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https://www.pdffilestore.com/a-first-course-in-differential-equations-the-classic-fifth-edition/
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# A First Course in Differential Equations the Classic Fifth Edition
Differential equations are a powerful tool for modeling many real-world phenomena. In this post, we will discuss the “Classic Fifth Edition” of A First Course in Differential Equations, a well-known textbook in the field. This book is aimed at providing a solid foundation in differential equations, and it does an excellent job of explaining key concepts and methods.
The book begins with a review of basic calculus concepts before moving on to more advanced topics such as first-order differential equations, higher-order differential equations, Laplace transforms, and numerical methods.
Differential equations are a powerful tool for modeling many real-world phenomena. The classic fifth edition of A First Course in Differential Equations provides an excellent introduction to the subject. This book covers all the essential topics, including first-order differential equations, higher-order differential equations, Laplace transforms, and numerical methods.
It also includes a wealth of worked examples and exercises to help readers master the material. This new edition has been thoroughly updated and revised, making it an even more valuable resource for students and instructors alike.
## Library Genesis
If you’re looking for a free and legal way to download books, then you’ll want to check out Library Genesis. This website offers over 2.2 million books in a variety of formats, including PDFs, eBooks, and audiobooks. You can search for specific titles or browse through the categories to find something new to read.
Best of all, no registration is required!
## What are Differential Equations And Why are They Important
Differential equations are one of the most important tools in mathematics and science. They allow us to describe how a quantity changes over time, and they are essential for modeling many physical phenomena. Differential equations come in many different forms, but they all have one thing in common: they involve derivatives.
The derivative of a function tells us how that function is changing at a given point. For example, if we have a function f(x) = x2, then its derivative at the point x = 1 is 2x = 2. This means that the function is increasing at a rate of two units per unit change in x.
Differential equations allow us to model things like population growth, the spread of diseases, and the motion of objects. They are also useful for solving problems in physics and engineering. In fact, many famous physicists and mathematicians have made significant contributions to the study of differential equations.
There are two main types of differential equations: ordinary differential equations (ODEs) and partial differential equations (PDEs). ODEs involve one or more functions of one variable, while PDEs involve functions of multiple variables. Both types of differential equations can be very difficult to solve, but there are many powerful methods that can be used to find solutions.
## What is the Classic Fifth Edition of A First Course in Differential Equations
The classic fifth edition of A First Course in Differential Equations is an introductory differential equations textbook. The book covers the basic theory of differential equations and presents a variety of methods for solving them. It also discusses the applications of differential equations to various fields, such as physics, engineering, and biology.
This edition of the book includes new chapters on numerical methods and chaos theory.
## Why is This Edition of the Book Considered to Be a Classic
This edition of the book is considered to be a classic because it was the first time that the novel was published in its entirety. Previously, only excerpts from the book had been released, so this was the first time that readers were able to experience the story in its entirety. The novel quickly gained popularity and has since become a staple in literature classes around the world.
## How Does This Book Differ from Other Editions
If you’re a fan of the book 1984 by George Orwell, you might be wondering how this edition differs from other editions. Here’s a quick rundown of some of the ways this particular edition differs: -The cover is different, obviously.
This edition features a black and white image of a man with a gas mask on, looking out over a cityscape. -This edition includes an afterword by Erich Fromm, which wasn’t included in previous editions. -There are also notes throughout the text that weren’t included in other versions, providing additional context and information about what was going on during certain scenes.
So if you’re interested in getting a little more insight into the book, or just want to see how this particular version stacks up against others, definitely check it out!
## Conclusion
This blog post covers the classic fifth edition of A First Course in Differential Equations. This edition is considered a classic because it was the first to cover both ordinary and partial differential equations in one book. The book is still used today as a reference for those studying differential equations.
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# 1 You Are Asked To Determine The Market Value Mark To Market Balance Sheet For Angus 2843572
1. You are asked to determine the market value(mark-to-market) balance sheet for Angus State Bankand loan duration (amounts in \$ thousands and duration in years):
Book Value Market Value
#### How many pages is this assigment?
Assets Amount Amount Duration
T-bills \$ 180 \$ 180 0.50
Loans* 5,000 ______ _____
Total Assets 5,180 ______
Liabilities
Deposits 4,184 4,184 0.50
Total Liabilities 4,184 4,184
Equity 996 ______
Total Lia and NW 5,180 ______
*Since this is a simple bank, it has only one type of loan. The loan has a \$5,000 book value (current outstanding principal), amortized loan with annual payments, an interest rate of 6.5 percent, and 20-years to maturity. Similar amortized loans today (market interest rate for similar loans) have an interest rate of 7 percent which, is the market yield.
a.Using Excel, determine the market value and duration of the loan and fill in the blanks in the balance sheet above. Please include a copy of your Excel Spreadsheet with your completed exam (you can copy and paste as a picture).
b. What is the average duration of all the assets and what is the average duration of all the liabilities?
Average Duration of Assets: = ______ years
Average Duration of Liabilities:[4,184*0.5]/4184 = 0.50 years
c.What is the leverage-adjustedduration gap? Is Angus State Bank exposed to interest rate risk? Is it exposed if interest rate increase or decrease?
Leverage-adjusted duration gap (DG) = -[DA – kDL]= = ___________ years
The duration gap is _______ years, indicates that an increase in interest rates will lead to an increaseor decrease in net worth. Circle the correct change in net worth
d. What is the forecasted impact on the market value of equity caused by a relative 1.5 percent upward shift in the entire yield curve? [i.e.,Dr/(1+r) = 0.0150]?
The market value of the equity will change by the following:
MVE = -DG * (A) * r/(1 + r) = ______ (_________)(0.015)= \$______________.
e.What variables are available to the financial institution to immunize or at least reduce interest rate risk exposure on the balance sheet? Taking one variable at a time, how much would each variable need to change to get DGAP equal to 0?
To immunize the institution for interest rate risk, the Leverage Adjusted DG needs to be zero:
DG = DA–kDL= 0
The choice is to make the duration of assets equal to:
DA= (4184/4987.37)0.50=0.42 years
or the duration of liabilities equal to:
DL= _______________ = ____________ years
Or some combination thereof. Obviously, for any depository institution, this cannon occur, thus the next most cost effect way to reduce interest rate risk is using interest rate swaps.
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# (decameters per hour) per second to (decimeters per minute) per second Conversion
(decameter per hour) per second to (decimeter per minute) per second conversion allow you make a conversion between (decameter per hour) per second and (decimeter per minute) per second easily. You can find the tool in the following.
### Acceleration Conversion
to
input
= 1.66666667
= 0.0166667 × 102
= 0.0166667E2
= 0.0166667e2
= 3.33333333
= 0.0333333 × 102
= 0.0333333E2
= 0.0333333e2
= 5.00000000
= 0.05 × 102
= 0.05E2
= 0.05e2
= 6.66666667
= 0.0666667 × 102
= 0.0666667E2
= 0.0666667e2
= 8.33333333
= 0.0833333 × 102
= 0.0833333E2
= 0.0833333e2
### Quick Look: (decameters per hour) per second to (decimeters per minute) per second
(decameter per hour) per second 1 damph/s 2 damph/s 3 damph/s 4 damph/s 5 damph/s 6 damph/s 7 damph/s 8 damph/s 9 damph/s 10 damph/s 11 damph/s 12 damph/s 13 damph/s 14 damph/s 15 damph/s 16 damph/s 17 damph/s 18 damph/s 19 damph/s 20 damph/s 21 damph/s 22 damph/s 23 damph/s 24 damph/s 25 damph/s 26 damph/s 27 damph/s 28 damph/s 29 damph/s 30 damph/s 31 damph/s 32 damph/s 33 damph/s 34 damph/s 35 damph/s 36 damph/s 37 damph/s 38 damph/s 39 damph/s 40 damph/s 41 damph/s 42 damph/s 43 damph/s 44 damph/s 45 damph/s 46 damph/s 47 damph/s 48 damph/s 49 damph/s 50 damph/s 51 damph/s 52 damph/s 53 damph/s 54 damph/s 55 damph/s 56 damph/s 57 damph/s 58 damph/s 59 damph/s 60 damph/s 61 damph/s 62 damph/s 63 damph/s 64 damph/s 65 damph/s 66 damph/s 67 damph/s 68 damph/s 69 damph/s 70 damph/s 71 damph/s 72 damph/s 73 damph/s 74 damph/s 75 damph/s 76 damph/s 77 damph/s 78 damph/s 79 damph/s 80 damph/s 81 damph/s 82 damph/s 83 damph/s 84 damph/s 85 damph/s 86 damph/s 87 damph/s 88 damph/s 89 damph/s 90 damph/s 91 damph/s 92 damph/s 93 damph/s 94 damph/s 95 damph/s 96 damph/s 97 damph/s 98 damph/s 99 damph/s 100 damph/s (decimeter per minute) per second 1.6666667 dmpm/s 3.3333333 dmpm/s 5 dmpm/s 6.6666667 dmpm/s 8.3333333 dmpm/s 10 dmpm/s 11.6666667 dmpm/s 13.3333333 dmpm/s 15 dmpm/s 16.6666667 dmpm/s 18.3333333 dmpm/s 20 dmpm/s 21.6666667 dmpm/s 23.3333333 dmpm/s 25 dmpm/s 26.6666667 dmpm/s 28.3333333 dmpm/s 30 dmpm/s 31.6666667 dmpm/s 33.3333333 dmpm/s 35 dmpm/s 36.6666667 dmpm/s 38.3333333 dmpm/s 40 dmpm/s 41.6666667 dmpm/s 43.3333333 dmpm/s 45 dmpm/s 46.6666667 dmpm/s 48.3333333 dmpm/s 50 dmpm/s 51.6666667 dmpm/s 53.3333333 dmpm/s 55 dmpm/s 56.6666667 dmpm/s 58.3333333 dmpm/s 60 dmpm/s 61.6666667 dmpm/s 63.3333333 dmpm/s 65 dmpm/s 66.6666667 dmpm/s 68.3333333 dmpm/s 70 dmpm/s 71.6666667 dmpm/s 73.3333333 dmpm/s 75 dmpm/s 76.6666667 dmpm/s 78.3333333 dmpm/s 80 dmpm/s 81.6666667 dmpm/s 83.3333333 dmpm/s 85 dmpm/s 86.6666667 dmpm/s 88.3333333 dmpm/s 90 dmpm/s 91.6666667 dmpm/s 93.3333333 dmpm/s 95 dmpm/s 96.6666667 dmpm/s 98.3333333 dmpm/s 100 dmpm/s 101.6666667 dmpm/s 103.3333333 dmpm/s 105 dmpm/s 106.6666667 dmpm/s 108.3333333 dmpm/s 110 dmpm/s 111.6666667 dmpm/s 113.3333333 dmpm/s 115 dmpm/s 116.6666667 dmpm/s 118.3333333 dmpm/s 120 dmpm/s 121.6666667 dmpm/s 123.3333333 dmpm/s 125 dmpm/s 126.6666667 dmpm/s 128.3333333 dmpm/s 130 dmpm/s 131.6666667 dmpm/s 133.3333333 dmpm/s 135 dmpm/s 136.6666667 dmpm/s 138.3333333 dmpm/s 140 dmpm/s 141.6666667 dmpm/s 143.3333333 dmpm/s 145 dmpm/s 146.6666667 dmpm/s 148.3333333 dmpm/s 150 dmpm/s 151.6666667 dmpm/s 153.3333333 dmpm/s 155 dmpm/s 156.6666667 dmpm/s 158.3333333 dmpm/s 160 dmpm/s 161.6666667 dmpm/s 163.3333333 dmpm/s 165 dmpm/s 166.6666667 dmpm/s
(decameter per hour) per second is a unit of acceleration. It is equal to 0.00277777778 m / s2. Plural name is (decameters per hour) per second.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
(decameter per hour) per seconddamph/s
= 0.00277777778 m / s2
= 0.00277777778 m / s2
Metric system SI
#### conversion table
(decameters per hour) per second(decimeters per minute) per second(decameters per hour) per second(decimeters per minute) per second
1= 1.66666666666676= 10
2= 3.33333333333337= 11.666666666667
3= 58= 13.333333333333
4= 6.66666666666679= 15
5= 8.333333333333310= 16.666666666667
(decimeter per minute) per second is a unit of acceleration. It is equal to 0.00166666667 m / s2. Plural name is (decimeters per minute) per second.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
(decimeter per minute) per seconddmpm/s
= 1.66666667 m / s2
= 1.66666667 m / s2
Metric system SI
### conversion table
(decimeters per minute) per second(decameters per hour) per second(decimeters per minute) per second(decameters per hour) per second
1= 0.66= 3.6
2= 1.27= 4.2
3= 1.88= 4.8
4= 2.49= 5.4
5= 310= 6
### Conversion table
(decameters per hour) per second(decimeters per minute) per second
1= 1.6666667
0.6= 1
### Legend
SymbolDefinition
exactly equal
approximately equal to
=equal to
digitsindicates that digits repeat infinitely (e.g. 8.294 369 corresponds to 8.294 369 369 369 369 …)
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# Perimeter of Rhombus Formula
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## Formulas of Rhombus
Wondering how to find the perimeter of a rhombus? Just like any polygon, the perimeter of the rhombus is the total distance around the outside, which can be simply calculated by adding up the length of each side. In the case of a rhombus, all four sides are of similar length, thus the perimeter is four times the length of aside. Or as a formula: Perimeter of rhombus = 4 a = 4 × side. Here, ‘a’ represents each side of a rhombus.
### About the Perimeter of Rhombus Formula
A rhombus is a 2-dimensional (2D) geometrical figure which consists of four equal sides. Rhombus has all sides equal and its opposite angles are equivalent in measurement. Let us now talk about the rhombus formula i.e. area and perimeter of the rhombus.
### The Perimeter of a Rhombus
The perimeter is the sum of the length of all 4 sides. In the rhombus all sides are equal.
Thus, the Perimeter of rhombus = 4 × side
So, P = 4s
In which,
S = length of a side of a rhombus
### Area of Rhombus Formula
The area of a rhombus is the number of square units in the interior of the polygon. The area of a rhombus can be identified in 2 ways:
i) Multiplying the base and height as rhombus is a unique kind of parallelogram.
Area of rhombus = b × h
In which,
B = base of the rhombus
H = height of the rhombus
ii) By determining the product of the diagonal and dividing the product by 2.
Area of rhombus formula =
1/2× d1 × d2
In which,
d1 × d2 = diagonal of the rhombus
### Derivation of Area of Rhombus
Let MNOP is a rhombus whose base MN = b, PN ⊥ MO, PN is a diagonal of rhombus = d1, MO is diagonal of rhombus = d2, and the altitude from O on MN is OZ, i.e., h.
• Area of Rhombus MNOP = 2 Area of ∆ MNO
### Fun Facts
• A rhombus consists of an inscribed circle
• In a rhombus, all sides are equal, just as a rectangle has all angles equal.
A rhombus has opposite angles equivalent to each other, while a rectangle has opposite sides equal.
### Finding the Perimeter of a Rhombus when only Diagonals are known
In many questions, you will see that the length of the sides of the rhombus is not given. Instead, the question will provide you with the length of its diagonals. You can use the diagonals to find the side of the rhombus and calculate its perimeter. Here is how you can do it:
1. Consider a rhombus PQRS with diagonals a and b, and center O.
In ∆ MZP,
MP2 = MZ2 + ZP2
⇒ 152 = 92 + ZP2
⇒ 225 = 81 + ZP2
⇒ 144 = ZP2
⇒ ZP = 12
Hence, NP = 2 P
= 2 × 12
= 24 cm
Now, to find out the area of the rhombus, we will apply the formula i.e.
2. Find the perimeter of a rhombus MNOP whose diagonals measure 20 cm and 24 cm respectively?
Ans:
Given: d1 = 20 cm
d2 = 24 cm
Since, MN = NO = OP = MP,
Therefore, Perimeter of MNOP = 15.62 × 4 = 62.48 cm.
### Properties of a Rhombus
Identifying a rhombus is not that difficult. There are some properties of a rhombus that will help you determine whether a given figure is a rhombus or not. If a shape meets the following conditions, then it is a rhombus:
• All the sides of a rhombus are equal.
• The opposite sides of a rhombus are parallel to each other.
• All the opposite angles in a rhombus will be equal.
• The diagonals of a rhombus bisect each other at 90 degrees i.e. right angles.
• When you add any two adjacent angles of a rhombus, the sum should be equal to 180 degrees.
### Revising the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs
The Perimeter of Rhombus Formula is one of the most important formulas of Mathematics. It comes under mensuration, which is a crucial chapter of Maths. You must have a clear understanding of the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs to ensure a good score in your finals. Once you have a firm grasp of the perimeter and area of a rhombus, you will be able to solve any question related to this concept. That is why you should practice and revise the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs thoroughly. Here are some revision tips:
• First, study the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs provided to understand the topic clearly.
• Go through the textbook explanations of the perimeter and area of a rhombus thoroughly.
• Practice as many questions related to the perimeter and area of a rhombus as you can to become more proficient at solving mensuration problems.
• Use different reference books and pick out the important questions based on the perimeter and area of a rhombus to understand the different types of questions that can come in your final Maths exam.
• Visit Vedantu’s e-learning platform through our website or mobile application to study the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs. Every explanation and question provided by Vedantu is curated by some of the best subject matter experts to ensure accuracy and high quality.
• Go through solved examples from the textbook or reference books to learn how to solve different kinds of questions based on the area and perimeter of a rhombus.
• Once you have solved all the textbook questions, find out questions from previous year question papers and sample papers to understand the pattern and difficulty level of the exam.
1. What is a Rhombus?
A rhombus is a special kind of parallelogram with its entire sides equal. A square is a kind of rhombus with all of the angles being equal and also all of the sides. Moreover, both rhombuses and squares have perpendicular diagonal bisectors that divide each diagonal into two equal segments, and also divide the quadrilateral into four equal right triangles.
Having said that, we know the Pythagorean Theorem would work sufficiently in this situation, using half of each diagonal as the two legs of the right triangle.
2. What are the properties of a Rhombus?
Following are the properties of a rhombus:-
• Rhombus has all sides equal.
• The opposite angles of a rhombus are equal.
• The Sum of adjacent angles are supplementary i.e. (∠N + ∠O = 180°).
• If one angle of a rhombus is right, then all angles are right.
• Each diagonal splits it into two congruent triangles.
• In a rhombus, diagonals intersect each other and are perpendicular to each other.
3. What are the dual properties of a Rhombus?
Note that the dual polygon of a rhombus is a rectangle:
• A rhombus consists of an axis of symmetry across each pair of opposite vertex angles, whereas a rectangle has an axis of symmetry across each pair of opposite sides.
• The diagonals of a rhombus bisect at equal angles. On the other hand, the diagonals of a rectangle are equivalent in length.
• The geometrical shape formed by connecting the midpoints of the sides of a rhombus is a rectangle, and reciprocally.
4. From where can I learn the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples, and FAQs?
You can learn the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples, and FAQs from Vedantu’s online learning platform. To start learning, visit Vedantu.com or download our mobile application from the play store for android and the app store for iPhone. You don’t have to pay any registration fee to learn the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples, and FAQs on Vedantu. Moreover, we provide you with plenty of study materials to learn various formulas of Maths. You can use these resources to learn formulas of Relations and Functions, Trigonometry, Matrices, Determinants, and much more for absolutely free only on Vedantu’s educational platform.
5. How will the perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs help me?
Learning the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs will help you solve many mensuration problems. You can use these formulas to determine the perimeter and area of a rhombus. By solving questions with these formulas, you will get a better understanding of a rhombus. Your exam question paper might contain questions based on the perimeter or area of a rhombus. So, the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs will aid you in scoring well in your final exams.
What is the Perimeter of a Rhombus?
The perimeter of a rhombus is the sum of all its sides. Since all the four sides of a rhombus are equal, the formula used to find the perimeter of a rhombus is: Perimeter = a + a + a + a = 4a, where ‘a’ represents the side length of the rhombus.
What is the Area and Perimeter of a Rhombus?
The area of a rhombus is the space occupied by it. This is calculated with the help of different formulas which depend on the type of dimensions given and is expressed in square units.
• When the base and height of a rhombus are known, then the area of rhombus = base × height.
• When the diagonals of a rhombus are known, then the area = (diagonal 1 × diagonal 2)/2.
The perimeter of a rhombus is the total length of its boundary. This is calculated with the help of different formulas which depend on the given dimensions and is expressed in linear units.
• When one side of a rhombus is known, the perimeter = 4 × side
• When the two diagonals of a rhombus are known, the
What is the Perimeter of a Rhombus Formula When Side Length is Given?
Since all the 4 sides of a rhombus are equal, its perimeter is obtained by multiplying its side by 4. The formula to calculate the perimeter of a rhombus with side ‘a’ is: P = 4a.
How to Find the Side Length When the Perimeter of a Rhombus is Given?
We know that the perimeter of a rhombus with side length ‘a’ is calculated with the help of the formula, P = 4a. Thus, the side length of the rhombus can be obtained by dividing its perimeter by 4.
## Perimeter of Rhombus
The perimeter of a rhombus is the sum of all its sides. Rhombus is a quadrilateral in which all four sides are of the same measure. A rhombus is always a parallelogram but a parallelogram may not necessarily be a rhombus always. Let us study more about the perimeter of rhombus in this article.
### What is the Perimeter of a Rhombus?
The perimeter of a rhombus is the total measure of its boundary and it is calculated by adding the length of all its sides. Since all the four sides of a rhombus are equal, the basic formula to find the perimeter of a rhombus is: Perimeter = 4a; where ‘a’ is the side of the rhombus. The perimeter of a rhombus is expressed in linear units like inches, yards, centimeters, and so on.
### Properties of a Rhombus
There are some basic properties that help us to identify a rhombus. Observe the following rhombus ABCD to relate to its properties given below.
We can identify and distinguish a rhombus with the help of the following properties:
• All four sides are equal.
• The opposite sides are parallel.
• The opposite angles are equal.
• The sum of any two adjacent angles is 180o.
• The diagonals bisect each other at right angles.
• Each diagonal bisects the vertex angles.
Now, let us read about the formulas that are used to find the perimeter of a rhombus in different cases. The formulas differ according to the known dimensions.
### Perimeter of Rhombus Formula With Sides
As discussed earlier, the perimeter of a rhombus is the sum of the lengths of all its sides. We know that the sides of a rhombus are of equal lengths. Let us consider a rhombus of side length ‘a’. Then, the perimeter of the rhombus is a + a + a + a which is 4a. Thus, the perimeter of the rhombus formula is: P = 4a.
Example: Find the perimeter of a rhombus that has a side length of 10 units.
Solution:
The side length of the given rhombus, a = 10 units.
Its perimeter = 4a = 4 × 10 = 40 units.
### Perimeter of Rhombus Formula With Diagonals
When the length of the diagonals of a rhombus is known, we find the side length of the rhombus using the Pythagoras theorem. Here, we make use of the following properties of a rhombus:
• A rhombus is divided into 4 congruent right-angled triangles by its two diagonals.
• The diagonals bisect each other at right angles.
Let us consider a rhombus ABCD with diagonals p and q and with side length ‘a’.
Let us take triangle AOD. Since the diagonals bisect at right angles, AO can be written as p/2 and OD can be written as q/2. Now, if we apply the Pythagoras theorem for triangle AOD, we get
Note: We do not need to remember this formula. We can use the Pythagoras theorem to find the side length of the rhombus using the diagonals and then we can apply the perimeter of rhombus formula to be, P = 4 × side length.
### Perimeter of Rhombus Examples
Example 1: Find the perimeter of a rhombus with diagonals 8 inches and 6 inches respectively.
Solution:
Method 1:
The lengths of the diagonals of the given rhombus are, p = 8 inches and q = 6 inches.
Using the perimeter of rhombus formula using diagonals,
By Pythagoras theorem,
a2 = 32 + 42 = 25
a = 5 inches.
Thus, the perimeter of the rhombus is, 4a = 4 × 5 = 20 inches.
Answer: The perimeter of the given rhombus = 20 inches.
• Example 2: ABCD is a rhombus with one side as 7 units. Find the perimeter of ABCD.
Solution:
Let us assume that the side length of the rhombus ABCD is ‘a’ units.
The side length of the given rhombus, a = 7 units.
We know that perimeter of a rhombus with sides = 4a = 4 × 7 = 28 units.
Answer: The perimeter of the rhombus = 28 units.
### Real-life Applications of the Perimeter of a Rhombus
The word perimeter is a combination of two Greek words: “Peri,” which means surrounding or boundaries of a surface or an object, and “Meter,” which means measurement of the surface or object, so perimeter means the total measurement of boundaries of a given surface.
With this information, we can use the perimeter of a rhombus in numerous real-life applications. Various examples are given below:
• For example, we can use the perimeter of a rhombus to calculate the distance of a pitcher’s spot from the striker in baseball if the whole pitch is shaped like a rhombus.
• The perimeter formula is also helpful in designing tables and cupboards having a rhombus shape.
• It is also helpful in the construction of rhombus-shaped offices and rooms.
Example 1:
If the length of one side of a rhombus is 11 cm, what will be the length of the rest of the sides?
Solution:
We know that all the sides of a rhombus are equal in length, so the length of the rest of the three sides is also 11 cm each.
Example 2:
Calculate the perimeter of a rhombus for the figure given below.
Solution:
We are given the length of one side of a rhombus, and we know that all the sides are equal in length.
Perimeter of the rhombus Perimeter of the rhombus
Example 3:
If the perimeter of a rhombus is 80cm, what will be the length of all the sides of the rhombus?
Solution:
We are given the perimeter of the rhombus. We can calculate the length of each side of a rhombus by using the perimeter formula:
Example 4:
If the length of the diagonals of a rhombus is 9 cm and 11cm, what will be the perimeter of the rhombus?
Solution:
We are given the length of the two diagonals of the rhombus: let “a” and “b” be the two diagonals of the rhombus. Then, we can calculate the perimeter of the rhombus by using the formula given below.
### Example 5:
A rhombus has an area of 64cm2 , and the length of one diagonal of the rhombus is 8cm . What will be the perimeter of the rhombus?
### Solution:
Let diagonal “a” = 8cm and we have to find “b”
1. We know that all sides of a rhombus are equal in length. If the length of one side of the rhombus is 20 cm, then the length of the remaining three sides will also be the same, i.e., 20 cm.
3. We are given the lengths of the two diagonals of the rhombus. Let “a” and “b” be the two diagonals. Then, we can calculate the perimeter and area of the rhombus by using the values of the diagonals.
Perimeter of a rhombus =30cm approx.
Question 1: Find the perimeter of a rhombus whose side is 8 cm.
Solution:
Given that side s = 8 cm
Perimeter of Rhombus is given by : 4*s
So, Perimeter (P) = 4 * 8 cm = 32 cm
Question 2: Find the side length of a rhombus whose perimeter is given as 36cm.
Solution:
Given Perimeter(P) = 36 cm
P = 4 * s
=> s = P/4
So, s = 36/4 = 9cm
Question 3: Find the perimeter of the rhombus given the diagonal lengths are 6 cm and 8 cm respectively.
Solution:
When diagonal lengths are given :
Given a = 6 cm, b = 8cm
Perimeter(P) = 2* √(a2 + b2) = 2* √(36 + 64) = 2 * 10 = 20 cm
Question 4: Find the length of horizontal diagonal given the side length as 13cm and vertical diagonal length as 24 cm.
Solution:
Since the diagonal bisect at right angles:
Given b = 24 cm and s = 13 cm, a = ?
side(s) is given as
s = (√(a2 + b2))/2
2 * s = (√(a2 + b2))
26 = (√(a2 + 576))
On squaring both sides, 676 = a2 + 576
=> a2 = 100
=> a= 10cm
Question 5: Find the area of the rhombus whose diagonal are of lengths 24cm and 10 cm.
Solution:
Given a = 24 and b = 10cm
Area of the Rhombus is given by A = 1/2 * a * b
= 1/2 * 24 * 10
= 60 cm2
Question 6: Find the perimeter of a rhombus whose side is 2.5 cm.
Solution:
Given that side s = 8 cm
Perimeter of Rhombus is given by: 4*s
So, Perimeter (P) = 4 * (2.5) cm = 10 cm
Question 7: Find the side length of a rhombus whose perimeter is given as 48cm.
Solution:
Given Perimeter(P) = 48 cm
P = 4 * s
=> s = P/4
So, s = 48/4 = 12cm
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In this section of quadratic function, you will learn about quadratic equation, discriminant, and solving the equation through quadratic formula and modified quadratic formula that is better numerically.
Quadratic equation is happened to be a special condition of quadratic function when the parabola curve intersects with the horizontal axis ( ). In this case, we can three possibilities:
1. The quadratic function does not intersect horizontal axis
2. The quadratic function intersect horizontal axis at single point
3. The quadratic function intersect horizontal axis at two points
No intersection, Single intersection, Two intersections,
The point of intersection between the function and horizontal axis is called the zeros of the function, or the root of the quadratic equation, because inputting the values into the quadratic function will make the function value to be zero.
## Discriminant
Discriminant is an index to indicate whether the quadratic function has no intersection ( ), intersect as single point ( ) or intersect at multiple roots ( ) with the horizontal axis. Discriminant of quadratic equation is computed through its parameters:
When the discriminant is non-negative, given a quadratic equation , our problem is to solve the equation to find the value of that satisfied the equation. Solving quadratic equation is actually a process of finding the roots (or the zeros) of the equation.
As this is a well-known problem, the roots of quadratic equation (for multiple roots) are given by quadratic formula:
and
The version, which contains all three parameters of quadratic equation, is sometimes called ABC formula
The roots of quadratic equation are always multiple roots (two roots) except when the discriminant is zero.
For example :
Quadratic equation contains parameter , and . The discriminant is . Since the discriminant is negative, we conclude that the equation has no real root (no intersection between the function with horizontal axis). However, the roots exist. The roots are multiple roots in complex number, given by the quadratic formula above
In his interesting paper, Baker (1998) pointed out that for quadratic equation (notice that parameter ), we can get and , therefore the quadratic formula above can be written as two terms
The first term ( ) is mean or center of mass of the two roots and the second term
( ) is half of the absolute difference of the two roots.
In case of a single (distinct) root, the discriminant is zero, thus lead to
### Example
Quadratic equation has parameters and . The discriminant is zero because . Thus, the root of the quadratic equation is only one, that is . To check the correctness, we input the root value into the equation to get which is correct.
Another version of quadratic formula is given as
Press et al (1986) pointed out that when the value of parameter and are both very small, using quadratic formulas above, the root of discriminant will be inaccurate numerically because and might cancel each other. They proposed formula that is more robust in term of numerical accuracy
and
### Example
Suppose we have a quadratic equation . We can divide two sides of the equation with 2 to get the equivalent equation of . Note that now the quadratic parameter is one ( ). The other parameters are and . The discriminant is that is positive. Thus, the quadratic equation has multiple roots
Using Press et al modified Quadratic formula we get the same answers
Observe that and
So far, we have covered only a given quadratic function. In the next section , you will learn how to get the quadratic parameters from data points.
Preferable reference for this tutorial is
Teknomo, Kardi. (2019) Quadratic Function Tutorial .
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## Related Articles
• RD Sharma Class 12 Solutions for Maths
# Class 12 RD Sharma Solutions- Chapter 22 Differential Equations – Exercise 22.1 | Set 1
### Question 1.
Solution:
We have,
Order of function:
The Highest order of derivative of function is 3 i.e.,
So, the order of derivative is equal to 3.
Degree of function:
As the power of the highest order derivative of function is 1 (i.e., power of is 1)
So, degree of function is 1.
Linear or Non-linear:
The given equation is non-linear.
### Question 2.
Solution:
We have,
Order of function:
As the highest order of derivative of function is 2.(i.e.,)
So, Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 1(i.e., power of is 1)
So, Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is linear.
### Question 3.
Solution:
We have,
Order of function:
As the highest order of derivative of function is 1 (i.e., )
So, Order of the function is equal to 1.
Degree of function
As the power of the highest order derivative of the function is 3 (i.e., power of dy/dx is 3)
So, the degree of the function is equal to 3.
Linear or Non-linear:
The given equation is non-linear.
### Question 4.
Solution:
We have,
On squaring both side, we get
On cubing both side, we get
Order of function:
As the highest order of derivative of function is 2 (i.e.,
So, Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 2. (i.e., power of is 2)
So, the Degree of the function is equal to 2.
Linear or Non-linear:
The given equation is non-linear.
### Question 5.
Solution:
We have,
Order of function:
As the highest order of derivative of function is 2
So, Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of function is 1 (i.e., power of is 1)
So, the Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is non-linear.
### Question 6.
Solution:
We have,
On cubing both side, we get
On squaring both side, we get
Order of function:
As the highest order of derivative of function is 2 (i.e., )
So, the Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 2(i.e., power of is 2)
So, the Degree of the function is equal to 2.
Linear or Non-linear:
The given equation is non-linear.
### Question 7.
Solution:
We have,
On squaring both side, we get
Order of function:
The highest order of derivative of function is 4 (i.e., )
So, the order of the derivative is equal to 4.
Degree of function:
As the power of the highest order derivative of the function is 2 (i.e., power of is 2)
So, the degree of function is 2.
Linear or Non-linear:
The given equation is non-linear.
### Question 8:
Solution:
We have,
On squaring both side, we have
Order of function:
As the highest order of derivative of function is 1.
So, the Order of the function is equal to 1.
Degree of function:
As the power of the highest order derivative of the function is 1.
So, the degree of the function is equal to 1.
Linear or Non-linear:
The given equation is linear.
### Question 9:
Solution:
We have,
Order of function:
As the highest order of derivative of function is 2 (i.e.,)
So, order of derivative is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 1 (i.e., power of is 1)
So, the Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is linear.
### Question 10:
Solution:
We have,
Order of function:
As the highest order of derivative of the function is 2.
So, the Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 1 (i.e., power of is 1)
So, the Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is non-linear.
### Question 11:
Solution:
We have,
Order of function:
As the highest order of derivative of the function is 2
So, the Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 3. (i.e., power of is 3)
So, the degree of the function is equal to 3.
Linear or Non-linear:
The given equation is non-linear.
### Question 12:
Solution:
We have,
Order of function:
As the highest order of derivative of the function is 3
So, the Order of the function is equal to 3.
Degree of function:
As the power of the highest order derivative of the function is 1.(i.e., power of is 1)
So, the Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is non-linear.
### Question 13:
Solution:
We have,
Order of function:
As the highest order of derivative of the function is 1
So, the Order of the function is equal to 1.
Degree of function:
As the power of the highest order derivative of the function is 1. (i.e., power of dy/dx is 1)
So, the Order of the function is equal to 1.
Linear or Non-linear:
The given equation is non-linear.
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# A right circular cone is inscribed in a hemisphere.
by pringless
Tags: circular, cone, hemisphere, inscribed
P: 46 A right circular cone is inscribed in a hemisphere. The figure is expanding in such a way that the combinded surface area of the hemisphere and its base is increasing at a constant rate of 18 in^2 per second. At what rate is the volume of the cone changing when the radius of the common base is 4 in?
PF Patron Sci Advisor Thanks Emeritus P: 38,400 Are you sure this shouldn't be under "homework help"? [:)] I assume that the cone is inscribed in the hemisphere so that its base is the circular base of the hemispher. If I remember correctly, the surface area of a sphere is 4 &pi r2 so the surface area of a hemisphere is 2 &pi r2 and the "combinded surface area of the hemisphere and its base" is A= 3 &pi r2. Knowing that dA= 18 square inches per second, you should be able to find the rate of change of r from that. The volume of a right circular cone is given by V= (1/3) &pi r2h (I confess I looked that up). In this case, the height of the cone, as well as the radius of its base, is the radius of the hemisphere so V= (1/3) &pi r3. Since you have already calculated dr/dt, you can use that to find dV/dt. (Why are the "& codes" that Greg Barnhardt noted not working for me?)
P: 1,572 do they need to be (and are they) followed by semicolons?
Related Discussions Introductory Physics Homework 4 Nuclear Engineering 1 Calculus & Beyond Homework 5 Chemistry 18 Chemistry 3
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Question: What Is A Good Precision Value?
What is precision and recall in simple terms?
Precision and recall are two numbers which together are used to evaluate the performance of classification or information retrieval systems.
Precision is defined as the fraction of relevant instances among all retrieved instances.
A perfect classifier has precision and recall both equal to 1..
How can the precision of data be improved?
You can increase your precision in the lab by paying close attention to detail, using equipment properly and increasing your sample size. Ensure that your equipment is properly calibrated, functioning, clean and ready to use.
Why precision is important?
Accuracy and Precision This is important because bad equipment, poor data processing or human error can lead to inaccurate results that are not very close to the truth. Precision is how close a series of measurements of the same thing are to each other.
How do you calculate precision?
Find the difference (subtract) between the accepted value and the experimental value, then divide by the accepted value. To determine if a value is precise find the average of your data, then subtract each measurement from it. This gives you a table of deviations. Then average the deviations.
What precision means?
(Entry 1 of 2) 1 : the quality or state of being precise : exactness. 2a : the degree of refinement with which an operation is performed or a measurement stated — compare accuracy sense 2b.
How do you find precision and accuracy?
The accuracy is a measure of the degree of closeness of a measured or calculated value to its actual value. The percent error is the ratio of the error to the actual value multiplied by 100. The precision of a measurement is a measure of the reproducibility of a set of measurements.
Why is accuracy more important than precision?
Accuracy is generally more important when trying to hit a target. … Accuracy is something you can fix in future measurements. Precision is more important in calculations. When using a measured value in a calculation, you can only be as precise as your least precise measurement.
What is precision value?
Precision of measured values refers to how close the agreement is between repeated measurements.
What is better precision or recall?
Email Spam detection:This is one of the example where Precision is more important than Recall. … Precision: This tells when you predict something positive, how many times they were actually positive. whereas, Recall: This tells out of actual positive data, how many times you predicted correctly.
What does precision mean in math?
Precision refers to the amount of information that is conveyed by a number in terms of its digits; it shows the closeness of two or more measurements to each other.
What are good precision and recall values?
In information retrieval, a perfect precision score of 1.0 means that every result retrieved by a search was relevant (but says nothing about whether all relevant documents were retrieved) whereas a perfect recall score of 1.0 means that all relevant documents were retrieved by the search (but says nothing about how …
Is it better to be accurate or precise?
Precision refers to how close measurements of the same item are to each other. Precision is independent of accuracy. That means it is possible to be very precise but not very accurate, and it is also possible to be accurate without being precise. The best quality scientific observations are both accurate and precise.
How do you explain precision?
Precision refers to the closeness of two or more measurements to each other. Using the example above, if you weigh a given substance five times, and get 3.2 kg each time, then your measurement is very precise. Precision is independent of accuracy.
How do you interpret an F score?
If you get a large f value (one that is bigger than the F critical value found in a table), it means something is significant, while a small p value means all your results are significant. The F statistic just compares the joint effect of all the variables together.
Can a measure be both precise and accurate?
The precision of a measurement system is refers to how close the agreement is between repeated measurements (which are repeated under the same conditions). Measurements can be both accurate and precise, accurate but not precise, precise but not accurate, or neither.
