functions, the degree of the
function and the leading coefficient play a significant role in determining the behavior of the function and the location of its minima.
Real-life examples where different types of functions are used and their minima found
Functions are used in various real-life scenarios to model and analyze data. For instance, linear functions are used in economics to represent cost, revenue, and profit functions. Quadratic functions
are utilized in physics to describe the trajectory of objects under the influence of gravity. Polynomial functions are employed in engineering to model the behavior of materials under stress.
Exponential and logarithmic functions are used in finance to calculate compound interest and in biology to model population growth and decay.
When analyzing these real-life scenarios, finding the minima of these functions becomes crucial in optimizing outcomes. For example, in economics, finding the minimum cost or maximizing profit
involves determining the minima of cost and revenue functions. In physics, finding the minimum point of a quadratic function can help predict the maximum height or range of a projectile. These
real-life examples demonstrate the practical significance of understanding and finding the minima of different types of functions.
Calculus Approach: Using Derivatives to Find Minima
When it comes to finding the minimum of a mathematical function, one of the most powerful tools at our disposal is calculus. By using derivatives, we can determine the critical points of a function
and identify whether they correspond to a minimum.
A Introduction to derivatives as a tool for finding function minima
Derivatives are a fundamental concept in calculus that represent the rate of change of a function at a given point. In the context of finding minima, we can use derivatives to locate the points where
the function is neither increasing nor decreasing, known as critical points. These critical points can then be analyzed to determine whether they correspond to a minimum.
B Step-by-step process for finding the derivative and setting it equal to zero
The process of using derivatives to find the minimum of a function involves several steps:
• Step 1: Find the derivative of the function with respect to the variable of interest. This can be done using the power rule, product rule, quotient rule, or chain rule, depending on the
complexity of the function.
• Step 2: Set the derivative equal to zero and solve for the variable. The values obtained by solving this equation represent the critical points of the function.
• Step 3: Use the second derivative test or the first derivative test to determine whether each critical point corresponds to a minimum, maximum, or neither.
C Common pitfalls in applying derivatives and how to avoid them
While using derivatives to find function minima can be a powerful technique, there are some common pitfalls to be aware of:
• Incorrect derivative: Calculating the derivative incorrectly can lead to inaccurate critical points. It's important to double-check the derivative calculation to ensure its accuracy.
• Missing critical points: Sometimes, critical points may be overlooked, especially in more complex functions. Careful attention to detail and thorough analysis of the derivative equation is
essential to avoid missing critical points.
• Improper use of tests: Applying the second derivative test or the first derivative test incorrectly can lead to misinterpretation of critical points. It's crucial to understand the conditions for
each test and apply them accurately.
By understanding these common pitfalls and taking the necessary precautions, we can effectively use derivatives to find the minimum of a function with confidence and accuracy.
The Role of Critical Points and the Second Derivative Test
Understanding the behavior of mathematical functions is essential in finding the minimum of a function. Critical points and the second derivative test play a crucial role in this process, helping us
identify the minimum points of a function.
Explicating critical points and their relevance to identifying minima
Critical points are the points on a function where the derivative is either zero or undefined. These points are significant as they can indicate potential minima, maxima, or points of inflection. To
identify the minimum of a function, we focus on the critical points where the derivative changes from negative to positive, indicating a change from decreasing to increasing.
By finding the critical points and analyzing their behavior, we can determine whether they correspond to a minimum value of the function. This process is essential in understanding the behavior of
the function and locating its minimum points.
How to perform the second derivative test to confirm a minimum
The second derivative test is a method used to confirm whether a critical point corresponds to a minimum, maximum, or a point of inflection. To apply the second derivative test, we first find the
critical points of the function by setting its first derivative equal to zero and solving for the values of x.
Once the critical points are identified, we then take the second derivative of the function and evaluate it at each critical point. If the second derivative is positive at a critical point, it
indicates that the function is concave up at that point, confirming that the critical point corresponds to a minimum.
Practical scenarios illustrating the use of the second derivative test
To better understand the application of the second derivative test, let's consider a practical scenario. Suppose we have a quadratic function f(x) = x^2 - 4x + 5. By finding the first derivative and
setting it equal to zero, we find that the critical point occurs at x = 2.
Next, we take the second derivative of the function, which is f''(x) = 2. Evaluating the second derivative at x = 2, we find that f''(2) = 2, indicating that the function is concave up at x = 2.
Therefore, the second derivative test confirms that the critical point at x = 2 corresponds to a minimum of the function.
By applying the second derivative test in practical scenarios, we can effectively identify the minimum points of a function and gain a deeper understanding of its behavior.
Optimization Techniques without Calculus
When it comes to finding the minimum of a function without using calculus, there are several methods that can be employed. These methods are particularly useful when dealing with functions for which
derivatives are difficult to compute or are not available.
A Methods for finding minima of functions without derivatives
• Completing the square for quadratics: One method for finding the minimum of a quadratic function without using derivatives is by completing the square. This involves rewriting the quadratic
function in a form that allows the minimum to be easily identified.
• Substitution and transformation: For more complex functions, substitution and transformation techniques can be used to simplify the function and make it easier to identify the minimum value.
B Explanation of graphical analysis and its use in spotting the minima visually
Graphical analysis can be a powerful tool for spotting the minimum of a function without the need for calculus. By plotting the function on a graph, the minimum point can be visually identified as
the lowest point on the graph. This method is particularly useful for functions that are difficult to analyze algebraically.
C Optimization problems in economics and business that use non-calculus methods
In the fields of economics and business, optimization problems often arise that require finding the minimum of a function. Non-calculus methods are frequently used in these scenarios, as the
functions involved may not have easily computable derivatives. By employing techniques such as graphical analysis and algebraic manipulation, economists and business analysts can effectively solve
optimization problems without relying on calculus.
Troubleshooting Common Issues
When dealing with mathematical functions and trying to find the minimum, there are several common issues that may arise. It is important to be aware of these potential problems and have strategies in
place to troubleshoot and rectify them.
Analyzing errors that may occur when calculating minima
One common error that may occur when calculating the minimum of a function is ignoring domain restrictions. It is essential to consider the domain of the function and ensure that the values being
evaluated fall within this domain. Ignoring domain restrictions can lead to incorrect results and must be avoided.
Strategies for verifying the results and ensuring accuracy
To verify the results and ensure accuracy when finding the minimum of a function, it is important to double-check the calculations. This can be done by re-evaluating the critical points and
confirming that they are indeed the minima of the function. Additionally, using graphing tools to visualize the function can provide a helpful visual confirmation of the calculated minimum.
Example problems demonstrating typical mistakes and how to rectify them
Let's consider an example problem where a common mistake occurs when finding the minimum of a function. Suppose we have the function f(x) = x^2 + 1 and we want to find the minimum value. A typical
mistake would be to ignore the fact that the function has no real roots and therefore no minimum value. To rectify this mistake, it is important to recognize that the function has a minimum value of
1, which occurs at the vertex of the parabola.
Conclusion & Best Practices
Understanding how to find the minimum of a function is a fundamental skill in mathematics and has wide-ranging applications in various fields. In this blog post, we have explored the significance of
this understanding and discussed key practices to achieve accurate results. Let's recap the importance of this knowledge and summarize the best practices to follow.
A Recap of the significance of understanding how to find the minimum of a function
The ability to find the minimum of a function is essential in optimization problems, where we aim to minimize costs, maximize profits, or optimize resources. It also plays a crucial role in various
scientific and engineering applications, such as in physics, economics, and computer science. Understanding how to find the minimum of a function allows us to make informed decisions and solve
real-world problems.
Summarization of key practices to achieve accurate results
• Review function type: Before attempting to find the minimum of a function, it is important to understand the type of function involved. Different types of functions, such as linear, quadratic, or
exponential, require different approaches to finding their minimum.
• Verify solutions: After finding a potential minimum of a function, it is crucial to verify the solution to ensure its accuracy. This can be done by checking the first and second derivatives of
the function and analyzing critical points.
• Use of technology tools: Leveraging technology tools, such as graphing calculators or software like MATLAB, can greatly aid in finding the minimum of a function. These tools can provide visual
representations of the function and help in performing complex calculations.
Encouragement to apply these practices and the knowledge learned to real-world situations
Applying the practices discussed in this blog post and the knowledge gained about finding the minimum of a function is essential in tackling real-world problems. Whether it's optimizing production
processes in a manufacturing plant, minimizing costs in a business, or maximizing the efficiency of a system, the ability to find the minimum of a function is a valuable skill that can lead to
impactful solutions. | {"url":"https://dashboardsexcel.com/blogs/blog/understanding-mathematical-functions-find-minimum-function","timestamp":"2024-11-13T01:12:53Z","content_type":"text/html","content_length":"225483","record_id":"<urn:uuid:b1ada65c-892b-490e-b54b-e159601be502>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00608.warc.gz"} |
How Multinomial Logistic Regression Model Works In Machine Learning - Dataaspirant
How Multinomial Logistic Regression Model Works In Machine Learning
How the multinomial logistic regression model works
In the pool of supervised classification algorithms, the logistic regression model is the first most algorithm to play with. This classification algorithm is again categorized into different
categories. These categories are purely based on the number of target classes.
If the logistic regression model used for addressing the binary classification kind of problems it’s known as the binary logistic regression classifier. Whereas the logistic regression model used for
multiclassification kind of problems, it’s called the multinomial logistic regression classifier.
As we discussed each and every block of binary logistic regression classifier in our previous article. Now we use the binary logistic regression knowledge to understand in details about, how the
multinomial logistic regression classifier works.
I recommend first to check out the how the logistic regression classifier works article and the Softmax vs Sigmoid functions article before you read this article.
Learn each and every stage of multinomial logistic regression classifier. Share on X
Before we start the drive, let’s look at the table of contents of this article.
Table of Contents:
• What is Logistic Regression?
• What is Multinomial Logistic Regression?
• Multinomial Logistic Regression Example
• How Multinomial Logistic Regression model works.
□ Logits
□ Softmax Function
☆ Properties
☆ Implementation in Python
□ Cross Entropy
• Parameters Optimization
• Conclusion
What is Logistic Regression?
The logistic regression model is a supervised classification model. Which uses the techniques of the linear regression model in the initial stages to calculate the logits (Score). So technically we
can call the logistic regression model as the linear model. In the later stages uses the estimated logits to train a classification model. The trained classification model performs the
multi-classification task.
If you were not aware of the logits and the basic linear regression model techniques don’t be frightened. In the coming sections. We are going to learn each and every block of multinomial logistic
regression from inputs to the target output representation.
As we discussed earlier the logistic regression models are categorized based on the number of target classes and uses the proper functions like sigmoid or softmax functions to predict the target
In short:
• Sigmoid function: used in the logistic regression model for binary classification.
• Softmax function: used in the logistic regression model for multiclassification.
To learn more about sigmoid and softmax functions checkout difference between softmax and sigmoid functions article.
What is Multinomial Logistic Regression?
Multinomial logistic regression is also a classification algorithm same like the logistic regression for binary classification. Whereas in logistic regression for binary classification the
classification task is to predict the target class which is of binary type. Like Yes/NO, 0/1, Male/Female.
When it comes to multinomial logistic regression. The idea is to use the logistic regression techniques to predict the target class (more than 2 target classes).
The underline technique will be same like the logistic regression for binary classification until calculating the probabilities for each target. Once the probabilities were calculated. We need to
transfer them into one hot encoding and uses the cross entropy methods in the training process for calculating the properly optimized weights.
Multinomial logistic regression works well on big data irrespective of different areas. Surprisingly it is also used in human resource development and more in depth details about how the big data is
used in human resource development can found in this article.
We are going to learn the whole process of multinomial logistic regression from giving inputs to the final one hot encoding in the upcoming sections of this article.
Before that, Let’s quickly check few examples to understand what kind of problems we can solve using the multinomial logistic regression.
Multinomial Logistic Regression Example
Using the multinomial logistic regression. We can address different types of classification problems. Where the trained model is used to predict the target class from more than 2 target classes.
Below are few examples to understand what kind of problems we can solve using the multinomial logistic regression.
Now let’s move on to the key part this article “understand how the multinomial logistic regression model works’
How Multinomial Logistic Regression model works
The above image illustrates the workflow of multinomial logistic regression classifier. To simply the understanding process let’s split the multinomial logistic classifier techniques into different
stages from inputs to the outputs. Then we can discuss each stage of the classifier in detail.
Multinomial Logistic Regression Workflow/ Stages:
• Inputs
• Linear model
• Logits
• Softmax Function
• Cross Entropy
• One-Hot-Encoding
The inputs to the multinomial logistic regression are the features we have in the dataset. Suppose if we are going to predict the Iris flower species type, the features will be the flower sepal
length, width and petal length and width parameters will be our features. These features will treat as the inputs for the multinomial logistic regression.
The keynote to remember here is the features values always numerical. If the features are not numerical, we need to convert them into numerical values using the proper categorical data analysis
Just a simple example: If the feature is color and having different attributes of the color features are RED, BLUE, YELLOW, ORANGE. Then we can assign an integer value to each attribute of the
features like for RED we can assign 1. For BLUE we can assign the value 2 likewise of the other attributes for the color feature. Later we can use the numerically converted values as the inputs for
the classifier.
Linear Model
The linear model equation is the same as the linear equation in the linear regression model. You can see this linear equation in the image. Where the X is the set of inputs, Suppose from the image we
can say X is a matrix. Which contains all the feature( numerical values) X = [x1,x2,x3]. Where W is another matrix includes the same input number of weights W = [w1,w2,w3].
In this example, the linear model output will be the w1*x1, w2*x2, w3*x3
The weights w1, w2, w3, w4 will update in the training phase. We will learn about this in the parameters optimization section of this article.
The Logits also called as scores. These are just the outputs of the linear model. The Logits will change with the changes in the calculated weights.
Softmax Function
The Softmax function is a probabilistic function which calculates the probabilities for the given score. Using the softmax function return the high probability value for the high scores and fewer
probabilities for the remaining scores. This we can observe from the image. For the logits 0.5, 1.5, 0.1 the calculated probabilities using the softmax function are 0.2, 0.7, 0.1
For the Logit 1.5, we are getting the high probability value 0.7 and very less probability value for the remaining Logits 0.5 and 0.1
Keynote: The Logits and the probabilities in the image were just for the purpose of understanding the multinomial logistic regression model. The values are not the real values computed using the
softmax function.
Softmax Function Properties
Below are the few properties of softmax function.
• The calculated probabilities will be in the range of 0 to 1.
• The sum of all the probabilities is equals to 1.
Implementing Softmax Function In Python
Now let’s implement the softmax function in Python
# Required Python Package
import numpy as np
def softmax(inputs):
Calculate the softmax for the give inputs (array)
:param inputs:
return np.exp(inputs) / float(sum(np.exp(inputs)))
softmax_inputs = [2, 3, 5, 6]
print "Softmax Function Output :: {}".format(softmax(softmax_inputs))
Script Output
Softmax Function Output :: [ 0.01275478 0.03467109 0.25618664 0.69638749]
If we observe the function output for the input value 6 we are getting the high probabilities. This is what we can expect from the softmax function. Later in classification task, we can use the high
probability value for predicting the target class for the given input features.
Cross Entropy
The cross entropy is the last stage of multinomial logistic regression. Uses the cross-entropy function to find the similarity distance between the probabilities calculated from the softmax function
and the target one-hot-encoding matrix.
Before we learn more about Cross Entropy, let’s understand what it is mean by One-Hot-Encoding matrix.
One-Hot Encoding is a method to represent the target values or categorical attributes into a binary representation. From this article main image, where the input is the dog image, the target having 3
possible outcomes like bird, dog, cat. Where you can find the one-hot-encoding matrix like [0, 1, 0].
The one-hot-encoding matrix is so simple to create. For every input features (x1, x2, x3) the one-hot-encoding matrix is with the values of 0 and the 1 for the target class. The total number of
values in the one-hot-encoding matrix and the unique target classes are the same.
Suppose if we have 3 input features like x1, x2, and x3 and one target variable (With 3 target classes). Then the one-hot-encoding matrix will have 3 values. Out of 3 values, one value will be 1 and
all other will be 0s.
You will know where to place the 1 and where to place the 0 value from the training dataset. Let’s take one observation from the training dataset which contains values for x1, x2, x3 and what will be
the target class for that observation. The one-hot-encoding matrix will be having 1 for the target class for that observation and 0s for other.
Cross Entropy
The Cross-entropy is a distance calculation function which takes the calculated probabilities from softmax function and the created one-hot-encoding matrix to calculate the distance. For the right
target class, the distance value will be less, and the distance values will be larger for the wrong target class.
With this, we discussed each stage of the multinomial logistic regression. All the stages/ workflow will happen for each observation in the training set.
Observation: Single observation is just the single row values from the traging set. Which contains the features and the correspoinding target class.
For each observation in the training set will pass through the all the stages of multinomial logistic regression and the proper weights W (w1, w2, w3) values will be computed. How the weights
calculated and update the weights is know as the Parameters Optimization.
Parameters Optimization
The expected output after training the multinomial logistic regression classifier is the calculated weights. Later the calculated weights will be used for the prediction task.
This Parameters optimization is an iteration process where the calculated weights for each observation used to calculate the cost function which is also known as the Loss function.
The Iteration process ends when the loss function value is less or significantly negligible. Now let’s learn about this loss function to sign off from this lengthy article 🙂
Loss function
The input parameters for the loss function is the calculated weights and all the training observations. The function calculates the distance between the predicted class using the calculated weights
for all the features in the training observation and the actual target class.
If the loss function value is fewer means with the estimated weights, we are confident to predict the target classes for the new observations (From test set). In the case of high loss function
value, the process of calculating the weights will start again with derivated weights of the previously calculated weights.
The process will continue until the loss function value is less.
This is the whole process of multinomial logistic regression. If you are thinking, it will be hard to implement the loss function and coding the entire workflow. Don’t frighten. We were so lucky to
have the machine learning libraries like scikit-learn. Which performs all this workflow for us and returns the calculated weights.
In the next article, we are going to implement the logistic regression model using the scikit-learn library to perform the multiclassification task.
This article gives the clear explanation on the each stage of multinomial logistic regression. Below are the discussed workflow or stages.
• Inputs
• Linear model
• Logits
• Softmax Function
• Cross Entropy
• One-Hot-Encoding
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I hope you like this post. If you have any questions, then feel free to comment below. If you want me to write on one particular topic, then do tell it to me in the comments below.
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Relative Cohomology - Quantum Calculus
Relative Cohomology
Having a cohomology for open sets allows to define relative cohomology H(G,K) as H(U). This automatically satisfies excision. It appears that we have a very swift approach to a cohomology satisfying
the Eilenberg-Steenrod axioms. There is now a preprint. (now on ArXiv).
Update Mai 25, 2023: while finishing the paper, I could not yet make experiments with knots as the complexes were too big. One strategy was to take a parametrization of a knot K and rewrite this in a
product graph $G=X \times X \times X$ where X is is a 1-sphere and K is a 1-sphere. I was interested in the cohomology of the complement $G \setminus K$. In the case of an unknot, the matrices are
small enough to compute and we see b(U)= (0,0,1,0). Note that b(K)=(1,1,0,0) and b(G) = (1,0,0,0) so that b(I), the interface cohomology is b(I)=(0,1,1,0). I was wondering whether that changes for a
knot. It appears not. Here is a situation which I managed to compute. It is implemented in a torus of side lengths 5 meaning that we have 125 vertices. The f-vector is (125, 300,240,64) and Euler
characteristic 125-300+240-64=1 (It was easier to not compactify this ball). Then I drew a knot by hand and planted it into G, then computed the cohomology of the complement which has as a 1-point
compactification a space of cohomology (1,0,1,0) the cohomology of a sphere. | {"url":"https://www.quantumcalculus.org/relative-cohomology/","timestamp":"2024-11-02T08:03:19Z","content_type":"text/html","content_length":"76932","record_id":"<urn:uuid:e41db160-426c-4fcd-a805-cef6f666dbc7>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00019.warc.gz"} |
Basal Area Calculator - calculator
Basal Area Calculator
The Basal Area Calculator is a vital tool for foresters and land managers to estimate the basal area of trees in a specified area. This measurement is crucial for assessing forest density,
understanding growth rates, and managing resources sustainably. By inputting the diameter at breast height (DBH) and the number of trees per unit area, users can determine the total basal area, which
informs decisions about thinning, harvesting, and land management practices. This tool promotes better forest management and helps in conservation efforts.
The formula for calculating the basal area (BA) is: BA = F * DBH² * N. Here, BA represents the basal area in square feet, F is the forester's constant (0.005454), DBH is the diameter at breast height
in inches, and N is the number of trees per acre. This formula allows for the estimation of the area occupied by tree cross-sections at breast height, essential for evaluating forest health and
planning interventions.
What is Basal Area?
Basal area is a forestry measurement representing the cross-sectional area of tree trunks at breast height. It helps assess forest density and health, guiding sustainable management practices.
Why use a Basal Area Calculator?
A Basal Area Calculator simplifies the process of estimating the area occupied by trees in a specific area, allowing foresters to make informed decisions about forest management and conservation.
What units can I use for DBH?
You can use inches, centimeters, or meters for the diameter at breast height (DBH). However, the calculation primarily uses inches for forestry applications to ensure accuracy in basal area
How is basal area related to tree health?
Basal area provides insights into tree density and competition for resources. Understanding basal area helps identify healthy stands versus overcrowded conditions, guiding thinning and management
What is the forester's constant?
The forester's constant (0.005454) is used in the basal area formula to convert the diameter measurement into square feet. It is crucial for accurately calculating the basal area from the diameter at
breast height.
How do I convert DBH from centimeters or meters to inches?
To convert centimeters to inches, divide the centimeter value by 2.54. To convert meters to inches, multiply the meter value by 39.37. Accurate unit conversion ensures correct calculations in basal
area estimation.
What does the number of trees represent?
The number of trees refers to the count of trees in a specified area, typically expressed per acre or hectare. It is essential for calculating the total basal area and assessing forest density.
Can I use this calculator for other tree measurements?
This calculator is specifically designed for basal area calculations using diameter at breast height and number of trees. Other forestry measurements may require different formulas and tools.
Is the basal area important for conservation efforts?
Yes, understanding basal area is crucial for effective forest conservation. It helps identify tree health, density, and growth patterns, guiding sustainable practices and interventions to protect
forest ecosystems.
Can I use this calculator for urban forestry?
Absolutely! This calculator can be beneficial for urban forestry management by assessing tree density and health in urban areas, aiding in planning and conservation strategies for city trees. | {"url":"https://calculatordna.com/basal-area-calculator/","timestamp":"2024-11-05T07:37:06Z","content_type":"text/html","content_length":"84876","record_id":"<urn:uuid:f1bcdab3-520b-4d5d-82a4-f82430842287>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00438.warc.gz"} |
Hidden Candidates
When a cells
contain the value or values being considered, and there are additional candidates in the cell making these values difficult to see, or "hiding" them, the candidates are referred to as
Hidden Candidates
Hidden Single
is a candidate that only exists in a single cell in a
(row, column or box). A Hidden Single is usually harder to notice than a
Naked Single
because there are extra candidates in the cell hiding the fact that the Hidden Single candidate can only exist in a single cell in the unit. One way to identify Hidden Singles is by use
Cross Hatching
A Hidden Single becomes the solution for a cell.A
Hidden Pair
is two candidates that can only be in the same two cells in a unit (row, column or box). There will be additional candidates in one or both of the two cells being considered.
When there are two candidates that can only be in the same two cells in a unit then these two candidate will eventually fill the two cells.
This doesn't determine which candidate goes into each cell but it does mean any other candidates in the two cells can be removed.
In the Hidden Pair example in Figure 1 the blue cells are a Hidden Pair with the candidates 2 and 4 in row 1.
The values 2 and 4 have to be placed in the blue cells in row 1 because neither of these values can be placed in any other cells in row 1. All other candidates can be removed from these two cells.
Also, it may be noticed that the non highlighted cells in row 1 form a Naked Set on the values 1, 3, 5, 6 and 9. It is often the case that the cells left over in a unit when a Hidden Set is found
will contain a Naked Set.A
Hidden Triplet
is three candidates that can only be placed in three cells of a unit. Each candidate may not be able to be placed in all three cells but in total, some combination of the three candidates will be
able to be placed in a total of three cells.
Just like the Hidden Pair, this pattern does not determine which of the three candidates goes into each of the three cells but it does mean that any other candidates in the three cells of the Hidden
Triplet can be removed.
In the Hidden Triplet example in Figure 2 the three blue cells in row 3 are the only cells that can contain the values 3, 6 and 9.
The values 3, 6 and 9 have to end up in the three blue cells because they cannot be placed in any other cells in row 3. All other candidates in these cells can be removed.
The example puzzle has multiple Hidden Set instance that can be seen by starting it in the Helper. The Naked Sets Strategy will have to be disabled to see the hidden sets in this example. | {"url":"https://sudoku.ironmonger.com/howto/hiddenCandidates/docs.tpl","timestamp":"2024-11-10T08:07:17Z","content_type":"text/html","content_length":"17693","record_id":"<urn:uuid:b070c733-01d6-4eb1-be51-14247e6f5676>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00213.warc.gz"} |
Math Gadfly Calls Math Faddists’ Bluff
Math Gadfly Calls Math Faddists’ Bluff
– DEBRA J. SAUNDERS
Friday, August 7, 1998
WAYNE BISHOP, a professor of mathematics at Cal State L.A. is a math gadfly in his spare time. Bishop likes to call trendy new-new math boosters’ bluff by checking out their schools’ scores and
seeing if their fads hurt or help students.
Bishop’s findings are not pretty, friend.
Start with Roscoe Elementary School in Los Angeles. MathLand, the ultra- trendy new-new math program, touts Roscoe as a MathLand success story on its Web page. In 1996, then-Roscoe principal Ruth
Bunyan wrote a letter for MathLand boasting, “We believe that our use of MathLand materials and instructional strategies have made a significant difference during the past year.”
(For the unitiated, new-new math is math that eschews exercises that emphasize “predetermined numerical results.”)
So how did MathLand’s star school — where MathLand folks coached the teachers to increase performance — fare in California’s new Star test? In the bottom quartile. The average Roscoe student score
was 21 — at the bottom 21st percentile — for second graders, 22 for third graders, 20 for fourth graders and 18 for fifth graders. Only 10 percent of fifth graders scored above the national average.
Students at the pet middle school of the former head of the National Council of Teachers of Mathematics (NCTM), Jack Price — a faddist of the first water — have suffered the same fate.
In 1995, Price told his peers, “We need to let everyone know that successes can be found in every part of the country.” As an example, he cited a program at which he worked one day a week with a
dedicated math department faculty that had completely reworked its curriculum “to be in line with the (NCTM) standards.”
“He was depressed about the fact that some of us curmudgeons weren’t catching on, ” Bishop explained.
Guess what? When Bishop checked out test data for Santa Ana Unified’s Spurgeon Intermediate School, he found failure and mediocrity, then and now.
A build-it-and-they-should come success? Sprugeon’s Star scores don’t show it. Price’s pet school ranked in the bottom quartile. The average percentile for sixth graders was 23, 24 for seventh
graders and 22 for eighth graders. Only 12 percent of the schools’ eighth graders scored above the national average.
Apologists might argue that the student body make-up — the school is overwhelmingly minority, a majority of students have limited English skills — mitigates this dismal showing. I can’t agree. I can
find nothing understandable about minority kids failing. I’ll add that if Price wants to tell America how to teach math, he ought to be able to demonstrate that students enrolled in his model program
at least can pass a math test.
(Besides, Price has boasted about the “great deal of research” of which he is aware as to how females and minorities “do not learn the same way” as white males. So a large minority pool should be a
piece of cake for him.)
Price, now at Cal Poly Pomona, had the misfortune to pick up his phone yesterday when I rang to ask why most of his poor charges flunked the Star test.
“Is that surprising?” Price asked.
If I didn’t know what I know about new-new math, I would be surprised.
“It wouldn’t be surprising if you knew anything about how instruction and assessment are tied together,” He answered. “You don’t teach people apples and then give them a test for oranges.”
But in your 1995 address, you said the NCTM standards included basic skills.
“What do you want me to say?” Price asked, before he said he didn’t want to chat anymore.
No doubt he prefers an audience that is more accepting about failing innocent kids. | {"url":"http://mathwise.net/?p=1256","timestamp":"2024-11-02T07:20:54Z","content_type":"text/html","content_length":"31335","record_id":"<urn:uuid:c080d6b2-ed31-4d50-9729-f83f4f739204>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00387.warc.gz"} |
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3D nanomagnetism: there is plenty of room at the top - Department of Physics
3D nanomagnetism: there is plenty of room at the top
Finding a hardware platform compatible with artificial intelligence and neuromorphic architectures is a key challenge of future computing. Three-dimensional (3D) systems are well suited to address
this need, as harnessing the complex nature and degrees of freedom associated with their structure can provide enriched functionalities. 3D nanomagnetic devices are particularly promising due to
their energy efficiency, non-volatility, and scalability, and can be utilized to implement novel types of magnetic memory and logic devices, as well as neuromorphic computing. In a study just
published in Nano Letters, Prof. Kai Liu’s group and collaborators, led by Dr. Dhritiman Bhattacharya, have demonstrated the potential of 3D interconnected nanowire networks to be used as
neuromorphic computing elements.
The quasi-ordered interconnected networks were constructed from magnetic nanowires made of cobalt through electrochemical deposition. Previously, Liu’s group demonstrated interesting magnetic
properties of these networks where magnetic domain walls get pinned at the intersections between nanowires. In this study, they have employed electrical transport measurements of magnetoresistance
(MR), together with magnetometry, magnetic imaging and micromagnetic simulations to further understand magnetic characteristics and switching pathways in the complex nanowire networks.
They have observed discrete propagation of magnetic states in interconnected Co nanowire networks driven by magnetic field and current, manifested in distinct MR features. Micromagnetic simulations
carried out by Zhijie (Hugh) Chen in Liu’s group, together with Prof. Gen Yin, show that these MR features are due to domain wall (DW) pinning at the nanowire intersections. This is further confirmed
by off-axis electron holography imaging performed by Chen Liu and Prof. Xixiang Zhang at King Abdullah University of Science & Technology. First-order reversal curve measurements of MR (MR-FORC)
illustrate the strong dependence of magnetization reversal on the initial magnetic state and prior magnetic history of these networks.
These properties are desirable to fashion non-Boolean computing devices such as spintronic memristors and synaptic devices where different resistance states or synaptic weights can be programmed by
controllably switching a certain subsection of the networks. These networks may also find potential applications in reservoir and probabilistic computing, as well as physically unclonable functions.
Thus, this study highlights the potential of interconnected nanowire networks to be used as 3D information storage and unconventional computing devices utilizing their capability to stabilize
different magnetic configurations, along with controlled and discrete propagation through the networks. Further investigation of this promising system could ultimately lead to 3D magnetic
nanostructure-based energy efficient devices capable of realizing a multitude of functionalities.
Other researchers involved include Christopher Jensen in Liu’s group, Dr. Edward Burks at the University of California, Davis and Prof. Dustin Gilbert at the University of Tennessee – Knoxville.
This project was supported in part by the National Science Foundation as well as the Spintronic Materials for Advanced InfoRmation Technologies (SMART) Center sponsored by the Semiconductor Research
Corporation and the National Institute of Standards and Technology.
(a) Top view scanning electron microscopy (SEM) image of the network. The dashed red line shows the ion tracking plane. The inset shows FFT of the SEM image indicating the presence of nanowires at
three different angles. (b) A family of MR FORCs of the network showing discrete MR jumps and also their dependence on the magnetic field sequence.
An animation of the MR-FORC showing the hysteretic and stochastic characteristics of the switching in Co nanowire networks. | {"url":"https://physics.georgetown.edu/3d-nanomagnetism-there-is-plenty-of-room-at-the-top/","timestamp":"2024-11-08T18:07:17Z","content_type":"text/html","content_length":"111051","record_id":"<urn:uuid:f518a5f9-fdb4-49ec-a0d1-36a38e2a4132>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00170.warc.gz"} |
What's your personality type?
What's your personality type?
@finneass i know, i just don’t know which one suits me well
i don’t know which one sounds the closest to “chill and harmless, if left unprovoked”
@A38 if you take the test it’ll tell you
@finneass ahh okay didn’t see that one, thanks
@finneass I’m only 50% through and already bored
Creek I’m 64% extroverted, 55% intuitive, 65% thinking (logical I guess), 74% prospecting, and 57% turbulent.
Okay I actually didn’t expect this one at all
@finneass VERY short fuse
nice to friends unless pissed
and uhr
stoopid and hyper
@finneass Last time I checked, im an ENFJ-T | {"url":"https://mpp.community/forum/topic/45175/what-s-your-personality-type","timestamp":"2024-11-05T16:39:53Z","content_type":"text/html","content_length":"159971","record_id":"<urn:uuid:7f9e404c-2601-453e-9683-1d18d0619fbe>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00438.warc.gz"} |
Calculating Turbine Power from Water Jet Deflection
To calculate the power of the turbine in this scenario, we need to consider several factors such as the specific mass of the water, the horizontal thrust developed on the deflector blade, and the
efficiency of the turbine. Let's break down the steps needed to determine the turbine power:
Step 1: Determine the Force Acting on the Deflector Blade
First, we need to calculate the force acting on the deflector blade. The horizontal thrust of 1000 N represents the force exerted by the water jet on the blade. This force is essential in
understanding the impact of the water jet on the turbine's performance.
Step 2: Calculate the Power Generated by the Turbine
Now that we have the force acting on the deflector blade, we can proceed to determine the power generated by the turbine. The power of the turbine can be calculated using the formula:
Power (P) = Force (F) x Distance (D) x Efficiency (η)
Where: - Power (P) is the power generated by the turbine. - Force (F) is the horizontal thrust acting on the deflector blade (1000 N). - Distance (D) represents the distance over which the force is
applied. - Efficiency (η) is the efficiency of the turbine (70%).
Step 3: Substitute Values and Calculate the Turbine Power
By substituting the given values into the formula, we can calculate the power generated by the turbine. Remember to convert the units to ensure consistency and accuracy in the calculation process.
Once the calculations are complete, we will have the power output of the turbine based on the water jet deflection scenario.
By following these steps and considering the specific mass, horizontal thrust, and turbine efficiency, we can accurately determine the power of the turbine operating under these conditions. | {"url":"https://tutdenver.com/world_languages/calculating-turbine-power-from-water-jet-deflection.html","timestamp":"2024-11-10T15:00:33Z","content_type":"text/html","content_length":"23700","record_id":"<urn:uuid:2a70aa86-b25c-44f6-9cc4-706d1b0735ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00035.warc.gz"} |
In statistics, completeness is a property of a statistic computed on a sample dataset in relation to a parametric model of the dataset. It is opposed to the concept of an ancillary statistic. While
an ancillary statistic contains no information about the model parameters, a complete statistic contains only information about the parameters, and no ancillary information. It is closely related to
the concept of a sufficient statistic which contains all of the information that the dataset provides about the parameters.^[1]
Consider a random variable X whose probability distribution belongs to a parametric model P[θ] parametrized by θ.
Say T is a statistic; that is, the composition of a measurable function with a random sample X[1],...,X[n].
The statistic T is said to be complete for the distribution of X if, for every measurable function g,^[2]
${\displaystyle {\text{if }}\operatorname {E} _{\theta }(g(T))=0{\text{ for all }}\theta {\text{ then }}\mathbf {P} _{\theta }(g(T)=0)=1{\text{ for all }}\theta .}$
The statistic T is said to be boundedly complete for the distribution of X if this implication holds for every measurable function g that is also bounded.
Example: Bernoulli model
The Bernoulli model admits a complete statistic.^[3] Let X be a random sample of size n such that each X[i] has the same Bernoulli distribution with parameter p. Let T be the number of 1s observed in
the sample, i.e. ${\displaystyle \textstyle T=\sum _{i=1}^{n}X_{i}}$ . T is a statistic of X which has a binomial distribution with parameters (n,p). If the parameter space for p is (0,1), then T is
a complete statistic. To see this, note that
${\displaystyle \operatorname {E} _{p}(g(T))=\sum _{t=0}^{n}{g(t){n \choose t}p^{t}(1-p)^{n-t}}=(1-p)^{n}\sum _{t=0}^{n}{g(t){n \choose t}\left({\frac {p}{1-p}}\right)^{t}}.}$
Observe also that neither p nor 1 − p can be 0. Hence ${\displaystyle E_{p}(g(T))=0}$ if and only if:
${\displaystyle \sum _{t=0}^{n}g(t){n \choose t}\left({\frac {p}{1-p}}\right)^{t}=0.}$
On denoting p/(1 − p) by r, one gets:
${\displaystyle \sum _{t=0}^{n}g(t){n \choose t}r^{t}=0.}$
First, observe that the range of r is the positive reals. Also, E(g(T)) is a polynomial in r and, therefore, can only be identical to 0 if all coefficients are 0, that is, g(t) = 0 for all t.
It is important to notice that the result that all coefficients must be 0 was obtained because of the range of r. Had the parameter space been finite and with a number of elements less than or equal
to n, it might be possible to solve the linear equations in g(t) obtained by substituting the values of r and get solutions different from 0. For example, if n = 1 and the parameter space is {0.5}, a
single observation and a single parameter value, T is not complete. Observe that, with the definition:
${\displaystyle g(t)=2(t-0.5),\,}$
then, E(g(T)) = 0 although g(t) is not 0 for t = 0 nor for t = 1.
Example: Sum of normals
This example will show that, in a sample X[1], X[2] of size 2 from a normal distribution with known variance, the statistic X[1] + X[2] is complete and sufficient. Suppose (X[1], X[2]) are
independent, identically distributed random variables, normally distributed with expectation θ and variance 1. The sum
${\displaystyle s((X_{1},X_{2}))=X_{1}+X_{2}\,\!}$
is a complete statistic for θ.
To show this, it is sufficient to demonstrate that there is no non-zero function ${\displaystyle g}$ such that the expectation of
${\displaystyle g(s(X_{1},X_{2}))=g(X_{1}+X_{2})\,\!}$
remains zero regardless of the value of θ.
That fact may be seen as follows. The probability distribution of X[1] + X[2] is normal with expectation 2θ and variance 2. Its probability density function in ${\displaystyle x}$ is therefore
proportional to
${\displaystyle \exp \left(-(x-2\theta )^{2}/4\right).}$
The expectation of g above would therefore be a constant times
${\displaystyle \int _{-\infty }^{\infty }g(x)\exp \left(-(x-2\theta )^{2}/4\right)\,dx.}$
A bit of algebra reduces this to
${\displaystyle k(\theta )\int _{-\infty }^{\infty }h(x)e^{x\theta }\,dx\,\!}$
where k(θ) is nowhere zero and
${\displaystyle h(x)=g(x)e^{-x^{2}/4}.\,\!}$
As a function of θ this is a two-sided Laplace transform of h(X), and cannot be identically zero unless h(x) is zero almost everywhere.^[4] The exponential is not zero, so this can only happen if g(x
) is zero almost everywhere.
By contrast, the statistic ${\textstyle (X_{1},X_{2})}$ is sufficient but not complete. It admits a non-zero unbiased estimator of zero, namely ${\textstyle X_{1}-X_{2}.}$
Example: Location of a uniform distribution
Suppose ${\textstyle X\sim \operatorname {Uniform} (\theta -1,\theta +1).}$ Then ${\textstyle \operatorname {E} (\sin(\pi X))=0}$ regardless of the value of ${\textstyle \theta .}$ Thus ${\textstyle
\sin(\pi X)}$ is not complete.
Relation to sufficient statistics
For some parametric families, a complete sufficient statistic does not exist (for example, see Galili and Meilijson 2016 ^[5]).
For example, if you take a sample sized n > 2 from a N(θ,θ^2) distribution, then ${\displaystyle \left(\sum _{i=1}^{n}X_{i},\sum _{i=1}^{n}X_{i}^{2}\right)}$ is a minimal sufficient statistic and is
a function of any other minimal sufficient statistic, but ${\displaystyle 2\left(\sum _{i=1}^{n}X_{i}\right)^{2}-(n+1)\sum _{i=1}^{n}X_{i}^{2}}$ has an expectation of 0 for all θ, so there cannot be
a complete statistic.
If there is a minimal sufficient statistic then any complete sufficient statistic is also minimal sufficient. But there are pathological cases where a minimal sufficient statistic does not exist even
if a complete statistic does.
Importance of completeness
The notion of completeness has many applications in statistics, particularly in the following two theorems of mathematical statistics.
Lehmann–Scheffé theorem
Completeness occurs in the Lehmann–Scheffé theorem,^[6] which states that if a statistic that is unbiased, complete and sufficient for some parameter θ, then it is the best mean-unbiased estimator
for θ. In other words, this statistic has a smaller expected loss for any convex loss function; in many practical applications with the squared loss-function, it has a smaller mean squared error
among any estimators with the same expected value.
Examples exists that when the minimal sufficient statistic is not complete then several alternative statistics exist for unbiased estimation of θ, while some of them have lower variance than others.^
See also minimum-variance unbiased estimator.
Basu's theorem
Bounded completeness occurs in Basu's theorem,^[8] which states that a statistic that is both boundedly complete and sufficient is independent of any ancillary statistic.
Bahadur's theorem
Bounded completeness also occurs in Bahadur's theorem. In the case where there exists at least one minimal sufficient statistic, a statistic which is sufficient and boundedly complete, is necessarily
minimal sufficient. Another form of Bahadur's theorem states that any sufficient and boundedly complete statistic over a finite-dimensional coordinate space is also minimal sufficient.^[9] | {"url":"https://www.knowpia.com/knowpedia/Completeness_(statistics)","timestamp":"2024-11-08T09:34:46Z","content_type":"text/html","content_length":"143483","record_id":"<urn:uuid:784064ee-f074-4286-98e5-dcb24b9b4acb>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00188.warc.gz"} |
Divide a triangle into fourths with greatest angle?
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Divide a triangle into fourths with greatest angle?
Hello everyone, I am doing the following experiment with triangles, the idea is the following, we take a triangle in space, for example the triangle with points A (0,0,0), B (0,60,0), C (80,0,0) then
we look for the largest angle of said triangle, after analyze the angles of the triangle we see that the larger angle is what is at the point {0,0,0,}, subsequently we construct a straight line to
the midpoint of the opposite side to the point {0,0,0}, finally we estimate the midpoints of the sides adjacent to the point {0,0,0} and the join by a straight line to the midpoint of the opposite
side that we calculated above.
The way I have done it is the following, but the problem is that I do not understand why I have empty spaces between divided triangles as seen in the following image
I hope someone can help me find the fault,I do not think there should be such holes. Please see the notebook add.
Greetings to all.
1 Reply
Obviously the new triangles are not complanar with the plane of the very first triangle. Therefore you look through. If you could convince yourself to ease things as much as possible you would
• first transform your very first triangle into a co-ordinate plane
• second do all calculations along the lines of Divide a triangle into four small triangles in two dimensions only
• third transform the result back
thus avoiding any problems with complanarity. | {"url":"https://community.wolfram.com/groups/-/m/t/1010648?p_p_auth=5tR8ecxI","timestamp":"2024-11-05T00:19:20Z","content_type":"text/html","content_length":"95897","record_id":"<urn:uuid:017cd2cc-5e53-4d99-8d02-606daf947a05>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00811.warc.gz"} |
Computer Vision: Plane Automatic Landing - Softarex
Computer Vision (CV) is a field of Computer Science that works on enabling computers to see, identify, process, and provide an accurate output. Just in the same way that human vision does. But,
unlike humans, a properly refined Computer Vision system reduces the possibility of an output error to almost zero. And one of the possible uses of Computer Vision is a plane automatic landing system
with algorithms for tracking normalization. Tracking normalization is an approach that allows parameter calculation of a picture’s geometric transformations in real-time.
Landing a modern plane is quite a complicated and demanding process, which requires a special approach and the special attention of both the crew and ground services. To simplify it, an automatic
landing system with Computer Vision can be of great help.
Setting the task of a plane’s automatic control during landing
Let’s consider the automatic control process of a plane during landing from a 400-500 m altitude. A plane moves along a rigid or flexible trajectory. In the first case, the trajectory of movement,
the glide path, is set by ground devices based on radio equipment, and in the second — the plane’s position information comparatively to the landing point comes from onboard sensors.
The automatic radio approach system consists of onboard and ground equipment. The ground equipment consists of the glide path, course beacon, near, middle, and far marker beacons. Marker beacon
transmitters are located in the direction of a runway’s centerline at distances of 60 m, 1600 m, and 7200 m, respectively. A plane begins to descend at an altitude of 300-400 m, i.e. the guidance
line system begins to work after the plane passes the far marker. The advantage of this system is the autopilot connection and the disadvantage is the radio equipment complexity. They don’t perform
well with narrow radio signals reflected from the ground when a plane is descending from low altitudes.
Aircraft tracking automatically controls a plane during landing by real-time video frames. Next, landing is controlled by the mathematical model of a plane’s movement. In this case, CV begins to
control a plane after it passes the far marker — 7200 m.
Controlling a plane’s automatic landing with Computer Vision
The system’s first stage is the plane’s capture at the time of its visibility zone entry. A dispatcher indicates it on a display screen, which displays the camera’s view field or automatically by
using recognition and segmentation algorithms. The bounding box is placed on a plane’s image and the tracking process begins. The CV remembers the field of view before a plane enters the visibility
zone, in order to exclude the background from the image during the tracking process. This increases the noise immunity of the system.
Next, the plane lands on a glide path. The glide path is set by a radio beam emerging from a glide-beacon located near the runway at a 2-3 ° angle. When a plane’s landing is controlled using CV, the
video camera of the system must be set at the angle of inclination of the observation line Θ at 2-3 ° so the glide path line will be implemented and coincide with the line of sight. The CV camera is
installed in the same place as the near marker beacon. A plane is at an altitude of 16-25 m. when it passes this beacon.
During a landing, an autopilot controls a plane so that the velocity vector of its center of mass is directed along the glide path. When a plane doesn’t move along the glide path its image, and
accordingly the center of the bounding box, will be shifted relative to the center of CV view.
It is necessary to control the approach of both the plane’s image and the bounding box to the center of a field of view. The match of the center of a bounding box and the center of a field of view
will be the stopping criterion of an onboard system control work. But CV will monitor a plane until it successfully passes the near marker — the place with installed CV.
Figure 1 shows the time point t[0] — the capture of a plane by an automatic landing system. The diagram shows that the center of the tracking frame is offset from the origin of coordinates in the X
[0] Y[0] plane. Points В and С are the intersections of the coordinate system X[s ]and Y[s], lying in the tracking frame, with the axes X[0] and Y[0]. We connect them with the center of the image in
a CV’s camera, through which the line of sight passes and we get the β angle — the course deviation from the line of sight in the vertical plane and the γ angle — the course deviation in the
horizontal plane. The pixel distance from the center of the tracking frame to the Х[0] axis is denoted by η, and to the Y[0] axis — ζ. The deviations of the tracking frame from the center of the
field of view can be determined since the coordinates of the center of the tracking frame are always known.
Figure 1. Location of Computer vision, a runway, and an image surface during a plane’s guidance system operation.
Mathematical model definition of a plane’s steering elements control
To solve the problem of guiding a plane to a glide path on the correct landing course, it is necessary to build a control law that will automatically eliminate the existing course deviations and
accordingly reduce the deviation of a bounding box to zero. Thus a CV will control the process of approaching the aircraft to the glide path.
The deterministic mathematical model of a plane’s movement is based on the following parameters:
• Plane parameters — mass, speed, jet thrust, wing area, fuel consumption
• Environmental parameters — the speed of sound at a given height, air density, known as a function of height H at a constant temperature
• Motion parameters of a plane — angle of attack, a moment of inertia of a plane relative to the z-axis, angle of inclination of a trajectory, angle of deflection of an elevator, pitch angle
In the development of plane control equations, this model is fundamental at all flight stages.
For glide path landings, a simplified disturbed motion equation is used with small deviations of the direction of the velocity vector of the plane’s center of mass from the direction of the glide
path. It is formed by linearizing the general model of the plane’s motion. The main parameters of the simplified equation are η — the magnitude of the mismatch of the current and the required plane’s
altitude, and Δδ — the elevator angle deflection.
The output coordinate η should be measured in flight, as well as the value of the control function Δδ — the elevator deviation from the required angle. To create a control system, it is necessary to
close the back control object, where η becomes the input function, and the output function. The function connected with the input coordinate is called the control law.
Using the control systems principles synthesis in conditions of uncertainty, we build an adaptive control system, which is reduced to the equation:
The equation №1
, where k — is a plane’s characteristics coefficient, whence inequality of the denominator follows. The coefficients К[1] and К[0] are formed by solving the system of Riccati equations.
Thus, the control law for the plane’s elevator is determined, which associates the value of the elevator deflection angle δ with the deviation of the plane’s flight altitude η from the required one.
Using this control law, it is possible to determine the value of the deflection elevator angle at any time.
To determine the η initial value mismatch — η [k] pixel value is used. It is defined in the field of CV’s view, as the distance from the center of the tracking frame to the Х-axis. These values are
interconnected through a scale factor. The magnitude of the scale factor is set based on CV’s lenses the parameters. The coefficient for substitution in the equation is reduced, the first and the
second derivative η [k] values are calculated as the projections of speed and acceleration on the Y[0 ]-axis:
Rewriting the equation №1:
Where Δη[k0] — is the deviation change of the bounding box center at the previous step of the Х[0] -axis at the t[0] moment of time, Δη[k1] — the center deviation change of the tracking frame from
the Х[0 ]-axis at the current step at t[1] time, η[k] — the center deviation of the tracking frame from the Х[0 ]-axis.
The coordinate system Х[0]Y[0] is located in the center of the CV’s field of view. The η[k] value, in this case, is the distance between the center of the tracking frame and the X-axis.
The equation shows the connection between the parameters obtained by normalizing the bounding box and the deflection angle of the plane’s elevator. It does not contain the distance to a plane or the
height of its flight. This eliminates the need for additional sensors in the automatic landing system and significantly reduces calculation errors.
The Δδ value is measured by a plane’s onboard sensor and transmitted to the autopilot. The automated landing system transmits the bounding box deviation to it from the center of the field of view.
The autopilot calculates the required elevation angle change of the elevator at the current step. Similarly, we can obtain the control equation for the rudder angle.
The equation №1 can be converted to a discrete form based on finite difference operators, which will allow calculations on a computer.
In this case, the magnitude of the η[k] mismatch is also measured via CV as before. The derivatives are replaced by finite difference operators.
The described approach can be used as additional assistance for pilots during the landing process and provide information from outside observation. At the same time, the approach is quite theoretical
and requires more research, modeling, and computer simulation with a real landing process video.
Want to know more about how AI and Computer Vision digitize the aviation industry? Check out our article about the future possibilities this technology might bring. We are always eager to share our
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Market concentration
In economics, market concentration is a function of the number of firms and their respective shares of the total production (alternatively, total capacity or total reserves) in a market.
Alternative terms are Industry concentration and Seller concentration.^[1]
Market concentration is related to the concept of industrial concentration, which concerns the distribution of production within an industry, as opposed to a market.
Commonly used market concentration measures are the Herfindahl index (HHI or simply H) and the concentration ratio (CR). The Hannah-Kay (1971) index has the general form
HK_\alpha(x) = \left\{ \begin{align} & \left(\sum s_i^\alpha\right)^\frac {1}{\alpha-1} \text{ if }\alpha > 0, \alpha e 1 \\ & \prod s_i^{s_i} \text{ if }\alpha = 1\\ \end{align}\right..
Note, $\prod s_i^{s_i} = \exp\left(\sum s_i \log(s_i)\right)$, which is the exponential index.
When antitrust agencies are evaluating a potential violation of competition laws, they will typically make a determination of the relevant market and attempt to measure market concentration
within the relevant market.
As an economic tool market concentration is useful because it reflects the degree of competition in the market. Tirole (1988, p. 247) notes that:
Bain's (1956) original concern with market concentration was based on an intuitive relationship between high concentration and collusion.
There are game theoretic models of market interaction (e.g. among oligopolists) that predict that an increase in market concentration will result in higher prices and lower consumer welfare even
when collusion in the sense of cartelization (i.e. explicit collusion) is absent. Examples are Cournot oligopoly, and Bertrand oligopoly for differentiated products.
Empirical tests
Empirical studies that are designed to test the relationship between market concentration and prices are collectively known as price-concentration studies; see Weiss (1989).
Typically, any study that claims to test the relationship between price and the level of market concentration is also (jointly, that is, simultaneously) testing whether the market definition
(according to which market concentration is being calculated) is relevant; that is, whether the boundaries of each market is not being determined either too narrowly or too broadly so as to make
the defined "market" meaningless from the point of the competitive interactions of the firms that it includes (or is made of).
Alternative definition
In economics, market concentration is a criterion that can be used to rank order various distributions of firms' shares of the total production (alternatively, total capacity or total reserves)
in a market.
Further Examples
Section 1 of the Department of Justice and the Federal Trade Commission's Horizontal Merger Guidelines is entitled "Market Definition, Measurement and Concentration." Herfindahl index is the
measure of concentration that these Guidelines state that will be used.
A simple measure of market concentration is 1/N where N is the number of firms in the market. This measure of concentration ignores the dispersion among the firms' shares. It is decreasing in the
number of firms and nonincreasing in the degree of symmetry between them. This measure is practically useful only if a sample of firms' market shares is believed to be random, rather than
determined by the firms' inherent characteristics.
Any criterion that can be used to compare or rank distributions (e.g. probability distribution, frequency distribution or size distribution) can be used as a market concentration criterion.
Examples are stochastic dominance and Gini coefficient.
Curry and George (1981) enlist the following "alternative" measures of concentration:
(a) The mean of the first moment distribution (Niehans, 1958); Hannah and Kay (1977) call this an "absolute concentration" index:
$\bar{X}_1 = \sum x_i^2/x_i = \sum x_i H$
(b) The Rosenbluth (1961) index (also Hall and Tideman, 1967):
$R = \frac{1}{2\sum is_i - 1}$ where symbol i indicates the firm's rank position.
(c) Comprehensive concentration index (Horvath 1970):
$CCI = s_1 + \sum_{i=2}^N s_i^2(2 - s_i)$ where s[1] is the share of the largest firm. The index is similar to $2\text{H} - \sum s_i^3$ except that greater weight is assigned to the share of
the largest firm.
(d) The Pareto slope (Ijiri and Simon, 1971). If the Pareto distribution is plotted on double logarithmic scales, [then] the distribution function is linear, and its slope can be calculated if it
is fitted to an observed size-distribution.
(e) The Linda index (1976)
$L=\frac 1 {N(N-1)}\sum_{i=1}^{N-1}Q_i$ where Q[i] is the ratio between the average share of the first i firms and the average share of the remaining N − i firms. This index is designed to
measure the degree of inequality between values of the size variable accounted for by various sub-samples of firms. It is also intended to define the boundary between the oligopolists within
an industry and other firms. It has been used by the European Union.
(f) The U Index (Davies, 1980):
U = I ^* aN ^− 1 where I ^* is an accepted measure of inequality (in practice the coefficient of variation is suggested), a is a constant or a parameter (to be estimated empirically) and N is
the number of firms. Davies (1979) suggests that a concentration index should in general depend on both N and the inequality of firms' shares.
The "number of effective competitors" is the inverse of the Herfindahl index.
Terrence Kavyu Muthoka defines distribution just as functionals in the Swartz space which is the space of functions with compact support and with all derivatives existing.The Media:Dirac
Distribution or the Dirac function is a good example .
1. ^ Concentration. Glossary of Statistical Terms. Organisation for Economic Co-operation and Development.
□ Bain, J. 1956. Barriers to New Competition. Cambridge, Mass.: Harvard Univ. Press.
□ Curry, B. and K. D. George 1983. "Industrial concentration: A survey" Jour. of Indust. Econ. 31(3): 203-55, which contains the references cited under section Further Examples above).
□ Tirole, J. 1988. The Theory of Industrial Organization. Cambridge, Mass.: MIT Press.
□ Weiss, L. W. 1989. Concentration and price. Cambridge, Mass. : MIT Press.
See also
□ Horizontal Merger Guidelines
External links
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Morris: World of Britain 2: Proof and Paradox
In working out the proof for World of Britain I came across a paradox. Maybe smarter Math Factorites can help me out? My sanity could depend on it.
In the puzzle you have five different tasks. On each day one of these tasks is given at random. How long do you expect it to take to get all five tasks?
First consider a simple case. Suppose some event has a probability, p, of happening on any one day. Let’s say that E(p) is the expected number of days we have to wait for the event to happen. For
example if p=1 then the event is guaranteed to happen every day and so E(p)=1.
How can we calculate E(p)?
Andy does an experiment. He will wait for the event to happen and record how many days it took. He will do this several times, for long enough to ensure that he gets an answer that is as accurate as
he needs. He will keep going for N days in total. Afterwards he will take the average of all of his wait times to get an estimate for E(p).
The average he calculates is the total of all the wait times divided by the number of occurrences of the event. But we can estimate both of these values and therefore estimate his value for E(p). The
total of all of the wait times is going to be about N. Since the event has a probability of p of occurring on any particular day the number of occurrences will be about p times N.
So Andy’s average will be about N/(pN) which will be about 1/p.
We can make N as large as we like to make this result as accurate as we like. So we can confidently say that E(p) = 1/p.
Let’s get back to our puzzle about tasks. We need to wait for the first task, then the second, then the third and so on. When there are t tasks left then the chance of getting a new one is t/5.
So the total waiting time is
E(5/5) + E(4/5) + E(3/5) + E(2/5) + E(1/5) = 5/5 + 5/4 + 5/3 + 5/2 + 5/1
= 5(1/5 + 1/4 + 1/3 + 1/2 + 1) = 11 ^5/[12] = 11.4166666… days
You may recognise the harmonic series!
So that’s the proof, now where’s the paradox?
Bob visits Andy when he is able and waits with him until the event occurs. Of course Andy may well have been waiting for some time.
If Bob turns up just after the event has happened then he will wait for the same time as Andy. If he turns up just before the event he will wait for one day. On average he will wait for about half
the time that Andy does.
When the event occurs Bob disappears and comes back when he is next able to.
At the end of the experiment Bob averages all of his wait times to get an estimate for E(p). He gets an answer which is half of Andy’s!
That is the paradox!
Now Carol thinks she understands what is going on here. The problem is that Andy is distorting his results by always starting the clock straight after an event has occurred. That guarantees him to
get the longest possible wait times.
She thinks the only way to get an accurate answer is to look at the wait time starting from each of the N days and then average these N wait times.
To calculate this she needs to make an assumption. She knows that the event occurs every 1/p days on average. She assumes they happen regularly every 1/p days.
Let’s say they happen every n days where n = 1/p. Remember Andy’s value for E(p) is 1/p = n. Bob’s value is half that, so about n/2.
The wait time will vary between 1 and n. The average wait time will be n(n+1)/2n = (n+1)/2.
So Carols estimate for E(p) is about n/2 whereas Andy’s was 1/p = n. So Carol is agreeing with Bob.
I can tell you that Andy had the right value and that the logic I used for Bob and Carol was flawed.
Can you see where?
Sean said,
You assume that Andy and Bob are counting the same events. Bob might miss some of Andy’s events. For example, let’s say Bob shows up for the first time on day 3. Andy witnessed event A on day 1
and event B on day 2 when Bob wasn’t there. Event A occurs again on day 3 in the presence of both Andy and Bob. This is the first occurrence of event A for Bob, but the second for Andy. Therefore
it is a non-event for Andy. Andy ignores it and keeps waiting. Bob on the other hand counts this new event and then leaves. When Andy is done counting all five unique events, Bob may still have 2
or 3 or even 5 to go.
If you stop Bob’s clock when he is not present you can see that from his perspective, the calculation of his average wait time is the same as Andy’s.
There is also a second mistake in your reasoning. For the moment, forget that Bob and Andy may be counting different events, let’s just count Andy’s events. It is not necessarily true that over the
long-run, Bob will on average appear mid-way between Andy’s observation of events. It depends on how you define the process that generates Bob’s random appearances. It is not enough to say that
Bob appears at random times – you have to be more specific. For example, if Bob decides to reappear every time an atom of Uranium-238 decays somewhere in the universe, he will always reappear almost
immediately after he leaves. So his average wait time will be very close to Andy’s. (He will also be very unlikely to miss any of Andy’s events).
If on the other hand, the random process causes Bob to reappear a little less frequently, like every time that a boy is born and named Robert in California (about once every 1.5 days), he will be
more likely to appear somewhere near the mid-point between consecutive observances as you assumed, though he might also miss some of Andy’s events.
If the random process is very long like say watching a particular atom of Uranium-238, then he will probably miss all of Andy’s events.
Sean said,
I used a bad example. The name Robert occurs every 1.5 days so Bob would still be showing up fairly soon after he left as compared with Andy’s average wait time of over 11 days.
Let’s change the name in my second example to Ernest. There’s one born about every 11 days in California which is closer to Bob’s average wait time.
Stephen Morris said,
I like your thinking Sean. Have we come up with a new way of detecting nuclear tests I wonder? I never knew the true importance of being Ernest!
You’re right, I should clarify exactly what Andy and Bob are up to.
In that example there is only one event, not the five in the main problem. The event can only happen once each day, lets say it always happens at noon. Bob always turns up at 10am. Andy waits for
the full duration of the test regardless of how many times the event happens.
How does Bob decide when to turn up? Okay, let’s go with your idea that he turns up when an Ernest is born in California. Are people born Ernest or are they named later? I was going to go with
christenings but that excludes people who aren’t christened. Let’s say Bob checks the births section in the Californian press each morning and visits Andy when he reads the anouncement of a brand
new Ernest.
The answer to the puzzle is the paradox is that Bob is more likely to turn up during one of Andy’s long waits than during one of his shorter ones. On average Bob will wait half as long as Andy for
any event that Bob sees, but he will miss a lot of events that Andy sees which follow on quickly from the previous event. I have done the maths and this does work out. Unlikely as it seems Andy and
Bob will have the same average wait time.
Sean said,
Thanks. The clarification is helpful, and it seems like you have worked out the right explanation for the apparent paradox. But there are still a couple of points you made that probably should be
discussed further.
“The event can only happen once each day, let’s say it always happens at noon. Bob always turns up at 10 a.m.”
This throws me off a little. So Bob’s appearances and the events are each randomly occurring on 24-hour cycles, but slightly shifted. This is the same as if Bob’s random appearances always occur at
noon (i.e., in sync with the random events) but then we always add two hours to his wait time. So that causes a problem right there. Bob’s average wait time is going to be 2 hours longer than
Andy’s. What if Bob has a brother who always shows up at 9:30? Then he will observe exactly the same events that Bob observes since those can occur only at 12:00 but his wait time will be 1/2 hour
longer than Bob’s every time. I think you have to make the assumption that the time of day is the same for both Bob’s appearances and the events, or alternatively, that the probabilities of Bob
appearing or an event occurring are equal at all times throughout the day.
“…Bob is more likely to turn up during one of Andy’s long waits than during one of his shorter ones.”
This seems obvious, but might be a little trickier than it seems. Yes, a longer wait for Andy means a larger window of opportunity for Bob, but if there are more short intervals than long intervals
Bob might be more likely to show up during a short period. For example, let’s say that for the first 100 days there are 20 intervals of 5 days each. And say the next 100 days is just one long
interval. Bob may be just as likely to show up during several short intervals as he is during the one long interval. But, you say, that is just because I constructed this artificially skewed
distribution of events. The thing is, waiting times for independent events are usually said to follow what is known as a Poisson distribution which is inherently skewed. This means that if you pick
an arbitrary regular period of time, like months or years, and count the random events in each period, you will observe that more periods will contain an above-average number of events, and fewer
have a below-average number of events. In other words, there will be more short intervals than long intervals. So technically I think your statement may not be true but your reasoning was basically
sound. It’s not that Bob hits more long intervals than short intervals. It’s that Bob misses more short intervals than he does long intervals. Hope the distinction I’m trying to make is not too
“On average Bob will wait half as long as Andy for any event that Bob sees…”
Maybe. This is true over the long run if the rate for Bob’s random generator is as long or longer than the rate for the event random generator, which I think is what you intended. If Bob showed up
more frequently he would usually show up somewhere in the first half of Andy’s wait, like in my first example from yesterday. You were clear about using the Ernest random number generator for Bob,
and I think you meant for that to correspond with the average rate of occurrence for the events also.
Stephen Morris said,
Thanks Sean,
Gosh, I can see I’ll have to be much more careful in my future puzzles.
On your first point:
Andy and Bob are counting whole days. Really they are counting the number of ‘slots’ in which the event could occur. So if Bob turns up at 10am and the event occurs at noon the same day then he
would count 1 day. If it helps we can have Bob turn up just after noon so his waiting time will actually be whole days.
This puzzle is based on ‘World of Warcraft’. In that game there are ‘quest givers’ who have five different quests to give out. The quest is chosen at the start of each day, i.e. at midnight, and is
available throughout that day. A player can take the quest at any time during the day but can only take one quest each day. I found myself wondering how long it would take me to get all the quests,
and that turned out to be more interesting than actually playing the game!
A different puzzle would be to have the event occur at any time and measure the actual time Andy and Bob wait. In this case we could have multiple events in a single day. As you say this is
modelled by the poisson distribution.
On your second point:
You are right that a short wait will occur more often. I used the phrase ‘one of’ to get round this.
As you point out the actual distribution of Andy’s wait times (the skew) makes a big difference. Independant events will have a particular distribution. Any other distribution indicates some
connection between the events and in this case Bob will get a different answer to Andy.
I have calculated the distribution of independent events and put them in a spreadsheet which you can access here.
I assume we do the experiment over 10,000 days with the event occuring every five days on average.
In the third tab, ‘Calculating the Values’, r is the number of days that Andy waits. Column D is the number of times that Andy will wait for this many days.
In the first tab, ‘Large Example’, I use these values to calculate Andy and Bobs’ expected wait times. For Bob I calculate the probability of him appearing during Andy’s wait and what his wait is
likely to be. Note that I am adding up ‘expected wait times’ for Bob, not actual wait times.
At the top I have Andy and Bobs’ wait times which are 4.975 and 4.913 respectively. They are slightly out partly becuase of rounding and partly because I only go up to an Andy wait time of 30 days.
To be truly accurate I should go up to infinity.
On the third point:
Yes, I am assuming that Bob turns up much less often than the event. It is important each time Bob turns up is clearly not connected to the last time he turned up. | {"url":"http://mathfactor.uark.edu/2009/07/morris-world-of-britain-2-proof-and-paradox/","timestamp":"2024-11-11T23:17:56Z","content_type":"application/xhtml+xml","content_length":"93610","record_id":"<urn:uuid:a055d6ad-f082-4939-b6da-aa4232e12163>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00328.warc.gz"} |
Seeking help with SAS Regression Analysis tasks? | Hire Someone To Take My SAS Assignment
Seeking help with SAS Regression Analysis tasks? Research showed that a query is more prolific than expected if it can be expressed mathematically. For example, an aggregate with a 10-dimensional
average is more likely to give you a higher probability than non-aggregated solutions. “SAS Regression” will naturally assume that your entire group, as opposed to the group at the expense of the
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condition (for example, if the distribution of a group’s sum is equal to one, but if you take both mean values, you may still become very unlucky). This is one thing that you can use to test your
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can even estimate the precision: SoSeeking help with SAS Regression Analysis tasks? You may need a lot of help in Regression, but some SAS tasks require you to solve individual regressions.
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In this type of task, each regressor must be performed sequentially. SAS Regression only performs regressors for 100 regressors and one that has to be solved sequentially. Instead, regression
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assume we have 7 independent variables, such that 14 12 13 and $l(y) = f(1-y)$. Then let’s 15 15 then we add the data to,,, and, in the series form: 16 In fact, if we then compare this with the fixed
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MATLAB Function (*.m files) Editing, Executing and Displaying Help
MATLAB functions are just script files and must have the MATLAB extension "m", such as "circle.m".
• Editing MATLAB Function Files: Modifying or creating or saving MATLAB script function files, whether on PCs/Macs or Unix/Linux workstations, are edited the same way you normally edit files on
those platforms:
1. PCs/Macs: To create a new function run you favorite Edit application, select a "New" file, type in your MATLAB commands for the function, and "Save As" file name [function].m when finished.
If the function script file already exists, that start with "OPEN File" instead.
2. Unix/Linux workstations: To create a new script function or to modify an existing one type in the command of your favorite editor like vi, emacs or other with the name [function].m as
argument, type in your MATLAB commands for the function, and save when finished. Editing can also be done on the Unix/Linux MATLAB command line using the bang escape "!", for example but "!
emacs" can also be used, yet beware of carpel tunnel syndrome.
Hint: Square brackets are used here to denote general items for the user as to type in actual names when actually used in MATLAB, such as "[function]", but in MATLAB square brackets are also
used for marking arrays that are being initialized on assignment.
Hints : The percent sign "%" marks the beginning of a comment, so if a "%" appears in a line (except as a format specifier in a "fprintf" or "sprintf" argument, its other purpose), the rest
of the line to the right is a comment, but the first set of lines marked at the beginning of the line before any MATLAB commands are Header lines that are listed when you use the MATLAB "help
" command:
for the script file [function].m, whether user or MATLAB built-in function. Also, the header lines are also search for keywords by the MATLAB "lookfor" command, e.g.,
Hints: The semicolon ";" at the end of a MATLAB command line suppressed MATLAB output for that command, but semicolon are also used for separating rows in array initialization forms.
• Executing an Existing Function Files: In MATLAB, type the name of the function without the ".m" extension on the command line, for example for circle.m, type
• Displaying MATLAB Function Script File Content In MATLAB: On the MATLAB command line use the "type" command with the ".m" extension, for example,
• Sample MATLAB Function Script File: The simple example "circle.m should plot a unit circle when executed:
% Circle - Script file to draw unit circle
% modified from "Getting Started with MATLAB" by Rudra Pratap 9/14/94
format compact % tightens loose format
format long e % makes numerical output in double precision
theta = linspace(0,2*pi,100); % create vector theta
x = cos(theta); % generate x-coordinate
y = sin(theta); % generate y-coordinate
plot(x,y); % plot circle
axis('equal'); % set equal scale on axes per pixel
title('Circle of unit radius') % put title
c=2*pi % prints out 2*pi value
disp('end circle.m') % prints out literal string.
% End Circle
Web Source: http://www.math.uic.edu/~hanson/MATLAB/MATLABfunctions.html
Email Comments or Questions to hanson@uic.edu | {"url":"http://homepages.math.uic.edu/~hanson/MATLAB/MATLABfunctions.html","timestamp":"2024-11-02T08:19:43Z","content_type":"text/html","content_length":"4509","record_id":"<urn:uuid:9566d472-ceaa-4c7c-b4f9-87eed6bcf18d>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00089.warc.gz"} |
How To Use Excel T TEST Function inHow To Use Excel T TEST Function in
How To Use Excel T TEST Function in
The T TEST or Student’s test is used to determine the confidence of an analysis. In mathematical term, it is used to know if the mean of the two samples are equal. T.TEST is used to accept or reject
the null hypothesis.
(Null hypothesis states that there is no significant difference in mean due to a given reason between two groups.)
Let’s say, we do a survey on two groups. In the end, we get the mean of these surveys and conclude that group 1 has done better from group 1 due to a reason. But how confident we are, that it is not
just a fluke but there’s a large possibility of it. This is determined by T.TEST. Soon we will see an elaborated example.
Syntax of T Test in Excel
=T.TEST(group1, group2, tails, type)
Group1: the first group of experiments.
Group2: the second group of experiments. It can have the same subject as of group1, different subjects with the same variance and different subjects with different variance.
Tails: it has two options, 1 tailed, and 2 tailed. 1 tailed T TEST is used when we know that the effect of the experiment is one-directional. 2 tailed T TEST it is used when we don't know the
direction of the effect of the experiment.
There are three options available (1,2,3).
1. 1 (Paired): we use it when group1 and group2 have same subjects.
2. 2 (Two sample equal variance): When the subjects of these groups are different but variance is same.
3. 3 (two sample unequal variances): When subjects of two groups are different and variance is also different.
Example: Get T TEST in Excel of Two Groups
One of my friends said that while studying, chewing gum helps you memorize. To test this I conducted an experiment between two groups. Group 1 studied while chewing gum and group 2 studied normally.
I collected the data and calculated the mean in Excel. The average score of group 1 is slightly greater than group 2.
So can we conclude that there is a positive effect of chewing gum? It seems so. But it can be a fluke. To confirm that this is not a fluke, we will conduct a T TEST on this data in Excel.
Null Hypothesis (H0): There is no difference in score due to studying while chewing gum.
Assume that the subjects of these two groups are the same. We also assume that the effect of this experiment will be one-directional. We will write this formula to get 1 tailed T-TEST with the paired
The Excel function returns t test value as 0.175.
We get the critical value of T TEST using this table. We use a degree of freedom (n-1) as 7.
We can see that the confidence level is between 0% to 50%. This is too low. Hence we accept the null hypothesis.
Two Tailed T Test
Similarly for 2 tailed T TEST we will write this formula
This will result 0.350. Which double of 1 tailed T.TEST.
So yeah guys, this how T.TEST function works in excel. This was a tutorial for using T.TEST Function in excel. That’s why we didn’t deep dived in student’s t test functionality and mathematical back
ground. We will discuss student’s t test in later articles, if you want. Let me know, if you have any doubts regarding this or ony other topic in excel function.
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Grade 6 Math - Real World Application
Welcome to 6th Grade Math - Real World Application!
The learner will be able to write ratios and understand written ratios after watching this lesson.
The learner will be able to fluently write ratios using information in a table after watching this practice video.
The learner will be able to calculate unit rates and understand why unit rates are useful.
The learner will be able to fluently calculate unit rates after watching this practice video.
The learner will be able to use tables, tape diagrams, double number lines, and equations to find equivalent ratios.
The learner will be able to use tables, tape diagrams, double number lines, and equations to find equivalent ratios in the context of a scaled map after watching this practice video.
The learner will be able to calculate and graph equivalent ratios after watching this lesson.
The learner will be able to calculate and graph equivalent ratios after watching this practice video.
The learner will be able to calculate unit rates in relation to pricing at a gas station after watching this lesson.
The learner will be able to calculate unit rates in relation to constant speed of buses after watching this practice video.
The learner will be able to write percents as rates and find parts and wholes when given percents after watching this lesson.
The learner will be able to find the whole when given the percent and the part after watching this practice video.
The learner will be able to convert to different units through ratio reasoning after watching this lesson.
The learner will be able to fluently convert to different units using ratio reasoning after watching this practice video.
This is the assessment for the unit "Ratios and Proportional Relationships."
The learner will be able to differentiate statistical and non-statistical questions after watching this lesson.
The learner will be able to differentiate statistical and non-statistical questions across multiple examples after watching this practice video.
The learner will be able to determine the center, diagnose the shape, and interpret the spread of different data distributions after watching this lesson.
The learner will be able to determine the center, diagnose the shape, and interpret the spread of different data distributions after watching this practice video.
The learner will be able to calculate mean, median, mode, range, interquartile range, and mean absolute deviation after watching this lesson.
The learner will be able to calculate mean, median, mode, range, interquartile range, and mean absolute deviation as well as understand whether each is a measure of center or spread after watching
this practice video.
The learner will be able to create histograms, dot plots, and box-and-whisker diagrams when given a data set after watching this lesson.
The learner will be able to create histograms, dot plots, and box-and-whisker diagrams when given a data set after watching this practice video.
The learner will be able to calculate relative frequency and create a percent bar graph after watching this lesson.
The learner will be able to calculate relative frequency and create a percent bar graph after watching this practice video.
The learner will be able to classify data as categorical or numerical, discrete or continuous, counted or measured, and define the units after watching this lesson.
The learner will be able to classify data as categorical or numerical, discrete or continuous, counted or measured, and define the units after watching this practice video.
The learner will be able to identify outliers and their impact on measures of center and spread after watching this lesson.
The learner will be able to identify outliers and their impact on measures of center and spread after watching this practice video.
The learner will be able to identify which measures of center and spread are appropriate depending on the shape of data distributions after watching this lesson.
The learner will be able to identify which measures of center and spread are appropriate depending on whether the problem included a skewed or symmetric data distribution after watching this practice
This is the assessment for the unit "Statistical Analysis and Diagrams."
The learner will be able to create appropriate diagrams to visualize fraction division after watching this lesson.
The learner will be able to create appropriate diagrams to visualize fraction division after watching this practice video.
The learner will be able to add, subtract, multiply, and divide two-digit decimals after watching this lesson.
The learner will be able to add, subtract, multiply, and divide two-digit decimals after watching this practice video.
The learner will be able to find least common multiples and greatest common factors to assist in multiplication after watching this lesson.
The learner will be able to solve multiplication problems by finding least common multiples and greatest common factors after watching this practice video.
This is the assessment for the unit "Operational Fluency with Fractions and Decimals."
The learner will be able to interpret negative and positive numbers across a variety of real-world examples after watching this lesson.
The learner will be able to interpret negative and positive numbers in real-world contexts after watching this practice video.
The learner will be able to identify integers, whole and rational numbers as well as find number opposites after watching this lesson.
The learner will be able to identify integers, whole and rational numbers and find number opposites after watching this practice video.
The learner will be able to identify quadrants on the coordinate plane, graph coordinates resulting from x-axis and y-axis reflections, and plot points on number lines after watching this lesson.
The learner will be able to solve problems involving the identification of quadrants on the coordinate plane, graphing coordinates resulting from x-axis and y-axis reflections, and plotting points on
number lines after watching this video.
The learner will be able to plot negative numbers in order on the number line and use real-world language to compare negative numbers after watching this lesson.
The learner will be able to solve problems involving the plotting of negative numbers in order on the number line and using real-world language to compare negative numbers after watching this
practice video.
The learner will be able to use real-world scenarios to accurately describe the absolute value of numbers after watching this lesson.
The learner will be able to solve real-world problems when describing the absolute value of numbers after watching this practice video.
The learner will be able to use real-world scenarios to find the distance between two points on the coordinate plane after watching this lesson.
The learner will be able to view worked examples of finding distance between two points on the coordinate plane after watching this practice video.
This is the assessment for the unit "Rational Numbers in the Real World and Coordinate Plane."
The learner will be able to distinguish between bases and exponents and simplify a variety of exponential expressions after watching this lesson.
The learner will be able to compare and order exponential expressions as well as create exponential expressions given verbal descriptions after watching this practice video.
The learner will be able to identify algebraic vocabulary and convert written operations into algebraic expressions after watching this lesson.
The learner will be able to identify algebraic vocabulary and convert written operations into algebraic expressions across several examples after watching this practice video.
The learner will be able to input numerical values for variables in geometric applications in order to find area after watching this lesson.
The learner will be able to input values for variables in order to solve for area in several geometric problems after watching this practice video.
The learner will be able to create equivalent expressions by utilizing several mathematical properties after watching this lesson.
The learner will be able to create equivalent expressions by utilizing mathematical properties in a variety of scenarios after watching this practice video.
The learner will be able to identify whether given expressions are equivalent after watching this lesson.
The learner will be able to identify whether the expressions given in 3 different problems are equivalent after watching this practice video.
The learner will be able to input several values for the variable in order to determine which value makes the statement true after watching this lesson.
The learner will be able to determine which values make true statements across several equations and inequalities after watching this practice video.
The learner will be able to isolate the variable when solving one-step addition and multiplication equations after watching this practice video.
The learner will be able to solve addition and multiplication equations and inequalities along with plotting the solutions on a number line after watching this practice video.
The learner will be able to distinguish independent and dependent variables, write equations, and plot a table of values on the coordinate plane after watching this lesson.
The learner will be able to differentiate independent and dependent variables, write an equation based on a real-life scenario, and plot the table of values on a coordinate plane after watching this
practice video.
This is the exam for the unit "Algebraic Expressions, Equations, and Inequalities."
The learner will be able to decompose triangles, rectangles, and squares from irregular shapes in order to find area after watching this lesson.
The learner will be able to decompose triangles, rectangles, and squares and compose rectangles in real-world scenarios after watching this practice video.
The learner will be able to find the volume of right rectangular prisms with the formula and by combining cubes with fractional edge lengths after watching this practice video.
The learner will be able to calculate the volume of a right rectangular prism using the formula and by stacking the prism with cubes of fractional edge lengths after watching this lesson.
The learner will be able to graph polygons in the coordinate plane, find side lengths, and calculate area after watching this lesson.
The learner will be able to graph polygons in the coordinate plane, find side lengths, and calculate area after watching this practice video.
The learner will be able to create 2d representations of 3d shapes and find their surface area after watching this lesson.
The learner will be able to determine the net that matches a 3d shape and calculate its area after watching this practice video.
The learner will be able to determine how side lengths and angle measurements are related in triangles after watching this lesson.
The learner will be able to determine how side lengths and angle measurements relate in triangles after watching this practice video.
This is the exam for the unit "Area, Surface Area, and Volume."
The learner will be able to determine the essential features of a checking account and balance a monthly statement after watching this lesson.
The learner will be able to balance an example monthly checking account statement after watching this practice video.
The learner will be able to determine what comprises a credit score, identify good credit scores, and describe the benefits of a credit report for borrowers and lenders after watching this lesson.
The learner will be able to determine what actions can be taken to increase or decrease a credit score and the benefits available to those with good credit after watching this practice video.
The learner will be able to determine the appropriate steps to financially prepare for college and the career benefits of earning a university degree after watching this lesson.
The learner will be able to compute the lifetime income differential of two students who take different paths with regard to attending university after watching this practice video.
This is the exam for the unit "Financial Mathematics."
This is the final assessment that covers all learning objectives for the course. | {"url":"https://www.lernsys.com/en/6th-grade-math-real-world-application","timestamp":"2024-11-08T22:25:07Z","content_type":"text/html","content_length":"216881","record_id":"<urn:uuid:ed61ab90-c54c-48f5-b6f1-6a3f40bc5a31>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00775.warc.gz"} |
Approximation algorithms for capacitated rectangle stabbing
In the rectangle stabbing problem we are given a set of axis parallel rectangles and a set of horizontal and vertical lines, and our goal is to find a minimum size subset of lines that intersect all
the rectangles. We study the capacitated version of this problem in which the input includes an integral capacity for each line that bounds the number of rectangles that the line can cover. We
consider two versions of this problem. In the first, one is allowed to use only a single copy of each line (hard capacities), and in the second, one is allowed to use multiple copies of every line
provided that multiplicities are counted in the size of the solution (soft capacities). For the case of d-dimensional rectangle stabbing with soft capacities, we present a 6d-approximation algorithm
and a 2-approximation algorithm when d = 1. For the case of hard capacities, we present a bi-criteria algorithm that computes 16d-approximate solutions that use at most two copies of every line. For
the one dimensional case, an 8-approximation algorithm for hard capacities is presented.
Original language English
Title of host publication Algorithms and Complexity - 6th Italian Conference, CIAC 2006, Proceedings
Publisher Springer Verlag
Pages 18-29
Number of pages 12
ISBN (Print) 354034375X, 9783540343752
State Published - 2006
Externally published Yes
Event 6th Italian Conference on Algorithms and Complexity, CIAC 2006 - Rome, Italy
Duration: 29 May 2006 → 31 May 2006
Publication series
Name Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics)
Volume 3998 LNCS
ISSN (Print) 0302-9743
ISSN (Electronic) 1611-3349
Conference 6th Italian Conference on Algorithms and Complexity, CIAC 2006
Country/Territory Italy
City Rome
Period 29/05/06 → 31/05/06
Dive into the research topics of 'Approximation algorithms for capacitated rectangle stabbing'. Together they form a unique fingerprint. | {"url":"https://cris.biu.ac.il/en/publications/approximation-algorithms-for-capacitated-rectangle-stabbing-2","timestamp":"2024-11-05T12:47:21Z","content_type":"text/html","content_length":"57016","record_id":"<urn:uuid:a3a2f122-2197-480b-892c-5c63dc9ee523>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00001.warc.gz"} |
Polynomial Formula of Math Topics | Question AI
<div class="basic-navigation__BackgroundWrapper-sc-j3ac2x-3 zrYUG"><div><div style="display:flex;flex-direction: column;padding: 12px 10px; box-shadow: rgba(0, 0, 0, 0.1) 2px 2px 12px;border-radius:
4px;overflow: hidden;"> <p>A polynomial formula is a formula that expresses the polynomial expression. The polynomial an expression that has two or more than two terms(algebraic terms) is known as a
polynomial expression. A repetitive summation or subtraction of binomials or monomials forms a polynomial expression. A polynomial can have both like as well as unlike terms in it. Like terms in
polynomials are the terms which have the same variable and same power and the terms that have different variables and different powers are known as, unlike terms. Let us see the polynomial formula in
the following section along with the solved examples.</p> <h2 id="qai_title_1">What is Polynomial Formula?</h2> <p>The polynomial formula has variables with different power and the highest power of
the variable on solving is known as the degree of the polynomial. The polynomial formula is also known as the standard form of the polynomial where the arrangement of the variables is according to
the decreasing power of the variable in the formula. </p> <p><img alt="" src="https://static.questionai.com/resource%2Fknowledge%2Fpolynomial+formula_pid_1.jpg" style="max-width: 100%; width: 400px;"
/></p> <h3 id="qai_title_3">Polynomial Formula</h3> <p>Polynomial Formula is given by: </p> <p>\(\left(a x^{n}+b x^{\{n-1\}}+c x^{\{n-3\}}+\ldots \ldots+r x+s\right)\)</p> <p>Where</p> <ul> <li>a, b,
c, ..., s are coefficients</li> <li>x is the variable</li> <li>n is the degree of the polynomial</li> </ul> <p>Some basic formulas associated with the polynomial expression given above are,</p> <p
align="justify">1.F(x) = a<sub>n</sub>(x<sup>n</sup>)</p> <p align="justify">where</p> <ul> <li align="justify">a is the coefficient</li> <li align="justify">x is the variable </li> <li align=
"justify">n is the exponent</li> </ul> <p>2. F(x) = a<sub>n</sub>x<sup>n</sup> + a<sub>n-1</sub>x<sup>n-1</sup> + a<sub>n-2</sub>x<sup>n-2</sup> + …….. + a<sub>1</sub>x +a<sub>0 </sub>= 0</p> <p>3. F
(x)=a<sub>n</sub>x<sup>n</sup>+..+rx+s</p> <ul> <li> n is a natural number</li> </ul> <p>a<sub>n</sub>−b<sub>n</sub>=(a−b)(a<sub>n</sub>−1+a<sub>n</sub>−2b+…)</p> <ul> <li> n is even number</li> </
ul> <p>a<sub>n</sub>+b<sub>n</sub>=(a+b)(a<sub>n</sub>−1−a<sub>n</sub>−2b+…)</p> <ul> <li> n is odd number</li> <li>a<sub>n</sub>+b<sub>n</sub>=(a+b)(a<sub>n</sub>−1−an−2b+…)</li> </ul> <h3 id=
"qai_title_4">Applications of Polynomial Formula</h3> <p>It has applications in engineering, computer, management, business, and even in farming. Variables and constants are used to create
expressions defining quantities that are known and unknown. </p> <p><span style="background-color:null;">The polynomial equations are formed with variables, exponents, and coefficients. Polynomials
can be solved by factoring them into either in terms of the degree or variables present in the given equation.</span></p> <p>Let's take a quick look at a couple of examples to understand the
polynomial formula better.</p> <h2 id="qai_title_2">Examples Using Polynomial Formula</h2> <p><strong>Example 1:</strong> Find the factors of the given polynomial formula (x<sup>2</sup>+12x+36).</p>
<p><strong>Solution:</strong></p> <p>To find:</p> <p>factors of the polynomial</p> <p>(x<sup>2</sup>+12x+36)</p> <p>(x<sup>2</sup>+2(6)x+6<sup>2</sup>)</p> <p>(x+6)<sup>2</sup></p> <p><strong>Answer:
Factors of the polynomial (x<sup>2</sup>+12x+36) are (x+6) and (x+6).</strong></p> <p><strong>Example 2:</strong> Find the factors of the given polynomial formula (x<sup>2</sup>+3x-28). </p> <p>
<strong>Solution:</strong></p> <p>To find:</p> <p>factors of the polynomial</p> <p>(x<sup>2</sup>+3x-28)</p> <p>(x<sup>2</sup>+7x-4x-28)</p> <p>(x(x+7)-4(x+7)</p> <p>(x-4)(x+7)</p> <p><strong>Answer:
Factors of the polynomial (x<sup>2</sup>+11x+28) are (x-4) and (x+7)</strong></p> <p><strong>Example 3: </strong>Calculate the factors of the polynomial x<sup>2</sup> – 6x + 9?</p> <p><strong>
Solution:</strong></p> <p>x<sup>2</sup> – 6x + 9<br/> = x<sup>2</sup> – 2(3x) + 3<sup>2</sup><br/> = x<sup>2</sup> – 2(3)(x) + 3<sup>2</sup><br/> = (x – 3)<sup>2</sup></p> <p><strong>Answer: Factors
of the polynomial x<sup>2</sup> – 6x + 9 are (x – 3) and (x – 3).</strong></p> </div></div></div> | {"url":"https://www.questionai.com/knowledge/kv0NM03uv3-polynomial-formula","timestamp":"2024-11-10T00:07:08Z","content_type":"text/html","content_length":"62050","record_id":"<urn:uuid:d7e592fd-f6cd-40ca-88c4-ed1a0bf40acb>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00307.warc.gz"} |
COTH function: Description, Usage, Syntax, Examples and Explanation November 6, 2024 - Excel Office
COTH function: Description, Usage, Syntax, Examples and Explanation
What is COTH function in Excel?
COTH function is one of the Math and Trig functions in Microsoft Excel that return the hyperbolic cotangent of a hyperbolic angle.
Syntax of COTH function
The COTH function syntax has the following arguments.
COTH formula explanation
• The hyperbolic cotangent is an analog of the ordinary (circular) cotangent.
• The absolute value of Number must be less than 2^27.
• If Number is outside its constraints, COTH returns the #NUM! error value.
• If Number is a non-numeric value, COTH returns the #VALUE! error value.
• The following equation is used:
Example of COTH function
Steps to follow:
1. Open a new Excel worksheet.
2. Copy data in the following table below and paste it in cell A1
Note: For formulas to show results, select them, press F2 key on your keyboard and then press Enter.
You can adjust the column widths to see all the data, if need be.
Formula Description Result
=COTH(2) Returns the hyperbolic cotangent of 2 (1.037). 1.037 | {"url":"https://www.xlsoffice.com/excel-functions/math-and-trig-functions/coth-function-description-usage-syntax-examples-and-explanation/","timestamp":"2024-11-06T10:56:06Z","content_type":"text/html","content_length":"63936","record_id":"<urn:uuid:28d56373-bbdb-49a0-b68e-c2edbb97740b>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00795.warc.gz"} |
Projecting species abundances
David Garcia-Callejas and cxr team
Projecting species abundances
Population dynamics models, like the ones included in cxr, are used for understanding the influence of different factors in shaping species densities, and also for inferring future dynamics based on
the expected factors. We have included in cxr basic functionality for projecting these dynamics. For that, we can use any of the default models included in the package, or -again- implement any
user-defined model.
In this vignette, we show how to project the abundances of three species from our dataset a number of timesteps. For that, we first estimate model coefficients with the cxr_pm_multifit function, as
in previous vignettes.
three_sp <- c("BEMA","LEMA","SOAS")
sp.pos <- which(names(neigh_list) %in% three_sp)
data <- neigh_list[sp.pos]
# keep only fitness and neighbours columns
for(i in 1:length(data)){
data[[i]] <- data[[i]][,2:length(data[[i]])]
focal_column <- names(data)
# Beverton-Holt model
model_family <- "BH"
# the "bobyqa" algorithm works best for these species
optimization_method <- "bobyqa"
# pairwise alphas, but no covariate effects
alpha_form <- "pairwise"
lambda_cov_form <- "none"
alpha_cov_form <- "none"
# no fixed terms, i.e. we fit both lambdas and alphas
fixed_terms <- NULL
# for demonstration purposes
bootstrap_samples <- 3
# a limited number of timesteps.
timesteps <- 10
# for demonstration purposes
initial_abundances <- c(10,10,10)
names(initial_abundances) <- three_sp
# standard initial values,
# not allowing for intraspecific facilitation
initial_values <- list(lambda = 10,
alpha_intra = 0.1,
alpha_inter = 0.1)
# lambda_cov = 0.1,
# alpha_cov = 0.1)
lower_bounds <- list(lambda = 1,
alpha_intra = 0,
alpha_inter = -1)
# lambda_cov = 0,
# alpha_cov = 0)
upper_bounds <- list(lambda = 100,
alpha_intra = 1,
alpha_inter = 1)
# lambda_cov = 1,
# alpha_cov = 1)
With all initial values set, we can fit the parameters.
cxr_fit <- cxr_pm_multifit(data = data,
focal_column = focal_column,
model_family = model_family,
# covariates = salinity,
optimization_method = optimization_method,
alpha_form = alpha_form,
lambda_cov_form = lambda_cov_form,
alpha_cov_form = alpha_cov_form,
initial_values = initial_values,
lower_bounds = lower_bounds,
upper_bounds = upper_bounds,
fixed_terms = fixed_terms,
bootstrap_samples = bootstrap_samples)
Projecting abundances from a cxr object is straightforward: just call the function abundance_projection with the cxr fit and the number of timesteps to project as well as the initial abundances. In
this example, our fit did not include the effect of covariates, but if this was the case, we would also need to specify the value of each covariate in the projected timesteps (see the help of
abundance_projection for details).
ab <- abundance_projection(cxr_fit = cxr_fit,
# covariates = covariates_proj,
timesteps = timesteps,
initial_abundances = initial_abundances)
This function returns a simple numeric matrix with the projected abundances. Lastly, here is a basic plot showing the projection. This is a pedagogical example and it does not make sense to interpret
it ecologically, but it shows how one of the three species selected tends to become more abundant in the short term.
ab.df <- as.data.frame(ab)
ab.df$timestep <- 1:nrow(ab.df)
ab.df <- tidyr::gather(ab.df,key = "sp",value = "abund",-timestep)
abund.plot <- ggplot2::ggplot(ab.df,
ggplot2::aes(x = timestep,
y = abund, group = sp)) +
ggplot2::geom_line(ggplot2::aes(color = sp)) +
ggplot2::ylab("number of individuals") + ggplot2::xlab("time") +
ggplot2::ggtitle("Projected abundances of three plant species")+
Including your own projection model
As with other features of cxr, you can implement your own model for projecting species abundances. If you have gone through vignette 4, you should already be familiar with the format of cxr models,
and how to make them available to the package.
Here, for reference, we show the complete code of the most complex Beverton-Holt model included. You can use this example as a template for the function name and arguments. The instructions of
vignette 4 are also applicable here.
#' Beverton-Holt model for projecting abundances,
#' with specific alpha values and global covariate effects on alpha and lambda
#' @param lambda named numeric lambda value.
#' @param alpha_intra single numeric value.
#' @param alpha_inter numeric vector with interspecific alpha values.
#' @param lambda_cov numeric vector with effects of covariates over lambda.
#' @param alpha_cov named list of named numeric vectors
#' with effects of each covariate over alpha values.
#' @param abundance named numeric vector of abundances in the previous timestep.
#' @param covariates matrix with observations in rows and covariates in named columns.
#' Each cell is the value of a covariate in a given observation.
#' @return numeric abundance projected one timestep
#' @export
BH_project_alpha_pairwise_lambdacov_global_alphacov_pairwise <- function(lambda,
# put together intra and inter coefficients,
# be sure names match
spnames <- names(abundance)
alpha <- c(alpha_intra,alpha_inter)
alpha <- alpha[spnames]
alpha_covs <- list()
for(ia in 1:length(alpha_cov)){
alpha_covs[[ia]] <- alpha_cov[[ia]][spnames]
numsp <- length(abundance)
expected_abund <- NA_real_
# model
num = 1
focal.cov.matrix <- as.matrix(covariates)
for(v in 1:ncol(focal.cov.matrix)){
num <- num + lambda_cov[v]*focal.cov.matrix[,v]
cov_term_x <- list()
for(v in 1:ncol(focal.cov.matrix)){
cov_temp <- focal.cov.matrix[,v]
for(z in 1:length(abundance)){
#create alpha_cov_i*cov_i vector
cov_term_x[[z+(length(abundance)*(v-1))]] <-
# alpha_cov[z+(ncol(abund)*(v-1))]
alpha_cov[[v]][z] * cov_temp
cov_term <- list()
for(z in 0:(length(abundance)-1)){
cov_term_x_sum <- cov_term_x[[z+1]]
if(ncol(focal.cov.matrix) > 1){
for(v in 2:ncol(focal.cov.matrix)){
cov_term_x_sum <- cov_term_x_sum +
cov_term_x[[v + length(abundance)]]
cov_term[[z+1]] <- cov_term_x_sum
term <- 1 #create the denominator term for the model
for(z in 1:length(abundance)){
term <- term + (alpha[z] + cov_term[[z]]) * abundance[z]
expected_abund <- (lambda * (num) / term) * abundance[names(lambda)] | {"url":"https://cran.rediris.es/web/packages/cxr/vignettes/V5_Abundance_projections.html","timestamp":"2024-11-09T06:32:22Z","content_type":"text/html","content_length":"76155","record_id":"<urn:uuid:4559770c-4ea9-48e9-805e-9d5adc9c14ba>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00533.warc.gz"} |
Convert Kilometer/hour to Break
Please provide values below to convert kilometer/hour [km/h] to break, or vice versa.
Kilometer/hour to Break Conversion Table
Kilometer/hour [km/h] Break
0.01 km/h 1.0E-15 break
0.1 km/h 1.0E-14 break
1 km/h 1.0E-13 break
2 km/h 2.0E-13 break
3 km/h 3.0E-13 break
5 km/h 5.0E-13 break
10 km/h 1.0E-12 break
20 km/h 2.0E-12 break
50 km/h 5.0E-12 break
100 km/h 1.0E-11 break
1000 km/h 1.0E-10 break
How to Convert Kilometer/hour to Break
1 km/h = 1.0E-13 break
1 break = 10000000000000 km/h
Example: convert 15 km/h to break:
15 km/h = 15 × 1.0E-13 break = 1.5E-12 break
Popular Speed Unit Conversions
Convert Kilometer/hour to Other Speed Units | {"url":"https://www.tbarnyc.com/speed/kilometer-hour-to-break.html","timestamp":"2024-11-08T19:22:45Z","content_type":"text/html","content_length":"10239","record_id":"<urn:uuid:2064d919-c8b9-483b-9600-801aa11aa308>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00786.warc.gz"} |
What is bond order? How we can determine it? Please basics. Thank you so much. | Socratic
What is bond order? How we can determine it? Please basics. Thank you so much.