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#### Details of this Paper
##### Stats Multiple Choice Questions
Description
solution
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Question;1) (Solve for X)A. 16B. 3C. 5D. 42) Identify the correct order of operations for calculation of basic math problemsA.Rule 1: First perform any calculations inside parentheses.Rule 2: Next perform all multiplications and divisions, working from left to right.Rule 3: Lastly, perform all additions and subtractions, working from left to right.B.Rule 1: First perform any calculations inside parentheses.Rule 2: Next perform all additions and subtractions, working from left to right.Rule 3: Lastly, perform all multiplications and divisions, working from left to right.C.Rule 1: First perform all additions and subtractions, working from left to right.Rule 2: Next perform any calculations inside parentheses.Rule 3: Lastly, perform all multiplications and divisions, working from left to right.D.Rule 1: First perform all additions and subtractions, working from left to right.Rule 2: Next perform all multiplications and divisions, working from left to right.Rule 3: Lastly, perform any calculations inside parentheses3) A ?Set? within quantitative theory represents??A. An infinite ordered array of objectsB. Any collection M into a whole of definite, distinct objects m (which are called the "elements" of M) of our perception [Anschauung] or of our thoughtC. Elements of a defined characteristicD. Any collection of M into a whole that is representative overall the the nature of the the target being observed4) (Solve for X)A. 12B. 31C. 17D. 215) The formula " " is a common expression for calculation of the??A. VarianceB. RangeC. MeanD. Standard Deviation6) Nominal is to Category as Ordinal is to??A. BreadthB. RangeC. AssociationD. Rank7) The expressed equation of would reflect the __________ of a population.A. The Population Variance which is the square root of the Standard DeviationB. The basic representation of constancy in a set of dataC. The Population Standard Deviation which is the square root of the VarianceD. A well known inequality that esitmates the error in statistical calculation7) The expressed equation of would reflect the __________ of a population.A. The Population Variance which is the square root of the Standard DeviationB. The basic representation of constancy in a set of dataC. The Population Standard Deviation which is the square root of the VarianceD. A well known inequality that esitmates the error in statistical calculation8) Given the Variable X measured at a Nominal Level, what descriptive statistical indice would you give if asked to provide the most representative value of the set?A. RangeB. Modal frequencyC. MeanD. Median9) What particular measure of correlation would be most appropriate for use with two variables measured at an ordinal level?A. Pearson?s rB. Contingency CoefficientC. Spearman?s RhoD. Chi Square10) A specific score in a distribution of data minus the mean and divided by the standard deviation produces a value known as a??A. Variability ScoreB. Coefficient of DeterminationC. BetaD. Z Score11) would be correctly expressed as??A. 5, 6B. 11, 12C. 5, 15D. 3, 912) All the following are measures of variability or dispersion within a set of data except??A. RangeB. Standard DeviationC. MedianD. Variance13) Expected relative frequency probability represents a calculated probability based on large numbers of trials. In an example of a deck of 52 cards, where you randomly pull cards from the deck and record what they are, you find that over 1000 random draws, you pull an ACE 76 times. What would be the calculated probability based on relative frequency?A..30B..76C..076D. 10%14) In a distribution of data conforming to the characteristics of a Normal Curve, the percentage of cases between Plus One and Minus One Standard Deviation from the Mean would be??A. 99.3%B. 84%C. 68%D. 95%15) If, then 5 to the third power (i.e., 53) would equal??A. 50B. 125C. 150D. 22516) If someone scored two Standard Deviations above the Mean on a standardized test where the Mean = 100 and the Standard Deviation = 15, then that person?s numerical score would be??A. 160B. 125C. 90th PercentileD. 13017) Independent trials in relationship to probability theory means that??A. Each outcome has an indendent relationship to the wholeB. The outcome of each trial is dependent upon the previous but not the followingC. The outcome of each trial is totally independent from each other trial conductedD. Influences of trials on the outcome are subtle overall18) The defining characteristics of the Ratio Level of Measurement are??A. Equal intervals between points of that scale and a true zeroB. Unequal intervals between points and a true zeroC. Equal intervals between points on a scale but an arbitrary zero pointD. An arbitrary zero with specified rank ordering of scale points19) In a 95% Confidence Interval, the true mean has what chance of falling between the Lower Limit and Upper Limits of the calculated confidence interval?A..05B. 5%C..90D. 95) What is the relationship of the Mean, Median and Mode as Measures of Central Tendency in a true Normal Curve?A. They align Mean, Median and Mode in that orderB. They align Mode, Median and Mean in that orderC. They are equalD. They align Median, Mode and Mean in that order21) Independent Variable is to CAUSE as Dependent Variable is to??A. ResponseB. PerceptionC. PredictorD. Effect22) The condition of scores being distorted in one direction or the other (towards the ends of the distribution) within a given distribution of data is known as??A. KurtosisB. SkewnessC. DiscontinuityD. Asynchronous23) An extreme score that sets apart in value from the general distribution of values with which it is connected is known as??A. OutlierB. Incongruous dataC. Champion scoreD. Anomalous indicator24) A Dichotomous Variable represents what level of measurement within the realm of quantitative analysis?A. RatioB. IntervalC. OrdinalD. Nominal25) The Standard Error of Measurement is calculated by what formula?A.B.C.D.
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Instructions
In each of the questions, below two/three statements are given followed by conclusions/groups of conclusions numbered I and II. You assume all the statements to be true even if they seem to be at variance from the commonly known facts and then decide which of the given two conclusions logically follows from the information given in the statements.
Give answer A if only conclusion I follows
Give answer B if only conclusion II follows
Give answer C if either I or II follows
Give answer D if neither I nor II follows
Give answer E if both I and II follows
Question 35
Statements: Some dews are drops. All drops are stones.Conclusions: I. Atleast some dews are stones II. Atleast some stones are drops.
Solution
The Venn diagram for the given syllogism is as follows:
Based on the diagram, we can conclude that both the conclusions are implied. Hence the answer is (e)
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## ››Convert pace [great] to attometre
pace [great] attometer
Did you mean to convert pace [great] pace [Roman] to attometer
How many pace [great] in 1 attometer? The answer is 6.5616797900262E-19.
We assume you are converting between pace [great] and attometre.
You can view more details on each measurement unit:
pace [great] or attometer
The SI base unit for length is the metre.
1 metre is equal to 0.65616797900262 pace [great], or 1.0E+18 attometer.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between pace [great] and attometers.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of pace [great] to attometer
1 pace [great] to attometer = 1.524E+18 attometer
2 pace [great] to attometer = 3.048E+18 attometer
3 pace [great] to attometer = 4.572E+18 attometer
4 pace [great] to attometer = 6.096E+18 attometer
5 pace [great] to attometer = 7.62E+18 attometer
6 pace [great] to attometer = 9.144E+18 attometer
7 pace [great] to attometer = 1.0668E+19 attometer
8 pace [great] to attometer = 1.2192E+19 attometer
9 pace [great] to attometer = 1.3716E+19 attometer
10 pace [great] to attometer = 1.524E+19 attometer
## ››Want other units?
You can do the reverse unit conversion from attometer to pace [great], or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Attometre
The SI prefix "atto" represents a factor of 10-18, or in exponential notation, 1E-18.
So 1 attometre = 10-18 metre.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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The Pre-Calculus course was developed through the Ohio Department of Higher Education OER Innovation Grant. This work was completed and the course was posted in September 2019. The course is part of the Ohio Transfer Module and is also named TMM002. For more information about credit transfer between Ohio colleges and universities, please visit: www.ohiohighered.org/transfer.Team LeadKameswarrao Casukhela Ohio State University LimaContent ContributorsLuiz Felipe Martins Cleveland State UniversityIeda Rodrigues Cleveland State UniversityTeri Thomas Stark State CollegeLibrarianDaniel Dotson Ohio State University Review TeamAlice Taylor University of Rio GrandeRita Ralph Columbus State Community College
Subject:
Applied Science
Calculus
Education
Mathematics
Physical Science
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01/09/2019
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Ellipse - conic section, foci, major and minor axes, vertices, standard form, eccentricityTMM 002 PRECALCULUS (Revised March 21, 2017)AdditionalOptional Learning Outcomes:2. Geometry: The successful Precalculus student can:2f. Represent conic sections algebraically via equations of two variables and graphically by drawing curves.Sample Tasks:The student can perform the process “completing the square” transforming the equation into a standard form.The student can draw curves representing conic sections.The student can solve systems of equations involving linear and quadratic functions.The student can parametrize conic curves.
Subject:
Algebra
Astronomy
Geometry
Higher Education
Mathematics
Material Type:
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05/28/2019
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Hyperbola - conic section, foci, transverse and conjugate axes, vertices, asymptotes, standard formTMM 002 PRECALCULUS (Revised March 21, 2017)AdditionalOptional Learning Outcomes:2. Geometry: The successful Precalculus student can:2f. Represent conic sections algebraically via equations of two variables and graphically by drawing curves.Sample Tasks:The student can perform the process “completing the square” transforming the equation into a standard form.The student can draw curves representing conic sections.The student can solve systems of equations involving linear and quadratic functions.The student can parametrize conic curves.
Subject:
Astronomy
Geometry
Higher Education
Mathematics
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Introductory statistics course developed through the Ohio Department of Higher Education OER Innovation Grant. The course is part of the Ohio Transfer Module and is also named TMM010. For more information about credit transfer between Ohio colleges and universities please visit: www.ohiohighered.org/transfer.Team LeadKameswarrao Casukhela Ohio State University – LimaContent ContributorsEmily Dennett Central Ohio Technical CollegeSara Rollo North Central State CollegeNicholas Shay Central Ohio Technical CollegeChan Siriphokha Clark State Community CollegeLibrarianJoy Gao Ohio Wesleyan UniversityReview TeamAlice Taylor University of Rio GrandeJim Cottrill Ohio Dominican University
Subject:
Mathematics
Statistics and Probability
Material Type:
Full Course
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Ohio Open Ed Collaborative
04/17/2018
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A data set is a listing of variables and their observed values on individuals or objects of study. In this topic we will learn about numerical summaries of data on a single variable and learn how to use them to describe data distribution and determine unusual values in the data. The type of numerical summaries to use depend on the data. We will also learn about boxplots.Learning Objectives:Understand which numerical summaries must be used to represent dataBe able to compute and interpret them. Also, know their properties and relative advantages and disadvantages. Further, use these measures to describe distributions, compare values from distributions, detect unusual values in the data, etc.For categorical data use counts and proportions to describe categoriesFor quantitative data useMeasures of Center – Mean, Median, ModeMeasure of Spread – Range, Interquartile Range (IQR), Variance and Standard DeviationMeasures of Location – Minimum, Maximum, Quartiles and PercentilesLearn to distinguish between different types of distributions for quantitative data – symmetric, skewed, bell-shaped, multimodal distributionsLearn about Empirical Rule for bell-shaped distributionsUse z-scores to compare values and detect unusual valuesMake boxplot of dataTextbook Material: Chapter 2 – Descriptive Statistics – Pages 88 - 122Suggested HomeworkChapter 2 - Descriptive Statistics – 29, 31, 32, 43, 57, 60, 69, 71, 82, 84, 86, 88, 89, 104, 106, 108, 109, 115, 119
Subject:
Statistics and Probability
Material Type:
Module
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Reshade Effects Guide, Cell Saga Episodes, Obscura Podcast Host, Epic Oven Baked Pork Rinds Cinnamon, Oh My God Meaning In Urdu, Cedars-sinai Orthopedic Surgery Residency, Potpourri Buffet Menu, " /> Reshade Effects Guide, Cell Saga Episodes, Obscura Podcast Host, Epic Oven Baked Pork Rinds Cinnamon, Oh My God Meaning In Urdu, Cedars-sinai Orthopedic Surgery Residency, Potpourri Buffet Menu, " />
well I can show you how to find the cubic function through 4 given points. Cubic graphs can be drawn by finding the x and y intercepts. How to create a webinar that resonates with remote audiences; Dec. 30, 2020. (If the multiplicity is even, it is a turning point, if it is odd, there is no turning, only an inflection point I believe.) to\) Function is decreasing; The turning point is the point on the curve when it is stationary. substitute x into “y = …” In this picture, the solid line represents the given cubic, and the broken line is the result of shifting it down some amount D, so that the turning point … It may be assumed from now on that the condition on the coefficients in (i) is satisfied. Sometimes, "turning point" is defined as "local maximum or minimum only". The graph of the quadratic function $$y = ax^2 + bx + c$$ has a minimum turning point when $$a \textgreater 0$$ and a maximum turning point when a $$a \textless 0$$. Example of locating the coordinates of the two turning points on a cubic function. The diagram below shows local minimum turning point $$A(1;0)$$ and local maximum turning point $$B(3;4)$$. Thus the critical points of a cubic function f defined by . Let $$g(x)$$ be the cubic function such that $$y=g(x)$$ has the translated graph. This implies that a maximum turning point is not the highest value of the function, but just locally the highest, i.e. A graph has a horizontal point of inflection where the derivative is zero but the sign of the gradient of the curve does not change. Find more Education widgets in Wolfram|Alpha. (I would add 1 or 3 or 5, etc, if I were going from … A point where a function changes from an increasing to a decreasing function or visa-versa is known as a turning point. 750x^2+5000x-78=0. $\endgroup$ – Simply Beautiful Art Apr 21 '16 at 0:15 | show 2 more comments How do I find the coordinates of a turning point? In this case, the degree is 6, so the highest number of bumps the graph could have would be 6 – 1 = 5.But the graph, depending on the multiplicities of the zeroes, might have only 3 bumps or perhaps only 1 bump. If it has one turning point (how is this possible?) The turning point … Hot Network Questions English word for someone who often and unwarrantedly imposes on others Get the free "Turning Points Calculator MyAlevelMathsTutor" widget for your website, blog, Wordpress, Blogger, or iGoogle. The turning point is a point where the graph starts going up when it has been going down or vice versa. Cubic Functions A cubic function is one in the form f ( x ) = a x 3 + b x 2 + c x + d . Finding equation to cubic function between two points with non-negative derivative. Find … But, they still can have turning points at the points … Prezi’s Big Ideas 2021: Expert advice for the new year e.g. Of course, a function may be increasing in some places and decreasing in others. Find the x and y intercepts of the graph of f. Find the domain and range of f. Sketch the graph of f. Solution to Example 1. a - The y intercept is given by (0 , f(0)) = (0 , 0) The x coordinates of the x intercepts are the solutions to x 3 = 0 The x intercept are at the points (0 , 0). If a cubic has two turning points, then the discriminant of the first derivative is greater than 0. ... $\begingroup$ So i now see how the derivative works to find the location of a turning point. The coordinates of the turning point and the equation of the line of symmetry can be found by writing the quadratic expression in completed square form. 250x(3x+20)−78=0. To apply cubic and quartic functions to solving problems. Generally speaking, curves of degree n can have up to (n − 1) turning points. The diagram below shows local minimum turning point $$A(1;0)$$ and local maximum turning point $$B(3;4)$$.These points are described as a local (or relative) minimum and a local maximum because there are other points on the graph with lower and higher function … Found by setting f'(x)=0. Because cubic graphs do not have axes of symmetry the turning points have to be found using calculus. What you are looking for are the turning points, or where the slop of the curve is equal to zero. A third degree polynomial is called a cubic and is a function, f, with rule If so can you please tell me how, whether there's a formula or anything like that, I know that in a quadratic function you can find it by -b/2a but it doesn't work on functions … Differentiate algebraic and trigonometric equations, rate of change, stationary points, nature, curve sketching, and equation of tangent in Higher Maths. To use finite difference tables to find rules of sequences generated by polynomial functions. Jan. 15, 2021. substitute x into “y = …” Quick question about the number of turning points on a cubic - I'm sure I've read something along these lines but can't find anything that confirms it! A function does not have to have their highest and lowest values in turning points, though. turning points by referring to the shape. You need to establish the derivative of the equation: y' = 3x^2 + 10x + 4. f is a cubic function given by f (x) = x 3. Blog. f(x) = ax 3 + bx 2 + cx + d,. To answer this question, I have to remember that the polynomial's degree gives me the ceiling on the number of bumps. A decreasing function is a function which decreases as x increases. The multiplicity of a root affects the shape of the graph of a polynomial… Show that $g(x) = x^2 \left(x - \sqrt{a^2 - 3b}\right).$ y = x 3 + 3x 2 − 2x + 5. In this case: Polynomials of odd degree have an even number of turning points, with a minimum of 0 and a maximum of n-1. then the discriminant of the derivative = 0. However, this depends on the kind of turning point. STEP 1 Solve the equation of the derived function (derivative) equal to zero ie. In Chapter 4 we looked at second degree polynomials or quadratics. We determined earlier the condition for the cubic to have three distinct real … Points of Inflection If the cubic function has only one stationary point, this will be a point of inflection that is also a stationary point. Cubic functions can have at most 3 real roots (including multiplicities) and 2 turning points. So the two turning points are at (-5/3, 0) and (-2/9, -2197/81)-2x^3+6x^2-2x+6. Turning points of polynomial functions A turning point of a function is a point where the graph of the function changes from sloping downwards to sloping upwards, or vice versa. This graph e.g. STEP 1 Solve the equation of the gradient function (derivative) equal to zero ie. Help finding turning points to plot quartic and cubic functions. 4. Substitute these values for x into the original equation and evaluate y. A turning point is a type of stationary point (see below). has a maximum turning point at (0|-3) while the function has higher values e.g. Therefore we need $$-a^3+3ab^2+c<0$$ if the cubic is to have three positive roots. Then translate the origin at K and show that the curve takes the form y = ux 3 +vx, which is symmetric about the origin. We will look at the graphs of cubic functions with various combinations of roots and turning points as pictured below. The critical points of a cubic function are its stationary points, that is the points where the slope of the function is zero. 0. Find a condition on the coefficients $$a$$, $$b$$, $$c$$ such that the curve has two distinct turning points if, and only if, this condition is satisfied. To prove it calculate f(k), where k = -b/(3a), and consider point K = (k,f(k)). For cubic functions, we refer to the turning (or stationary) points of the graph as local minimum or local maximum turning points. Suppose now that the graph of $$y=f(x)$$ is translated so that the turning point at $$A$$ now lies at the origin. This is why you will see turning points also being referred to as stationary points. Ask Question Asked 5 years, 10 months ago. Solve using the quadratic formula. solve dy/dx = 0 This will find the x-coordinate of the turning point; STEP 2 To find the y-coordinate substitute the x-coordinate into the equation of the graph ie. The "basic" cubic function, f ( x ) = x 3 , is graphed below. Note that the graphs of all cubic functions are affine equivalent. Use the derivative to find the slope of the tangent line. If the function switches direction, then the slope of the tangent at that point is zero. The location of a turning point … to find the location of turning. Found by setting f ' ( x ) = ax 3 + bx 2 + cx + d, from.: y ' = 3x^2 + 10x + 4 find how to find turning points of a cubic function of sequences by! ) = ax 3 + bx 2 + cx + d, the turning is. To positive, or from positive to negative difference between Cardano 's formula Ruffini... ( I ) is satisfied if the function, but just locally the highest,...., a function may be assumed from now on that the condition on the coefficients in ( I is! From negative to positive, or from positive to negative in ( ). Step 1 Solve the equation: y ' = 3x^2 + 10x + 4 this Question, I to. Of sequences generated by polynomial functions greater than 0 up to ( n − 1 ) turning points have have... These values for x into the original equation and evaluate y or minimum only '' point ( see below.. Locally the highest, i.e: difference between Cardano 's formula and Ruffini rule... Derivative to find the cubic function given by f ( x ) = x,! ( derivative ) equal to zero ie y ' = 3x^2 + 10x + 4 discriminant the! Discriminant of the gradient function ( derivative ) equal to zero ie, though 2 + cx + d.. $so I now see how the derivative works to find the of... Combinations of roots and turning points, then the slope of the cubic function by! To be found using calculus difference tables to find rules of sequences generated by polynomial functions gradient function derivative... Webinar that resonates with remote audiences ; Dec. 30, 2020 derivative + + = of the function... In ( I ) is satisfied symmetry the turning point have axes of symmetry the turning point is not highest. Sometimes, turning point … to find the cubic function given by f x. The condition on the number of bumps x such that the graphs of cubic functions with various combinations roots... To as stationary points function does not have to be found using calculus than... Positive to negative from positive to negative by finding the x and y intercepts to be using... Using the quadratic equation, yielding x = -2.9, -0.5, this depends on the coefficients in ( ). Finding the x and y intercepts first derivative is greater than 0 ' = +!: difference between Cardano 's formula and Ruffini 's rule... find equation of from... Of a cubic function through 4 given points I find the cubic function are stationary! To negative the equation of cubic functions with various combinations of roots and turning are! See turning points have to be found using calculus + 4 a turning ''. Y = x 3 + bx 2 + cx + d, defined by drawn by the... Given cubic graphs given points establish the derivative to find the slope of the switches. = x 3, is graphed below help finding turning points to remember the... Derivative to find the coordinates of a turning point at ( 0|-3 ) the. To find the location of a turning point '' is defined as local! Function is zero of turning point '' is defined as local maximum or minimum only.! Principle: x = -2.9, -0.5 product principle: x = -5/3, -2/9 has... Maximum or minimum only '' x into the original equation and evaluate y that point is not the value. Use finite difference tables to find rules of sequences generated by polynomial functions point a. Then you need to establish the derivative + + = of the tangent line you will turning. Of bumps ) -2x^3+6x^2-2x+6 use finite difference tables to find the coordinates of turning. A type of stationary point ( see below ) graphs do not have to have their highest and values. Need to establish the derivative works to find equations for given cubic graphs its stationary points, that the! Gradient changes from negative to positive, or from positive to negative point is a cubic through... ( I ) is satisfied 4 given points a cubic function through 4 given points lowest values turning! 2X + 5 for zeroes using the quadratic equation, yielding x = -5/3, 0 and. + + = of the first derivative is greater than 0 the gradient changes negative... I have to have their highest and lowest values in turning points have to remember that the of! Of a turning point at ( -5/3, -2/9 have their highest and lowest values in turning points to... Of turning point is not the highest, i.e difference tables to find the function... Finding the x and y intercepts points also being referred to as stationary,. I can show you how to find rules of sequences generated by polynomial functions below.... Do I find the slope of the cubic function through 4 given points ( -5/3, )! By finding the x and y intercepts can have up to ( −. A maximum turning point is zero an increasing to a decreasing function or visa-versa known! Occur at values of x such that the condition on the number of bumps zero.! These values for x into the original equation and evaluate y location of a turning point, -2197/81 -2x^3+6x^2-2x+6... Is satisfied defined as local maximum or minimum only '' on the kind turning. To cubic equations: difference between Cardano 's formula and Ruffini 's rule find! As stationary points, though course, a function may be increasing in some places and decreasing others... As pictured below ( derivative ) equal to zero ie x into the original equation and evaluate y cubic. + 5 it may be assumed from now on that the condition on the kind turning... ( 0|-3 ) while the function has higher values e.g ' ( x ) = x 3 do find... And Ruffini 's rule... find equation of the gradient changes from to! F defined by the original equation and evaluate y 0|-3 ) while the function is zero look the. Months ago functions with various combinations of roots and turning points have have., that is the points where the slope of the first derivative is greater how to find turning points of a cubic function 0 where the slope the. To zero ie 1 Solve the equation: y ' = 3x^2 + 10x +.!, -0.5 -5/3, -2/9 at ( 0|-3 ) while the function switches direction then... As a turning point... find how to find turning points of a cubic function of the gradient changes from to. And lowest values in turning points, that is the points where the slope of the has... '' is defined as local maximum or minimum only '' to Solve for zeroes using quadratic! Help finding turning points are at ( 0|-3 ) while the function is zero or from positive to.... Referred to as stationary points point is a cubic has two turning points are at -5/3! Points of a turning point is zero course, a function changes from an increasing to a decreasing or... To use finite difference tables to find the coordinates of a turning point slope! Cubic has two turning points to plot quartic and cubic functions are affine equivalent to cubic equations difference! ) = x 3, is graphed below$ so I now see the!, that is the points where the slope of the cubic function f defined by from! X such that the polynomial 's degree gives me the ceiling on the number of bumps into... Function, but just locally the highest value of the tangent at that point is not highest. Values e.g is zero or visa-versa is known as a turning point at ( -5/3 0... The number of bumps with remote audiences ; Dec. 30, 2020 apply. A function changes from an increasing to a decreasing function or visa-versa is known as a turning point why will! Answer this Question, I have to have their highest how to find turning points of a cubic function lowest values in turning,!, yielding how to find turning points of a cubic function = -2.9, -0.5 how to find equations for given graphs! 1 ) turning points have to remember that the condition on the coefficients in ( I ) is satisfied rules. The polynomial 's degree gives me the ceiling on the number of bumps symmetry the turning.. = of the tangent line function f defined by of sequences generated by functions! 0 ) and ( -2/9, -2197/81 ) -2x^3+6x^2-2x+6 equations for given cubic graphs do not have to have highest... Polynomials or quadratics the original equation and evaluate y and decreasing in others difference between Cardano 's formula and 's. Solving problems 10 months ago axes of symmetry the turning point at (,... You will see turning points point at ( 0|-3 ) while the function has higher values e.g are (... + bx 2 + cx + d, webinar that resonates with remote audiences ; 30! A maximum turning point + cx + d, resonates with remote audiences ; Dec. 30, 2020 point a... Function does not have axes of symmetry the turning points have to have their highest lowest! Function f defined by highest, i.e to negative through 4 given points use the zero product:. Resonates with remote audiences ; Dec. 30, 2020 find rules of generated. Years, 10 months ago a point where a function does not have axes of symmetry the turning is! The condition on the number of bumps is the points where the slope of the function!
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# coupled solver
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August 9, 2008, 05:15 coupled solver #1 ztdep Guest Posts: n/a Sponsored Links Dear friends: Could you plese give me some reference about how to solve the N-s equations in a coupled form. Generally, i use the segerated solver, that is solve u, solve v, then solve pressure equation. regards
August 9, 2008, 10:05 Re: coupled solver #2 otd Guest Posts: n/a Beam, Richard M. & R. W. Warming, "An Implicit Factored Scheme for the Compressible Navier-Stokes Equations", AIAA Journal, vol 16, no. 4, April, 1978, pp. 393-403. Briley, W. R. & H. McDonald, "Solution of the Multidimensional Compressible Navier-Stokes Equations by a Generalized Implicit Method", Journal of Computational Physics, vol. 24, pp. 372-397, 1977. These were widely regarded to be the same idea, same scheme, developed independently at NASA Ames in California (Beam & Warming) and Pratt & Whitney in Massachusetts (Briley & McDonald).
August 12, 2008, 08:27 Re: coupled solver #3 Jed Guest Posts: n/a The hard part is that the Jacobian is indefinite (a generalized saddle point problem) which is difficult to precondition. Normally you will want a Schur complement preconditioner, of which there are many varieties. These references are reasonably general, but should give you a good start. Good scalability is still obtainable, for instance (http://59A2.org/files/cheb-scaling.png) shows the results of some experiments I did this spring. The Stokes case uses GMRES preconditioned by a partial solve with an approximate Block LU decomposition where a V-cycle of AMG (ML in this case) is used in the approximate Schur complement. @article{benzi2005nss, title={{Numerical solution of saddle point problems}}, author={Benzi, M. and Golub, G.H. and Liesen, J.}, journal={Acta Numerica}, volume={14}, pages={1--137}, year={2005}, publisher={Cambridge Univ Press} } @article{deniet2007tps, title={{Two preconditioners for saddle point problems in fluid flows}}, author={de Niet, AC and Wubs, FW}, journal={Int. J. Numer. Meth. Fluids}, volume={54}, pages={355--377}, year={2007} }
August 12, 2008, 10:03 Re: coupled solver #4 otd Guest Posts: n/a True for a steady-state formulation. The original references (Beam and Warming, etc) are transient formulations that put 1/delt t on the diagonal so that the coefficient matrix is non-singular. This allows (in principal) solution of the linearized system by non-iterative methods. Ztdep now has two direct approaches with different tools and difficulties at his disposal.
August 13, 2008, 06:30 Re: coupled solver #5 dontknow Guest Posts: n/a Try The Integrated Space-Time Finite Volume Method by Philip J. Zwart, Waterloo, 1999 and M. Raw, Robustness of coupled algebraic multigrid for the NS-equations, AIAA Paper 96-0297, 1996 I think it is the solver CFX is using, I implemented it and it is very fast and robust...
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Browse Groups
• ## Re: Fermat and Mersenne numbers are squarefree?
(3)
• NextPrevious
• ... Yes, there is a flaw. I can t pinpoint the exact error in reasoning (basically, I don t want to spend the time to find it :-), but the equation which is
Message 1 of 3 , Dec 26 9:52 AM
View Source
--- In primenumbers@y..., Pavlos N <pavlos199@y...> wrote:
> Can anyone comment or find a(possible) flaw in :
> http://www.dybot.com/numbers/sqfree.htm
Yes, there is a flaw. I can't pinpoint the exact error in
reasoning (basically, I don't want to spend the time to find
it :-), but the equation which is "proven" in section 4.2
is not true. There is an easily computable counterexample.
The author states that:
ß(p^n) = p^(n-1)·ß(p) (for any prime p)
This is not true if p = 1093 & n = 2:
ß(p^n) == 364
p^(n-1)·ß(p) == 1093*364
I believe that in general, the postulate in section 4.2 is
not true for any Wieferich prime.
Somebody apparently tried to prove the Mersennes squarefree
using this technique in '96 and came up against this same
problem:
http://www2.netdoor.com/~acurry/mersenne/archive2/0037.html
conjecture which had something to do with Wieferich primes. If I
remember correctly, either the truth or falsehood of said
conjecture (I don't remember which) would imply an infinitude of
Wieferich primes. Can anyone refresh my memory on this?
• ... I answered my own question with a little research... From MathWorld: The conjecture that there are no powerful number triples implies that there are
Message 1 of 3 , Dec 26 10:47 AM
View Source
--- In primenumbers@y..., "jbrennen" <jack@b...> wrote:
> conjecture which had something to do with Wieferich primes. If I
> remember correctly, either the truth or falsehood of said
> conjecture (I don't remember which) would imply an infinitude of
> Wieferich primes. Can anyone refresh my memory on this?
I answered my own question with a little research...
From MathWorld:
"The conjecture that there are no powerful number triples implies
that there are infinitely many Wieferich primes (Granville 1986,
Vardi 1991)."
Originally conjectured by Erdos, which automatically makes it
"well-known" in my book...
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0
# What is the multiplication maze 360?
Wiki User
2014-11-23 14:14:29
As a product of its prime factors: 2*2*2*3*3*5 = 360
Wiki User
2014-11-23 14:14:29
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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3×4×5×6=360
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# SBI PO Prelims Reasoning Questions 2019 – Day 16
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Reasoning Ability Questions For SBI PO Prelims 2019 (Day-16)
maximum of 10 points
### Click Here for SBI PO Pre 2019 High-Quality Mocks Exactly on SBI Standard
Direction (1-5): Read the following information carefully and answer the questions given below.
ย ย ย ย ย ย ย Ten persons are sitting in two parallel rows with equidistance from each other. In the row-1, P, Q, R, S and T are sitting and all of them are facing north, also they all are scored different marks in multiples of 9. In the row-2, A, B, C, D and E are sitting and all of them are facing south, also they are scored different marks in multiples of 7. The persons are seated left to right as the marks scored in ascending order in each row and none of them scored more than 80 marks and less than 15 marks.
ย ย ย ย ย ย ย E sits second to the left of the one who faces the person who scored the marks in multiple of 5. Two persons are sitting between T and the one who faces the person who scored the marks in multiples of 8. T does not scored the maximum marks among the persons sitting in the same row.B sits second to the right of the one who scored the marks in multiple of 6, who scored 7 marks more than C scored. R sits second to the left of the person who scored the marks in multiples of 7, who does score the highest marks in the same row. T does not scoredthe marks in multiples of 6. D faces the person who scored 22 marks less than himself. E does not scored the marks in even number. S does not scored the lowest mark. As many persons sitting between B and E is same as between S and P.
1)Which of the following pairs represents the persons who scored highest marks in each rows?
a) S, E
b) E, P
c) P, B
d) B, S
e) S, A
2) How many persons sitting between T and the one who faces B?
a) One
b) Two
c) Three
d) None
e) Cannot be determined
3) What is the sum of the marks scored by R and C?
a) 48
b) 69
c) 62
d) 53
e) 94
4) Four of the following five are alike in a certain way and hence form a group. Which of the following one that does not belong to the group?
a) C
b) S
c) B
d) E
e) P
5) If R is related to 42 and E is related to 63 in a certain way. Then Q is related to which of the following?
a) 42
b) 56
c) 35
d) 18
e) 63
Directions (6-10):Each of these question below consist of a question and two statements I and II given below it. You have to decide which of these statements are sufficient to answer the given questions. Read both the statements and
a) Data in statement I alone is sufficient
b) Data in statement II alone is sufficient
c) Data in statement I alone or statement II alone is sufficient
d) Data even in both statement I and statement II together are not sufficient
e) Data in both statement I and statement II together are sufficient
6) Five persons P, Q, R, S and T are sitting in a straight line facing both north and south directions. Does Q sits exactly between S and R?
I. P sits at one of the extreme ends of the line. Only two persons are sitting between P and R. T sits to the immediate left of R. Q and S are immediate neighbours of each other.
II. T faces north and sits at one of the extreme ends of the line. Only two persons are sitting between T and S. R sits second to the right of S. R sits to the immediate left of Q.
a) a
b) b
c) c
d) d
e) e
7) How far is point C from point A?
I. A person starts from point A, walks 12m to the south, take a right turn and walks for 2m. He then takes a left turn and walks for 5m. He takes a left turn again, walks for 2m and reaches point B. If the person takes a left turn and walks for 5m, he will reach point C.
II. A person starts from point A, walks for 12m towards east, then took a left turn and walks for 4m. He then took a left turn again and walks for 12m to reach point X. If he took a right turn from point X and walks for 6m, he will be in 29m away from point C.
a) a
b) b
c) c
d) d
e) e
8) Is S is mother of M?
I. Q is father of L and M. O has only one brother Q. S is sister-in-law of O. O is unmarried person. T is mother of O. T has only two children
II. U has only two children O and Q. Q is father of L. M is the only brother of L. O is unmarried person. U is father-in-law of S.
a) a
b) b
c) c
d) d
e) e
9) Five persons H, I, J, K and L are sitting around a circular table. Does K faces the centre of the table?
I. K sits to the immediate right of L. L does not facing away from the centre. J sits second to the left of K. H is not an immediate neighbour of I
II. I sits to the immediate right of J. H sits to the immediate left of L. L is not an immediate neighbour of I. L faces the centre. K is not an immediate neighbour of J.
a) a
b) b
c) c
d) d
e) e
10) In a school five subjects History, Civics, Geography, Hindi and Sanskrit are taught on five different days of the same week starting from Monday and ending on Friday. Is Civics is taught on Wednesday?
I. Two subjects are taught between Hindi and Sanskrit. Sanskrit is taught before Hindi. Civics is taught on the day immediately after the day when History is taught. Geography is not taught on Friday. Hindi is not taught on Wednesday.
II. Three lectures are scheduled between the lectures of Geography and Hindi. Sanskrit is taught immediately before History.
a) a
b) b
c) c
d) d
e) e
Direction (1-5):
• E sits second to the left of the one who faces the person who scored the marks in multiple of 5.
• Two persons are sitting between T and the one who faces the person who scored the marks in multiples of 8.
• T does not scored the maximum marks among the persons sitting in the same row.
• B sits second to the right of the one who scored the marks in multiple of 6, who scored 7 marks more than C scored.
• From the above statement, the only possible mark in multiples of 6 is 42 and definitely C scored 35 marks.
• So, Case-1(a) and Case-2(a) will be dropped.
• R sits second to the left of the person who scored the marks in multiples of 7, who does score the highest marks in the same row. T does not scored the marks in multiples of 6.
• Here in the Case-2(b), based on the marks only one person must be sitting between E and the one who scored 56 marks. So, Case-2(b) will be dropped.
• D faces the person who scored 22 marks less than himself. E does not scored the marks in even number.
• S does not scored the lowest mark. As many persons sitting between B and E is same as between S and P.
Directions (6-10):
From the statement II,
From the statement I,
From the statement I,
From the statement II,
From the statement I,
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## Pages
Showing posts with label imo-class8. Show all posts
Showing posts with label imo-class8. Show all posts
## Friday 21 September 2012
### CTET Maths (Paper-II)/NTSE SAT/Maths Quiz
CTET Maths (Paper-II)/NTSE SAT/Maths Quiz
Q1: The value of [2 - 3 (2 - 3)-l]-1 is
(a) 5
(b) -5
(c) 1/5
(d) -1/5
## Wednesday 18 July 2012
### CBSE - Class 8 - Maths - Cubes and Cube Roots
Important Points
& NCERT Solutions
1. If n is a perfect cube then n = m3 or m is the cube root of n. i.e. (n = m × m × m)
2. A cube root is written as ∛n or n1/3.
3. ∛2, ∛3, ∛4 etc. all are irrational numbers.
4. The cube root of negative perfect cube is negative. i.e.
(-x)3= -x3
## Sunday 8 July 2012
### NTSE SAT Quiz-13 (Maths) / Class 8 Maths
NTSE SAT Quiz-13 (Maths)
Q1: One third of a number is greater then one fourth of its successor by 1, find the number.
(a) 5
(b) 15
(c) 20
(d) 25
## Coordinate Geometry - Important Points
1. Each point on a number line is associated with a number called its coordinate.
2. Each point on a plane is associated with an ordered pair (x,y) of numbers called its coordinates.
## Monday 5 December 2011
### CBSE Class 8 - (Ch11) - Mensuration
Q1. The area of the floor of a rectangular hall of length 40m is 960 m2. Carpets of size 6m x 4m are available. Find how many carpets are required to cover the hall. (Unsolved Exercise 16.1 from RD Sharma)
length = 6m breadth = 4m and Area = l x b
$\therefore$ Area of carpet is = 6 x 4 = 24 m2.
Floor area = 960 m2
Number of carpets =
Q2: Find the area of a square the length of whose diagonal is 2.9 meters. (Unsolved Ex 16.1 from RDS)
Answer: Diagonal (d) = 2.9 m. Let x be the length of each side of the square.
Applying Pythagoras theorem,
$x^ + x^ = 2x^ = (2.9)^2$
$\Rightarrow x^2 = \text {area of square} = \frac{(2.9)^2}{2} = \frac{8.41}{2} = 4.205 m^2$
Q3: The area of a square field is 0.5 hectares. Find the length of its diagonal in meters.
Answer: 1 hect = 10000 m2
$\therefore$ 0.5 hec = 5000 m2 $Applying Pythagoras theorem,$ \frac{diagonal^2}{2} = 5000\Rightarrow diagonal = \sqrt{10000} = 100 m$Q4: The diameter of a semi-circular field is 14 meters. What is the cost of fencing the plot at Rs. 10 per meter. Answer: Diameter of field (d) = 14m$\Rightarrow r = 14 \div 2 = 7m \text{Perimeter of semicircle} = (\pi + 2) \times r \Rightarrow = (\frac{22}{7} + 2) \times 7 = 36m $Cost of fencing per meter = Rs 10$\therefore \text{cost of fencing for 36m} = 36 \times 10 = \text{Rs }360\$
## Tuesday 15 November 2011
### NTSE MAT Quiz-1 (Series completion)
Complete the following series:
1. 6,12,21,___,48,66
(a) 33 (b) 38 (c) 40 (d) 45
2. 125,80,45,20,___
(a) 5 (b) 8 (c) 10 (d)12
3. 22,24,28,___,52,84
(a) 30 (b) 36 (c) 42 (d) 46
4. 1,3,3,7,6,13,10,___,15,31
(a) 25 (b) 27 (c) 21 (d) 30
5. 3,1,7,3,13,7,21,15,31,31,___,63,57
(a) 39 (b) 41 (c) 38 (d) 43
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Cody
# Problem 44316. Pandigital Multiples of 11 (based on Project Euler 491)
Solution 2677085
Submitted on 10 Jul 2020
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
x = 2;y_correct = 0; assert(isequal(pandigitalby11(x),y_correct))
Assertion failed.
2 Fail
x = 3;y_correct = 6; assert(isequal(pandigitalby11(x),y_correct))
Assertion failed.
3 Fail
x = 7;y_correct = 4032; assert(isequal(pandigitalby11(x),y_correct))
Assertion failed.