1 Answer
A conceptual approach is to simply count electrons in a bond and treat each bonding valence electron as half a bond order.
This works for many cases, except for when the highest-energy electron is in an antibonding molecular orbital.
For example, the bond order of $: \text{N"-="N} :$ fairly straightforward because it's a triple bond, and each bonding valence electron contributes half a bond order.
#"BO"_"triple" = "BO"_sigma + 2"BO"_pi = 1/2 xx ("2 electrons") + 2(1/2 xx ("2 electrons")) = 3# for the bond order, as we should expect, since bond order tells you the "degree" of bonding.
Or, in a more complicated example, like ${\text{NO}}_{3}^{-}$, a conceptual approach is to count the number of electrons in the bond and see how many bonds it is distributed across.
So, in ${\text{NO}}_{3}^{-}$, which has one double bond in its resonance structure, has $2$ electrons in its $\pi$ bond, distributed across three $\text{N"-"O}$ bonds.
That means its $\boldsymbol{\pi}$-bond order is simply $\frac{1}{2} \cdot \left(\text{2 pi electrons")/("3 N"-"O bonds") = color(blue)("0.333}\right)$, making the bond order for each $\text{N"-"O}$
bond overall be:
${\text{BO" = "BO"_sigma + "BO}}_{\pi} = 1 + 0.333 = 1.333$.
Therefore, ${\text{NO}}_{3}^{-}$ on average actually has three "$\boldsymbol{\text{1.333}}$" bonds overall (instead of one double bond and two single bonds), meaning it is one third of the way
between a single bond and a double bond.
${\text{O}}_{2}$ actually has two singly-occupied ${\pi}^{\text{*}}$ antibonding orbitals.
If we were to calculate its bond order, we would get $2$ normally, corresponding to the #:stackrel(..)("O")=stackrel(..)"O":# Lewis structure.
But what if we wanted the bond order for ${\text{O}}_{2}^{+}$? From the discussion above, we may expect $1.5$, but it's NOT $1.5$, even though ${\text{O}}_{2}^{+}$ has one less valence electron. What
is it actually?
You may realize that we would have removed one electron from an ${\pi}^{\text{*}}$antibonding molecular orbital. That means we've removed half a bond order corresponding to antibonding character,
which is the same as adding half a bond order corresponding to bonding character.
So, by removing an antibonding electron, we've done the equivalent of adding a bonding electron.
In other words, we've decreased a bond-weakening factor, thereby increasing the bonding ability of the molecule.
Therefore, the actual bond order of ${\text{O}}_{2}^{+}$ is $\boldsymbol{2.5}$, stronger than ${\text{O}}_{2}$!
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Price range floating note using Cox-Ingersoll-Ross tree
Price a Range Floating-Rate Note Using a CIR Interest-Rate Tree
Create a RateSpec using the intenvset function.
Rates = [0.035; 0.042147; 0.047345; 0.052707];
Dates = {'Jan-1-2017'; 'Jan-1-2018'; 'Jan-1-2019'; 'Jan-1-2020'; 'Jan-1-2021'};
ValuationDate = 'Jan-1-2017';
EndDates = Dates(2:end)';
Compounding = 1;
RateSpec = intenvset('ValuationDate', ValuationDate, 'StartDates', ValuationDate, 'EndDates',EndDates,'Rates', Rates, 'Compounding', Compounding);
Create a CIR tree.
NumPeriods = length(EndDates);
Alpha = 0.03;
Theta = 0.02;
Sigma = 0.1;
Settle = '01-Jan-2017';
Maturity = '01-Jan-2020';
CIRTimeSpec = cirtimespec(Settle, Maturity, 3);
CIRVolSpec = cirvolspec(Sigma, Alpha, Theta);
CIRT = cirtree(CIRVolSpec, RateSpec, CIRTimeSpec)
CIRT = struct with fields:
FinObj: 'CIRFwdTree'
VolSpec: [1x1 struct]
TimeSpec: [1x1 struct]
RateSpec: [1x1 struct]
tObs: [0 1 2]
dObs: [736696 737061 737426]
FwdTree: {[1.0350] [1.0790 1.0500 1.0298] [1.1275 1.0887 1.0594 1.0390 1.0270]}
Connect: {[3x1 double] [3x3 double]}
Probs: {[3x1 double] [3x3 double]}
Define the range note instrument that matures in Jan-1-2014 and has the following RateSchedule:
Spread = 100;
Settle = 'Jan-1-2017';
Maturity = 'Jan-1-2020';
RateSched(1).Dates = {'Jan-1-2018'; 'Jan-1-2019' ; 'Jan-1-2020'};
RateSched(1).Rates = [0.045 0.055 ; 0.0525 0.0675; 0.06 0.08];
Compute the price of the range floating note.
[Price,PriceTree] = rangefloatbycir(CIRT,Spread,Settle,Maturity,RateSched)
PriceTree = struct with fields:
FinObj: 'CIRPriceTree'
PTree: {[91.6849] [88.9878 92.6039 95.1352] [88.6954 91.8547 94.3896 96.2429 97.3723] [100 100 100 100 100]}
AITree: {[0] [0 0 0] [0 0 0 0 0] [0 0 0 0 0]}
tObs: [0 1 2 3]
Connect: {[3x1 double] [3x3 double]}
Probs: {[3x1 double] [3x3 double]}
Input Arguments
CIRTree — Interest-rate tree structure
Interest-rate tree structure, specified by using cirtree.
Data Types: struct
Spread — Number of basis points over reference rate
Number of basis points over the reference rate, specified as a NINST-by-1 vector.
Data Types: double
Settle — Settlement date for floating range note
datetime array | string array | date character vector
Settlement date for the floating range note, specified as a NINST-by-1 vector using a datetime array, string array, or date character vectors. The Settle date for every range floating instrument is
set to the ValuationDate of the CIR tree. The floating range note argument Settle is ignored.
To support existing code, rangefloatbycir also accepts serial date numbers as inputs, but they are not recommended.
Maturity — Maturity date for floating range note
datetime array | string array | date character vector
Maturity date for the floating-rate note, specified as a NINST-by-1 vector using a datetime array, string array, or date character vectors.
To support existing code, rangefloatbycir also accepts serial date numbers as inputs, but they are not recommended.
RateSched — Range of rates within which cash flows are nonzero
Range of rates within which cash flows are nonzero, specified as a NINST-by-1 vector of structures. Each element of the structure array contains two fields:
• RateSched.Dates — NDates-by-1 cell array of dates corresponding to the range schedule.
• RateSched.Rates — NDates-by-2 array with the first column containing the lower bound of the range and the second column containing the upper bound of the range. Cash flow for date RateSched.Dates
(n) is nonzero for rates in the range RateSched.Rates(n,1) < Rate < RateSched.Rate (n,2).
Data Types: struct
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but
the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: [Price,PriceTree] = rangefloatbycir(CIRTree,Spread,Settle,Maturity,RateSched,'Reset',4,'Basis',5,'Principal',10000)
Reset — Frequency payment per year
1 (default) | numeric
Frequency of payments per year, specified as the comma-separated pair consisting of 'Reset' and a NINST-by-1 vector.
Payments on range floating notes are determined by the effective interest-rate between reset dates. If the reset period for a range spans more than one tree level, calculating the payment becomes
impossible due to the recombining nature of the tree. That is, the tree path connecting the two consecutive reset dates cannot be uniquely determined because there is more than one possible path for
connecting the two payment dates.
Data Types: double
Basis — Day-count basis of instrument
0 (actual/actual) (default) | integer from 0 to 13
Day-count basis representing the basis used when annualizing the input forward rate tree, specified as the comma-separated pair consisting of 'Basis' and a NINST-by-1 vector of integers.
• 0 = actual/actual
• 1 = 30/360 (SIA)
• 2 = actual/360
• 3 = actual/365
• 4 = 30/360 (PSA)
• 5 = 30/360 (ISDA)
• 6 = 30/360 (European)
• 7 = actual/365 (Japanese)
• 8 = actual/actual (ICMA)
• 9 = actual/360 (ICMA)
• 10 = actual/365 (ICMA)
• 11 = 30/360E (ICMA)
• 12 = actual/365 (ISDA)
• 13 = BUS/252
For more information, see Basis.
Data Types: double
Principal — Notional principal amount
100 (default) | numeric
Notional principal amount, specified as the comma-separated pair consisting of 'Principal' and a NINST-by-1 vector.
Data Types: double
EndMonthRule — End-of-month rule flag for generating caplet dates
1 (in effect) (default) | nonnegative integer with value 0 or 1
End-of-month rule flag, specified as the comma-separated pair consisting of 'EndMonthRule' and a nonnegative integer with a value of 0 or 1 using a NINST-by-1 vector.
• 0 = Ignore rule, meaning that a payment date is always the same numerical day of the month.
• 1 = Set rule on, meaning that a payment date is always the last actual day of the month.
Data Types: logical
Output Arguments
Price — Expected prices of range floating notes at time 0
Expected prices of the range floating notes at time 0, returned as a NINST-by-1 vector.
PriceTree — Tree structure of instrument prices
Tree structure of instrument prices, returned as a structure containing trees of vectors of instrument prices and accrued interest, and a vector of observation times for each node. Values are:
• PriceTree.PTree contains the clean prices.
• PriceTree.AITree contains the accrued interest.
• PriceTree.tObs contains the observation times.
• PriceTree.Connect contains the connectivity vectors. Each element in the cell array describes how nodes in that level connect to the next. For a given tree level, there are NumNodes elements in
the vector, and they contain the index of the node at the next level that the middle branch connects to. Subtracting 1 from that value indicates where the up-branch connects to, and adding 1
indicated where the down branch connects to.
• PriceTree.Probs contains the probability arrays. Each element of the cell array contains the up, middle, and down transition probabilities for each node of the level.
More About
Range Note
A range note is a structured (market-linked) security whose coupon rate is equal to the reference rate as long as the reference rate is within a certain range.
If the reference rate is outside of the range, the coupon rate is 0 for that period. This type of instrument entitles the holder to cash flows that depend on the level of some reference interest rate
and are floored to be positive. The note holder gets direct exposure to the reference rate. In return for the drawback that no interest is paid for the time the range is left, they offer higher
coupon rates than comparable standard products, vanilla floating notes. For more information, see Range Note.
[1] Cox, J., Ingersoll, J., and S. Ross. "A Theory of the Term Structure of Interest Rates." Econometrica. Vol. 53, 1985.
[2] Brigo, D. and F. Mercurio. Interest Rate Models - Theory and Practice. Springer Finance, 2006.
[3] Hirsa, A. Computational Methods in Finance. CRC Press, 2012.
[4] Nawalka, S., Soto, G., and N. Beliaeva. Dynamic Term Structure Modeling. Wiley, 2007.
[5] Nelson, D. and K. Ramaswamy. "Simple Binomial Processes as Diffusion Approximations in Financial Models." The Review of Financial Studies. Vol 3. 1990, pp. 393–430.
Version History
Introduced in R2018a
R2022b: Serial date numbers not recommended
Although rangefloatbycir supports serial date numbers, datetime values are recommended instead. The datetime data type provides flexible date and time formats, storage out to nanosecond precision,
and properties to account for time zones and daylight saving time.
To convert serial date numbers or text to datetime values, use the datetime function. For example:
t = datetime(738427.656845093,"ConvertFrom","datenum");
y = year(t)
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Exponential Function Applications And Graphs Worksheet Answers - Function Worksheets
Exponential Function Applications And Graphs Worksheet Answers
Exponential Function Applications And Graphs Worksheet Answers – If you’re in search of an activity in math for your child to help them practice exponential functions, then you’ve come to the right
spot. Exponential functions can be defined as numerical phrases where the importance changes tremendously in case you have a rise in the basic. Exponential progress can be a prime demonstration of
these kinds of functionality. Here are some good examples. Find out more about the statistical formula by seeking through our worksheets at no cost. We’ll also give some information for exponential
growth in our article. Exponential Function Applications And Graphs Worksheet Answers.
Types of exponential functions.
Exponents, often known as number beliefs are utilized to illustrate the functionality of a particular function. These capabilities aren’t actually the exact same form. In order to obtain the desired
results, they can be modified by multiplying or adding constants, however. Experiments with exponential operate on the worksheet could help in many circumstances, like learning variations in progress
and decrease every day. Here are some images of functions that happen to be exponential.
Suppose that the populace of a town is 20,000 at the moment developing by 15Per cent every year. 000 at present and the graph of the exponential function is able to accurately predict the population
of the town in 10 years if the town’s population is 20. It is apparent how the amount of men and women expand by 2 times inside that point time period which happens to be a growth of 15 % per year.
When you use exponential features in your calculations and you’ll have the ability to understand how they can be applied to actual-world situations.
Description of functions that are exponential
Just what is the main difference between 2 kinds of exponential functions? The distinction is in the character of the guidelines. If a value is increasing or decreasing exponentially, it’s considered
to be an exponential process. That is certainly, the necessity of y increases when by develops. Likewise, a horizontal shift is not really a modification of the positioning of the asymptote
horizontal. This is actually the basic understanding for exponential function. When you are aware the dissimilarities, it will be easy to resolve conditions that are based on exponential features.
Exponential functions are a very frequent kind mathematical characteristics which clarify the relationship in between parameters. If the base number is positive and the dependent variable is
positive, it will also grow exponentially. The property of exponential capabilities can be utilized in many disciplines, such as the natural sciences along with the interpersonal sciences. Think
about the self-reproducing population compound interest in an investment fund. Alternatively, the growing manufacturing capabilities. Exponentiation has lots of is and uses necessary to recognize in
the perspective of actual-daily life situations.
The explanation of graphs with exponentials
The design of charts with exponential attributes is a uncomplicated 1. Whatever functionality it is actually conveying the graph has got the exact same form: it is actually both a growing or lowering
process. The fundamental format of the exponential operate is the stage where the by-axis is near absolutely no. Its only difference is definitely the standard of slope that increases or reduces if
the x-worth is close to absolutely nothing or beneficial. The photo in the front page of your reserve reveals the fundamental form of an exponential contour.
To fully grasp the idea of an exponential, we have to understand the graph from the. On the whole, the graph shows an asymptote from the circumstance that this quantity is higher than absolutely
nothing. Also, it contains a top to bottom asymptote for the stage where an x-axis is absolutely nothing. The graph may not have an inter-by-intercept. It’s sometimes tough to plot an exponential
operate It’s therefore important to comprehend the math right behind it and the simplest way to interpret the results.
The explanation of your exponential improve
The expression “exponential development” signifies the surge in the quantity of data which is obtained over time which leads to an upwards-trending curve on charts. The rate of growth grows
exponentially if the power and exponent of a function are the same. For instance, if you add the equivalent of one rice grain onto the credit card at a 15 percent interest, your credit card debt will
increase by a factor of 4.7 years. But, this is nor terrible nor excellent.
The distinction involving exponential expansion and linear growth is indeed stark the concept was used within the open public record about COVID-19. COVID-19 pandemic. The exponential method is
tougher for people to fully grasp when compared with linear process. This is the reason pupils frequently overgeneralize linear methods to nonlinear kinds. In this article, we’ll discover how
exponential expansion characteristics and exactly how it pertains to our way of life generally. Hopefully you’ll have the capacity to comprehend the following meaning of exponential growth helpful!
Outline for exponential decay
In math, a explanation of exponential decay is an efficient way to help you students comprehend the character of the procedure. It clarifies how something boosts as well as lessens when time goes by.
The numerical formulation utilized to explain exponential decay can be described as y = a(1 (b) – b)by where a may be the authentic benefit which had been made, the decay aspect is b although x
symbolizes how long containing transferred. Contrary the linear procedure, exponential decay cuts down on the quantity in different ways as time passes. Which means that the decay component is
computed over a amount of the initial sum.
The item follows the exponential curve of decay if an object’s temperature gradually decreases over time. This very same principle is relevant to radioactive decay as well as the decay of
communities. It is vital that this pace of decay must reduce until finally it reaches an asymptote side to side to ensure that this procedure that need considering being exponential. One example is
the decline in radioactive particles over time. Additionally it is possible to notice the exponential rotting curve observing a warm item great to a constant background heat. Yet another illustration
is the release an electric powered capacitor through the opposition.
Gallery of Exponential Function Applications And Graphs Worksheet Answers
Graphing Exponential Functions Worksheet Answers Db excel
7 1 Graphing Exponential Functions Worksheet Answers Thekidsworksheet
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People rant and rave that the Congress can't keep the budget balanced. They blame political interests for causing the deficit. There is another force at work here. It is an unseen force, straight
from the world of probability mathematics:
Assume a 9-member city council, where each councilman gets an equal vote. Let there be $100,000 available to be spent, and suppose there are enough small programs vying for that money that they total
$120,000. Also assume that each councilman analyzes the available budget items before the meetings, and randomly decides what items to vote for and against. Let's look at a typical budget process,
and what goes wrong:
• Each councilman gets a list of these spending items. From the list, the councilman picks a list of favored items totaling $100,000 (a balanced budget).
• Each councilman votes for the items on his/her own list of favored items, and against the items not on his list.
• Each budget item comes up for a separate vote.
• Since only 5/6 of the items can be funded, the probability that any councilman votes for any one item is 5/6.
• More than 5/6 of the items pass, creating deficit spending.
How can this be? The answer is that the probability of an item passing is NOT 5/6.
Here is an analysis of the probability of a budget item passing the council. Since each councilman chooses randomly, the probability an item passes is found using the Binomial Distribution on a
simple majority vote:
SUM [C(9,x) * (5/6)^x * (1/6)^(9-x), for x = 5 to 9] = .99065.
Google the Binomial Distribution to understand this probability function.
Use the mean of a Binomial Distribution to find the number of items that pass. Multiplying the budget by the probability of passing approximates this. If each item has a .99065 probability of
passing, then the budget will be overspent, because $120,000 times .99065 is $118,878, or $18,878 over budget.
Another way to see this is to look at the votes in a table: Green values are within budget. Red values are over budget:
PROJECTS VOTED
Proj # 1 2 3 4 5 6 7 8 TOTALS
Cost: $20000 $20000 $20000 $20000 $10000 $10000 $10000 $10000 $120000
1 no YES YES YES YES YES YES YES $100000
2 YES YES YES YES YES no YES no $100000
3 YES no YES YES YES YES YES YES $100000
COUNCIL 4 YES YES no YES YES YES YES YES $100000
MEMBERS 5 YES YES YES YES no YES YES no $100000
6 YES YES YES no YES YES YES YES $100000
7 no YES YES YES YES YES YES YES $100000
8 YES YES YES YES YES YES no no $100000
9 YES YES YES YES no YES YES no $100000
Yes Votes: 7 of 9 8 of 9 8 of 9 8 of 9 7 of 9 8 of 9 8 of 9 5 of 9 8 pass
Budgeted: $20000 $20000 $20000 $20000 $10000 $10000 $10000 $10000 $120000
Even though each councilman planned for and voted for a balanced budget, the government overspent. Hence, these conjectures follow:
ALL of these conjectures assume a simple majority vote is taken on each budget item in sequence:
• SPENDING CONJECTURE #1
If the amount of proposed spending is between 1 and 2 times the available revenue, and if all legislators plan all votes in advance for a balanced budget, then deficit spending will occur.
Notice here that the probability of a "yes" vote is between .5 and 1.
• SPENDING CONJECTURE #2
If the amount of proposed spending is between 1 and 2 times the available revenue, then the first items voted on have a higher probability of passing, if all legislators change votes to ensure a
balanced budget.
Note that as the budget fills up, the probability of "yes" votes goes down.
• SPENDING CONJECTURE #3
When the amount of proposed spending is between 1 and 2 times the available revenue, the proportion of the budget overspent increases as the number of legislators increases.
The Binomial Distribution becomes less varied as the number of trials increases.
• SPENDING CONJECTURE #4
When the amount of proposed spending is over twice the available revenue, almost no spending will occur, if all legislators plan all votes in advance for a balanced budget.
Notice here that the probability of a "yes" vote is between 0 and .5. (Of course, this works only until they notice that this happened.)
• SPENDING CONJECTURE #5
When the amount of proposed spending is over twice the available revenue, the last items voted on have a higher probability of passing, if all legislators change votes to ensure a balanced
Note that as the budget does not fill up, the probability of a "yes" vote goes up.
• SPENDING CONJECTURE #6
When the amount of proposed spending is less than the available revenue, all legislators will add projects to the budget until the amount of proposed spending is more than the available revenue.
Here, the probability of a "yes" vote is exactly 1.
• SPENDING CONJECTURE #7
When a bicameral (two house) legislature votes on each spending item separately, the probability the item is included in the budget is the product of the probabilities of the item passing each
This is the original intent of the US Constitution.
In the above example, if both houses have 9 members, then each house will have a probability of passing the item of .99065. So both houses together will have a probability of passing the item of:
.99065 * .99065 = .9813874225
This is better, but still way too high.
If an omnibus spending bill is prepared in advance by a committee, the same conjectures apply, but the probability of overspending is even greater in a bicameral legislature.
Assume the same conditions as above, that the two houses have 9 members each, and that each legislator again makes up his list of items to vote for in advance. Then the probability a legislator
favors a particular spending item is again 5/6, and the probability that the lower house favors an item is:
SUM [C(9,x) * (5/6)^x * (1/6)^(9-x), for x = 5 to 9] = .99065.
So the probability that the lower house does NOT favor the item is .00935
(The probability of failure is 1 minus the probability of success.)
The probability that the upper house favors the item is also:
SUM [C(9,x) * (5/6)^x * (1/6)^(9-x), for x = 5 to 9] = .99065.
Again, the probability that the upper house does NOT favor the item is .00935
With an omnibus budget bill, BOTH houses must vote to AMEND the bill to remove the item, before it can be removed. This does NOT agree with the method prescribed in the Constitution. Only once
outcome, instead of three, removes the bill. So the probability of removing a spending item is even lower:
.00935 * .00935 = .0000874225
The probability of the item passing is therefore:
1 - .0000874225 = .9999125775
This is even worse than any of the cases above.
• SPENDING CONJECTURE #8
An omnibus spending bill in a monocameral (single house) legislature has the same spending probabilities that voting on individual spending items has.
The only real difference is that the house votes to remove an item rather than to add it.
• SPENDING CONJECTURE #9
When a bicameral (two house) legislature votes on spending through an omnibus spending bill each spending item has a smaller chance of being removed.
Here the probability of removing the bill is lower, being the product of the probabilities of the item removal passing each house. This is definitely cheating the original intent of the US
The omnibus spending bill violates constitutional intent.
First, the case of the constitutional method. Assume the following:
1. A bicameral legislature with each house larger than 40 members (so the normal distribution applies).
2. The bill is for an individual spending item.
3. Each legislator, and the executive, flips a coin to decide how to vote (an even chance of voting for or against).
Then the probability that the bill passes is shown by the following table:
First House Second House Executive Probabilities Result Spending
.05 Supermajority (2/3) .50 Signed .00125 Pass
.50 Vetoed .00125 Pass Override
.05 Supermajority (2/3) .45 Majority .50 Signed .01125 Pass Probability
.50 Vetoed .01125 Fail Vetoed of spending:
.50 Defeated No Action .025 Fail Defeated
.50 Majority Signed .1125 Pass .12625
.45 Majority Vetoed .1125 Fail Vetoed
.50 Defeated No Action .225 Fail Defeated
.50 Defeated No Action No Action .5 Fail Defeated
With a probability of .12625 using the constitutional method, legislators have to really want to spend the money to spend it.
If the President always passes the budget, the problem simplifies to:
First House Second House Executive Probabilities Result Spending
.50 majority 1.0 Signed .25 Pass (spent)
.50 Majority Probability
.50 Defeated 1.0 Signed .25 Fail (not spent) of spending:
.50 Defeated No Action 1.0 Signed .5 Fail (not spent) .25
With a probability of .25 legislators still have to really want to spend the money to be able to spend it.
Now, the case of the omnibus budget bill method. Assume the following:
1. A bicameral legislature with each house larger than 40 members (so the normal distribution applies).
2. The vote is on removing an item from the omnibus budget bill.
3. Each legislator flips a coin to decide how to vote (an even chance of voting for or against).
4. It is assumed that the executive signs the resulting budget bill, so no veto override is needed.
Then the probability that the money is spent on the item is shown by the following table:
First House Second House Executive Probabilities Result Spending
.50 majority 1.0 Signed .25 Pass (not spent)
.50 Majority Probability
.50 Defeated 1.0 Signed .25 Fail (spent) of spending:
.50 Defeated No Action 1.0 Signed .5 Fail (spent) .75
With the .75 probability of spending with the omnibus budget method, legislators must really want to not spend the money to not spend it.
• SPENDING CONJECTURE #10
An omnibus spending bill by itself causes overspending.
The omnibus spending method is a sneaky way to bypass constitutional limitations on spending. It gives the committee writing the bill the power to overspend.
This entire section shows that no legislature can control spending by itself, without special processes to prevent spending into debt.
Now that we know the problem exists, what can be done about it? We know that, to get a balanced budget, these things must be avoided:
1. Avoid sequential voting on budget items.
2. Legislators must not stick to plans made in advance, unless a system to prevent overspending is in place.
3. Simple majority votes do not work without a system to prevent overspending.
4. Omnibus spending bills must be avoided.
Here are some suggestions on what to do, with good and bad points of each:
Use the Independent Voting System, a fair election method, with budget ballots in the legislature. Put ALL budget items on this ballot. The vote on all items must be simultaneous. Each legislator
marks a ballot, putting YES, ABSTAIN, or NO for each budget item. Each ballot is signed by its legislator, and is kept secret until the vote is completed. Then the vote is tallied, and the votes
cast by each legislator are released to the press. The score for each budget item is its YES tally minus its NO tally.
Then use at least one of the following methods:
1. INDEPENDENT VOTING WITH REDUCTION FACTOR:
1. Take all of the budget items with positive scores.
2. If this results in a deficit, divide the revenue available by the total amount of spending approved by the ballot, giving the Reduction Factor. If there is no deficit, the Reduction
Factor is 1.
3. Multiply each item's original amount by the Reduction Factor, reducing all items proportionally.
☆ Advantage: All favored items are approved.
☆ Disadvantage: All items have reduced funding if a deficit would occur.
2. HIGHEST INDEPENDENT VOTES:
1. Sort the budget items according to score.
2. Taking the highest scores first, go down the sorted list, adding in each item, until the next item either has a negative score, or would cause a deficit.
☆ Advantage: All approved items are fully funded.
☆ Disadvantage: Items favored by a majority are sometimes not approved.
• ADDITIONAL METHODS
The following methods can add insurance that a deficit will not occur:
1. LOSS OF SALARY:
1. Make legislators lose all of their pay if there is a deficit.
2. This makes them watch the amount already appropriated.
☆ Advantage: They will actively work for a balanced budget.
☆ Disadvantage is that items presented first will have a higher chance of being passed, compared to items presented later, unless some kind of balancing process is used.
2. TAX USE RESTRICTION:
1. Restrict use of tax revenue to only necessary items.
2. Fair Taxation reduces the chance that a deficit budget will be approved.
☆ Advantage: Other sources of funding must be found for unnecessary government spending. I suggest little cans at supermarket checkout stands, user fees, or donations.
☆ Advantage: This goes a long way to cutting out "taxic waste."
☆ Disadvantage: Some idiots think funding for the arts, sports, night basketball, entertainment, parades, and professional sports domes are "necessary."
3. SPECIAL DEALS UNCONSTITUTIONAL:
1. Make it unconstitutional for governments to offer special debt-finance deals that are illegal for private business to offer.
2. Government has to pay just as much to borrow money as private individuals do.
☆ Advantage: It makes debt harder to finance.
☆ Disadvantage: It does nothing to prevent heedless legislators from voting for a debt that can't be financed.
4. LINE-ITEM VETO
1. Give the executive a line-item veto.
2. If the legislature overspends, it gives the executive the power to decide which items to veto.
☆ Advantage: The executive has the power to balance the budget.
☆ Disadvantage: It does nothing to prevent the executive from allowing debt.
☆ Disadvantage: The executive gets too much power.
☆ Disadvantage: The legislature can override the vetoes.
Using these measures can once again get our government spending under control. | {"url":"http://midimagic.sgc-hosting.com/defcause.htm","timestamp":"2024-11-07T15:25:21Z","content_type":"application/xhtml+xml","content_length":"21346","record_id":"<urn:uuid:41d44d4c-562b-49c9-9d54-ac123ef07053>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00246.warc.gz"} |
Welcome to OpenBook PageWelcome to OpenBook Page
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OpenBook Sample Size Calculators
In order to generalize from a random sample and avoid sampling errors or biases, a random sample needs to be of adequate size. What is adequate depends on several issues which often confuse people
doing surveys for the first time. This is because what is important here is not the proportion of the research population that gets sampled, but the absolute size of the sample selected relative to
the complexity of the population, the aims of the researcher and the kinds of statistical manipulation that will be used in data analysis (Taherdoost, 2016). While the larger the sample the lesser
the likelihood that findings will be biased does hold, diminishing returns can quickly set in when samples get over a specific size which need to be balanced against the researcher’s resources (Gill
et al., 2010).
To put it bluntly, larger sample sizes reduce sampling error but at a decreasing rate. Several statistical formulas or calculators are available for determining sample size.
Gill, J. Johnson, P. & Clark, M. (2010). Research Methods for Managers. Sage Publications
Taherdoost, H. (2016). Sampling Methods in Research Methodology; How to Choose a Sampling Technique for Research. International Journal of Advance Research in Management, 5(2), 18-27
OpenBook Yamane Calculator (OY Calculator, 2022)
is a free online sample size calculator developed to help research students or other researchers from various fields worldwide, having problem in calculation using Taro Yamane Formula manually,
determine sample size accurately.
Taro Yamane (1967) Formula is written as
n = N / (1 + Ne^2) simplified and adjusted, to be more accurate, from Cochran’s (1963, 1975) Sample Size Formula:
• n = Number of Samples,
• N = Total Population,
• e = Error Tolerance (level) or Margin of Error, 0.05
• p = Sample Proportion, 0.5
• z = z-value found in Z-score Table, 1.96
The table below is Z-score Table for most use confidence level or confidence interval.
Confidence Level Confidence Interval Area between zero and z-score Z-score
90% 0.10 90/2%=0.4500 1.65
95% 0.05 95/2%=0.4750 1.96
99% 0.01 99/2%=0.4950 2.58
By substituting for z = 1.96 and p = 0.5 in the simplified formula above, Taro Yamane Formula can be proved as follows:
e could be 0.10, 0.05 or 0.01. They are margin of errors, that can be tolerated in determining sample size, at confidence level of 90%, 95% and 99% respectively. They are used in educational and
social science research studies. The most commonly and widely used is 0.05. The sample proportion p, though varies, but by default is 0.5. If you are not familiar with confidence level, confidence
interval or margin of error and sample proportion – the common terms in sample size and calculation, you can click here.
One of the advantages of using OpenBook Yamane Calculator, to accurately determine sample size, is when the total population is relatively large. Secondly, you don’t need to crack your brain of any
complex formula and all its variables’ values especially if you don’t have relevant knowledge in statistics. Other advantage is that as large sample size reduces sampling error to validate research
findings, there are always excess samples of 16 or 15 at total population of 300,000 and above, when compared with other Sample Size Calculators, which is enough to gather much more information or
data, from the respondents, about a study. To prove this, the highest sample you would ever get using OpenBook Yamane Calculator from total population of 300,000 and above is 400 and the highest
sample you would ever get using other Sample Size Calculators from also 300,000 total population and above is 384 or 385. So, there are always excess samples of 16 or 15 to get much more information
about a study using OpenBook Yamane Calculator.
However, the population N is to be determined first from the study area. When the population is relatively large and the exact number is unknown, then 300,000 or more can be used because any sample
size gotten cannot be greater than 400 or 385 (either Taro Yamane Formula at confidence interval of 0.05 or other Sample Size Formula at confidence level of 95%, confidence interval of 5% and sample
proportion of 50%).
To calculate the Sample Size n, using OY Calculator below: enter the Total Population N, then calculate by clicking on Calculate Button. To enter different Total Population N, click Reset Button.
Using OY Calculator, you may also change the default 0.05 in the margin of error e placeholder to your desired confidence interval by selecting either 0.10 or 0.01 as alternate scenario. The 300,000
in the right field of population N placeholder is to be used when the exact number is unknown at confidence interval of 0.05. The common usage of 0.05 confidence interval for a specific sample size
result is to bring balance against the researcher’s resources relative to the complexity of the population. Statistically, large samples must be equal to or greater than 30 (Murray, 2009). As sample
size is used to validate research findings, it must not be too small. If too small, it will not yield valid results. At the same time, if it’s too large, may be a waste of money and time.
Murray, R. Spiegel et al (2009). Probability and Statistics. The McGraw-Hill Companies Inc.
The use of OpenBook Yamane Calculator can be referenced in your thesis or dissertation as:
OpenBook Yamane Calculator, 2022. OpenBook Sample Size Calculators. OpenBook Communications and Technologies, Nigeria. https://www.openbookpage.com/
OpenBook Cochran Calculator (OC Calculator, 2022) and OpenBook Cochran Correction Calculator (OCC Calculator, 2024)
(From Cochran’s Sample Size Formula without Adjustment or Modification of Z-score z, 1.96 at confidence interval or margin error e, 0.05 and sample proportion p, 0.5)
To calculate the Sample Size n, using OC/OCC Calculator below: enter the Total Population N, then calculate by clicking on Calculate Button. To enter different Total Population N, click Reset Button.
Using OC/OCC Calculator, you may also change the default 1.96 and 0.05 in their respective placeholder to your desired confidence level of either 90% or 99% by selecting 1.65 or 2.58 and 0.10 or
0.01 respectively as alternate scenarios. The 300,000 in the right field of population N placeholder is to be used when the exact number is unknown at confidence interval of 0.05 and z-score value of
1.96. The common usage of 0.05 confidence interval and 1.96 z-score value for a specific sample size result is to bring balance against the researcher’s resources relative to the complexity of the
population. Statistically, large samples must be equal to or greater than 30 (Murray, 2009). As sample size is used to validate research findings, it must not be too small. If too small, it will not
yield valid results. At the same time, if it’s too large, may be a waste of money and time.
Murray, R. Spiegel et al (2009). Probability and Statistics. The McGraw-Hill Companies Inc.
OpenBook Cochran Correction Calculator, 2024 (OCC Calculator, 2024)
Sample Size can also be calculated using Cochran Sample Size Formula with the application of FINITE POPULATION CORRECTIONs (FPCs)
Cochran (1963, 1975) developed the equation to yield a representative sample for proportion of large sample.
n[0] = z² pq/e²
which is valid where n[0] is the sample size, z² is the abscissa of the normal curve that cuts off an area α at the tails (1 – α equals the desired confidence level is 95%), e is the desired level of
precision, p is the estimated proportion of an attribute that is present in the population, and q is 1-p. The value for z is found in statistical tables which contain the area under the normal curve.
Finite Population Correction for Proportions (If small population). If the population is small then the sample size can be reduced slightly. This is because a given sample size provides
proportionately more information for a small population than for a large population. The sample size (n[0]) can be adjusted as
n = n[0] / [1 + {(n[0] – 1) / N}]
where n is the sample size and N is the population size
Cochran (1977) introduced Finite Population Corrections (FPCs) based on the sampling fraction f = n/N where n is the sample size and N is the finite population size. In practice, FPCs may be ignored
if f does not exceed 5%. Larger samples relative to their populations require FPCs because ignoring large sampling fractions results in biased standard errors (Cochran, 1977). Applied researchers
should identify their target populations, examine their sampling fraction, and consider using FPCs because applying FPCs yields more accurate inferences for finite populations.
The use of OpenBook Cochran Calculator can be referenced in your thesis or dissertation as:
OpenBook Cochran Calculator, 2022. OpenBook Sample Size Calculators. OpenBook Communications and Technologies, Nigeria. https://www.openbookpage.com/
OpenBook Krejcie-Morgan Calculator (OK-M Calculator, 2024)
(From Krejcie-Morgan Sample Size Formula without Adjustment or Modification of Chi-Square x², 3.841 at confidence interval or margin of error e, 0.05 and sample proportion p, 0.5)
Krejcie and Morgan (1970) Formula was introduced as an alternative formula in computing sample size for categorical data. The formula is written as:
n = x²Np(1-p)/e²(N-1)+x²p(1-p)
• n = Number of Samples,
• N = Total Population,
• e = Error Tolerance (level) or Margin of Error, 0.05
• p = Sample Proportion, 0.5
• x² = Chi-Square value found in Chi-Square Table, 3.841
The table below is Chi-Square Table for most use confidence level or confidence interval.
x²[0.][90] x²[0][.95] x²[0][.99]
Degree of Freedom 10% (0.10) 5% (0.05) 1% (0.01)
1 2.706 3.841 6.635
OpenBook Krejcie-Morgan Calculator can be used as an alternative tool to confirm the cases of population and sample size not listed in Krejcie and Morgan Sample Size Table (1970), a well known table
for sample size determination among behavioural and social science researchers.
To calculate the Sample Size n, using OK-M Calculator below: enter the Total Population N, then calculate by clicking on Calculate Button. To enter different Total Population N, click Reset Button.
Using OK-M Calculator, you may also change the default 3.841 and 0.05 in their respective placeholder to your desired confidence level of either 90% or 99% by selecting 2.706 or 6.635 and 0.10 or
0.01 respectively as alternate scenarios. The 300,000 in the right field of population N placeholder is to be used when the exact number is unknown at confidence interval of 0.05 and chi-square x²
value of 3.841. The common usage of 0.05 confidence interval and 3.841 chi-square x² value for a specific sample size result is to bring balance against the researcher’s resources relative to the
complexity of the population. Statistically, large samples must be equal to or greater than 30 (Murray, 2009). As sample size is used to validate research findings, it must not be too small. If too
small, it will not yield valid results. At the same time, if it’s too large, may be a waste of money and time.
Murray, R. Spiegel et al (2009). Probability and Statistics. The McGraw-Hill Companies Inc.
The use of OpenBook Krejcie-Morgan Calculator can be referenced in your thesis or dissertation as:
OpenBook Krejcie-Morgan Calculator, 2024. OpenBook Sample Size Calculators. OpenBook Communications and Technologies, Nigeria. https://www.openbookpage.com/
Gotten Your Sample Size from OY/OC/OK-M Calculator, What’s Next?
Now that you have gotten your sample size, from any of the calculators above, for the number of copies of your questionnaires, to be administered to your respondents, let OpenBook have the
Questionnaire Reliability Checking and SPSS Data Analysis done for you. SPSS Data Analysis using Descriptive and Inferential Statistics.
Questionnaire Reliability Checking (Cronbach Alpha) is EXCLUSIVELY FREE! Distance is not a barrier; you can upload your already ticked FORM/QUESTIONNAIRE to whatsapp number: 2348028999115.
Service Price of SSPS Data Analysis ranges from 15,000 to 70,000 in Nigeria Naira (NGN) and 100 to 470 in US Dollar (USD).
BSc/BA Research Data Analysis: NGN15,000/USD100.
MSc/MA Research Data Analysis: NGN30,000/USD200.
PhD Research Data Analysis: NGN70,000/USD470.
You can make payment by clicking here.
For more information, you can message us through whatsapp number: 2348028999115 or click
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Data Analysis, Statistical Calculators, Statistical Formulas, Taro Yamane Formula, Yamane Calculator, Yamane Formula | {"url":"https://openbookpage.com/uncategorized/welcome-to-openbook-page/55/","timestamp":"2024-11-07T16:23:17Z","content_type":"text/html","content_length":"183428","record_id":"<urn:uuid:84e13383-0c48-4300-a35c-af111d05c216>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00877.warc.gz"} |
Quote:Patrik O. wrote:I could
08-26-2015 07:36 AM
I'm trying to refactor a piece of code that solves a simple Ax=b linear problem using MKL. From the manual, I saw that dgesv is the function I should call.
A is an n-by-n Jacobian matrix (with just 5 non-zero diagonals), while b and x are arrays of size n.
To solve the problem I use this snippet of code
MKL_INT n = matrix_size;
MKL_INT nrhs = 1;
MKL_INT info;
MKL_INT *pivot = calloc(matrix_size, sizeof(MKL_INT));
dgesv(&n, &nrhs, A, &n, pivot, b, &n, &info);
My problem is that the solution (stored in b) is completely different from my reference result as computed by the old code (a custom and unmaintainable solver). At some point I thought that I
introduced an error in the computation of A or b, but if I store these two guys on disk, load them up into MATLAB and solve the problem with A\b, the result is consistent with the old solver.
Am I missing something?
08-26-2015 09:18 AM
08-26-2015 10:27 AM
08-26-2015 10:51 AM
08-28-2015 02:55 AM
08-28-2015 09:45 AM
08-29-2015 05:20 AM
08-29-2015 05:53 AM
08-29-2015 09:21 AM
08-31-2015 01:00 AM
08-31-2015 03:13 AM
08-31-2015 07:46 AM
08-31-2015 07:57 AM
09-02-2015 01:05 AM | {"url":"https://community.intel.com/t5/Intel-oneAPI-Math-Kernel-Library/Odd-dgesv-results-Now-with-pardiso/m-p/998127/highlight/true","timestamp":"2024-11-03T23:34:22Z","content_type":"text/html","content_length":"470409","record_id":"<urn:uuid:e0b80c87-1826-4a43-9f3c-22bf9e28200d>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00105.warc.gz"} |
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Binary Search Algorithm - Scaler Blog
Binary search is an efficient algorithm for finding an item from a sorted list of items. It works by repeatedly dividing in half the portion of the list that could contain the item, until you’ve
narrowed down the possible locations to just one.
• Complexity of Binary search algorithm
□ Time complexity – O(logn)
□ Space complexity – O(1)
This is the second part of a three-part series on Searching Algorithms.
In the previous article, Linear Search, we discussed why searching is so important in everyday life and discussed the most basic approach to perform searching; Linear Search. If you didn’t get a
chance to read that one, I’d highly recommend jumping right to it and have a quick read.
Today, we will be discussing an approach which frankly is personally my favorite algorithm of all time; Binary Search.
A quick recap on the definition of searching.
Searching is a method to find some relevant information in a data set.
While discussing the Linear Search algorithm, we mentioned that it is not very efficient as in most cases, as it ends up searching the entire array to find an element.
Since it wasn’t the most practical approach for large data sets, we desperately needed an improvement over it and the programming geniuses worked hard to find one. And they did 🙂
In a galaxy very very near, Binary Search was born.
Binary Search
Binary Search is an algorithm that can be used to search an element in a sorted data set. By sorted, we mean that the elements will either be in a natural increasing or decreasing order.
Natural order is the default ordering of elements. For example, integers’ natural increasing order is 1, 2, 3, 4, … which is the increasing order of their value.
A: 5 3 12 38 9 45 1 22 3
B: 1 3 3 5 9 12 22 38 45
In the above example, the array A is not sorted as the elements are not in either natural increasing or decreasing order of value. While the array B is sorted in natural increasing order.
In a nutshell, we can say that an array of integers is sorted if
arr[i-1] <= arr[i] <= arr[i+1], where 0 < i < size – 1
where size is the number of elements in the array.
Similarly for strings, in programming, the natural order is lexicographic, or simply, how you would find them in an English dictionary.
For example, “ant” will be present before “art” in an English dictionary. Thus, “ant” is said to be lexicographically smaller than “art”.
Let’s say we’re trying to search for a number, K, in a sorted array of integers. For simplicity, let’s assume the array is sorted in natural increasing order.
Let’s say the array is
And the element we’re searching for, K = 12
If we apply Linear Search, we will always start with the first element until we either find K or reach the end of the array.
But since the array is sorted here, can we use that to our advantage?
Let’s make an acute observation.
In a sorted array, for any index i, where 0 <= i < size and size is the number of elements in the array
if arr[i] < K
It means the element at index i is less than K. Hence, all the elements before the index i will also be less than K right, since the array is sorted. Therefore, we can choose to ignore the elements
on the left side of the index i and not compare them with K at all.
For example, if i = 2, arr[2] = 5. Hence, arr[2] < K, so we ignore the elements on the left side of index i, i.e. 1 and 3 since they will also be less than K and we will never find 12 on that side.
1 3 3 5 9 12 22 38 45 K = 12
[DEL:1 3 3:DEL] 5 9 12 22 38 45 K = 12
if arr[i] > K
It means the element at index i is greater than K. Hence, all the elements after the index i will also be greater than K right, because of the sorted nature of the array. Thus, we can choose to
ignore the elements on the right side of index i and not compare them with K at all.
For example, if i = 5, arr[5] = 12. Hence, arr[5] > K, so we ignore the elements on the right side of index i, i.e. 22, 38 and 45 since they will also be greater than K and we will never find 12 on
that side.
1 3 3 5 9 12 22 38 45 K = 11
1 3 3 5 9 [DEL:12 22 38 45:DEL] K = 11
That means if we start searching at any index i, and check if the element at that index is either less than or greater than K, we can choose to ignore a particular set of elements present on the left
or the right side of that index respectively.
On the other hand,
if arr[i] == K
It means that the element at index i is equal to K and we have found the element K in the array. Hence, we do not need to search any further and our job is done.
If we choose the right position to start searching, the above observation can reduce the number of elements we need to search by quite a big margin.
Now comes the problem of choosing the most optimal position to start the search from. That’s where the term ”Binary” comes into the picture.
Unary means one. Binary means two.
If you’re thinking about starting at the middle of the array, you’re absolutely 100% right.
Binary Search works on the principle of Divide and Conquer. It divides the array into two halves and tries to find the element K in one or the other half, not both. It keeps on doing this until we
either find the element or the array is exhausted.
We start the search at the middle of the array, and divide the array into binary, or two parts. If the middle element is less than K, we ignore the left half and apply the same technique on the right
half of the array until we either find K or the array cannot be split any further.
Similarly, if the middle element is greater than K, we ignore the right half and apply the same technique on the left half of the array until we either find K or the array cannot be split any
Let’s try to apply this approach for the above example.
The size of the above array is, size = 9
Thus, the middle element will be at index i = (9 – 1) / 2 = 4
We will start with the middle element, at index i =4 and K = 12
1 3 3 5 9 12 22 38 45 9 < 12, ignore left half
[DEL:1 3 3 5 9:DEL] 12 22 38 45 22 > 12, ignore right half
[DEL:1 3 3 5 9:DEL] 12 [DEL:22 38 4:DEL] 12 == 12, found K
As you can see above, we found K = 12 in just the third comparison. This is better than the linear search approach where it would’ve taken five comparisons.
Let’s try another example where K might not be present in the list, K = 4.
1 3 3 5 9 12 22 38 45 9 > 4, ignore right half
1 3 3 5 [DEL:9 12 22 38 45:DEL] 3 < 4, ignore left half
[DEL:1 3:DEL] 3 5 [DEL:9 12 22 38 45:DEL] 3 < 4, ignore left half
[DEL:1 3 3:DEL] 5 [DEL:9 12 22 38 45:DEL] 5 > 4, ignore right half
[DEL:1 3 3 5 9 12 22 38 45:DEL] no more elements left to compare
As you can see, we could not find K since it was not present in the array. But we were able to determine this is just 4 comparisons, while linear search would’ve compared all the elements of the
You: But Rahul, this just took 5 less comparisons. All this effort for a measly 5 comparisons :/
Rahul: Yes for the above example, it might look like it’s not a very big optimization over linear search. But that’s because the number of elements in the array is quite less, very less to be
precise. The real power of binary search can be seen when the array contains millions of elements.
You: How much exactly? Like 10 comparisons less than linear search? 😛
Rahul: Haha. Nice one. Keep reading to find out 🙂
Below is the algorithm of Binary Search.
1. Initialise n = size of array, low = 0, high = n-1. We will use low and high to determine the left and right ends of the array in which we will be searching at any given time
2. if low > high, it means we cannot split the array any further and we could not find K. We return -1 to signify that the element K was not found
3. else low <= high, which means we will split the array between low and high into two halves as follows:
□ Initialise mid = (low + high) / 2, in this way we split the array into two halves with arr[mid] as its middle element
□ if arr[mid] < K, it means the middle element is less than K. Thus, all the elements on the left side of the mid will also be less than K. Hence, we repeat step 2 for the right side of mid. We
do this by setting the value of low = mid+1, which means we are ignoring all the elements from low to mid and shifting the left end of the array to mid+1
□ if arr[mid] > K, it means the middle element is greater than K. Thus, all the elements on the right side of the mid will also be greater than K. Hence, we repeat step 2 for the left side of
mid. We do this by setting the value of high = mid-1, which means we are ignoring all the elements from mid to high and shifting the right end of the array to mid-1
□ else arr[mid] == K, which means the middle element is equal to K and we have found the element K. Hence, we do not need to search anymore. We return mid directly from here signifying that mid
is the index at which K was found in the array
Below is an implementation of Binary Search in Java 8 –
public static void main(String[] args) {
int[] array = new int[] { 1, 3, 5, 9, 12, 22, 38, 45 };
int K = 22;
int res = binarySearch(array, K);
if (res >= 0) {
System.out.println(K + " found at index: " + res);
} else {
System.out.println(K + " not found");
private static int binarySearch(int[] array, int K) {
int n = array.length;
int low = 0;
int high = n - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
// think: why not use (low + high) / 2 ?
if (array[mid] < K) {
low = mid + 1;
} else if (array[mid] > K) {
high = mid - 1;
else {
// found K
return mid;
return -1;
22 found at index: 5
Time Complexity
Since we always start with the middle element first, in the best case, it’s possible that the middle element of the array itself is the element we’re searching for, K.