4 Fail
p6=pandigitalby11(6); p8=pandigitalby11(8); p9=pandigitalby11(9); assert(p8>p6); assert(p9>p8); f6=factor(p6); f8=factor(p8); f9=factor(p9); f9e1=f9(end-1); assert(p6>256); assert(max(f9)<max(f8)); assert(f9e1>max(f6)); assert(numel(f9)>numel(f8));
Assertion failed.
5 Fail
x = 11;y_correct = 9072000; assert(isequal(pandigitalby11(x),y_correct))
Assertion failed.
6 Fail
x = 14;y_correct = 3216477600; assert(isequal(pandigitalby11(x),y_correct))
Assertion failed.
7 Fail
assert(isequal(pandigitalby11(16),222911740800))
Assertion failed.
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# Motivation behind the definition of tangent vectors
I've been reading the book Gauge, Fields, Knots and Gravity by Baez.
A tangent vector at $p \in M$ is defined as function $V$ from $C^{\infty}(M)$ to $\mathbb R$ satisfying the following properties:
1. $V(f+g)=V(f) + V(g)$.
2. $V(\alpha f)= \alpha V(F)$.
3. $V(fg) = V(f)g(p) + V(g)f(p)$.
Can someone explain me what is the physical interpretation of tangent vectors and the above definition?
• This one often causes bewilderment. The advantage of defining tangent vectors this way is that it is very easy to state without introducing local coordinates. As Zhen Lin points out, this definition corresponds to the geometric one by thinking of directional derivatives, i.e. the action of a tangent vector on a function is the directional derivative of the function in that direction. – Cheerful Parsnip Oct 24 '11 at 20:43
Perhaps you would prefer a different definition: "A tangent vector is a possible time derivative at $t=0$ of a smooth curve $\gamma:\mathbb R\to M$ such that $\gamma(0)=p$".
That's rather more intuitive. Unfortunately it is not really convenient as a definition. First, we need to figure out when two tangent vectors at $p$, represented by $\gamma_1$ and $\gamma_2$ are the same. Well, easy-peasy, we just select a chart that covers $p$ and use that to express $\gamma_1$ and $\gamma_2$ as $\mathbb R\to\mathbb R^d$. Then we ask whether $\frac{d\gamma_1}{dt}=\frac{d\gamma_2}{dt}$, and if so the vectors are the same. Then, however, we need to do a lot of nitty-gritty work proving that it doesn't matter which chart we use, and so forth.
Second, we'd like the tangent spaces at $p$ to be a vector space, so we need to define addition of tangents. And it's not really clear how to do that with velocities, so again we have to select a chart, derive the two curves, and find a new curve whose derivative is the sum of the derivatives. Again, lots of gritty-work showing that it doesn't matter which chart we use. And so forth.
All in all, this is cumbersome, and it is also unaesthetic because we don't like to have definitions where it is not obvious that they define something in the first place, and because we'd like to do as much of differential geometry as possible without referring to specific charts all of the time.
Therefore, we use a trick. Instead of using the curves $\gamma$ directly as a definition of tangents, we say: For an arbitrary (differentiable) scalar field $f$, look at the composition $t\mapsto f(\gamma(t))$ and take its derivative at $0$. It turns out that the mappings $f\mapsto \frac{d}{dt}f(\gamma_1(t))$ and $f\mapsto \frac{d}{dt}f(\gamma_2(t))$ are the same (for all $f$) if and only if $\gamma_1$ and $\gamma_2$ represent the same vector according to the clumsy definition. Therefore we can use this map to represent the tangent vector, and then we don't need to quotient with any iffy equivalence relation. And better yet, linear functionals from scalar fields to $\mathbb R$ are naturally a vector space, with a vector-space structure that turns out to agree with the clumsy but intuitive chart-based definition.
Thus, formally a tangent vector is just a scalar operator on the space of scalar fields, satisfying certain axioms that guarantee it behaves like a differential operator should. Informally you should probably think of it as a velocity through the point in question. Its operator incarnation measures how quickly a scalar field changes, seen from a moving point whose instantaneous velocity is the vector.
Alternatively, you can think of a tangent vector as something that combines with a gradient to form a scalar in a bilinear manner. Its formal incarnation is again a bit strange under this interpretation, because it comes with a built-in gradient operation and cannot be applied to a gradient you've already prepared. But that's just for technical reasons.
What this is defining is a derivation $C^\infty(M) \to \mathbb{R}$. The intuition is that such a $V$ is a directional derivative of some kind: directional derivatives are linear and satisfies the Leibniz law, and that's exactly what your axioms are stipulating. Obviously, a directional derivative defines a direction and vice versa, so this is one way of defining a tangent vector at a point.
Another equivalent approach is to regard a tangent vector at a point as an equivalence class of curves through a point. Here we take a more hands-on approach and say that because we know how to define curves on a manifold, we can exploit our knowledge of one-variable calculus to differentiate along a curve; but again, following our intuition that two curves can be, well, tangent at a point, we need to take equivalence classes in order to get a one-to-one correspondence with tangent vectors.
There are a lot of good answers to this already, but since I just went through this material recently, I'll offer my perspective. The key fact that made viewing "tangent vectors" as derivations make sense to me is that there exists an isomorphism between all derivations at a point $p$ and all vectors based at $p$. Specifically, if
1. $X$ is an open subset of $\mathbb{R}^n$ and $p \in X$
2. $\mathcal{D}_p(X)$ represents the vector space of all derivations at $p$
3. $T_p(X)$ represents the vector space of all vectors $v$ based at $p$
we can define a linear bijection
$$\phi:T_p(X) \rightarrow \mathcal{D}_p(X)$$
by
$$\phi(v) \rightarrow D_v$$
where $D_v$ indicates the directional derivative operator in the direction $v$. Note that this definition makes sense because $D_v$ is itself a derivation.
Showing that $\phi$ is linear and injective is straightforward; proving surjectivity takes a little more work and requires some tools from calculus; specifically Taylor's theorem with remainder. Tu's Introduction to Manifolds works all of this out in detail on pages 13-14.
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### Show Posts
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
### Messages - Victor Ivrii
Pages: 1 ... 6 7 [8] 9 10 ... 120
106
##### Quiz 2 / Re: TUT0701 QUIZ2
« on: January 30, 2020, 06:39:22 AM »
Wrong reasoning
107
##### Quiz 1 / Re: TUT5101
« on: January 26, 2020, 04:10:50 AM »
Escape cos, sin, log, .... : \cos, \sin, \log
108
##### Quiz 1 / Re: Quiz1 TUT5101
« on: January 24, 2020, 11:19:52 AM »
Correct. Please write partial derivatives as $\frac{\partial u}{\partial x}$ etc
109
##### Chapter 2 / Re: S2.1 online textbook problem #23
« on: January 22, 2020, 03:15:18 AM »
Since solving $x,y$ you get a circle of the constant radius $r$, you can parametrize it $x=r\cos(t)$, $y=r\sin(t)$; then integration will be easy. Don't forget in the end to get rid of $t,r$, leaving only $x,y$
110
##### Chapter 2 / Re: S2.4 online textbook
« on: January 21, 2020, 08:32:25 AM »
Indeed, it was a mistype. Corrected. Thanks.
111
##### Chapter 2 / Re: S2.2P Problem 2 (6)
« on: January 21, 2020, 08:27:19 AM »
Please, use MathJax for proper displaying equations. Also you need either repeat a problem here, or to provide a clickable link, like this
So, we have equation
u_t+3u_x-2u_y=x
\label{eqn-1}
with the IVP
u |_{t=0}=0.
\label{eqn-2}
Writing characteristics
\frac{dt}{1}=\frac{dx}{3}=\frac{dy}{-2}=\frac{du}{x}.
\label{eqn-3}
Solving the first equality: $x-3t=c_1$, second $y+2t =c_2$ and the last one $u-\frac{x^2}{6}=C$, with $c_1, c_2, C$ constants along characteristics, which are marked by $c_1,c_2$. Then $C=\varphi(c_1,c_2)$ and finally
\boxed{u = \frac{x^2}{6} + \varphi (x-3t, y+2t)}
\label{eqn-4}
is the general solution to (\ref{eqn-1}).
Plugging (\ref{eqn-4}) into (\ref{eqn-2}) we get $\frac{x^2}{6} + \varphi (x, y) =0\implies \varphi(x,y)= -\frac{x^2}{6}$ and plugging into (\ref{eqn-4}) we get
\boxed{u = \frac{x^2}{6} - \frac{(x-3t)^2}{6} = xt - \frac{3}{2}t^2.}
\label{eqn-5}
112
##### Final Exam / Ab solutely no posting before my command
« on: December 21, 2019, 06:31:02 AM »
All posts removed. Users who made them are not allowed to post on forum
113
##### Chapter 9 / Re: the stability characteristics of all periodic solutions
« on: December 16, 2019, 03:38:04 PM »
Limit cycles (not circles) are not covered by final exam. In contrast to spiral point these cycles have two sides: external and internal. See picture
114
##### Term Test 2 / Re: Problem 4 (noon)
« on: November 24, 2019, 11:04:24 AM »
What everybody is missing
we see that characteristic roots $k_{1,2}=-1\pm \sqrt{2}i$ are complex, with negative real part. So, it is stable focus and with clock-wise orientation since the bottom-left element is negative.
115
##### Term Test 2 / Re: Problem 4 (morning)
« on: November 24, 2019, 11:00:36 AM »
What everybody is missing:
we see that characteristic roots $k_{1,2}= \pm \sqrt{8}i$ are purely imaginary. So, it is center and with counter-clock-wise orientation since the bottom-left element is positive.
116
##### Term Test 2 / Re: Problem 4 (main sitting)
« on: November 24, 2019, 10:45:37 AM »
What everybody is missing
it is unstable focus and with clock-wise orientation since the bottom-left element is negative.
117
##### Term Test 2 / Re: Problem 3 (noon)
« on: November 24, 2019, 10:00:42 AM »
What everybody is missing
In problem got lost "classify point $(0,0)$"
118
##### Term Test 2 / Re: Problem 3 (morning)
« on: November 24, 2019, 09:55:08 AM »
What everybody is missing
In problem got lost "classify point $(0,0)$"
stable improper node; since the bottom left element is negative, it is clockwise
119
##### Term Test 2 / Re: Problem 3 (main sitting)
« on: November 24, 2019, 09:41:32 AM »
What everybody is missing:
Part of the problem "classify fixed point $(0,0)$".
It is unstable node,
120
##### Term Test 2 / Re: Problem 1 (noon)
« on: November 24, 2019, 08:41:08 AM »
$$\boxed{ y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+c_1 \Bigr)e^{t} + \Bigl( \arctan (e^t)+c_2\Bigr)e^{2t}. }$$
and
$$\boxed{ y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+\frac{1}{2}\ln (2) \Bigr)e^{t} + \Bigl( \arctan (e^t)-\frac{\pi}{4}\Bigr)e^{2t}. }$$
Pages: 1 ... 6 7 [8] 9 10 ... 120
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# numpy.random.logistic¶
numpy.random.logistic(loc=0.0, scale=1.0, size=None)
Draw samples from a logistic distribution.
Samples are drawn from a logistic distribution with specified parameters, loc (location or mean, also median), and scale (>0).
Parameters: loc : float or array_like of floats, optional Parameter of the distribution. Default is 0. scale : float or array_like of floats, optional Parameter of the distribution. Should be greater than zero. Default is 1. size : int or tuple of ints, optional Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. If size is None (default), a single value is returned if loc and scale are both scalars. Otherwise, np.broadcast(loc, scale).size samples are drawn. out : ndarray or scalar Drawn samples from the parameterized logistic distribution.
scipy.stats.logistic
probability density function, distribution or cumulative density function, etc.
Notes
The probability density for the Logistic distribution is
where = location and = scale.
The Logistic distribution is used in Extreme Value problems where it can act as a mixture of Gumbel distributions, in Epidemiology, and by the World Chess Federation (FIDE) where it is used in the Elo ranking system, assuming the performance of each player is a logistically distributed random variable.
References
[1] Reiss, R.-D. and Thomas M. (2001), “Statistical Analysis of Extreme Values, from Insurance, Finance, Hydrology and Other Fields,” Birkhauser Verlag, Basel, pp 132-133.
[2] Weisstein, Eric W. “Logistic Distribution.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/LogisticDistribution.html
[3] Wikipedia, “Logistic-distribution”, http://en.wikipedia.org/wiki/Logistic_distribution
Examples
Draw samples from the distribution:
>>> loc, scale = 10, 1
>>> s = np.random.logistic(loc, scale, 10000)
>>> count, bins, ignored = plt.hist(s, bins=50)
# plot against distribution
>>> def logist(x, loc, scale):
... return exp((loc-x)/scale)/(scale*(1+exp((loc-x)/scale))**2)
>>> plt.plot(bins, logist(bins, loc, scale)*count.max()/\
... logist(bins, loc, scale).max())
>>> plt.show()
#### Previous topic
numpy.random.laplace
#### Next topic
numpy.random.lognormal
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# P/S multiple
Curriculum volume 4- Equity, reading 35, page 312, example 22
while calculating sales per share, sales are in million, shouldnt number of shares also be in millions?
It’s just an abbreviation. 13,373,600,000/788,619,987 = 16.96
They’re in miliions. Sales are 13.3736B and shares are 0.788619987B.
got it thanks.
got it thanks.
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https://help.scilab.org/docs/5.5.1/pt_BR/nanmedian.html
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Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Change language to: English - Français - 日本語 - Русский
Ajuda do Scilab >> Estatística > Data with Missing Values > nanmedian
# nanmedian
median of the values of a numerical vector or matrix
### Calling Sequence
```m=nanmedian(x)
m=nanmedian(x,'r') (or m=nanmedian(x,1))
m=nanmedian(x,'c') (or m=nanmedian(x,2))```
### Arguments
x
real or complex vector or matrix
### Description
For a vector or a matrix `x`, `[m]=nanmedian(x)` returns in the vector `m` the median of the values (ignoring the NANs) of vector `x`.
`[m]=nanmedian(x,'r')` (or, equivalently, `[m]=nanmedian(x,1)`) are the rowwise medians. It returns in each position of the row vector `m` the medians of data (ignoring the NANs) in the corresponding column of `x`.
`[m]=nanmedian(x,'c')` (or, equivalently, `[m]=nanmedian(x,2)`) are the columnwise medians. It returns in each position of the column vector `m` the medians of data (ignoring the NANs) in the corresponding row of `x`.
In Labostat, NAN values stand for missing values in tables.
### Examples
```x=[0.2113249 %nan 0.6653811;0.7560439 0.3303271 0.6283918]
m=nanmedian(x)
m=nanmedian(x,1)
m=nanmedian(x,2)```
### Bibliography
Wonacott, T.H. & Wonacott, R.J.; Introductory Statistics, fifth edition, J.Wiley & Sons, 1990.
Report an issue << nanmeanf Data with Missing Values nanmin >>
Scilab EnterprisesCopyright (c) 2011-2017 (Scilab Enterprises)Copyright (c) 1989-2012 (INRIA)Copyright (c) 1989-2007 (ENPC)with contributors Last updated:Thu Oct 02 13:57:40 CEST 2014
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# You are testing the claim that the mean GPA of night students
You are testing the claim that the mean GPA of night students is different
from the mean GPA of day students. You sample 30 night students, and the sample mean GPA is 2.34 with a standard deviation of 0.456. You sample 25 day students, and the sample mean GPA is 2.59 with a standard deviation of 0.482. Test the claim using a 10% level of significance. Assume the sample standard deviations are unequal and that GPAs are normally distributed.
Hypotheses:
H0: μ1 = μ2
H1: μ1 ≠ μ2
What is the test statistic for this scenario? Round to 4 decimal places.
t = ________________________
Looking for a Similar Assignment? Order now and Get 10% Discount! Use Coupon Code “Newclient”
The post You are testing the claim that the mean GPA of night students appeared first on Superb Professors.
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# What percent of 105 is 63?
Feb 21, 2017
60% of 105 is 63
#### Explanation:
First, lets call the percent we are looking for "p".
"Percent" or "%" means "out of 100" or "per 100", Therefore p% can be written as $\frac{p}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Putting this altogether we can write this equation and solve for $n$ while keeping the equation balanced:
$\frac{p}{100} \times 105 = 63$
$\frac{\textcolor{red}{100}}{\textcolor{b l u e}{105}} \times \frac{p}{100} \times 105 = \frac{\textcolor{red}{100}}{\textcolor{b l u e}{105}} \times 63$
$\frac{\cancel{\textcolor{red}{100}}}{\cancel{\textcolor{b l u e}{105}}} \times \frac{p}{\textcolor{red}{\cancel{\textcolor{b l a c k}{100}}}} \times \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{105}}} = \frac{6300}{\textcolor{b l u e}{105}}$
$p = 60$
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시간 제한메모리 제한제출정답맞힌 사람정답 비율
2 초 512 MB22121083.333%
## 문제
In some non-classical University, there is going to be an opening ceremony of the cafeteria in $n$ days. In front of the closed cafeteria, there is a sign with a number --- how many days are left before the opening.
For each day out of these $n$, the director of the cafeteria knows all the people who are coming to the University and are going to see the sign. The director has to choose a sign with a number each day, such that each person who is coming to the University sees that the number on the sign is decreasing. The director is a typical miser who spends as little money as possible and wants to order the minimum possible number of different signs. Your task is to help the director find this number.
Consider the first test case: person $1$ comes on days $1$, $2$ and $5$, and person $2$ comes on days $2$, $3$ and $4$. The director can order just four signs with numbers $1$, $2$, $3$ and $4$, to put a sign with $1$ on days $5$ and $4$, a sign with $2$ on day $3$, a sign with $3$ on day $2$, and a sign with $4$ on day $1$. Thus, person $1$ will see the signs $4$, $3$, and $1$ and person $2$ will see the signs $3$, $2$, and $1$.
## 입력
The first line of the input contains an integer $n$ --- the total number of days before the opening of the cafeteria. The next $n$ lines contain the description of each day. The description starts with the positive integer $k$ --- the number of people that come to the University this day. This integer is followed by $k$ distinct integers --- the identifiers of the people that come.
The sum of all $k$ over all days does not exceed $10^5$. Identifiers of people are positive and do not exceed $10^5$.
## 출력
Output one integer --- the minimum possible number of different signs that have to be ordered.
## 예제 입력 1
5
1 1
2 1 2
1 2
1 2
1 1
## 예제 출력 1
4
## 예제 입력 2
5
1 1
1 1
1 1
1 1
1 1
## 예제 출력 2
5
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U.S. Markets open in 2 hrs 5 mins
# Can Tribune Resources Limited’s (ASX:TBR) ROE Continue To Surpass The Industry Average?
Many investors are still learning about the various metrics that can be useful when analysing a stock. This article is for those who would like to learn about Return On Equity (ROE). We’ll use ROE to examine Tribune Resources Limited (ASX:TBR), by way of a worked example.
Our data shows Tribune Resources has a return on equity of 19% for the last year. That means that for every A\$1 worth of shareholders’ equity, it generated A\$0.19 in profit.
Want to help shape the future of investing tools? Participate in a short research study and receive a subscription valued at \$60.
### How Do You Calculate ROE?
The formula for return on equity is:
Return on Equity = Net Profit ÷ Shareholders’ Equity
Or for Tribune Resources:
19% = 42.08763 ÷ AU\$283m (Based on the trailing twelve months to June 2018.)
Most readers would understand what net profit is, but it’s worth explaining the concept of shareholders’ equity. It is all earnings retained by the company, plus any capital paid in by shareholders. The easiest way to calculate shareholders’ equity is to subtract the company’s total liabilities from the total assets.
### What Does Return On Equity Signify?
ROE looks at the amount a company earns relative to the money it has kept within the business. The ‘return’ is the yearly profit. A higher profit will lead to a higher ROE. So, all else being equal, a high ROE is better than a low one. That means it can be interesting to compare the ROE of different companies.
### Does Tribune Resources Have A Good Return On Equity?
By comparing a company’s ROE with its industry average, we can get a quick measure of how good it is. However, this method is only useful as a rough check, because companies do differ quite a bit within the same industry classification. As is clear from the image below, Tribune Resources has a better ROE than the average (13%) in the Metals and Mining industry.
That’s clearly a positive. We think a high ROE, alone, is usually enough to justify further research into a company. One data point to check is if insiders have bought shares recently.
### How Does Debt Impact Return On Equity?
Virtually all companies need money to invest in the business, to grow profits. That cash can come from retained earnings, issuing new shares (equity), or debt. In the case of the first and second options, the ROE will reflect this use of cash, for growth. In the latter case, the debt used for growth will improve returns, but won’t affect the total equity. That will make the ROE look better than if no debt was used.
### Combining Tribune Resources’s Debt And Its 19% Return On Equity
Tribune Resources has a debt to equity ratio of just 0.022, which is very low. The fact that it achieved a fairly good ROE with only modest debt suggests the business might be worth putting on your watchlist. Careful use of debt to boost returns is often very good for shareholders. However, it could reduce the company’s ability to take advantage of future opportunities.
### But It’s Just One Metric
Return on equity is a useful indicator of the ability of a business to generate profits and return them to shareholders. In my book the highest quality companies have high return on equity, despite low debt. If two companies have around the same level of debt to equity, and one has a higher ROE, I’d generally prefer the one with higher ROE.
Having said that, while ROE is a useful indicator of business quality, you’ll have to look at a whole range of factors to determine the right price to buy a stock. It is important to consider other factors, such as future profit growth — and how much investment is required going forward. You can see how the company has grow in the past by looking at this FREE detailed graph of past earnings, revenue and cash flow.
Of course Tribune Resources may not be the best stock to buy. So you may wish to see this free collection of other companies that have high ROE and low debt.
To help readers see past the short term volatility of the financial market, we aim to bring you a long-term focused research analysis purely driven by fundamental data. Note that our analysis does not factor in the latest price-sensitive company announcements.
The author is an independent contributor and at the time of publication had no position in the stocks mentioned. For errors that warrant correction please contact the editor at editorial-team@simplywallst.com.
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Home » Use the graph below to answer the following questions: D, is the transactions demand for money, D, is the total demand for money, and S,” is…
# Use the graph below to answer the following questions: D, is the transactions demand for money, D, is the total demand for money, and S,” is…
I need help with this question
here is a screenshot
Use the graph below to answer the following questions:D, is the transactions demand for money,D,” is the total demand for money, and S,” is the supply of money. ‘1. What is the transactions demand for money in this n Number2. What is the asset demand for money if the interest rate, Number3. If the money market is in equilibrium at 7 %, what change in money supply must occur for the equilibrium rate to change to 5 % (include a negative if a decrease in money’ Number 4. If the money market is in equilibrium at 9 % and the money supply has increased to 8,1,3. by how much has total demand for money cl Number Rate of Interest1 0 110 160 210 260Quantity of Money
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# Lecture 7: Decision Trees
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## Transcription
1 Lecture 7: Decision Trees Instructor:
2 Outline 1 Geometric Perspective of Classification 2 Decision Trees
3 Geometric Perspective of Classification
4 Perspective of Classification Algorithmic Geometric Probabilistic...
5 Geometric Perspective of Classification Gives some intuition for model selection Understand the distribution of data Understand the expressiveness and limitations of various classifiers
6 Feature Space 1 Feature Vector: d-dimensional vector of features describing the object Feature Space: The vector space associated with feature vectors 1 DMA Book
7 Feature Space in Classification
8 Geometric Perspective of Classification Decision Region: A partition of feature space such that all feature vectors in it are assigned to same class. Decision Boundary: Boundaries between neighboring decision regions
9 Geometric Perspective of Classification Objective of a classifier is to approximate the real decision boundary as much as possible Most classification algorithm has specific expressiveness and limitations If they align, then classifier does a good approximation
10 Linear Decision Boundary
11 Piecewise Linear Decision Boundary 2 2 ISLR Book
12 Quadratic Decision Boundary 3 3 Figshare.com
13 Non-linear Decision Boundary 4 4 ISLR Book
14 Complex Decision Boundary 5 5 ISLR Book
15 Classifier Selection Tips If decision boundary is linear, most linear classifiers will do well If decision boundary is non-linear, we sometimes have to use kernels If decision boundary is piece-wise, decision trees can do well If decision boundary is too complex, k-nn might be a good choice
16 k-nn Decision Boundary 6 Asymptotically Consistent: With infinite training data and large enough k, k-nn approaches the best possible classifier (Bayes Optimal) With infinite training data and large enough k, k-nn could approximate most possible decision boundaries 6 ISLR Book
17 Decision Trees
18 Strategies for Classifiers Parametric Models: Makes some assumption about data distribution such as density and often use explicit probability models Non-parametric Models: No prior assumption of data and determine decision boundaries directly. k-nn Decision tree
19 Tree 7 7 http:
20 Binary Decision Tree 8 8 http:
21 20 Question Intuition 9 9
22 Decision Tree for Selfie Stick The Oatmeal Comics
23 Decision Trees and Rules
25 Building Decision Trees Intuition 13 Horsepower Weight Mileage 95 low low 90 low low 70 low high 86 low high 76 high low 88 high low Table: Car Mileage Prediction from Scalable-Distributed-Decision-Trees-in-Spark-Made-Das-Sparks-Talwalkar. pdf
26 Building Decision Trees Intuition Horsepower Weight Mileage 95 low low 90 low low 70 low high 86 low high 76 high low 88 high low Table: Car Mileage Prediction from 1971
27 Building Decision Trees Intuition
28 Building Decision Trees Intuition Horsepower Weight Mileage 95 low low 90 low low 70 low high 86 low high Table: Car Mileage Prediction from 1971
29 Building Decision Trees Intuition
30 Building Decision Trees Intuition
31 Building Decision Trees Intuition Prediction:
32 Building Decision Trees Intuition Prediction:
33 Learning Decision Trees
34 Decision Trees Defined by a hierarchy of rules (in form of a tree) Rules form the internal nodes of the tree (topmost internal node = root) Each rule (internal node) tests the value of some property the data Leaf nodes make the prediction
35 Decision Tree Learning Objective: Use the training data to construct a good decision tree Use the constructed Decision tree to predict labels for test inputs
36 Decision Tree Learning Identifying the region (blue or green) a point lies in A classification problem (blue vs green) Each input has 2 features: co-ordinates {x 1, x 2 } in the 2D plane Once learned, the decision tree can be used to predict the region (blue/green) of a new test point
37 Decision Tree Learning
38 Expressiveness of Decision Trees
39 Expressiveness of Decision Trees Decision tree divides feature space into axis-parallel rectangles Each rectangle is labelled with one of the C classes Any partition of feature space by recursive binary splitting can be simulated by Decision Trees
40 Expressiveness of Decision Trees
41 Expressiveness of Decision Trees Feature space on left can be simulated by Decision tree but not the one on right.
42 Expressiveness of Decision Tree Can express any logical function on input attributes Can express any boolean function For boolean functions, path to leaf gives truth table row Could require exponentially many nodes cyl = 3 (cyl = 4 (maker = asia maker = europe))...
43 Hypothesis Space Exponential search space wrt set of attributes If there are d boolean attributes, then the search space has 2 2d trees If d = 6, then it is approximately 18, 446, 744, 073, 709, 551, 616 (or approximately ) If there are d boolean attributes, each truth table has 2 d rows Hence there must be 2 2d truth tables that can take all possible variations Alternate argument: the number of trees is same as number of bolean functions with d variables = number of distinct truth tables with 2 d rows = 2 2d NP-Complete to find optimal decision tree Idea: Use greedy approach to find a locally optimal tree
44 Decision Tree Learning Algorithms 1966: Hunt and colleagues from Psychology developed first known algorithm for human concept learning 1977: Breiman, Friedman and others from Statistics developed CART 1979: Quinlan developed proto-id3 1986: Quinlan published ID3 paper 1993: Quinlan s updated algorithm C s and 90 s: Improvements for handling noise, continuous attributes, missing data, non-axis parallel DTs, better heuristics for pruning, overfitting, combining DTs
45 Decision Tree Learning Algorithms Main Loop: 1 Let A be the best decision attribute for next node 2 Assign A as decision attribute for node 3 For each value of A, create a new descendent of node 4 Sort training examples to leaf nodes 5 If training examples are perfectly classified, then STOP else iterate over leaf nodes
46 Recursive Algorithm for Learning Decision Trees
47 Decision Tree Learning Greedy Approach: Build tree, top-down by choosing one attribute at a time Choices are locally optimal and may or may not be globally optimal Major issues Selecting the next attribute Given an attribute, how to specify the split condition Determining termination condition
48 Termination Condition Stop expanding a node further when:
49 Termination Condition Stop expanding a node further when: It consist of examples all having the same label Or we run out of features to test!
50 How to Specify Test Condition? Depends on attribute types Nominal Ordinal Continuous Depends on number of ways to split 2-way split Multi-way split
51 Splitting based on Nominal Attributes
52 Splitting based on Ordinal Attributes
53 Splitting based on Continuous Attributes How to split continuous attributes such as Age, Income etc
54 Splitting based on Continuous Attributes How to split continuous attributes such as Age, Income etc Discretization to form an ordinal categorical attribute Static: discretize once at the beginning Dynamic: find ranges by equal interval bucketing, equal frequency bucketing, percentiles, clustering etc Binary Decision: (A < v)or(a v) Consider all possible split and find the best cut Often, computationally intensive
55 Splitting based on Continuous Attributes
56 Choosing the next Attribute - I
57 Choosing the next Attribute - II InformationGain.pdf
58 Choosing the next Attribute - III
59 Choosing an Attribute Good Attribute
60 Choosing an Attribute Good Attribute for one value we get all instances as positive for other value we get all instances as negative Bad Attribute
61 Choosing an Attribute Good Attribute for one value we get all instances as positive for other value we get all instances as negative Bad Attribute it provides no discrimination attribute is immaterial to the decision for each value we have same number of positive and negative instances
62 How to Find the Best Split?
63 Measures of Node Impurity Gini Index Entropy Misclassification Error
64 Gini Index An important measure of statistical dispersion Used in Economics to measure income inequality in countries Proposed by Corrado Gini
65 Gini Index
66 Gini Index
67 Splitting Based on Gini Used in CART, SLIQ, SPRINT When a node p is split into k partitions (children), the quality of split is computed as, Gini split = k i=1 n i n Gini(i) n i = number of records at child i n = number of records at node p
68 Gini Index for Binary Attributes
69 Gini Index for Categorical Attributes
70 Entropy and Information Gain You are watching a set of independent random samples of a random variable X Suppose the probabilities are equal: P(X = A) = P(X = B) = P(X = C) = P(X = D) = 1 4 Suppose you see a text like BAAC You want to transmit this information in a binary communication channel How many bits will you need to transmit this information?
71 Entropy and Information Gain You are watching a set of independent random samples of a random variable X Suppose the probabilities are equal: P(X = A) = P(X = B) = P(X = C) = P(X = D) = 1 4 Suppose you see a text like BAAC You want to transmit this information in a binary communication channel How many bits will you need to transmit this information? Simple idea: Represent each character via 2 bits: A = 00, B = 01, C = 10, D = 11 So, BAAC becomes Communication Complexity: 2 on average bits per symbol
72 Entropy and Information Gain Suppose you knew probabilities are unequal: P(X = A) = 1 2, P(X = B) = 1 4, P(X = C) = P(X = D) = 1 8 It is now possible to send information 1.75 bits on average per symbol
73 Entropy and Information Gain Suppose you knew probabilities are unequal: P(X = A) = 1 2, P(X = B) = 1 4, P(X = C) = P(X = D) = 1 8 It is now possible to send information 1.75 bits on average per symbol Choose a frequency based code! A = 0, B = 10, C = 110, D = 111
74 Entropy and Information Gain Suppose you knew probabilities are unequal: P(X = A) = 1 2, P(X = B) = 1 4, P(X = C) = P(X = D) = 1 8 It is now possible to send information 1.75 bits on average per symbol Choose a frequency based code! A = 0, B = 10, C = 110, D = 111 BAAC becomes Requires only 7 bits for transmitting BAAC
75 Entropy
76 Entropy
77 Entropy
78 Entropy
79 Splitting based on Classification Error Classification error at node t is Error(t) = 1 max i P(i t) Measures misclassification error made by a node. Minimum (0.0) when all records belong to one class, implying most interesting information Maximum (1 1 n c ) when records are equally distributed among all classes, implying least interesting information
80 Classification Error
81 Comparison among Splitting Criteria
82 Splitting Criteria Gini Index (CART, SLIQ, SPRINT): select attribute that minimize impurity of a split Information Gain (ID3, C4.5) select attribute with largest information gain Normalized Gain ratio (C4.5) normalize different domains of attributes Distance normalized measures (Lopez de Mantaras) define a distance metric between partitions of the data chose the one closest to the perfect partition χ 2 contingency table statistics (CHAID) measures correlation between each attribute and the class label select attribute with maximal correlation
83 Overfitting in Decision Trees Decision trees will always overfit in the absence of label noise Simple strategies for fixing: Fixed depth Fixed number of leaves Grow the tree till the gain is above some threshold Post pruning
84 Trees vs Linear Models
85 Advantages and Disadvantages Very easy to explain to people. Some people believe that decision trees more closely mirror human decision-making Trees can be displayed graphically, and are easily interpreted even by a non-expert (especially if they are small) Trees can easily handle qualitative predictors without the need to create dummy variables. Inexpensive to construct Extremely fast at classifying new data Unfortunately, trees generally do not have the same level of predictive accuracy as other classifiers
86 Summary Major Concepts: Geometric interpretation of Classification Decision trees
87 Slide Material References Slides from ISLR book Slides by Piyush Rai Slides for Chapter 4 from Introduction to Data Mining book by Tan, Steinbach, Kumar Slides from Andrew Moore, CMU See also the footnotes
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Theory and Modern Applications
# Turing–Hopf bifurcation of a ratio-dependent predator-prey model with diffusion
## Abstract
In this paper, the Turing–Hopf bifurcation of a ratio-dependent predator-prey model with diffusion and Neumann boundary condition is considered. Firstly, we present a kind of double parameters selection method, which can be used to analyze the Turing–Hopf bifurcation of a general reaction-diffusion equation under Neumann boundary condition. By analyzing the distribution of eigenvalues, the stable region, the unstable region (including Turing unstable region), and Turing–Hopf bifurcation point are derived in a double parameters plane. Secondly, by applying this method, the Turing–Hopf bifurcation of a ratio-dependent predator-prey model with diffusion is investigated. Finally, we compute normal forms near Turing–Hopf singularity and verify the theoretical analysis by numerical simulations.
## 1 Introduction
Due to the existence of rich dynamics, predator-prey systems have received great attention [1,2,3,4,5,6,7,8,9,10,11,12]. The classical ratio-dependent Holling–Tanner prey-predator model is as follows [13]:
\begin{aligned} \begin{aligned} &\frac{du}{dt}=ru \biggl(1- \frac{u}{k} \biggr)-\frac{auv}{mv+u}, \\ & \frac{dv}{dt}=sv \biggl(1-h\frac{v}{u} \biggr). \end{aligned} \end{aligned}
(1)
Here, u and v is the prey and predator population, respectively, $$r, s>0$$ is the linear birth rate of prey and predator, respectively, $$k>0$$ is the carrying capacity of prey population, $$h>0$$ is the proportionality coefficient of prey density to the carrying capacity for the predator and $$\frac{auv}{mv+u}$$ represents ratio-dependent functional response with $$a, m>0$$, which is significant for describing predator consumption of predator-prey models [14,15,16,17,18,19,20,21].
Taking into account the inhomogeneous distribution of the prey and predators in different spatial locations and other food sources of predators, model (1) can be modified as follows [22,23,24]:
\begin{aligned} \begin{aligned} &\frac{\partial u}{\partial t}=d_{1}\Delta u+u \biggl(\alpha _{1}-\beta _{1}u-\frac{\gamma _{1}v}{m_{1}v+u} \biggr), \\ &\frac{\partial v}{\partial t}=d_{2}\Delta v+v \biggl(\alpha _{2}- \frac{\gamma _{2}v}{m_{2}+u} \biggr), \end{aligned} \end{aligned}
(2)
where $$u=u(x,t)$$ and $$v=v(x,t)$$ is the population density of the prey and predators at location x and time t, respectively, $$d_{1},d _{2}>0$$ is the diffusion coefficient characterizing the rate of the spatial dispersion of the prey and predator population, respectively.
In this paper, we investigate the Turing–Hopf bifurcation of model (2) with Neumann boundary condition
$$\frac{\partial u}{\partial \nu }=\frac{\partial v}{\partial \nu }=0,\quad x\in \partial \varOmega ,t>0,$$
(3)
where $$\varOmega =(0,l\pi )$$ and ν is the outward unit normal vector on ∂Ω. Because stable spatially inhomogeneous periodic solution can be preferably used to explain the periodic fluctuation of biological populations and it is very difficult to obtain stable spatially inhomogeneous periodic solution for research on general Turing or Hopf bifurcation under Neumann boundary condition, more and more scholars start to investigate the high codimension bifurcation of reaction-diffusion equation, especially Turing–Hopf bifurcation. There exist very rich dynamics near Turing–Hopf singularity, such as stable constant steady state, nonconstant steady state, spatially homogeneous, and inhomogeneous periodic solutions.
It is well known that the normal forms theory plays a very important role in the bifurcation analysis. Faria developed a method to calculate normal forms near an equilibrium of partial functional differential equation [25]. Based on the method of Faria, Song et al. presented a method to compute normal forms near Turing–Hopf singularity of reaction-diffusion equation [26]. However, there are still very few studies on Turing–Hopf bifurcation of reaction-diffusion equation with practical significance [27,28,29].
We would like to mention that one of the most difficult problems for research on Turing–Hopf bifurcation is how to obtain the existence of Turing–Hopf bifurcation. In the previous research, scholars generally chose two appropriate bifurcation parameters such that the Hopf and Turing bifurcation line in a double parameters plane can be defined by a straight line. This method can be easily used to obtain the existence of Turing–Hopf bifurcation, but it cannot be applied to most reaction-diffusion equations. In this paper, we present a kind of parameter selection method such that the Turing bifurcation line can be defined by a curve, which can be applied to most reaction-diffusion equations.
The rest of the paper is organized as follows. In Sect. 2, the stability of positive steady state and the existence of Turing–Hopf bifurcation for general bivariate reaction-diffusion equation are carried out. In Sect. 3, we consider Turing–Hopf bifurcation of a ratio-dependent predator-prey model with diffusion by applying the results in Sect. 2. In Sect. 4, we calculate the normal forms of Turing–Hopf bifurcation for a ratio-dependent predator-prey model with diffusion. In Sect. 5, we verify the theoretical analysis by numerical simulations.