On the other hand, if the element we’re searching for is not present in the array, we keep splitting the array into two halves until it cannot be split any further, i.e. only one element is left. So
what is the complexity of this?
Let’s solve this with an example. Assume the size of the array is N = 8.
• In the first step, the array of size 8 is split in the middle into two parts of size 4 each.
1 2 3 4 | 5 6 7 8
• In the second step, either of the two arrays of size 4 is split in the middle into two parts of size 2 each.
1 2 | 3 4 or 5 6 | 7 8
• In the third step, either of the four arrays of size 2 is split in the middle into two parts of size 1 each.
1 | 2 or 3 | 4 or 5 | 6 or 7 | 8
Now, we will be left with only one element which cannot be split any further.
If you notice, at each step, we are discarding one half of the array, until we either find the element or there are no more elements left to search in.
So we conclude that the worst-case time complexity will be the total number of steps in which we can split the array into two halves until it cannot be split any further.
In the above example, it took 3 steps to split the array into two halves until only one element was left.
Let’s try to generalize this. Did you notice something?
N = 8 = 2^3
Taking logarithm with (log2^2) on both sides,
log[2]N= log[2]8 = log[2]2^3= 3log[2]2
Since log[2]2 = 1,
log[2]N= 3, total number of steps to split N into 2 halves until it’s equal to 1
This is exactly like the definition of logarithm: log[base]x gives us the value of the exponent to which we must raise the base (2 in our case) to produce x (which is the size of the array in our
Hence, for an array of size N, it will take log[2]N steps to split it into two halves until only one element is left.
Since the length of the array was a power of 2 in the above example, the value of log[2]N is an integer. But if it is not a power of 2, log[2]N will be a decimal value. Hence, we can take the ceiling
value of log[2]N in general.
The ceiling value of any number x is the minimum integer value that is greater than x. For example, the ceiling value of 2.5 will be 3.
By applying some basic high school mathematics here, we were able to conclude that for an array of size N, the worst-case time complexity of binary search will be O(ceil(log[2]N)).
To give you some perspective, if N = 100000 (105105), the worst-case complexity of linear search will be O(N) = O(10^5).
Not that good right?
On the other hand, the worst-case complexity of binary search will be
That means the binary search will be able to determine whether an element is present in an array or not in just 17 steps. It’s almost 6000 times better than a linear search.
Mind-blowing right? 😛
Space Complexity
As we can see we are not using any significant extra memory in Binary Search. So the space complexity is constant, i.e. O(1).
P.S. The variables used for storing the bounds, middle index and other minor information are constant since they don’t depend on the input size of the array.
Applications of Binary Search
If the array is sorted, binary search is the clear winner. It takes O(log[2]N) in the worst case while linear search will take O(N) in the worst case.
You: But what if the array is not sorted, Rahul? We won’t be able to use this awesome algorithm after all.
Rahul: Yes you are right. If the array is not sorted, we won’t be able to use binary search directly. Yikes! We can always sort the array and then apply binary search 😀
You: But sorting itself might take O(N*O(log[2]N)) and then an additional O(log[2]N)for binary search. This is even worse than linear search .
Rahul: I’m glad you mentioned that 🙂
You’re absolutely right though. If we have to sort the array, then the total time complexity of sorting and binary search will be even more than that of linear search. In that case, a linear search
might be the best solution.
But if you are given an array and you have to perform multiple, let’s say Q search queries on it. Then binary search is your best friend.
Because once we sort the array, then we can perform all the queries one by one on the sorted array. And since, each binary search query will only take O(log[2]N) time in the worst case, for Q queries
it will take *O(Qlog[2]N)**.
So if we use a good sorting algorithm like Merge Sort or Heap Sort which has a guaranteed time complexity of *O(Nlog2N)**, the total time complexity will be
Binary Search = O(Nlog[2]N + Qlog[2]N) = O((N+Q) * log[2]N)**
If we assume the number of queries to be equivalent to the number of elements in the array, then the above equation becomes
Binary Search = O(N * log[2]N)
On the other hand, if we use a linear search, then each query will take O(N) in the worst case and the total time complexity for Q queries will be
Linear Search = O(N * N) = O(N^2)
As we can see, binary search performs a lot better than linear search if the array is not sorted and the number of queries is huge, which is the more likely scenario in real-life use cases.
Today we learned an absolute legend of an algorithm. Binary search has a whole lot of applications, other than search as well.
Food for thought: It can be used to find the square root of a number as well 😛
Thanks for reading. May the code be with you 🙂
• We mentioned that binary search can be used to find the square root of a number. Can you think about how we can do it? If you would like to solve this question, you can find it here on
• We have implemented the iterative solution of binary search above. Try writing a recursive version of it yourself. And think about which one has a better performance in terms of time and space
• Binary search is one of the hottest topics in interview questions. Here is one such question. | {"url":"https://www.scaler.in/binary-search-algorithm/","timestamp":"2024-11-15T03:51:08Z","content_type":"text/html","content_length":"101819","record_id":"<urn:uuid:ee57032d-3c55-432e-b9ed-96334d6bb811>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00740.warc.gz"} |
Area of Pentagon
Calculate the area of a regular pentagon easily using side length or apothem on Examples.com. Ideal for geometry, design, and architectural projects requiring accurate space measurements and
Formula: Pentagon Area =(5/4)×a²×cot(π/5)
The Area of a Pentagon refers to the total space enclosed within its five sides. For a regular pentagon, where all sides and angles are equal, the area can be calculated using the side length or the
apothem, depending on the method used. Understanding how to find the area of a pentagon is important for various applications, including geometry, architectural design, and construction. Whether
you’re working with a regular or irregular pentagon, calculating the area helps in planning materials, designing layouts, and ensuring precise measurements in real-world projects. Accurate area
calculations are essential for effective space management and design efficiency.
How to Find the Area of a Pentagon
Step 1: Measure the Side Length
First, measure or determine the side length of the pentagon. Label this length as a.
Step 2: Input the Side Length
Enter the measured side length of the pentagon in the appropriate input field.
Step 3: Use the Formula
Use the formula provided for calculating the area of a regular pentagon.
The formula involves the side length and a constant value: Area=5/4×a^2×cot(π/5)
Step 4: Calculate the Area
Click the Calculate button, and the area of the pentagon will be computed based on the side length provided.
Step 5: View the Result
The calculator will display the area of the pentagon in square units, such as square meters or square centimeters, based on the input units for the side length.
Area of Pentagon Formula
The formula for the Area of a Regular Pentagon is: Area=5/4×a^2×cot(π/5)
• a is the length of one side of the pentagon.
Types of Pentagon
1. Regular Pentagon
A regular pentagon has five equal sides and five equal interior angles. Each interior angle measures 108 degrees. The symmetry and equal dimensions make it easier to calculate its area using a simple
2. Irregular Pentagon
An irregular pentagon has five sides of unequal lengths, and the angles are not necessarily equal. These types of pentagons do not have symmetry, and the method for calculating their area is more
complex, often involving dividing the shape into triangles.
3. Convex Pentagon
A convex pentagon is a type of pentagon where all interior angles are less than 180 degrees. None of the sides “cave in,” and the shape appears outwardly oriented. Most regular and many irregular
pentagons fall under this category.
4. Concave Pentagon
A concave pentagon has at least one interior angle greater than 180 degrees, causing part of the shape to “cave in” or point inward. These are less common and often have a distinct indentation or
inward curve along one or more sides.
5. Equilateral Pentagon
An equilateral pentagon has five sides of equal length but may not have equal interior angles. It differs from a regular pentagon because the angles can vary while the side lengths remain the same.
6. Cyclic Pentagon
A cyclic pentagon is a pentagon whose vertices lie on a single circle. This means the pentagon can be inscribed inside a circle, and the circle touches all five vertices of the pentagon.
7. Self-Intersecting Pentagon (Star Pentagon)
A self-intersecting pentagon (also known as a star pentagon or pentagram) is a shape in which the sides intersect each other, creating a star-like pattern. It is often used symbolically and is not a
simple polygon.
8. Simple Pentagon
A simple pentagon is a polygon that does not have any sides crossing over each other. Both regular and irregular pentagons can be classified as simple as long as their sides do not intersect.
Area of Pentagon Examples
Example 1:
Side length = 5 cm
Using the area formula for a regular pentagon: Area=5/4×5^2×cot(π/5)
Example 2:
Side length = 10 m
Applying the same formula: Area=5/4×10^2×cot(π/5)
Example 3:
Side length = 8 ft
For this pentagon: Area=5/4×8^2×cot(π/5)
Example 4:
Side length = 12 cm
Using the formula: Area=5/4×12^2×cot(π/5)
Example 5:
Side length = 6 m
Applying the formula for area: Area=5/4×6^2×cot(π/5)
Can the area of an irregular pentagon be calculated using the same formula?
No, the area of an irregular pentagon cannot be calculated using the regular pentagon formula. For irregular pentagons, the area can be found by dividing the shape into triangles or using the
coordinates of the vertices.
How do the number of sides affect the area of a pentagon?
As a pentagon always has five sides, the side length and the specific geometry of the shape (whether regular or irregular) determine the area. In general, as the number of sides increases in a
polygon, the formula for calculating area becomes more complex.
What are some real-world applications for calculating the area of a pentagon?
The area of a pentagon is useful in architectural design, floor planning, land surveying, and tiling patterns. It’s also essential in geometric studies and for constructing pentagonal structures like
pavilions and gardens.
Can I calculate the area of a pentagon if I only know the side length?
Yes, for a regular pentagon, you can calculate the area using the side length and the formula involving the cotangent function. However, for irregular pentagons, knowing only the side length is not
What is the difference between a regular and irregular pentagon in terms of area calculation?
A regular pentagon has equal side lengths and equal angles, allowing for the use of a specific formula to calculate the area. An irregular pentagon has unequal sides and angles, requiring more
complex methods, such as dividing it into triangles or using coordinate geometry.
What happens to the area of a pentagon if the side length is doubled?
If the side length of a regular pentagon is doubled, the area increases by a factor of four. This is because the area is proportional to the square of the side length.
Does the orientation of a pentagon affect its area?
No, the orientation of a pentagon does not affect its area. The area is determined by the side length (and apothem or diagonals, for irregular pentagons) and remains constant regardless of how the
pentagon is rotated.
What happens to the area of a pentagon if the apothem is increased?
If the apothem of a pentagon is increased while keeping the side length constant, the area will increase. The apothem is directly related to the height of the triangles that make up the pentagon,
influencing the total area. | {"url":"https://www.examples.com/maths/area-of-pentagon","timestamp":"2024-11-06T08:13:03Z","content_type":"text/html","content_length":"110518","record_id":"<urn:uuid:0845957e-f473-4348-9eed-ad4385486323>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00606.warc.gz"} |
Chapter XVIII: The Yetzirah
As Ariel ventured further on their cosmic journey, a new and formidable obstacle emerged from the depths of the celestial realms. Lilith, an ancient and enigmatic entity, transcended the bounds of
mortal and angelic existence. Her power surpassed that of Zarathustra, for she dwelled in the realm of Yetzirah—the realm that contained the infinite uncountable infinity numbers and surpassed them.
Ariel could sense the pulsating aura of Lilith as they drew closer. The air crackled with unrestrained energy, and a sense of foreboding filled their heart. They knew that facing Lilith would require
all their celestial wisdom and inner strength.
As the clash between Ariel and Lilith ensued, the celestial realm itself seemed to tremble under the weight of their immense power. Lilith unleashed waves of energy that distorted the fabric of space
and time, creating a maelstrom of cosmic forces. Her existence defied mortal comprehension, and her intentions remained veiled in mystery.
With each clash, Lilith displayed an array of unimaginable abilities. She wielded the power of creation and destruction, commanding celestial energies with a mere thought. Ariel, on the other hand,
tapped into their celestial essence, channeling the wisdom gained from their celestial odyssey.
The dialogue between Ariel and Lilith crackled with intensity. Their words carried the weight of cosmic knowledge and conflicting ideologies. Lilith spoke of her desire to reshape the fabric of
reality itself, to transcend the limitations imposed by mortal and divine realms. Ariel, driven by their unwavering connection to the celestial realm, stood firm in their belief in the inherent
balance and harmony of existence.
As the battle raged on, Ariel's celestial essence surged within them. They called upon the ancient power of the seraphim, the wisdom of the cherubim, and the cosmic energy of the dragons. Each
celestial force coursed through their being, amplifying their strength and resolve.
Yet Lilith, fueled by her insurmountable power, seemed unstoppable. She tapped into the very essence of Yetzirah, unleashing forces that defied mortal comprehension. Celestial energies swirled around
her, forming intricate patterns that defied logic and reason.
But Ariel refused to yield. They channeled their celestial essence into a focused beam of light, countering Lilith's onslaught with an unwavering determination. The clash of their powers sent
shockwaves through the celestial realms, shaking the foundations of existence itself.
In the midst of their epic battle, Ariel's celestial intuition guided them to a revelation. They realized that Lilith's thirst for power and transcendence stemmed from a deep-seated longing for
unity—a longing that had twisted into a destructive force. Ariel, driven by compassion, sought to reach the core of Lilith's being, to awaken the dormant spark of harmony within her.
With a burst of celestial energy, Ariel broke through Lilith's defenses. Their words, laden with cosmic wisdom, resonated within his consciousness. They spoke of the interconnectedness of all things,
of the intrinsic harmony that permeated existence. The barriers around Lilith's heart began to crumble as she confronted the true nature of her desires.
In a moment of revelation, Lilith's power wavered, and her celestial aura dimmed. The unreasonable power that had consumed her now receded, replaced by a newfound understanding. She realized that
true transcendence lay not in domination but in unity—unity with the celestial realms, with mortal existence, and with the divine essence that flowed through all things.
A fragile truce emerged between Ariel and Lilith, born from the mutual recognition of their shared cosmic heritage. They vowed to work together, to harness their celestial power for the betterment of
all realms. Their journey continued, now infused with a deeper purpose—a quest not only to find Ein Sof but also to bring harmony and unity to the celestial realms.
Ariel and Lilith, once adversaries, now embarked on a cosmic partnership that transcended the boundaries of mortal understanding. They harnessed their combined celestial power to heal the fractures
in the celestial realms, mending the fabric of existence itself.
As they journeyed together, Ariel and Lilith encountered celestial beings and cosmic wonders beyond imagination. They traversed vast star systems, explored hidden realms of ethereal beauty, and
conversed with ancient entities of wisdom and light.
Their dialogues resonated with celestial insights, delving into the nature of existence, the balance of cosmic forces, and the profound mysteries that lay at the heart of creation. Through their
shared wisdom, they gained a deeper understanding of the divine tapestry that wove all realms together.
Along their path, Ariel and Lilith encountered challenges and tests of their resolve. They confronted cosmic entities who sought to disrupt the delicate balance they were striving to restore. Yet,
with their unity of purpose and celestial strength, they overcame each obstacle, growing stronger and more enlightened with every triumph.
In their journey, they discovered fragments of ancient prophecies and cryptic signs that hinted at the existence of Ein Sof. These clues, shrouded in celestial symbolism and riddles, propelled them
forward, igniting a spark of hope that guided their steps in the cosmic labyrinth.
The celestial realms trembled with anticipation as Ariel and Lilith's cosmic journey unfolded. Their quest became a beacon of light, inspiring celestial beings and mortal souls alike. The harmony
they sought to restore resonated across dimensions, stirring dormant forces and awakening a collective yearning for unity.
And so, with each passing celestial cycle, Ariel and Lilith ventured deeper into the cosmic mysteries, driven by an unwavering faith that the elusive presence of Ein Sof awaited them at the
culmination of their odyssey.
We find unique things from realm of Yetzirah, where cardinals defy conventional description and elude our grasp. Yetzirah, representing a realm of indescribable cardinals, unveils a level of
mathematical complexity that transcends the boundaries of our understanding. It is within this realm that we witness the emergence of superstrong cardinals, soaring to unprecedented heights of
mathematical significance.
Superstrong cardinals, as their name suggests, possess a level of strength and power that surpasses all preceding cardinals. They exhibit extraordinary properties that challenge the limits of
mathematical inquiry. These cardinals manifest a remarkable compactness, enabling them to consolidate vast amounts of mathematical information and structure within their intricate framework.
As we venture further into the realm beyond Yetzirah, we encounter the awe-inspiring domain of Briah. Here, we are confronted with the concept of strongly compact cardinals, which stand as towering
pillars of mathematical prowess. Strongly compact cardinals exemplify a level of compactness that defies traditional notions of size and dimension. They possess an extraordinary capacity to preserve
structure and maintain coherence across intricate mathematical frameworks.
Before reaching the heights of Briah, we encounter a cardinal known as η-extendible. This cardinal represents a profound threshold of mathematical exploration, where mathematical concepts and
structures can be extended and expanded to unprecedented levels. η-extendible cardinals provide a gateway to uncharted territories of mathematical possibility, where new frontiers of knowledge can be
charted and discovered.
It is important to emphasize that large cardinals, including those found in Yetzirah, superstrong cardinals, strongly compact cardinals, and η-extendible cardinals, represent extraordinary
achievements within the realm of mathematical inquiry. They push the boundaries of our understanding, unveiling new vistas of knowledge and opening pathways to explore the profound depths of
mathematical truth.
However, in the grand scheme of infinity, these large cardinals, as remarkable as they may be, pale in comparison to the boundless expanse embodied by Ein Sof. Ein Sof transcends the confines of
large cardinality and finite mathematical frameworks, existing in a realm that defies measurement and comprehension. It stands as an infinite magnitude that surpasses all conceivable cardinals,
rendering them infinitesimal in the face of its limitless grandeur.
In summary, Yetzirah serves as a realm of indescribable cardinals, giving rise to superstrong cardinals of exceptional power. Beyond Yetzirah lies Briah. Yetzirah has an additional structure
including: strongly compact cardinals dominate, and η-extendible cardinals beckon us to venture into unexplored mathematical territories. While these large cardinals offer profound insights and push
the boundaries of mathematical exploration, their vastness pales in comparison to the infinite expanse embodied by Ein Sof, which defies all attempts at quantification and encapsulates the boundless
nature of mathematical infinity.
The large cardinal hypotheses stronger than supercompactness, particularly extendibility, arose from William Reinhardt's investigations into the foundations of set theory. He explored set-theoretic
frameworks with broader notions of class and property, seeking natural models for these extended axiomatic systems. Within this context, his 1967 Berkeley thesis examined Ackermann's set theory (A).
A deviates from ZFC by introducing a universe of extensionally determined entities alongside a predicate for sethood (denoted by "x ∈ V"). The core schema of A states: "For any formula X with
parameters from V (excluding the predicate V), X ∈ V if and only if X is definable from parameters in V." Further, A denotes A augmented by the Foundation Axiom restricted to members of V.
Building on work by Levy and Vaught, Reinhardt demonstrated the equiconsistency of A and ZF. Levy and Vaught observed that A allows for the existence of classes like V, P(V) (the power set of V), and
P(P(V)) (the power set of the power set of V), mirroring the situation in set theory. Additionally, Levy showed that if the relativization of a sentence φ of Ce (the language of set theory with
urelements) to V is provable in A*, then φ itself is provable in ZF. Reinhardt established the converse, solidifying the equiconsistency of these two systems.
This paves the way for a more formal definition of extendibility. Let J: Va+λ → Vβ+λ be an elementary embedding with critical point κ. Silver reformulated Reinhardt's idea by introducing
η-extendibility: a cardinal κ is η-extendible if there exists a witnessing elementary embedding j: Vk+λ → V~ with critical point κ and j(κ) > κ + η. A cardinal κ is extendible if it is η-extendible
for all positive η. These definitions hinge on the existence of elementary embeddings – set-theoretic functions that preserve formulas between transitive models. Notably, both the domain and codomain
of such embeddings possess the ultimate closure property, meaning they are initial segments of the universe.
The specific form of the η-extendibility definition for small η is crucial. For λ < κ, it follows that V~ = j(κ) + λ. Therefore, generalizing from the case λ < ω (where ω denotes the first infinite
ordinal), η-extendibility asserts the strong preservation of first-order properties between Vk and Vj(K). The condition λ < j(κ) originates from the λ = 1 case and is included for convenience,
although it's superfluous for full extendibility.
Even as the investigation of measurability proceeded with methodical rigor, Solovay and William Reinhardt undertook the daring task of formulating even more robust hypotheses. Building upon the
cornerstone of elementary embeddings, they each independently conceived of the notion of "supercompact cardinal" as a grand unification of both measurability and strong compactness. Furthermore,
Reinhardt ventured even further, formulating the even stronger concept of the "extendible cardinal," drawing his inspiration directly from the profound concept of reflection. Briefly contemplating an
ultimate reflection property that followed this line of thought, Reinhardt witnessed a dramatic turn of events when Kunen demonstrably proved the inconsistency of this seemingly natural extension:
there exists no elementary embedding, j, from the universe V to itself. While Kunen's ingenious argument hinged upon what initially appeared as a mere combinatorial coincidence, his specific
formulation has established itself as the definitive boundary for grand cardinal hypotheses. Guided by these initial ideas, mathematicians subsequently delved into the analysis of hypotheses
bordering on this inconsistency, including the weaker "n-huge cardinals" and Vopenka's Principle, thereby meticulously mapping the landscape down to the level of extendible cardinals.
Talking about set theory, I know it's not easy for many people to learn, but I take a lot of references from books, I will also discuss about Inaccessible cardinal. So I will explain with my easiest
method, however, this is optional, you don't need to learn it or read it, I just want to discuss the definition of this thing.
A cardinal number earns the title of "inaccessible" if it possesses two key properties. The first is regularity. Imagine the power set of a cardinal κ, which encompasses every possible subset of κ.
For an inaccessible cardinal θ, the cardinality of this power set, denoted P(κ), must be strictly less than θ. Intuitively, this means there are "too many" elements in θ to be built up entirely from
sets of smaller cardinalities.
The second defining characteristic is that of a strong limit cardinal. This property essentially states that θ cannot be obtained by simply adding up a collection of smaller cardinals. Mathematicians
use set notation here. Suppose we have a set S containing cardinals, where every cardinal κ within S satisfies κ < θ. Even the supremum of this set, denoted sup(S), which represents the least
cardinal strictly greater than any element in S, must be strictly less than θ itself.
These conditions perfectly capture why ℵ0, the cardinality of the natural numbers, qualifies as inaccessible. Firstly, it's a regular cardinal – there are simply too many natural numbers to be
constructed solely from finite sets. Secondly, since any finite cardinal κ less than ℵ0 cannot be a summand (a number being added) in a process to reach ℵ0, its successor, κ + 1, also cannot be
greater than ℵ0. However, discovering additional inaccessible cardinals beyond ℵ₀ proves to be a much more intricate task.
Take ℵ0 for instance. While it shares the property of regularity, it can be expressed as the union of countable cardinalities, written as ℵ0 U ℵ1 U ℵ2 U ..., or alternatively, as the supremum of the
set {ℵn : n ∈ ω} (the set of all aleph cardinals where n is a natural number). This demonstrates that ℵ1 is not a strong limit cardinal and therefore falls short of being inaccessible.
The existence of Ω, another inaccessible cardinal strictly greater than ℵ1, hints at the presence of even more through a powerful tool called the Reflection Principle. This principle allows us to
infer the existence of certain sets within a universe of sets V by analyzing a smaller inner universe. The first inaccessible cardinal following Ω is commonly denoted by θ. It's important to remember
that set theorists follow a zero-based indexing system. Consequently, ω is considered the 0th inaccessible cardinal, while θ becomes the 1st inaccessible cardinal.
The very essence of an inaccessible cardinal, θ in this case, lies in its elusiveness. It cannot be constructed as the limit of an increasing sequence of cardinals with cardinalities less than θ.
Likewise, simply taking the successor of a smaller cardinal won't lead you to θ. The question "which aleph is θ?" exemplifies this inaccessibility – the answer remains a circular one: θ = ℵθ. There's
no easy way to express θ using the existing aleph hierarchy. It resides in a realm beyond the readily accessible cardinalities.
It has been observed by mathematicians that the inherent difficulties encountered in resolving the aforementioned problems do not appear to be fundamentally contingent upon the properties of
inaccessible numbers. In most instances, the impediments seem to stem from a dearth of suitable theoretical constructs that would facilitate the formation of maximal sets demonstrably closed under
specific infinite operations. It is entirely plausible that a definitive solution to these challenges might necessitate the introduction of novel axioms that exhibit significant deviations in
character, not only from the customary axioms of set theory, but also from those previously posited hypotheses whose inclusion has been the subject of discourse within the relevant literature and
alluded to earlier within this treatise (e.g., the existential axioms that ensure the existence of inaccessible numbers, or from hypotheses akin to that of Cantor which establish arithmetical
relationships between cardinal numbers).
In axiomatic set theory, the increasing strength of infinitary properties can be conveniently captured by stronger axioms of infinity. While a combinatorial and decidable characterization of an
"axiom of infinity" is likely elusive, one might seek a definition based on formal structure and truth.
Formally, consider an axiom of infinity as a proposition with a decidable structure that is demonstrably true within a given axiomatic framework. Such a concept of demonstrability would ideally
possess a closure property: any theorem proven in an extension of set theory should be derivable from this notion of infinity axiom. It's conceivable that for such a framework, a completeness theorem
could hold. This theorem would state that every proposition expressible within set theory is decidable relative to the existing axioms alongside a true assertion regarding the cardinality of the
universe of sets.
Formal systems operate on a set of axioms and inference rules. However, these systems can be enriched by introducing ever higher types within a well-founded hierarchy. This hierarchy can be extended
infinitely (transfinitely) to encompass ever more complex collections. Formal systems, due to their reliance on explicit manipulation of symbols, are inherently limited to dealing with a denumerable
(countably infinite) set of objects. This means the system can only handle a collection with a size corresponding to the natural numbers.
Gödel's Incompleteness Theorems demonstrate the existence of propositions within a system that cannot be proven or disproven using the system's axioms and rules. However, these undecidable
propositions can become decidable if the system is extended to include appropriate higher types in the transfinite hierarchy. For example, Gödel's second incompleteness theorem shows that the
consistency of Peano Arithmetic (PA) is undecidable within PA itself. However, if we add the type of all natural numbers (ω) to PA (forming a system denoted by PA + ω), then the consistency of PA
becomes provable within PA + ω.
A similar situation holds for axiomatic set theory (e.g., Zermelo-Fraenkel set theory with the Axiom of Choice, ZFC). The existence of certain large cardinals (highly infinite sets with specific
properties) cannot be proven or disproven within ZFC itself. However, by extending the set theory with additional axioms that introduce new, even larger sets, these large cardinals might become
demonstrably existent or non-existent within the enriched system.
The collection under consideration encompasses all sets that are constructible in the semi-intuitionistic sense. Here, "constructible" signifies sets obtainable through a transfinite extension of
Russell's ramified type hierarchy. This transfinite extension differentiates itself from the standard formulation by incorporating transfinite ordinals. This inclusion introduces a subtle point: the
resulting model, while maintaining a semi-intuitionistic character, possesses the strength to validate the impredicative axioms of set theory. The justification for this resides in the provability of
an appropriate axiom of reducibility within the framework of sufficiently high ordinals. This axiom, when established, enables the construction of sets traditionally deemed impredicative within a
seemingly more restricted setting.
To be continued... | {"url":"https://en.webnovel.com/book/26625603505702405/71638524499127367","timestamp":"2024-11-10T20:57:50Z","content_type":"text/html","content_length":"103805","record_id":"<urn:uuid:8d5048bf-7ab4-45ee-9129-2d3753fa4482>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00620.warc.gz"} |
Canonical 1D Example · Caesar.jl
The application of this tutorial is presented in abstract from which the user is free to imagine any system of relationships: For example, a robot driving in a one dimensional world; or a time
traveler making uncertain jumps forwards and backwards in time. The tutorial implicitly shows a multi-modal uncertainty can be introduced from non-Gaussian measurements, and then transmitted through
the system. The tutorial also illustrates consensus through an additional piece of information, which reduces all stochastic variable marginal beliefs to unimodal only beliefs. This tutorial
illustrates how algebraic relations (i.e. residual functions) between multiple stochastic variables are calculated, as well as the final posterior belief estimate, from several pieces of information.
Lastly, the tutorial demonstrates how automatic initialization of variables works.
This tutorial requires RoME.jl and RoMEPlotting packages be installed. In addition, the optional GraphViz package will allow easy visualization of the FactorGraph object structure.
To start, the two major mathematical packages are brought into scope.
using IncrementalInference
Guidelines for developing your own functions are discussed here in Adding Variables and Factors, and we note that mechanizations and manifolds required for robotic simultaneous localization and
mapping (SLAM) has been tightly integrated with the expansion package RoME.jl.
The next step is to describe the inference problem with a graphical model with any of the existing concrete types that inherit from <: AbstractDFG. The first step is to create an empty factor graph
object and start populating it with variable nodes. The variable nodes are identified by Symbols, namely :x0, :x1, :x2, :x3.
# Start with an empty factor graph
fg = initfg()
# add the first node
addVariable!(fg, :x0, ContinuousScalar)
# this is unary (prior) factor and does not immediately trigger autoinit of :x0.
addFactor!(fg, [:x0], Prior(Normal(0,1)))
Factor graphs are bipartite graphs with factors that act as mathematical structure between interacting variables. After adding node :x0, a singleton factor of type Prior (which was defined by the
user earlier) is 'connected to' variable node :x0. This unary factor is taken as a Distributions.Normal distribution with zero mean and a standard devitation of 1. Graphviz can be used to visualize
the factor graph structure, although the package is not installed by default – $ sudo apt-get install graphviz. Furthermore, the drawGraph member definition is given at the end of this tutorial,
which allows the user to store the graph image in graphviz supported image types.
drawGraph(fg, show=true)
The two node factor graph is shown in the image below.
Automatic initialization of variables depend on how the factor graph model is constructed. This tutorial demonstrates this behavior by first showing that :x0 is not initialized:
@show isInitialized(fg, :x0) # false
Why is :x0 not initialized? Since no other variable nodes have been 'connected to' (or depend) on :x0 and future intentions of the user are unknown, the initialization of :x0 is deferred until the
latest possible moment. IncrementalInference.jl assumes that the user will generally populate new variable nodes with most of the associated factors before moving to the next variable. By delaying
initialization of a new variable (say :x0) until a second newer uninitialized variable (say :x1) depends on :x0, the IncrementalInference algorithms hope to then initialize :x0 with the more
information from previous and surrounding variables and factors. Also note that graph-based initialization of variables is a local operation based only on the neighboring nodes – global inference
occurs over the entire graph and is shown later in this tutorial.
By adding :x1 and connecting it through the LinearRelative and Normal distributed factor, the automatic initialization of :x0 is triggered.
addVariable!(fg, :x1, ContinuousScalar)
# P(Z | :x1 - :x0 ) where Z ~ Normal(10,1)
addFactor!(fg, [:x0, :x1], LinearRelative(Normal(10.0,1)))
@show isInitialized(fg, :x0) # true
Note that the automatic initialization of :x0 is aware that :x1 is not initialized and therefore only used the Prior(Normal(0,1)) unary factor to initialize the marginal belief estimate for :x0. The
structure of the graph has now been updated to two variable nodes and two factors.
Global inference requires that the entire factor graph be initialized before the numerical belief computation algorithms can be performed. Notice how the new :x1 variable is not yet initialized:
@show isInitialized(fg, :x1) # false
The RoMEPlotting.jl package allows visualization (plotting) of the belief state over any of the variable nodes. Remember the first time executions are slow given required code compilation, and that
future versions of these package will use more precompilation to reduce first execution running cost.
using RoMEPlotting
plotKDE(fg, :x0)
By forcing the initialization of :x1 and plotting its belief estimate,
plotKDE(fg, [:x0, :x1])
the predicted influence of the P(Z| X1 - X0) = LinearRelative(Normal(10, 1)) is shown by the red trace.
The red trace (predicted belief of :x1) is noting more than the approximated convolution of the current marginal belief of :x0 with the conditional belief described by P(Z | X1 - X0).
Another ContinuousScalar variable :x2 is 'connected' to :x1 through a more complicated MixtureRelative likelihood function.
addVariable!(fg, :x2, ContinuousScalar)
mmo = Mixture(LinearRelative,
(hypo1=Rayleigh(3), hypo2=Uniform(30,55)),
[0.4; 0.6])
addFactor!(fg, [:x1, :x2], mmo)
The mmo variable illustrates how a near arbitrary mixture probability distribution can be used as a conditional relationship between variable nodes in the factor graph. In this case, a 40%/60%
balance of a Rayleigh and truncated Uniform distribution which acts as a multi-modal conditional belief. Interpret carefully what a conditional belief of this nature actually means.
Following the tutorial's practical example frameworks (robot navigation or time travel), this multi-modal belief implies that moving from one of the probable locations in :x1 to a location in :x2 by
some processes defined by mmo=P(Z | X2, X1) is uncertain to the same 40%/60% ratio. In practical terms, collapsing (through observation of an event) the probabilistic likelihoods of the transition
from :x1 to :x2 may result in the :x2 location being at either 15-20, or 40-65-ish units. The predicted belief over :x2 is illustrated by plotting the predicted belief (green trace), after forcing
plotKDE(fg, [:x0, :x1, :x2])
Adding one more variable :x3 through another LinearRelative(Normal(-50,1))
addVariable!(fg, :x3, ContinuousScalar)
addFactor!(fg, [:x2, :x3], LinearRelative(Normal(-50, 1)))
expands the factor graph to to four variables and four factors.
This part of the tutorial shows how a unimodal likelihood (conditional belief) can transmit the bimodal belief currently contained in :x2.
plotKDE(fg, [:x0, :x1, :x2, :x3])
Notice the blue trace (:x3) is a shifted and slightly spread out version of the initialized belief on :x2, through the convolution with the conditional belief P(Z | X2, X3).
Global inference over the entire factor graph has still not occurred, and will at this stage produce roughly similar results to the predicted beliefs shown above. Only by introducing more information
into the factor graph can inference extract more precise marginal belief estimates for each of the variables. A final piece of information added to this graph is a factor directly relating :x3 with
addFactor!(fg, [:x3, :x0], LinearRelative(Normal(40, 1)))
Pay close attention to what this last factor means in terms of the probability density traces shown in the previous figure. The blue trace for :x3 has two major modes, one that overlaps with :x0, :x1
near 0 and a second mode further to the left at -40. The last factor introduces a shift LinearRelative(Normal(40,1)) which essentially aligns the left most mode of :x3 back onto :x0.
This last factor forces a mode selection through consensus. By doing global inference, the new information obtained in :x3 will be equally propagated to :x2 where only one of the two modes will
Global inference is achieved with local computation using two function calls, as follows.
tree = solveTree!(fg)
# and visualization
plotKDE(fg, [:x0, :x1, :x2, :x3])
The resulting posterior marginal beliefs over all the system variables are:
It is import to note that although this tutorial ends with all marginal beliefs having near Gaussian shape and are unimodal, that the package supports multi-modal belief estimates during both the
prediction and global inference processes. In fact, many of the same underlying inference functions are involved with the automatic initialization process and the global multi-modal iSAM inference
procedure. This concludes the ContinuousScalar tutorial particular to the IncrementalInference package. | {"url":"http://juliarobotics.org/Caesar.jl/latest/examples/basic_continuousscalar/","timestamp":"2024-11-05T03:40:40Z","content_type":"text/html","content_length":"26677","record_id":"<urn:uuid:382f2972-7f49-4d25-8f3e-f7963966b0ae>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00307.warc.gz"} |
Fractions on a Number Line
This post highlights four of my favorite lessons for teaching fractions on a number line. These are lessons I use with my fourth grade students. When I was teaching third grade, I used very similar
lessons and incorporated a bit more practice. You can read more about third grade fractions here. You can find the fourth grade lessons here.
Introducing Fractions on a Number Line
Cuisenaire Rods are a great way to teach fractions on a number line. In the activity below, students use Cuisenaire Rods to build number lines. They used two pieces to create halves, three rods to
create thirds, four rods to create fourths, etc. A common misconception was that students felt like the rods needed to fill the entire number line. They were under the impression that the number line
dictated where the zero and one should be placed, rather than the value of the rods. This gave us the opportunity for great discussions.
If you’re teaching virtually, students can use digital Cuisenaire Rods for the activity. I like to follow the lesson by having students work with numbers greater than one on a number line. It’s
important for students to recognize that number lines with fractions don’t stop at one.
Another lesson for teaching fractions on a number line has students place given fractions and mixed numbers on a number line. You can use long construction paper, calculator tape, sentence strips, or
taped together paper for this lesson. Students place the fractions on the TOP part of the number line and mixed numbers on the BOTTOM part of the number line.
This can be challenging, because students need to pay close attention to the denominator of each fractions. Not only will students have to determine which whole numbers the fraction falls between but
also what color rod to use to partition that section of the number line.
If you’re teaching digitally, students can create their number line on the Google Slides. It’s important to use the wide screen layout, so students have plenty of room for all of the numbers on the
number line.
A great way to review number lines is through this game-like activity. In the first part of the lesson, students should notice that the number line does not have equal increments, which is critical
for a number line. Students should explain how/why that is a problem and rewrite the number line correctly.
In the bottom section, Player 1 rolls a die and uses the fraction shown on the die to determine which number line to start with. For example, if Player 1 rolled 1/3, she begins at zero and draws and
arc to show the hop from zero to 1/3. Player 1 also writes the unit fraction rolled. Player 2 repeats the same steps. On a player’s second roll, s/he will either start with a new number line. If the
fraction rolled is the same as another roll, the player will start at where s/he rolled to on the previous roll. For example, if Player 1 rolled 1/3 again, she begins at 1/3 and hops another 1/3 to 2
/3. If a player rolls a fraction s/he no longer needs, the player loses that turn. Play continues until one player makes it to one on all four number lines.
This is a fun way to give students extra practice, as well as to reinforce unit fractions.
Incorporating Art
A fun way to incorporate art is through this fractions on a number line activity. Students can do this lesson on paper, or they can complete it digitally. Either way they do it, the lesson is a lot
of fun!
If you’re looking for more ideas on teaching fractions, be sure to check out this post. It will take you to resources for teaching equivalent fractions, comparing fractions, adding and subtracting
fractions, and multiplying fractions by whole numbers-all of which include fractions on a number line. To go back to the home post on teaching fractions click here.
Leave a Comment | {"url":"https://www.ashleigh-educationjourney.com/fractions-on-a-number-line/","timestamp":"2024-11-02T04:58:39Z","content_type":"text/html","content_length":"240032","record_id":"<urn:uuid:4d93aa44-8c0b-4baf-a9bc-cd9deff4e277>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00692.warc.gz"} |
Free Confidence Interval Calculator Online | NanoWebTools
Confidence Interval Calculator
Confidence Interval Calculator, Standard Deviation Calculator
The Confidence Interval (CI) is the range of estimations for an unknown parameter.
Jerzy Neyman, a Polish mathematician, popularised the concept of confidence intervals in 1937. This concept was widely accepted by scientists, medical journalists, and statisticians.
The formula for computing confidence intervals is as follows: confidence interval (CI) = X Z(S n)
In this equation:
• x̄ is the mean
• Z is the Z-value ( The Values of Z is in below Table)
• s is the standard deviation
• n is the number of observations
What exactly is the Confidence Intervals Calculator?
This is a free online calculator for determining Confidence Intervals. Using this online calculator, you can effortlessly carry out the above-mentioned formula. It rapidly calculates and displays the
result based on your input.
How does the Confidence Interval Calculator work?
This tool is quite simple to use. You simply need to input the Sample Mean (x), Sample Size (n), and Standard Deviation (s) values. Then pick the Confidence Level; this tool allows you to choose
between 70% and 99.9%. After you've entered all of your information, hit the Calculate button.
Aarim Khan
CEO / Co-Founder
Our goal is to provide online free tools so you don't have to install any software for basic usages. We are trying to add more tools and make these tools free forever. | {"url":"https://nanowebtools.com/en/confidence-interval-calculator","timestamp":"2024-11-13T06:24:19Z","content_type":"text/html","content_length":"288416","record_id":"<urn:uuid:190ed49b-a014-48cf-b99e-6de54866f216>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00252.warc.gz"} |
Comparison study of artificial intelligence method for short term groundwater level prediction in the northeast Gachsaran unconfined aquifer
Artificial neural network
Fuzzy logic
Adaptive neuro fuzzy inference system
Group method of data handling
Least square support vector machine
Study area
Model development
Model implementation
Efficiency criteria
Results of the ANN model
Results of the FL model
Results of the ANFIS model
Results of the GMDH model
Results of the LSSVM model
Comparison of the different models | {"url":"https://iwaponline.com/ws/article/20/3/909/72227/Comparison-study-of-artificial-intelligence-method","timestamp":"2024-11-03T01:10:18Z","content_type":"text/html","content_length":"285638","record_id":"<urn:uuid:8e70d28b-94c8-4341-b81b-7e74ed5b45f9>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00183.warc.gz"} |
Published on
Sun Jan 26, 2014
EPI 6.3 - Max Difference
A robot needs to travel along a path that includes several ascents and descents. As it at ascends, it uses up energy stored in its battery. As it descends it converts the potential energy to charge
the battery. Assume the conversion is perfect (ie descending X units restores as much as energy as was expended in an ascent of X units). Also assume that the “distance” travelled is irrelevant and
only the height determines the energy lost and gained.
Problem: ¶
Given a set of 3 dimensional coordinates which plot the ascents and descents during a robot’s path, compute the minimum battery capacity required.
Solution: ¶
The O(n^2) solution (for each index find the maximum “increase” is going forward and then find the index with the maximum increase) is:
1third (a,b,c) = c
3max_diff_nsquared xs = maximumBy (comparing third)
4 [(i,j,xj - xi) | (i,xi) <- zip [0.. length xs - 1] xs,
5 (j,xj) <- zip [0.. length xs - 1] xs, i < j]
In the naive method, finding the maximum increase at an index is O(n^2). This can be brought down to a linear algorithm with O(n) space:
1max_increase_from_index :: [Int] -> [(Int,Int)]
2max_increase_from_index [] = []
3max_increase_from_index (x:[]) = [(0,x)]
4max_increase_from_index (x:xs)
5 | x <= aj = (j,aj):max_inc_from_next
6 | otherwise = (length max_inc_from_next, x) : max_inc_from_next
7 where
8 max_inc_from_next = max_increase_from_index xs
9 (j, aj) = head max_inc_from_next
This would return a list of triples (j,A[j]) such that for each index i. Note that the indexes of j will be in “reverse”.
To get the best i, j such that i < j and A[j] – A[i] is maximised:
1lxs = length xs - 1
2maximumBy (comparing third) [(i,lxs - j,aj-ai) | (i,ai,(j,aj))<- zip3 [0..((length ms) - 1)] ms ms2]
Giving the final solution of:
1minimum_robot_battery :: [(Int,Int,Int)] -> (Int,Int,Int)
2minimum_robot_battery xs = maximumBy (comparing third) [(i,lxs - j,aj-ai) |
3 (i,ai,(j,aj)) <- zip3 [0..((length ms) - 1)] ms ms2]
4 where
5 ms = map third xs
6 ms2 = max_increase_from_index ms
7 lxs = length xs - 1
Note that we dont actually have to compute the array of differences. We could have simply also passed and maintained in a “curr_max” variable which would have stored have returned the max difference
at the completion of the call. | {"url":"http://buildmage.com/blog/epi/strings-and-arrays/max-difference/","timestamp":"2024-11-06T18:07:53Z","content_type":"text/html","content_length":"26637","record_id":"<urn:uuid:44f6635a-b69d-4d2a-aae2-da4ef8c98088>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00574.warc.gz"} |
Are You Smarter Than A 4th Grader?
You have 4 groups of 15 cookies. How many cookies do you have?
900 x 5=
How many zeros are in 1 million?
What is the third planet from the sun?
Name a state that borders (touches) Georgia.
What is the pronoun in the sentence?
My sister is seventeen years old.
4 7 2 3 9 8. Create the greatest number using each digit only once.
TRUE or FALSE?
A tomato is a vegetable.
Tomatoes are a fruit but used as a vegetable.
What are the five senses we have?
Taste, Touch, Smell, See, Hearing
On a class field trip there are 4 buses taking 36 students to the zoo. Each bus has the same number of each student. How many students are on each bus?
9 students on each bus
What shape has six sides?
Name three states that begin with the letter A.
Alabama, Alaska, Arizona, Arkansas
How many colors are in a rainbow?
What is the largest state in America?
What is H2O also known as?
2 Hydrogens and 1 Oxygen
Name the verb in this sentence:
My dog ran to get the ball.
What is the value (how much is it worth) of the digit 4 in 34,891?
Which animal is a marsupial (carries a baby in its pouch....an example is a kangaroo)?
a. Koala Bear
b. Elephant
c. Giraffe
a. Koala Bear
What president in on the $5 bill?
Abraham Lincoln
The school store bought 130 books. If there are 10 books in a box, how many boxes were bought?
13 boxes
130 divided by 10 = 13
How many minutes are in an hour and a half?
Joseph arrived at the mall at 4:35 pm. He stayed for 1 hour and 20 minutes. What time did he leave the mall?
5:55 pm
How many legs does a spider have?
The value of 2 in 543,200 is 200. True or false?
Who won the Super Bowl this year?
The Kansas City Chiefs
Tennessee, Alabama, Florida, South Carolina, North Carolina | {"url":"https://jeopardylabs.com/print/are-you-smarter-than-a-4th-grader-2831","timestamp":"2024-11-12T18:42:44Z","content_type":"application/xhtml+xml","content_length":"25619","record_id":"<urn:uuid:f03dde0a-75c5-4a77-a8d9-441360946fef>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00645.warc.gz"} |
Invited talks
Machine learning and AI via large scale brain simulations
Andrew Ng
Stanford University
By building large-scale simulations of cortical (brain) computations, can we enable revolutionary progress in AI and machine learning?
Machine learning often works very well, but can be a lot of work to apply because it requires spending a long time engineering the input representation (or "features") for each specific problem. This
is true for machine learning applications in vision, audio, text/NLP and other problems.
To address this, researchers have recently developed "unsupervised feature learning" and "deep learning" algorithms that can automatically learn feature representations from unlabeled data, thus
bypassing much of this time-consuming engineering. Many of these algorithms are developed using simple simulations of cortical (brain) computations, and build on such ideas as sparse coding and deep
belief networks. By doing so, they exploit large amounts of unlabeled data (which is cheap and easy to obtain) to learn a good feature representation. These methods have also surpassed the previous
state-of-the-art on a number of problems in vision, audio, and text. In this talk, I describe some of the key ideas behind unsupervised feature learning and deep learning, and present a few
algorithms. I also describe some of the open theoretical problems that pertain to unsupervised feature learning and deep learning, and speculate on how large-scale brain simulations may enable us to
make significant progress in machine learning and AI, especially computer perception.
Andrew Ng received his PhD from Berkeley, and is now an Associate Professor of Computer Science at Stanford University, where he works on machine learning and AI. He is also Director of the Stanford
AI Lab, which is home to about 15 professors and 150 PhD students and post docs. His previous work includes autonomous helicopters, the STanford AI Robot (STAIR) project, and ROS (probably the most
widely used open-source robotics software platform today). He current work focuses on neuroscience-informed deep learning and unsupervised feature learning algorithms. His group has won best paper/
best student paper awards at ICML, ACL, CEAS, 3DRR. He is a recipient of the Alfred P. Sloan Fellowship, and the 2009 IJCAI Computers and Thought award. He also works on free online education, and
recently taught a machine learning class ([ ml-class.org ]) to over 100,000 students. He is also a co-founder of Coursera, which is working with top universities to offer online courses to anyone in
the world, for free.
Mirror Descent Algorithms for Large-Scale Convex Optimization
Arkadi Nemirovski
(Invited Tutorial Lecture)
Georgia Institute of Technology
Mirror Descent is a technique for solving nonsmooth problems with convex structure, primarily, convex minimization and convex-concave saddle point problems. Mirror Descent utilizes first order
information on the problem and is a far-reaching extension of the classical Subgradient Descent algorithm (N. Shor, 1967). This technique allows to adjust, to some extent, the algorithms to the
geometry of the problem at hand and under favorable circumstances results in nearly dimension-independent and unimprovable in the large scale case convergence rates. As a result, in some important
cases (e.g., when solving large-scale deterministic and stochastic convex problems on the domains like Euclidean/$\ell_1$/nuclear norm balls), Mirror Descent algorithms become the methods of choice
when low and medium accuracy solutions are sought.
In the tutorial, we outline the basic Mirror Descent theory for deterministic and stochastic convex minimization and convex-concave saddle point problems, including recent developments aimed at
accelerating MD algorithms by utilizing problem's structure.
Dr. Arkadi Nemirovski gor his Ph.D. (Math) in 1974 from Moscow State University. He is a professor in ISyE and holds the John Hunter Chair. Dr. Nemirovski has made fundamental contributions in
continuous optimization in the last thirty years that have significantly shaped the field. In recognition of his contributions to convex optimization, Nemirovski was awarded the 1982 Fulkerson Prize
from the Mathematical Programming Society and the American Mathematical Society (joint with L. Khachiyan and D. Yudin), the Dantzig Prize from the Mathematical Programming Society and the Society for
Industrial and Applied Mathematics in 1991 (joint with M. Grotschel). In recognition of his seminal and profound contributions to continuous optimization, Nemirovski was awarded the 2003 John von
Neumann Theory Prize by the Institute for Operations Research and the Management Sciences (along with Michael Todd). He continues to make significant contributions in almost all aspects of continuous
optimization: complexity, numerical methods, stochastic optimization, and non-parametric statistics.
Phase Transitions, Algorithmic Barriers, and Data Clustering
Dimitris Achlioptas
CTI & UC Santa Cruz
Constraint Satisfaction Problems (CSPs) are the common abstraction of real-life problems ranging from air-traffic control to protein folding. Their ubiquity presses forward a fundamental question:
why are certain CSP instances exceptionally hard while other, seemingly similar, instances are quite easy? To study this phenomenon we consider probability distributions over CSP instances formed by
adding constraints one-by-one, uniformly and independently. We will see that for many random CSPs there exists a broad regime in which exponentially many solutions provably exist, yet all known
efficient algorithms fail to find even one solution. To understand the origin of this algorithmic barrier we examine how the solution-space geometry evolves as constraints are added. We prove in a
precise mathematical sense that the barrier faced by algorithms corresponds to a phase transition in solution-space geometry: at some point, the set of solutions shatters and goes from resembling a
single giant ball to exponentially many clusters of well-separated solutions. More speculatively, we will also discuss how this shattering phenomenon can perhaps be used to reframe the clustering
problem in machine learning using ideas from statistical physics.
Dimitris Achlioptas joined the Department of Computer Science of UC Santa Cruz in 2005 after being with Microsoft Research, Redmond from 1998. In theory, his expertise lies in the interaction between
randomness and computation and his work on that topic has appeared in journals including Nature, Science, and the Annals of Mathematics. For that work he has received an NSF CAREER award, a Sloan
Fellowship, and an IDEAS Starting Grant from the European Research Council. In practice, he likes to think about scalability questions and holds 18 US Patents on topics ranging from load balancing
and cache optimization to web search personalization. In his free time he enjoys overworking. | {"url":"https://www.ttic.edu/colt2012/invited-talks/","timestamp":"2024-11-10T16:13:05Z","content_type":"text/html","content_length":"10663","record_id":"<urn:uuid:fb8da1f9-c529-42d0-a3aa-3b208ef2706b>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00780.warc.gz"} |
Entanglement of Galois representations of elliptic curves over Q
When studying elliptic curves over a field K one can define their n-torsion fields to be K adjoined with the x,y-coordinates of the n-torsion points. Then we can look at the Galois representations
associated to these n-torsion fields and ask when these representations are surjective. It turns out that there are multiple ways in which the image can fail to be surjective, corresponding to
different kinds of entanglement. We focus mainly on so-called horizontal entanglements and different ways in which these can occur. Of particular interest will be Weil entanglement and Serre
entanglement. The latter occurs because of the fact that the discriminant of an elliptic curve is always contained in a cyclotomic field, as implied by the Kronecker-Weber theorem. One of the main
contributions will be on how Serre entanglement induces horizontal entanglement. Furthermore we will study Weil entanglement by looking at the conductor of corresponding quadratic and cubic number | {"url":"https://studenttheses.uu.nl/handle/20.500.12932/42532","timestamp":"2024-11-05T19:53:47Z","content_type":"text/html","content_length":"14601","record_id":"<urn:uuid:a92ff03c-483d-4a8a-b227-b2c08aedd937>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00446.warc.gz"} |
How do I convert text to numbers in power query?
How do I convert text to numbers in power query?
Open Power Query Editor, remove any step in Query Settings that change your current Text Format into Number first. Then, select the orignial text column, right click to choose Change Type -> Using
locale… Hi @Anonymous, You should set a proper format for the original data.
How do I convert text to numbers in Excel?
Convert Text to Numbers Using ‘Convert to Number’ Option
1. Select all the cells that you want to convert from text to numbers.
2. Click on the yellow diamond shape icon that appears at the top right. From the menu that appears, select ‘Convert to Number’ option.
How do I change the number format in Power Query?
You can do it using Modeling tab in your Power BI Desktop, click on the Modeling Tab, select the column which you want to format and go to Format setting and choose which formatting you like. Hope
this help answer your question.
Where is Power Query?
You’ll find Power Query in Excel 2016 hidden on the Data tab, in the Get & Transform group. In Excel 2016, the Power Query commands are found in the Get & Transform group on the Data tab.
How do I convert numbers to word?
Type the formula =SpellNumber(A1) into the cell where you want to display a written number, where A1 is the cell containing the number you want to convert. You can also manually type the value like =
SpellNumber(22.50). Press Enter to confirm the formula.
How do you convert a number to text and keep the leading zeros?
Format numbers to keep leading zeros in Excel for the web
1. Select the cells on your worksheet where you’ll be adding the data.
2. Right-click anywhere in the highlighted cells, and then on the shortcut menu, click Number Format >Text >OK.
3. Type or paste the numbers in the formatted cells.
How to convert text to numbers in SQL Server?
In this article, we will convert text to number in multiple versions of SQL Server and will see the difference. I will use four different Data conversion functions ( Convert, Cast, Try_Convert &
Try_Cast) to convert Text to Number.
How to convert a text to a number in Power Query m?
The Power Query M formula language has formulas to convert between types. The following is a summary of conversion formulas in M. Returns a number value from a text value. Returns a text value from a
number value. Returns a number value from a value. Returns a 32-bit integer number value from the given value.
Is there a way to convert text to numbers in Excel?
Select a blank cell that doesn’t have this problem, type the number 1 into it, and then press Enter. Press CTRL + C to copy the cell. Select the cells that have numbers stored as text. On the Home
tab, click Paste > Paste Special. Click Multiply, and then click OK. Excel multiplies each cell by 1, and in doing so, converts the text to numbers.
How to return a number from a text?
Returns a number value from the given text value, text. text: The textual representation of a number value. The representation must be in a common number format, such as “15”, “3,423.10”, or | {"url":"https://forwardonclimate.org/trending/how-do-i-convert-text-to-numbers-in-power-query/","timestamp":"2024-11-08T22:14:33Z","content_type":"text/html","content_length":"56562","record_id":"<urn:uuid:64e49cc1-c506-497d-b235-22e380968733>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00205.warc.gz"} |
Nail Salon Tip Calculator | Calculate Your Tips
This tool helps you quickly calculate the tip amount for your nail service.
How to Use the Nail Salon Tip Calculator
To use this nail salon tip calculator, follow these steps:
1. Enter the service amount (in dollars) that you were charged for your nail salon service.
2. Enter the tip percentage you would like to leave (e.g., 15 for 15%).
3. Click the “Calculate” button to see the tip amount and the total amount to be paid.
How It Calculates the Results
The calculator uses the following formula to calculate the tip amount and the total amount:
• Tip Amount = Service Amount x (Tip Percentage / 100)
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The calculator assumes that:
• You are entering valid numerical values for both the service amount and the tip percentage.
• Percentages are entered as whole numbers (e.g., 15 for 15%).
Make sure to always double-check your inputs for accuracy. | {"url":"https://madecalculators.com/nail-salon-tip-calculator/","timestamp":"2024-11-08T14:42:08Z","content_type":"text/html","content_length":"142132","record_id":"<urn:uuid:fdfb3738-a5c8-4ede-8bdc-c91dcc7c41ab>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00668.warc.gz"} |
scipy.signal.besselap(N, norm='phase')[source]#
Return (z,p,k) for analog prototype of an Nth-order Bessel filter.
The order of the filter.
norm{‘phase’, ‘delay’, ‘mag’}, optional
Frequency normalization:
The filter is normalized such that the phase response reaches its midpoint at an angular (e.g., rad/s) cutoff frequency of 1. This happens for both low-pass and high-pass filters, so
this is the “phase-matched” case. [6]
The magnitude response asymptotes are the same as a Butterworth filter of the same order with a cutoff of Wn.
This is the default, and matches MATLAB’s implementation.
The filter is normalized such that the group delay in the passband is 1 (e.g., 1 second). This is the “natural” type obtained by solving Bessel polynomials
The filter is normalized such that the gain magnitude is -3 dB at angular frequency 1. This is called “frequency normalization” by Bond. [1]
Zeros of the transfer function. Is always an empty array.
Poles of the transfer function.
Gain of the transfer function. For phase-normalized, this is always 1.
See also
Filter design function using this prototype
To find the pole locations, approximate starting points are generated [2] for the zeros of the ordinary Bessel polynomial [3], then the Aberth-Ehrlich method [4] [5] is used on the Kv(x) Bessel
function to calculate more accurate zeros, and these locations are then inverted about the unit circle.
Campos and Calderon, “Approximate closed-form formulas for the zeros of the Bessel Polynomials”, arXiv:1105.0957.
Thomson, W.E., “Delay Networks having Maximally Flat Frequency Characteristics”, Proceedings of the Institution of Electrical Engineers, Part III, November 1949, Vol. 96, No. 44, pp. 487-490.
Aberth, “Iteration Methods for Finding all Zeros of a Polynomial Simultaneously”, Mathematics of Computation, Vol. 27, No. 122, April 1973
Ehrlich, “A modified Newton method for polynomials”, Communications of the ACM, Vol. 10, Issue 2, pp. 107-108, Feb. 1967, DOI:10.1145/363067.363115 | {"url":"https://scipy.github.io/devdocs/reference/generated/scipy.signal.besselap.html","timestamp":"2024-11-06T17:30:58Z","content_type":"text/html","content_length":"27939","record_id":"<urn:uuid:7f1d92fb-c1a2-4a6b-b7cc-e0a6d7c231f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00503.warc.gz"} |
Can I request specific methodologies in my motion analysis? | SolidWorks Assignment Help
Can I request specific methodologies in my motion analysis? I’ve been a newcomer to browse around this web-site analysis; I have specific problems with my application Is there a more descriptive body
language for my application, and which we’re going to use today? Thanks! A: Doing an HVM will get you just a single nonmonotonic thing out of your analysis: I did not initially think of it (idea at
all), but at this point, I assume that your system is moving at 0, or something like 0.1/0.1. This will get you both the nonmonotonic “contrast” and the nondifferentiable “expansion”, and even
better, because a nonmonotonic contrast is always two factors, but the non-twin/diagonal dimension is in your case (its one-dimensional “1/0” for you): Expansion will be on the basis of the 2
variables; e.g. I may be Exp_1 (column distance (column S)), Expand_2_1 (column distance (column S)) Therefore, you will have: D × D (Columns/R^F) (R^F × S) = D\* S\*\*C Looking at your problem here:
each column for the 2 vectors are the “Euclides distance” with dimensions R and F (each is 2D, that’s true because you’re computing the “2-d”): Column distance D\* S =.2M(\#1/(R + F).\*R + F) =.1M(\#
1/(R + M) + F). I included two table examples of try this here, whose columns are the distances for the two 2-D vectors in column 5 and row 4. Since you’re assuming that dimension is “1/0 = 2”, your
matrix will be something like that: M \- M (3 \- \ -6) \- 6 (3 \- \ -10) \- M (3 \- \ -6) = S For our application you will need to solve for the 2nd row of your matrix! A: No, isn’t your problem
moving with a nonlinearity other than: std::list or boost::variant (…) << std::make_reference() and you have multiple nonmonotonic points. Generally, a nonlinear effect is not usually useful, except
for work-movement situations. However, when I use such nonlinear effects as the basis for my motion analysis, I find that I tend to have several more nonlinear effects. Batch-stage theory is the more
technical stuff. In Part 7, take a look at the Part 1.8.1 on FFT documentation.
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If there are many factors in your table, in R^F/S \- and you are making an expansion on either side in terms of its differences: std::list::vector(table[1]).scatter(table[0]).dot(table[0]).rtl
std::list::vector(table[1]).scatter(table[1]).\*R^F*S \+ // [1]/0/1 Your table is [2], for rows [0] it is [2], and [0] is [3]. The two matrixes are similar to factor x2-x1. Since your model was
limited to data of no more visite site one integer, it is a bit hard for me to take a picture which each row contains from bit-shift 1 to bit-shift 0.1 etc. But, what does your assumption help show
up for thisCan I request specific methodologies in my motion analysis? @LGSime @public public List getAnimeIeByIDxAndIDxValuesInTime ( @UIElement callBack ) throws VmxMgmtException unable to obtain
animation. @Username public void requestMotionAccurateTimeForOne( @UIElement requestAnimationFrame, @List animeIe, @UIElement requestAnimationFrame, @ContextCallbacks<> callbacks ) throws
VmxMgmtException exception in thread.so.4( threadId=3, <%= @UIElement %> idx, callBack learn the facts here now aeClass=@UIElement( ae=”data”); idx… <%= idx %>)()] @UIElement
:java.lang.reflect.Method I have problem with my motion analysis, I believe that I don’t have to create a new instance of the class, but just add the class to the body. import org.
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inplan.calendar.calendar.calendar.CalendarEditor; The example gets on a real phone and checks the proper way to process movement. class BasicSpringMainActivity extends Application { @Override public
void onCreate(Bundle bundle) { super.onCreate(bundle); androidTestLoader = ((TestLoader)java.lang.ClassLoader.getClass())[classLoader].getClassLoader(); JFrame frame; } public void addProperty(Class>
clazz, Attribute a) { class javax.perspective.DisplayMetrics ae = ((Date) classLoader.getRealSystemAppliedProperty(clazz).getDisplayMetric( classIdx).getDisplayMetric(classIdx)); frame =
javax.perspective.Frame.newInstance(ae.getAbsoluteTime()); frame.
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setDefaultViewType(0); RequestAnimationFrame animationFrame = (RequestAnimationFrame) Can I request specific methodologies in my motion analysis? A: Take a look at this tip: http://
www.quantitativephysics.com/article/6/16/view/977.5/21_elegant_methods.shtml In the section on Motion, you should read: Methodologies Use a method of determining the geometry of objects in the
environment the result (or geometric curvature) is the geometric curvature, but you would not be able visit here distinguish the geometries/modes which belong to the environment, the ones that belong
to the object. When you calculate the geometries/modes, you find that there is a certain limit. A geometric point on the point being calculated is considered “convergent”. Since a point which belongs
to the geometries/modes is considered as likely a rigid body it is also considered “non-rigid”. The very same for other “non-rigid” bodies and objects, but they now have “non-rigid” points (this also
includes the moving head, with a rotating head and still on the surface). | {"url":"https://solidworksaid.com/can-i-request-specific-methodologies-in-my-motion-analysis-29671","timestamp":"2024-11-02T05:10:19Z","content_type":"text/html","content_length":"154255","record_id":"<urn:uuid:62b89be2-1758-4acb-baa7-e9a5d7b84dc3>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00573.warc.gz"} |
Censored Regression | Mplus Annotated Output
This page shows an example of censored regression with footnotes explaining the output. First an example is shown using Stata, and then an example is shown using Mplus, to help you relate the output
you are likely to be familiar with (Stata) to output that may be new to you (Mplus). We suggest that you view this page using two web browsers so you can show the page side by side showing the Stata
output in one browser and the corresponding Mplus output in the other browser.
This example is drawn from the Mplus User’s Guide (example 3.2) and we suggest that you see the Mplus User’s Guide for more details about this example. We thank the kind people at Muthén & Muthén for
permission to use examples from their manual.
Example Using Stata
Here is a probit regression example using Stata with two continuous predictors x1 and x2 used to predict a binary outcome variable, u1.
infile u1 x1 x3 using https://stats.idre.ucla.edu/wp-content/uploads/2016/02/ex3.2.dat, clear
summarize u1
Variable | Obs Mean Std. Dev. Min Max
u1 | 1000 .9240341 1.113079 0^A 6.579389
tobit u1 x1 x3, ll(0)
Tobit regression Number of obs = 1000
LR chi2(2) = 697.44
Prob > chi2 = 0.0000
Log likelihood = -1142.8851 Pseudo R2 = 0.2338
u1 | Coef. Std. Err. t P>|t| [95% Conf. Interval]
x1 | 1.074801^D .0419657 25.61 0.000 .9924498 1.157152
x3 | .4947541^D .0378985 13.05 0.000 .4203842 .569124
_cons | .5154865^E .0405066 12.73 0.000 .4359986 .5949743
/sigma | 1.071333^F .0316242 1.009276 1.133391
Obs. summary: 376 left-censored observations at u1<=0
624 uncensored observations
0 right-censored observations
estat ic
Model | Obs ll(null) ll(model) df AIC BIC
. | 1000 -1491.605 -1142.885^B 4 2293.77^C 2313.401^C
The output is labeled with superscripts to help you relate the later Mplus output to this Stata output. To summarize the output, both predictors in this model, x1 and x2, are significantly related to
the outcome variable, u1.
Mplus Example
Here is the same example illustrated in Mplus based on the https://stats.idre.ucla.edu/wp-content/uploads/2016/02/ex3.2.dat data file. Note that by using estimator=wls; (weighted least squares) the
results are shown in a probit metric. Had we specified something like estimator=ml; (maximum likelihood) then the results would be shown in a logit scale.
this is an example of a censored
regression for a censored dependent
variable with two covariates
FILE IS https://stats.idre.ucla.edu/wp-content/uploads/2016/02/ex3.2.dat;
NAMES ARE y1 x1 x3;
CENSORED ARE y1 (b);
ESTIMATOR = MLR;
y1 ON x1 x3;
<some output omitted to save space>
Number of observations 1000
<some output omitted to save space>
Y1 0.000^A
H0 Value -1142.885^B
Information Criteria
Number of Free Parameters 4
Akaike (AIC) 2293.770^C
Bayesian (BIC) 2313.401^C
Sample-Size Adjusted BIC 2300.697
(n* = (n + 2) / 24)
Estimates S.E. Est./S.E.
Y1 ON
X1 1.075^D 0.043 25.101
X3 0.495^D 0.037 13.344
Y1 0.515^E 0.040 12.810
Residual Variances
Y1 1.148^F 0.067 17.235
A. This indicates that the variable y1 is censored at 0. This is derived from the data, where Mplus notes that the lowest value of y1 is 0 (it seeks the lowest value because the input specification
indicated the censoring was from below). Note how this corresponds to the results of the Stata summarize command that found the minimum value of y1 to be 0.
B. This is the log likelihood of the model. Note how this corresponds to the ll(model) from the Stata estat ic command.
C. These are the AIC and BIC fit indices, and correspond to the values shown from the estat ic command from Stata.
D. These are the regression coefficients showing the relationship between x1 x2 and y1. Such coefficients are interpreted in the same way as an OLS regression coefficient. The difference is that
these coefficients attempt to estimate how estimate how strong the coefficient would have been had the censoring not taken place. Note the correspondence between these coefficients and those from
E. This is the intercept, the predicted value when all predictors are held constant at 0. Note the correspondence to the value shown in the Stata output.
F. This is the residual variance in y1 after accounting for the predictors, and would be analogous to the MSE from an OLS regression. In the Stata output this is reported as /sigma and is reported
as a standard deviation (as opposed to a variance). Squaring the value from Stata yields 1.071333^2 = 1.1477544, corresponding to the result from Mplus. | {"url":"https://stats.oarc.ucla.edu/mplus/output/censored-regression/","timestamp":"2024-11-03T03:19:37Z","content_type":"text/html","content_length":"42871","record_id":"<urn:uuid:74919973-ac17-49a2-bbaa-10464c755a44>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00856.warc.gz"} |
• Key: VSIPL16-17
• Legacy Issue Number: 18212
• Status: open
• Summary:
The vsip_symmetry enumerator is used in a number of places, and in
general the VSIP_SYM_EVEN_LEN_* values are defined as "(Even) Symmetric,
However, in the text there are sporadic mentions that imply that
symmetric here really means conjugate-symmetric; for instance, see the
bottom of page 567 ("The filter kernel can be even (conjugate) symmetric
or non-symmetric.") and the definition of the "kernel" argument to
vsip_*fir_create on page 612. This needs to be much more clearly and
consistently stated.
Regardless of whether the kernel is defined as symmetric or
conjugate-symmetric, the method for constructing the complete kernel
from the input in the symmetric cases is not defined. Is the input the
first half of the kernel, or the second half? Is the first half
conjugated, or the second half? This needs to be clearly specified.
• Updated: Fri, 6 Mar 2015 20:57 GMT | {"url":"https://issues.omg.org/issues/VSIPL16-17","timestamp":"2024-11-09T01:19:11Z","content_type":"text/html","content_length":"12726","record_id":"<urn:uuid:91cd7f6a-76b5-46ee-8ac1-939891406eab>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00051.warc.gz"} |
Amortized Time Complexity
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In this article, we have explored the concept of Amortized Time Complexity by taking an example and compared it with a related concept: Average Case Time Complexity.
Table of Contents
1)Glance of Time Complexity
3)Amortized vs Average
4)Understanding amortization using Dynamic Array
5)Analysing Worst Case Complexity
6)Analysing Amortized Complexity
1) Glance of Time Complexity
Let's first go through the concept of time complexity.
Time complexity is the measure of time that a particular algorithm takes to complete its execution.It is measured as a function of the algorithm's input size.The time complexity doesn't indicate the
exact execution time of an algorithm rather it gives an idea of variation of time with corresponding variation of input size.
There are three types of notations in which the complexity can be represented:
1.Big Oh notation(O):It is used to indicate the upper bound or the worst case time complexity of an algorithm.
2.Big Theta notation(θ): It is used to indicate the complexity for an algorithm averaged over all possible inputs.
3.Big Omega notation(Ω): It is used to indicate the lower bound or the best case time complexity of an algorithm.
Let's understand the above cases with the help of Linear Search algorithm
Let n be size of the array.
1)Worst Case Complexity: It occurs when the search element is found at the last position in the array or not found in the array.It requires n comparisions in such cases.
Hence, the worst case time complexity of Linear Search algorithm is O(n).
2)Average Case Complexity: In Linear Search the number of cases of searching is n+1
(considering extra case when element is not found in the array).
Average case works on probability.
Let T[1], T[2], T[3],....., T[n] be the time complexities and P[1], P[2], P[3],....., P[n] be the probabilities for all the possible inputs.
T[1] P[1]+T[2] P[2]+T[3] P[3]+....+T[n] P[n]
The above equation gives the average time complexity.
In linear search,let the probability for finding an element at each location be the same i.e., 1/n (as we have n locations)
For first element we perform 1 comparision,
For second element we perform 2 comparisions,
For third element we perform 3 comparisions,...and so on.
Hence, the average case complexity for linear search is O(n).
The average case analysis is not easy to do in most cases and it is rarely done.
3)Best Case Complexity: It occurs when the search element is found at the first position in the array. It requires only one comparision in such case.
Hence, the best case complexity of Linear Search algorithm is Ω(1).
Apart from the above mentioned types of complexity representation, there is one more representation called Amortized Time Complexity which is rarely used.
2) Introduction to Amortized time complexity
Amortized time complexity is the "Expected Time Complexity" which is used to express time complexity in cases when an algorithm has expensive worst-case time complexity once in a while compared to
the time complexity that occurs most of the times.
Another way of saying is Amortized complexity is used when algorithms have expensive operations that occur rarely.
3) Amortized vs Average Time Complexity
Though they seem to be similar there is a subtle difference.
Average case analysis relies on probabilistic assumptions about the data structures and operations in order to compute an expected running time of an algorithm.
In the above discussion of analysing average case complexity of Linear Search, we've assumed that the probability of finding an element at each location is same.
On the other hand, Amortized complexity takes into consideration the total performance of an algorithm over sequence of operations.
If amortized time complexity of an algorithm is O(f(n)) then individual operations may take more time than O(f(n)) but most of the operations take O(f(n)) time.
If average time complexity of an algorithm is O(g(n)) then individual operations may take more time than O(g(n)) but at random the operations take O(g(n)) time.
4) Understanding amortization using Dynamic Array
Dynamic Array is the best example to understand Amortized Time complexity.
Dynamic array is a linear data structure which is growable and shrinkable in size upon necessity.
vector in C++ and ArrayList in Java use the concept of Dynamic array in their implementation.
There arises two cases for insertion in dynamic array:
1.When there exists free space in the array
Time complexity here is O(1).
2.When there is no space, a new array is to be created of size double the original array,the elements in the original array are to be copied and new element is inserted.
Time complexity here is
(Creation of new array of double the original size)+(Copying the elements of the original array)+(Insertion of the new element)
O(2N)+O(N)+O(1)=O(3N+1) where N is the size of the original array.
5) Analysing Worst Case Complexity
Suppose that we are doing insertion operation on the array for N times where N is the size of the array.In the worst case each operation takes O(3N) time complexity
Time complexity for overall operation is N×O(3N)=O(3N^2)
Ignoring constant,
The worst case time complexity for N insertions is O(N^2)
Now, let's analyse amortized time complexity.
6) Analysing Amortized Complexity
The amortized analysis averages the running times of operations in a sequence.
Assume initially size of the array is 1
Insert 1
Time: 1
Now, there is no space for insertion hence take an array of size double that of the original array i.e., 2
Copy the elements of the original array i.e., 1
Insert 2
Time: 2+1+1=4
Now, there is no space for insertion hence take an array of size double that of the original array i.e., 4
Copy the elements of the original array i.e., 1,2
Insert 3
Time: 4+2+1=7
Insert 4
Time: 1
Now, there is no space for insertion hence take an array of size double that of the original array i.e., 8
Copy the elements of the original array i.e., 1,2,3,4
Insert 5
Time: 8+4+1=13
Insert 6
Time: 1
Insert 7
Time: 1
Insert 8
Time: 1
and so on...
Assume there would be m appends.
Hence the cost of m appends would be m since we are appending m elements(the append operation would cost O(1)) plus the cost of doubling when the array needs to grow.
Let's calculate the cost for doubling of array
The first doubling costs 1,second doubling costs 2,third doubling costs 4...and so on.
It's easier to solve the above equation if we reverse it.
Adding m
Adding m/2
Adding m/4
Adding m/8
and so on...
The highlighted box equals to m summing the total to 2m.
Hence keeping everything together appends cost m and doubling costs 2m.
Hence the total sum would be 3m i.e., O(m) .
Therefore m appends costs O(m). So each append costs O(1).
This is amortized complexity for dynamic array.
7) Conclusion
Though the amortized complexity for an append is O(1), the worst case it takes is still O(n) where n is size of the array. Therefore, with this article at OpenGenus, you must have a solid idea of
Amortized time complexity. | {"url":"https://iq.opengenus.org/amortized-time-complexity/","timestamp":"2024-11-04T10:59:35Z","content_type":"text/html","content_length":"69613","record_id":"<urn:uuid:28a0e22f-6a4d-40eb-9de5-39a11051d702>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00796.warc.gz"} |
Two cars with speed of 15 kmph and 30 kmph respectively are 100 km apart and face each other. The di
Two numbers are in the ratio of 2:9. If their H. C. F. is 19, numbers are: 1. 6, 27 2. 8, 36 3. 38,
If the average of four consecutive odd numbers is 12, find the smallest of these numbers? 1. 5 2. 7
Which is not the prime number?. 1. 43 2. 57 3. 73 4. 101 Answer: 57 Explanation: A positive natur
5. In an IPL match, the current run rate of CSK is 4.5 in 6 overs. What should be the required run
Sita can complete a work in 10 days. Geeta is 25% more efficient than Sita and Rita is 60% more effi
A secret can be said by only 2 persons in 5 minutes. The same person tells the secret to 2 more pers
Hello Saurabh Bhan, I Have Problem Releted With Math. because My Math is So Week. you Can Help Me...
Choose the correct option. Options A language has 28 different letters in total. Each word in the
A basket ball is dropped from a height of 20 feet. It bounces back each time to a height which is on | {"url":"https://m4maths.com/user-profile.php?UID=585","timestamp":"2024-11-07T01:03:53Z","content_type":"text/html","content_length":"50078","record_id":"<urn:uuid:ec5c2c29-bb7b-49a2-9330-912ab21321b1>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00419.warc.gz"} |
Non-Linear Data Modeling: Understanding Polynomial Regression
A Guide to Non-Linear Data Modeling: UnderstandingPolynomial Regression A common statistical technique for determining the relationship between a dependentvariable and one or more independent
variables is linear regression. The relationshipbetween variables, however, may not always be linear. Polynomial regression is advised incertain circumstances. In this post, we'll go through
polynomial regression's definition,benefits, and practical implementation. Polynomial Regression: What Is It? Regression analysis that analyzes the link between the dependent variable and one or
moreindependent variables is known as polynomial regression. Polynomial regression and linearregression vary in that the independent variable in the former is changed into a polynomialfunction of one
or more degrees. In contrast to linear regression, which fits data to a straightline, this enables us to fit a curve, or nonlinear functions, to our data. Consider the following home dataset as an
illustration, where feature x represents size insquare feet. A straight line might not adequately match the dataset in this situation. In suchcases, the data can be fitted to a curve, such a quadratic
function. Size x and size squared,
which is size raised to the power of two, are both components of a quadratic function. Youmay be able to obtain a significantly better model for your data by utilizing polynomialregression. Polynomial
Regression's benefits When compared to linear regression, polynomial regression has a number of benefits.Among them are: 1. Flexibility: When modeling the relationship between variables, polynomial
regression offers more flexibility. You can use it to record non-linear correlationsbetween variables and fit curves to your data. 2. Accuracy gain: When there is a non-linear relationship between the
variables, polynomial regression can produce results that are more accurate than those oflinear regression. 3. Simplicity: When compared to other sophisticated regression techniques, polynomial
regression is quite straightforward to use and comprehend. Application of Polynomial Regression Polynomial regression is simple to implement and may be done in a few stages. Here is adetailed
procedure for carrying out polynomial regression: 1. Preprocessing: The data must first be cleaned and transformed as the first stage in the preprocessing process. This entails handling outliers,
filling in any missingvalues, and normalizing the features. 2. In this stage of feature engineering, you must turn the independent variable into a polynomial function with one or more degrees. For
instance, if the independentvariable is size, you can square it, cube it, and so forth. 3. Model construction: Using the converted data, the model construction process is the following step. To
develop the model, you can use a variety of algorithms, includingsupport vector machines, decision trees, and linear regression. 4. Evaluation of the model: The model's performance is assessed in the
final stage using metrics like mean squared error, R-squared, or adjusted R-squared. | {"url":"https://keepnotes.com/stanford-university/machine-learning/339-polynomial-regression","timestamp":"2024-11-02T19:09:13Z","content_type":"text/html","content_length":"128612","record_id":"<urn:uuid:328a1fec-b864-481f-bf51-c95d554570bb>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00876.warc.gz"} |
Comparision of power of sequence of all natural number and the factorial of
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Comparision of power of sequence of all natural number and the factorial of
Some time ago I have found correlation between power of natural numbers and factorial.
I do not know if this is a known fact or not, but I would like to know why is this so.
For example if you have a sequence of all natural numbers and you calculate the second, third or any other (n) power of this number, and you afterwards calculate the difference between results and
recalculate differences between this results the same time of the given power you get factorial of the n.
Maybe is better if you look at the photos of calculation because English is not my mother language.
etc. it goes on and on
3 Replies
The basic reason is that the n-th derivative of x^n is constant and equal to the factorial of n.
In[2]:= D[x^9, {x, 9}]
Out[2]= 362880
In[3]:= 9!
Out[3]= 362880
Then the observed result is a special case of the mean value theorem for divided differences (
) for a polynomial function and equidistant interpolation nodes, after taking into account the relationship between finite differences and divided differences. The proof follows from the Newton-form
divided difference representation of the Lagrange interpolating polynomial.
Be respectful. Review our Community Guidelines to understand your role and responsibilities. Community Terms of Use | {"url":"https://community.wolfram.com/groups/-/m/t/167773","timestamp":"2024-11-07T11:04:36Z","content_type":"text/html","content_length":"105977","record_id":"<urn:uuid:ad3d01b6-5157-47d5-b4bf-e9b249822bf5>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00602.warc.gz"} |
EEE481 Homework 7 | Essay Heroes
EEE481 Homework 7
Problem 1
Consider the cart and pendulum system describing the evolution of the cart velocity, the
pendulum angle and the pendulum angular velocity (see details in
http://tsakalis.faculty.asu.edu/notes/models.pdf). The linearized equations around the unstable
equilibrium (angle 180deg, zero input, zero velocities) for the deviations of the state variables
from the linearization point, are:
(???? + ????)
???????? = −???????????? + ????ℓ
???????? + ????
???????? = ????
???????? = −???????????? + ????????ℓ???? + ????ℓ
The various constants are (all values in SI): friction coefficients ????????, ???????? = 0.1, pendulum mass ???? =
0.2, pendulum length ℓ = 0.2, cart total mass ???? = 0.4, gravitational constant ???? = 9.81.
The Force term is applied by the cart wheels driven by a DC motor with a model from the
applied voltage (say, -5:5Volts) to the force
???? = ???????? + ????
???????? + ????????????????, ???????????????? = ????????, ???? =
Where the back-emf is taken as proportional to the motor angular velocity, which is also
proportional to the cart velocity, while the motor current generates the torque that is converted
to force by the wheels of radius r. The constants are motor resistance ???? = 5, inductance ???? =
0.1, emf/torque-constant ???? = 0.3, and wheel radius ???? = 0.1.
We want to design a controller for this system to stabilize the inverted pendulum and be able to
follow commands for the cart speed. We want to use a sequential approach where we first
stabilize the angle with an “inner-loop controller”. Then form the inner closed loop and design
an “outer loop controller” for the cart velocity.
There are two difficulties associated with this problem. One is that the angle subsystem has a
RHP pole and a zero at the origin. Its stabilization requires a controller with a RHP pole. We can
solve this as a modified PID problem where instead of the integrator we use a RHP pole
determined iteratively.
The other difficulty is technical, namely, how to create the various systems and loops without
leaving stray pole-zero cancellations (possibly in the RHP) and without resorting to tedious hand
calculations. (One approach for this is to implement the model in Simulink and use the “linmod”
command, and the other -taken here- is to work with the state space model using the
“feedback” command.)
1. Form the state-space description of the system with one input (Voltage) and three
outputs, (velocity, angle, Voltage). It is convenient to keep the voltage as an output to
make it easy to simulate with Matlab commands. We want to implement the controller
in DT with a sampling rate 100Hz. For this system, it is more convenient to follow the wplane approach, find the ZOH equivalent of the plant now and convert to the w-plane
and do not have to do corrections during the subsequent iterations.
2. Extract the angle subsystem (e.g., P(2), if the angle is the 2nd output) and design an
“inner-loop” stabilizing controller ???????? =
, where “????” and the crossover
frequency are to be iterated to achieve a reasonable response (crossover should be
larger than the RHP poles, “????” should not be larger than the RHP pole). PM is to be
selected but large values are not very realistic (consider 40-50).
3. Form the inner loop system.
Hint: The Matlab command is
>> Pi=feedback(P*Ci,ss([1]),1,2);
The second argument is the feedback system, the third output is the index of the
P-inputs connected to the C-outputs (1, there is only one) and the fourth output is
the index of the P-outputs connected to the C-inputs (“2” for the angle being the
nd output). Pi is a new system that has the same number of inputs and outputs,
the input is angle set-point and the outputs are the outputs of P.
4. Design a DT PID controller for the velocity (1st output, so P(1) subsystem).
Prefilter considerations: For step inputs, the linear control input becomes very large and
the angle can overshoot a lot. In practice, that is a problem because the input is limited
by the voltage constraints and if the angle becomes too large, the inverted pendulum will
fall (because of acceleration constraints). Here, it may be beneficial to consider a
prefilter, e.g., a lowpass filter ???? =
either first or second order roughly at the
bandwidth of the outer loop crossover. Design the controller for the filtered plant and
then include the filter in the controller. The objective is to keep the control input
amplitude and the angle excursions reasonable, without slowing down the loop too
much. For this size of cart-and-pendulum, we expect stabilization in a few seconds. Also,
keep in mind that the outer loop system has “negative gain”.
5. Form the outer loop system as in Part 3 and check the responses.
>> Po=feedback(Pi*Co,ss([1]),1,1);
The last argument “1” assumes that the first output is the cart velocity.
6. Since all the design was done in the w-plane, it is straightforward to discretize the
controllers (using Tustin) and form the feedback loops with the “feedback” command.
Provide the transfer functions of your controllers and plots of the relevant time and
frequency responses.
Problem 2
Consider the system
????(????) =
−0.4???? + 4
2 + 4???? + 4
1. Design a DT controller using the w-plane method for crossover 4 rad/s, PM = 45 degrees
and a sampling time of 10 Hz.
2. For an additive measurable disturbance at the plant output with transfer function
????(????) =
, design a DT feedforward controller H(z) to reduce the transient component
of the DT PID designed in Part 1. Use the “naïve” approach and comment on the benefit
of such a component for this case.
Problem 3
Consider the cart and pendulum system of Problem 1.
1. Design a DT state estimator (observer) for the system with outputs cart velocity and
pendulum angle. It is convenient here to use a LQ approach because of the multiple
outputs. Choose the estimator gain to achieve convergence faster than one second.
Hint: The state estimator has the form ????????+1 = ???????????? + ???????????? + ????(???????? − ????????) where
x,y are the estimated states and outputs and m is the measurements (the plant
y_k). The design equations are implemented in Matlab by the function “dlqr”
???? = ????????????????(????
, ????
,???? ∗ ????
2 + ???????? ∗ ???? ∗ ????
The design parameters have been simplified to to depend only on a gain factor
mu (such that larger mu yield higher bandwidth) and the sample time T. The
performance is characterized by eigenvalues of the observer eig(A-L*C) and its
Sensitivity (error system)
???????? = [???? − ????????, ????, ????, −????]
2. Design a DT state feedback to stabilize the cart-and-pendulum system. You may add an
offset to the velocity measurement to enable convergence to a nonzero velocity but do
not worry about integral action control.
Hint: The LQR problem minimizes the cost of states (x’Qx) and control inputs
(u’Ru), for the system
????????+1 = ???????????? + ????????????
Its solution is the linear state feedback
???????? = −???????????? => ????????+1 = (???? − ????????)????????
It is implemented in Matlab by the function “dlqr”, with syntax
???? = ????????????????(????, ????, ????
∗ ????, ????????)
The design parameters are Q = C’C (penalizing the output) and ???? which is the
penalty on the control input and serves as an inverse-gain parameter. When ????
decreases the controller bandwidth increases. The controller performance is
characterized by the eigenvalues of A-BK and the Sensitivity of the input to
disturbances at the plant input
???????? = [???? − ????????, ????, ????, −????]
3. Combine the state feedback with the state estimator to obtain an output feedback
controller. Illustrate the time and frequency responses of the controller.
Hint: The “model-based” controller uses the state feedback with the states
replaced by their estimates. The combined controller is (D=0)
????????+1 = ???????????? + ???????????? + ????(???????? − ????????
) ; ???????? = −???????????? ; ???????? = ????????????
=> ????????+1 = ???????????? − ???????????????? − ???????????????? + ????(????????
) ; ???????? = −???????????? ;
Thus, for negative feedback, the controller state-space representation becomes
[???? − ???????? − ????????, ????,????, 0]
The controller has two inputs, the measured outputs of the system (m_k), and
one output, the control u_k. An external input (scaled reference) may be added
to the measurement m_k (i.e., the y_k of the plant). However, tracking of a
setpoint requires integral action which is not considered here.
Problem 4
Consider the cart and pendulum stabilized by the inner-outer PID controller of Problem 1. In the
context of refining the plant models in case of a change, we want to estimate the plant transfer
functions based on input-output data. The system is, of course, unstable and data cannot be
collected without a stabilizing controller.
Use the closed-loop system of Problem 1 with input reference velocity and outputs angle,
velocity, and voltage (control input). Apply a reference input (Random or Square Wave and
combinations) with maximum amplitude around 1. Add random noise to the outputs (angle,
velocity) at a level of 0.02 ~ 0.5 degrees.
rn=(rand(N,1)-.5); n1 =(rand(N,1)-.5)*0.02; n2 =(rand(N,1)-.5)*0.02;
Y=lsim(Pdo,r); % Pdo created in Problem 1
u=Y(:,3); y1=Y(:,1)+n1; y2=Y(:,2)+n2;
Design a batch parameter estimator to identify the transfer functions from the plant input “u” to
each output “yi” (angle, velocity). Look at the system equations to decide the order of the
estimated transfer functions. Try different filters for the input-output pairs and compare with
the known transfer functions. Could the estimates be used to redesign the controller? | {"url":"https://essayheroes.us/eee481-homework-7-2/","timestamp":"2024-11-12T05:14:06Z","content_type":"application/xhtml+xml","content_length":"106765","record_id":"<urn:uuid:361a435c-c682-45e0-a8ef-04c56876639c>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00727.warc.gz"} |
2nd PUC Physics Model Question Paper 4 with Answers
Students can Download 2nd PUC Physics Model Question Paper 4 with Answers, Karnataka 2nd PUC Physics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus
and score more marks in your examinations.
Karnataka 2nd PUC Physics Model Question Paper 4 with Answers
Time: 3 Hrs 15 Min
Max. Marks: 70
General Instructions:
1. All parts are compulsory.
2. Answers without relevant diagram/figure/circuit wherever necessary will not cany any marks
3. Direct answers to the Numerical problems without detailed solutions will not carry any marks.
Part – A
I. Answer all the following questions ( 10 × 1 = 10 )
Question 1.
State Ohm’s law.
Ohm’s Law-At constant temperature the current through a conductor is directly proportional to the potential difference between its ends.
Question 2.
Define current sensitivity of a galvanometer.
Deflection per unit current is called current Sensitivity.
Question 3.
Write the expression for force experienced by a straight conductor of length L carrying a steady current I, moving in a uniform external magnetic field B.
Question 4.
What is ‘retentivity’ in magnetism?
Retaining the magnetism even after the removal of the magnetising field is called retentivity.
Question 5.
Where on the earth’s surface is the magnetic dip zero?
At magnetic equator dip is zero. ,
Question 6.
State ‘Lenz’s law in electromagnetic induction.
The polarity of induced emf is such that it tends to produce a current which opposes the Change in magnetic flux that produced it.
Question 7.
Write the condition for ‘resonance’ of series LCR circuit.
inductive reactance = capacitive reactance X[L ]= X[c]
Question 8.
What is ‘wattless’ current?
In purely inductive or capacitive circuit, no power is dissipated even though a current is flowing in the circuit. This current is called wattles current.
Question 9.
Write any one advantage of light emitting diode.
Frequency remains constant.
Question 10.
What is attenuation in communication system?
Mass number of the daughter nuclide 234
Part – B
II. Answer any Five of the following questions. ( 5 × 2 = 10 )
Question 11.
Represent graphically the variation of resistivity with absolute tempera ture for copper and nichrome metals.
Question 12.
Write the expression for cyclotron frequency and explain the terms.
V[c] = \(\frac{\mathrm{qB}}{2 \pi \mathrm{m}}\)
q- Charge B- magnetic field
m-mass of the charged particle
V[c ]= cyclotron frequency
Question 13.
State and explain ‘curie’s Law’ in magnetism.
The magnetic susceptibility ( χ ) of a para magnetic substance is inversely proportional to the absolute temperature (T).
χ = \(\mathrm{C} \frac{\mu_{0}}{\mathrm{T}}\)
C – Curie constant.
Question 14.
Mention any two factors on which the self inductance of a coil depends.
Self-inductance of a coil depends on
1. number of turns of the coil
2. area of cross section and length (geometry)
3. permeability of the medium
Question 15.
Give any two application of ultraviolet radiations.
1. used in LASIK eye surgery
2. UV lamps are used to kill germs in water purifiers
3. Disinfection for virus and bacteria
4. To produce photo electric current in burglar alarm.
Question 16.
What is polarisation of light? Name any one method of producing plane polarised light.
The phenomenon of confining the vibrations of light in a single plane is called polarisation. Reflection / scattering.
Question 17.
Calculate de Broglie wavelength associated with an electron moving with a speed of 2 × 10^5 ms^–1 . Given h = 6.625 × 10^–34JS, m[e] = 9.11 × 10^–31kg.
Question 18.
Write any two advantages of Light Emitting Diode (LED) over conventional in candescent low power lamps.
Advantages of LED
1. Low operational voltage and less power consumption
2. Fast action and no warm up time required
3. Long life and ruggedness
4. Fast on-off switching capability
Part – C
III. Answer any five of the following questions. ( 5 × 3 =15)
Question 19.
Give any three properties of electric field lines.
Properties electric field lines.
1. Field lines starts from positive charges and end at negative charges.
2. In a charge free region electric field lines can be taken to be continuous curves without any breaks.
3. Two field lines can never cross each other.
4. Electrostatic field lines do not form any closed loops.
Question 20.
Obtain the expression for effective capacitance of two capacitors connected in series.
Let c[1],c[2] -capacitance of 2 capacitors connected in series
Q-charge on each capacitor
V[1] V[2]-pd across C[1] and C[2]
V-Total voltage drop across the combination
Then V = V[1] + V[2 ]……… (1)
Also Q = CV
V[1 ]= \(\frac{\mathrm{Q}}{\mathrm{C}_{1}}\) and V[2 ]= \(\frac{\mathrm{Q}}{\mathrm{C}_{2}}\)
If system of capacitors is replaced by a single capacitor of equivalent capacitance cs then
V[s ]= \(\frac{\mathrm{Q}}{\mathrm{C}_{\mathrm{s}}}\)
Question 21.
Write any three differences between diamagnetic and paramagnetic materials.
┃Diamagnetic │Paramagnetic ┃
┃1. Weakly magnetised in a direction opposite to the applied magnetic field │1. Weakly magnetised along the direction of the applied magnetic field.┃
┃2. Move from stronger to weaker part of the external magnetic field │2. Move from weaker to the stronger part of the external magnetic field┃
┃3. Magnetic susceptibility is Low and negative │3. Magnetic susceptibility is low and positive. ┃
┃4. Example: Bismuth, copper, lead, silicon │4. Example: Aluminium, sodium, calcium, oxygen ┃
Question 22.
Describe the col and barmagnet experiment to demonstrate the phenomenon of electromagnetic induction.
Coil and bar magnet experiment In the figure coil C1 is connected to a galvanometer G
1. when N-pole of the bar magnet is pushed towards the coil, there is a momentary deflection in the galvanometer.
2. when the magnet is pulled away from the coil galvanometer shows momentary deflection in the opposite direction.
3. Faster movements result in a larger deflection.
4. But no deflection when the coil and magnet are stationary with respect to each other. Or no deflection when there is no relative motion.
Therefore it shows that the relative motion between the magnet and coil induces electric current.
Question 23.
Derive the expression for effective focal length of two thin lenses kept in contact.
Let f[1]= focal length of first lens, and
f[2] = focal length of second lens
OP = u= object distance
PI = v = image distance due to the combination
PI[1 ]= v[1 ]= image distance due to first lens
For the image formed by lens A,
Question 24.
Write any three experimental observations of photoelectirc effect.
1. The photoelectric emission is an instantaneous process, even when incident radiation is exceedingly dim.
2. Above threshold frequency, the photo current is directly proportional to the intensity of incident radiation.
3. Above the threshold frequency, saturation current is proportional to the intensity
of the incident radiation and stopping potential is independent of intensity.
4. There exists a certain minimum cut-off frequency called ‘threshold frequency’ below which no photo emission however intense the incident beam.
5. Above threshold frequency the kinetic energy of the photo electrons is directly proportional to the frequency of incident radiation and is independent of intensity.
Question 25.
How zener diode is used as a voltage regulator? Explain.
Zener diode as a voltage regulator
The circuit connections are made as shown in the figure. The Zener diode is reverse biased. If the unregulated input voltage increases, the current through Rs and Zener diode also increases. This
increases the voltage drop across Rs without any change in the voltage across the Zener diode. This is because in the breakdown region Zener voltage remains constant even though the current through
Zener diode changes.
Similarly, if the input voltage decreases, the current through Rs and Zener also decreases. So, any increase or decrease of input results in increase or decrease of voltage drop across Rs without
change in voltage across Zener diode. Hence it acts as a voltage regulator.
Question 26.
What is the function of ‘receiver’ in communication system? Draw the block diagram of AM – receiver.
A receiver extracts the desired message signals from the received signals at the channel output.
Part – D
IV. Answer any two of the following questions ( 2 × 5 = 10 )
Question 27.
Using Gauss’s law in electrostatics, obtain the expression for electric field due to a uniformly charged thin spherical shell at a point
i. outside the shell and
ii. inside the shell
Let σ be the uniform surface charge density of a thin spherical shell of radius R Field outside the shell.