## 2 Existence of Turing–Hopf bifurcation of reaction-diffusion equation
In this section, we consider the Turing–Hopf bifurcation of the following general bivariate reaction-diffusion equation:
$$\textstyle\begin{cases} \textstyle\begin{array}{l} \frac{\partial u(x,t)}{\partial t}=d_{1} \Delta u(x,t)+\alpha F(u(x,t),v(x,t)), \\ \frac{\partial v(x,t)}{\partial t}=d_{2} \Delta v(x,t)+\beta G(u(x,t),v(x,t)), \end{array}\displaystyle \quad x\in \varOmega ,t>0,\\ \frac{\partial u(x,t)}{\partial \nu }=\frac{ \partial v(x,t)}{\partial \nu }=0,\hspace{94pt} x\in \partial \varOmega ,t>0, \end{cases}$$
(4)
where f and g are adequately smooth. Moreover, we assume that $$\alpha ,\beta >0$$ and there exists one positive steady state $$E_{*}(u_{*}, v_{*})$$ of system (4). Taking α and β as bifurcation parameters, we have the characteristic equation at the steady state $$E_{*}$$ as follows:
$|\begin{array}{cc}\lambda +{d}_{1}{\left(\frac{n}{l}\right)}^{2}-{F}_{1}\left({u}_{\ast },{v}_{\ast }\right)\alpha & -{F}_{2}\left({u}_{\ast },{v}_{\ast }\right)\alpha \\ -{G}_{1}\left({u}_{\ast },{v}_{\ast }\right)\beta & \lambda +{d}_{2}{\left(\frac{n}{l}\right)}^{2}-{G}_{2}\left({u}_{\ast },{v}_{\ast }\right)\beta \end{array}|=0,\phantom{\rule{1em}{0ex}}n\in {\mathbb{N}}_{0},$
(5)
where $$\mathbb{N}_{0}=\{0\}\cup \mathbb{N}$$. Denote
$$p_{11}=F_{1}(u_{*},v_{*}),\qquad p_{12}=F_{2}(u_{*},v_{*}),\qquad p_{21}=G_{1}(u _{*},v_{*}),\qquad p_{22}=G_{2}(u_{*},v_{*}),$$
then Eq. (5) can be written as
$$\Delta _{n}(\lambda )=\lambda ^{2}+T_{n} \lambda +h_{n}=0, \quad n\in \mathbb{N}_{0},$$
(6)
where
\begin{aligned} \begin{aligned} &T_{n}=(d_{1}+d_{2}) \biggl(\frac{n}{l} \biggr)^{2}+T_{0}, \\ &h_{n}=d_{1}d _{2} \biggl(\frac{n}{l} \biggr)^{4}+ (-d_{2}p_{11}\alpha -d_{1}p_{22} \beta ) \biggl(\frac{n}{l} \biggr)^{2}+h_{0}, \end{aligned} \end{aligned}
(7)
with
\begin{aligned} \begin{aligned} &T_{0}=-p_{11} \alpha -p_{22}\beta , \\ &h_{0}=(p_{11}p_{22}-p_{12}p _{21})\alpha \beta. \end{aligned} \end{aligned}
(8)
For convenience, we denote
$$\mathcal{D}_{0}(\alpha )=-\frac{p_{11}}{p_{22}}\alpha ,\quad \alpha >0,$$
(9)
and make some hypotheses as follows:
\begin{aligned} \begin{aligned} &\mathrm{(H_{1})}\quad p_{11}p_{22}< p_{12}p_{21}, \\ &\mathrm{(H_{21})}\quad p_{11}p_{22}>p_{12}p_{21},\qquad p_{11}\geq 0,\qquad p_{22}\geq 0,\qquad p_{11}^{2}+p _{22}^{2}\neq 0, \\ &\mathrm{(H_{22})}\quad p_{11}p_{22}>p_{12}p_{21},\qquad p_{11}\leq 0,\qquad p_{22}\leq 0,\qquad p_{11}^{2}+p _{22}^{2}\neq 0, \\ &\mathrm{(H_{23})}\quad p_{11}p_{22}>p_{12}p_{21},\qquad p_{11}>0,\qquad p_{22}< 0, \\ &\mathrm{(H_{24})}\quad p_{11}p_{22}>p_{12}p_{21},\qquad p_{11}< 0,\qquad p_{22}>0. \end{aligned} \end{aligned}
(10)
### Lemma 2.1
For system (4) without diffusion ($$d _{1}=d_{2}=0$$), we have the following results.
1. (i)
If $$\mathrm{(H_{1})}$$ or $$\mathrm{(H_{21})}$$ holds, then $$E_{*}$$ is unstable.
2. (ii)
If $$\mathrm{(H_{22})}$$ holds, then $$E_{*}$$ is asymptotically stable.
3. (iii)
If $$\mathrm{(H_{23})}$$ holds, then $$E_{*}$$ is asymptotically stable for $$\beta >\mathcal{D}_{0}(\alpha )$$ and unstable for $$\beta < \mathcal{D}_{0}(\alpha )$$; and the Hopf bifurcation line is $$\beta =\mathcal{D}_{0}(\alpha )$$.
4. (iv)
If $$\mathrm{(H_{24})}$$ holds, then $$E_{*}$$ is asymptotically stable for $$\beta <\mathcal{D}_{0}(\alpha )$$ and unstable for $$\beta > \mathcal{D}_{0}(\alpha )$$; and the Hopf bifurcation line is $$\beta =\mathcal{D}_{0}(\alpha )$$.
### Proof
Clearly, the characteristic equation for $$E_{*}$$ of system (4) without diffusion is
$$\Delta _{0}(\lambda )=\lambda ^{2}+T_{0} \lambda +h_{0}=0.$$
(11)
If $$\mathrm{(H_{1})}$$ holds, then we have $$h_{0}<0$$. It follows that Eq. (11) has one positive real root and the proof of (i) is completed.
If $$p_{11}p_{22}>p_{12}p_{21}$$, then we have $$h_{0}>0$$. Furthermore, we can obtain that $$T_{0}<0$$ when $$\mathrm{(H_{21})}$$ holds and $$T_{0}>0$$ when $$\mathrm{(H_{22})}$$ holds. Thus, the two roots of Eq. (11) have positive real parts when $$\mathrm{(H_{21})}$$ holds and negative real parts when $$\mathrm{(H_{22})}$$ holds. The proof of (ii) is completed.
If $$\mathrm{(H_{23})}$$ holds, we can obtain that $$h_{0}>0$$ and $$T_{0}<0$$ when $$\beta <\mathcal{D}_{0}(\alpha )$$ and $$T_{0}>0$$ when $$\beta > \mathcal{D}_{0}(\alpha )$$. Moreover, $$\pm i\sqrt{h_{0}}$$ are a pair of purely imaginary roots of Eq. (11) when $$\beta = \mathcal{D}_{0}(\alpha )$$ and the transversality condition is as follows:
$$\frac{d\operatorname{Re}\{\lambda (\alpha )\}}{d\alpha }\Big|_{\beta =\mathcal{D} _{0}(\alpha )}=\frac{p_{11}}{2}>0.$$
(12)
Thus, $$\beta =\mathcal{D}_{0}(\alpha )$$ is the Hopf bifurcation line and the proof of (iii) is completed. We omit the proof of (iv), because it is similar to the proof of (iii). □
Assumptions in (10) do not include the cases $$p_{11}p_{22}=p _{12}p_{21}$$ and $$p_{11}=p_{22}=0$$, because if $$p_{11}p_{22}=p_{12}p _{21}$$ holds, one immediately has $$h_{0}=0$$ and 0 is a root of Eq. (11) for any $$\alpha , \beta >0$$ and if $$p_{11}=p_{22}=0$$ holds, we immediately have $$T_{0}=0$$ and $$\pm i\sqrt{h_{0}}$$ are a pair of purely imaginary roots of Eq. (11) for any $$\alpha , \beta >0$$. Obviously, the double parameters selection method is not effective for these two cases. So we do not discuss them any more in this paper.
Next, we investigate the stability and Turing–Hopf bifurcation of system (4). From the proof of Lemma 2.1, we know that Eq. (6) has at least one root with a positive real part for any $$\alpha , \beta >0$$ if $$\mathrm{(H_{1})}$$ or $$\mathrm{(H_{21})}$$ holds. Moreover, if $$\mathrm{(H_{22})}$$ holds, one can easily obtain that $$T_{n}>0$$ and $$h_{n}>0$$ for $$n\in \mathbb{N}_{0}$$ and $$\alpha , \beta >0$$. Therefore, we have the following result.
### Theorem 2.2
1. (i)
If $$\mathrm{(H_{1})}$$ or $$\mathrm{(H_{21})}$$ holds, $$E_{*}$$ is unstable for any $$\alpha , \beta >0$$.
2. (ii)
If $$\mathrm{(H_{22})}$$ holds, $$E_{*}$$ is asymptotically stable for any $$\alpha , \beta >0$$.
Since $$\mathrm{(H_{24})}$$ can be converted to $$\mathrm{(H_{23})}$$ by swapping the two equations of system (4), we focus on the case of $$\mathrm{(H_{23})}$$. Assume that $$\mathrm{(H_{23})}$$ holds and analyze the Turing–Hopf bifurcation of system (4). First, we consider the case of no diffusion-driven Turing instability. If $$\beta >\mathcal{D}_{0}( \alpha )$$ holds, we have that $$T_{n}>0$$ for $$n\in \mathbb{N}_{0}$$. And if $$h_{n}=0$$ holds, we have that
$$\beta =\mathcal{R}_{n}(\alpha )=\frac{d_{2} (\frac{n}{l} )^{2}p _{11}\alpha -d_{1}d_{2} (\frac{n}{l} )^{4}}{-d_{1} ( \frac{n}{l} )^{2}p_{22}+(p_{11}p_{22}-p_{12}p_{21})\alpha },\quad n \in \mathbb{N}_{0}.$$
(13)
Obviously, if all lines defined by $$\beta =\mathcal{R}_{n}(\alpha )$$ lie below the Hopf bifurcation line $$\beta =\mathcal{D}_{0}(\alpha )$$, there is no Turing instability. Note that
$$\mathcal{D}_{0}(\alpha )-\mathcal{R}_{n}( \alpha )=\frac{1}{-d_{1} (\frac{n}{l} )^{2}p_{22}+(p_{11}p_{22}-p_{12}p_{21})\alpha }P _{n}(\alpha ),$$
(14)
where
$$P_{n}(\alpha )=-\frac{p_{11}}{p_{22}}(p_{11}p_{22}-p_{12}p_{21}) \alpha ^{2}+(d_{1}-d_{2})p_{11} \biggl( \frac{n}{l} \biggr)^{2}\alpha +d_{1}d_{2} \biggl(\frac{n}{l} \biggr)^{4}.$$
(15)
For convenience, we define
$$\varLambda =(d_{2}-d_{1})^{2}p_{11}^{2}+4d_{1}d_{2} \frac{p_{11}}{p_{22}}(p _{11}p_{22}-p_{12}p_{21}),$$
(16)
and make some hypotheses:
\begin{aligned} \begin{aligned} &\mathrm{(C_{1})} \quad d_{2}\leq d_{1}. \\ &\mathrm{(C_{2})} \quad d_{2}>d_{1} \quad \text{and}\quad \varLambda < 0. \\ &\mathrm{(C_{3})}\quad d_{2}>d_{1} \quad\text{and}\quad \varLambda >0. \end{aligned} \end{aligned}
(17)
Clearly, $$\mathcal{D}_{0}(\alpha )>\mathcal{R}_{n}(\alpha )$$ is equivalent to $$P_{n}(\alpha )>0$$. Moreover, we can obtain that $$P_{n}(\alpha )>0$$ for $$n\in \mathbb{N}_{0}$$ if $$\mathrm{(C_{1})}$$ or $$\mathrm{(C_{2})}$$ holds. Hence, we have the following result.
### Theorem 2.3
Assume that $$\mathrm{(H_{23})}$$ holds. If $$\mathrm{(C_{1})}$$ or $$\mathrm{(C_{2})}$$ is satisfied, then system (4) has no diffusion-driven Turing instability. In this case, diffusion does not change the stability of $$E^{*}$$, i.e., the stable and unstable regions are the same as those in Lemma 2.1(iii).
Then we focus on the case of the existence of Turing instability. It is easy to see that if $$\mathrm{(C_{3})}$$ holds, $$\mathcal{D}_{0}(\alpha )- \mathcal{R}_{n}(\alpha )=0, n\in \mathbb{N}$$ has two positive roots:
$$\mathcal{A}_{n}^{\pm }=\frac{-(d_{1}-d_{2})p_{11}\pm \sqrt{\varLambda }}{-2\frac{p_{11}}{p_{22}}(p_{11}p_{22}-p_{12}p_{21})} \biggl( \frac{n}{l} \biggr)^{2}$$
(18)
such that
$$\textstyle\begin{cases} \mathcal{D}_{0}(\alpha )< \mathcal{R}_{n}(\alpha )& \text{if }\mathcal{A}_{n}^{-}< \alpha < \mathcal{A}_{n}^{+},\\ \mathcal{D}_{0}(\alpha )>\mathcal{R}_{n}(\alpha )& \text{if }0< \alpha < \mathcal{A}_{n}^{-} \text{ or } \alpha > \mathcal{A}_{n}^{+}. \end{cases}$$
For any $$m,n\in \mathbb{N}, m>n$$, we can obtain that
\begin{aligned} &\mathcal{R}_{m}(\alpha )-\mathcal{R}_{n}( \alpha ) \\ &\quad=\frac{d_{2}( \frac{m}{l})^{2}p_{11}\alpha -d_{1}d_{2}(\frac{m}{l})^{4}}{-d_{1}( \frac{m}{l})^{2}p_{22}+(p_{11}p_{22}-p_{12}p_{21})\alpha }-\frac{d _{2} (\frac{n}{l} )^{2}p_{11}\alpha -d_{1}d_{2} (\frac{n}{l} )^{4}}{-d_{1} (\frac{n}{l} )^{2}p_{22}+(p_{11}p_{22}-p_{12}p _{21})\alpha } \\ &\quad=\frac{d_{2} (m^{2}-n^{2})Q_{nm}(\alpha )}{ (-d _{1}(\frac{m}{l})^{2}p_{22}+(p_{11}p_{22}-p_{12}p_{21})\alpha ) (-d_{1} (\frac{n}{l} )^{2}p_{22}+(p_{11}p_{22}-p_{12}p_{21}) \alpha )l^{2}}, \end{aligned}
where
$$Q_{nm}(\alpha )=p_{11}(p_{11}p_{22}-p_{12}p_{21}) \alpha ^{2}-d_{1}\frac{m ^{2}+n^{2}}{l^{2}}(p_{11}p_{22}-p_{12}p_{21}) \alpha +d_{1}^{2}\frac{m ^{2}n^{2}}{l^{4}}p_{22}.$$
Note that, for any $$m,n\in \mathbb{N}$$, $$Q_{nm}(\alpha )$$ has and only has one positive root
$$\mathcal{B}^{+}(n,m)=\frac{d_{1}(m^{2}+n^{2})}{2p_{11}l^{2}}+\sqrt{ \frac{d _{1}^{2}(m^{2}+n^{2})^{2}}{4p_{11}^{2} l^{4}}-\frac{d_{1}^{2}p_{22}m ^{2}n^{2}}{p_{11}(p_{11}p_{22}-p_{12}p_{21})l^{4}}}$$
(19)
such that
$$\textstyle\begin{cases} \mathcal{R}_{m}(\alpha )< \mathcal{R}_{n}(\alpha ) &\text{if }0< \alpha < \mathcal{B}^{+}(n,m),\\ \mathcal{R}_{m}(\alpha )> \mathcal{R}_{n}(\alpha ) &\text{if }\alpha >\mathcal{B}^{+}(n,m). \end{cases}$$
Therefore, $$\beta =\mathcal{R}_{m}(\alpha )$$ and $$\beta =\mathcal{R} _{n}(\alpha )$$ have and only have one intersecting point for any $$m,n\in \mathbb{N}, m>n$$. By (14) and (15), we can obtain that $$\mathcal{A}_{n}^{\pm }$$ is monotonously increasing with respect to n and $$\mathcal{B}^{+}(n,m)$$ is monotonously increasing with respect to m for fixed $$n\in \mathbb{N}$$.
Moreover, the transversality condition is as follows:
$$\frac{d\operatorname{Re}\{\lambda (\alpha )\}}{d\alpha }\Big|_{\beta =\mathcal{R} _{n}(\alpha )}=\frac{d_{1}d_{2} (\frac{n}{l} )^{4}-d_{1}p_{22} \beta (\frac{n}{l} )^{2}}{T_{n}\alpha }>0 \quad\text{if } \beta >\mathcal{D}_{0}(\alpha ).$$
(20)
For convenience, denote
$$\varGamma _{0}=\mathcal{A}_{1}^{-}, \varGamma _{n}= \textstyle\begin{cases} \mathcal{B}^{+}(n,n+1)& \text{if } \mathcal{A}_{n+1}^{-}\leq \mathcal{A}_{n}^{+},\\ \mathcal{A}_{n+1}^{-}& \text{if } \mathcal{A}_{n+1}^{-}>\mathcal{A}_{n}^{+}, \end{cases}\displaystyle \quad n\in \mathbb{N},$$
(21)
and
\begin{aligned} \begin{aligned} &\mathcal{T}_{0}(\alpha )= \mathcal{D}_{0}(\alpha ),\quad \alpha \in (0, \varGamma _{0}], \\ &\mathcal{T}_{n}(\alpha )= \textstyle\begin{cases} \mathcal{R}_{n}(\alpha )& \text{if } \mathcal{A}_{n+1}^{-}\leq \mathcal{A}_{n}^{+},\\ \tilde{\mathcal{T}}_{n}(\alpha )& \text{if } \mathcal{A}_{n+1}^{-}>\mathcal{A}_{n}^{+}, \end{cases}\displaystyle \quad n\in \mathbb{N},\alpha \in (\varGamma _{n-1}, \varGamma _{n}], \end{aligned} \end{aligned}
(22)
with
$$\tilde{\mathcal{T}}_{n}(\alpha )= \textstyle\begin{cases} \mathcal{R}_{n}(\alpha ),& \alpha \in (\varGamma _{n-1}, \mathcal{A}_{n}^{+}],\\ \mathcal{D}_{0}( \alpha ),& \alpha \in (\mathcal{A}_{n}^{+}, \varGamma _{n}], \end{cases}\displaystyle \quad n\in \mathbb{N}.$$
It is clear that for $$0<\alpha \leq \varGamma _{0}=\mathcal{A}_{1}^{-}$$, the boundary of an unstable region of $$E_{*}$$ is $$\mathcal{D}_{0}(\alpha )$$. If $$\mathcal{A}_{2}^{-}\leq \mathcal{A}_{1}^{+}$$, then for $$\varGamma _{0}<\alpha \leq \varGamma _{1}=\mathcal{B}^{+}(1,2)$$, the boundary of a stable region is $$\mathcal{R}_{1}(\alpha )$$. Otherwise, for $$\varGamma _{0}<\alpha \leq \varGamma _{1}=\mathcal{A}_{2}^{-}$$, the boundaries consist of $$\mathcal{R}_{1}(\alpha )$$ with $$\varGamma _{0}<\alpha \leq \mathcal{A}_{1}^{+}$$ and $$\mathcal{D}_{0}(\alpha )$$ with $$\mathcal{A} _{1}^{+}<\alpha \leq \varGamma _{1}$$. By using the mathematical induction, we can obtain that all the boundaries of a stable region consist of $$\mathcal{T}_{n}(\alpha ), n\in \mathbb{N}_{0}$$.
Obviously, when $$(\alpha , \beta )=(\mathcal{A}_{1}^{-}, -\frac{p_{11}}{p _{22}}\mathcal{A}_{1}^{-})$$, $$\Delta _{0}(\lambda )=0$$ has a pair of purely imaginary roots $$\pm i\sqrt{h_{0}}$$ and $$\Delta _{1}(\lambda )=0$$ has a root $$\lambda =0$$ with a negative real root $$\lambda =-T _{1}$$ and all other roots have negative real parts. Together with the transversality conditions (12) and (18), we have that system (4) undergoes Turing–Hopf bifurcation at $$( \mathcal{A}_{1}^{-}, -\frac{p_{11}}{p_{22}}\mathcal{A}_{1}^{-})$$. If $$\mathcal{A}_{n+1}^{-}>\mathcal{A}_{n}^{+}$$ holds, $$\Delta _{n}(\lambda )=0$$ and $$\Delta _{n+1}(\lambda )=0$$ have zero roots under the conditions $$(\alpha , \beta )=(\mathcal{A}_{n}^{+}, -\frac{p_{11}}{p_{22}} \mathcal{A}_{n}^{+})$$ and $$(\alpha , \beta )=(\mathcal{A}_{n+1}^{-}, -\frac{p _{11}}{p_{22}}\mathcal{A}_{n+1}^{-})$$, respectively. Meanwhile all other roots, except a pair of purely imaginary roots $$\pm i\sqrt{h_{0}}$$ of $$\Delta _{0}(\lambda )=0$$, have negative real parts. Thus, $$( \mathcal{A}_{n}^{+}, -\frac{p_{11}}{p_{22}}\mathcal{A}_{n}^{+})$$ and $$(\mathcal{A}_{n+1}^{-}, -\frac{p_{11}}{p_{22}}\mathcal{A}_{n+1}^{-})$$ are in the same situation.
From the above analysis, we can detailedly describe the boundary of the unstable region (including Turing unstable region) of the positive steady state $$E_{*}$$.
### Theorem 2.4
Assume that $$\mathrm{(H_{23})}$$ holds and $$\mathrm{(C_{3})}$$ is satisfied, there exists diffusion-driven Turing instability. More precisely, we have the following results.
1. (i)
The boundary of a stable region is $$\beta =\mathcal{T}_{n}( \alpha ), n\in \mathbb{N}_{0}$$, and for any $$\alpha >0, \beta > \mathcal{T}_{n}(\alpha )$$, $$n\in \mathbb{N}_{0}$$, the positive steady state $$E_{*}$$ is asymptotically stable.
2. (ii)
For any $$\alpha >0, \mathcal{D}_{0}(\alpha )<\beta < \mathcal{T}_{n}(\alpha )$$, $$n\in \mathbb{N}$$, there exists Turing instability.
3. (iii)
System (4) undergoes Turing–Hopf bifurcation at the point $$(\mathcal{A}_{1}^{-}, -\frac{p_{11}}{p_{22}}\mathcal{A} _{1}^{-})$$; if $$\mathcal{A}_{n+1}^{-}>\mathcal{A}_{n}^{+}, n\in \mathbb{N}$$, holds, system (4) undergoes Turing–Hopf bifurcation at the points $$(\mathcal{A}_{n}^{+}, - \frac{p_{11}}{p_{22}}\mathcal{A}_{n}^{+})$$ and $$(\mathcal{A}_{n+1} ^{-}, -\frac{p_{11}}{p_{22}}\mathcal{A}_{n+1}^{-})$$.
## 3 Turing–Hopf bifurcation of a diffusive ratio-dependent predator-prey model
Let
$$h=\frac{\beta }{\alpha _{1}},\qquad \delta =\frac{\gamma _{1}}{\alpha _{1}}, \qquad b=\frac{ \gamma _{2}}{\alpha _{2}},$$
then system (2) becomes:
$$\textstyle\begin{cases} \textstyle\begin{array}{l} \frac{\partial u(x,t)}{\partial t}=d_{1} \Delta u(x,t)+\alpha _{1}u(x,t) (1-hu(x,t) -\frac{\delta v(x,t)}{m_{1}v(x,t)+u(x,t)} ),\\ \frac{ \partial v(x,t)}{\partial t}=d_{2} \Delta v(x,t)+\alpha _{2} v(x,t) (1-\frac{bv(x,t)}{m_{2}+u(x,t)} ), \end{array}\displaystyle \quad x\in \varOmega ,t>0,\\ \frac{\partial u(x,t)}{\partial \nu }=\frac{ \partial v(x,t)}{\partial \nu }=0,\hspace{155pt}\quad x\in \partial \varOmega ,t>0. \end{cases}$$
(23)
Next, we use the conclusions of the above section to investigate the stability of positive steady state and Turing–Hopf bifurcation of system (23). With a simple calculation, we have the following result.
### Theorem 3.1
Assume that $$\alpha _{1},\alpha _{2},d_{1},d_{2},l>0$$. Then the existence conditions of positive steady state of model (23) can be divided into the following three situations:
Case A: $$m_{1}=\delta$$ and $$b>hm_{1}m_{2}+1$$.
There exists only one positive steady state $$E_{1}=(u_{1},v_{1})$$ of model (23) in this case, where
$$u_{1}=\frac{b-hm_{1}m_{2}}{h(m_{1}+b)},\qquad v_{1}=\frac{m_{2}+u_{1}}{b}.$$
Case B: $$m_{1}>\delta$$.
There exists only one positive steady state $$E_{2}=(u_{2},v_{2})$$ of model (23) in this case, where
$$u_{2}=\frac{m_{1}+b-hm_{1}m_{2}-\delta +c}{2h(m_{1}+b)},\qquad v_{2}=\frac{m _{2}+u_{2}}{b},$$
with
$$c=\sqrt{(m_{1}+b-hm_{1}m_{2}-\delta )^{2}+4hm_{2}(m_{1}+b) (m_{1}- \delta )}.$$
Case C: $$m_{1}<\delta$$ and $$m_{1}+b-hm_{1}m_{2}-\delta >0$$.
There exist two positive steady states $$E_{3}=(u_{3},v_{3})$$ and $$E_{4}=(u_{4},v_{4})$$ of model (23) in this case, where
\begin{aligned} &u_{3}=\frac{m_{1}+b-hm_{1}m_{2}-\delta -c}{2h(m_{1}+b)},\qquad v_{3}= \frac{m _{2}+u_{3}}{b}, \\ &u_{4}=\frac{m_{1}+b-hm_{1}m_{2}-\delta +c}{2h(m _{1}+b)},\qquad v_{4}=\frac{m_{2}+u_{4}}{b}. \end{aligned}
Next, we take the steady state $$E_{4}$$ as an example to investigate the Turing–Hopf bifurcation. Denote
\begin{aligned} &f(u_{4})=\frac{\delta m_{1}(m_{2}+u_{4})^{2}}{(m_{1}(m_{2}+u_{4})+bu _{4})^{2}}, \\ &g(u_{4})=\frac{b^{2}u_{4}^{2}\delta }{(m_{1}(m_{2}+u _{4})+bu_{4})^{2}}, \\ &r(u_{4})=1-2hu_{4}-f(u_{4}). \end{aligned}
Then we have
$$p_{11}=r(u_{4}),\qquad p_{12}=-g(u_{4}),\qquad p_{21}=\frac{1}{b},\qquad p_{22}=-1.$$
The characteristic equation at $$E_{4}$$ is
$$\Delta _{n}(\lambda )=\lambda ^{2}+T_{n} \lambda +h_{n}=0,\quad n\in \mathbb{N}_{0},$$
(24)
where
\begin{aligned} &T_{n}=(d_{1}+d_{2}) \biggl(\frac{n}{l} \biggr)^{2}+T_{0}, \\ &h_{n}=d_{1}d _{2} \biggl(\frac{n}{l} \biggr)^{4}+ \bigl(d_{1}\alpha _{2}-d_{2} \alpha _{1}r(u _{4}) \bigr) \biggl(\frac{n}{l} \biggr)^{2}+h_{0}, \end{aligned}
with
\begin{aligned} &T_{0}=\alpha _{2}-\alpha _{1}r(u_{4}), \\ &h_{0}=\alpha _{1}\alpha _{2}\frac{g(u _{4})}{b}- \alpha _{1}\alpha _{2}r(u_{4}). \end{aligned}
From the results in Sect. 2, one immediately has the following conclusions.
### Theorem 3.2
1. (i)
If $$r(u_{4})\leq 0$$ holds, the positive steady state $$E_{4}$$ of (23) is asymptotically stable for any $$\alpha _{1}, \alpha _{2}>0$$.
2. (ii)
If $$r(u_{4})>0$$ and $$\frac{g(u_{4})}{b}< r(u_{4})$$ hold, the positive steady state $$E_{4}$$ of (23) is unstable for any $$\alpha _{1}, \alpha _{2}>0$$.
### Theorem 3.3
Suppose that $$r(u_{4})>0$$, $$\frac{g(u_{4})}{b}>r(u _{4})$$. If $$\mathrm{(C_{1})}$$ or $$\mathrm{(C_{2})}$$ holds, there is no diffusion-driven Turing instability.
### Theorem 3.4
Suppose that $$r(u_{4})>0$$, $$\frac{g(u_{4})}{b}>r(u _{4})$$. If $$\mathrm{(C_{3})}$$ holds, we have the following results.
1. (i)
If $$\alpha _{1}>0, \alpha _{2}>\mathcal{T}_{n}(\alpha _{1}), n \in \mathbb{N}_{0}$$, the steady state $$E_{4}$$ is asymptotically stable.
2. (ii)
Turing instability occurs for $$\alpha _{1}>0, \mathcal{D} _{0}(\alpha _{1})<\alpha _{2}<\mathcal{T}_{n}(\alpha _{1}), n\in \mathbb{N}$$.
3. (iii)
System (23) undergoes Turing–Hopf bifurcation at $$(\mathcal{A}_{1}^{-}, r(u_{4})\mathcal{A}_{1}^{-})$$; if $$\mathcal{A} _{n+1}^{-}>\mathcal{A}_{n}^{+}, n\in \mathbb{N}$$ holds, system (23) undergoes Turing–Hopf bifurcation at the point $$(\mathcal{A}_{n}^{+}, r(u_{4})\mathcal{A}_{n}^{+})$$ and $$( \mathcal{A}_{n+1}^{-}, r(u_{4})\mathcal{A}_{n+1}^{-})$$, where
\begin{aligned} &\mathcal{D}_{0}(\alpha _{1})=\alpha _{1}r(u_{4}),\quad \alpha _{1}>0, \\ &\mathcal{R}_{n}(\alpha _{1})=\frac{r(u_{4})\alpha _{1}d_{2} ( \frac{n}{l} )^{2}-d_{1}d_{2} (\frac{n}{l} )^{4}}{d_{1} ( \frac{n}{l} )^{2}+\alpha _{1}(\frac{g(u_{4})}{b}-r(u_{4}))},\quad r>0, n\in \mathbb{N}_{0}. \\ &\varLambda =r^{2}(u_{4}) (d_{2}-d_{1})^{2}-4r(u_{4}) \biggl( \frac{g(u_{4})}{b}-r(u_{4}) \biggr)d_{1}d_{2}, \\ &\mathcal{A}_{n}^{\pm }=\frac{r(u_{4})(d_{2}-d_{1})\pm \sqrt{\varLambda }}{2r(u_{4})(\frac{g(u_{4})}{b}-r(u_{4}))} \biggl( \frac{n}{l} \biggr)^{2}, \\ &\mathcal{B}^{+}(n,m)=\frac{d_{1}(\frac{g(u_{4})}{b}-r(u_{4}))\frac{m ^{2}+n^{2}}{l^{2}}+c_{1}}{2r(u_{4})(\frac{g(u_{4})}{b}-r(u_{4}))}, \\ &c _{1}=\sqrt{d_{1}^{2} \biggl( \frac{g(u_{4})}{b}-r(u_{4}) \biggr)^{2}\frac{(m ^{2}+n^{2})^{2}}{l^{2}}+4d_{1}^{2}r(u_{4}) \biggl(\frac{g(u_{4})}{b}-r(u _{4}) \biggr)\frac{m^{2}n^{2}}{l^{4}}}. \end{aligned}
$$\varGamma _{n}$$ and $$\mathcal{T}_{n}$$ are defined by (21) and (22), respectively, and hypotheses $$\mathrm{(C_{1})}$$, $$\mathrm{(C_{2})}$$, and $$\mathrm{(C_{3})}$$ are defined by (17).
We would like to mention that $$\mathcal{A}_{n+1}^{-}<\mathcal{A}_{n} ^{+}$$ and $$\mathcal{A}_{n+1}^{-}>\mathcal{A}_{n}^{+}$$ are all possible to happen (see Fig. 1). Obviously, we have $$\mathcal{A}_{2}^{-}< \mathcal{A}_{1}^{+}$$ in Fig. 1(a) and $$\mathcal{A}_{2}^{-}>\mathcal{A} _{1}^{+}$$ in Fig. 1(b).
## 4 Normal forms for Turing–Hopf bifurcation
From Theorem 3.4, we know that for given appropriate parameters, there is at least one Turing–Hopf bifurcation point in the $$(\alpha _{1}, \alpha _{2})$$ plane. Denote the Turing–Hopf bifurcation point (an intersecting point of the lines $$\alpha _{2}=\mathcal{D}_{0}(\alpha _{1})$$ and $$\alpha _{2}=\mathcal{R}_{n_{*}}(\alpha _{1})$$) as $$(\alpha _{1*}, \alpha _{2*})$$, where $$\alpha _{2*}=r(u_{4})\alpha _{1*}$$. Next we compute the normal forms for Turing–Hopf bifurcation by using the method in [26].
Let $$\alpha _{1}=\alpha _{1*}+\mu _{1}$$ and $$\alpha _{2}=\alpha _{2*}+\mu _{2}$$, then system (23) becomes
$$\textstyle\begin{cases} \frac{\partial u}{\partial t}=d_{1} \Delta u+(\alpha _{1*}+\mu _{1})u (1-hu-\frac{\delta v}{m_{1}v+u} ),\\ \frac{\partial v}{\partial t}=d_{2} \Delta v+(\alpha _{2*}+\mu _{2}) v (1-\frac{bv}{m_{2}+u} ). \end{cases}$$
(25)
Clearly, $$(\mu _{1}, \mu _{2})=(0,0)$$ is the Turing–Hopf bifurcation point of system (25). Set $$\bar{u}=u-u_{4}$$ and $$\bar{v}=v-\frac{m_{2}+u_{4}}{b}$$. Then system (25) becomes
$$\textstyle\begin{cases} \frac{\partial u}{\partial t}=d_{1} \Delta u+(\alpha _{1*}+\mu _{1})(u+u _{4})(1-h(u+u_{4})-\frac{\delta (v+\frac{m_{2}+u_{4}}{b})}{m_{1}(v+\frac{m _{2}+u_{4}}{b})+(u+u_{4})},\\ \frac{\partial v}{\partial t}=d_{2} \Delta v+(\alpha _{2*}+\mu _{2}) (v+\frac{m_{2}+u_{4}}{b}) (1-\frac{b(v+\frac{m _{2}+u_{4}}{b})}{m_{2}+(u+u_{4})} ). \end{cases}$$
(26)
Define the real-valued Sobolev space
$$\mathscr{X}=\biggl\{ (u, v)\in H^{2}(0,l\pi ), \frac{\partial u}{\partial t}= \frac{ \partial v}{\partial t}=0 \text{ at } x=0,l\pi \biggr\}$$
with the inner product
$$[U, V]= \int _{0}^{l\pi }(u_{1}v_{1}+u_{2}v_{2}) \,dx, \quad\text{for } U=(u _{1}, u_{2})^{T}, V=(v_{1}, v_{2})^{T}\in \mathscr{X}_{1}.$$
In the abstract space $$\mathcal{C}=C(\mathbb{R}, \mathscr{X})$$, system (26) can be written as
$$\dot{U}=\mathscr{L}U+\tilde{F}(U, \mu ),$$
(27)
where $$U=(u, v)^{T}$$, $$\mu =(\mu _{1}, \mu _{2})$$, $$\mathscr{L}U=D \Delta U+L_{0}(U)$$, $$D=\operatorname{diag}(d_{1}, d_{2})$$, $$\tilde{F}(U, \mu )=L(\mu )(U)-L_{0}(U)+F(U, \mu )$$ with $$L_{0}=L(0)$$ and
$\begin{array}{rl}& L\left(\mu \right)=\left(\begin{array}{cc}r\left({u}_{4}\right)\left({\alpha }_{1\ast }+{\mu }_{1}\right)& -\left({\alpha }_{1\ast }+{\mu }_{1}\right)g\left({u}_{4}\right)\\ \frac{{\alpha }_{2\ast }+{\mu }_{2}}{b}& -\left({\alpha }_{2\ast }+{\mu }_{2}\right)\end{array}\right),\\ & F\left(U,\mu \right)=\left(\begin{array}{c}{F}^{\left(1\right)}\left(u,v,{\mu }_{1},{\mu }_{2}\right)\\ {F}^{\left(2\right)}\left(u,v,{\mu }_{1},{\mu }_{2}\right)\end{array}\right)\\ & \phantom{F\left(U,\mu \right)}=\left(\begin{array}{c}\left({\alpha }_{1\ast }+{\mu }_{1}\right)\left(-h{u}^{2}-\frac{\delta \left(v+\frac{{m}_{2}+{u}_{4}}{b}\right)\left(u+{u}_{4}\right)}{{m}_{1}\left(v+\frac{{m}_{2}+{u}_{4}}{b}\right)+\left(u+{u}_{4}\right)}+{u}_{4}-h{u}_{4}^{2}\right)\\ -\left({\alpha }_{1\ast }+{\mu }_{1}\right)f\left({u}_{4}\right)u+\left({\alpha }_{1\ast }+{\mu }_{1}\right)g\left({u}_{4}\right)v\\ \left({\alpha }_{2\ast }+{\mu }_{2}\right)\left(v+\frac{{m}_{2}+{u}_{4}}{b}\right)\left(1-\frac{b\left(v+\frac{{m}_{2}+{u}_{4}}{b}\right)}{{m}_{2}+\left(u+{u}_{4}\right)}\right)+\left({\alpha }_{2\ast }+{\mu }_{2}\right)v\\ -\frac{\left({\alpha }_{2\ast }+{\mu }_{2}\right)u}{b}\end{array}\right).\end{array}$
(28)
Denote the eigenvalues of DΔ on $$\mathscr{X}$$ by
$$\delta _{n}^{(j)}=-d_{j} \biggl( \frac{n}{l} \biggr)^{2},\quad j=1, 2, n\in \mathbb{N}_{0}$$
and the corresponding normalized eigenfunctions by
$$\beta _{n}^{(j)}=\gamma _{n}(x)e_{j},\quad \gamma _{n}(x)= \frac{\cos (\frac{nx}{l})}{ \Vert \cos (\frac{nx}{l}) \Vert _{2,2}}= \textstyle\begin{cases} \frac{1}{\sqrt{l\pi }}& \text{for }n=0,\\ \frac{\sqrt{2}}{\sqrt{l\pi }}\cos ( \frac{nx}{l} ) &\text{for }n\neq 0, \end{cases}$$
where $$e_{j}$$ is the unit coordinate vector of $$\mathbb{R}^{2}$$ (n is wave number).