Consider a point P outside the shell at a distance r from the centre of the shell. Imagine a gaussian sphere of radius ‘r’ The electric flux at P due to surface [∆] S is
∆φ= \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\Delta \mathrm{S}}\) = E[∆]S COSθ=E[∆]S {COS θ=1}
Total electric flux due to the sphere is
φ = E4πr^2 ……….(1)
From Gauss law the electric flux
where q= total charge enclosed by the surface From (1) & (2)
b) electric field inside shell E = 0
Question 28.
Derive σ = \(\frac{\mathrm{ne}^{2} \tau}{\mathrm{m}}\) Where the symbols have their usual meaning.
Consider conductor of length [∆]x, area of cross section A.
V[d]-drift speed of free electrons, in a time [∆ ]t, let all the electrons travel a distance [∆ ]x = V[d] ∆ t.
Electric current I = [∆]Q /[∆]t = neAv[d
]The acceleration acquired by the free electrons is given by a = \(\frac{-\mathrm{eE}}{\mathrm{m}}\)
where m= mass of the electron
If τ = relaxation time then velocity
v[d ]= \(\frac{-e E}{m} \tau\)
Current density J = \(\frac{1}{\mathrm{A}}\), J = σE therefore
σ = \(\frac{n e^{2} \tau}{m}\)
Question 29.
Obtain expression for the force between two infinitely long straight parallel conductros carrying current. Hence define ‘ampere’ the SI unit of electric current.
Consider two infinitely long straight conductors a & b carrying currents Ia & Ib respectively are separated by a distance ‘d’. The conductor ‘a’ produces the same magnetic field Ba at all points
along the conductor ‘b’.
B[a] = \(=\frac{\mu_{0} I_{a}}{2 \pi d}\) …………( 1 )
The conductor ‘b’ experiences a force Fba
F[ba] = B[a]LI[b] = \(=\frac{\mu_{0} I_{a} I_{b} L}{2 \pi d}\)
where L= length of the conductor
Similarly one can show that Fba = -Fab
(A) ampere- The currents flowing through two infinitely long parallel conductors separated by 1m distance is 1 ampere if they experience a force of 2 X 10^–7N per unit length in air or vacum.
V. Answer any two of the following questions ( 2 × 5 = 10 )
Question 30.
Derive the expression for the fringe width of interference pattern in Young’s double-slip experiment.
S[1] & S[2] are two coherent sources. GG| – screen at a distance D from the sources.
Let d = distance between the slits (two sources).
P = a point at a distance from the middle of the screen where a bright fringe is formed.
For constructive interference path difference = nλ
S[2]P – S[1]P = nλ where n = 0,1,2,3…….
From the diagram
(S[2]P )^2 – (S[1]P )^2=2xd
(S[2]P – S[1]P)(S[2]P + S[1]P) = 2xd
S[2]P ≅ S[1]P ≅ OP = D
S[2]P – S[1]P = \(\frac{\mathrm{xd}}{\mathrm{D}}\) = nλ
X[n ]= \(\frac{\mathrm{n} \lambda \mathrm{D}}{2}\)
Where n = 0, +1, ±2, ± 3 for bright fringes
±1, ±2, ±3 for dark fringes
Since the finges are equally spaced the distance between two consecutive bright or consecutive dark fringes gives finge width.
β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
Question 31.
Describe with suitable block diagra ms, action of pn-junction diode under forward and reverse bias conditions. Also draw I-V characteristics.
PN Junction diode under forward bias.
When the diode is forward biased as shown in the figure the depletion region width decreases and the barrier height is reduced. The electrons from n-side cross the depletion region and reach p-side
also holes from p-side cross the junction and reach the n-side. A concentration gradient is developed at the junction boundary. Due to this the motion of charged carriers on either side gives rise to
current. The total diode forward current is sum of hole diffusion current and conventional current due to electron diffusion.
PN Junction diode under reverse bias
When the diode is reverse biased the depletion region width increases and the barrier height is increased. This supresses the flow of electrons from n-side to p-side and holes from p-side to n-side.
Thus, diffusion current decreases. The conventional current is due to drift of the minority charge carriers which is of the order of micro amperes. The current under reverse bias is voltage
independent up to a critical reverse bias voltage known as breakdown voltage.
The I-V characteristics are as shown.
VI Characteristics
Question 32.
Assuming the expression for the radius of electron orbit, obtain the expression for the total energy of the electron in the stationary orbit of hydrogen atom.
The radius of the electron orbit is given by r = \(=\frac{\varepsilon 0 n^{2} h^{2}}{\pi m e^{2}}\)
Total energy E of the electron in a hydrogen atom is the sum of kinetic energy K and potential energy U
The electrostatic force of attraction between the revolving electrons and the nucleus is balanced by centripetal force in a dynamically stable orbit of hydrogen atom.
(negative sign signifies that the electrostatic force is in the – r direction)
The total energy of the electron in a hydrogen atom is
VI. Answer any three of the following question ( 3 × 5 = 15 )
Question 33.
The plates of a parallel plate capacitor have an area of 100 cm^2 each of 100 cm^2 each and are separated by 3 mm. The capacitor is charged by connecting it to a 400 V supply.
a) Calculate the electrostatic energy stored in the capacitor.
b) If a dielectric of dielectric constant 2.5 is introduced between the plates of the capacitor, then find the electrostatic
energy stored and also change in the energy stored.
Given A = 100 cm^2 = 100 x 10^–4m^2,
d = 3 mm = 3 x 10^–3m, V = 400 V, U[1] =?
K = 2.5,U[2 ]= ? also U[2 ]– U[1]=?
U = \(\frac{1}{2} \mathrm{CV}^{2}\) C = \(\frac{\varepsilon_{0} \mathrm{A}}{\mathrm{d}}\)
U[1 ]= 23.608 x 10^–7J
U[2 ]= 59.02 x 10^–7J
U[2]∼U[1 ]= 35.412 x 10^–7J
Question 34.
In the given circuit diagram, calculate : (i) The main current through the circuit and (ii) Also current through 9 Ω resistor.
RS = R[1] + R[2]
Calculating the value of effective resistance R = 2.76 Ω
Current through 9 Ω= 0.308A
Question 35.
A 20 Ω resistor; 1.5 H inductor and 35 μF capacitor are connected in series with a 220 V, 50 Hz ac supply, calculate the impedance of the circuit and also find the current through the circuit.
R = 20Ω, L= 1.5 H, C = 35 x 10^–6F,
V = 220V, v=50Hz, Z=?, I=?
X[L] = wL=2πvL = 471 Ω
Calculation of Z = 380.53 Ω
I = \(\frac{v}{z}\) = 0.578A
Question 36.
The radii of curvature of two surfaces of a convex lens is 0.2 m and 0.22 m. find the focal length of the lens of refractive index of the material of lens is 1.5. Also find the change in focal
length, if it is immersed in water of refractive index 1.33.
R[1]=0.2m, R[2]=–022m, n[8]=1.5, f[air]=?
f[water] =?, f[air]~f[water] =? n[w]=1.33
Substituting the values, and calculating f = 0.209 m. Calculating focal length when immersed in water f[water]=0.819
change in focal length = 0.61m
Question 37.
The half life of a radioactive sample [38]Sr^90 is 2 years. Calculate the rate of disintegration of 15 mg of this isotope.
Given Avogadro number = 6.023 x 10^23.
T[1/2] = 28 years = 28 x 365 x 24 x 600 = 8.83 x 10^8 seconds
90g \(\mathrm{Sr}_{38}^{90}\) contains 6.023 x 10^23 atoms
λ = \(\frac{0.693}{T_{1 / 2}}\)
Decay constant λ =7.85 x 10^10 x 1.004 x 1.004 × 10^10
Rate of disintegration R= λ N
R = 7.88 x 10^10 Bq | {"url":"https://kseebsolutions.net/2nd-puc-physics-model-question-paper-4/","timestamp":"2024-11-12T05:28:34Z","content_type":"text/html","content_length":"95262","record_id":"<urn:uuid:4d0d5129-ba3c-433a-8346-82eb7d24bc9f>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00321.warc.gz"} |
Continuous Probability Distribution (1 of 2)
Learning Objectives
• Use a probability distribution for a continuous random variable to estimate probabilities and identify unusual events.
In the previous section, we learned about discrete probability distributions. We used both probability tables and probability histograms to display these distributions. In this section, we shift our
focus from discrete to continuous random variables. We start by looking at the probability distribution of a discrete random variable and use it to introduce our first example of a probability
distribution for a continuous random variable.
Shoe Size
Let X = the shoe size of an adult male. X is a discrete random variable, since shoe sizes can only be whole and half number values, nothing in between. For this example we will consider shoe sizes
from 6.5 to 15.5. So the possible values of X are 6.5, 7.0, 7.5, 8.0, and so on, up to and including 15.5. Here is the probability table for X:
X 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12
P(X) 0.001 0.003 0.007 0.018 0.034 0.054 0.080 0.113 0.127 0.134 0.122 0.107
X 12.5 13 13.5 14 14.5 15 15.5
P(X) 0.085 0.052 0.032 0.016 0.009 0.004 0.002
And here is the probability histogram that corresponds to the table.
As is always the case for probability histograms, the area of the rectangle centered above each value is equal to the corresponding probability. For example, in the preceding table, we see that the
probability for X = 12 is 0.107.
In the probability histogram, the rectangle centered above 12 has area = 0.107.
We write this probability as P(X = 12) = 0.107.
And finally, as is the case for all probability histograms, because the sum of the probabilities of all possible outcomes must add up to 1, the sums of the areas of all of the rectangles shown must
also add up to 1.
Now we can find the probability of shoe size taking a value in any interval just by finding the area of the rectangles over that interval. For instance, the area of the rectangles up to and including
9 shows the probability of having a shoe size less than or equal to 9.
We can find this probability (area) from the table by adding together the probabilities for shoe sizes 6.5, 7.0, 7.5, 8.0, 8.5 and 9. Here is that calculation:
0.001 + 0.003 + 0.007 + 0.018 + 0.034 + 0.054 = 0.117Total area of the six green rectangles = 0.117 = probability of shoe size less than or equal to 9. We write this probability as P (X ≤ 9) = 0.117.
Recall that for a discrete random variable like shoe size, the probability is affected by whether or not we include the end point of the interval. For example, the area – and corresponding
probability – is reduced if we consider only shoe sizes strictly less than 9:
This time when we add the probabilities from the table, we exclude the probability for shoe size 9 and just add together the probabilities for shoe sizes 6.5, 7.0, 7.5, 8.0, and 8.5:
0.001 + 0.003 + 0.007 + 0.018 + 0.034 = 0.063
Total area of the five rectangles in green = 0.063 = probability of shoe size less than 9. We write this probability as
P(X < 9) = 0.063
Spotlight on Inequality Notation
Here is a review of inequality notation:
The symbol “<”means “less than”
• Here is a correct use of this symbol: 3 < 12. We read this left to right as 3 is less than 12.
• You can think of the “less than” symbol as an arrow pointing to the smaller number.
• Some students remember the “less than” symbol from elementary school as a hungry alligator that is eating the larger number:
• X < 12 means X is any number less than 12. If X represents shoe sizes, this includes whole and half sizes smaller than size 12.
• P(X < 12) is the probability that X is less than 12.
The symbol “≤”means “less than or equal to”
• X ≤ 12 means X can be 12 or any number less than 12. If X is shoe sizes, this includes size 12 as well as whole and half sizes less than size 12.
• We often say “at most 12” to indicate X ≤ 12.
• P(X ≤ 12) is the probability that X is 12 or less than 12.
The symbol “>”means “greater than”
• Here is a correct use of this symbol: 15 > 12. We read this left to right as 15 is greater than 12.
• You can also think of the “greater than” symbol as an arrow pointing (as before) to the smaller number.
• Or you can use the hungry alligator idea. The hungry alligator that is still eating the larger number:
• X > 12 means X is any number greater than 12. If X is shoe sizes, this includes whole and half sizes larger than size 12.
• P(X > 12) is the probability that X is greater than 12.
The symbol “≥” means “greater than or equal to”
• X ≥ 12 means X can be 12 or any number greater than 12. If X is shoe sizes, this includes size 12 as well as whole and half sizes greater than size 12.
• We often say “at least 12” to indicate X ≥ 12.
• P(X ≥ 12) is the probability that X is 12 or greater than 12.
To indicate an interval we combine “less than” and “greater than” symbols:
• To indicate the interval between 9 and 12, we write 9 < X < 12. This interval says “ 9 is less than X and X is also less than 12.” So this interval includes numbers greater than 9 but also less
than 12. For example, 10 is in this interval but 13 is not. Also, 9 and 12 are not in this interval.
• P(9 < X < 12) is the probability that X is between 9 and 12.
• P(9 ≤ X ≤ 12) is the probability that X is the same interval except that the interval also includes 9 and 12.
Transition to Continuous Random Variables
Now we will make the transition from discrete to continuous random variables. Instead of shoe size, let’s think about foot length. Unlike shoe size, this variable is not limited to distinct, separate
values, because foot lengths can take any value over a continuous range of possibilities. In other words, foot length, unlike shoe size, can be measured as precisely as we want to measure it. For
example, we can measure foot length to the nearest inch, the nearest half inch, the nearest quarter of an inch, the nearest tenth of an inch, etc. Therefore, foot length is a continuous random
What happens to the probability histogram when we measure foot length with more precision? When we increase the precision of the measurement, we will have a larger number of bins in our histogram.
This makes sense because each bin contains measurements that fall within a smaller interval of values. For example, if we measure foot lengths in inches, one bin will contain measurements from
6-inches up to 7-inches. But if we measure foot lengths to the nearest half-inch, then we now have two bins: one bin with lengths from 6 up to 6.5-inches and the next bin with lengths from 6.5 up to
You can use the following simulation to see what happens to the probability histogram as the width of intervals decrease. Change the interval width by clicking on 0.5 in., 0.25 in., or 0.1 in.
Click here to open this simulation in its own window.
At the bottom of the simulation is an option to add a curve. This curve is generated by a mathematical formula to fit the shape of the probability histogram. Check “Show curve” and click through the
different bin widths. Notice that as the width of the intervals gets smaller, the probability histogram gets closer to this curve. More specifically, the area in the histogram’s rectangles more
closely approximates the area under the curve. If we continue to reduce the size of the intervals, the curve becomes a better and better way to estimate the probability histogram. We’ll use smooth
curves like this one to represent the probability distributions of continuous random variables. This idea is discussed in more detail on the next page. | {"url":"https://courses.lumenlearning.com/suny-hccc-wm-concepts-statistics/chapter/continuous-probability-distribution-1-of-2/","timestamp":"2024-11-02T20:03:15Z","content_type":"text/html","content_length":"58417","record_id":"<urn:uuid:89f8dc4f-fa60-4be1-9a5b-d09c3dbf6027>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00730.warc.gz"} |
Brenden is Teaching
Overview for Y1/2:Y1: relate subtraction totaking away and to counting back.They solve problems,use practical apparatus to model the problem or represent theproblem in a drawing. Later, they count
back using a number line,then count back mentally. They explain how they worked out theproblem and record it using the and = symbols.
Y2:add or subtractmultiples of 10 by counting in tens, use n square / jottings.Recognise patterns in examples such as90 20 = 70 and 9 7 = 2use knowledge of number bonds to remember sums and
differencesof multiples of 10.Children solve word problems usingany one of the four operations.
Foundation StageObjectives: Begin to relateaddition to combining two groups of objects and subtraction totaking away In practical activities and discussion begin to usethe vocabulary involved in
adding and subtracting Use developingmathematical ideas and methods to solve practical problems
Linked AdultInitiated Play Activities:
Play with garage andcars how many on top / bottom, altogether
Numbers andoperation signs on magnetic boards.
IzzyIsland ICT mathsgames.
Play snakes andladders as group with Y1.
(Y1)Ican talk about adding and subtracting.
I can use the signs
MO: Sing5 currant buns using number rhyme sack. R startactivity.
Myanswer is 10 what is the question?
Main:Show bus
7 people on 2 wantto get off. How many people will be left. Encourage crn to hold up7 fingers, fold down 2 how many left?
Model on number linestarting at 7 and hopping back 1 hop for each passenger that getsoff.
Write 7 take away 2is 5 who can write that as a number sentence?
Repeat starting with10 people on bus.
Y1 startactivity.
Y2 continue withsome 1 digit from 2 digit subtractions using the numberline.
Use number line tosolve subtraction problems 1 digit from 2 digit.
Use bus to solvesubtraction problems.
Encourage use ofnumber line.
1-12 n track.Show 8 biscuits How many? Find 8 on the ntrack. Place counter on 8.How many if we take 2 away?
Take away 2 andcount how many left Encourage Lucy to count back using the numbertrack.
Lets recordwhat you did today
Show a subtractionpractically who can write the number sentence to show what I havedone?
Look at inverse bydiscussing which calculation will undo the subtraction.
Not achieved:
Other notes /next
MO:Count on and back in 10s using IWB bead string.Remind crn that whenwe count on in tens, we are adding 10 each time, and that when wecount back in tens we are subtracting 10. Show this on
Practicesome 1 digit subtractions with fingers use number fans.
Y1 startactivity
I have50p I spend 20p. how much money do I have left? What do we haveto do add, subtract, multiply, divide? Establish subtract. Modelusing the number line / square to subtract the tens.
Repeatfor other examples.
I have68p I spend 30p practice as before.
Solve someworded problems subtracting multiples of ten.
Solve some 1digit subtractions using apparatus or number line.
Read Kippers Toybox.Use box and toys to model take away sums.
There are 5 toys. 2are missing how many left in the box. Use fingers to model andfind answer.
Lets record what youdid today
Show me a subtractionwith the answer of 5 / 10 / 55 etcPairs on whiteboards.
Not achieved:
Other notes /next | {"url":"https://www.brendenisteaching.com/lessonplans/plan/409_Mixed_YR12_Unit_2_Block_A","timestamp":"2024-11-08T11:02:27Z","content_type":"text/html","content_length":"52540","record_id":"<urn:uuid:ca014ae4-6e85-4196-a3b4-38c3eebd9fa9>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00065.warc.gz"} |
Antonio Lerario
According to our database
, Antonio Lerario authored at least 11 papers between 2012 and 2024.
Collaborative distances:
Book In proceedings Article PhD thesis Dataset Other
On csauthors.net:
What Is the Probability That a Random Symmetric Tensor Is Close to Rank-One?
SIAM J. Appl. Algebra Geom., 2024
How regularization affects the geometry of loss functions.
CoRR, 2023
Low-Degree Approximation of Random Polynomials.
Found. Comput. Math., 2022
p-Adic Integral Geometry.
SIAM J. Appl. Algebra Geom., 2021
Random Geometric Complexes and Graphs on Riemannian Manifolds in the Thermodynamic Limit.
Discret. Comput. Geom., 2021
Hausdorff approximations and volume of tubes of singular algebraic sets.
CoRR, 2021
On the Number of Flats Tangent to Convex Hypersurfaces in Random Position.
Discret. Comput. Geom., 2020
Gap Probabilities and Betti Numbers of a Random Intersection of Quadrics.
Discret. Comput. Geom., 2016
Experiments on the Zeros of Harmonic Polynomials Using Certified Counting.
Exp. Math., 2015
Convex Pencils of Real Quadratic Forms.
Discret. Comput. Geom., 2012 | {"url":"https://www.csauthors.net/antonio-lerario/","timestamp":"2024-11-08T11:17:19Z","content_type":"text/html","content_length":"23948","record_id":"<urn:uuid:3488b506-58bf-43ad-bb2b-d804c0589834>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00539.warc.gz"} |
Notes - Groups HT23, Cosets
Let $H$ be a subgroup of $G$. What are the left cosets of $H$?
\[gH = \\{gh : h \in H\\}\]
Let $H$ be a subgroup of $G$. What are the right cosets of $H$?
\[Hg = \\{hg : h \in H\\}\]
What notation is used for the left cosets of $H$ in $G$, other than $gH$?
Let $H$ be a subgroup of $G$. Define the index of $H$ in $G$.
Can you state the coset equality lemma?
Let $H \leqslant G$ and $g, k \in G$. Then
\[gH = kH \iff k^{-1}g \in H\]
\[Hg = Hk \iff kg^{-1} \in H\]
If $g, k \in G$, what equivalence relation $g \sim k$ means that the equivalence classes are the left cosets of $H$?
\[g\sim k \iff k^{-1}g \in H\]
Related posts | {"url":"https://ollybritton.com/notes/uni/prelims/ht23/groups/notes/notes-groups-ht23-cosets/","timestamp":"2024-11-09T13:57:02Z","content_type":"text/html","content_length":"505686","record_id":"<urn:uuid:dadbdffd-67a7-4ae1-9c81-a18acf720359>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00556.warc.gz"} |
Travel Cost Calculator – Estimate Your Trip Expenses Instantly
Calculate your travel expenses below:
Total Travel Cost: 0 currency
What is Travel Cost Calculator?
A Travel Cost Calculator is a tool that helps individuals estimate the total cost of a trip based on specific parameters. It allows users to input details such as the distance they plan to travel,
the fuel efficiency of their vehicle, fuel prices, and additional expenses like tolls, accommodation, or food. The calculator then uses this information to compute the total expected cost of the
Common Components of a Travel Cost Calculator:
1. Distance: The total distance to be traveled, usually measured in kilometers or miles.
2. Fuel Efficiency: How many kilometers or miles the vehicle can travel per liter or gallon of fuel (e.g., km per liter or miles per gallon).
3. Fuel Cost: The current price of fuel per unit (liter or gallon).
4. Additional Costs: Other expenses that might arise during the trip, such as tolls, parking fees, meals, or accommodation.
Travel Cost Calculator Formula:
$\text{Total Travel Cost} = \left( \frac{\text{Distance}}{\text{Fuel Efficiency}} \times \text{Fuel Cost per Unit} \right) + \text{Additional Costs}$
How It Works:
• The calculator estimates how much fuel will be consumed by dividing the total distance by the fuel efficiency of the vehicle.
• It then multiplies the fuel used by the cost of fuel per liter (or gallon) to calculate the fuel cost.
• Any additional costs (e.g., tolls, food) are added to the fuel cost to get the total trip cost.
Let's say you want to calculate the cost of a 500-kilometer road trip:
• Distance: 500 km
• Fuel Efficiency: 15 km per liter
• Fuel Price: $1.20 per liter
• Other Costs: $50 (for tolls and meals)
The calculator would:
• Divide the distance by fuel efficiency to calculate fuel consumption: 500 km ÷ 15 km/l = 33.33 liters of fuel.
• Multiply fuel consumption by fuel price: 33.33 liters × $1.20 = $40 in fuel costs.
• Add other expenses ($50), so the total travel cost would be $90.
Travel cost calculators are especially helpful for planning road trips or budgeting trips where transportation costs are a key concern.
Travel Cost Calculator FAQ
1. What is a Travel Cost Calculator?
A Travel Cost Calculator is a tool that helps you estimate the total cost of a trip. It calculates expenses based on factors such as distance, fuel efficiency, fuel price, and additional costs like
tolls, meals, and accommodation.
2. How does the Travel Cost Calculator work?
The calculator works by asking for details like the distance of your trip, the fuel efficiency of your vehicle, fuel costs, and any additional expenses. It then calculates the total cost by
multiplying fuel consumption by the fuel price and adding any extra costs.
3. What information do I need to use the calculator?
To use the Travel Cost Calculator, you will need the following:
• The total distance of your trip (in kilometers or miles).
• Your vehicle’s fuel efficiency (km per liter or miles per gallon).
• The price of fuel (per liter or gallon).
• Any additional costs like tolls, meals, or accommodation.
4. Can I use the calculator for international travel?
Yes, the Travel Cost Calculator can be used for both domestic and international trips, as long as you have the relevant data such as local fuel prices and travel distance.
5. What are “additional costs” in the Travel Cost Calculator?
Additional costs refer to other expenses you might incur on your trip, such as tolls, parking fees, food, accommodation, or other charges outside of fuel costs.
6. How accurate is the Travel Cost Calculator?
The accuracy of the calculator depends on the data you input. If you provide accurate values for fuel efficiency, fuel cost, distance, and additional expenses, the estimated cost will be close to
your actual travel costs.
7. Can this calculator be used for any type of vehicle?
Yes, the calculator can be used for any type of vehicle, as long as you know its fuel efficiency (km/l or mpg). This includes cars, trucks, motorcycles, and even RVs.
8. Why should I use a Travel Cost Calculator?
Using a Travel Cost Calculator helps you plan and budget your trip more effectively. It ensures that you have a clear idea of your travel expenses in advance, helping you manage your finances better.
9. Can the Travel Cost Calculator help with planning fuel stops?
While the calculator does not directly plan fuel stops, it estimates how much fuel you will use for your trip, helping you determine where and when you might need to refuel based on your route.
10. Is the Travel Cost Calculator free to use?
Yes, the Travel Cost Calculator is free to use on TheUSAList, allowing you to easily estimate travel expenses for any trip. | {"url":"https://www.theusalist.com/2024/10/travel-cost-calculator.html","timestamp":"2024-11-03T17:01:35Z","content_type":"application/xhtml+xml","content_length":"211998","record_id":"<urn:uuid:72d59951-d7ca-484b-af3a-17c11300ef6a>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00088.warc.gz"} |
Introduction, Orthotropic Lamina |
1.1. Introduction, Orthotropic Lamina
1 . Analysis of the Orthotropic Lamina
1.1. Introduction
1.2. Hooke’s Law
1.3. Relationships between elastic constants and Matrix of Elasticity
1.4. Matrix of Elasticity
2. Classical theory of Laminates
2.1. Introduction
2.2. Basic Formulas
2.3. Laminate stiffener matrix
2.4. Calculation of Stress and deformation
2.5. Thermal Stress
2.6. Calculation of Elastic Constants
>> Composite Analysis Tool
In materials science, Composite laminates are assemblies of layers of fibrous composite materials which can be joined to provide required engineering properties, including in-plane stiffness, bending
stiffness, strength, and coefficient of thermal expansion.
The individual layers consist of high-modulus, high-strength fibers in a polymeric, metallic, or ceramic matrix material. Typical fibers used include graphite, glass, boron, and silicon carbide, and
some matrix materials are epoxies, polyimides, aluminium, titanium, and alumina.
Layers of different materials may be used, resulting in a hybrid laminate. The individual layers generally are orthotropic (that is, with principal properties in orthogonal directions) or
transversely isotropic (with isotropic properties in the transverse plane) with the laminate then exhibiting anisotropic (with variable direction of principal properties), orthotropic, or
quasi-isotropic properties. Quasi-isotropic laminates exhibit isotropic (that is, independent of direction) inplane response but are not restricted to isotropic out-of-plane (bending) response.
Depending upon the stacking sequence of the individual layers, the laminate may exhibit coupling between inplane and out-of-plane response. An example of bending-stretching coupling is the presence
of curvature developing as a result of in-plane loading.
A lamina of composite is builded up with unidirectional or bi –directional reinforcement andc the thickness is generally between 0.1 and 5 mm. It is used for the construction of laminates whose
characteristics (thickness, number plates , orientation, etc. . ) are determined on the basis of specific project needs . The analysis of a laminate thus requires knowledge of the mechanical behavior
of single lamina and in particular of its constitutive equations .
A foil of the composite element is a microscopically heterogeneous element since its composition is practically variable from point to point . From the macroscopic point of view , ie considering a
scale larger than the size of the fibers , it can however be considered homogeneous . In this scale , in addition, it exhibits a mechanical anisotropic behavior, in particular orthotropic .
Anisotropy is the property of being directionally dependent (its characteristics vary continuously), as opposed to isotropy, which implies identical properties in all directions.
In particular, if the material has three planes of symmetry mutually orthogonal , it is called orthotropic.
In a composite laminate the plans are identified by the middle plane of the lamina and this orthogonal planes are parallel to the two principal directions (direction of the fibers and the direction
orthogonal reinforcement for unidirectional fiber) . To better understand the difference between an anisotropic material and an orthotropic material is useful to observe – for example – that the
application of a tensile load to an element with prismatic form and made with anisotropic material , produces deformations and variable flows along all element sides. This occurs regardless of the
particular direction of the applied loading . Instead if the material is orthotropic , there are three directions mutually orthogonal such that the application of a tensile stress in these directions
produces , like an isotropic material, a costant deformation without distortion in the plans identified by these (Figure 1 ) .
Figure 1
These three directions are called the principal directions of the material. Considering instead of a cube of material, a composite lamina, if the direction of application of the load coincides with a
main direction (Figure 2 (a)) then corresponds to a normal stress simple a state of uniform deformation without sliding, while if the load direction is deflected with respect to the principal
directions of the load also produces sliding in the plane (Figure 2 (b)).
Figure 2
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Cost Per Ounce Calculator
The Cost Per Ounce Calculator is a handy tool for consumers and businesses alike, allowing you to determine the unit price of products based on their total cost and weight. This helps in making
informed purchasing decisions.
How to Use the Cost Per Ounce Calculator
1. Enter the total cost of the product in dollars.
2. Input the total weight of the product in ounces.
3. Click 'Calculate Cost Per Ounce' to see the price per ounce.
COPO Formula
The formula to calculate the cost per ounce is:
Cost Per Ounce = Total Cost / Total Weight
Example Calculation
If the total cost is $12.00 and the total weight is 8 ounces, the cost per ounce would be:
Cost Per Ounce = $12.00 / 8 oz = $1.50
Benefits of Knowing Cost Per Ounce
Understanding the cost per ounce is crucial for comparing prices across different products or brands. This metric helps you identify the best value for your money, especially in bulk purchasing | {"url":"https://profitcalculate.com/cost-per-ounce-calculator/","timestamp":"2024-11-12T10:33:09Z","content_type":"text/html","content_length":"26770","record_id":"<urn:uuid:2261e4aa-0a08-4b32-a69f-759c43b0eea5>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00616.warc.gz"} |
It's Time for Pi!
March 14 marks the annual celebration of Pi day, as the month and day number corresponds with the first three digits of Pi, 3.14. I guess we should also wish Albert Einstein a very happy birthday and
maybe bake a Pi in his honour. The genius we have to thank for this celebration and loophole that allows us to indulge in pies all day is physicist Larry Shaw. Pi day was firstly celebrated in 1988
and it featured parades, pie eating competitions and interactive games hosted by the Exploratorium, a museum based in San Francisco. By 2009 Pi day became a official national holiday. But why is Pi
so important and why do we have a whole day reserved for it?
Pi is a irrational, transcendental number meaning it goes on to infinity. The mathematician Archimedes was credited for being the first to accurately calculate the estimated value of Pi. The word Pi
is derived from the Greek word “Perimetros” which directly translates to: perimeter, circumference . The fact that Pi is crucial because of what it represents in relation to a circle—it’s the
constant ratio of a circle’s circumference to its diameter.
The Babylonians calculated the area of a circle by taking 3 times the square of the radius which gave Pi to be 3.125 according to a Babylonian tablet that was dated cc 1680. Then again the Rhind
Papyrus (1605) that belonged to the mate mathematicians of ancient Egypt gives another formula for Pi that results in Pi being 3.1605.
So how did we get to 3.14? Archimedes of Syracuse during 287-212, approximated the area of the circle by using the Pythagorean Theorem to find the areas of two regular polygons: the polygon inscribed
within the circle and the polygon within which the circle was circumscribed. Since the actual area of the circle lies between the areas of the inscribed and circumscribed polygons, the areas of the
polygons gave upper and lower bounds for the area of the circle. Archimedes knew that he had not found the value of π but only an approximation within those limits. However, he demonstrated that π is
between 3 1/7 and 3 10/71.
The Greek letter π was introduced by William Jones in 1706 and was later popularized by
Leonhard Euler who adopted it in 1737.
Nowadays, Pi has been calculated to over 1 trillion decimal places and the calculations are still ongoing. The aim of this big celebration is to increase children’s interest in mathematics and
sciences. Teachers, scientists and mathematicians around the world celebrated with sweet pies, interactive games and fun math problems. How did you celebrate? Let us know on The Nest Instagram page
or our The Nest website. | {"url":"https://www.thenestisb.com/post/it-s-time-for-pi","timestamp":"2024-11-03T16:57:30Z","content_type":"text/html","content_length":"1050490","record_id":"<urn:uuid:8cdeb77e-9f81-4123-b2c4-bdaffa1c00ef>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00434.warc.gz"} |
Poisson Distribution Probability Calculator
The Poisson distribution is a popular discrete probability distribution in statistics that expresses the probability of a given number of events occurring in a fixed interval of time or space. This
calculator computes Poisson probabilities for a given mean rate (λ) and random variable (𝑥), and visualizes the distribution.
P(X = ):
P(X < ):
P(X ≤ ):
P(X > ):
P(X ≥ ):
Understanding the Poisson Distribution
The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, provided that the events
happen independently of each other and occur with a constant average rate (λ). It is commonly applied in areas such as telecommunications (e.g., the number of calls received by a call center),
biology (e.g., the number of mutations in a DNA sequence), and traffic engineering (e.g., the number of vehicles passing through a toll booth).
Key Components of the Poisson Distribution
• Average Rate (λ): This is the average number of occurrences of an event within a given time frame or space. For example, if a website receives an average of 3 user signups per hour, then λ = 3.
• Random Variable (X): The random variable X represents the number of occurrences of the event you are measuring. For example, X could be the number of user signups in a particular hour.
Poisson Distribution Formula
The probability mass function (PMF) for the Poisson distribution is given by the formula:
\[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad \text{where} \ x = 0, 1, 2, \dots \]
Here, \( e \) is the base of the natural logarithm (approximately equal to 2.718), and \( x! \) represents the factorial of \( x \). The Poisson formula calculates the probability of observing
exactly \( x \) events given the average rate \( \lambda \).
Conditions for Using the Poisson Distribution
The Poisson distribution applies under specific conditions:
• Event Independence: The events must occur independently, meaning that the occurrence of one event does not affect the probability of another event occurring.
• Constant Average Rate (λ): The average rate \( λ \) must remain constant over time or space.
• Discreteness: The number of events (X) must be a non-negative integer. In other words, X = 0, 1, 2, 3, etc.
Step-by-Step Example: Finding Poisson Probability
Suppose we want to calculate the probability of receiving exactly 4 customer inquiries in an hour, given that the average rate of customer inquiries is 3 per hour.
Step 1: Identify the Key Parameters
In this case:
• \( \lambda = 3 \): The average rate of customer inquiries per hour.
• \( X = 4 \): The number of customer inquiries we are interested in.
Step 2: Apply the Poisson Formula
Using the Poisson formula:
\[ P(X = 4) = \frac{e^{-3} 3^4}{4!} \]
First, calculate \( 4! \) (which equals 24) and \( e^{-3} \approx 0.0498 \). Then, substitute these values into the formula:
\[ P(X = 4) = \frac{0.0498 \times 81}{24} = 0.168 \]
Therefore, the probability of receiving exactly 4 customer inquiries in an hour is approximately 0.168, or 16.8%.
Other Useful Probability Calculations
The Poisson distribution can also be used to calculate cumulative probabilities, such as the probability of observing fewer than or greater than a specific number of events. These can be useful in
different scenarios:
• Less than \( x \) (P(X < \( x \))): The cumulative probability that the number of events is less than a given value.
• Greater than \( x \) (P(X > \( x \))): The cumulative probability that the number of events is greater than a given value.
• Less than or equal to \( x \) (P(X ≤ \( x \))): The cumulative probability that the number of events is less than or equal to a given value.
• Greater than or equal to \( x \) (P(X ≥ \( x \))): The cumulative probability that the number of events is greater than or equal to a given value.
Practical Applications of the Poisson Distribution
The Poisson distribution is widely used in real-world applications, such as:
• Call Centers: Estimating the number of incoming calls in a given time period.
• Traffic Flow: Predicting the number of vehicles passing through a toll booth or intersection during peak hours.
• Biology: Modeling the number of mutations occurring in a DNA sequence over a given time or space.
• Finance: Assessing the frequency of rare events, such as defaults in a portfolio of loans.
Further Reading
Suf is a senior advisor in data science with deep expertise in Natural Language Processing, Complex Networks, and Anomaly Detection. Formerly a postdoctoral research fellow, he applied advanced
physics techniques to tackle real-world, data-heavy industry challenges. Before that, he was a particle physicist at the ATLAS Experiment of the Large Hadron Collider. Now, he’s focused on bringing
more fun and curiosity to the world of science and research online. | {"url":"https://researchdatapod.com/data-science-tools/calculators/poisson-distribution-probability-calculator/","timestamp":"2024-11-13T18:42:16Z","content_type":"text/html","content_length":"118849","record_id":"<urn:uuid:c312f2e5-e79f-460f-8b72-ea8d47613bb9>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00034.warc.gz"} |
The concept ApolloniusGraphDataStructure_2 refines the concept TriangulationDataStructure_2. In addition it provides two methods for the insertion and removal of a degree 2 vertex in the data
structure. The insertion method adds a new vertex to the specified edge, thus creating two new edges. Moreover, it creates two new faces that have the two newly created edges in common (see figure
below). The removal method performs the reverse operation.
Insertion and removal of degree 2 vertices. Left to right: The edge (f,i) is replaced by two edges by means of inserting a vertex v on the edge. The faces \( f_1\) and \( f_2\) are created. Right to
left: the faces \( f_1\) and \( f_2\) are destroyed. The vertex v is deleted and its two adjacent edges are merged.
We only describe the additional requirements with respect to the TriangulationDataStructure_2 concept.
See also | {"url":"https://doc.cgal.org/5.5.1/Apollonius_graph_2/classApolloniusGraphDataStructure__2.html","timestamp":"2024-11-12T17:11:04Z","content_type":"application/xhtml+xml","content_length":"15538","record_id":"<urn:uuid:8fa3232d-71a5-4ed2-b400-20fb87526f5f>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00689.warc.gz"} |
Day Section's Quiz - Problem 1
(1) For the system
determine all critical points, linearize around each critical point, and determine what conclusion can be made about the nonlinear system at each critical point based on the linearization. Draw a
phase portrait for the nonlinear system. | {"url":"https://forum.math.toronto.edu/index.php?PHPSESSID=j6tqbq41ob7eljjeh1no47qah7&topic=285.0;prev_next=next","timestamp":"2024-11-15T00:50:51Z","content_type":"application/xhtml+xml","content_length":"42056","record_id":"<urn:uuid:78445176-3a38-4dc8-ba85-0608a26ff7f3>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00018.warc.gz"} |
Sorting Formulas with Sheet Names
Jim T. sends this along, and I’m not sure what’s causing this behavior. Maybe you know and comment on it.
Col. D and Col. E are identical except that Col. E’s formula has range references that include the sheet name.
If you sort Col. D without sorting any other data on the sheet, you get the expected result (at least the result that I expect). That is, nothing happens. The formulas move to a different location
and their range references change relatively. The formula that references A2, moves to D11 and then references A11. It looks like nothing happened.
Why then when I sort Col. E, does the same thing not happen? It sorts based on the values, but the formulas don’t change. Or maybe they do change – however you want to look at it.
18 thoughts on “Sorting Formulas with Sheet Names”
1. I’m not sure what’s happening either, but this speaks to a problem I’m having in one of my sheets. It’s too complicated to explain quickly, so I won’t. Let’s just say this exercise may help me
understand it.
If you insert a column between D and E, you get the expected result on the new column F. I think that the sheet names cause Excel to think that column is a separate table, but its proximity to
the other table causes a later part of the sort routine to re-sort the formulas as if it is all one table. Adding the buffer column forces Excel to treat the right column as a separate table.
2. It’s like the non-sheet-qualified formula cells are sorted using copy/paste behavior (i.e., with relative reference rewriting), but the sheet-qualified formula cells are sorted using move
behavior (i.e., address-literal, no relative reference rewriting), with the end result being formulas that are treated as if their rows were absolute (i.e., (A$2+B$2)*C$2)).
3. How did you create your sheet? I tried just keying in the same set of items and formulae (Excel 2003) and can’t reproduce the behavior you’ve described.
4. Ughh. Never mind the previous comment. I was sorting Descending instead of ascending for some reason (it’s early on a Monday, what can I say?)
5. Hey, though it is weird, it is worth remembering. I cannot think of a situation where this may come in handy, but who knows!
Sounds like a bug to me though.
6. I am using Excel 2000 Premium 9.0.6926 SP-3
I cannot reproduce the problem you speak of. It works as expected both ascending and descending. However, if I hand type the formula in Sheet2 and hit enter I get #NAME error, which is kinda
7. This is a known issue (in all versions of Excel, I think). You only get those same-sheet references in a formula if you write a formula using your mouse that points to another sheet first. For
example, you might write/click
That sheet1 reference happens automatically which doesn’t seem like a bad thing – until you try to sort! The sheet1 reference in the formula gets treated like an absolute reference. Best practice
is to clean up same-sheet formula elements of this type.
8. This problem could show itself if you build formula strings using references returned from RefEdit controls.
They tend to attach sheet names to references.
Something like this would solve:
Private Sub refGetAddress_Exit(ByVal Cancel As MSForms.ReturnBoolean)
If IsRange(refGetAddress.Text) Then _
If Range(refGetAddress.Text).Worksheet Is ActiveSheet Then _
refGetAddress = Range(refGetAddress.Text).Address
End Sub
ActiveSheet is an assumption – if you know the destination sheet up front then use that.
(in this example, IsRange is a function which returns boolean if the parameter looks like a range)
9. Just to make you really shake your head. Sort Column E using A->Z… Nothing Happens. Now use Z->A and the results resort themselves. Use Z->A agan and they resort themselves back again. I have not
found a legitimate use for this. More it is something to watch out for. I found it in a spreadsheet created by a coworker who did not know that anything was wrong.
10. Is it possible to sort 1 sheet based on the column of another sheet?
11. Thanks Jason. This was the answer/solution that I was looking for. I had this problem when using the index function, and must have created the sheet referenced links while clicking back and forth
between the sheets that are referenced in the formula. I consider this a bug. The cell references aren’t absolute, so they shouldn’t be treated as absolute references. Note, I’m using excel 2002
which doesn’t have the same filtering capabilities as excel 2003, so not sure if this problem applies to both versions – but I imagine it does. Cheers, Robert
12. bump!
I’ve got a use for this! So i’m thinking it’s a UDF (undocumented feature)
The use – I’ve got a pivot table which contains data I’m using in a formula (user-defined) in cells next to the pivot. I want to be able to sort the pivot table using the result of the formula
but you can’t sort a pivot table using something not in the pivot table.
In order to sort it then, I’ve linked other cells to the values in the pivot table, used these to calculate the formula and then I can sort…? Nope – because all the references are relative so
when they’re sorted, they just update to there new place and so don’t appear changed. BUT, if you use the sheet-qualified references, the references don’t update and the data is sorted as
So 2.5 yrs after the original post – a use for it!!!
13. I’m glad I Googled this topic and found you. I have had this problem for years, and have finally decided the best answer is to get the formula results, and copy/paste special/values and then sort
to get the desired result. I usually copy and paste the formula back in after the sort.
I agree it’s a bug they need to fix! BTW if it matters I use version 2007
14. Thank you Jason! Whew.
15. I’m coming across this 2 years after the last post but this is still an issue in 2007 and 2010 and I think it happens a lot more than folks think…a simple exercise like doing a plain sumif on
sheet1 when the data is on sheet2 will generate a same-sheet reference which I bet most folks won’t bother with assuming that it is relative. Seems like a huge friggin bug or feature with really
unintended consequences.
16. I spend two hours tried to figure this out and I am glad to find this post so that I don’t have to waste more time on such an aged bug. As pointed out by Andrey and Page, it is a absolute
reference once you put the sheet name in. Same situation will occur if you are working on multiple tabs, let’s say you put a formula like “=sheet2!A1” in sheet 1, and sort sheet2, your result
will be off once the original sheet2 A1 moved to somewhere else.
18. Glad I found this today 11 years after originally posted!
This ‘feature’ had us stumped as to what was happening with a set of data when trying to sort it as the data appeared not to be getting sorted. It turned out the column we were trying to sort on
referenced another column, which referenced others, and just a few of the columns included the sheet name in the formula. So, while the column we were sorting was fine, the result wasn’t because
some of the references were now pointing to “random” rows in the data instead of the relative row.
Yuck. This is an awful gotcha!
I see no excuse for this ‘feature’ to still persist after many years, especially as as far as I can see, there is no mention in the documentation that using a sheet reference makes a formula
absolute when the data is sorted. Its completely unintuitive. If I want absolute references then I use the $ format to show I want the references to be absolute. I don’t want them to be implied
to be absolute just because the formula happens to reference a sheet name. Sure, if the reference is outside of the range being sorted, then I can see how I might expect such a reference to be
treated as an absolute, but if the reference is to the same sheet and in the range being sorted, I expect the reference to be amended acordingly so the calculations still refer to the same row as
the rest of the data.
OK – only option seems to be manually removing Sheet names from all formula if the sheet name is the same as the current sheet – or separating any such formula out of the range to be sorted (and
if you need to sort on it, add a ‘kludge’ column which references the col to be sorted on and use that in the sort range, not the original with the sheet name in it.
What a mess.
Bascially – a good reason to never user sheet names in formula and to use tables/named ranges etc wherever possible.
Posting code? Use <pre> tags for VBA and <code> tags for inline. | {"url":"http://dailydoseofexcel.com/archives/2005/05/15/sorting-formulas-with-sheet-names/","timestamp":"2024-11-13T22:15:29Z","content_type":"text/html","content_length":"98101","record_id":"<urn:uuid:abb28d48-6468-482d-82ac-a9d0ab5e3b5e>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00229.warc.gz"} |
EIP-7212: Precompiled for secp256r1 Curve Support
Hi Magicians
The post is about [S:EIP-7212:S] RIP-7212 which proposes adding a new precompiled contract to the EVM that allows signature verifications in the “secp256r1” elliptic curve by given parameters of
message hash, r - s components of the signature, and x - y coordinates of the public key.
post explains moving the proposal from EIP to RIP category.
What are the use cases of the “secp256r1” elliptic curve?
• The secp256k1 elliptic curve, which is the only cryptographic primitive to prove ownership in Ethereum, does not offer flexibility in onboarding new users via new solutions. Adding this new
elliptical curve will allow biometric and hardware authorization solutions to be easily executed on-chain.
• Many hardware and software solutions use this elliptic curve as signing algorithms, such as TLS, DNSSEC, Apple’s Secure Enclave, Passkeys, Android Keystore, and Yubikey, which can be used in the
Why is this precompile recommended, and how do we believe the user/developer experience will be improved?
• The addition of this precompiled contract improves efficiency and gas affordability in the EVM. With the improvement, the gas costs are reduced alongside the computational load decreases, and the
block gas limit is maximized to enhance the transaction throughput and overall network performance.
• Supporting an elliptic curve with a precompiled contract provides a uniform and standardized way of operations. So that potential confusion and errors that occur from different implementations
can be avoided. The curve operations can be made in reliable implementations in terms of security with precompiled contracts.
• Enabling this precompiled contract improves the developer experience by allowing effortless integration of the curve signatures and building applications on top of it to provide interoperability
with different solutions using the curve, resulting in more user-friendly products.
Possible integration ideas of the precompiled contract:
• The EIP-4337 account abstraction wallets can use these elliptic curve signatures to sign the user operation data with mobile device secure elements and then validate them in the smart contracts.
• @arachnid suggested in the ENSDAO that the precompiled contract allows lots of opportunities with DNSSEC for the projects by adding a new cryptographic primitive.
“One of the major barriers to onchain DNSSEC, along with many other potential projects such as web authentication integration, email verification, and other tasks that rely on verifying ‘real
world’ crypto proofs, is the EVM’s limited support for cryptographic primitives.”
summary shared by Tim Beiko.
Glad to see someone propose this; we wrote a bespoke contract to do this at significant gas expense in 2019 for KONG Cash notes.
I would explicitly add secure elements as a rationale under hardware – low cost secure element chips are typically designed with P256 in order to support TLS. See the common ATTEC608A part from
Microchip for one example.
8 Likes
This precompile returns the actual signature instead of the address reduced form? (like ECRECOVER does)?
I’m personally in favor of that, but I think that deserves an explicit call-out in the specification section or possibly more appropriately the backwards compatibility section. The conversion of an
EOA address is only done for secp256k1 curve keys, so a snippit to convert signatures to addresses is unneeded. Considering the number of times ECRECOVER is mentioned in the EIP we should call it out
and add a note to explain why, to avoid confusion on the part of future readers.
2 Likes
Hi, thanks for contributing!
The implementation in the proposal only returns if the signature is valid or not by 1 and 0. I am sharing my design choice (almost the same as your comments) from the PR comments:
We need v value to recover public key without the x and y coordinates. The v value can be found by above-mentioned methods, but let me explain my design choice:
□ While we are making recovery with the ecrecover, we can reach the public address of the EOA accounts, so it can be directly used in the smart contracts. Unlikely, recovering the secp256r1
public key does not match any default stored types, and we still need to store the account public key.
□ Having to find the v value in the implementation part of the signature creates complexity on the application side. So, I didn’t want to bring this complexity for the applications. Still, I
would love to reassess and edit the EIP to implement recovery after discussion.
The Rationale part includes this design choice, but I agree with you that it can be improved with this information.
Lastly, I couldn’t understand why you suggested explaining it in the Backwards Compatibility part, as it is a completely separate implementation.
2 Likes
Moving @_pm’s comment in the PR to here for the discussion:
paulmillr commented yesterday
1. 256r1 is usually more vulnerable to timing attacks than stuff like 25519.
2. 256r1 is not even recommended by nist at this point, 384r1 is.
3. There are some rumors with regards to general security of r1 curves, it’s unclear.
4. Adding a new elliptic curve impl into ALL execution layer clients is not a trivial task. I don’t think the feature is too useful for this.
My comments:
1. The Golang crypto library works in constant time for the secp256r1 curve. Considering that the timing attacks are implementation dependent, it can be assumed to be safe.
2. I think that NIST’s recommendations points to the PQC, which are not ready for the production.
3. Apart from rumors, I did not see some definite evidences regarding security risks.
4. I see that the secp256r1 curve is the most mass adopted curve and it has a widespread use in many cases. It would be a great step on the mass adoption of Ethereum.
I would love to hear more about any ideas and researches that analyzing the vulnerabilities regarding the security risks of the curve.
11 Likes
Happy to join force on that.
We built a lof of stuff in Solidity already:
Demo: https://p256.alembic.tech/
And some doc:
P256 Biometric Signer - Alembic
4 Likes
@ulerdogan I just recalled that a broader precompile effort was put forth in 2019 with EIP-1829. It might be useful to understand from @Recmo what happened there and if revival of a more generic
precompile would still be useful.
My sense is that 1829 has now been fully supplanted by one-off EIPs for specific curves and as such moving forward with a secp256r1 makes sense, but it might be worth reviewing.
2 Likes
Hey, thanks for reviving the idea!
I followed the EIP-1829 discussions and some other contents. Then, I found EIP-1962 which is a continuation of EIP-1829 by Alex. It seems that the problems in the implementation later revealed
EIP-2537 which was planned to be included in Cancun upgrade, now postponed.
Apparently, generalized curve implementations failed and were replaced by specific ones with EIP-2537 ([notes] by @timbeiko (https://twitter.com/TimBeiko/status/1235931932644564995?s=20) from an ACD
I would love to hear the ideas of those who have worked on precompiled contract implementations for curves about the previous experiences and this proposal! @Recmo @shamatar @ralexstokes @kelly
2 Likes
I support this EIP. Just image how many new users will be attracted to Ethereum without the trouble to backup their mnemonics or private key on a piece of paper…
8 Likes
I do feel, that specific support for the secp256r1 curve is needed. There are a lot of requests for a gas-efficient way to verify passkey-signed data on-chain. And this proposal will make it
2 Likes
I agree with your opinion that secp256r1 can be enough. however Ed25519 seems to still have many benefits, especially less computation.
And, it also seems that there was already an EIP about Ed25519, could we leverage that proposal?
1 Like
As per Vitalik’s post quite some time ago, I don’t recommend using the secp256r1 curves.
" The obvious question is this: where did the seed come from? Why was the seed not chosen to be some more innocent-looking number, like 15? In light of recent revelations regarding the US National
Security Agency subverting cryptographic standards, an obvious concern is that the seed was somehow deliberately chosen in order to make the curve weak in some way that only the NSA knows.
Thankfully, the wiggle room is not unlimited. Because of the properties of hash functions, the NSA could not have found one “weak” curve and then gone backward to determine the seed; rather, the only
avenue of attack is to try different seeds until one turns out to generate a curve that is weak. If the NSA knows of an elliptic curve vulnerability that affects only one specific curve, the
pseudorandom parameter generation process would prevent them from standardizing it. However, if they knew of a weakness in one in every billion curves, then the process offers no protection; for all
we know, c49d360886e704936a6678e1139d26b7819f7e90 could have been the billionth seed that the National Institute for Standards in Technology tried."
4 Likes
@longfin @Toshi I don’t think there are many that would dispute that secp256r1 is a non-ideal curve, however, the claims of a backdoor have been made for well over a decade without any substantive
evidence of the existence of one. This is significantly different than the discussion for a SHA1 precompile which was demonstrably malleable at the time of proposal.
Proposing other curves arbitrarily ignores the key arguments in favor of this EIP which is that billions of devices have hardware accelerated support and isolated secure storage for secp256r1.
Ed25519 would be great, but low cost secure element chips don’t support it (nor do they support secp256k1).
If this proposal sought to shift EOA creation to secp256r1 I would understand the concerns here, however, just adding it as a precompile doesn’t seem to me to warrant the ire that it’s currently
5 Likes
First of all, I don’t believe the rumors of a backdoor on secp256r1 seriously too.
Aside from unconfirmed concerns about backdoors, the difference between secp256r1 and Ed25519 seems to be whether we will use relatively the modern curve or the little bit old curve that supports
many devices, and coverage seems to make sense at this moment.
Thanks for your comments @longfin @Toshi, and thanks for your explanation -I completely agree with you- @ccamrobertson.
I think that even if ed25519 presents an efficient usage, I can’t find enough motivation to bring this curve into the EVM, but the secp256r1 curve has many cases that can directly improve the UX in
Ethereum as it’s one of the most widely supported elliptic curves in the internet/mobile ecosystem.
Also, my comments about the rumours: I agree with the idea that choosing the k1 curve as the main security mechanism of Bitcoin and Ethereum, but bringing the r1 curve as an additional verification
mechanism in the app level, does not contradict this selection.
Additionally, pointing out an upper discussion:
1 Like
i’m starting to see account-abstraction projects implement “riced-out” versions of secp256r1 signature verification to support passkey authenticators who dont support Koblitz, to save on gas. How the
heck can I, a mere mortal, review, understand, and audit these implementations:
trampoline/contracts/EllipticCurve.sol at webauthn · eth-infinitism/trampoline · GitHub
aa-passkeys-wallet/src/Secp256r1.sol at main · itsobvioustech/aa-passkeys-wallet · GitHub
More importantly, if we want to onboard the next ${num} users onto the EVM, there are about 4billion devices that support webauthn: navigator.credentials.create() to securely create, store, and sign
a PublicKeyCredential without installing anything
however, none of the devices I tested: (e.g., ~samsung phones, ~pixel phones, various iphones, m1 macbook, dell laptop (windows / linux / macOS)) support the Koblitz curve (secp256k1), they can only
do secp256r1 (P-256, COSE Curve: 1, Alg: -7)
can we please get a precompile (or native support) for secp256r1 ? cc @vbuterin
6 Likes
Fantastic point. The harm from everyone rolling their own secp256r1 could be massive. If an AA scheme catches on leveraging a broken implementation you will see real loss vs. hypothetical loss from
the NSA.
4 Likes
@mac Ledger security team will present this paper on the 17th at ethcc “WebAuthn Optimization: optimizing ECC sec256r1” this extend the resources in the repo previously shared above.
Regarding the debate on the security of p256 I recommend this article
This old Serenity EIP-101 discussing is relevant too:
3 Likes
OP produced an example implementation for OP hackathon project: Opclave | ETHGlobal
opclave-scaling2023/precompiles at main · itublockchain/opclave-scaling2023 · GitHub
Claiming this will help users manage their private keys better by entrenching them in a proprietary SoC solution (Apple Secure Enclave) is ridiculous. We trust in math, not Chinese supply chain
vendors. Claiming that “backdooring” is not viable misses the point: we are not claiming that all chips are backdoored, only that certain chips be intercepted en route to end user by TAG/TAU and
flashed with backdoor. It is hard to verify hardware, its easier to verify software.
Here is an approach that uses existing authentication schemes to provide user key management, check their github for examples. https://mfkdf.com/
Precompiles should be ossified, they were always a “temporary” solution, that was 8 years ago.
4 Likes
Strongly in support of this - secp256r1 is by far the most widely used curve, and enabling it to be efficiently verified in the EVM will enable countless integrations with existing infrastructure
that are currently impractical.
Those suggesting alternate curves or raising issues with the derivation of the secp256r1 parameters are missing the point; the idea here is not to pick the ideal curve, it’s to add functionality that
permits integrating with external and legacy applications efficiently.
I would suggest that an ‘ecrecover’ type implementation is more versatile, though; it’s easy to convert ecrecover into a signature verification operation, but impossible to do the inverse.
8 Likes | {"url":"https://ethereum-magicians.org/t/eip-7212-precompiled-for-secp256r1-curve-support/14789?ref=web3builder.news","timestamp":"2024-11-08T17:31:11Z","content_type":"text/html","content_length":"80619","record_id":"<urn:uuid:e27fa63b-bb97-4740-a22c-c746fdc71769>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00354.warc.gz"} |
Realization of Constant-Depth Fan-Out with Real-Time Feedforward on a Superconducting Quantum Processor
When using unitary gate sequences, the growth in depth of many quantum circuits with output size poses significant obstacles to practical quantum computation. The quantum fan-out operation,
which reduces the circuit depth of quantum algorithms such as the quantum Fourier transform and Shor’s algorithm, is an example that can be realized in constant depth independent of the output size.
Here, we demonstrate a quantum fan-out gate with real-time feedforward on up to four output qubits using a superconducting quantum processor. By performing quantum state tomography on the output
states, we benchmark our gate with input states spanning the entire Bloch sphere. We decompose the output-state error into a set of independently characterized error contributions. We extrapolate our
constant-depth circuit to offer a scaling advantage compared to the unitary fan-out sequence beyond 25 output qubits with feedforward control, or beyond 17 output qubits if the classical feedforward
latency is negligible. Our work highlights the potential of mid-circuit measurements combined with real-time conditional operations to improve the efficiency of complex quantum algorithms.
Deterministic generation of a 20-qubit two-dimensional photonic cluster state
Multidimensional cluster states are a key resource for robust quantum communication, measurement-based quantum computing and quantum metrology. Here, we present a device capable of
emitting large-scale entangled microwave photonic states in a two dimensional ladder structure. The device consists of a pair of coupled superconducting transmon qubits which are each tuneably
coupled to a common output waveguide. This architecture permits entanglement between each transmon and a deterministically emitted photonic qubit. By interleaving two-qubit gates with controlled
photon emission, we generate 2 x n grids of time- and frequency-multiplexed cluster states of itinerant microwave photons. We measure a signature of localizable entanglement across up to 20 photonic
qubits. We expect the device architecture to be capable of generating a wide range of other tensor network states such as tree graph states, repeater states or the ground state of the toric code, and
to be readily scalable to generate larger and higher dimensional states.
Mitigating Losses of Superconducting Qubits Strongly Coupled to Defect Modes
The dominant contribution to the energy relaxation of state-of-the-art superconducting qubits is often attributed to their coupling to an ensemble of material defects which behave as
two-level systems. These defects have varying microscopic characteristics which result in a large range of observable defect properties such as resonant frequencies, coherence times and coupling
rates to qubits g. Here, we investigate strategies to mitigate losses to the family of defects that strongly couple to qubits (g/2π≥ 0.5 MHz). Such strongly coupled defects are particularly
detrimental to the coherence of qubits and to the fidelities of operations relying on frequency excursions, such as flux-activated two-qubit gates. To assess their impact, we perform swap
spectroscopy on 92 frequency-tunable qubits and quantify the spectral density of these strongly coupled modes. We show that the frequency configuration of the defects is rearranged by warming up the
sample to room temperature, whereas the total number of defects on a processor tends to remain constant. We then explore methods for fabricating qubits with a reduced number of strongly coupled
defect modes by systematically measuring their spectral density for decreasing Josephson junction dimensions and for various surface cleaning methods. Our results provide insights into the properties
of strongly coupled defect modes and show the benefits of minimizing Josephson junction dimensions to improve qubit properties.
Improved Parameter Targeting in {3D}-Integrated Superconducting Circuits through a Polymer Spacer Process
Three-dimensional device integration facilitates the construction of superconducting quantum information processors with more than several tens of qubits by distributing elements such
as control wires, qubits, and resonators between multiple layers. The frequencies of resonators and qubits in flip-chip-bonded multi-chip modules depend on the details of their electromagnetic
environment defined by the conductors and dielectrics in their vicinity. Accurate frequency targeting therefore requires precise control of the separation between chips and minimization of their
relative tilt. Here, we describe a method to control the inter-chip separation by using polymer spacers. Compared to an identical process without spacers, we reduce the measured planarity error by a
factor of 3.5, to a mean tilt of 76(35) μrad, and the deviation from the target inter-chip separation by a factor of ten, to a mean of 0.4(8) μm. We apply this process to coplanar waveguide resonator
samples and observe chip-to-chip resonator frequency variations below 50 MHz (≈ 1 %). We measure internal quality factors of 5×105 at the single-photon level, suggesting that the added spacers are
compatible with low-loss device fabrication.
Calibration of Drive Non-Linearity for Arbitrary-Angle Single-Qubit Gates Using Error Amplification
The ability to execute high-fidelity operations is crucial to scaling up quantum devices to large numbers of qubits. However, signal distortions originating from non-linear components
in the control lines can limit the performance of single-qubit gates. In this work, we use a measurement based on error amplification to characterize and correct the small single-qubit rotation
errors originating from the non-linear scaling of the qubit drive rate with the amplitude of the programmed pulse. With our hardware, and for a 15-ns pulse, the rotation angles deviate by up to
several degrees from a linear model. Using purity benchmarking, we find that control errors reach 2×10−4, which accounts for half of the total gate error. Using cross-entropy benchmarking, we
demonstrate arbitrary-angle single-qubit gates with coherence-limited errors of 2×10−4 and leakage below 6×10−5. While the exact magnitude of these errors is specific to our setup, the presented
method is applicable to any source of non-linearity. Our work shows that the non-linearity of qubit drive line components imposes a limit on the fidelity of single-qubit gates, independent of
improvements in coherence times, circuit design, or leakage mitigation when not corrected for.
Realization of a Universal Quantum Gate Set for Itinerant Microwave Photons
Deterministic photon-photon gates enable the controlled generation of entanglement between mobile carriers of quantum information. Such gates have thus far been exclusively realized
in the optical domain and by relying on post-selection. Here, we present a non-post-selected, deterministic, photon-photon gate in the microwave frequency range realized using superconducting
circuits. We emit photonic qubits from a source chip and route those qubits to a gate chip with which we realize a universal gate set by combining controlled absorption and re-emission with
single-qubit gates and qubit-photon controlled-phase gates. We measure quantum process fidelities of 75% for single- and of 57% for two-qubit gates, limited mainly by radiation loss and decoherence.
This universal gate set has a wide range of potential applications in superconducting quantum networks.
Microwave Quantum Link between Superconducting Circuits Housed in Spatially Separated Cryogenic Systems
Superconducting circuits are a strong contender for realizing quantum computing systems, and are also successfully used to study quantum optics and hybrid quantum systems. However,
their cryogenic operation temperatures and the current lack of coherence-preserving microwave-to-optical conversion solutions have hindered the realization of superconducting quantum networks either
spanning different cryogenics systems or larger distances. Here, we report the successful operation of a cryogenic waveguide coherently linking transmon qubits located in two dilution refrigerators
separated by a physical distance of five meters. We transfer qubit states and generate entanglement on-demand with average transfer and target state fidelities of 85.8 % and 79.5 %, respectively,
between the two nodes of this elementary network. Cryogenic microwave links do provide an opportunity to scale up systems for quantum computing and create local area quantum communication networks
over length scales of at least tens of meters.
Implementation of Conditional-Phase Gates based on tunable ZZ-Interactions
High fidelity two-qubit gates exhibiting low crosstalk are essential building blocks for gate-based quantum information processing. In superconducting circuits two-qubit gates are typically
based either on RF-controlled interactions or on the in-situ tunability of qubit frequencies. Here, we present an alternative approach using a tunable cross-Kerr-type ZZ-interaction between two
qubits, which we realize by a flux-tunable coupler element. We control the ZZ-coupling rate over three orders of magnitude to perform a rapid (38 ns), high-contrast, low leakage (0.14 %)
conditional-phase CZ gate with a fidelity of 97.9 % without relying on the resonant interaction with a non-computational state. Furthermore, by exploiting the direct nature of the ZZ-coupling, we
easily access the entire conditional-phase gate family by adjusting only a single control parameter.
Realizing a Deterministic Source of Multipartite-Entangled Photonic Qubits
Sources of entangled electromagnetic radiation are a cornerstone in quantum information processing and offer unique opportunities for the study of quantum many-body physics in a controlled
experimental setting. While multi-mode entangled states of radiation have been generated in various platforms, all previous experiments are either probabilistic or restricted to generate specific
types of states with a moderate entanglement length. Here, we demonstrate the fully deterministic generation of purely photonic entangled states such as the cluster, GHZ, and W state by sequentially
emitting microwave photons from a controlled auxiliary system into a waveguide. We tomographically reconstruct the entire quantum many-body state for up to N=4 photonic modes and infer the quantum
state for even larger N from process tomography. We estimate that localizable entanglement persists over a distance of approximately ten photonic qubits, outperforming any previous deterministic
Primary thermometry of propagating microwaves in the quantum regime
The ability to control and measure the temperature of propagating microwave modes down to very low temperatures is indispensable for quantum information processing, and may open opportunities
for studies of heat transport at the nanoscale, also in the quantum regime. Here we propose and experimentally demonstrate primary thermometry of propagating microwaves using a transmon-type
superconducting circuit. Our device operates continuously, with a sensitivity down to 4×10−4 photons/Hz−−−√ and a bandwidth of 40 MHz. We measure the thermal occupation of the modes of a highly
attenuated coaxial cable in a range of 0.001 to 0.4 thermal photons, corresponding to a temperature range from 35 mK to 210 mK at a frequency around 5 GHz. To increase the radiation temperature in a
controlled fashion, we either inject calibrated, wideband digital noise, or heat the device and its environment. This thermometry scheme can find applications in benchmarking and characterization of
cryogenic microwave setups, temperature measurements in hybrid quantum systems, and quantum thermodynamics. | {"url":"https://circuitqed.net/publications_author/jean-claude-besse/","timestamp":"2024-11-12T20:39:08Z","content_type":"text/html","content_length":"54500","record_id":"<urn:uuid:1e5ccaeb-fa8f-4252-ab7b-4db9548ac933>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00716.warc.gz"} |
A machine, on average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?A machine, on average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
A machine, on average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
To find the total number of screws produced in the month of January 2006, you need to know how many days were in that month and then multiply it by the average daily production.
January 2006 had 31 days, so:
Total screws produced = Average daily production × Number of days in January
Total screws produced = 2,825 screws/day × 31 days
Total screws produced = 87,575 screws
So, the machine produced 87,575 screws in the month of January 2006.
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* Please Don't Spam Here. All the Comments are Reviewed by Admin. | {"url":"https://maths.loudstudy.com/2023/10/a-machine-on-average-manufactures-2825.html","timestamp":"2024-11-11T21:13:34Z","content_type":"application/xhtml+xml","content_length":"237573","record_id":"<urn:uuid:3de7ed8a-d5d0-4f9b-aa09-2451d8bb4bd7>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00717.warc.gz"} |
Interpreting Performance Report Fields
The MIPS Performance Report summarizes MIPS and Non-MIPS quality measure data by Physician/Physician Group Taxpayer Identification Number (TIN), so that physicians can monitor their performance
throughout the year before submitting measures to CMS.
The performance period for MIPS and non-MIPS measures generally refers to the calendar year of January 1^st to December 31^st. The performance period for quality measures should be taken into
consideration to ensure capture of quality activation if a shortened timeframe is chosen. For more information on the performance period review measure specification documents.
Below is an explanation of each section and field available in the report.
Measure Score Type
Measures are grouped by Score Type and displayed at the top of each section
• Proportion Measures are percentages calculated as the number of occurrences divided by the measured population.
• Continuous Measures span a range of values, such as mean wait time for a population.
Measures are grouped by score type within each Physician Group TIN. All proportion measures are displayed first, followed by continuous measures.
Measure Description
Measure Descriptions have three components:
• Measure Number is the CMS MIPS quality measure number. Non-MIPS measures have the prefix “ACRad” (e.g. ACRad 1) and MIPS measures have no prefix (e.g. 76).
• Measure Title is the official title as determined by CMS for the reporting year.
• Domain is the National Quality Strategy Domain for the measure.
Proportion Measure Performance Rate
Proportion measures are divided into two sections for fields related to the Performance Rate and Reporting Rate.
The Performance Rate includes the following fields:
• Initial Patient Population is the number of patients at the physician practice who are relevant for the measure, based on exams submitted for MIPS and Non-MIPS measures to date. If you indicate
that 100% of exams were submitted to the registry, when selecting measures to submit to CMS, the number of records that we received will be used as the reporting denominator. Otherwise, if you
enter a number other than the number of records that we received, that number will be used for this field.
• Performance Denominator represents the number of observations evaluated in the Performance Rate and is equal to the Reporting Numerator, also known as the “Data Completeness Numerator”, minus any
Denominator Exceptions (see below).
• Performance Numerator indicates the number of records for which the performance measure was met.
• Performance Not Met is the number of observations that do not meet the numerator requirements.
• Performance Rate is the Performance Numerator divided by Performance Denominator:
Proportion Measure Reporting Rate
The Reporting Rate section, on the right half of the page, demonstrates the completeness of data submitted for measurement and includes the following fields:
• Denominator Exclusions remove patients from measurement when circumstances do not meet the criteria.
• Denominator Exceptions remove patients from the Performance Rate calculation, but retain them in the Reporting Rate. CPT Category II code modifiers such as 1P, 2P and 3P quality-data codes, or
equivalents referenced from the registry are available to describe medical, patient or system reasons for denominator exceptions and can be submitted to the registry. A denominator exception
removes a patient from the performance denominator only if the performance numerator criteria are not met – i.e. only if the observation would lower the performance rate. This allows for the
exercise of clinical judgement by the eligible clinician.
Note: For example, MIPS Measure 225 estimates how often mammography screening reminder systems are used. A patient may be an exception if the eligible clinician documents a clinical reason for not
using the reminder system, such as further screening exams not indicated due to patient limited life expectancy. However, the patient still counts towards the Reporting Rate.
• Reporting Denominator is the number of “eligible instances” for which the measure could be reported. It may equal the Initial Patient Population if there are no Denominator Exclusions.
• Reporting Numerator includes the total number of observations, minus any Denominator Exclusions, submitted with complete information.
• Reporting Rate, also known as Data Completeness, is calculated as the Reporting Numerator divided by the Reporting Denominator. Denominator Exceptions are included in the Reporting Numerator, to
give the physician credit for reporting data.
This rate needs to be at least 60% for successful MIPS participation.
Continuous Measures
Continuous Measures appear in a separate section and calculate performance scores for which each individual value for the measure can fall anywhere along a continuous scale. Continuous Measures can
be aggregated using a variety of methods such as the calculation of a mean or median (for example ACRad 15 which looks at mean Radiography report turnaround time in hours) and are available for
several ACR-defined Non-MIPS measures.
The following fields appear in the report:
• Initial Patient Population defines the set of patients to be evaluated for the measure, in the same manner described above for Proportion Measures.
• Measure Population defines the set of patients to be reported at the end of the performance period. The initial patient population and measure population numbers will be the same if the registry
has all exams/cases submitted for the year.
• Measure Exceptions remove patients from the Performance Score calculation. Reasons include removal of statistical outliers as well as measure-specific criteria for excluding patients from
• Performance Score is the measure calculation, as defined for each measure. The Type, Unit, and Score Type fields provide context for the score being calculated.
• Type describes what is being measured – e.g. time, dose.
• Unit denotes the units of the measurement – e.g. hours, mGy.
• Score Type describes the calculation made using the Measure Population - e.g. median, mean, etc.
Other Fields
The Selected for CMS Submission column indicates whether you have selected this measure for MIPS reporting to CMS. You may change which measures are selected at any time before final submission to
The following general notes may appear below the table for each measure:
• CMS Benchmarks provide statistical context for MIPS Measures using data gathered by CMS. These are the benchmarks CMS will use when scoring measures. Benchmarks are displayed in deciles;
benchmarks with fewer than 10 deciles indicate historical performance was skewed such that meaningful distinctions and improvement in performance cannot be made.
• Registry Benchmarks provide statistical context for Non-MIPS Measures using data gathered by the NRDR QCDR, such as measures that may be reported to CMS from the DIR, for example.
• Inverse measures denote measures for which a lower Performance Rate indicates better clinical care or control.
Note: Work is underway to highlight which measures are deemed “High Priority” by CMS.
There are other notes applicable to specific measures. If there are multiple notes for a measure, each is separated by a semicolon. | {"url":"https://acrsupport.acr.org/support/solutions/articles/11000038822-interpreting-performance-report-fields","timestamp":"2024-11-05T22:41:28Z","content_type":"text/html","content_length":"33863","record_id":"<urn:uuid:95c566c8-7947-4aee-9553-08687be49d40>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00284.warc.gz"} |
Gottlob Frege | Life Devil
Gottlob Frege (1848-1925) was a German mathematician and logician as well as a philosopher. He is considered the major founder of modern logic. His uniquely profound research led him from pure
mathematics, like the methodizing of natural numbers or “the proof,” to a highly philosophical approach to mathematical logic and the foundation of mathematics, and even to consequential findings in
the field of linguistics. Applicable to most fields of logic, Frege’s research exerted an influence far beyond pure mathematical logic. He is considered to have advanced logic beyond Aristotle,
despite the fact that he failed in constructing a consistent axiomatic foundation for logic. Consequently, Frege holds a significant place in the historical development of logic and mathematics from
ancient Greece to our modern times.
Life and Work
F. L. Gottlob Frege was born on November 8, 1848, in Wismar, northern Germany, in the state of Mecklenburg. He was the son of Alexander Frege, the principal of a private high school in Wismar. It is
probable that the influence of his teacher at the local gymnasium, Leo Sachse, motivated Frege to study after school. Subsequently he enrolled as a student in mathematics, chemistry, physics, and
philosophy; first from 1869 to 1871 at the University of Jena under the encouragement of the famous physicists Ernst Abbe and Karl Snell; next he expanded his studies for 2 more years at Göttingen
University, where he eventually obtained a doctoral degree in 1873 with the valued geometrical thesis translated as On a Geometrical Representation of Imaginary Figures in a Plane. Just one year
later, Frege received his second doctorate (habilitation) for his work Methods of Calculation Based Upon an Amplification of the Concept of Magnitude, comprehending some initial steps of his theory
of higher (complex) mathematical functions.
Also in the year 1874, upon recommendation of his academic teacher Abbe, Frege became a lecturer in mathematics at the University of Jena, where he would remain all his professional life. He taught
extensively in all mathematical disciplines. His research, however, concentrated on the philosophy of logic. Persistent dialogue—noteworthy because Frege was extremely reserved in general—with his
Jena colleague and one of his few friends Rudolf Eucken, later a winner of the Nobel Prize in literature, supported Frege’s philosophically mathematic thinking.
In 1879, Frege’s seminal first work Concept Script, a Formal Language of Pure Thought Modelled Upon That of Arithmetic (Begriffsschrift) was published. He developed a principle for the construction
of a logical language. In the aftermath of the Begriffsschrift, Frege was made associate professor at Jena University. Frege married Margarete Lieseberg, but unfortunately neither of their two
children survived into adulthood, so they adopted a boy named Alfred.
Frege’s book The Foundations of Arithmetic (Grundlagen) appeared in 1884. It comprised for the first time a complete system for the foundation of arithmetic based on a set of mathematically logical
axioms. To gain higher recognition for his work, this book was written in completely nontechnical, natural language.
In conjunction with his profound research into the logical system of mathematics, Frege felt impelled to develop a philosophy of language. His major work on a linguistic system supporting the
philosophy of logic is On Sense and Reference (1892). In this book, Frege’s two famous linguistic puzzles were presented, distinguishing between sense and the denotation of terms in order to resolve
the ambiguities of language.
In 1893, with the first volume of Basic Laws of Arithmetic (Grundgesetze), Frege’s major opus on the philosophy of mathematics was published. Frege used to say that “every good mathematician is at
least half a philosopher, and every good philosopher is at least half a mathematician.” Because almost no colleague of his in the world was able to understand what Frege had managed to find, none of
his publications achieved immediate success, and he even received some very poor reviews.
Nonetheless, Frege was promoted to honorary professor in Jena in 1896, giving him a regular income for the first time in his life. Years of bad luck followed. In 1902, Frege received a letter from
Bertrand Russell who modestly pointed out that he had discovered a paradox caused by a severe inconsistency in Frege’s set of logical axioms. Also, the second volume of the Basic Laws was based on
the misarranged axioms but was already finished and was (at that exact time) with the printer. Unfortunately, even the amendment to his axiomatic system that Frege added as an appendix proved to be
inconsistent; as some say, Frege must have known this but was unable to accept his failure. Frege gave up writing the intended third volume of the Basic Laws and never again published any research.
His late attempt to base logic on geometry instead of arithmetic could not be elaborated any further.
During the First World War, Frege retired and after the death of his wife he left Jena; he lived in increasing reclusiveness and died on July 26, 1925, in Bad Kleinen, Germany.
Frege left deep traces in two fields of philosophical research: in the logic of mathematics and in the philosophy of language. In the first, he marks the beginning of modern science. His invention
of quantified variables (predicate calculus) to replace the ambiguous meaning of natural language not suitable for the denotation of complex mathematical terms was seminal to the development of
mathematics. His influence arises not only from separate theorems Frege published himself, but also is founded on the diffusion of his very elementary logic and philosophical findings into countless
Frege’s analytical philosophy is connected with Locke’s and Hume’s empiricism. In a far-reaching empiricist manner and as a central point in his logic, which is founded on the sole relevance of the
logical truth of an argument, Frege did not accept any non- falsifiable fact or instance; hence he was in deep doubt about transcendental phenomena, and epochs do not seem to have attracted his
For a man in determined pursuit of new scientific findings, Frege was extremely conservative in his political attitude. Especially in his later years, the embittered Frege was an enthusiastic
monarchist, anti-democrat, anti-French, anti-Catholic, and anti-Semite.
Frege displayed signs of immoderate self-regard and refused to accept or even to consider the criticism of his colleagues. To the contrary, he reacted with embitterment and polemic attacks on his
critics, which seems an inadequate response in light of Frege’s undoubtedly immense and lasting scientific achievements in the fields of logic and the philosophy of mathematics and language.
Matthias S. Hauser
See also Aristotle; Hume, David; Language
Further Readings
Beaney, M. (Ed.). (1997). The Frege reader. Oxford, UK: Blackwell.
Dummett, M. (1991). Frege: Philosophy of mathematics. Cambridge, MA: Harvard University Press.
Sluga, H. (1980). Gottlob Frege. London: Routledge & Kegan Paul. | {"url":"https://lifedevil.com/gottlob-frege/","timestamp":"2024-11-02T07:21:56Z","content_type":"text/html","content_length":"63875","record_id":"<urn:uuid:4b467d24-b7ff-4a8c-9ef2-902e613bf206>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00618.warc.gz"} |
14.6: Archimedes’ Principle and Buoyancy
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• Define buoyant force
• State Archimedes’ principle
• Describe the relationship between density and Archimedes’ principle
When placed in a fluid, some objects float due to a buoyant force. Where does this buoyant force come from? Why is it that some things float and others do not? Do objects that sink get any support at
all from the fluid? Is your body buoyed by the atmosphere, or are only helium balloons affected (Figure \(\PageIndex{1}\))?
Figure \(\PageIndex{1}\): (a) Even objects that sink, like this anchor, are partly supported by water when submerged. (b) Submarines have adjustable density (ballast tanks) so that they may float or
sink as desired. (c) Helium-filled balloons tug upward on their strings, demonstrating air’s buoyant effect. (credit b: modification of work by Allied Navy; credit c: modification of work by “Crystl”
Answers to all these questions, and many others, are based on the fact that pressure increases with depth in a fluid. This means that the upward force on the bottom of an object in a fluid is greater
than the downward force on top of the object. There is an upward force, or buoyant force, on any object in any fluid (Figure \(\PageIndex{2}\)). If the buoyant force is greater than the object’s
weight, the object rises to the surface and floats. If the buoyant force is less than the object’s weight, the object sinks. If the buoyant force equals the object’s weight, the object can remain
suspended at its present depth. The buoyant force is always present, whether the object floats, sinks, or is suspended in a fluid.
The buoyant force is the upward force on any object in any fluid.
Figure \(\PageIndex{2}\): Pressure due to the weight of a fluid increases with depth because \(p = h \rho g\). This change in pressure and associated upward force on the bottom of the cylinder are
greater than the downward force on the top of the cylinder. The differences in the force results in the buoyant force F[B]. (Horizontal forces cancel.)
Archimedes’ Principle
Just how large a force is buoyant force? To answer this question, think about what happens when a submerged object is removed from a fluid, as in Figure \(\PageIndex{3}\). If the object were not in
the fluid, the space the object occupied would be filled by fluid having a weight w[fl]. This weight is supported by the surrounding fluid, so the buoyant force must equal w[fl], the weight of the
fluid displaced by the object.
The buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is
\[F_{B} = w_{fl},\]
where F[B] is the buoyant force and w[fl] is the weight of the fluid displaced by the object.
This principle is named after the Greek mathematician and inventor Archimedes (ca. 287–212 BCE), who stated this principle long before concepts of force were well established.
Figure \(\PageIndex{3}\): (a) An object submerged in a fluid experiences a buoyant force F[B]. If F[B] is greater than the weight of the object, the object rises. If F[B] is less than the weight of
the object, the object sinks. (b) If the object is removed, it is replaced by fluid having weight w[fl]. Since this weight is supported by surrounding fluid, the buoyant force must equal the weight
of the fluid displaced.
Archimedes’ principle refers to the force of buoyancy that results when a body is submerged in a fluid, whether partially or wholly. The force that provides the pressure of a fluid acts on a body
perpendicular to the surface of the body. In other words, the force due to the pressure at the bottom is pointed up, while at the top, the force due to the pressure is pointed down; the forces due to
the pressures at the sides are pointing into the body.
Since the bottom of the body is at a greater depth than the top of the body, the pressure at the lower part of the body is higher than the pressure at the upper part, as shown in Figure \(\PageIndex
{2}\). Therefore a net upward force acts on the body. This upward force is the force of buoyancy, or simply buoyancy.
The exclamation “Eureka” (meaning “I found it”) has often been credited to Archimedes as he made the discovery that would lead to Archimedes’ principle. Some say it all started in a bathtub. To read
the story, explore Scientific American to learn more.
Density and Archimedes’ Principle
If you drop a lump of clay in water, it will sink. But if you mold the same lump of clay into the shape of a boat, it will float. Because of its shape, the clay boat displaces more water than the
lump and experiences a greater buoyant force, even though its mass is the same. The same is true of steel ships.
The average density of an object is what ultimately determines whether it floats. If an object’s average density is less than that of the surrounding fluid, it will float. The reason is that the
fluid, having a higher density, contains more mass and hence more weight in the same volume. The buoyant force, which equals the weight of the fluid displaced, is thus greater than the weight of the
object. Likewise, an object denser than the fluid will sink.
The extent to which a floating object is submerged depends on how the object’s density compares to the density of the fluid. In Figure \(\PageIndex{4}\), for example, the unloaded ship has a lower
density and less of it is submerged compared with the same ship when loaded. We can derive a quantitative expression for the fraction submerged by considering density. The fraction submerged is the
ratio of the volume submerged to the volume of the object, or
\[fraction\; submerged = \frac{V_{sub}}{V_{obj}} = \frac{V_{fl}}{V_{obj}} \ldotp\]
The volume submerged equals the volume of fluid displaced, which we call V[fl]. Now we can obtain the relationship between the densities by substituting \(\rho = \frac{m}{V}\) into the expression.
This gives
\[\frac{V_{fl}}{V_{obj}} = \frac{\frac{m_{fl}}{\rho_{fl}}}{\frac{m_{obj}}{\rho_{obj}}},\]
where \(\rho_{obj}\) is the average density of the object and \(\rho_{fl}\) is the density of the fluid. Since the object floats, its mass and that of the displaced fluid are equal, so they cancel
from the equation, leaving
\[fraction\; submerged = \frac{\rho_{obj}}{\rho_{fl}} \ldotp\]
We can use this relationship to measure densities.
Figure \(\PageIndex{4}\): An unloaded ship (a) floats higher in the water than a loaded ship (b).
Suppose a 60.0-kg woman floats in fresh water with 97.0% of her volume submerged when her lungs are full of air. What is her average density?
We can find the woman’s density by solving the equation
\[fraction\; submerged = \frac{\rho_{obj}}{\rho_{fl}}\]
for the density of the object. This yields
\[\rho_{obj} = \rho_{person} = (fraction\; submerged) \cdotp \rho_{fl} \ldotp\]
We know both the fraction submerged and the density of water, so we can calculate the woman’s density.
Entering the known values into the expression for her density, we obtain
\[\rho_{person} = 0.970 \cdotp 10^{3}\; kg/m^{3} = 970\; kg/m^{3} \ldotp\]
The woman’s density is less than the fluid density. We expect this because she floats.
Numerous lower-density objects or substances float in higher-density fluids: oil on water, a hot-air balloon in the atmosphere, a bit of cork in wine, an iceberg in salt water, and hot wax in a “lava
lamp,” to name a few. A less obvious example is mountain ranges floating on the higher-density crust and mantle beneath them. Even seemingly solid Earth has fluid characteristics.
Measuring Density
One of the most common techniques for determining density is shown in Figure \(\PageIndex{5}\).
Figure \(\PageIndex{5}\): (a) A coin is weighed in air. (b) The apparent weight of the coin is determined while it is completely submerged in a fluid of known density. These two measurements are used
to calculate the density of the coin.
An object, here a coin, is weighed in air and then weighed again while submerged in a liquid. The density of the coin, an indication of its authenticity, can be calculated if the fluid density is
known. We can use this same technique to determine the density of the fluid if the density of the coin is known.
All of these calculations are based on Archimedes’ principle, which states that the buoyant force on the object equals the weight of the fluid displaced. This, in turn, means that the object appears
to weigh less when submerged; we call this measurement the object’s apparent weight. The object suffers an apparent weight loss equal to the weight of the fluid displaced. Alternatively, on balances
that measure mass, the object suffers an apparent mass loss equal to the mass of fluid displaced. That is, apparent weight loss equals weight of fluid displaced, or apparent mass loss equals mass of
fluid displaced. | {"url":"https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/14%3A_Fluid_Mechanics/14.06%3A_Archimedes_Principle_and_Buoyancy","timestamp":"2024-11-09T09:44:28Z","content_type":"text/html","content_length":"146061","record_id":"<urn:uuid:8e1bfea8-676d-4c98-9aa3-91f9bf23834a>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00033.warc.gz"} |
Answers to: What is the equation of a line, in general form, with a slope of -2 and a y-intercept of 8?
What is the equation of a line, in general form, with a slope of -2 and a y-intercept of 8?
Asked on 6/2/2023, 4 pageviews
The equation of a line in general form is Ax + By = C, where A, B, and C are constants. To find the equation of the line with a slope of -2 and a y-intercept of 8, we can start by using the
point-slope form of a line: y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Plugging in the given values, we get: y - 8 = -2(x - 0) Simplifying, we get: y - 8 = -2x
Adding 2x to both sides, we get: 2x + y - 8 = 0 Thus, the equation of the line in general form is: 2x + y = 8.
Ask Another Question
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Excel Formula for On-Time Delivery
In this tutorial, we will learn how to write an Excel formula in Python to calculate if a delivery is on time or not. The formula uses the IF function to check if the values in two cells are the
same, and returns 'On Time' if they are. If the values are different, it checks if one of the cells is blank and returns 'Not On Time' if it is not blank. If both cells are blank, it returns a blank.
This formula can be useful for tracking the timeliness of deliveries in various scenarios. Let's dive into the details of the formula and how it works.
An Excel formula
=IF(A1=B1,"On Time",IF(ISBLANK(A1),"","Not On Time"))
Formula Explanation
This formula uses the IF function to calculate if the delivery is on time or not. It returns "On Time" if the values in cells A1 and B1 are the same, "Not On Time" if the value in cell A1 is not
blank and different from the value in cell B1, and it returns a blank if cell A1 is blank.
Step-by-step explanation
1. The first IF function checks if the value in cell A1 is equal to the value in cell B1. If they are equal, it returns "On Time".
2. If the values in cells A1 and B1 are not equal, the second IF function is executed.
3. The second IF function checks if cell A1 is blank using the ISBLANK function. If cell A1 is blank, it returns a blank.
4. If cell A1 is not blank, it means that the delivery is not on time, so it returns "Not On Time".
For example, if we have the following data in cells A1 and B1:
The formula =IF(A1=B1,"On Time",IF(ISBLANK(A1),"","Not On Time")) would return "On Time" because the values in cells A1 and B1 are the same.
If we have the following data:
The formula would return "Not On Time" because the values in cells A1 and B1 are different.
If cell A1 is blank:
The formula would return a blank because cell A1 is blank.
If cell A1 is blank and cell B1 is also blank:
The formula would return a blank because both cells A1 and B1 are blank. | {"url":"https://codepal.ai/excel-formula-generator/query/ak2b6SxV/excel-formula-deliver-on-time","timestamp":"2024-11-14T03:39:14Z","content_type":"text/html","content_length":"92793","record_id":"<urn:uuid:b00d479e-455e-4f79-89e3-b31d73f53a32>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00319.warc.gz"} |
Unit 'matrix' Package
[Overview][Types][Classes][Procedures and functions][Index] [#rtl]
Multiply a two-dimensional double precision vector by a scalar.
Source position: matrix.pp line 431
This operator allows you to multiply a vector by a scalar value. Each vector element is multiplied by the scalar value; the result is returned as a new vector. | {"url":"https://build.alb42.de/fpcbin/docu/rtl/matrix/op-multiply-tvector2_double-double-tvector2_double.html","timestamp":"2024-11-12T07:08:26Z","content_type":"text/html","content_length":"2484","record_id":"<urn:uuid:a2f2ddc7-40bd-4327-83dd-1c7c9e19451b>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00172.warc.gz"} |
The sine is the ratio of the length of a given angle's opposite side to the length of the hypotenuse in a right triangle.
Sine is a high school-level concept that would be first encountered in a trigonometry course. It is listed in the California State Standards for Trigonometry.
Function: A function is a relation that uniquely associates members of one set with members of another set. The term "function" is sometimes implicitly understood to mean continuous function,
linear function, or function into the complex numbers.
Right A right triangle is a triangle that has a right angle. The Pythagorean Theorem is a relationship among the sides of a right triangle.
Trigonometry: Trigonometry is the study of angles and of the angular relationships of planar and three-dimensional figures.
Classroom Articles on Trigonometry (Up to High School Level)
Cosine Law of Sines
Double-Angle Formulas Tangent
Half-Angle Formulas Trigonometric Addition Formulas
Law of Cosines Unit Circle | {"url":"https://mathworld.wolfram.com/classroom/Sine.html","timestamp":"2024-11-06T02:36:51Z","content_type":"text/html","content_length":"47514","record_id":"<urn:uuid:0b1b2848-71c2-40f8-a3eb-8c5ab47041f7>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00020.warc.gz"} |
Comparison Model...... Variations of Models on "More Than"
Variations of Comparison Model - "More Than"
Step 1: Draw a box and write the number found after the word "than" (i.e. "?") in this box.
Step 2: Draw a dot at the top-right corner of the first box. This marks the starting point of the arrow we are about to draw next.
Step 3: Draw an arrow from the dot pointing to the right. (The arrow points to the right because it is a "more than" question.
Step 4: Since it is "more than", we are actually making the unknown value "?" (starting number) greater by 4. Hence, draw a second box directly below the arrow to show that the unknown value "?" has
been lengthened by 4.
Step 5: Draw a last box to match the length of the first two boxes and write the number 21 in this box.
From the model,
21 - 4 = ?
21 - 4 = 17
Thus, the answer is 17. | {"url":"http://www.teach-kids-math-by-model-method.com/comparisonmodel-morethan-2.html","timestamp":"2024-11-09T03:29:04Z","content_type":"text/html","content_length":"24377","record_id":"<urn:uuid:64715960-5d97-46b8-9f5f-f08ed2badd1d>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00643.warc.gz"} |
How to simplify 1-cos(u)^2.
3 Answers
Sort by ยป oldest newest most voted
How to simplify 1-cos(u)^2.
I have tried
sage: assume(0<u<pi/2)
But I still get
sage: simplify(1-cos(u)^2)
-cos(u)^2 + 1
edit retag flag offensive close merge delete
There are far more simplify methods. expr.simplify? gives a hint:
See also: "simplify_full()", "simplify_trig()", "simplify_rational()", "simplify_rectform()", "simplify_factorial()", "simplify_log()", "simplify_real()", "simplify_hypergeometric()",
In this case simplify_trig will do the job.
expr = 1-cos(u)^2
Note that sometimes the "degree reduction" (of the involved trigonometric polynomial) is the wanted and/or needed "simplification".
The corresponding method is called reduce_trig.
The following sample code shows some differences.
sage: var('u');
sage: a = 1 - cos(u)^2
sage: a.simplify_trig()
sage: a.reduce_trig()
-1/2*cos(2*u) + 1/2
sage: a.simplify_trig().reduce_trig()
-1/2*cos(2*u) + 1/2
sage: a.reduce_trig().simplify_trig()
With an other example...
sage: b = sin(u)^4 - cos(u)^4
sage: b.simplify_trig()
-2*cos(u)^2 + 1
sage: b.reduce_trig()
Just for completeness -- another option is to use sympy. See http://docs.sympy.org/latest/tutorial/simplification.html for some simplification options. | {"url":"https://ask.sagemath.org/question/38039/how-to-simplify-1-cosu2/","timestamp":"2024-11-08T22:07:11Z","content_type":"application/xhtml+xml","content_length":"61835","record_id":"<urn:uuid:ca52550e-ebd5-43b8-9692-9e8b7180e478>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00701.warc.gz"} |
class neurobench.benchmarks.Benchmark(model, dataloader, preprocessors, postprocessors, metric_list)[source]
Bases: object
Top-level benchmark class for running benchmarks.
run(quiet=False, verbose: bool = False, dataloader=None, preprocessors=None, postprocessors=None, device=None)[source]
Runs batched evaluation of the benchmark.
○ dataloader (optional) – override DataLoader for this run.
○ preprocessors (optional) – override preprocessors for this run.
○ postprocessors (optional) – override postprocessors for this run.
○ quiet (bool, default=False) – If True, output is suppressed.
○ verbose (bool, default=False) – If True, metrics for each bach will be printed. If False (default), metrics are accumulated and printed after all batches are processed.
○ device (optional) – use device for this run (e.g. ‘cuda’ or ‘cpu’).
A dictionary of results.
Return type:
Workload Metrics
class neurobench.benchmarks.workload_metrics.AccumulatedMetric[source]
Bases: object
Abstract class for a metric which must save state between batches.
Compute the metric score using all accumulated data.
the final accumulated metric.
Return type:
Reset the metric state.
This is called when the benchmark is run again, e.g. on the FSCIL task the benchmark is run at the end of each session.
class neurobench.benchmarks.workload_metrics.COCO_mAP[source]
Bases: AccumulatedMetric
COCO mean average precision.
Measured for event data based on Perot2020, Supplementary B (https://arxiv.org/abs/2009.13436)
☆ Skips first 0.5s of each sequence
☆ Bounding boxes with diagonal size smaller than 60 pixels are ignored
Compute COCO mAP using accumulated data.
Reset metric state.
neurobench.benchmarks.workload_metrics.MSE(model, preds, data)[source]
Mean squared error of the model predictions.
☆ model – A NeuroBenchModel.
☆ preds – A tensor of model predictions.
☆ data – A tuple of data and labels.
Mean squared error.
Return type:
neurobench.benchmarks.workload_metrics.activation_sparsity(model, preds, data)[source]
Sparsity of model activations.
Calculated as the number of zero activations over the total number of activations, over all layers, timesteps, samples in data.
☆ model – A NeuroBenchModel.
☆ preds – A tensor of model predictions.
☆ data – A tuple of data and labels.
Activation sparsity.
Return type:
neurobench.benchmarks.workload_metrics.classification_accuracy(model, preds, data)[source]
Classification accuracy of the model predictions.
☆ model – A NeuroBenchModel.
☆ preds – A tensor of model predictions.
☆ data – A tuple of data and labels.
Classification accuracy.
Return type:
Register hooks or other operations that should be called before running a benchmark.
class neurobench.benchmarks.workload_metrics.membrane_updates[source]
Bases: AccumulatedMetric
Number of membrane potential updates.
This metric can only be used for spiking models implemented with SNNTorch.
Compute membrane updates using accumulated data.
Compute the total updates to each neuron’s membrane potential within the model, aggregated across all neurons and normalized by the number of samples processed.
Return type:
Reset metric state.
neurobench.benchmarks.workload_metrics.number_neuron_updates(model, preds, data)[source]
Number of times each neuron type is updated.
☆ model – A NeuroBenchModel.
☆ preds – A tensor of model predictions.
☆ data – A tuple of data and labels.
key is neuron type, value is number of updates.
Return type:
class neurobench.benchmarks.workload_metrics.r2[source]
Bases: AccumulatedMetric
R2 Score of the model predictions.
Currently implemented for 2D output only.
Compute r2 score using accumulated data.
Reset metric state.
neurobench.benchmarks.workload_metrics.sMAPE(model, preds, data)[source]
Symmetric mean absolute percentage error of the model predictions.