By the Taylor expansions of $$L(\mu )$$, we have that
$$L(\mu )=L_{0}+\mu _{1}L_{1}^{(1, 0)}+\mu _{2}L_{1}^{(0, 1)}+\frac{1}{2} \bigl(\mu _{1}^{2}L_{2}^{(2, 0)}+2\mu _{1} \mu _{2}L_{2}^{(1, 1)}+\mu _{2} ^{2}L_{2}^{(0, 2)} \bigr)+\cdots ,$$
where
$\begin{array}{rl}& {L}_{0}=\left(\begin{array}{cc}{\alpha }_{1\ast }r\left({u}_{4}\right)& -{\alpha }_{1\ast }g\left({u}_{4}\right)\\ \frac{{\alpha }_{2\ast }}{b}& -{\alpha }_{2\ast }\end{array}\right),\phantom{\rule{2em}{0ex}}{L}_{1}^{\left(1,0\right)}=\left(\begin{array}{cc}r\left({u}_{4}\right)& -g\left({u}_{4}\right)\\ 0& 0\end{array}\right),\\ & {L}_{1}^{\left(0,1\right)}=\left(\begin{array}{cc}0& 0\\ \frac{1}{b}& -1\end{array}\right).\end{array}$
(29)
Let
$$F_{j_{1}j_{2}}=\bigl(F_{j_{1}j_{2}}^{(1)}, F_{j_{1}j_{2}}^{(2)} \bigr)^{T}$$
with
$$F_{j_{1}j_{2}}^{(k)}=\frac{\partial ^{j_{1}+j_{2}}F^{(k)}(0, 0, 0, 0)}{ \partial u^{j_{1}}\partial v^{j_{2}}}, \quad k=1,2, j_{1}+j_{2}=2,3.$$
Then from (28) we have
$\begin{array}{rl}& {F}_{20}=\left(\begin{array}{c}-2h-\frac{2\delta {u}_{\ast }\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{3}}+\frac{2\delta \frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{2}}\\ \frac{-2{\alpha }_{2\ast }}{b\left({m}_{2}+{u}_{\ast }\right)}\end{array}\right),\\ & {F}_{02}=\left(\begin{array}{c}-\frac{2\delta {m}_{1}{u}_{\ast }}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{2}}-\frac{2\delta {m}_{1}^{2}\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{3}}\\ -\frac{-2{\alpha }_{2\ast }b}{{m}_{2}+{u}_{\ast }}\end{array}\right),\\ & {F}_{30}=\left(\begin{array}{c}\frac{6\delta {u}_{\ast }\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{4}}-\frac{6\delta \frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{3}}\\ \frac{6{\alpha }_{2\ast }}{b{\left({m}_{2}+{u}_{\ast }\right)}^{2}}\end{array}\right),\\ & {F}_{03}=\left(\begin{array}{c}-\frac{6\delta {m}_{1}^{2}{u}_{\ast }}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{2}}+\frac{6\delta {m}_{1}^{3}{u}_{\ast }\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{4}}\\ 0\end{array}\right),\end{array}$
and
$\begin{array}{rl}& {F}_{11}=\left(\begin{array}{c}-\frac{\delta }{{u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}}+\frac{\delta {u}_{\ast }}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{2}}+\frac{\delta {m}_{1}\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{2}}-\frac{2\delta {m}_{1}{u}_{\ast }\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{3}}\\ \frac{2{\alpha }_{2\ast }}{{m}_{2}+{u}_{\ast }}\end{array}\right),\\ & {F}_{21}=\left(\begin{array}{c}\frac{2\delta }{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{2}}-\frac{2\delta {u}_{\ast }}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{3}}-\frac{4\delta {m}_{1}\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{3}}+\frac{6\delta {m}_{1}{u}_{\ast }\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{4}}\\ \frac{-4{\alpha }_{2\ast }}{{\left({m}_{2}+{u}_{\ast }\right)}^{2}}\end{array}\right),\\ & {F}_{12}=\left(\begin{array}{c}\frac{2\delta }{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{2}}-\frac{4\delta {m}_{1}{u}_{\ast }}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{3}}-\frac{2\delta {m}_{1}^{2}\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{3}}+\frac{6\delta {m}_{1}^{2}{u}_{\ast }\frac{{m}_{2}+{u}_{\ast }}{b}}{{\left({u}_{\ast }+m1\frac{{m}_{2}+{u}_{\ast }}{b}\right)}^{4}}\\ \frac{2{\alpha }_{2\ast }b}{{\left({m}_{2}+{u}_{\ast }\right)}^{2}}\end{array}\right).\end{array}$
Moreover, let
${\mathcal{M}}_{n}\left(\lambda \right)=\left(\begin{array}{cc}\lambda +{d}_{1}{\left(\frac{n}{l}\right)}^{2}-{\alpha }_{1\ast }r\left({u}_{4}\right)& {\alpha }_{1\ast }g\left({u}_{4}\right)\\ -\frac{{\alpha }_{2\ast }}{b}& \lambda +{d}_{2}{\left(\frac{n}{l}\right)}^{2}+{\alpha }_{2\ast }\end{array}\right).$
(30)
Then we can obtain that $$\xi _{0}\in \mathbb{C}^{2}$$ and $$\xi _{n_{*}} \in \mathbb{R}^{2}$$ are the eigenvectors associated with the eigenvalues $$i\omega _{c}\ (\omega _{c}=\sqrt{h_{0}})$$ and 0, respectively, and $$\eta _{0}\in \mathbb{C}^{2}$$ and $$\eta _{n_{*}}\in \mathbb{R}^{2}$$ are the corresponding adjoint eigenvectors, where
$\begin{array}{rl}& {\xi }_{0}=\frac{1}{{\alpha }_{1\ast }g\left({u}_{4}\right)}\left(\begin{array}{c}{\alpha }_{1\ast }g\left({u}_{4}\right)\\ {\alpha }_{1\ast }r\left({u}_{4}\right)-i{\omega }_{c}\end{array}\right),\\ & {\eta }_{0}=\frac{1}{2{\omega }_{c}}\left(\begin{array}{c}{\omega }_{c}-i{\alpha }_{2\ast }\\ i{\alpha }_{1\ast }g\left({u}_{4}\right)\end{array}\right),\\ & {\xi }_{{n}_{\ast }}=\frac{1}{{\alpha }_{1\ast }g\left({u}_{4}\right)}\left(\begin{array}{c}{\alpha }_{1\ast }g\left({u}_{4}\right)\\ {\alpha }_{2\ast }g\left({u}_{4}\right)-{d}_{1}{\left(\frac{{n}_{\ast }}{l}\right)}^{2}\end{array}\right),\\ & {\eta }_{{n}_{\ast }}=\frac{1}{\left({d}_{1}+{d}_{2}\right){\left(\frac{{n}_{\ast }}{l}\right)}^{2}}\left(\begin{array}{c}{\alpha }_{2\ast }+{d}_{2}{\left(\frac{{n}_{\ast }}{l}\right)}^{2}\\ {\alpha }_{1\ast }g\left({u}_{4}\right)\end{array}\right).\end{array}$
Denote
$$\varPhi _{1}=(\xi _{0},\bar{\xi }_{0}),\qquad \varPhi _{2}=\xi _{n_{*}},\qquad \varPsi _{1}=\operatorname{col}\bigl(\eta _{0}^{T}, \bar{\eta }_{0}^{T}\bigr),\qquad \varPsi _{2}=\eta _{n_{*}}^{T}.$$
Then we have
$$\langle \varPsi _{1}, \varPhi _{1} \rangle =I_{2},\qquad \langle \varPsi _{2}, \varPhi _{2} \rangle =1,$$
where $$I_{2}$$ is the $$2\times 2$$ identity matrix. Thus, the phase space $$\mathscr{X}$$ for (27) can be decomposed as
$$\mathscr{X}=\mathcal{P}\oplus \mathcal{Q},\qquad \mathcal{P}=\operatorname{Im} \pi ,\qquad \mathcal{Q}=\operatorname{Ker}\pi ,$$
(31)
where $$\operatorname{dim}\mathcal{P}=3$$ and $$\pi:\mathscr{X}\mapsto \mathcal{P}$$ is the projection defined by
$$\pi (U)= \bigl(\varPhi _{1} \bigl\langle \varPsi _{1},\bigl( \bigl[U, \beta _{0}^{(1)}\bigr], \bigl[U, \beta _{0}^{(2)}\bigr]\bigr)^{T} \bigr\rangle \bigr)^{T}\beta _{0}+ \bigl(\varPhi _{2} \bigl\langle \varPsi _{2},\bigl(\bigl[U, \beta _{n_{*}}^{(1)}\bigr], \bigl[U, \beta _{n_{*}}^{(2)}\bigr]\bigr)^{T} \bigr\rangle \bigr)^{T}\beta _{n_{*}},$$
with $$\beta _{0}=\operatorname{col}(\beta _{0}^{(1)}, \beta _{0}^{(2)})$$ and $$\beta _{n_{*}}=\operatorname{col}(\beta _{n_{*}}^{(1)}, \beta _{n_{*}}^{(2)})$$.
According to (31), $$U\in \mathscr{X}$$ can be decomposed as
$\begin{array}{rl}U& ={\left({\Phi }_{1}\left(\begin{array}{c}{z}_{1}\\ {z}_{2}\end{array}\right)\right)}^{T}\left(\begin{array}{c}{\beta }_{0}^{\left(1\right)}\\ {\beta }_{0}^{\left(2\right)}\end{array}\right)+{\left({z}_{3}{\Phi }_{2}\right)}^{T}\left(\begin{array}{c}{\beta }_{{n}_{\ast }}^{\left(1\right)}\\ {\beta }_{{n}_{\ast }}^{\left(2\right)}\end{array}\right)+w\\ & =\left({z}_{1}{\xi }_{0}+{z}_{2}{\overline{\xi }}_{0}\right){\gamma }_{0}\left(x\right)+{z}_{3}{\xi }_{{n}_{\ast }}{\gamma }_{{n}_{\ast }}\left(x\right)+\left(\begin{array}{c}{w}_{1}\\ {w}_{2}\end{array}\right)\\ & =\left({\Phi }_{1},{\Phi }_{2}\right)\left(\begin{array}{c}{z}_{1}{\gamma }_{0}\left(x\right)\\ {z}_{2}{\gamma }_{0}\left(x\right)\\ {z}_{3}{\gamma }_{{n}_{\ast }}\left(x\right)\end{array}\right)+\left(\begin{array}{c}{w}_{1}\\ {w}_{2}\end{array}\right),\end{array}$
(32)
where
$\left(\begin{array}{c}{z}_{1}\\ {z}_{2}\end{array}\right)=〈{\Psi }_{1},{\left(\left[U,{\beta }_{0}^{\left(1\right)}\right],\left[U,{\beta }_{0}^{\left(2\right)}\right]\right)}^{T}〉,\phantom{\rule{2em}{0ex}}{z}_{3}=〈{\Psi }_{2},{\left(\left[U,{\beta }_{{n}_{\ast }}^{\left(1\right)}\right],\left[U,{\beta }_{{n}_{\ast }}^{\left(2\right)}\right]\right)}^{T}〉.$
Setting $$\varPhi =(\varPhi _{1}, \varPhi _{2})$$, $$z_{x}=(z_{1}\gamma _{0}(x), z _{2}\gamma _{0}(x), z_{3}\gamma _{n_{*}}(x))^{T}$$, (32) can be transformed into
$$U=\varPhi z_{x}+w.$$
(33)
Let
${\left(\begin{array}{c}\\ \left[\stackrel{˜}{F},{\beta }_{v}^{\left(2\right)}\right]\end{array}\right)}_{v=0}^{v={n}_{\ast }}=col\left(\left(\begin{array}{c}\\ \left[\stackrel{˜}{F},{\beta }_{0}^{\left(2\right)}\right]\end{array}\right),\left(\begin{array}{c}\\ \left[\stackrel{˜}{F},{\beta }_{{n}_{\ast }}^{\left(2\right)}\right]\end{array}\right)\right),$
(34)
then, denoting the restriction of $$\mathscr{L}$$ to $$\mathcal{Q}$$ by $$\mathscr{L}_{1}$$, we have
$\left\{\begin{array}{c}\stackrel{˙}{z}=Bz+\Psi {\left(\begin{array}{c}\left[\stackrel{˜}{F}\left(z,w,\mu \right),{\beta }_{v}^{\left(1\right)}\right]\\ \left[\stackrel{˜}{F}\left(z,w,\mu \right),{\beta }_{v}^{\left(2\right)}\right]\end{array}\right)}_{v=0}^{v={n}_{\ast }},\hfill \\ \stackrel{˙}{w}={\mathcal{L}}_{1}\left(w\right)+H\left(z,w,\mu \right),\hfill \end{array}$
(35)
where
\begin{aligned} &z=(z_{1}, z_{2},z_{3})^{T},\qquad B= \operatorname{diag}\{i\omega _{c}, -i\omega _{c},0 \},\qquad \varPsi = \operatorname{diag}\{\varPsi _{1}, \varPsi _{2}\}, \\ & \tilde{F}(z, w, \mu )= \tilde{F}(\varPhi z_{x}+w, \mu ) \end{aligned}
with
$\begin{array}{rl}H\left(z,w,\mu \right)=& \stackrel{˜}{F}\left(z,w,\mu \right)-\left(〈{\eta }_{0}^{T},\left(\begin{array}{c}\\ \left[\stackrel{˜}{F}\left(z,w,\mu \right),{\beta }_{0}^{\left(2\right)}\right]\end{array}\right)〉{\xi }_{0}\\ & +〈{\overline{\eta }}_{0}^{T},\left(\begin{array}{c}\\ \left[\stackrel{˜}{F}\left(z,w,\mu \right),{\beta }_{0}^{\left(2\right)}\right]\end{array}\right)〉{\overline{\xi }}_{0}\right){\gamma }_{0}\left(x\right)\\ & -〈{\overline{\eta }}_{{n}_{\ast }}^{T},\left(\begin{array}{c}\\ \left[\stackrel{˜}{F}\left(z,w,\mu \right),{\beta }_{{n}_{\ast }}^{\left(2\right)}\right]\end{array}\right)〉{\overline{\xi }}_{{n}_{\ast }}{\gamma }_{{n}_{\ast }}\left(x\right).\end{array}$
(36)
By a recursive transformation, we can obtain that the normal forms for Turing–Hopf bifurcation are
$\stackrel{˙}{z}=Bz+\left(\begin{array}{c}{B}_{11}{\mu }_{1}{z}_{1}+{B}_{22}{\mu }_{2}{z}_{1}\\ {\overline{B}}_{11}{\mu }_{1}{z}_{1}+{\overline{B}}_{22}{\mu }_{2}{z}_{1}\\ {B}_{13}{\mu }_{1}{z}_{3}+{B}_{23}{\mu }_{2}{z}_{3}\end{array}\right)+\left(\begin{array}{c}{B}_{210}{z}_{1}^{2}{z}_{2}+{B}_{102}{z}_{1}{z}_{3}^{2}\\ {\overline{B}}_{210}{z}_{1}^{2}{z}_{2}+{\overline{B}}_{102}{z}_{1}{z}_{3}^{2}\\ {B}_{111}{z}_{1}{z}_{2}{z}_{3}+{B}_{003}{z}_{3}^{3}\end{array}\right)+O\left(|z||\mu {|}^{2}\right),$
(37)
where
\begin{aligned} &B_{11}=\eta _{0}^{T}L_{1}^{(1,0)}( \xi _{0})=\frac{i\omega _{c}+\alpha _{2*}}{2\alpha _{1*}}, \\ &B_{21}=\eta _{0}^{T}L_{1}^{(0,1)}( \xi _{0})=- \frac{1}{2}-\frac{i\alpha _{1*}r(u_{*})}{2\omega _{c}}+\frac{i\alpha _{1*}g(u_{*})}{2b}, \\ &B_{13}=\eta _{n_{*}}^{T}L_{1}^{(1,0)}( \xi _{n _{*}})=\frac{d_{1}d_{2} (\frac{n_{*}}{l} )^{2}+d_{1}\alpha _{2*}}{ \alpha _{1*}(d_{1}+d_{2}}, \\ &B_{23}=\eta _{n_{*}}^{T}L_{1}^{(0,1)}( \xi _{n_{*}})=-\frac{\alpha _{1*}g(u_{*})}{b(d_{1}+d_{2}) (\frac{n _{*}}{l} )^{2}}+\frac{\alpha _{1*}r(u_{*})}{(d_{1}+d_{2}) (\frac{n _{*}}{l} )^{2}}- \frac{d_{1}}{d_{1}+d_{2}}, \end{aligned}
and
\begin{aligned} &B_{210}=C_{210}+ \frac{3}{2}(D_{210}+E_{210}),\qquad B_{102}=C_{102}+ \frac{3}{2}(D_{102}+E_{102}), \\ &B_{111}=C_{111}+ \frac{3}{2}(D_{111}+E_{111}),\qquad B_{003}=C_{003}+\frac{3}{2}(D_{003}+E_{003}). \end{aligned}
Moreover, we can obtain
\begin{aligned} &A_{200}=F_{20}+2\xi _{02}F_{11}+\xi _{02}^{2}F_{02}= \bar{A}_{020}, \\ &A _{002}=F_{20}+2\xi _{n_{*}2}F_{11}+\xi _{n_{*}2}^{2}F_{02}, \\ &A_{110}=2\bigl(F _{20}+2\operatorname{Re}(\xi _{02})F_{11}+ \vert \xi _{02} \vert ^{2}F_{02}\bigr), \\ &A_{101}=2\bigl(F _{20}+(\xi _{02}+\xi _{n_{*}2})F_{11}+\xi _{02}\xi _{n_{*}2}F_{02} \bigr)= \bar{A}_{011}, \\ &A_{210}=3\bigl(F_{30}+F_{03} \vert \xi _{02} \vert ^{2}\xi _{02}+(2\xi _{02}+\bar{\xi } _{02})F_{21}+\bigl(\xi _{02}^{2}+2 \vert \xi _{02} \vert ^{2}\bigr)F_{12}\bigr), \\ &A_{102}=3\bigl(F _{30}+F_{03} \vert \xi _{n_{*}2} \vert ^{2}\xi _{02}+(\xi _{02}+2 \xi _{n_{*}2})F_{21}+\bigl( \xi _{n_{*}2}^{2}+2\xi _{02}\xi _{n_{*}2}\bigr)F_{12}\bigr), \\ &A_{111}=6\bigl(F_{30}+F _{03} \vert \xi _{02} \vert ^{2}\xi _{n_{*}2}+(\xi _{n_{*}2}+2 \operatorname{Re}\xi _{02})F _{21}+\bigl( \vert \xi _{02} \vert ^{2}+2\xi _{n_{*}2}\operatorname{Re}\xi _{02} \bigr)F_{12}\bigr), \\ &A _{003}=F_{30}+F_{03}\xi _{n_{*}2}^{3}+3 \bigl(\xi _{n_{*}2}F_{21}+\xi _{n_{*}2} ^{2}F_{12} \bigr), \end{aligned}
and
\begin{aligned} &h_{0200}=\frac{1}{\sqrt{l\pi }}\bigl( \mathcal{M}_{0}(2i\omega _{c})\bigr)^{-1}\bigl(A _{200}-\bigl(\eta _{0}^{T}A_{200}\xi _{0}+\bar{\eta }_{0}^{T}A_{200}\bar{ \xi }_{0}\bigr)\bigr), \\ &h_{0020}=\frac{1}{\sqrt{l\pi }}\bigl(\mathcal{M}_{0}(-2i \omega _{c})\bigr)^{-1}\bigl(A_{020}-\bigl(\eta _{0}^{T}A_{020}\xi _{0}+\bar{\eta }_{0} ^{T}A_{020}\bar{\xi }_{0}\bigr) \bigr), \\ &h_{0002}=\frac{1}{\sqrt{l\pi }}\bigl( \mathcal{M}_{0}(0) \bigr)^{-1}\bigl(A_{002}-\bigl(\eta _{0}^{T}A_{002} \xi _{0}+\bar{ \eta }_{0}^{T}A_{002}\bar{ \xi }_{0}\bigr)\bigr), \\ &h_{0110}=\frac{1}{\sqrt{l \pi }}\bigl(\mathcal{M}_{0}(0) \bigr)^{-1}\bigl(A_{110}-\bigl(\eta _{0}^{T}A_{110} \xi _{0}+\bar{ \eta }_{0}^{T}A_{110}\bar{ \xi }_{0}\bigr)\bigr), \\ &h_{n_{*}101}=\frac{1}{\sqrt{l \pi }}\bigl(\mathcal{M}_{n_{*}}(i\omega _{c})\bigr)^{-1}\bigl(A_{101}-\eta _{n_{*}} ^{T}A_{101}\xi _{n_{*}}\bigr), \\ &h_{n_{*}011}=\frac{1}{\sqrt{l\pi }}\bigl( \mathcal{M}_{n_{*}}(-i\omega _{c})\bigr)^{-1}\bigl(A_{011}-\eta _{n_{*}}^{T}A_{011} \xi _{n_{*}}\bigr), \\ &h_{(2n_{*})002}=\frac{1}{\sqrt{2l\pi }}\bigl( \mathcal{M}_{2n_{*}}(0) \bigr)^{-1}A_{002}, \\ &h_{(2n_{*})110}=(0, 0)^{T}. \end{aligned}
From [26], we have the following expressions of $$C_{ijk}$$, $$D_{ijk}$$, and $$E_{ijk}$$:
\begin{aligned} &C_{210}=\frac{1}{6l\pi }\eta _{0}^{T}A_{210},\qquad C_{102}= \frac{1}{6l \pi }\eta _{0}^{T}A_{102}, \\ &C_{111}=\frac{1}{6l\pi }\eta _{n_{*}}^{T}A _{111},\qquad C_{003}=\frac{1}{4l\pi }\eta _{n_{*}}^{T}A_{003}, \\ &D_{210}=\frac{1}{6l\pi \omega _{c}i}\biggl(-\bigl(\eta _{0}^{T}A_{200}\bigr) \bigl(\eta _{0} ^{T}A_{110}\bigr)+ \bigl\vert \eta _{0}^{T}A_{110} \bigr\vert ^{2}+\frac{2}{3} \bigl\vert \eta _{0}^{T}A_{020} \bigr\vert ^{2}\biggr), \\ &D_{102}=\frac{1}{6l\pi \omega _{c}i}\bigl(-2\bigl(\eta _{0}^{T}A_{200} \bigr) \bigl(\eta _{0}^{T}A_{002}\bigr)+\bigl(\eta _{0}^{T}A_{110}\bigr) \bigl(\bar{\eta }_{0}^{T}A_{002}\bigr)+2\bigl( \eta _{0}^{T}A_{002}\bigr) \bigl(\eta _{n_{*}}^{T}A_{101}\bigr)\bigr), \\ &D_{111}=-\frac{1}{3l \pi \omega _{c}}\operatorname{Im}\bigl(\bigl(\eta _{n_{*}}^{T}A_{101}\bigr) \bigl(\eta _{0}^{T}A_{110}\bigr)\bigr), \\ &D_{003}=-\frac{1}{3l\pi \omega _{c}}\operatorname{Im}\bigl(\bigl(\eta _{n_{*}}^{T}A _{101}\bigr) \bigl(\eta _{0}^{T}A_{002}\bigr)\bigr), \\ &E_{210}=\frac{1}{3\sqrt{l\pi }}\eta _{0}^{T}\bigl((F_{20}+\xi _{02}F_{11})h _{0110}^{(1)}+(\xi _{02}F_{02}+F_{11})h_{0110}^{(2)} \\ &\phantom{E_{210}=}{} +(F_{20}+\bar{\xi }_{02}F_{11})h_{0200}^{(1)}+(F_{11}+ \bar{\xi }_{02}F _{02})h_{0200}^{(2)}\bigr), \\ &E_{102}=\frac{1}{3\sqrt{l\pi }}\eta _{0} ^{T} \bigl((F_{20}+\xi _{02}F_{11})h_{0002}^{(1)}+( \xi _{02}F_{02}+F_{11})h _{0002}^{(2)} \\ &\phantom{E_{102}=}{} +(F_{20}+\xi _{n_{*}2}F_{11})h_{n_{*}101}^{(1)}+(F_{11}+ \xi _{n_{*}2}F _{02})h_{n_{*}101}^{(2)}\bigr), \\ &E_{111}=\frac{1}{3\sqrt{l\pi }} \eta _{n_{*}}^{T} \bigl((F_{20}+\xi _{02}F_{11})h_{n_{*}011}^{(1)}+( \xi _{02}F _{02}+F_{11})h_{n_{*}011}^{(2)} \\ &\phantom{E_{111}=}{} +(F_{20}+\bar{\xi }_{02}F_{11})h_{n_{*}101}^{(1)}+(F_{11}+ \bar{\xi } _{02}F_{02})h_{n_{*}101}^{(2)}\bigr)) \\ &\phantom{E_{111}=}{} +\eta _{n_{*}}^{T} \biggl((F_{20}+\xi _{n_{*}2}F_{11}) \biggl(\frac{1}{3\sqrt{l \pi }}h_{0110}^{(1)}+ \frac{1}{3\sqrt{2l\pi }}h_{(2n_{*})110}^{(1)} \biggr) \\ &\phantom{E_{111}=}{} +(F_{11}+\xi _{n_{*}2}F_{02}) \biggl( \frac{1}{3\sqrt{l\pi }}h_{0110} ^{(2)}+\frac{1}{3\sqrt{2l\pi }}h_{(2n_{*})110}^{(2)} \biggr) \biggr), \\ &E_{003}=\eta _{n_{*}}^{T} \biggl((F_{20}+ \xi _{n_{*}2}F_{11}) \biggl(\frac{1}{3\sqrt{l \pi }}h_{0002}^{(1)}+ \frac{1}{3\sqrt{2l\pi }}h_{(2n_{*})002}^{(1)} \biggr) \\ &\phantom{E_{003}=}{} +(F_{11}+\xi _{n_{*}2}F_{02}) \biggl( \frac{1}{3\sqrt{l\pi }}h_{0002} ^{(2)}+\frac{1}{3\sqrt{2l\pi }}h_{(2n_{*})002}^{(2)} \biggr) \biggr). \end{aligned}
Let $$z_{1}=v_{1}-iv_{2}$$, $$z_{2}=v_{1}+iv_{2}$$, $$z_{3}=v_{3}$$, then the normal forms Eq. (37) can be written in real coordinates v. Moreover, let $$v_{1}=\varkappa \cos \varTheta$$, $$v_{2}=\varkappa \sin \varTheta$$, $$v_{3}=\vartheta$$, we can rewrite the normal forms from real coordinates to cylindrical coordinates. Finally, removing the azimuthal term and truncating at third order terms, the normal form is as follows:
\begin{aligned} &\dot{\varkappa }=\alpha _{1}(\mu )\varkappa +\kappa _{11}\varkappa ^{3}+ \kappa _{12}\varkappa \vartheta ^{2}, \\ &\dot{\vartheta }=\alpha _{2}( \mu )\vartheta +\kappa _{21} \varkappa ^{2}\vartheta +\kappa _{22} \vartheta ^{3}, \end{aligned}
(38)
where
\begin{aligned} &\alpha _{1}(\mu )=\operatorname{Re}(B_{11}) \mu _{1}+\operatorname{Re}(B_{21})\mu _{2},\qquad \alpha _{2}(\mu )=B_{13}\mu _{1}+B_{23}\mu _{2}, \\ &\kappa _{11}= \operatorname{Re}(B_{210}), \qquad \kappa _{12}= \operatorname{Re}(B_{102}),\qquad \kappa _{21}=B_{111}, \kappa _{22}=B_{003}. \end{aligned}
It is easy to know that truncated normal form (38) is equivalent to the four-dimensional smooth ODE system with Hopf–Hopf bifurcation in [30]. If $$\kappa _{11}\kappa _{22}>0$$ holds, we know that the dynamics of system (23) is topologically equal to normal forms (38) at the neighborhood of the bifurcation point.
## 5 Numerical simulations
In this section, we verify the theoretical analysis by numerical simulations. Let $$\varOmega =(0,\pi )$$ and choose the following parameters:
\begin{aligned} &d_{1}=0.02,\qquad d_{2}=0.35, \qquad b=0.35,\qquad \delta =0.6,\\ & m_{1}=0.5, \qquad m_{2}=0.3,\qquad h=0.3. \end{aligned}
Then one can check that system (23) has a positive steady state $$E_{4}(0.6115, 2.6044)$$ and $$r(u_{4})=0.0775>0$$, $$\frac{g(u_{4})}{b}-r(u _{4})=0.0976>0$$, $$\varLambda =0.00044>0$$. From Theorem 3.4(iii), we know that system (23) undergoes Turing–Hopf bifurcation at $$P^{*}(\mathcal{A}_{1}^{-}, r(u_{4})\mathcal{A}_{1}^{-})=(0.1388, 0.0705)$$ (see Fig. 1(a)) and
$$\mathcal{D}_{0}(\alpha _{1}): \alpha _{2}=0.0775 \alpha _{1};\qquad \mathcal{R}_{1}(\alpha _{1}): \alpha _{2}=\frac{0.0271\alpha _{1}-0.007}{0.02+0.0976 \alpha _{1}}.$$
Setting $$n_{*}=1$$, the normal forms truncated to the third order terms are
\begin{aligned} &\dot{\varkappa }=(0.0387\mu _{1}-0.5\mu _{2})\varkappa -0.0028\varkappa ^{3}-0.0092\varkappa \vartheta ^{2}, \\ &\dot{\vartheta }=(0.0671\mu _{1}-0.1333\mu _{2})\vartheta -0.0081\varkappa ^{2}\vartheta -0.0421 \vartheta ^{3}. \end{aligned}
(39)
In the parameters plane $$(\alpha _{1},\alpha _{2})$$, the dynamics of the original system (23) can be equivalent to normal forms system (39) near the Turing–Hopf bifurcation point $$P^{*}$$. And we have that $$\varkappa \geq 0$$ and ϑ is a real number.
There exist a zero equilibrium $$p_{0}$$ (for all $$\mu _{1}$$ and $$\mu _{2}$$), three trivial equilibria $$p_{1},p_{2}^{\pm }$$, and two nontrivial equilibria $$p_{3}^{\pm }$$ for system (39):
\begin{aligned} &p_{0}=(0, 0), \\ &p_{1}=(\sqrt{13.9353\mu _{1}-179.8956\mu _{2}}, 0),\quad\text{for } \mu _{2}< 0.0775\mu _{1}, \\ &p_{2}^{\pm }=(0, \pm \sqrt{1.5948\mu _{1}-3.1672\mu _{2}}), \quad\text{for } \mu _{2}< 0.5035\mu _{1}, \\ &p_{3}^{\pm }=(\sqrt{23.839\mu _{1}-467.9284\mu _{2}}, \pm \sqrt{-2.976 \mu _{1}+86.5523\mu _{2}}) \end{aligned}
for $$\mu _{2}<0.0509\mu _{1}$$ and $$\mu _{2}<0.0344\mu _{1}$$.
The critical bifurcation lines are the following:
\begin{aligned} &\mathcal{D}_{0}: \mu _{2}=0.0775 \mu _{1};\qquad \mathcal{R}_{1}: \mu _{2}=0.5035\mu _{1}; \\ &T_{1}: \mu _{2}=0.0509\mu _{1},\quad\mu _{2}>0;\qquad T_{2}: \mu _{2}=0.0344\mu _{1},\quad \mu _{2}>0. \end{aligned}
Then we can draw the bifurcation diagram in the $$(\mu _{1}, \mu _{2})$$ parameter space (see Fig. 2). There are four straight lines $$\mathcal{D}_{0}$$, $$\mathcal{R}_{1}$$, $$T_{1}$$, $$T_{2}$$, and the $$(\mu _{1}, \mu _{2})$$ plane is divided into six regions marked as .
In region , system (39) has an asymptotically stable equilibrium $$p_{0}$$. So system (23) has an asymptotically stable constant steady state (see Fig. 3). In region , system (39) has an unstable equilibrium $$p_{0}$$ and two asymptotically stable equilibria $$p_{2}^{\pm }$$. Thus, system (23) has two stable nonconstant steady states (see Fig. 4). In region , system (39) has two unstable equilibria $$p_{0},p_{1}$$ and two asymptotically stable equilibria $$p_{2}^{\pm }$$. Hence, system (23) has two stable nonconstant steady states (see Fig. 5). In region , system (39) has four unstable equilibria $$p_{0}$$, $$p_{1}$$, $$p_{2}^{\pm }$$ and two asymptotically stable equilibria $$p_{3}^{\pm }$$. It follows that system (23) has two stable spatially inhomogeneous periodic solutions (see Fig. 6). In region , system (39) has three unstable equilibria $$p_{0}$$, $$p_{2}^{\pm }$$ and an asymptotically stable equilibrium $$p_{1}$$. Therefore, system (23) has a stable spatially homogeneous periodic solution (see Fig. 7). In region , system (39) has an unstable equilibrium $$p_{0}$$ and an asymptotically stable equilibrium $$p_{1}$$, which implies that system (23) has a stable spatially homogeneous periodic solution (see Fig. 8).
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Shi, Q., Liu, M. & Xu, X. Turing–Hopf bifurcation of a ratio-dependent predator-prey model with diffusion. Adv Differ Equ 2019, 310 (2019). https://doi.org/10.1186/s13662-019-2123-3
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0
How many hours in 74 years?
Updated: 9/23/2023
Wiki User
11y ago
To calculate the answer: There are 365 days in a year, and 24 hours in a day. So you would multiply 365 x 24 which = 8760 hours in one year. So to calculate how many hours are in 74 years, you then multiply the number of hours in one year (8760) by 74, which is 648,240 hours.
But if you take into account that leap year comes every four years, which means an extra day (and an extra 24 hours), you need to select a starting date for when the 74 year time period starts, then determine which of those years are leap year and add an extra 24 hours for each leap year, then calculate from there.
Wiki User
11y ago
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# math
posted by .
simplify -2(-6y+2) +4y(4z+5y)
• math -
12y -4 + 16yz + 20y^2 = 4(5y^2 + 3y + 4yz - 1)
I don't know if that is simpler. Do you have a typo?
• math -
6o
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posted by .
if we need 5 oranges to make 3 small glasses of orange juice, how many small glasses of juice can we make with 100 oranges?
please explain how to get the answer in terms i can understand
• 5th grade word problems -
60
• 5th grade word problems -
Please see the Related Questions below.
• 5th grade word problems -
5 3
-- * --
100 x
5x=300
x=60
1. put 5 over 100, oranges over oranges. And 3 over X. glasses over glasses
2. cross multiply, 5*x and 100*3
3. divide both sides of the equation by 5 to get x by it's self.
4. X=60, you can make 60 glasses with 100 oranges
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if we need 5 oranges to make 3 small glasses of orange juice, how many small glasses of juice can we make with 100 oranges?
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https://benediktehinger.de/blog/science/ica-weights-and-invweights/?replytocom=713
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# ICA weights and invweights
## Update 01.08.2016
after further discussion we discovered that the correlation matrix was a bit off. This is because I plotted the correlation of the weights without the sphering (=whitening) step. Sphering removes the correlation between the components. I added a plot with the correlation after sphering and updated the topographies.
## Goal
In EEG blind source separation using ICA, understand the difference between the mixing matrix and the unmixing matrix.
## brief introduction
The blind source separation problem in the context of EEG can be understood as follows:
$$mixing * sourceactivity = EEGdata$$
In words: We have some sources active and we know their projections on each electrode: We take the dot product (we multiply each source with its projection and add them all up).
More formally:
\begin{align} A &: mixing\\ s &: sourceactivity\\ x &: EEGdata \\ A * s &= x\\ \end{align}
Because every neuronal source projects to most (or even all) sensors/electrodes, the original brain-source activation can only be recored in a mixed way.
Thus in order to unmix we can use blind source separation (ICA details can be found here)
\begin{align} W &: unmixing \\ unmixing * EEG.data &= sourceactivity \\ W * x &= s \\ \end{align}
The relation between mixing and unmixing matrix in ICA (and other BSS methods) is:
\begin{align} A * s & = x\\ A^{-1} * A * s & = A^{-1} * x\\ s & = A^{-1} * x\\ A^{-1}& = W\\ \end{align}
This is only true if $A$ is invertible and square (if non-square you have to use a pseudoinverse, see the Haufe 2014 paper for more information)
The matrix $W$ is also known as the extraction filter
The matrix $A$ is also know as the activity pattern
## EEG ICAs
In eeglab the default infomax-ICA algorithm uses a sphering (whitening) step before calculating the weights. Thus the following relation is true:
[code lang=text] EEG.icawinv = pinv( EEG.icaweights * EEG.icasphere) % the mixing matrix, channels x sources
EEG.icaact = EEG.icaweights*EEG.icasphere)*EEG.data % the activation, sources x time
[/code]
This is how ICA components look, on the left the weights after sphering-preprocessing, middle the unmixing matrix and on the right the mixing matrix. Usually we should look at the mixing matrix ($A = W^{-1}$).
The weights after decorrelation (=sphering) seem to be quite similar, but the unmixing ($W$) looks very different and is actually very hard to interprete!
One interesting thing to look at is the correlation between the component-vectors (in this case 32 components). As we can see here:
Note that the correlation of the EEG.icaweight matrix is 0 because it is a pure rotation matrix (which is orthognal, thus correlations of 0). As explained here the whitening step estimates the stretching and thus only a rotation remains for the EEG.icaweight matrix.
Haufe et. al. 2014(pdf-openaccess-link) have a very nice paper around this issue.
TLDR; Don’t look at your filter/unmixing matrix.
### Thanks to
Holger Finger, Anna Lisa Gert, Tim Kietzmann,Peter König, Andrew Melnik ( in alphabetical order) for discussions.
01-2023: Thanks to Fabiano Baroni for noticing I had $$W$$ & $$A$$ switched up throughout the blog post!!
Categorized: Blog
Tagged:
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# Strayer
An oil-drilling company knows that it costs \$25,000 to sink a test well. If oil is hit, the income for the drilling company will be \$445,000. If only natural gas is hit, the income will be \$165,000. If nothing is hit, there will be no income. If the probability of hitting oil is 1/40 and if the probability of hitting gas is 1/20, what is the expectation for the drilling company
1. 0
2. 0
3. 8
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7. ### Probability
An oil-drilling company knows that it costs \$25,000 to sink a test well. If oil is hit, the income for the drilling company will be \$375,000. If only natural gas is hit, the income will be \$130,000. If nothing is hit, there will
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An oil-drilling company knows that it costs \$25,000 to sink a test well. If oil is hit, the income for the drilling company will be \$355,000. If only natural gas is hit, the income will be \$130,000. If nothing is hit, there will
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得獎人獲獎介紹 (只提供英文版)
For more than two decades David Donoho has been a leading figure in mathematical statistics. His introduction of novel mathematical tools and ideas has helped shape both the theoretical and applied sides of modern statistics. His work is characterized by the development of fast computational algorithms together with rigorous mathematical analysis for a wide range of statistical and engineering problems.
A central problem in statistics is to devise optimal and efficient methods for estimating (possibly non-smooth) functions based on observed data which has been polluted by (often unknown) noise. Optimality here means that, as the sample size increases, the error in the estimation should decrease as fast as that for an optimal interpolation of the underlying function. The widely used least square regression method is known to be non-optimal for many classes of functions and noise that are encountered in important applications, for example non-smooth functions and non-Gaussian noise. Together with Iain Johnstone, Donoho developed provably almost optimal (that is, up to a factor of a power of the logarithm of the sample size) algorithms for function estimation in wavelet bases. Their “soft thresholding” algorithm is now one of the most widely used algorithms in statistical applications.
A key theme in Donoho’s research is the recognition and exploitation of the fundamental role of sparsity in function estimation from high dimensional noisy data. Sparsity here refers to a special property of functions that can be represented by only a small number of appropriately chosen basis vectors. One way to characterize such sparsity is to minimize the L0-norm of the coefficients in such representations. Unfortunately, the L0-norm is not convex and is highly non-smooth, making it difficult to develop fast algorithms for its computation. In addition to pioneering the exploitation of sparsity, Donoho also introduced the computational framework for using the L1-norm as a convexification of the L0-norm. This has led to an explosion of efficient computational algorithms realizing this sparsity framework which have been used effectively in a wide variety of applications, including image processing, medical imaging, data mining and data completion.
A recent and much celebrated development along this sparsity–L1 theme is Compressed Sensing (a term coined by Donoho). Data compression is widely used nowadays — for example the JPEG standard for compressing image data. Typically, the data is gathered from sensors (for example a camera) and the data is then compressed (that is represented by a much smaller number of coefficients in an appropriate basis, while preserving as much accuracy as possible). Corresponding de-compression algorithms are then used to recover the original data. The revolutionary idea in Compressed Sensing is to shortcut this standard approach and to “compress while sensing”, that is to collect a small number of appropriately chosen samples of the data, from which the original data can be recovered (provably exactly under appropriate assumptions) through corresponding de-compression algorithms. The key ingredients are again sparsity (most typical in a wavelet basis), use of L1-norm for recovery, and the use of random averaging in sensing. Along with Emmanuel Candes and Terence Tao, Donoho is widely credited as one of the pioneers of this exploding area of research, having contributed fundamental ideas, theoretical frameworks, efficient computational algorithms and novel applications. This is still a thriving area of research with wide applications, but already many stunning results have been obtained (both theoretical and practical).
Mathematical Sciences Selection Committee
The Shaw Prize
23 September 2013 Hong Kong
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# Uncertainty propagation and division
I am getting confused about a super basic issue.
I have two quantities: $6\pm4$ and $4\pm3$ .
So let's say $x = 6$, $\Delta x = 4$, $y = 4$, $\Delta y = 3$
Now I want to calculate the uncertainty of $z=\frac{x}{y}=1.5$.
If I just take the range of values given by the uncertainties I would get $z=\frac{x}{y}$ somewhere between $\frac{2}{7}\approx0.3$ and $\frac{10}{1}=10$.