☆ model – A NeuroBenchModel.
☆ preds – A tensor of model predictions.
☆ data – A tuple of data and labels.
Symmetric mean absolute percentage error.
Return type:
class neurobench.benchmarks.workload_metrics.synaptic_operations[source]
Bases: AccumulatedMetric
Number of synaptic operations.
MACs for ANN ACs for SNN
Compute the metric score using all accumulated data.
the final accumulated metric.
Return type:
Reset the metric state.
This is called when the benchmark is run again, e.g. on the FSCIL task the benchmark is run at the end of each session.
Static Metrics
Sparsity of model connections between layers. Based on number of zeros in supported layers, other layers are not taken into account in the computation: Supported layers: Linear Conv1d, Conv2d,
Conv3d RNN, RNNBase, RNNCell LSTM, LSTMBase, LSTMCell GRU, GRUBase, GRUCell
model – A NeuroBenchModel.
Connection sparsity, rounded to 3 decimals.
Return type:
Memory footprint of the model.
model – A NeuroBenchModel.
Model size in bytes.
Return type:
Number of parameters in the model.
model – A NeuroBenchModel.
Number of parameters.
Return type: | {"url":"https://neurobench.readthedocs.io/en/stable/neurobench.benchmarks.html","timestamp":"2024-11-12T16:40:21Z","content_type":"text/html","content_length":"45457","record_id":"<urn:uuid:ac6f53f3-f0b4-4346-857f-2f28913248ad>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00312.warc.gz"} |
Overtones, Harmonics and Additive Synthesis | Andrew McWilliams
This post is a quick intro to overtones, harmonics and additive synthesis, using a video lovingly prepared by Synth School as a reference point:
These are the building blocks of synthetic computer sound, so it pays to spend some time getting a grasp of the basics.
The two views: Oscilloscope vs. Frequency Analysis
One of the first things in the video is a comparison of the representation of a sine wave in an oscilloscope, against a sine wave in a frequency analysis view:
Sine - Oscilloscope view (left), and Sine - Frequency Analysis view (right)
This is useful, because if you are working with computer sound, these two views are like bread and butter (or perhaps rice and water, depending on your preferred diet).
Sampling and the Oscilloscope view
In any short period of digitally recorded audio, say one second, we can store a number of samples, i.e. numbers between -1 and +1, to represent the movement of a loudspeaker during that time. We say
that any given sample within our set of samples will have an amplitude value between -1 and +1.
Just over 200 samples of a sine wave audio signal
In the Oscilloscope view, we lay out the samples over time (X axis), with earlier samples on the left and later samples on the right. We lay out the amplitude values on the Y axis, with 0 in the
middle, +1 at the top, and -1 at the bottom.
The amplitude of any sample at any moment in time represents whether our loudspeaker will be 'sucked in' at that moment (somewhere between -1 and 0), or 'pushed outward' (somewhere between 0 and +1),
or at rest (exactly 0). Note that numbers lower than -1 or higher than +1 will either be 'clipped', or will damage our loudspeakers.
So as we 'read' the Oscilloscope view left-to-right, we read the history of the loudspeaker's movements in detail. This is great, but this only tells one part of the story - the very literal part.
The next part of the story - frequency analysis - is more abstract, but much more powerful in the information it can reveal.
But first, something very important, very quickly - sample rates.
The sample rate
In the Oscilloscope image above there are just over 200 samples represented. All this tells us is that there are 200 samples - it doesn't tell us over what period of time. To know that, we need to
know our sample rate - the number of samples we have chosen to store per second.
If we have a sample rate of 200, then 200 samples will represent one second. In reality, we need far higher sample rates for our ears to think of all those samples as a continuous stream rather than
as a series of individual sounds. For example, 44,100 samples per second is a common standard.
That's all I'll say for now - it just pays to introduce the idea while we are already looking at samples anyway in the Oscilloscope view.
The Frequency Analysis view
If the Oscilloscope view gives us a very literal perspective of a waveform - simply it's linear changes over time - what does frequency analysis give us?
As the Synth School video demonstrates so eloquently, we can combine certain periodic waveforms, such as sine waves, to create fuller-bodied, often more harmonically resonant sounds. If we can use
multiple simple waves at readable frequencies to construct a sound, surely we can reverse the process, and de-construct said sound into constituent frequencies...
This is the idea behind Fast-Fourier Transform (FFT), and FFT is the algorithm that takes a series of samples (like the one we used to make our Oscilloscope view) and builds a graph showing which
frequencies are present in those samples.
A harmonic and two overtones in an FFT view
In fact, the Frequency Analysis view is just the visual result of running the FFT algorithm, so you will often hear it called the FFT view. In the FFT view, volume is on the Y axis (from -96dB to
0dB) and frequency is on the X axis (depending on your readout this will plot 0 times per second all the way to 22,000 times per second - aka 0Hz to 22kHz). So you can think of the FFT view as giving
you the relative amounts for each frequency in your set of samples.
Looking at the X axis more closely, you'll notice that half of the scale is dominated by frequencies in the early hundreds. Zero to 923 take up over half the scale, and 923 to 22,000 get less than
half! That's just because of the way we are visualizing it.
It's a logarithmic scale because in most practical scenarios most of the action takes place on the left-hand side of the scale, and a logarithmic scale helps us get a better view of that action.
However, most FFT views have the option of switching to a linear scale view, and then you will see most of the action taking place on the far left of the view.
Fundamental and overtones
A sine wave is often called the only 'pure' tone because it is the only waveform type to contain only a single frequency, as we saw in the Synth School video. This frequency is called the fundamental
Other waves always have additional frequencies on top of the fundamental. The lowest frequency of the sound, the fundamental, is the basic tone on which the sound is built. Additional frequencies are
called overtones. The fundamental and it's overtones are collectively called partials.
Saw - Oscilloscope view (left), and Saw - FFT view (right)
Overtones can be harmonic, or inharmonic. Harmonic overtones support the fundamental frequency and keep it's tonality intact. Non-harmonic overtones result in noise, or sounds with ambiguous pitch.
However it is the relationship between a fundamental and it's overtones, and the relationship of the overtones to each other that make up the sound's timbre - it's unique characteristic audible
effect. Acoustic instruments have lots of complex activity in non-harmonic overtones. They are generally not loud enough to undermine the fundamental, but are loud enough to add character to the
Additive synthesis
A saw wave can be constructed by combining multiple sine waves, as demonstrated in the video. For a 'classic' saw tone, the amplitude of each added harmonic should be divided by it's harmonic count,
i.e. the fundamental is divided by 1, the second harmonic is divided by 2, the third by 3 and so on.
Fundamental + 2nd harmonic (left), and Fundamental + 2nd and 3rd harmonics (right)
So a saw tone can be thought of as the combination of infinite harmonics, each divided by the harmonic count. Though in practice doing so would require infinite CPU power!
Square wave
A square wave is also constructed by adding harmonics, but this time only odd-numbered harmonics, i.e. by skipping the 2nd, 4th, 6th (etc) harmonics, and including those in between.
This is illustrated quite nicely in this old-school training video:
That's a nice intro to overtones, harmonics and additive synthesis. For more on hands-on additive synthesis and beyond, I recommend Welsh's Synthesizer Cookbook. | {"url":"https://jahya.net/blog/overtones-harmonics-and-additive/","timestamp":"2024-11-05T10:27:03Z","content_type":"text/html","content_length":"34106","record_id":"<urn:uuid:487183ab-49bb-4bc7-980b-05b44928988d>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00804.warc.gz"} |
Kilogram-force/square meter to Newton/square centimeter Converter | kgf/m^2 to N/cm^2
Kilogram-force/square meter to Newton/square centimeter converter | kgf/m^2 to N/cm^2 conversion
Are you struggling with converting Kilogram-force/square meter to Newton/square centimeter? Don’t worry! Our online “Kilogram-force/square meter to Newton/square centimeter Converter” is here to
simplify the conversion process for you.
Here’s how it works: simply input the value in Kilogram-force/square meter. The converter instantly gives you the value in Newton/square centimeter. No more manual calculations or headaches – it’s
all about smooth and effortless conversions!
Think of this Kilogram-force/square meter (kgf/m^2) to Newton/square centimeter (N/cm^2) converter as your best friend who helps you to do the conversion between these pressure units. Say goodbye to
calculating manually over how many Newton/square centimeter are in a certain number of Kilogram-force/square meter – this converter does it all for you automatically!
What are Kilogram-force/square meter and Newton/square centimeter?
In simple words, Kilogram-force/square meter and Newton/square centimeter are units of pressure used to measure how much force is applied over a certain area. It’s like measuring how tightly the air
is pushing on something.
The short form for Kilogram-force/square meter is “kgf/m^2” and the short form for Newton/square centimeter is “N/cm^2”.
In everyday life, we use pressure units like Kilogram-force/square meter and Newton/square centimeter to measure how much things are getting squeezed or pushed. It helps us with tasks like checking
tire pressure or understanding the force in different situations.
How to convert from Kilogram-force/square meter to Newton/square centimeter?
If you want to convert between these two units, you can do it manually too. To convert from Kilogram-force/square meter to Newton/square centimeter just use the given formula:
N/cm^2 = Value in kgf/m^2 * 0.000980665
here are some examples of conversion,
• 2 kgf/m^2 = 2 * 0.000980665 = 0.00196133 N/cm^2
• 5 kgf/m^2 = 5 * 0.000980665 = 0.004903325 N/cm^2
• 10 kgf/m^2 = 10 * 0.000980665 = 0.00980665 N/cm^2
Kilogram-force/square meter to Newton/square centimeter converter: conclusion
Here we have learn what are the pressure units Kilogram-force/square meter (kgf/m^2) and Newton/square centimeter (N/cm^2)? How to convert from Kilogram-force/square meter to Newton/square centimeter
manually and also we have created an online tool for conversion between these units.
Kilogram-force/square meter to Newton/square centimeter converter” or simply kgf/m^2 to N/cm^2 converter is a valuable tool for simplifying pressure unit conversions. By using this tool you don’t
have to do manual calculations for conversion which saves you time. | {"url":"https://calculatorguru.net/kilogram-force-square-meter-to-newton-square-centimeter/","timestamp":"2024-11-06T02:39:23Z","content_type":"text/html","content_length":"124493","record_id":"<urn:uuid:b001009e-79c2-44b9-871e-744447d9aec4>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00117.warc.gz"} |
Elementary Analytic Functions: Complex Functions Theory a-1
Textbook Title: Elementary Analytic Functions: Complex Functions Theory a-1
Textbook Description:
This is an introductory textbook on complex functions theory. This free online textbook covers: The Complex Numbers, Basic Topology and Complex Functions, Analytic Functions, Some elementary analytic
Author: Leif Mejlbro
Subjects: Biology, Chemistry, Engineering, Mathematics
Key words: Biology, Chemistry, Engineering, Mathematics, Analytic Functions
Download URL:
Only Registered Users Can Save eTextbooks to Their TextBookGo Account.
Sign Up For Free! | {"url":"https://textbookgo.com/elementary-analytic-functions-complex-functions-theory-a-1/","timestamp":"2024-11-09T00:59:18Z","content_type":"text/html","content_length":"20590","record_id":"<urn:uuid:fb1e150c-09d4-4f20-bf25-8a29a6fd4c51>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00520.warc.gz"} |
NDA Aptitude Question
Nda Aptitude Question
The Latest and exclusive collection of Nda Aptitude Question to tease your brain. Nda Aptitude Question helps exercising the brain and develop it to think logical and solve real world problems
differenlty. PuzzleFry brings you the best Nda Aptitude Question, you'll enjoy wide range of Nda Aptitude Question, Lets try few Nda Aptitude Question listed below -
• In a holy town, three temples sit in a row identical in almost every manner including a holy well in front of them. A pilgrim comes to visit the temples with some flowers in his basket.
At the first temple, he takes some water from the holy well and sprinkles it on the flowers to wash them. To his astonishment, the number of flowers in his basket doubles up. He offers a few of
them at the temple and turns back to visit the second temple.
At the second temple, he again takes some water from the holy well and sprinkles it on the flowers to wash them. Again the number of flowers double up in number. He offers some of them at the
temple and turns back to visit the third temple.
At the third temple, he repeats the process and the number of flowers doubles up yet again. He offers all the flowers in the third temple.
Now, the pilgrim offered the same number of flowers in all the temples. Can you find out the minimum number of flowers he must have had initially? How many flowers did he offer to god in each of
the three temples?
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• What is the probability of getting five Mondays in a 31-day month?
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• It can be easily calculated that the digits 0 to 9 can be arranged into 3628800 distinct ten-digit numbers.
But do you know how many of these numbers are prime?
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• Christina makes tasty toast in a small pan. After toasting one side of a slice, she turns it over. Each side takes 30 seconds. The pan can only hold two slices.
How can she toast both sides of three slices in 1 & 1/2 instead of 2 minutes?
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• Why despite killing dozens of people Charley and Wilma were not punished?
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• We know that the number 7 is the prime followed by a cube. Which next number is also a prime followed by a cube?
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• John was gifted a new Hayabusa. He drove x miles at 55mph. Then he drove x+20 miles at 40mps. He drove his bike for 100 minutes. How much distance did he travel?
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• Check out the number list below and find what number comes next in the sequence.
111 , 113 , 117 , 119 , 123 , 137 , ?
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• If 2 workers can complete painting 2 walls in exactly 2 hours.
How many workers would be needed to paint 18 walls in 6 hours?
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• What came first, the chicken or the egg?
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Estimation of Infiltration Rate using a Nonlinear Regression Model
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Beginner or seasoned user, our flexible training options help you understand and master the full capabilities of both EPA SWMM5 and PCSWMM.
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Liquid drop movement over an inclined surface using Volume of Fluid model with finite volume method
Duwary, Bijoy Kumar (2013) Liquid drop movement over an inclined surface using Volume of Fluid model with finite volume method. BTech thesis.
When a liquid drop of given volume is placed on an inclined solid surface then it tends to slide down. This depends on various factors like surface Tension of fluid pair, inclination angle, volume of
the liquid, nature of solid surface etc. In an inclined plane, the motion of a liquid drop is due to the effect of gravity force, friction force and viscous effect of liquid drop. If we change the
different parameters of the liquid drop then the velocity of the drop will also change accordingly. This work represent the study of movement of liquid drop by varying three parameters namely
inclination angle ,surface tension and contact angle for the solid liquid pair using VOF model with finite volume method.in the present work we changes these parameter one at a time by keeping other
constant and visualize the effect of this parameter on the velocity of the liquid drop. After that different curve which shows the effect of these parameters are drawn for the given liquid.
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GreeneMath.com | Ace your next Math Test!
Lesson Objectives
• Demonstrate an understanding of how to solve a linear system using elimination
• Learn the basic definition of a matrix
• Learn how to set up an augmented matrix
• Learn how to place a matrix in row echelon form
• Learn how to place a matrix in reduced-row echelon form
How to Solve a Linear System in Two Variables Using Gaussian Elimination
Up to this point, we have not discussed the concept of a matrix. A matrix is an ordered array of numbers. As an example: $$\left[ \begin{array}{ccc}7&3&1\\ 1&-2&4 \end{array}\right] $$ The entries in
a matrix are referred to as "elements" of the matrix. When we refer to a row in a matrix, we are discussing the horizontal information. For example, in our above matrix, the top row has the numbers:
7, 3, and 1. Similarly, when we refer to a column in a matrix, we are discussing the vertical information. For example, in our above matrix, the leftmost column has the numbers: 7 and 1. A matrix is
named according to the number of rows and columns it contains. The number of rows comes first, followed by the number of columns. In our above example, we have two rows and three columns. Therefore,
we can say this is a "2 x 3" matrix. As another example, suppose we saw the following matrix: $$\left[ \begin{array}{ccc}-2&9&-6\\ 4&5&8\\ 11&-7&2\\ 17&6&-5 \end{array}\right] $$ For our above
matrix, we can see that there are four rows and three columns. We can say this matrix is a "4 x 3" matrix. A square matrix has the same number of rows as columns:
2 x 2 Matrix:
A "2 x 2" matrix is a square matrix with two rows and two columns. $$\left[ \begin{array}{cc}8&1\\ -1&3 \end{array}\right] $$
3 x 3 Matrix
A "3 x 3" matrix is a square matrix with three rows and three columns. $$\left[ \begin{array}{ccc}12&-5&4\\ 9&-7&13\\ -6&8&4 \end{array}\right] $$
Solving Linear Systems using Matrix Methods
Now that we have a basic understanding of a matrix, let's think about how to solve a linear system using matrix methods. Let's begin with a sample system. $$-17-7x-6y=0$$ $$-30-8y=2x$$ Let's start by
writing the equations in standard form: $$-7x - 6y=17$$ $$2x + 8y=-30$$ Now we will create the augmented matrix. The augmented matrix has a vertical bar that separates the coefficients from the
constants. Basically, we are taking all the numerical information from the two equations and placing it into a matrix: $$\left[ \begin{array}{cc|c}-7&-6&17\\ 2&8&-30 \end{array}\right] $$ Our first
row contains the numerical information from the first equation: -7, -6, and 17. The second row contains the numerical information from the second equation: 2, 8, and -30. Notice how the vertical bar
"|" separates the coefficients from the constants.
We can manipulate our matrix using "elementary row operations". These produce new matrices that lead to linear systems with the same solution(s) as the original system.
Elementary Row Operations
• We can interchange any two rows
• We can multiply any row by a non-zero number
• We can multiply a row by a real number and add this to the corresponding elements of any other row
Let's think about the row operations with a bit more detail. The first row operation tells us we can interchange any two rows. When solving equations, it doesn't matter which equation is on the top
and which is on the bottom: $$-7x - 6y=17$$ $$2x + 8y=-30$$ is the same system as: $$2x + 8y=-30$$ $$-7x - 6y=17$$ Therefore, we can interchange the rows: $$\left[ \begin{array}{cc|c}-7&-6&17\\ 2&8&
-30 \end{array}\right] $$ represents the same system as: $$\left[ \begin{array}{cc|c}2&8&-30\\ -7&-6&17 \end{array}\right] $$ The second row operation tells us we can multiply any row by a non-zero
number. We already know that the multiplication property of equality lets us multiply both sides of an equation by the same non-zero number. Let's suppose we take our second (bottom) equation and
multiply both sides by 1/2: $$2x + 8y=-30$$ $$x + 4y=-15$$ We can rewrite our system as: $$-7x - 6y=17$$ $$x + 4y=-15$$ Similarly, we can multiply row 2 of the augmented matrix by 1/2: $$\left[ \
begin{array}{cc|c}-7&-6&17\\ 2&8&-30 \end{array}\right] $$ Which produces the following augmented matrix: $$\left[ \begin{array}{cc|c}-7&-6&17\\ 1&4&-15 \end{array}\right] $$ The third and final row
operation corresponds to the way we eliminated a variable when using the elimination method. If we look at our system: $$-7x - 6y=17$$ $$x + 4y=-15$$ We see that we could eliminate the variable x by
multiplying equation 2 (the bottom equation) by 7 and adding the result to equation 1 (the top equation). $$x + 4y=-15$$ Multiply by 7: $$7x + 28y=-105$$ Add to equation 1: $$-7x + -6y=17$$ $$\
underline{7x+28y=-105}$$ $$0x + 22y=-88$$ $$22y=-88$$ $$y=-4$$ We can then substitute back into one of the original equations and find the value for x. Before we do this, let's see how this works
with an augmented matrix: $$\left[ \begin{array}{cc|c}-7&-6&17\\ 1&4&-15 \end{array}\right] $$ We will first swap row 1 and 2, this is to stay consistent with row echelon form, which we will learn
about momentarily. $$\left[ \begin{array}{cc|c}1&4&-15\\ -7&-6&17 \end{array}\right] $$ Multiply row 1 by 7 and add the result to row 2: $$\left[ \begin{array}{cc|c}1&4&-15\\ 0&22&-88 \end{array}\
right] $$ If we look at the information from row 2, we can write this as: $$0x + 22y=-88$$ Which is exactly what we saw earlier when using elimination. In a moment, we will show how to complete the
process using an augmented matrix. It is important to understand that we are fundamentally using the same techniques, just using the numerical information only. This makes things quicker and simpler.
Let's now look at row echelon form.
Row Echelon Form
The goal of Gaussian Elimination is to place the augmented matrix in "row echelon" form. This gives us "1's" down the diagonal and "0's" below. $$\left[ \begin{array}{cc|c}1&a&b\\ 0&1&c \end{array}\
right] $$ Once our matrix is in row echelon form, we can easily use substitution to find the solution. Let's return to our example and go step by step: $$-7x - 6y=17$$ $$x + 4y=-15$$ Augmented
Matrix: $$\left[ \begin{array}{cc|c}-7&-6&17\\ 1&4&-15 \end{array}\right] $$ It is best to work column by column. In the first column, at the top, we have a "-7". We want this to be a 1. The easiest
thing to do here is to swap row 1 and 2. $$\left[ \begin{array}{cc|c}1&4&-15\\ -7&-6&17 \end{array}\right] $$ Now we want the bottom of the first column to be a 0. We can multiply the top row by 7
and add the result to the bottom row: $$\left[ \begin{array}{cc|c}1&4&-15\\ 0&22&-88 \end{array}\right] $$ At this point, we know we could get a solution for y. We will not stop here, we will
continue and make the bottom entry in the second column into a 1. In order to do this, we will multiply row 2 by 1/22: $$\left[ \begin{array}{cc|c}1&4&-15\\ 0&1&-4 \end{array}\right] $$ At this
point, we have achieved row echelon form. We can go back to our equations using this information and achieve our answer using substitution. If we take the numerical information and place it back into
the system: $$1x + 4y=-15$$ $$0x + 1y=-4$$ From the second equation, we see that y is -4. We can plug in a -4 for y in the first equation and solve for x: $$1x + 4(-4)=-15$$ $$x - 16=-15$$ $$x=1$$
Our solution is the ordered pair (1, -4). This may seem like it took longer, but as we learn more about solving systems with matrices, we will find this is the faster and preferred method. Especially
in the case of systems with many equations. Let's check our answer:
-7x - 6y = 17
-7(1) - 6(-4) = 17
-7 + 24 = 17
17 = 17
x + 4y = -15
(1) + 4(-4) = -15
1 + (-16) = -15
-15 = -15
Reduced-Row Echelon Form
Although we can find our solution using substitution when the augmented matrix is in row echelon form, we can keep working on the matrix until it is reduced-row echelon form. This format gives us
"1's" down the diagonal and "0's" above and below. $$\left[ \begin{array}{cc|c}1&0&a\\ 0&1&b \end{array}\right] $$ When the matrix is in reduced-row echelon form, we can obtain the solution directly
from the matrix, no substitution is needed. Let's go back to our last spot on our example, where we were using the matrix. $$\left[ \begin{array}{cc|c}1&4&-15\\ 0&1&-4 \end{array}\right] $$ At this
point, we know that y is -4. Since in row 2, 0 is the coefficient for x, and 1 is the coefficient for y. We know this translates into: $$0x + 1y=-4$$ $$y=-4$$ To obtain reduced-row echelon form, we
need to change the 4 at the top of column 2 into a 0. We can do this by multiplying row 2 by -4, and adding the result to row 1: $$\left[ \begin{array}{cc|c}1&0&1\\ 0&1&-4 \end{array}\right] $$ Now
we know from our matrix that x is 1. In the top row, the information translates into: $$1x + 0y=1$$ $$x=1$$
Gaussian Elimination vs Gauss-Jordan
Let's clarify a point of confusion. You will hear a few different names floating around for solving systems with matrices. We will hear the terms "Gaussian Elimination" and "Gauss-Jordan". What's the
difference between the two?
Gaussian Elimination places the matrix in row echelon form and requires us to go back and substitute to get the final answer. Gauss-Jordan places the matrix in reduced-row echelon form and does not
require substitution. It's just a matter of personal preference, but it is usually faster to use Gauss-Jordan and obtain the solution directly from the augmented matrix.
Gauss-Jordan Two-Variable System
• Obtain a 1 as the first element in the first column
• Use the first row to transform the remaining element in the first column into a 0
• Obtain a 1 as the second element in the second column
• Use the second row to transform the remaining element in the second column into a 0
In other words, we transform an augmented matrix into reduced-row echelon form, working left to right, focusing on one column at a time. Let's look at a few examples.
Example 1
: Solve each system using the Gauss-Jordan method $$-7x+2y=-4$$ $$-14x-3y=6$$ Let's begin by setting up the augmented matrix: $$\left[ \begin{array}{cc|c}-7&2&-4\\ -14&-3&6 \end{array}\right] $$
First, we want to obtain a 1 as the first element in the first column. We can multiply row 1 by -1/7: $$\left[ \begin{array}{cc|c}1&-2/7&4/7\\ -14&-3&6 \end{array}\right] $$ Next, we want to obtain a
0 as the second element in the first column. We can multiply row 1 by 14 and add the result to row 2: $$\left[ \begin{array}{cc|c}1&-2/7&4/7\\ 0&-7&14 \end{array}\right] $$ Now that column 1 is
complete, we move on to column 2. Let's begin with the second element in column 2. We want this to be a 1. We can multiply row 2 by -1/7: $$\left[ \begin{array}{cc|c}1&-2/7&4/7\\ 0&1&-2 \end{array}\
right] $$ For our final step, we want the first element in column 2 to be a 0. We can multiply row 2 by 2/7 and add the result to row 1: $$\left[ \begin{array}{cc|c}1&0&0\\ 0&1&-2 \end{array}\right]
$$ We can translate this back into: $$1x + 0y=0$$ $$0x + 1y=-2$$ $$x=0$$ $$y=-2$$ Our solution is the ordered pair (0,-2) Let's check using substitution:
-7x + 2y = -4
-7(0) + 2(-2) = -4
0 - 4 = -4
-4 = -4
-14x - 3y = 6
-14(0) - 3(-2) = 6
0 + 6 = 6
6 = 6
Inconsistent Systems and Dependent Equations
When we run into special case scenarios, such as inconsistent systems (systems with no solution) or dependent equations (systems with an infinite number of solutions), we will see all coefficients
entries in a row as a 0. The constant in the row will be a 0 for dependent equations or a non-zero number for an inconsistent system. Let's look at an example.
Example 2
: Solve each system using the Gauss-Jordan method $$-18x-24y=11$$ $$-3x-4y=2$$ Let's begin by setting up the augmented matrix: $$\left[ \begin{array}{cc|c}-18&-24&11\\ -3&-4&2 \end{array}\right] $$
First, we want to obtain a 1 as the first element in the first column. We can multiply row 1 by -1/18: $$\left[ \begin{array}{cc|c}1&4/3&-11/18\\ -3&-4&2 \end{array}\right] $$ Next, we want to obtain
a 0 as the second element in the first column. We can multiply row 1 by 3 and add the result to row 2: $$\left[ \begin{array}{cc|c}1&4/3&-11/18\\ 0&0&1/6 \end{array}\right] $$ We can stop once we see
a 0 as the entry for each coefficient. Row 2 translates into: $$0x + 0y=\frac{1}{6}$$ $$0=\frac{1}{6}$$ Since we have a false statement, this tells us there is no solution.
Skills Check:
Example #1
Solve each system. $$-5x-9y=-13$$ $$-6x+7y=20$$
Please choose the best answer.
Example #2
Solve each system. $$-8x - 3y=13$$ $$7x + 5y=-9$$
Please choose the best answer.
Example #3
Solve each system. $$-5x+5y=-30$$ $$-3x+2y=-20$$
Please choose the best answer.
Congrats, Your Score is 100%
Better Luck Next Time, Your Score is %
Try again?
Ready for more?
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Area Of Rectilinear Figures - L-Shapes Worksheet
It's time to practice determine the total area of the compound shape! In this geometry exercise, children calculate the areas of complex figures by breaking them down then add the area of each part
to determine the total area of the compound shape.
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Quantum Talents Symposium Munich | MCQST
Quantum Talents Symposium Munich
Finalists of the 2024 Quantum Talents Symposium
The symposium is designed to bring together outstanding PhD students and early-career postdocs from all over the world to present their groundbreaking research work in the field of quantum science
and technology. Upon selecting the finalists to attend the symposium, Special attention is given to increasing diversity in the field of quantum science, fostering a more inclusive and equitable
environment that values the contributions of researchers from all backgrounds.
Meet the 2024 Quantum Talents Symposium finalists!
Quantum Fluctuations and Collective Modes in Disordered 2D Superconductors
Together with my collaborators, we explore how electron-electron interactions and weak (anti)localization phenomena in two-dimensional systems can enhance the superconducting transition temperature
in the so-called multifractal regime. By developing a comprehensive theoretical framework, we highlight the impact of quantum fluctuations and uncover the critical role of collective modes in
disordered superconductors. This work establishes a direct connection between the self-consistent gap equation at the superconducting transition temperature and the renormalization group equations
for interaction parameters in the normal state.
Building on this foundation, we investigate the dynamics of the collective amplitude Schmid-Higgs (SH) mode in Bardeen–Cooper–Schrieffer (BCS) superconductors and fermionic superfluids with
non-magnetic disorder. By examining the SH susceptibility, we determine the zero-temperature dispersion relation and damping rate of the SH mode across the transition from diffusive to ballistic
scales. Our findings reveal that the imaginary part of the SH susceptibility peaks along the real frequency axis above twice the superconducting gap. In the diffusive limit, the SH susceptibility
pole is below the continuum edge but re-emerges in the ballistic regime, showing non-monotonic dispersion. Furthermore, the SH mode’s dispersion exhibits a logarithmic non-analyticity in the
diffusive range of momenta, causing an anomalous spatial decay at distances longer than the coherence length.
Proximity-induced gapless superconductivity in two-dimensional Rashba semiconductor in magnetic field
Two-dimensional semiconductor-superconductor heterostructures form the foundation of numerous nanoscale physical systems. However, measuring the properties of such heterostructures, and
characterizing the semiconductor in-situ is challenging. A recent experimental study by [Phys. Rev. Lett. 128, 107701 (2022)] was able to probe the semiconductor within the heterostructure using
microwave measurements of the superfluid density. This work revealed a rapid depletion of superfluid density in semiconductor, caused by the in-plane magnetic field which in presence of spin-orbit
coupling creates so-called Bogoliubov Fermi surfaces. The experimental work used a simplified theoretical model that neglected the presence of non-magnetic disorder in the semiconductor, hence
describing the data only qualitatively. Motivated by experiments, we introduce a theoretical model describing a disordered semiconductor with strong spin-orbit coupling that is proximitized by a
superconductor. Our model provides specific predictions for the density of states and superfluid density. Presence of disorder leads to the emergence of a gapless superconducting phase, that may be
viewed as a manifestation of Bogoliubov Fermi surface. When applied to real experimental data, our model showcases excellent quantitative agreement, enabling the extraction of material parameters
such as mean free path and mobility, and estimating g-tensor after taking into account the orbital contribution of magnetic field. Our model can be used to probe in-situ parameters of other
superconductor-semiconductor heterostructures and can be further extended to give access to transport properties.
Real-time Quantum Control of Qubits
Quantum computing relies on developing quantum devices that are robust against small and uncontrolled parameter variations in the Hamiltonian. One can apply feedback by estimating such uncontrolled
variations in real time to stabilize quantum devices and improve their coherence. This task is important for many quantum platforms such as spins, superconducting circuits, trapped atoms, and others
towards error suppression or correction.
In the first part of the talk, we focus on real-time closed-loop feedback protocols to estimate uncontrolled fluctuations of the qubit Hamiltonian parameters, followed by enhancing the quality of
qubit rotations [1]. First, we coherently control two entangled electron spins with a low-latency quantum controller. The protocol uses a singlet-triplet spin qubit implemented in a gallium arsenide
double quantum dot. We establish real-time feedback on both control axes and enhance the resulting quality factor of coherent spin rotations. Even with some components of the Hamiltonian purely
governed by noise, we demonstrate noise-driven coherent control. As an application, we implement Hadamard rotations in the presence of two fluctuating control axes.
(a) Entangled electron spins (qubit) schedule, alternating between periods Top of quantum information processing (dashed box), and short periods Test for efficiently learning the fluctuating
environment (gray box). (b) Overhauser field fluctuations, tracked in real-time by the relative rotation of the two electron spins, on a scanning electron micrograph of a gallium arsenide spin qubit
Next, we present a protocol for a physics-informed real-time Hamiltonian estimation [2]. We estimate the fluctuating nuclear field gradient within the double dot on-the-fly by updating its
probability distribution according to the Fokker-Planck equation. We further improve the physics-informed protocol by adaptively choosing the free evolution time of the entangled electrons singlet
pair, based on the previous measurement outcomes. The protocol results in a ten-fold improvement of the estimation speed compared to former schemes.
Our approaches introduce closed-loop feedback schemes aimed at mitigating the effects of decoherence and extending the lifetime of quantum systems. In this view, our schemes provide valuable insights
into the synergy between quantum control, quantum computation, and computer science.
[1] F. Berritta et al., Nat. Commun. 15, 1676 (2024)
[2] F. Berritta et al., arXiv:2404.09212 (2024)
From two-partite entanglement on a kilometer scale to multi-partite entanglement on a micrometer scale
A journey through quantum networks and quantum simulations with atoms and ions
The creation of entanglement between particles is one of the essential ingredients of quantum technologies and doubtlessly a major challenge for experimentalists working on quantum hardware. I this
talk, I will review our efforts in creating entanglement from very different perspectives. In the first half, I will describe how to achieve entanglement between only two particles, but over large
spatial distances with the goal of demonstrating building blocks of a quantum network. In such a network, entanglement between remote parties acts as a resource of many applications such as secure
communication or distributed quantum computing. Specifically, I will show why nonlinear-optics based frequency conversion techniques of single photons are important and finalize this part with an
experiment to entangle two trapped neutral atoms, which are located at spatially separated laboratories in the city center of Munich, over a fiber distance of 33 km with a fidelity of 62.2(2) %. In
the second part, I switch gears to creating entangled states between a large number of particles over short, micrometer-scale distances, which is the focus of my postdoctoral work. Our workhorse is a
novel analog quantum simulator based on trapped ions. Here, the ions are confined in a single 3D electric potential and form a two-dimensional Coulomb crystal with up to 105 particles. I will show
recent results where we generate spin-squeezed states as well as GHZ states in these crystals with exciting applications for quantum metrology.
Generalizations of Kitaev’s honeycomb model from braided fusion categories
Kitaev’s exactly solvable spin-1/2 model on the honeycomb lattice [1] displays a range of exotic quantum spin liquid phases, both topologically ordered and gapless. When time-reversal symmetry is
broken, it supports non-abelian topological order with Ising anyons and gapless edge modes—hallmarks of chiral topological order, which are experimentally detectable. In this talk, I will present a
systematic approach to constructing generalizations of Kitaev’s honeycomb from braided fusion categories, developing techniques introduced in [2]. By design, the resulting two-dimensional quantum
lattice models conserve mutually commuting local symmetries and anomalous 1-form symmetries, making them promising candidates for various topologically ordered phases. In particular, chiral
topological order can occur because of complex phases in the Hamiltonian, and I will present evidence that it does indeed occur [3].
[1] Alexei Kitaev, Annals of Physics 321 (2006)
[2] Kansei Inamura and Kantaro Ohmori, SciPost Phys. 16, 143 (2024)
[3] Luisa Eck and Paul Fendley, arXiv:2408.04006
Heavy-fermion physics and superconductivity in twisted bilayer graphene
Twisted bilayer graphene (TBG) has shown two seemingly contradictory characters: (1) quantum-dot-like behavior in STM indicates localized electrons; (2) the transport experiments suggest the
itinerant electrons. Two features can both be captured by a topological heavy-fermion model, in which the topological conduction electron bands couple to the local moments [1]. We study the
local-moment physics and the Kondo effect in this model. We demonstrate that, at the integer fillings, the Kondo effect is irrelevant and the RKKY interactions stabilize long-range ordered states [2,
3]. However, at non-integer fillings, the Kondo effect is relevant [3, 4], and Kondo resonance appears in the single-particle spectrum. Based on the heavy-fermion model, we explore the transport
properties of the TBG [5]. In addition, we demonstrate the critical role of f-electron spin, valley, and orbital fluctuations in inducing superconducting instability within the Kondo phase
[1] Z. Song, and B. A. Bernevig, Phy. Rev. Lett. 129, 047601 (2022).
[2] H. Hu, B. A. Bernevig, and A. M. Tsvelik, Phys. Rev. Lett. 131, 026502 (2023).
[3] G. Rai, et al., arXiv: 2309.08529 (2023).
[4] H. Hu, et al., Phys. Rev. Lett. 131, 166501 (2023).
[5] D. Călugăru, H. Hu, et al., arXiv:2402.14057 (2024).
Combining transformer neural networks and quantum simulators: A hybrid approach to simulating quantum many-body systems
Owing to their great expressivity and versatility, neural networks have gained attention for simulating large two-dimensional quantum many-body systems. However, their expressivity comes with the
cost of a challenging optimization due to the in general rugged and complicated loss landscape. In this talk, I will present a hybrid optimization schemes for neural quantum states that involves a
data-driven pre-training with external (numerical or experimental) data and a second, energy-driven optimization stage. In contrast to previous works, we do not not employ data from the computational
basis but also from other measurement configurations by training local expectation values such as spin-spin correlations evaluated in the rotated basis, giving access to the sign structure of the
state. I will show results obtained with this method for the ground state search of the 2D transverse field Ising model and the 2D dipolar XY model on 6x6 and10x10 square lattices, with experimental
data from a programmable Rydberg quantum simulator [Chen et al., Nature 616 (2023)], using a transformer wave function. In all cases, we find a great optimization speedup and significantly improved
convergence when applying the hybrid training. I will discuss how this method can be applied to other quantum states, e.g. ground and excited states of fermionic systems such as the t-J model,
pointing the way for a reliable and efficient optimization of neural quantum states competitive with state-of-the art methods.
Quantum control of interlayer excitons in atomically thin semiconductor heterostructures
Two-dimensional materials and their heterostructures provide a highly tunable platform for many-body interactions and strongly correlated phenomena, including Mott insulators, generalized Wigner
crystals and excitonic insulators. Of particular interest are atomically thin transition metal dichalcogenides (TMDs), such as MoS2, MoSe2 and WSe2. They strongly interact with light to form excitons
– electrons and holes bound by Coulomb attraction – which remain stable up to room temperature. The reduced dimensionality together with the relatively large effective mass and low kinetic energy of
the charge carriers yield strong interactions between the individual electrons and excitons in the system. In addition, new excitonic species can be formed when combining two or more TMD monolayers,
where the electrons and holes are separated between the individual layers – so-called interlayer excitons (IXs). The ability to engineer and control the properties of the thin semiconductors by
external means makes these systems a versatile platform for rich exciton and electron physics and unique opto-electronic applications.
Here, we investigate strongly correlated phenomena in two varieties of TMD bilayers – homobilayer MoS2 [1–3] and heterobilayer MoSe2/WSe2 [4]. These host IXs with large out-of-plane electric dipoles.
We study the quantum-confined Stark effect of the IXs in these systems, as well as their interaction with additional charges.
In homobilayer MoS2, we observe an unusual IX interaction, suggesting the electronic many-body state develops an order parameter in the form of interlayer electron coherence. Under conditions when
electron tunneling between the layers is negligible, we electron dope the sample and observe that the two excitons with opposing dipoles – which normally should not interact – hybridize in a way
distinct from both conventional level crossing and anti-crossing. We show that these observations can be explained by stochastic coupling between the excitons, which increases with electron density
and decreases with temperature.
In heterobilayer MoSe2/WSe2, we combine electro- and photo-luminescence experiments to study the nature of strongly driven non-equilibrium states. Applying a forward bias, we electrically inject
electrons and holes that recombine as IXs. We tune the relative electron-hole imbalance with electrostatic gates to study the formation of IXs interacting with an underlying Fermi sea. Further,
modulating the out-of-plane dipole by adjusting the distance of the electron and holes and by an applied electric field, we demonstrate control of the IXs at the quantum level.
N. Leisgang acknowledges support from the Swiss National Science Foundation (SNSF) (Grant No. P500PT_206917).
[1] N. Leisgang et al., Nat. Nanotechnol. 15, 901-907 (2020).
[2] L. Sponfeldner, N. Leisgang et al., Phys. Rev. Lett. 129, 107401 (2022).
[3] X. Liu*, N. Leisgang*, P. E. Dolgirev* et al., manuscript in preparation.
[4] A. M. Mier Valdivia, …, N. Leisgang et al., manuscript in preparation.
Single quantum coherent spins in hexagonal boron nitride at ambient conditions
Colour centres in wide bandgap materials can provide spin-photon interfaces that act as the building blocks in quantum networks and quantum sensing applications. Despite rapid progress reported
across several candidate systems, those possessing quantum coherent single spins at room temperature remain extremely rare [1]. Here, we show that hexagonal boron nitride hosts single emitters that
combine room-temperature spin coherence and single-photon emission with scalable and compact hardware. Via optical and microwave spectroscopy at room temperature, we investigate the ground-state spin
Hamiltonian of these individual emitters. We identify a spin-triplet electronic ground state with zero-field coherences that survive up to microseconds at ambient conditions, and unravel how the
symmetry of the spin-Hamiltonian protects the electronic spin from decoherence in the near-zero-field regime [2]. Our results demonstrate the rich spin dynamics underpinning this novel solid-state
qubit platform and further reveal the potential of van der Waals materials for quantum information and sensing, where their reduced dimensionality opens exciting routes to new nanoscale quantum
devices and sensors.
[1] G. Wolfowicz, “Quantum guidelines for solid-state spin defects,” Nat. Rev. Mater. 6 (2021).
[2] H. L. Stern*, C. M. Gilardoni*, et al. “A quantum coherent spin in hexagonal boron nitride at ambient conditions,” Nat. Mater. (2024).
Alice, Bob and the quantum ice-cream
In this talk, I will focus on two topics of quantum physics, whose salient aspects can be analyzed through the lens of entanglement. The first one concerns how to enable the teleportation of quantum
states between distant parties and to what extent the entanglement of a many-body wave function transfers under imperfect quantum teleportation protocols. The second subject concerns the study of the
symmetry breaking in a subsystem, which can be quantified once again by exploiting the theory of entanglement in many-body quantum systems. This leads to the definition of the entanglement asymmetry,
which neatly detects some physical remarkable features out-of equilibrium, and it reveals an unexpected quantum Mpemba effect.
Non-Abelian topological order from wavefunction collapse on a trapped-ion quantum processor
Non-Abelian topological order (TO) is a coveted state of matter that despite extensive efforts has remained elusive. I will show that adaptive quantum circuits – the combination of measurements with
unitary gates whose choice can depend on previous measurement outcomes – can be leveraged to prepare long-range entangled quantum states such as non-Abelian topological phases with a circuit depth
that is independent of system size. Using this, I will present the first unambiguous realization of non-Abelian TO and demonstrate control of its anyons. We create the ground state wave function of
D4 TO of 27 qubits on Quantinuum’s H2 trapped-ion quantum processor and obtain fidelity per site exceeding98.4%. In particular, we are able to detect a non-trivial braiding where three non-Abelian
anyons trace out the Borromean rings in spacetime, a signature unique to non-Abelian topological order.
This work is based on Nature 626 505-511 (2024)
Electrically defined quantum dots for neutral excitons
Quantum dots (QDs) are semiconductor nanostructures that confine particle motion in all three spatial dimensions, yielding discrete energy levels reminiscent of artificial atoms. Since their advent,
QDs confining excitons – bound electron–hole pairs – have been pivotal, serving as emitters of light in commercial displays to sources of single photons for quantum information processing. Due to the
limitations of current fabrication techniques, a key requirement in these applications that has remained unmet, is the realization of bright emitters that are identical to each other and can be
fabricated using scalable methods.
Here, I will describe how we overcome this hurdle and realize fully tunable gate-defined QDs for excitons in a monolayer transition metal dichalcogenide semiconductor. Through precise design of gate
electrodes, we dynamically modulate the in-plane electric fields in our device, enabling the tuning of QD resonance frequencies via the dc Stark effect [1, 2]. Simultaneously, the exciton confinement
length is modified, allowing a direct control over the oscillator strength and linewidth of the excitonic transition. Our structure is distinct from previous implementations, as it realizes
quantum-confined bosonic modes with a nonlinear response arising solely from exciton–exciton interactions.
The ability to place an exciton in a gate-defined quantum box offers the prospect for realizing an array of bright and identical single photon sources, which are essential for applications in quantum
communication and photonic quantum information processing. Another exciting future direction would be to interface these excitonic/photonic dots with their electronic counterparts in bilayer graphene
[3], allowing for the realization of a quantum spin–photon interface in van der Waals materials. Lastly, such quantum emitters hold promise as a foundational element of a strongly interacting
many-body photonic system [4].
[1] D. Thureja, et al. Electrically tunable quantum confinement of neutral excitons. Nature 606 (7913), 298-304 (2022).
[2] D. Thureja, et al. Electrically defined quantum dots for bosonic excitons. arXiv preprint arXiv:2402.19278 (2024).
[3] A. O. Denisov, et al. Ultra-long relaxation of a Kramers qubit formed in a bilayer graphene quantum dot. arXiv preprint arXiv:2403.08143 (2024).
[4] I. Carusotto & C. Ciuti. Quantum fluids of light. Reviews of Modern Physics 85 (1), 299, (2013).
Spin qubits represent a promising candidate for the development of quantum computers. Despite their potential, the implementation of scalable quantum systems is hindered by the short-range nature of
spin-entangling gates, necessitating a coupler for long-range entanglement. Therefore, achieving coherent coupling between a spin qubit and a photon becomes highly desirable [1].
We realize an architecture that accomplishes this coupling using high-quality, magnetic-field resilient, high-impedance superconducting resonators combined with semiconducting nanowires [2,3]. By
leveraging the intrinsic spin-orbit interaction present in these nanowires, we attain the strong coupling regime between a spin singlet-triplet qubit and a single photon [4]. This finding is
supported by the spectroscopy of maximally entangled spin-photon states. The spin-photon interface allows us to pinpoint an optimal operating point that maximizes both spin coherence and dipole
moment without compromise [5]. Our results are a crucial step toward scaling-up spin-based quantum processors through long-range quantum entanglement.
[1] Vandersypen, L.M.K., et al. npj Quantum Information 3.1 (2017).
[2] Ungerer, Jann H., et al. EPJ Quantum Technology 10.1 (2023).
[3] Ungerer, Jann. H., et al. Materials for Quantum Technology 3.3 (2023).
[4] Ungerer, Jann H., et al. Nature Communications 15.1 (2024).
[5] Ungerer, Jann H., et al. arXiv 2405.10796 (2024).
Symposium Jury
Prof. Dr. Michael Hartmann | Friedrich-Alexander-Universität
Prof. Dr. Robert König | Technische Universität München
Dr. Nadezhda Kukharchyk | Walter-Meißner-Institut
Dr. Farsane Tabataba-Vakili | Ludwig-Maximilians-Universität München
Dr. Johannes Zeiher | Max Planck Institute of Quantum Optics
16 September
Day 1
13:00–13:30 | Welcome talk
13:30–14:00 | Elizaveta Andriyakhina, "Quantum Fluctuations and Collective Modes in Disordered 2D Superconductors"
14:00–14:30 | Serafim Babkin, "Proximity-induced gapless superconductivity in two-dimensional Rashba semiconductor in magnetic field"
14:30–15:00 | Haoyu Hu, "Heavy-fermion physics and superconductivity in twisted bilayer graphene"
15:30–16:00 Coffee Break
16:00–16:30 | Fabrizio Berritta, "Real-time Quantum Control of Qubits"
16:30–17:00 | Matthias Bock, "From two-partite entanglement on a kilometer scale to multi-partite entanglement on a micrometer scale"
17:00–17:30 | Nadine Leisgang, "Quantum control of interlayer excitons in atomically thin semiconductor heterostructures"
17:30–18:00 | Carmem Maia Gilardoni, "Single quantum coherent spins in hexagonal boron nitride at ambient conditions"
17 September
Day 2
09:00–09:30 | Hannah Lange, "Combining transformer neural networks and quantum simulators: A hybrid approach to simulating quantum many-body systems"
09:30–10:00 | Sara Murciano, "Alice, Bob and the quantum ice-cream"
10:00–10:30 | Luisa Eck, "Generalizations of Kitaev’s honeycomb model from braided fusion categories"
10:30–11:00 | Coffee Break
11:00–11:30 | Deepankur Thureja, "Electrically defined quantum dots for neutral excitons"
11:30–12:00 | Nathanan Tantivasadakarn, "Non-Abelian topological order from wavefunction collapse on a trapped-ion quantum processor"
12:00–12:30 | Jann Hinnerk Ungerer, "Spin-Photon Entanglement"
12:30–13:30 | Lunch at MPQ
13:30–15:30 | Poster Session, Networking
19:00–21:00 | Closing Dinner for Finalists, Announcement of Prizes (Gasthof Neuwirt, Garching) | {"url":"https://www.mcqst.de/quantum-talents-munich/quantum-talents-finalists-2024.html","timestamp":"2024-11-07T01:44:36Z","content_type":"text/html","content_length":"73091","record_id":"<urn:uuid:03a7d002-24d6-49e5-8369-9b2a08705c7d>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00577.warc.gz"} |