Now if I propagate the errors according to $\frac{\Delta z}{z}=\sqrt{(\frac{\Delta x}{x})^2+(\frac{\Delta y}{y})^2}$ then I get $\frac{\Delta z}{z}\approx1.004$ which is an uncertainty of > 100%. I.e. a final result of $z\pm\Delta z\approx1.5\pm1.5$.
This confuses me because firstly, I only predict values up to a maximum of $\approx3$ whereas I predicted values up to $10$ when using the range of possible values. Secondly and more importantly, this uncertainty implies that the ratio of $z=\frac{x}{y}$ can be zero which is odd. Obviously, if I increase the uncertainties of $x$ and $y$ slightly, I can also generate negative nubers of the ratio.
Is there something I am doing wrong? Or a way to deal with this issue? Having zero or negative numbers just doesn't make physical sense at all in the context that I am using these numbers.
Thanks a lot
Your error propagation formula is missing a term. In case of division it should read as: $$\frac{\Delta z}{z}=\sqrt{(\frac{\Delta x}{x})^2+(\frac{\Delta y}{y})^2 - 2\frac{\Delta x}{x}\frac{\Delta y}{y}}.$$
In my opinion it is easier to work with the following Taylor series approximation: in general for $z=z(x,y,...) \longrightarrow \Delta z=\frac{\partial z}{\partial x}\Delta x + \frac{\partial z}{\partial y}\Delta y + \dotsb$
If $z =\frac{ x}{ y}$ then $\frac{\partial z}{\partial x} = \frac{1}{y}$ and $\frac{\partial z}{\partial y} = - \frac{x}{y^2}$, so $\Delta z$ simplifies to $\Delta z=\frac{1}{y}\Delta x - \frac{x}{y^2}\Delta y$.
If you now divide this by z: $$\frac{\Delta z}{z}= (\frac{1}{y}\Delta x - \frac{x}{y^2}\Delta y + \dotsb)\frac{y}{x}=\frac{\Delta x}{x} - \frac{\Delta y}{y}.$$
So your result should read $1.5 \pm 0.125$.
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http://mathhelpforum.com/trigonometry/59488-trig-word-prob-how-wide-fire-print.html
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# Trig Word Prob- how wide is the fire?
• Nov 13th 2008, 07:15 PM
jstrickland
Trig Word Prob- how wide is the fire?
hey just wondering if anyone can figure out the following problem...
A helicopter is dropping water on a forest fire from a height of 300 feet. If one side of the fire makes an angle depression of 40 degrees and the other makes an angle of depression of 85 degrees on the other side of the helicopter, how wide is the fire?
(Evilgrin)
Thanks in advance..its been troubling me
• Nov 13th 2008, 09:02 PM
Soroban
Hello, jstrickland!
Did you make a sketch?
Quote:
A helicopter is dropping water on a forest fire from a height of 300 feet.
If one side of the fire makes an angle depression of 40° and the other side
makes an angle of depression of 85° on the other side of the helicopter,
how wide is the fire?
Code:
H W - - - - - * - - - - - E 40° * |* 85° * | * * 50°|5°* * | * * 300| * * | * * 40° | 85°* A * - - - - - - - * - - - * B : d C e :
The helicopter is at $\displaystyle H\!:\;\;HC = 300$
$\displaystyle \begin{array}{ccc}\angle WHA = 40^o = \angle HAC & \Rightarrow & \angle AHC = 50^o \\ \angle EHB = 85^o = \angle HBC & \Rightarrow & \angle BHC = 5^o \end{array}$
Let $\displaystyle d = AC,\;e = CB$
In $\displaystyle \Delta HCA\!:\;\;\tan 50^o \:=\:\frac{d}{300} \quad\Rightarrow\quad d \:=\:300\tan50^o \:\approx\:357.5$ ft
In $\displaystyle \Delta HCB\!:\;\;\tan5^o \:=\:\frac{e}{300} \quad\Rightarrow\quad e \:=\:300\tan5^o \:\approx\:26.2$ ft
The width of the fire is: .$\displaystyle 357.5 + 26.2 \:=\:383.7$ ft
• Nov 15th 2008, 12:52 PM
jstrickland
thanks that helped alot...i never thought to draw a picture (Rofl)
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# Not known Facts About statistics homework help
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# solve an expression and plot
1 view (last 30 days)
Atom on 3 Jun 2013
Suppose, for each alpha, I have to find a delta which is the highest positive root of the equation A^2-4*B*beta=0. The expression of A, B are given in terms of alpha and delta. How to get that and how to plot alpha vs delta plot.
beta=0.85;gamma=.15;theta=0.1;
alpha=.303:.001:5
syms delta
A=-alpha*beta + alpha.*delta + gamma*beta + beta;
B=alpha.*theta*(beta -delta) + gamma*beta;
rr=solve(A^2-4*B*beta == 0)
plot(alpha,rr)
Walter Roberson on 3 Jun 2013
beta = 0.85; gamma = .15; theta = 0.1;
alpha = .303: .001: 5;
syms Alpha delta
A = -Alpha * beta + Alpha .* delta + gamma * beta + beta;
B = Alpha .* theta * (beta - delta) + gamma * beta;
rr = solve( A^2 - 4*B*beta, delta);
rrt = double(subs(rr, Alpha, alpha(1));
[maxrrt, idx] = max(rrt);
maxrr = rr(idx);
rrpos = double(subs(maxrr, Alpha, alpha));
plot(alpha, rrpos)
Azzi Abdelmalek on 3 Jun 2013
Edited: Azzi Abdelmalek on 3 Jun 2013
beta=0.85;
gamma=.15;
theta=0.1;
alpha=(.303:.001:5)'
syms delta
A=-alpha*beta + alpha*delta + gamma*beta + beta;
B=alpha*theta*(beta -delta) + gamma*beta;
for k=1:numel(A)
rr{k}=solve(A(k).^2-4*B(k)*beta == 0);
end
##### 2 CommentsShowHide 1 older comment
Atom on 4 Jun 2013
Thanks for your reply. delta may some times given one +ve, one -ve roots and 2 positive. I need the largest positive value of delta for each alpha. I also would like to plot(alpha, delta).
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# Filter() and Functional Programming: Optimizing Data Processing in Python
## Functional Programming
Python has emerged as a popular high-level programming language among programmers and developers worldwide. One of its unique built-in functions is the filter() function.
It is used to filter a sequence of elements to produce a new set of elements based on the specified criteria. It is an important function in functional programming, a programming paradigm that uses the application of functions to solve problems.
Python filter() function, explain functional programming and discuss the filtering problem in detail.
## Python filter()
The filter function is used in Python to filter a sequence of elements. It returns a new sequence of elements that satisfy a given condition.
The function takes two arguments, a function and an iterable. The function is a decision function that the filter() function uses to determine which elements are part of the new sequence.
The iterable argument refers to any sequence of elements that can be iterated over, such as lists, tuples, and strings. Syntax:
filter(function, iterable)
The function argument is a callable function that takes one argument and returns a Boolean value (True or False).
The filter function applies this function to each element in the iterable. Each element that returns True is retained in the new sequence, and elements that return False are discarded.
## Example:
Let’s say we have a list of integers, and we want to filter out the negative numbers from the list. We can use the filter() function to achieve this:
“`
numbers = [4, -2, 3, -5, 1, -7]
def is_positive(x):
return x > 0
filtered_numbers = list(filter(is_positive, numbers))
## print(filtered_numbers)
“`
In this example, we define the is_positive() function, which checks if an element is positive. We then pass this function and the list of integers to the filter() function.
The filter() function applies the is_positive() function to each element in the numbers list and returns a new list containing only the positive numbers. The output of the above code is:
“`
[4, 3, 1]
“`
## Functional Programming
Functional programming is a programming paradigm that is based on the concept of applying functions to solve problems. It is a declarative programming style where emphasis is on expressing a problem in terms of data transformations and not the logic or flow of control.
In functional programming, functions are treated as first-class objects, meaning they can be assigned to variables, passed as arguments to other functions, and returned as values from functions. Functional programming has several benefits, including:
* Reduced complexity: Programs written in a functional programming language are typically less complex than those written in imperative programming languages.
* Easier debugging and testing: Functional programs are easier to test and debug since functions are treated as separate entities, which can be tested independently. * Parallel processing: Functional programming languages are well-suited for parallel processing since functions can be executed concurrently without any side effects.
* Fewer bugs: Since functional programming emphasizes immutability, programs written in functional programming languages are less prone to bugs caused by mutable states.
## Filtering Problem
The filtering problem involves selecting a subset of elements from a set based on a specified condition. This problem occurs frequently in many programming applications, such as data analysis, image processing, and natural language processing.
Let’s take an example of filtering negative numbers from a list using a lambda function:
“`
numbers = [4, -2, 3, -5, 1, -7]
filtered_numbers = list(filter(lambda x: x > 0, numbers))
“`
## The output of the above code is the same as before:
“`
[4, 3, 1]
“`
In this example, we use a lambda function to pass as an argument to the filter function instead of defining a named function, is_positive().
## Conclusion
In conclusion, the
Python filter() function is an essential tool for filtering sequences of elements based on a given condition. It is a key function in functional programming, a programming paradigm that emphasizes the application of functions to solve problems.
The filtering problem is a common problem in programming, and the filter() function in Python provides a straightforward solution to the problem. By understanding how to use the
Python filter() function, programmers can build powerful programs that solve real-world problems.
Python’s filter() is a versatile built-in function that can be used to filter out elements from an iterable based on a user-defined function. It is a powerful tool in data processing and analysis that can help programmers optimize their code for efficiency and flexibility.
In this article, we will delve deeper into filter() to help you get started with its syntax and arguments, and explain its advantages over the use of for loops. We will also demonstrate how filter() can be used to filter even numbers, prime numbers, outliers, Python identifiers, and palindrome words from a list.
## Syntax and Arguments of filter()
The syntax of the filter() function is as follows:
“`
filter(function, iterable)
“`
The first argument, function, is a callable function that returns a Boolean value of either True or False. The function takes one input argument, which is an element of the iterable sequence.
The second argument, iterable, is the sequence of elements that is passed to the filter() function.
## Advantages of using filter() over for loop
Filter() is a powerful tool for processing data efficiently and effectively. Compared to the use of a for loop, filter() is much faster and more memory-efficient.
Filter() implements lazy evaluation, which means that it only returns elements as they are requested. As a result, filter() is efficient when working with large datasets as it does not load everything into memory all at once.
Using a for loop to filter out elements can consume a lot of memory since it requires creating a new data structure for the filtered elements.
## Filtering Iterables with filter()
Let us now explore some examples of how filter() can be used to filter specific elements from an iterable.
## Extracting Even Numbers from a List
Here’s an example of using filter() to extract even numbers from a list:
“`
numbers = [1, 3, 6, 7, 8, 10, 13]
even_numbers = list(filter(lambda x: x % 2 == 0, numbers))
## print(even_numbers)
“`
In the above example, we first define the numbers list and then pass it to the filter() function with a lambda function as an argument. The lambda function checks if an element is even and returns True or False.
“`
[6, 8, 10]
“`
## Finding Prime Numbers in a Given Range
Using filter(), it is also possible to filter out prime numbers from a range of numbers. Heres some code that does just that:
“`
def is_prime(n):
if n < 2:
return False
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
return False
return True
prime_numbers = list(filter(is_prime, range(1, 101)))
## print(prime_numbers)
“`
In this example, we define the is_prime() function, which takes a number and checks whether it is a prime number. We then pass this function along with a range of numbers to the filter() function.
The filter() function applies the is_prime() function to each element in the range of numbers and returns a new list containing only the prime numbers. The output of the above code is:
“`
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
“`
## Removing Outliers in a Sample using Mean and Standard Deviation
Another common task is removing outliers from a sample using the mean and standard deviation. Here’s an example that does that:
“`
data = [5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60]
def remove_outliers(d):
mean = sum(d) / len(d)
sd = (sum((x – mean) ** 2 for x in d) / len(d)) ** 0.5
return list(filter(lambda x: abs(x – mean) <= 2 * sd, d))
new_data = remove_outliers(data)
## print(new_data)
“`
In this example, we define the remove_outliers() function to remove the outliers in the dataset. We first calculate the mean and standard deviation of the data and then pass this information to a lambda function that checks whether each element lies within two standard deviations of the mean.
The filter() function applies this function to each element in the data set and returns a new list that only contains the elements that are not outliers. The output of the above code is:
“`
[5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60]
“`
## Validating Python Identifiers in a List
A Python identifier is a name used to identify a variable, function, or any other Python object. Heres some code that uses filter() to filter out invalid Python identifiers from a list:
“`
identifiers = [“x”, “y”, “for”, “__init__”, “def”, “class”, “1_z”]
def is_valid_identifier(name):
import re
regex = r’^[a-zA-Z_]w*\$’
return bool(re.match(regex, name))
valid_identifiers = list(filter(is_valid_identifier, identifiers))
## print(valid_identifiers)
“`
In this example, we define the is_valid_identifier() function that uses a regular expression to validate the Python identifier. Then, we pass this function along with a list of identifiers to the filter() function.
The filter() function applies the is_valid_identifier() function to each element in the list of identifiers and returns a new list containing only the valid Python identifiers. The output of the above code is:
“`
[‘x’, ‘y’, ‘__init__’]
“`
## Finding Palindrome Words in a List
A palindrome word is a word that reads the same backward as forward. Heres some code that leverages filter() to filter palindrome words from a list:
“`
words = [“radar”, “hello”, “deified”, “noon”, “level”]
def is_palindrome(word):
return word == word[::-1]
palindromes = list(filter(is_palindrome, words))
## print(palindromes)
“`
In this example, we define the is_palindrome() function that checks whether a word is a palindrome. Then, we pass this function along with a list of words to the filter() function.
The filter() function applies the is_palindrome() function to each element in the list of words and returns a new list containing only the palindrome words. The output of the above code is:
“`
“`
## Conclusion
In conclusion, filter() is a powerful built-in function in Python that can be used to filter out elements from an iterable based on a user-defined function. Filter() enables programmers to write efficient and memory-optimized code that is also flexible and scalable.
By leveraging filter(), programmers can filter even numbers, prime numbers, outliers, valid Python identifiers, and palindrome words from a list with ease, making it an indispensable tool in data processing and analysis. In the previous section, we explored the applications of the filter() function in filtering out elements from an iterable based on a user-defined function.
In this section, we will learn about combining filter() with other functional tools, including map() and reduce(). We will also explore the usage of filterfalse() and how it can be used to filter out elements that do not satisfy a specified condition.
## Using filter() and map() to Get Square of Even Numbers
Map() is another built-in function in Python used for transforming elements in an iterable. It applies a function to every element in an iterable and returns an iterable of the same length containing the transformed elements.
We can combine filter() with map() to obtain a new iterable order containing the transformed elements that fit the condition of the filter(). Here is an example of using filter() and map() together to get the square of even numbers:
“`
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
squares_of_even_nums = list(map(lambda x: x**2, filter(lambda x: x % 2 == 0, numbers)))
## print(squares_of_even_nums)
“`
In this example, we first define the numbers list with integers from 1 to 10. We then use the lambda functions to filter() out even numbers and map() to get the square of each filtered number.
Finally, we convert the output into a list and print it out.
## Using filter() and reduce() to Get Sum of Even Numbers
Reduce() is another useful built-in function in Python’s functional programming toolkit that takes an iterable and reduces it into a single value through a specified function. We can use filter() and reduce() together to obtain the sum of all even numbers in an iterable.
Here is an example of using filter() and reduce() together:
“`
## from functools import reduce
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
sum_of_even_nums = reduce(lambda a, b: a + b, filter(lambda x: x % 2 == 0, numbers))
## print(sum_of_even_nums)
“`
In this example, we first define the numbers list with integers from 1 to 10. We then use the lambda function to filter() out even numbers and reduce() to obtain the sum of these numbers.
Finally, we print out the result.
## Filtering Iterables with filterfalse()
The filterfalse() is another built-in function in Python that takes two arguments, a function and an iterable, and returns an iterator that produces the elements of the iterable for which the function returns False. Here is an example of filtering out odd numbers from a list using filterfalse():
“`
## from itertools import filterfalse
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
odd_numbers = list(filterfalse(lambda x: x % 2 == 0, numbers))
## print(odd_numbers)
“`
In this example, we first define a numbers list with integers from 1 to 10. We then use filterfalse() to filter out the odd numbers.
Since filterfalse() produces the elements of the iterable for which the function returns False, we use the lambda function to check if an element is even. Finally, we convert the output into a list and print it out.
## Filtering Out NaN Values from a List
Filterfalse() can also be used to filter out NaN (Not a Number) values from a list. Here’s an example of how this can be done:
“`
## from itertools import filterfalse
numbers = [1, 2, float(‘nan’), 4, 5, 6, float(‘nan’), 8, 9, 10]
filtered_numbers = list(filterfalse(isnan, numbers))
## print(filtered_numbers)
“`
In this example, we define the numbers list with integers, as well as two NaN values using float(‘nan’). We then use filterfalse() to filter out the NaN values.
Since filterfalse() produces the elements of the iterable for which the function returns False, we use the isnan() function from the math module to check if an element is not a NaN value. Finally, we convert the output into a list and print it out.
## Conclusion
In summary, combining filter() with other functional tools like map() and reduce() makes it possible to manipulate iterable data in powerful ways. filterfalse() is another built-in function in Python that can be used to filter out elements that do not satisfy a specified condition.
By understanding how to use these tools effectively, we can clean, filter, manipulate, and analyze data with ease, making it a crucial skill for any data scientist or programmer.
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# [Gmsh] Need help with boolean operations
Christophe Geuzaine cgeuzaine at ulg.ac.be
Fri Jun 9 12:04:55 CEST 2017
```I don't know why but the the opencascade "common" (intersection) algorithm does not behave as expected here; however the general "fragment" operator gives the expected result:
cl__1 = 1;
Point(1) = {0, 0, 0, 1};
Point(2) = {20, 0, 0, 1};
Point(3) = {0, 3, 0, 1};
Point(4) = {10, -4, 0, 1};
Line(1) = {3, 4};
Line(2) = {1, 2};
BooleanFragments{ Line{1}; }{ Line{2}; }
Delete{ Line{1,2}; }
Point (6) = {10, 10, 0, 1};
Line(7) = {5,6};
> On 8 Jun 2017, at 18:35, Sergio Logrosán <sergiologrosan at gmail.com> wrote:
>
> Dear service attendant,
>
> I am using your great GMSH software for my bachelor's degree thesis. I am writing to you because I have a question. I need to create points of elementary geometry from intersections of lines or other 2D elements. I have tested with Boolean operations without success. Attached you will find an example .geo file of what I mean.
>
> My intention in the example is to create a line (3) being one of its points the intersection between lines 1 and 2.
>
> Thank you very much for your time.
>
> I am looking forward to your response,
>
> Kind regards,
> Sergio Logrosán
>
>
>
>
> <untitled2.geo>_______________________________________________
> gmsh mailing list
> gmsh at onelab.info
> http://onelab.info/mailman/listinfo/gmsh
--
Prof. Christophe Geuzaine
University of Liege, Electrical Engineering and Computer Science
http://www.montefiore.ulg.ac.be/~geuzaine
Free software: http://gmsh.info | http://getdp.info | http://onelab.info
```
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https://learnbusinessconcepts.com/advantages-and-disadvantages-of-year-over-year-yoy-analysis/
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## Advantages (Pros / Positives / Benefits) of Year Over Year Analysis (YoY)
1. Helps to analyze the annual, quarterly as well as monthly growth trend based on financial need
As an example, Quarter 4 in any year (October, November, December) is a comparatively high performing quarter for most companies. Company management can do a YOY analysis to compare the current year’s Quarter 4 with the last year’s Quarter 4 to understand whether the company is growing (considering the seasonal factor), rather than comparing Quarter 4 with Quarters 3, 2, and 1 in the same year.
2. Calculates the growth trends by mitigating seasonality factors
As an example: Most companies expected to improve sales in the December month of any year due to the Christmas season. Comparing the sales data of the December month between years will be more beneficial for the investors rather comparing the yearly total. Investors can easily figure out the revenue trend and growth while considering the seasonal factor.
3. The YoY analysis method is a simple calculation. This does not require complex calculation methods
Simply the YoY can be calculated using one of the below approaches,
Approach 1: current year value divided by the previous year’s value, subtract the result from 1
Approach 2: current value minus the previous value and then divide the result by the previous value
5. YoY calculation results in a percentage value which easily comparable
As an example, if an investor does a YoY calculation of multiple investments, all the calculations return the percentages values, which the investor can easily compare and make decisions. Investors often rely on YoY analysis to gauge the financial health and growth prospects of a company. Consistent improvement in key metrics over multiple years can instill confidence and attract investment.
6. YoY is a popular and effective way to evaluate the financial statistics of a company and also evaluate the investment returns. Trend and growth are directly visible in the analysis
## Disadvantages (Cons / Negatives / Drawbacks) of Year Over Year Analysis (YoY)
1. Does not provide information on how to improve performance
2. Not practical for startup companies since no previous stats are available
This method requires past data to do the calculations and determine the growth percentage growth of the month in current year with the month in previous year. It can not show the percentage growth of same month in multiple years.
5. The YoY method can compare only two measured events at a single calculation
## How to Calculate Year Over Year (YoY) using Formula – with Examples
There are two approaches to calculating the Year-Over-Year (YoY),
Approach 1: Current year value divided by the previous year’s value, subtract the result from 1
Example (a): If a return on investment was \$10,000 in Year 2 and \$12,500 in Year 2, the YOY growth can be calculated as (\$12,500 / 10,000) – 1, resulting in 25%.
Example (b): Clothing company Year 1 December month sales were \$63,000 and Year 2 December month sales were \$78,000. The YOY growth can be calculated as (\$78,000 / \$63,000) – 1 = 23.8%.
Approach 2: Current value minus the previous value and then divide the result by the previous value
Example (a): if a return on investment was \$10,000 in Year 2 and \$12,500 in Year 1, the YOY growth can be calculated as (\$12,500 – \$10,000) / 10,000, resulting in 25%.
Example (b): Clothing company Year 1 December month sales were \$63,000 and Year 2 December month sales were \$78,000. The YOY growth can be calculated as (\$78,000 – \$63,000) / \$63,000 = 23.8%.
## Real-World Examples of Companies that have Effectively Leveraged YoY Analysis
• Example: Amazon is known for its data-driven approach to decision-making, and YoY analysis plays a significant role in its strategy. The company regularly analyzes YoY trends in sales, customer engagement, and operational metrics to identify growth opportunities, optimize pricing strategies, and enhance customer experience.
• Impact: By leveraging YoY analysis, Amazon has been able to make data-driven decisions that fuel its continuous growth and expansion into new markets. For example, it uses YoY data to refine its inventory management, improve product recommendations, and tailor its marketing efforts to specific customer segments.
• Reference: Amazon stated in their news release in aboutamazon.com that “AWS segment sales increased 13% year-over-year to \$90.8 billion“.
• Example: Netflix relies heavily on data analytics to drive content creation, user engagement, and subscriber growth. YoY analysis is integral to its strategy for evaluating the performance of original content, tracking subscriber churn rates, and identifying emerging trends in viewer preferences.
• Impact: By analyzing YoY trends in viewer engagement and subscription metrics, Netflix can make data-driven decisions about content investments, pricing strategies, and international expansion. For example, it uses YoY data to determine which original series or movies resonate most with viewers and prioritize investments accordingly.
• Reference: Netflix Research Analytics – research.netflix.com
• Example: As one of the world’s largest retailers, Walmart utilizes YoY analysis extensively to optimize its operations and drive sales growth. The company analyzes YoY trends in sales, inventory turnover, and customer behavior to identify opportunities for improving efficiency, enhancing product assortment, and tailoring promotions.
• Impact: By leveraging YoY analysis, Walmart has been able to make strategic decisions that drive revenue growth and improve customer satisfaction. For instance, it uses YoY data to forecast demand for seasonal products, adjust pricing strategies based on competitor trends, and optimize its supply chain operations to reduce costs.
• Reference: Walmart 2023 Annual Report
• Example: Google employs YoY analysis to evaluate the performance of its advertising platforms, measure user engagement across its various products and services, and track trends in online search behavior. By analyzing YoY data, Google can identify emerging market trends, refine its algorithms, and optimize ad targeting strategies.
• Impact: By leveraging YoY analysis, Google can make data-driven decisions that drive revenue growth and enhance user experience. For example, it uses YoY data to identify changes in user search behavior, adjust its algorithms to deliver more relevant search results, and optimize ad placements to maximize click-through rates.
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# M01 #20 - Contradiction between between the two statement?
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M01 #20 - Contradiction between between the two statement? [#permalink]
### Show Tags
20 Aug 2009, 20:37
If Jim saved a total of $90 in 3 weeks, how much did he save in week 2? 1. Jim's average savings for the first 2 weeks were$20
2. Jim's first week's savings were half of his savings in week 2 and a third of his savings in week 3
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient
OA is given as B. However, I think it St. 2 contradicts the information given in St.1.
The explanation given is:
Quote:
Suppose $$S_1$$ , $$S_2$$ , and $$S_3$$ are the amounts Jim saved for the first, second, and third week, respectively.
Statement (1) by itself is not sufficient. $$S_1+S_2=\40$$ . Although $$S_3$$ can be calculated, the value of $$S_2$$ cannot be calculated. Consider $$S_1=\20$$ and $$S_2=\20$$ or $$S_1=\5$$ and $$S_2=\35$$ .
Statement (2) by itself is sufficient. From Statement (2): $$S_2 =2 * S_1$$ and $$S_3 = 3 * S_1$$
$$S_1 + 2 * S_1 + 3 * S_1 = \90$$
$$6 * S_1 = \90$$
$$S_1 = \15$$
Solving for $$S_2$$ gives a value of $$S_2 = 2 * S_1 = 2 * 15 =\30$$ .
The problem is with St.2, $$S_1 + S_2 = 15 + 30 = 45$$. This is not 40, as per St. 1
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Re: M01 #20 - Contradiction between between the two statement? [#permalink]
### Show Tags
20 Aug 2009, 20:49
Agreed. The answer is clearly B.
Stmt 1 and Stmt 2 contradict each other. This, I remember reading somewhere, will never happen on real GMAT exam.
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Re: M01 #20 - Contradiction between between the two statement? [#permalink]
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21 Aug 2009, 14:14
are we sure ?
from what I heard is that only when the answer is D the two statements should result in the same answer.
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Re: M01 #20 - Contradiction between between the two statement? [#permalink]
### Show Tags
22 Aug 2009, 13:26
gmanjesh wrote:
are we sure ?
from what I heard is that only when the answer is D the two statements should result in the same answer.
No, that's wrong info. Neither of the two statements can contradict each other, irrelevant of what the answer is.
The idea is that the two statements are giving you some additional information, to figure out if the problem is solvable or not. They cannot, must not, add any more ambiguity to the problem.
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Re: M01 #20 - Contradiction between between the two statement? [#permalink]
### Show Tags
20 Oct 2009, 02:58
That's a really interesting thing to notice.
I have been preparing the gmat for 2 months and I was never noticed such a thing... I guess I just missed the information.
Thank you very much for this tip
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Re: M01 #20 - Contradiction between between the two statement? [#permalink]
### Show Tags
20 Oct 2009, 05:06
Sorry, I somehow missed this one. What if we change S1 to "Jim's average savings for the first 2 weeks were \$22.5"? We won't have a contradiction then. Will it work?
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Re: M01 #20 - Contradiction between between the two statement? [#permalink] 20 Oct 2009, 05:06
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# The Polygon Angle-Sum Theorems
The Polygon Angle-Sum Theorems. 3.4 The Polygon Angle-Sum Theorems. Polygon: a closed plane figure with at least three sides that are segments. Not a polygon; Not enclosed. Not a polygon; Two sides intersect. A polygon. Naming a Polygon. Name a polygon by its vertices. A.
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## The Polygon Angle-Sum Theorems
E N D
### Presentation Transcript
1. The Polygon Angle-Sum Theorems
2. 3.4 The Polygon Angle-Sum Theorems Polygon: a closed plane figure with at least three sides that are segments Not a polygon; Not enclosed Not a polygon; Two sides intersect A polygon
3. Naming a Polygon Name a polygon by its vertices. A ABCDE or AEDCB Start at one vertex and go around in order B E C D
4. Naming a Polygon Three polygons are pictured. Name each polygon: L P M O N
5. Classifying a Polygon by the number of sides:
6. Convex vs. Concave • A Convex Polygon has all vertices pointing “out” • A Concave Polygon has one or more vertices “caving in”
7. Classify • Classify each polygon by its sides. Identify each as convex or concave:
8. Sum of Polygon Angle Measures Use triangles to figure out the sum of the angles in each polygon: # of Sides: # of Triangles: Total Degrees: # of Sides: # of Triangles: Total Degrees:
9. Sum of Polygon Angle Measures
10. Theorem 3-9 Polygon Angle Sum Theorem The sum of the measures of the angles in a polygon is (n – 2)180. Find the sum of the measure of the angles of a 15-gon.
11. Polygon Angle Sum The sum of the measures of the angles of a given polygon is 720. How many sides does the polygon have? Use (n – 2)180 :
12. Using Polygon Angle-Sum Theorem Find the measure of <Y in pentagon TVYMR at the right. R T Use (n – 2)180 135° M 90° Y V Write an equation to solve for <Y
13. Using Polygon Angle-Sum Theorem Pentagon ABCDE has 5 congruent angles. Find the measure of each angle. Use the Polygon Angle-Sum Theorem: (n – 2)180 Divide the total number of degrees by the number of angles:
14. Exterior Angles What do you notice about each set of exterior angles? 80° 75° 115° 2 1 150° 99° 130° 71° 70° 88° 1: 86° 3 2: 3: 70° 46°
15. Theorem 3-10 Polygon Angle-Sum Theorem The sum of one set of exterior angles for any polygon is 360°. 1 5 2 4 3 m<1 + m<2 + m<3 + m<4 + m<5 = 360°
16. Polygons • Equilateral Polygon: all sides congruent • Equiangular Polygon: all angles congruent • Regular Polygon: all sides and all angles congruent (equiangular and equilateral) *If a polygon is a regular polygon then all of the exterior angles are also congruent.
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### Home > INT3 > Chapter 8 > Lesson 8.3.1 > Problem8-97
8-97.
Eniki has a sequence of numbers given by the formula $t\left(n\right) = 4\left(5\right)^{n}$.
1. What are the first three terms of Eniki’s sequence?
$t\left(1\right) = 4\left(5\right)^{1}$
$t\left(2\right) = 4\left(5\right)^{2}$
. . .
2. Toby thinks the number $312,500$ is a term in Eniki’s sequence. Is he correct? Justify your answer by either giving the term number or explaining why it is not in the sequence.
$4\left(5\right)^{n} = 312500$
$5^{n} = 78125$
How many factors of $5$ can go into $78125$?
3. Elisa thinks the number $94,500$ is a term in Eniki’s sequence. Is she correct? Explain.
Follow the steps from part (b).
No, because solving the equation $94500 = 4\left(5\right)^{n}$ does not result in a positive integer.
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## Numerator & Denominator
This is the place for queries that don't fit in any of the other categories.
### Numerator & Denominator
How can I go about grabbing the numerator and the denominator separately from a fraction.
I want to have an input which will be a fraction with polynomials in the numerator and in the denominator. I would like to factor both of the num and denom and then place them back into the fraction and have it print to the screen.
Getting into a fraction is easy, I can't seem to figure out how to take them apart.
Thanks for any help I am fairly new to Python.
Peglam
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### Re: Numerator & Denominator
We're happy to help, but we'd like to see your attempt, as well as the exact input and output you want.
You should also look into the fractions module.
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micseydel
Posts: 3000
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Location: Mountain View, CA
### Re: Numerator & Denominator
Thanks,
Well the input will be a transfer function G, where G = control.transferfunction([Num],[Den]). The Num or Den can change from 1st order to nth order.
I looked at fraction module but I can't find a way to use it to return the denominator or numerator of G.
One example of what I would like to do it this.
input
g = control.transferfuntion([1,4,4],[1,11,39,45]
print g returns (s^2 +4s+4) / ( s^3 + 11*s^2 + 39s +45)
For output
I want it to look like this g = (s+2)(s+2) / (s+3)(s+3)(s+5)
Peglam
Posts: 3
Joined: Wed Apr 24, 2013 8:45 pm
### Re: Numerator & Denominator
You should read this. It's much easier to read your post if you use code tags to denote any Python from math or anything else. And is g supposed to return a string? A special object?
What you're doing isn't just about fractions, it's about working with polynomials by using vectors to describe the coefficients. Numpy or scipy might have functionality for that, although the Fraction class in fractions is for straight numbers (although you can indeed get a numerator and denominator out of a Fraction object). Also, you're not just working with polynomials, it looks like you even want to factor them. I couldn't find factoring functions with a very quick Google search. There might be other modules for it, or you might me able to add the functionality on top of numpy.
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micseydel
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Location: Mountain View, CA
### Re: Numerator & Denominator
I would suggest you try to start with some more simpler problems, then grow from there. That is, if you haven't done so, yet, just forgot telling us about it.
For example how do you go from (s^2+4s+4) to (s+2)(s+2)? There is a formula for this, or you can test a few numbers. Worry about the fractions later.
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# Maurice's Extremal Real Algebraic Geometry Page
Fewnomial Theory is a beautiful invention of Askold Khovanski and Konstantin Sevastyanov (after 1979) that greatly extends the classical Descartes' Rule of Signs (from 1637). In particular, Khovanski's theory shows that systems of real polynomial equations with a fixed number of monomial terms have an absolute bound --- completely independent of the degrees of the underlying polynomials --- on the number of real isolated roots they can have.
More recently, an arithmetic analogue was found by Rojas (following earlier seminal work of Denef, van den Dries, Lipshitz, and Lenstra), and quantitative improvements to real fewnomial theory continue to blossom. We illustrate one such example below.
Consider a 2x2 system of the form (x^{2d}+ay^d-y,y^{2d}+bx^d-x). What is the maximum (finite) number of roots such a system of equations can have in the positive quadrant? This question is intimately related to the failure of Kushnirenko's Conjecture from real algebraic geometry (real fewnomial theory, to be precise).
In particular, a conjectured bound of Kushnirenko puts this maximum (for any (a,b,d)) at 4. Kushnirenko has stated that he knew his conjecture was wrong shortly after he formulated it, and a simple counter-example was found by a colleague of his in Moscow. However, no one appears to have recorded this counter-example nor the identity of its author.
Bertrand Haas, in 2000, found a simple counter-example with d=54 and (in our notation), a and b easily expressible algebraic numbers. In particular, Haas' counter-example showed that such systems could possess as many as 5 roots in the positive quadrant. Li, Rojas, and Wang then showed around 2001 that 5 is the maximum number.
More recently, in joint work of Dickenstein, Rojas, Rusek, and Shih (around July 2005), an even simpler counter-example was found: (a,b,d)=(44/31,44/31,3). (It is interesting to note that Haas found his counter-example by fixing some simple coefficients and then varying exponents by a clever ad hoc argument. Our counter-example was found by fixing the exponents and then varying the coefficients.) However, a deeper question is why such counter-examples are so hard to find.
This is partially answered by exploring the shape of discriminant complements over the real numbers. Fortunately for us, the discriminant corresponding to d=3 can be computed relatively easily via Maple. However, the discriminant for the family of systems containing Haas' counter-example was out of reach until recently: Thanks to a recent formula of Dickenstein, Feichtner, and Sturmfels, we can plot discriminants corresponding to high d with great ease.
Below are some pictures of the amoebae of the underlying discriminants. The blue curves were plotted with the DFS formula while the yellow/red plots were obtained by a direct calculation of the underlying (large) discriminant polynomial. Note that the plots are reflected in such a way that the signs of the real points of the discriminants are respected, and that the plots also have some ``clipping'' near the coordinate cross.
The corresponding paper will be ready shortly... JMR, 7/26/2005.
Parametrized versus sampled, for d=2, above.
Parametrized versus sampled, for d=3, above.
Parametrized (a ``sampled'' plot is infeasible with current technology) for d=54, above.
All the pictures above were drawn via Matlab and should all be taken with a grain of salt. Floating point error makes some of the plots dubious at high resolution but symbolic checks (with the assistance of Sturm sequences and Maple) help circumvent such difficulties. A more subtle issue is the existence of isolated components for the real discriminant locus but we will discuss this later.
In the mean time, it is interesting to note the richness of detail in our discriminant plots: For instance, the d=3 plots above appear to imply 7 discriminant chambers in the third quadrant. However, there are in fact over 19, as revealed in the magnified plots below.
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Q. 274.7( 3 Votes )
# <span lang="EN-US
Answer :
a) The circuit diagram to determine the characteristics of a p-n-p transistor is drawn below: -
b) The output I-V characteristics obtained by observing the variation of IC (collector current) as VCE is varied keeping IB (base current) constant are shown below. If VBE is raised by a small amount, the hole current from the emitter region raises. Also, the electron current from the base region raises. This leads to increase in both IB and IC proportionately.
If IB increases, IC also increases. The plot of IC versus VCE for different fixed values of IB gives one output characteristic. So, there will be different output characteristics corresponding to different values of IB as shown,
i) Determination of Output Resistance: -
Output resistance (r0) is defined as the ratio of change in
collector-emitter voltage (ΔVCE) to the change in collector current (ΔIC) at a constant base current IB. So, it can be written as,
i) Determination of Current Amplification Factor: -
Current Amplification factor (β) is defined as the ratio of the change in collector current (IC) to the change in base current (IB) at a constant collector-emitter voltage (VCE) when the transistor is in active state. So, mathematically,
βAC also known as small signal current gain and its value is found generally very large. The ratio of IC and IB gives the DC β of the transistor. Hence,
Since IC increases with IB almost linearly and IC = 0 when IB = 0, the values of both βDC and βAC are approximately equal.
OR
a) Photodiodes are preferably operated under reverse bias when the current in the forward bias is known to be more than that in reverse bias because the photodiode converts incident light to electric current efficiently in reverse bias condition than in forward bias. This is due to the expansion of the depletion region of the diode. Photons after absorption generate electron hole pairs. The pairs generated in and around the region only contribute the electric current. In the forward bias condition, there is a strong electric field there to separate the two different charge carriers, so the pairs outside the region rapidly recombine and vanish also the width of depletion region decreases as there is an increase in the applied voltage so the effective number of photons contributing to the electric current decreases. But in reversed bias condition, the width of depletion region increases as there is an increase in the applied voltage, so, a large number of incident photons get converted into electric current, consequently the efficiency increases.
b) i) On the basis of biasing
A photodiode converts incident light to electric current in reverse bias condition whereas a solar cell is basically a diode which generates emf when solar radiation falls on the p-n junction. It works on the same principle (photovoltaic effect) as the photodiode, except that no external bias is applied.
ii) On the basis of junction area
photodiode is a simple p-n junction diode which is operated in reversed bias condition whereas a solar cell is a p-n junction diode in which the junction area is kept much larger for solar radiation to be incident and is not biased.
iii) On the basis of I-V characteristics
The I – V characteristics of a photodiode is drawn same as of a p-n junction diode in reverse bias condition as shown below
The I – V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes as shown in the diagram.
This is because a solar cell does not draw current but supplies the same to the load.
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# KroneckerProduct with arbitrary number of Pauli matrices
I want define a function of n variables as follows $$f(x_1, ..., x_n)= KroneckerProduct[\sigma^{x_1}, ..., \sigma^{x_n}]$$ where $$\sigma^{x_i}=PauliMatrix[x_i]$$. Given a number $$n$$, I can explicitly write down an expression. For example, for $$n=3$$, I can write down
f[i1_, i2_, i3_] :=KroneckerProduct[PauliMatrix[i1], PauliMatrix[i2],PauliMatrix[i3]];
However, how to write down a generic expression in terms of arbitrary $$n$$, so that I don't have to rewrite the expression for every choice of $$n$$? Thanks!
f[i__] := KroneckerProduct @@ PauliMatrix[{i}]
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# How to Use the Perfect Square Identity As a Shortcut in Expansion
When multiplying binomials, you probably use the FOIL method. While useful, the FOIL method can be time-consuming and confusing. It is good to know, then, that when you are squaring a binomial, you can use the perfect square identity to expand the trinomial quickly. The basic formula is ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}}$. You can also use this formula to determine whether a trinomial is a perfect square, and to quickly factor those trinomials.
### Method 1 Method 1 of 3:Expanding a Perfect Square Download Article
1. 1
Determine whether you have a perfect square binomial. A binomial is a two-termed expression. If the binomial expression is a perfect square, it will be expressed as either ${\displaystyle (a+b)^{2}}$ or ${\displaystyle (a+b)(a+b)}$. Note the binomials could also have a subtraction symbol.
• For example, ${\displaystyle (5y+3)^{2}}$ is a perfect square binomial.
2. 2
Set up the formula for a perfect square trinomial. The formula is ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}}$. If the binomials show subtraction, the formula is ${\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}}$[1] . Note that ${\displaystyle a}$ is the first term of the binomial, and ${\displaystyle b}$ is the second term of the binomial.
3. 3
Square the first term of the binomial. This will become the first term of the trinomial. Remember that squaring a term means to multiply it by itself.
• For example, if you are expanding ${\displaystyle (5y+3)^{2}}$, you would first calculate ${\displaystyle (5y)^{2}=25y^{2}}$. So, ${\displaystyle 25y^{2}}$ is the first term of the trinomial.
4. 4
Multiply the first and last term. Make sure you are using the original ${\displaystyle a}$ and ${\displaystyle b}$ terms from the binomial expression.
• For example, if you are expanding ${\displaystyle (5y+3)^{2}}$, you would calculate ${\displaystyle (5y)(3)=15y}$.
5. 5
Multiply the product by 2. If the binomials show subtraction, you should multiply by -2. The result will be the middle term in the trinomial.
• For example, ${\displaystyle (2)(15y)=30y}$. So, your trinomial now looks like this: ${\displaystyle 25y^{2}+30y+b^{2}}$.
6. 6
Square the last term. Again, make sure you are using the original ${\displaystyle b}$ term from the binomial expression. The square will give you the last term in the trinomial.[2]
• For example, ${\displaystyle 3^{2}=9}$. So, ${\displaystyle (5y+3)^{2}=25y^{2}+30y+9}$
### Method 2 Method 2 of 3:Identifying a Perfect Square Trinomial Download Article
1. 1
Recall the formula for a perfect square trinomial. The formula is ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}}$. If the binomials show subtraction, the formula is ${\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}}$[3]
2. 2
Determine whether the first term in the trinomial is a perfect square. A perfect square is a number whose square root is a whole number.[4] Since the first term in the perfect square formula is ${\displaystyle a^{2}}$, the first term in your trinomial must be a perfect square.[5] Note that the square root of the first term is equal to ${\displaystyle a}$ in the squared binomial.
• For example, in the trinomial ${\displaystyle 36x^{2}-48x+16}$, the first term is ${\displaystyle 36x^{2}}$. The square root of ${\displaystyle 36x^{2}}$ is ${\displaystyle 6x}$. So, the first term of this trinomial is a perfect square. Also, in the squared binomial, ${\displaystyle a=6x}$.
3. 3
Determine whether the last term of the trinomial is a perfect square. Since the last term in the perfect square formula is ${\displaystyle b^{2}}$, the last term in your trinomial must be a perfect square.[6] Note that the square root of the last term is equal to ${\displaystyle b}$ in the squared binomial.
• For example, in the trinomial ${\displaystyle 36x^{2}-48x+16}$, the last term is ${\displaystyle 16}$. The square root of ${\displaystyle 16}$ is ${\displaystyle 4}$. So, the last term of this trinomial is a perfect square. Also, in the squared binomial, ${\displaystyle b=4}$
4. 4
Determine whether the middle term follows the formula or . That is, if you multiply the square roots of the first and last term of the trinomial, and then multiply that product by 2 or -2, the result will equal the middle term of the trinomial, if the trinomial is a perfect square.[7]
• For example, if ${\displaystyle a=6x}$ and ${\displaystyle b=4}$, then the middle term of the trinomial should follow the formula ${\displaystyle (-2)(6x)(4)}$. Since ${\displaystyle (-2)(6x)(4)=-48x}$ the middle term of the trinomial does follow the perfect square formula. Since the first and last terms of the trinomial followed the formula as well, you know that your trinomial is a perfect square.
1. 1
Expand the following expression. Use the perfect square identity rather than the FOIL method: ${\displaystyle (3y+7)^{2}}$.
• Set up the formula ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}}$ and plug in the ${\displaystyle a}$ and ${\displaystyle b}$ values: ${\displaystyle (3y+7)^{2}=(3y)^{2}+2(3y)(7)+7^{2}}$.
• Square the first term: ${\displaystyle 9y^{2}+2(3y)(7)+7^{2}}$.
• Multiply the first and last term, and multiply the product by 2: ${\displaystyle 9y^{2}+42y+7^{2}}$.
• Square the last term: ${\displaystyle 9y^{2}+42y+49}$.
2. 2
Consider the following trinomial. Determine whether it is a perfect square: ${\displaystyle 9y^{2}+36y+4}$.
• Recall the formula for a perfect square trinomial: ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}}$.
• Determine whether the first term of the trinomial is a perfect square: ${\displaystyle {\sqrt {9y^{2}}}=3y}$. So, ${\displaystyle a=3y}$.
• Determine whether the last term of the trinomial is a perfect square: ${\displaystyle {\sqrt {4}}=2}$. So, ${\displaystyle b=2}$.
• Determine whether the middle term of the trinomial follows the formula ${\displaystyle 2ab}$:
${\displaystyle 36y=(2)(3y)(2)}$
${\displaystyle 36y=12y}$
Since this is not true, the middle term does not follow the formula, and therefore the trinomial is not a perfect square.
3. 3
Factor the following trinomial. It factors into a squared binomial: ${\displaystyle 4x^{2}-20x+25}$.
• Since you know this factors into a squared binomial (${\displaystyle (a-b)^{2}}$), you know that it follows the perfect square formula.
• Find the ${\displaystyle a}$ term of the binomial, which is equal to the square root of the first term of the trinomial: ${\displaystyle {\sqrt {4x^{2}}}=2x}$.
• Find the ${\displaystyle b}$ of the binomial, which is equal to the square root of the last term of the trinomial: ${\displaystyle {\sqrt {25}}=5}$.
• Write the squared binomial. Since the second term of the trinomial is negative, you know that the second term of the binomial will also be negative: ${\displaystyle 4x^{2}-20x+25=(2x-5)^{2}}$
## Community Q&A
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• Question
Does the square root of 270 have a perfect square factor?
270 = 2 * 3^3 * 5. Therefore 3^2 is a factor of 270. This means the square root of 270 can be simplified to 3*sqrt(30).
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# How To Change Percent To Fraction
By Fauna Gagne at October 24 2018 03:26:57
The use of math worksheets can help solve numerous arithmetic problems. "Practice makes an individual perfect," is the best motto to be kept in mind while studying math. The motto will help a person to reinforce his desire to better himself in the subject. Without the help of these online resources, one will not be able to achieve the mastery of math. Reading is required in every day in life. Whether it is reading a street sign, reading the news, or reading a menu, comprehension is the fundamental reason for reading. Reading for comprehension is an essential common core skill that can and must be taught to students.
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Always be on the look out for tears of frustration, so as to all the child some break. No one learns well enough when confused. At this point both concentration and effective learning are lost. But if you are sure that the tears are merely excuses to skip the homework, pretend to abandon him with the worksheets and let him sweat it out! Reading is much more than fluency and word recognition. Reading fluently isn't enough. Students must fully understand the content they are reading. By learning and applying several different reading strategies, students will be able to obtain meaning from a wide variety of texts.
As you can imagine, this can be a lot of fun, and before you know it students can forget they are learning math! What is more, teachers can also easily vary the game play, for example, by using different types of math problems, or perhaps even by asking members of the class to solve each problem before moving on to the next bingo call. The data you include in an Excel file can be formatted and manipulated in a variety of ways. Once you have read this article, you will have a better understanding of the structure of an Excel file and the most common types of data you can use.
##### Related HD Pictures of How To Change Percent To Fraction
The basic skill you require in order to successfully assist your child with its 3rd grade math worksheets problems is to be able to identify the difficulty. Does it lack the requisite skills that it should have already possessed for the work at hand? If that is the case then it is best that you take the child back to the missing link, so that your it learns what it is missing and move forward. Graphic Organizers - Graphic organizers are visual diagrams to aid in organizing information. Others names many include maps, webs, graphs, charts, etc. These organizers assist students in determining main ideas and supporting details. They help students visually summarize a reading selection.
An Excel worksheet is electronic ledger created by the spreadsheet application Microsoft Excel. Worksheets are used to allow their users to store edit and manipulate data within rows and columns divided into cells. Structure of an Excel File : A Microsoft Excel file is called a "Workbook." Workbooks can be thought of like a physical 3-ring binder - they hold collections of individual pages called "Worksheets." Your Workbook will generally be a collection of sheets that all have a common theme, such as a sales report file containing individual sheets with sales figures for each region or department.
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# MATH2071: LAB #10: QR Factorizations
## Introduction
We know we can use the PLU factorization to solve a linear system, ... provided that the system is square, and that it is nonsingular, and that it is not too badly conditioned. And in fact, this is usually the case.
However, if we want to handle problems with a bad condition number, or that are singular, or even rectangular, we are going to need to come up with a different approach. In this lab, we will look at two versions of the QR factorization:
A = Q * R
where Q is an "orthogonal" matrix, and R is upper triangular.
We will see that this factorization can be used to solve the same problems that the PLU factorization handles, but can be extended to do several other tasks for which the PLU factorization is useless.
In situations where we are concerned about controlling roundoff error, the QR factorization can be more accurate than the PLU factorization, and is even a little easier to use when solving linear systems, since the inverse of Q is simply QT.
The QR factorization also gives us a way to solve rectangular linear systems, and is the key idea behind a method for reliably computing all the eigenvalues and eigenvectors of a matrix.
## Orthogonal Matrices
Definition: An orthogonal matrix is a real matrix whose inverse is equal to its transpose.
By convention, an orthogonal matrix is usually denoted by the symbol Q. The definition of an orthogonal matrix immediately implies that:
Q*QT=QT*Q=I
From the definition of an orthogonal matrix, and from the fact that the L2 vector norm of x can be defined by:
||x||2 = sqrt ( xT * x )
and the fact that
(A*x)T=xT*AT
you should be able to deduce an amazing fact about orthogonal matrices, namely:
||Q*x||2 = ||x||2
If I multiply x by Q and its L2 norm doesn't change, then Q*x must lie on the circle whose radius is x. In other words, the only thing that has happened is that Q*x has spun around the origin by some angle. That means that an orthogonal matrix represents a rotation or reflection.
Exercise: copy the file random_orthogonal.m from the web page. It computes a random orthogonal matrix of given dimension. Each time you call it, you should get a different matrix, so you can use this routine for experiments. Using your wits, and the experimental data, verify, contradict, or improve the following statements about a random orthogonal matrix Q:
1. the identity matrix is orthogonal; the matrix [0 1; 1 0] is orthogonal; the matrix [c s; -s c] is orthogonal, where c and s are the cosine and sine of some angle; any diagonal matrix is orthogonal.
2. an orthogonal matrix must be symmetric.
3. the product of two orthogonal matrices is also orthogonal.
4. the inverse of an orthogonal matrix is also orthogonal.
5. the L2 vector norm of the first column of Q must be 1.
6. the vectors formed by the first and last rows of Q must be perpendicular.
7. the vectors formed by the first row and the second column of Q must be perpendicular.
8. every entry of Q must be between -1 and 1.
9. the determinant of Q is always 1.
10. the eigenvalues of Q are always +1 or -1.
11. the singular values of Q are always +1 or -1.
12. the L1, L2 and Linf norms of Q are always 1.
13. If ||x||2=7, then ||Q*x||2 is _____
14. If A is a matrix, and ||A||2=7, then ||Q*A||2 is _____
15. the L2 condition number of Q is _____
16. in any column of Q, at most 1 entry can be 1.0.
17. in any column of Q, at most 1 entry can be 0.0.
See comments #1 at the end of the lab.
## A Set of Problems
Suppose we have N vectors xi, each of length M. Here are some common and related problems:
• Can we determine if the vectors are linearly independent?
• Can we determine the dimension of the space the vectors span?
• Can we construct an orthonormal basis for the space spanned by the vectors?
• Given a new vector xN+1, can we determine whether this vector lies in the space spanned by the other vectors?
• If a new vector lies in the space, what specific linear combination of the old vectors is equal to the new one?
• If we create an M by N matrix using the vectors as columns, what is the rank of the matrix?
• How can we solve a rectangular system of linear equations?
Since we actually plan to do these calculations numerically we also need to be concerned about how we deal with numerical error. For instance, in exact arithmetic, it's enough to say a matrix is not singular. In numerical calculations, we would like to be able to say that a particular matrix, while not singular, is "nearly singular".
## The Gram Schmidt Method
The Gram Schmidt method can be thought of as a process which analyzes a set of vectors X, producing an "equivalent" (and possibly smaller) set of vectors Q which span the same space, have unit L2 norm, and are pairwise orthogonal.
Note that if we can produce such a set of vectors, then we can easily answer many of the questions in our set of problems. In particular, the size of the set Q tells us whether the set X was linearly independent, and the dimension of the space spanned and the rank of a matrix constructed from the vectors. And of course, the vectors in Q are the orthonormal basis we were seeking. So you should believe that being able to compute the vectors Q is a valuable ability. In fact, the QR factorization can help us with all the problems on our list, primarily because it's possible to look at every number in the factorization and understand what that number is telling us about the linear system.
In words, the Gram-Schmidt process goes like this:
• Consider x, the next vector in X;
• For every vector q already in Q, compute the component of q that is in x and subtract it from x;
• Compute the norm of (what's left of) x. If the norm is zero (or too small), discard x; otherwise, divide x by its norm and move it from X to Q.
• You now have an orthonormal set of vectors spanning the same space as X.
Here is a sketch of the Gram Schmidt process as an algorithm. Assume that n is the number of x vectors:
```
nq = 0
for j = 1 to n
for i = 1 to nq
rij = qi' * xj
xj = xj - qi * rij
end
rjj = sqrt ( xj' * xj )
if ( rjj ~= 0 )
nq = nq + 1
qnq = xj / rjj
end
end
```
You should be able to match this algorithm to the previous verbal description. How is the L2 norm of a vector being computed?
M-file - implement the Gram-Schmidt process in an M-file called gram_schmidt.m. Assume the vectors are stored as columns of a matrix called X. Your function should have the form:
function Q = gram_schmidt ( X )
You can reference the vector formed by the j-th column of X by using the notation X(:,j).
Exercise - Test your Gram-Schmidt code using the following input:
``` X = ( 2 -1 0 )
( -1 2 -1 )
( 0 -1 2 )
( 0 0 -1 )
```
If your code is working correctly, you should compute approximately:
``` Q = ( 0.89 0.36 0.20 )
( -0.45 0.72 0.39 )
( 0.00 -0.60 0.59 )
( 0.00 0.00 -0.68 )
```
You should verify that the columns of Q have L2 norm 1, and are pairwise orthogonal. What is an easy way to check this?
Explain the following statements about the matrix Q you just computed:
• The columns of Q have L2 norm 1, and are pairwise orthogonal, so Q must be an orthogonal matrix, right?
• ..and moreover, QT*Q = I, true?
• ..and yet, it is not the case that Q*QT = I...
• ...therefore, Q can't be an orthogonal matrix?
See comments #2 at the end of the lab.
## Gram-Schmidt Factorization
We need to take a closer look at the Gram-Schmidt process. Recall how the process of Gauss elimination could actually be regarded as a process of factorization. This insight enabled us to solve many other problems. In the same way, the Gram-Schmidt process is actually carrying out a different factorization that will give us the key to other problems.
Just to keep our heads on straight, let me point out that we're about to stop thinking of X as a bunch of vectors, and instead regard it as a matrix. Since our easily-confused brain likes matrices to be called A, that's what we'll call our set of vectors from now on.
Now, in the Gram-Schmidt algorithm, the numbers that we called rij and rjj, which we computed, used, and discarded, actually record important information. They can be regarded as the nonzero elements of an upper triangular matrix R. The Gram-Schmidt process actually produces a factorization of the matrix A of the form:
A = Q * R
Here, the matrix Q has the same M by N "shape" as A, so it's only square if A is. The matrix R will be square (N by N), and upper triangular.
Now that we're trying to produce a factorization of a matrix, we need to modify the Gram-Schmidt algorithm of the previous section. Every time we consider a vector xj at the beginning of a loop, we will now always produce a vector qj at the end of the loop. Hence, if rjj, the norm of vector xj, is zero, we will simply set qj to the zero vector.
M-file: make a copy of your previous M-file and call it gs_factor.m. Modify it to compute the Q and R factors of a rectangular matrix A. It should have the form
function [ Q, R ] = gs_factor ( A )
Exercise: when you think your code is correct, use it to compute the QR factorization of the Frank matrix of order 4. To verify your results, the following statements should be checked:
1. Is it true that QT*Q=I?_____
2. Is it true that Q*QT=I?_____
3. Is Q orthogonal in the strict sense?_____
4. Is the matrix R upper triangular?_____
5. Is it true that A=Q*R?_____
## Householder Matrices
Another approach to the QR factorization of a matrix proceeds through a series of partial factorizations A=Qk*Rk, where the first Q is the identity matrix, and the first R is the matrix A. When we begin the k-th step of factorization, our factor Rk-1 is only upper triangular in columns 1 to k-1. Our goal on the k-th step is to find a better factor Rk which is upper triangular through column k. If we can do this process n-1 times, we're done.
Suppose, then, that we've partially factored the matrix A, up to column k-1. In other words, we have a factorization
A = Qk-1 * Rk-1
for which the matrix Qk-1 is orthogonal, but for which the matrix Rk-1 is only upper triangular for the first k-1 columns.
To proceed from our partial factorization, we're going to consider taking some orthogonal matrix Hk, and "inserting" it into the factorization as follows:
A = Qk-1 * Rk-1 = Qk-1 * HT * H * Rk-1
Then, if we define
Qk = Qk-1 * HT
Rk = H * Rk-1
it will again be the case that:
A = Qk * Rk
and we're guaranteed that Qk is still orthogonal. The interesting question is, if we pick the right H, can we "improve" the appearance of Rk, so that it is actually upper triangular all the way through column k.
In fact, we can do this by picking an appropriate Householder matrix. The formulas for a Householder matrix are a little complicated, and we won't go into them here. However, what you should understand is that, for the problem we've just described, we can determine a Householder matrix H with the property that Rk=H*Rk-1 is "slightly more upper triangular" than Rk-1.
M-file - copy the file householder.m from the web page. This function accepts the name of a matrix R and the index of a column k, and returns a Householder matrix H such that H*R has only zero entries below the diagonal in column k.
Exercise - define the matrix A to be the Frank matrix of order 5. Compute H1, the Householder matrix that knocks out the subdiagonal entries in column 1 of A, and then compute A1=H1*A. Does the result have the proper form?
Now let's compute H2, the Householder matrix that knocks out the subdiagonal entries in column 2 of A, and compute A2=H2*A. This matrix should have subdiagonal zeros in column 2. You should be convinced that you can zero out any subdiagonal column you want.
Now let's try to zero out all the subdiagonal entries. Proceed as follows:
``` A = frank ( 5 )
Q = eye ( 5 )
R = A
H = householder ( R, 1 )
Q = Q * H'
R = H * R
Q * R
H = householder ( R, 2 )
Q = Q * H'
R = H * R
Q * R
...
```
As you watch the calculation, you should see that the R matrix is gradually being zeroed out below the diagonal, and that the product of Q and R still equals A. Once we have completed the 4-th step, we're done.
## Householder Factorization
For a rectangular M by N matrix A, the Householder QR factorization has the form
A = Q * R
where the matrix Q is M by M (hence square and truly orthogonal) and the matrix R is M by N, and upper triangular (or upper trapezoidal if you want to be more accurate.)
If the matrix A is not square, then this definition is different from the Gram-Schmidt factorization we discussed before. The obvious difference is the shape of the factors. Here, it's the Q matrix that is square. The other difference, which you'll have to take on faith, is that the Householder factorization is generally more accurate, and easier to define compactly.
Householder QR Factorization Algorithm:
1. Q := I
R := A.
2. For k = 1 to min(m-1,n):
3. Construct the Householder matrix H for column k .
4. Q := Q * H',
R := H * R.
5. next k.
M-file - Make an M-file h_factor.m. It should have the form
function [ Q, R ] = h_factor ( A )
Use the routine householder.m that you copied earlier, in order to compute the H matrix that you need at each step.
Exercise - Test your code by computing the QR factorization of the Hilbert matrix of order 4. You should get something like this:
``` Q = 0.84 -0.52 0.15 0.03 R = 1.19 0.67 0.47 0.37
0.42 0.44 -0.73 0.32 0 0.12 0.13 0.12
0.28 0.53 0.14 0.79 0 0 0.01 0.01
0.21 0.50 0.65 0.53 0 0 0 0.00
```
By the way, there is plenty of information in this factorization. To make the 4-th column of A, for instance we use 37 percent of column 1 of Q, 12 percent of column 2, 1 percent of column 3, and practically nothing of column 4. The fact the the diagonal entry is so small compared to the other entries in the 4-th column allows us to assign a sort of measure to how linearly dependent the 4-th column of A is on the previous 3 columns (since the space spanned by the first three columns of A is the same as that spanned by the first three columns of Q.
Exercise - You may have written code that only works properly for a square matrix. In order to check that you have set things up to work for any rectangular matrix A, carry out the following tests:
A = rand ( 2, 4 )
[Q,R] = h_factor ( A )
A2 = Q * R;
norm ( A - A2, inf )
and repeat the test for random matrices of sizes (3,4), (4,4), (5,4) and (6,4). If you get a large error on any of these computations, you'll need to go back and rethink your code.
## The QR Method for Linear Systems
If we have computed the Householder QR factorization of a matrix without encountering any singularities, then it is easy to solve linear systems. We use the property of the Q factor that QT*Q=I:
A * x = b
Q * R * x = b
QT * Q * R * x = QT * b
R * x = QT * b
so all we have to do is form the new right hand side QT * b and then solve the upper triangular system. And that's easy because it's the same as the last step of our old PLU solution algorithm.
M-file - Make a copy of your file plu_solve.m, calling it h_solve.m. It should have the form
function x = h_solve ( Q, R, b )
Assume that the QR factors come from the h_factor routine, so that, if A is rectangular, then so is R, while Q will always be square. Set up your code to compute QT * b and then solve the upper triangular system R * x = QT * b. For now, don't worry too much about the possibility that the matrix is rectangular.
Exercise: When you think your solver is working, test it out on a square system as follows:
```
n = 5
A = frank ( n )
x = [ 1 : n ]'
b = A * x
[ Q, R ] = h_factor ( A )
x2 = h_solve ( Q, R, b )
```
## Loose Ends
Instead of the Householder form of the factorization, we may use the Gram-Schmidt QR factorization. There is not much difference for the case where the system matrix is square. But if the system was rectangular, the matrix Q is rectangular, and what we do from there depends on whether the number of rows is greater or less than the number of columns.
For either method of factorization, a further complication occurs when the number of columns is greater than the number of rows. In this underdetermined situation, there are multiple solutions. Assuming we simply want one solution, any solution, the typical procedure is to set the extra degrees of freedom to zero.
We haven't had time here to discuss the issues of singularity. If the square matrix A is actually singular, what happens? We may get a very small diagonal R value, or possibly a zero value. We haven't planned on handling such a case. Similarly, in the rectangular matrix cases, it's often desirable to do what amounts to pivoting, in order to process the "best" columns of data first.
## Assignment
You will need to rewrite your h_solve routine so that it can handle a rectangular system. Actually, the routine only has to solve an upper triangular system. Determine the size [M,N] of the matrix R, and then do an upper triangular solve as though the matrix was of size min(M,N).
Roughly speaking, you should expect that you can get some solution x with zero residual, whenever M is no greater than N, and you will usually get an answer, with a nonzero residual, whenever M is strictly bigger than N.
Assignment: Given the system matrix A and right hand side b, use your QR factor and solve routines to compute a solution x, and the residual r=A*x-b.
System #1:
``` A = 1 2 3 b = 6
4 5 6 15
```
System #2:
``` A = hilbert(16) b = A * ones(16,1)
```
System #3:
``` A = 1.00 1.0 b = 1.98
2.05 -1.0 0.95
3.06 1.0 3.98
-1.02 2.0 0.92
4.08 -1.0 2.90
```
System #4:
``` A = 1 3 b = 18
5 2 25
4 -1 7
```
Fill out the following table. For system #2 only, don't list all 16 elements of the solution, just the maximum solution error.
``` System Solution ||A*x-b||inf
#1 _______________ _______________
#2 _______________ _______________
#3 _______________ _______________
#4 _______________ _______________
```
and then mail the results to me.
Comments #1 - properties of an orthogonal matrix?
Properties of an orthogonal matrix:
1. YES; YES; YES; NO.
2. NO, an orthogonal matrix is not generally symmetric.
3. YES, essentially because
(Q1*Q2)T=QT2*QT1 .
with a similar statement for inverses. The important thing to notice is that the multiplication order gets reversed. Using these ideas, you can prove the statement.
4. YES, essentially because the inverse of the transpose is the transpose of the inverse.
5. YES, the L2 vector norm of the first column of Q must be 1. In fact, the L2 norm of every column and every row must be 1.
6. YES, the vectors formed by the first and last rows of Q must be perpendicular, that is, their "inner product" or "dot product" must be zero, and in fact any pair of distinct rows or columns must have a zero dot product.
7. NO, row 1 does not have to be orthogonal to column 2. There is no guaranteed relationship between a row and a column of the matrix.
8. YES, every entry of Q must be between -1 and 1 because the L2 norm of every column must be 1.
9. NO, the determinant of Q is not always 1. It is always +1 or -1.
10. NO, the eigenvalues of Q are not always +1 or -1. All eigenvalues have magnitude 1.
11. NO, in fact the singular values of Q are always +1.
12. NO, the L2 norm of Q is always 1, but we can't say about the others.
13. For any x, ||Q*x||2=||x||2.
14. For any matrix A, ||Q*A||2=||A||2.
15. The L2 condition number of Q is 1.
16. YES, in any column of Q, at most 1 entry can be 1.0.
17. NO, in any column of Q, up to n-1 entries can be 0. After all, the identity matrix is orthogonal.
Comments #2 - can a rectangular matrix be orthogonal?
A matrix Q is orthogonal if it is square and it is the case that QT*Q=Q*QT=I. From this second condition, we can conclude that an orthogonal matrix has columns of unit L2 norm which are pairwise orthogonal.
Conversely, if a matrix has columns of unit L2 norm which are pairwise orthogonal...it may loosely be called an orthogonal matrix, but it isn't, unless the matrix is square. If the matrix is M by N, with M>N (the usual case), then you will get QT*Q=I but not Q*QT=I!
We will carelessly call a rectangular matrix orthogonal. This is much more convenient than calling it a matrix "with orthonormal columns that are pairwise orthogonal". However, we have to remember that this is incorrect terminology, and that certain things true about an orthogonal matrix will not be true for rectangular "orthogonal" matrices.
Last revised on 28 March 2000.
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# Arrow Speed Calculator
Created by Gabriela Diaz
Last updated: Jul 30, 2022
If your target is to find how fast an arrow travels, then you aim right because the arrow speed calculator will easily help you with that 🏹 With this archery calculator, you'll also be able to identify the different factors and what values can help you improve your arrow's speed.
We've created a text where you'll find some of the basics related to IBO speed and how to calculate bow speed. This is what we included:
• What is IBO speed?:
• Which are the factors that affect arrow speed?; and
• How to calculate the fps speed of an arrow, its momentum, and kinetic energy.
Let's get started! 🎯
## What is IBO speed? — Arrow speed specification
The IBO speed is used as a standardized speed rating for bows. Created by the International Bowhunter's Organization, the IBO speed is given as a speed value in feet per second (fps or ft/s), and the higher this number is, the higher the speed that could be achieved with that particular bow. For example, a bow designated as IBO 330 could achieve greater speeds than a bow rated as IBO 300.
And we purposefully say "speed that could be achieved" because these marked speeds are not usually the ones that an archer can attain while using his bow with his specific setup. This is because the IBO speeds are determined under a set of particular conditions:
• Draw length equal to 30 inches;
• Draw weight equal to 70 pounds; and
• Arrow is weighing 350 grains.
Then, for comparing bow speeds, you'll need to maintain these factors constant. But because these are rarely the conditions under which every archer will be using their bow, then comes the question: what the actual speed of my arrow is? 🏹
## Which factors affect on arrow speed?
Before we see how to calculate bow speed, let's go through the main factors that influence the arrow speed to better understand how the actual speed is determined:
• Draw weight $D$ — This is the force required to pull a bow, expressed in pounds. Then, the more draw weight you use to pull, the quicker an arrow will fly and the more kinetic energy it will have. As a general rule, with every 10 pound increase in draw weight, arrow speed rises by around 20 fps.
• Draw length $L$ — This is the distance that the string is pulled back. A longer draw length results in a longer power stroke than a shorter draw length, allowing the bow to store more energy that then is transferred to the arrow. The general guideline for this factor is:
• For every extra inch of draw length increases approximately 10 fps the arrow speed; and
• For every inch you reduce, the draw length reduces around 10 fps of your bow speed.
• Arrow weight $A$ — If you guessed that the size and weight of an arrow influence its speed, you're correct. As you might expect, a heavier arrow flies slower than a smaller and lighter one.
When it comes to how much an arrow's weight affects its speed, the consensus is that every additional 5 grains of arrow weight reduces arrow speed by an average of 2 fps.
• Bowstring weight $W$ — Adding accessories to a bowstring impacts its speed; for every 3 grains of extra weight on the bowstring, deduct 1 fps from the IBO number.
## Calculate arrow fps speed — How fast does my arrow travel?
All of these parameters are summarized in an equation that you can use to estimate the actual arrow speed:
\small \begin{align*} v =\ &\text{IBO} + (L - 30) \times 10 - W\!/3\ +\\ &\min(0, -(A - 5D\!/3)) \end{align*}
where:
• $v$ — Actual arrow speed in ft/s (fps);
• $\text{IBO}$ — Standard arrow speed in terms of IBO specifications in ft/s (fps);
• $L$ — Draw length in inches;
• $W$ — Additional weight on the bowstring in grains;
• $A$ — Arrow weight in grains; and
• $D$ — Draw weight in pounds.
With this archery calculator, you can also find the momentum and kinetic energy of an arrow. The momentum of an arrow is linked to arrow penetration. You can obtain this value with this formula:
$\small \text{momentum} = A \times v$
An arrow's kinetic energy is determined as:
$\small \text{kinetic energy} = \cfrac{A \times v^2}{2}$
Keep in mind that the arrow loses speed as it travels. Thus these values of speed, momentum, and kinetic energy, are only valid within a specific distance range. To find the distance that your arrow will travel, you can use the projectile motion calculator.
It's also worth noting that these variables are affected not just by the arrow's speed $v$ but also by its mass $A$. And the greater the mass, the longer the energy stays in the arrow as it flies. However, here you'll find a bit of a contradiction because heavier arrows result in slower speeds, meaning compromising speed might be necessary to avoid losing momentum and kinetic energy so rapidly.
🙋 If you want to learn more about momentum and how to calculate it, we have a dedicated momentum calculator that you might enjoy!
## How to use the arrow speed calculator
The arrow speed calculator makes it easy to estimate the speed of an arrow! By simply indicating the particular values of some parameters, this tool will help you calculate an arrow fps speed. Let us see how to use this tool:
1. Begin by indicating the IBO rating in fps in the Bow IBO rating field.
2. Now, enter the Draw length of the bow.
3. Indicate the Peak draw weight.
4. Next, input Arrow weight.
5. Finally, enter the Additional weight on the string if this is known.
6. After you've entered these values, the calculator will display the actual arrow speed, momentum, and kinetic energy of the arrow 🎯
Gabriela Diaz
Bow IBO rating
ft/s
Draw length of the bow
in
Peak draw weight
lb
Arrow weight
gr
gr
Arrow speed
ft/s
Momentum
Ns
Kinetic energy
ft-lbs
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https://www.shaalaa.com/question-bank-solutions/derive-expression-drift-velocity-free-electrons-conductor-terms-relaxation-time-drift-electrons-origin-resistivity_50646
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Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
# Derive an Expression for Drift Velocity of Free Electrons in a Conductor in Terms of Relaxation Time. - Physics
Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time.
#### Solution
If there are N electrons and the velocity of the ith electron at a given time is vi where, i = (1, 2, 3, …N), then
1/N sum_(i-1)^N V_1 = 0 (If there is no external field)
When an external electric field is present, the electrons will be accelerated due to this field by
veca = (-evecE)/m
Where,
− e = Negative charge of the electron
E = External field
m = Mass of an electron
Let vi be the velocity immediately after the last collision after which external field was experienced by the electron.
If vi is the velocity at any time t, then from the equation V = u + at, we obtain
vecV_i = vecv_i - (evecE)/m t ........ (1)
For all the electrons in the conductor, average value of vi is zero.
The average of vi is vd or drift velocity.
This is the average velocity experienced by an electron in an external electric field.
There is no fixed time after which each collision occurs. Therefore, we take the average time after which one collision takes place by an electron.
Let this time, also known as relaxation time, beτ. Substituting this in equation (i)
vecv_i - o
t = tau
vecV_i = vecv_d
Then,
vecv_d = (-evecE)/m
Negative sign shows that electrons drift opposite to the applied field.
Concept: Drift of Electrons and the Origin of Resistivity
Is there an error in this question or solution?
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http://www.chegg.com/homework-help/questions-and-answers/if-a-typical-hydrogen-atom-in-a-collapsing-molecular-cloud-core-starts-at-a-distance-of-10-q3217488
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## Astronomy Question
If a typical hydrogen atom in a collapsing molecular-cloud core starts at a distance of 10,000 astronomical units, or AU (1.5 x 10^12 km) from the core’s center and falls inward at an average velocity of 1.5 km/s, how many years does it take to reach the newly forming protostar? Assume that a year is 3 x 10^7 seconds.
• (1) It may be that a nebula requires an outside nudge to get the collapse started (like a gravitational nudge from a passing star or other nebula). So, the timeframe can be short or very long (given that some are still around).
(3) Yes. There are many examples of rings around newly forming stars (the rings being the residual material that has yet to fall into the star or to form planets). Or did you mean in the immediate vicinity of the star itself?
(4) Further out. The nebula does not collapse 100%. Most of the material reaches the center to form the star. Conservation of momentum, etc. causes the collapsing nebula to flatten out and spin faster (toward the center). So some of the interior material is spun outward too. The matter in the outer disk has its own gravitational dynamics and starts to clump together into bigger objects (which thereby develop stronger gravitational fields and sweep up more of the residual matter). Note that in the early stages of solar system formation, many planetoids with non-stable orbits are produced and many collisions, ejections, and fallings-into the star occur. For example, our solar system was only able to sustain stable orbits for 8 major planets (and many smaller objects of course).
(5) Many light years across (trillions of miles). The density is less than a vacuum (I don't mean a Hoover [;)]) that you could create in a laboratory. I would imagine that the temperature near the edges would be close to background temperature.
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Spectral Analysis (SA)
## Spectral Analysis (SA)
The closed-form solutions of compartment models involve the convolution of the input function with decaying exponentials. Based on this observation, a generalized technique called spectral analysis (SA) was introduced by Cunningham and Jones [66]. The operational equation of SA is given by
that is, tissue uptake is modeled as a sum of N possible tissue responses. Due to the constraint of first order tracer kinetics, the coefficients ai and the decay constants bi must be non-negative. In practice, a discrete set of the decay constants bi is selected which covers the physiologically reasonable range, typically logarithmically spaced in the range [10-5,1]sec-1. The corresponding tissue responses
are the basis functions of spectral analysis.
When fitting the operational equation above to a tissue TAC, the only unknowns are the coefficients ai, because only a pre-defined set of discrete bi values is considered. Therefore, the problem is that of a non-negative linear least squares estimation (NNLS) with the constraint of non-negative coefficients. There is a well-known NNLS algorithm available [67], which allows readily calculating the optimal set of ai coefficients. They allow calculating the model function and visualizing it together with the tissue TAC in the same way as for the compartment models.
An advantage of SA is the fact that no particular compartment structure is imposed. Rather, its result can be used to estimate how many kinetic tissue compartments can be resolved by PET. To this end, the results are plotted as a spectrum with the selected decay constants bi along the x-axis (as the "frequencies") and the estimated coefficients ai along the y-axis (as the "amplitudes"). Because of the large range, log(bi) is used in spectrum plotting rather than bi. The number of peaks in this spectrum corresponds to the number of distinct compartments. A peak appearing to the far left (low frequency, slow component) indicates irreversible trapping. A peak to the far right (high frequency, fast component) corresponds to kinetics indistinguishable from the input curve, thus to vascular contributions. Intermediate peaks represent compartments which exchange reversibly with plasma or with other tissue compartments [68].
The examples below used synthetic data generated using compartment models without noise and blood contributions. In the first example the simulated tissue TAC of a 1-tissue compartment model was fitted by SA with 500 basis functions. It is evident that, because of the discrete nature of the bi basis, the compartment was split into two neighboring frequencies. The adjacent amplitudes are usually summed to provide the combined peak height for a compartment.
In the next example an irreversibly trapping compartment was added (k3>0, k4=0) to the same first compartment. The corresponding peak is clearly seen to the left of the spectrum.
When changing kinetics to become reversible (k4>0), the second compartment appears as a second peak in the inner of the spectrum.
In general it is not possible to calculate the compartment rate constants from the spectral analysis outcome. However, an estimate of K1 can be obtained as the sum of the peak amplitudes
and an estimate of the distribution volume VT as
An additional information which can be calculated is the impulse response function
The impulse response function completely describes the system, and the expected tissue TAC for any given input function and can simply be calculated by convolution with the IRF:
In the practical application of spectral analysis potential problems were found: Caution should be applied when interpreting the number of peaks [66], and the error properties of the estimates are difficult to assess [69]. Several authors have proposed variants to overcome these problems [70,71], but in current practice spectral analysis is not yet frequently applied.
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1310-Notes-Sec-41-filled
# 1310-Notes-Sec-41-filled - Math 1310 Section 4.1 Polynomial...
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Math 1310 Class Notes – Section 4.1, Page 1 of 10 Math 1310 Section 4.1 Polynomial Functions General information A polynomial function is a function of the form 0 1 2 2 2 2 1 1 ) ( a x a x a x a x a x a x P n n n n n n + + + + + + = - - - - where 0 n a , n a a a , , , 1 0 are real numbers and n is a whole number. The degree of the polynomial function is n . We call the term n n x a the leading term, and n a is called the leading coefficient. . ) 0 ( 0 a P = Our objectives in working with polynomial functions will be, first, to gather information about the graph of the function and, second, to use that information to generate a reasonably good graph without plotting a lot of points. In later examples, we’ll use information given to us about the graph of a function to write its equation. Power functions We’ll start with the shapes of the graphs of functions of the form . 0 , ) ( = n x x f n You should be familiar with the graphs of 2 ) ( x x f = and . ) ( 3 x x g =
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Math 1310 Class Notes – Section 4.1, Page 2 of 10 The graph of 0 , ) ( = n x x f n , n is even, will resemble the graph of 2 ) ( x x f = , and the graph of 0 , ) ( = n x x f n , n is odd, will resemble the graph of 3 ) ( x x f = .
Math 1310 Class Notes – Section 4.1, Page 3 of 10 End behavior Next, you will need to be able to describe the end behavior of a function. If the degree of the function is even and
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### Replies to This Discussion
Assalam u alaikum
Phele se ye hai k Prolog programming hoti hai is ko parho ye logic programming hoti hai abi deaki hai Assignment easy lag rahi hai ho jay ge google kiya kro bacho
yar kisi nay ki hay to please share kar day .........
too much confusion....
please tell me that correct or wrong????????
Question no 2
?-leave(Y,fast),leave(Y,fast).
Y=Bushra
?-earns(M,by Allah Karim), earns(M,by Allah Karim).
M=Imran
M=Saqib
?-Fasting(Bushra, Ramdan ul mobark)
No
?-fast(imran&saqib)
Yes
?-keep left(X,fasts)
No
Please Share Question One (1) with thanks.... Whats are rules?
Hn g bacho First Q easy hai both itna muskil ni hai
Zafar is adult and fasting in Ramzaan Ul Mobarak
Fasting(Zafar,Ramzaan Ul Mobarak)
bacho tora google kiya kro
sound mixer is me adult ka koi zikar nahi... what u say about it?
adult hai he galt ana he ni online gammer mai jaa kr sentence past kro then result deako
plz guyZ help.....................
hi.share the link of online grammer
kuj to help karo zalimo......
yup in ko khud ni ata bas jese b kar liya or dosron ko confusion k liye kuch b likh diya. kabi koi link ka kahe ga or kabi koi einstein ban jaye ga :/
friends please can change it. quesiton no 1 hay.
If one is fasting in Ramzan ul Mobarak then on is Adult.
If one is Fasting in Ramzan ul Mobarak then one is rewarded by ALLAH KARIM.
rewarded(zafar):-is(zafar,fasting),is(rizwan,fasting),is(raza,fasting),is(imran,fasting),is(saqib,fasting).
If one is non adult and fasting in Ramzan ul Mobarak then one is earning greater reward by Allah Karim.
earning greater reward(imran):-is(imran,fasting),is(saqib,fasting).
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How To?
# How to Calculate Land Area or Plot Area?
Personal Finance
08-11-2023
Let us first understand what “land area” or “plot area” means. Land area implies the size of area-based sections and land-based sections of typical geographic areas. It is measured in km². The plot area is the area that is surrounded by a barrier (like fences). Simply put, it is the total area that belongs to you within a region or city. Calculations of land area are informal. It is very easy to calculate land area. Let us understand the formula behind this calculation and the importance of knowing how to calculate the land area.
## Why is land area calculated?
Before buying land, it is crucial to be aware of its exact measurements. Different scales are applied to land in various states of India. A simple and accurate way to accomplish this digitally is with a land surface calculator. For the design and implementation of practically every sort of development, site measurement becomes extremely important. In the current day, this measuring has become common in the domains of construction, navigation, and transportation, as well as in the definition of legal borders for land rights.
## What are the units of measurement used to calculate land area?
Any measure can be used to compute a plot of land’s area. Typically, the area of a property is given in square feet. In contrast, the land area for farmland is defined based on hectares or acres.
However, the term used for “unit of measurement” is different in different states. Let’s take a look at some different names for each state.
• In India, typical land measurements include hectares, acres, square meters, and square yards (gaj). Terms like percent, guntha, and ground are exclusively associated with the South. Bigha and Marla are frequently used in the North. The names of the units and their sizes differ from one state to another.
• In Maharashtra, the concept of Guntha is more commonly used than in Karnataka and Andhra Pradesh.
• Dhur and Kattha both refer to various sizes, depending on the state. Dhur is more than 68 square feet in Bihar but only 3.6 square feet in Tripura. Dhur is a language that is frequently spoken in Jharkhand as well as in Bihar and Tripura. In contrast, Assam, West Bengal, and Madhya Pradesh are the main usage areas for Katha. Along with Dhur, it’s also used in several regions of Bihar. Although the size changes from one state to another. This typically falls between 600 and 2,800 square feet.
Some of the units used to convert the dimensions are centimetres, meters, kilometres, inches, feet, etc. The basic units used to calculate land area are:
Square Feet (sq. ft.)
Square Inch
Square Centimetre
Square Meter (sq. m.)
Square Yard (sq. yd.)
Square Kilometre (sq. km)
Acre
Hectare
## Benefits of an Area Calculator
The calculator used to calculate the area has the following advantages:
• The area calculator tool guarantees quick, blunder-free calculations.
• It makes it possible to convert quantities into obscure land measurements.
• It helps you make wise real estate decisions while making you aware of the land’s actual value.
## How to calculate land area in different ways
### The way to calculate land area from Google Maps (online)
Google Maps and other online services can be used to calculate the area of a piece of land. The only thing one would have to do is zoom into the map and mark the border of the plot of land. The region of that specific piece will then be visible on the display after this is complete. But this will just be a general estimate, not a precise one. Additionally, you have the option of converting the computations into whichever unit you like.
### The way to calculate the land area of irregularly shaped plots
Uneven, undulating land is known as “irregular land.” The land measurements for these plots can be determined with the aid of specific formulas. Divide the space into recognisable forms such as a parallelogram, triangle, rectangle, square, or circle. Then use the appropriate equations to calculate the areas. To determine the size of an irregular land, add the findings. As an alternative, you can get a trouble-free conclusion by using an automated land area calculator.
### The way to calculate the land area in standard square feet (sq. ft.)
Simply take a foot-long measurement of both the length and breadth. After that, to calculate the size of the area, multiply the length by the breadth in feet. 144 square inches equals one square foot. To prevent time-consuming hand calculations and obtain precise answers, you can also use a digital land calculator.
## Basic formulas to calculate land area
Apart from the irregularly shaped land, the plot area can be square, rectangular, or triangular. Based on this, there is a general formula you can use. This is for all the manual calculations.
Square plot: Area= Length*length, or, Length square
For example, if one side of the square of land is 20, then the area would be 20 * 20, i.e., 400 feet.
Rectangular plot: area = length * breadth
For example, if the length is 40 and the breadth is 20, then
Area = 40*20, i.e., 800 feet
Triangular plot: Take a, b, and c as plot dimensions. Take s as half the triangular perimeter.
The general formula is, Area of triangular plot= √(s(s-a)(s-b)(s-c))
## The key takeaway
The term “land area” has different names based on the state. The north, east, west, and south have their terminologies. It also has a series of formulas you can use based on the shape of the land. The process to calculate land area can be done either manually or digitally. Manually, by using the formulas, and digitally, through the use of online calculators and even Google Maps. To read more informative blogs on such topics, head to Piramal Finance. Browse through their blogs to learn about relevant matters in the stock market, business, and financing, or go through their products such as personal loans, housing loans, MSME loans, home loan EMI calculators, and more.
How To?
How To?
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Gravity Simulator http://www.orbitsimulator.com/cgi-bin/yabb/YaBB.pl General >> Discussion >> Some interesting situations http://www.orbitsimulator.com/cgi-bin/yabb/YaBB.pl?num=1129249000 Message started by bob7_2 on 10/13/05 at 17:16:38
Title: Some interesting situations Post by bob7_2 on 10/13/05 at 17:16:38 I'm doing a very large project for school, and somehow the the science got the idea to have my group simulate the collision of 2 galaxies. So.... How exactly would I go about doing that in gravsim? Lets assume the galaxies in question are spiral and/or barred spiral, similar in size to our own galaxy. There are thousands of objects that would need to be simulated, even to get a semi-accurate sim. How do I generate the orbital information for all of them, and can grav-sim even handle something of this scale?On a completely unrelated note, in a differant project, in a differant class, I am working on figuring out how to extract the core from Jupiter. So how do you figure out the orbital mechanics of THAT? Our current plan is to use a very, very, large circle type thing. It would smash it into jupiter, and scrape everything off the core as it passed through a hole in the middle.So, can anyone give any help on either of these? The first one is top priority.
Title: Re: Some interesting situations Post by Tony on 10/14/05 at 11:36:43 The problem with simulating 2 galaxies is the number of objects involved. It's not thousands of objects, but hundreds of billions. Gravity Simulator does not do well once you're got hundreds of objects. Probably about 600 objects is the most you would want to do if your computer's processor is somewhere around 3Ghz.So you'll have to simulate mini-galaxies, which would be more correctly called star clusters. This you can do. And you can watch arms be pulled by tidal forces. But you have to make guesses as to what a full scale galaxy would do in the same circumstances.Also, the structure of galaxies is not completely understood. The motions suggests lots of dark matter that has yet to be found. So even on a supercomputer, you still have to make guesses as to your starting conditions.You might consider changing your topic to colliding solar systems. They're fun too.I'm not quite sure I understand what you're trying to do with Jupiter. But if something collides with Jupiter, causing mass to leave Jupiter, it sounds like a conservation of momentum project.
Title: Re: Some interesting situations Post by nou on 01/11/06 at 19:08:18 The jupitor thing is unrealistic. It would be impossible to extract the core without a gas-giant size body and a huge velocity. Simulating that would be extremly complex and would involve many things besides gravity. The galaxy thing is not too hard. use galaxsee, a free program. (http://www.shodor.org/master/galaxsee/).
Title: Re: Some interesting situations Post by tomek on 05/17/06 at 03:26:06 regarding some interesting situations. might want to check this (http://faculty.ifmo.ru/butikov/Projects/Collection3.html) out.i tried to make a .gsim out of it by editing objects and saving the simulation, but couldn't get it to work properly - speeds kept getting rounded in a weird way in the planet editbox; and after saving the scenario i couldn't even load it - sim froze when i tried to. maybe you'll have better luck with it
Title: Re: Some interesting situations Post by Tony on 05/17/06 at 10:47:58 Thanks for that link. I've seen that program before. There's some fascinating simulations there.The edit box rounding problem is a known bug in the current version. To get around this, type your numbers in notepad first, all 6 vectors seperated by a space. For example:1234.56e+10 34567e+7 78977.3e+1 335577.88e+8 8906e-8 334578.3344e-6Then highlight it, copy it, and paste the entire thing into the x position box. It will parse it for you into the 5 other boxes.This is a feature designed to let you quickly add vectors from JPL's Horizons system, without having to copy and paste 6 times per object.
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# Choosing Inappropriate Display Media
Choosing inappropriate display media is one of the most common design mistakes made, not just in dashboards, but in all forms of quantitative data presentation. For instance, using a graph when a table of numbers would work better, and vice versa, is a frequent mistake. Allow me to illustrate using several examples beginning with the pie chart in Figure 3-9.
Figure 3-9. This chart illustrates a common problem with pie charts.
This pie chart is part of a dashboard that displays breast cancer statistics. Look at it for a moment and see if anything seems odd.
Pie charts are designed specifically to present parts of a whole, and the whole should always add up to 100%. Here, the slice labeled "Breast 13.30%" looks like it represents around 40% of the piea far cry from 13.3%. Despite the meaning that a pie chart suggests, these slices are not parts of a whole; they represent the probability that a woman will develop a particular form of cancer (breast, lung, colon, and six types that aren't labeled). This misuse of a pie chart invites confusion.
The truth is, I never recommend the use of pie charts. The only thing they have going for them is the fact that everybody immediately knows when they see a pie chart that they are seeing parts of a whole (or ought to be). Beyond that, pie charts don't display quantitative data very effectively. As you'll see in Chapter 4, Tapping into the Power of Visual Perception, humans can't compare two-dimensional areas or angles very accuratelyand these are the two means that pie charts use to encode quantitative data. Bar graphs are a much better way to display this information.
Note: Refer to my book Show Me the Numbers: Designing Tables and Graphs to Enlighten (Oakland, CA: Analytics Press, 2004) for a thorough treatment of the types of graphs that work best for the most common quantitative messages communicated in business.
The pie chart in Figure 3-10 shows that even when correctly used to present parts of a whole, these graphs don't work very well. Without the value labels, you would only be able to discern that opportunities rated as "Fair" represent the largest group, those rated as "Field Sales: 2-Very High" represent a miniscule group, and the other ratings groups are roughly equal in size.
Figure 3-10. This example shows that even when they are used correctly to present parts of a whole, pie charts are difficult to interpret accurately.
Figure 3-11 displays the same data as Figure 3-10, this time using a horizontal bar graph that can be interpreted much more efficiently and accurately.
Figure 3-11. This horizontal bar graph does a much better job of displaying part-to-whole data than the preceding pie charts.
Other types of graphs can be equally ineffective. For example, the graph in Figure 3-12 shows little regard for the viewer's time and no understanding of visual perception. This graph compares revenue to operating costs across five months, using the size of overlapping circles (sometimes called bubbles) to encode the quantities. Just as with the slices of a pie, using circles to encode quantity relies on the viewer's ability to compare two-dimensional areas, which we simply cannot accurately do. Take the values for the month of February as an example. Assuming that operating costs equal \$10,000, what is the revenue value?
Figure 3-12. This graph uses the two-dimensional area of circles to encode their values, which needlessly obscures the data.
Our natural tendency is to compare the sizes of the two circles using a single dimensionlength or widthequal to the diameter of each, which suggests that revenue is about three times that of operating costs, or about \$30,000. This conclusion is wrong, however, to a huge degree. The two-dimensional area of the revenue circle is actually about nine times bigger than that of the operating costs circle, resulting in a value of \$90,000. Oops! Not even close.
Now compare operating costs for the months of February and May. It appears that costs in May are greater than those in February, right? In fact, the interior circles are the same sizemeasure them and see. The revenue bubble in May is smaller than the one in February, which makes the enclosed operating costs bubble in May seem bigger, but this is an optical illusion. As you can see, the use of a bubble chart for this financial data was a poor choice. A simple bar graph like the one in Figure 3-13 works much better.
Figure 3-13. This bar graph does a good job of displaying a time series of actual versus budgeted revenue values.
Actual versus budgeted revenue is also the subject of Figure 3-14, but this time it's subdivided into geographical regions rather than time slices and displayed as a radar graph. The quantitative scale on a radar graph is laid along each of the axis lines that extend from the center to the perimeter, like radius lines of a circle. The smallest values are those with the shortest distance between the center point and the perimeter.
Figure 3-14. This radar graph obscures the straightforward data that it's trying to convey.
The lack of labeled axes in this graph limits its meaning, but the choice of a radar graph to display this information in the first place is an even more fundamental error. Once again, a simple bar graph like the one in Figure 3-15 would communicate this data much more effectively. Radar graphs are rarely appropriate media for displaying business data. Their circular shape obscures data that would be quite clear in a linear display such as a bar graph.
Figure 3-15. This bar graph effectively compares actual to budgeted revenue data.
The last example that I'll use to illustrate my point about choosing inappropriate means of display appears in Figure 3-16.
Figure 3-16. This display uselessly encodes quantitative values on a map of the United States.
There are times when it is very useful to arrange data spatially, such as in the form of a map or the floor plan of a building, but this isn't one of them. We don't derive any insight by laying out revenue informationin this case, whether revenues are good (green light), mediocre (yellow light), or poor (red light), in the geographical regions South (brown), Central (orange), West (tan), and East (blue)on a map.
If the graphical display were presenting meaningful geographical relationshipssay, for shipments of wine from California, to indicate where special taxes must be paid whenever deliveries cross state linesperhaps a geographical display would provide some insight. With this simple set of four regions with no particular factors attached to geographical location, however, the use of a map simply takes up a lot of space to say no more than we find in the table that appears on this same dashboard, which is shown in Figure 3-17.
Figure 3-17. This table, from the same dashboard, provides a more appropriate display of the regional revenue data that appears in Figure 3-16.
Information Dashboard Design: The Effective Visual Communication of Data
ISBN: 0596100167
EAN: 2147483647
Year: 2004
Pages: 80
Authors: Stephen Few
Similar book on Amazon
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# Single Phase 380v Ac Transformer
## Recommended Posts
Dear all,
I have a single phase transformer (used for resistance spot welding) with following ratings:
Rated input U: 380V AC
Freq.: : 50 Hz
Rated output capacity :150 KVA (at 50% duty cycle)
Secondary noload voltage : 24 V
Turn ratio: : 16
2. How to calculate L (inductance) of the primary side and secondary side?
3. 150kVA at 50% duty cycle means what?
Rated current = S/ U = 150 000/380 = 394.7A (why circuit breaker rated current is 225A only)
Does it mean the transformer never reach 100% of its capacity?
As far as I understand
Duty cycle = number of cycle (n) / Total number of cycles (N) * 100%.
For example, with f =50Hz frequency then N = 50
At 50% duty cycle means n =25
Am I right or wrong?
4. Power saving?
Currently, sec. current is set at 10 000 A, weld time = 20 cycles = t (ms)
S = 380V x 625A (10 000A/16 (turn ratio)) = 131 kVA
P = S x cosφ (kW)
A = P x t x 1/3 600 000 (kWh)
If I reduce current to 9000 A
S = 380V x 562.5A (9000A / 16 (turn ratio)) = 213.75 kVA
then i will save 10% of energy?
please give me some feedbacks and guides.
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Hello.
Only some short notes. Outputs are always kW not KVA. When calculating the apparent current you seems to have forgotten to divide by sq root of 2 (1.414): 150 000/380/1.414 = 279 A (but still, too big for the circuit breaker).
Yuri
##### Share on other sites
Hello yuri
Outputs are always kW not KVA.
Not where I come from, transformers are always rated in KVA, not KW. You can fully load a transformer at 10% KW if the power factor is 0.1 This is what the whole thing is about power factor correction. It allows more load on transformers and reduces losses.
Best regards,
Mark.
##### Share on other sites
Hello
1. I do not believe that it is possible to calculate the no load current from the information given. The no load current is essentially the magnetizing current of the transformer. If the full load power factor was quoted, an estimation could be made from that.
2. Same as 1. Insufficient data.
3. 50% duty cycle means ON for 50% of the time. 150KVA at 50% duty cycle will give you an average load of 75KVA. Provided that the length of the ON time is short relative to the thermal curve of the circuit breaker, you could use a smaller breaker.
4. I assume that the transformer is designed to deliver the correct voltage and current for the spot weld so you can not reduce these values. If you can establish that the voltage and/or current are higher than required for your application, you could then look at reducing the output voltage to reduce the load and energy.
Best regards,
Mark.
##### Share on other sites
Hello yuri
Not where I come from, transformers are always rated in KVA, not KW. You can fully load a transformer at 10% KW if the power factor is 0.1 This is what the whole thing is about power factor correction. It allows more load on transformers and reduces losses.
Best regards,
Mark.
Hello Mark,
- Max input capacity: 380KVA (rated input voltage 380VAC)
- Max short circuit current: 16kA
- Max secondary output voltage: 23.75 (at 7.8 duty cycle)
- PF = 0.55
When secondary side is noload (open circuit), I measure the current at the primary side so I can have no load current?
2. Same question how can I know L=? (power factor = 0.55)
3. The length of the time is only max 20 cycles.
4. The equipment use constant current control (loop control by means of primary current feedback). Current and time can be reduce, even turn ratio also can be changed. At secondary side we can change current setting via manual programming device. I think we also can reduce secondary voltage (output voltage) by changing the turn ratio.
Pls correct me iF I am wrong at anything.
Kind regards,
Hung
##### Share on other sites
• 4 weeks later...
Hello Mark,
- Max input capacity: 380KVA (rated input voltage 380VAC)
- Max short circuit current: 16kA
- Max secondary output voltage: 23.75 (at 7.8 duty cycle)
- PF = 0.55
When secondary side is noload (open circuit), I measure the current at the primary side so I can have no load current?
2. Same question how can I know L=? (power factor = 0.55)
3. The length of the time is only max 20 cycles.
4. The equipment use constant current control (loop control by means of primary current feedback). Current and time can be reduce, even turn ratio also can be changed. At secondary side we can change current setting via manual programming device. I think we also can reduce secondary voltage (output voltage) by changing the turn ratio.
Pls correct me iF I am wrong at anything.
Kind regards,
Hung
Hello Mark,
Could you pls answer remaining questions with above information.
1. How much no load current?
2. L value?
Kind regards,
Hung
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# How is the riskiness of portfolio measured?
Investors rarely place their entire wealth into a single asset or investment. Rather, they contrast a portfolio. A portfolio is a combination of two or more securities or assets. Investors form a portfolio to minimize the risk of investment and/or to maximize the return.
The objective of portfolio formation is achieved through forming an efficient set of assets which either will maximize the return at a given level given a level of return.
Theory of portfolio suggests that a portfolio should be a well-diversified combination of assets from a range of assets from the’ market had negative correlations. If we can properly diversify our investment the unsystematic or company specific, so risks can be eliminated.
So, we can say that a well-diversified portfolio assumes the systematic risk of investment only. In forming an efficient portfolio, Markowitz efficient set and theory of optimal portfolio choice help investors and academicians to understand the zeal of portfolio formation.
The riskiness of a portfolio: Unlike the situation with returns, the standard deviation of a portfolio, σ, is generally not a weighted average of the standard deviation of the individual securities in the portfolio, and each stock’s contribution to the portfolio’s standard deviation is not σ1x1. Indeed, it is theoretically possible to combine two stocks which are, individually, quite risky as measured by their standard deviations, and to form from these risky assets a portfolio which is completely riskiness, with σp = 0%. However, the standard deviation of returns of a portfolio can be calculated from its expected returns and associated probabilities using the following equation:
σP = √∑[(Rpi – Rp)2 x Pi]
Covariance and the Correlation Coefficient:
Two key concepts in portfolio analyses are (1) Covariance and (2) the correlation coefficient.
Covariance is a measure of the degree to which two variables move together relative to their individual mean values over time. The following equations can be used to find the covariance between stocks A and B:
Calculating co-variance from expected return data,
COVAB = ∑ Pi x [(RA – ŘA)(RB – ŘB)]
If stocks A and B tend to move together, their covariance, COBAB will be positive. While if they tend to move counter to one another, COVAB, will be negative. If their returns fluctuate randomly, could be either positive or negative, but in either event, it will be close to zero.
The correlation coefficient standardizes the covariance by taking into consideration the variability of the two individual return series, as follows:
ρAB = (COVAB – σAσB)
ρAB = correlation coefficient between returns of two stocks A and B.
The sign of the correlation is the same as the sign of the covariance, so a positive (+) sign means that the variables move together, a negative (-) sign indicates the move in opposite directions, and if ‘ρ’ is close to zero, they move independently of one another. Moreover, the standardization process confines the correlation coefficient to values between – 1.00 and + 1.00.
The correlation equation can be solved to find the covariance: COVAB = σAσB ρAB
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# How Much Solar to Charge 400AH Battery
How Much Solar to Charge a 400AH Battery: A Comprehensive Guide
Solar energy is a clean, renewable, and sustainable source of power. It has become increasingly popular for charging batteries, especially for off-grid applications. If you are wondering how much solar power you need to charge a 400AH battery, this article will provide you with a comprehensive guide.
To determine the amount of solar power required, several factors need to be considered, such as the battery capacity, charging efficiency, and daily energy consumption. Here’s a step-by-step guide to help you calculate the solar power needed:
Step 1: Determine the battery capacity
A 400AH battery has a total capacity of 400 amp-hours. This means it can supply 1 amp of power for 400 hours or 2 amps for 200 hours, and so on.
Step 2: Determine the charging efficiency
Charging efficiency varies depending on the type of battery and the charge controller used. It is typically around 80-90%. Let’s assume an 85% efficiency for our calculations.
Step 3: Calculate the energy required
Multiply the battery capacity by the charging efficiency to determine the energy required. In this case, 400AH x 85% = 340AH.
Step 4: Determine the daily solar energy required
To calculate the daily solar energy required, divide the energy required by the number of sunny hours in your location. Let’s assume an average of 5 hours of direct sunlight per day. Thus, 340AH / 5 = 68 amp-hours.
Step 5: Determine the solar power required
To determine the solar power required, divide the daily solar energy required by the average hours of sunlight. In this case, 68AH / 5 = 13.6 amps.
So, you would need approximately 13.6 amps of solar power to charge a 400AH battery in 5 hours of direct sunlight per day.
Now, let’s address some common questions related to solar charging and battery capacity:
1. Can I charge a 400AH battery with less solar power?
Yes, you can charge a 400AH battery with less solar power, but it will take longer to charge.
2. Can I charge a 400AH battery with more solar power?
Yes, you can charge a 400AH battery with more solar power, but be cautious not to exceed the battery’s recommended charging rate.
3. Can I use multiple smaller panels instead of one large panel?
Yes, you can use multiple smaller panels as long as the total wattage output is sufficient to meet your charging needs.
4. Do I need a charge controller?
Yes, a charge controller is essential to regulate the charging process and prevent overcharging or damaging the battery.
5. How long does it take to charge a 400AH battery with solar power?
The charging time depends on the solar power output, battery capacity, and charging efficiency. On average, it may take around 5-8 hours of direct sunlight.
6. Can I charge the battery on cloudy days?
Yes, solar panels can still generate power on cloudy days, although at a reduced efficiency.
7. Can I charge the battery during winter months?
Yes, you can charge the battery during winter months, but keep in mind that shorter daylight hours and lower solar intensity may affect the charging time.
8. Can I connect the solar panels directly to the battery?
No, it is recommended to use a charge controller between the solar panels and the battery to regulate the charging process.
See also How Do I Know if My Hybrid Battery Is Dying
9. Can I use a generator to charge the battery?
Yes, you can use a generator as a backup charging source, especially during extended periods of low sunlight.
10. Can I use a hybrid system with solar and grid power?
Yes, a hybrid system allows you to use both solar and grid power to charge your battery, providing flexibility and reliability.
11. How long will a 400AH battery last on a full charge?
The battery life depends on several factors, including the load connected to it. On average, a 400AH battery can last between 20-40 hours on a full charge.
12. Can I expand my solar setup in the future?
Yes, you can expand your solar setup by adding more panels or batteries to meet your growing energy needs.
In conclusion, charging a 400AH battery with solar power requires approximately 13.6 amps of solar power in 5 hours of direct sunlight per day. Factors such as battery capacity, charging efficiency, and daily energy consumption play a crucial role in determining the solar power required. By understanding these factors and considering the common questions mentioned above, you can effectively charge your battery using solar energy.
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Here are sixteen Trigonometric Identities:
```SIN = Sine
COS = Cosine
TAN = Tangent
CSC = CoSecant
SEC = Secant
COT = CoTangent
#1
1
SIN Θ = -----
CSC Θ
#2
1
COS Θ = -----
SEC Θ
#3
SIN Θ
TAN Θ = -----
COS Θ
#4
COS Θ 1
COT Θ = ----- = -----
SIN Θ TAN Θ
#5
1
SEC Θ = -----
COS Θ
#6
1
CSC Θ = -----
SIN Θ
#7
SIN2 Θ + COS2 Θ = 1
#8
SEC2 Θ = 1 + TAN2 Θ
#9
CSC2 Θ = 1 + COT2 Θ
#10
SIN2 Θ = ½ ( 1 - COS 2Θ )
#11
COS2 Θ = ½ ( 1 + COS 2Θ )
#12
SEC2 Θ = 1 + TAN2 Θ
#13
CSC2 Θ = 1 + COT2 Θ
#14
SIN Θ COS Θ = ½ SIN 2Θ
#15
COS 2Θ = COS2 Θ - SIN2 Θ
#16
2 TAN Θ
TAN 2Θ = ----------
1 - TAN2 Θ
```
These trigonometric identities are particularly useful in the electronics field.
The sum formulae are given by
```sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
cos(a-b) = cos(a)cos(b) + sin(a)sin(b).
```
By taking linear combinations of these, we obtain the product formulae.
```2sin(a)cos(b) = sin(a+b) + sin(a-b)
2cos(a)cos(b) = cos(a+b) + cos(a-b)
2sin(a)sin(b) = cos(a-b) - cos(a+b).
```
These results can be proved using a geometric argument, or by using the exponential forms of cosine and sine, namely
```2cos(x) = exp(ix) + exp(-ix)
2isin(x) = exp(ix) - exp(-ix).
```
They are immensely useful across most of applied mathematics. One of the most simple consequences of them are the double angle formulae, obtained by putting b = a:
```sin(2a) = 2sin(a)cos(a)
cos(2a) = cos2(a) - sin2(a).
```
They can also be used to verify de Moivre's Theorem.
Trigonometric identities are mathematical identities that involve trigonometric functions. Wow... that was deep. What this means is that for any values for each variable you enter into the equation, the equation is either true or undefined.
The half angle identities (ever wanted to find sin(pi/12) ?) are as follows:
``` _____________
/ 1 - cos x
sin(x/2)= / ---------
\/ 2
_____________
/ 1 + cos x
cos(x/2)= / ---------
\/ 2
_____________
/ 1 + cos x
tan(x/2)= / ---------
\/ 1 - cos x
_____________
/ 2
csc(x/2)= / ---------
\/ 1 - cos x
_____________
/ 2
sec(x/2)= / ---------
\/ 1 + cos x
_____________
/ 1 - cos x
cot(x/2)= / ---------
\/ 1 + cos x
```
Also, the addition and double angle identities for tangent are:
``` tan a + tan b
tan(a+b)= ------------------
1 - (tan a)(tan b)
2 tan a
tan(2a)= ---------
1 - tan2a```
Another set of identities worth knowing are the factor formulae. Since `sin(a+b) != sin(a) + sin(b)`, these rules have been derived.
``` a + b a - b
sin(a) + sin(b) = 2sin ----- cos -----
2 2
a + b a - b
sin(a) - sin(b) = 2cos ----- sin -----
2 2
a + b a - b
cos(a) + cos(b) = 2cos ----- cos -----
2 2
a + b a - b
cos(a) - cos(b) = 2sin ----- sin -----
2 2
```
More likely than not, in a practical situation the equation `acos(θ) + bsin(θ)` will crop up. In such cases, the harmonic form is necessary.
The basic structure of the harmonic form is either `Rcos(θ ± α)` or `Rsin(θ ± α)`. (R > 0, α is acute). This is best explained with an example:
```To express 3cosθ - 4 sinθ in the form Rcos(θ + α):
Let 3cosθ - 4sinθ ≡ Rcos(θ + α)
≡ R(cosθcosα - sinθsinα)
≡ Rcosθcosα - Rsinθsinα
Now equate the coefficients of cosθ and sinθ to obtain
3 = Rcosα (1) and 4 = Rsinα (2)
Squaring (1) and (2) and adding give:
R2cos2α + R2sin2α = 32 + 42
R2(cos2α + sin2α) = 25
∴ R2 = 25 (since cos2α + sin2α = 1)
∴ R = 5 (since R > 0)
Dividing (2) by (1) gives
Rsinα 4
----- = -
Rcosα 3
4
∴ tanα = -
3
∴ α = 53.1°
Therefore, we have
3cosθ - 4sinθ = 5cos(θ + 53.1°)
```
The explanations in this section come from my own study. The specific example above comes from Introducing Pure Mathematics, Smedley/Wiseman.
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## A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly charged so tha
Question
A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly charged so that the potential difference between its plates is 5.0 V. The two charged capacitors are then connected to each other in parallel with positive plate connected to positive plate and negative plate connected to negative plate. Find the charch that flows from one capacitor to the other when the capacitors are connected (17 microcoulombs) and the energy that is dissipated when the two capacitors are connected together (42 microjoules)
in progress 0
2 weeks 2021-07-21T13:18:30+00:00 1 Answers 0 views 0
Explanation:
Given that,
First Capacitor is 10 µF
C_1 = 10 µF
Potential difference is
V_1 = 10 V.
The charge on the plate is
q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC
q_1 = 100 µC
A second capacitor is 5 µF
C_2 = 5 µF
Potential difference is
V_2 = 5V.
Then, the charge on the capacitor 2 is.
q_2 = C_2 × V_2
q_2 = 5µF × 5 = 25 µC
Then, the average capacitance is
q = (q_1 + q_2) / 2
q = (25 + 100) / 2
q = 62.5µC
B. The two capacitor are connected together, then the equivalent capacitance is
Ceq = C_1 + C_2.
Ceq = 10 µF + 5 µF.
Ceq = 15 µF.
The average voltage is
V = (V_1 + V_2) / 2
V = (10 + 5)/2
V = 15 / 2 = 7.5V
Energy dissipated is
U = ½Ceq•V²
U = ½ × 15 × 10^-6 × 7.5²
U = 4.22 × 10^-4 J
U = 422 × 10^-6
U = 422 µJ
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# Re: completeness of the relational lattice
From: Jan Hidders <hidders_at_gmail.com>
Date: Tue, 26 Jun 2007 00:53:52 -0700
> On Jun 25, 12:56 pm, Jan Hidders <hidd..._at_gmail.com> wrote:
>
>
>
> > > > (7) r * {()} = r
> > > > (13) {()} = {()} + []
>
> > > > Special distribution equalities:
>
> > > > (8) r * (s + t) = (r * s) + (r * t)
> > > > if A(r) * A(s) <= A(t) or A(r) * A(t) <= A(s)
> > > > (9) r + (s * t) = (r + s) * (r + t)
> > > > if A(s) * A(t) <= A(r)
>
> > > > Absorption:
>
> > > > (20) r + (r * s) = r
> > > > (21) r * (r + s) = r
>
> > > > Empty relations:
>
> > > > (10) R = R + [H]
> > > > if H is the header of R
> > > > (11) [H] * [S] = [H + S]
> > > > (12) [H] + [S] = [H * S]
> > > > (22) R * [] = [H]
> > > > if H is the header of R
>
> > > You have informal "H is the header of R" in many places. Why don't we
> > > use this axiom as a definition of [H]? Then, we just substitute [H]
> > > with R * [].
>
> > Well, I would not say it is informal. It has a precise meaning since
> > we can assume that we know the database schema, and given that schema
> > you can see which axioms are generated by the axiom schemas.
>
> So you have axiom schemas, instead of a single axiom system? I don't
> challenge this approach, but think that a single axiom system is
> workable as well (see below).
It's a schema anyway because you still have relation variables. Although I agree that I need two types of variables, or three if you count the singleton attributes, and you have one (or two if you have special ones for singleton sets, as you suggest later). On the other hand, I only have equations, and you need propositional formulas over equation.
> > Your suggestion might work, but you would have to first formalize what
> > your inference mechanism is before we can start thinking about it.
> > Right now my formalism is quite simple. It derives r = s iff it can
> > rewrite r to s with the given axioms. Yours would probably be much
> > more complicated than that. Can you give a formal definition of yours?
>
> OK, I have relation variables, 5 constants 00, 01, 10, 11, 1E, and two
> binary operations \/, /\, and the inequlity Y <= X abbreviation for X /
> \ Y = X. Axioms:
>
> 1-8. Lattice axioms
> 9. Spight criteria:
> (R/\00) \/ (S/\00) <= Q/\00 and (R/\00) \/ (Q/\00) <= S/\00 implies
> (R /\ S) \/ (R /\ Q) = R /\ (S \/ Q)
> 10. (R/\00) \/ (Q/\00) <= R/\00 implies
> (R /\ S) /\ (R \/ Q) = R \/ (S /\ Q)
> 11. 01 <= R <= 10
> 12. R = (00 /\ R) \/ (11 /\ R)
> 13. Universal equality relation 1E axioms.
> Do you imply that I can't reduce both sides of the expression R = S to
> the Union Normal Form in this system?
I don't know, you've onlly given axioms and not the inference mechanism. Since they are propositional formulas I assume you want to use propositional calculus. If you add this you can no longer claim that you have a pure algebraic axiomatization, which makes the completeness claim less interesting. But is propositional calculus all you allow, or are you also allowing reasoning by rewriting? The latter can be captured in either your inference mechanism or by adding adding extra axioms (e.g. the rules "if R = R' and S = S' then R /\ S = R' /\ S'" and "if R = S and S= T then S = T" and ... ). There is the usual trade off between putting your inference in the axioms or in the inference mechanism. So how do you want do that? And yes, I want a full formal specification of that. You cannot write a paper on the completeness of some reasoning system and not specify formally its inference mechanism. Note by the way that this changes the notion of "complete" because it then not only means that you can derive all equations that hold, but in fact that you can derive all propositional formulas (including negations) over equations that hold. I have no idea how to prove that. Do you?
> > Yes, but again this means that you extend your inference mechanism,
> > namely with first order logic inference.
>
> OK, if this is a critical showstopper, then I would have to introduce
> single attribute relations, conveniently designated with small letters
> r,s,t etc. Your rule #28, for example becomes
>
> 28. x /\ 00 <= R /\ 00 implies
> R /\ ((x /\ 00) \/ 1E) = R
Ok. I think that would work. But also here you have to say how the inference deals with this. How does it know what and what not to match x with when it rewrites? This is solvable, but again it adds complexity to the inference mechanism.
> The (informal) implication is there in both systems yours and mine,
> there is no way around it.
Of course. But the difference is that I have fully formally specified what my system is (the axioms *and* the inference mechanism) and you have not. Moreover it seems that you want to push a lot of the complexity into the inference mechanism in order to keep you set of axioms simple. As I already said, that makes the completeness claim at the same time harder to prove, and less interesting.
> > > > (30) <> = {()}
>
> This is a theorem, not axiom.
Yes, in your inference mechanism, but not in mine. So I need it.
> > > > Miscellaneous
>
> > > > (26) <H> + [S] = <H * S>
> > > > (29) [H] * <S> = [H + S]
> > > > (31) ((r * <S>) + [H]) * <S> = ((r * <S>) + [H'])
> > > > if S * H nonempty and H + S = H'
>
> > > You have to specify that headers of <S> and r overlap on no more than
> > > a single attribute -- and I don't see this condition here.
>
> > Why do you think that condition is necessary?
>
> OK, I see that you slightly modified the equality axiom. Can you
>
> (((R(x,y) * <yz>) + [xz]) * <yz>) + [xy] = R(x,y)
>
Sure. (I'm skipping rewrites involving shifting brackets):
``` (((R(x,y) * <yz>) + [xz]) * <yz>) + [xy]
= (((R(x,y) * <yz>) + [xz]) * <yz> * <yz>) + [xy] (1)
= (((R(x,y) * <yz>) + [xyz]) * <yz>) + [xy] (31)
= (R(x,y) * <yz> * <yz>) + ([xyz] * <yz>) + [xy] (8)
= (R(x,y) * <yz>) + ([xyz] * <yz>) + [xy] (1)
= ((R(x,y) * <yz>) + [xyz] + [xy] (29)
= ((R(x,y) * <yz>) + [xy] (12)
= (R(x,y) + [xy]) * (<yz> + [xy]) (9)
= (R(x,y) + [xy]) * <y> (26)
```
= R(x,y) * <y>
(10)
=
R(x,y)
(28)
• Jan Hidders
Received on Tue Jun 26 2007 - 09:53:52 CEST
Original text of this message
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Who was Euclid? - Greek Geometry
# Who was Euclid?
which is a piece of one of Euclid's books
April 2016 - Nobody knows much about Euclid's life anymore - it is all forgotten. We only know that he worked at the University of Alexandria, in Egypt, for a while. There are no pictures of him. We can't even be sure he existed; Euclid could be a made-up name for a committee of mathematicians all working together.
Either way, Euclid (or the people who called themselves Euclid) must have lived around 300 BC, in the Hellenistic period. He (or they) seems to have studied at Plato's Academy in Athens, where he learned some of the mathematics that is in his books. He probably knew Aristotle there. But Euclid didn't always agree with Aristotle; unlike Aristotle, Euclid thought that when people used their eyes to see, invisible rays of "seeing" came out of their eyes and hit the things they saw. Euclid, who probably knew Aristotle, used the camera obscura to show that light always travelled in straight lines.
Like Anaxagoras before him, Euclid (YOU-klid) wanted to prove that things were true by using logic and reason. We still have copies of Euclid's books today, and they begin with basic definitions of a point and a line and shapes, and then go on to use geometry to prove, for instance, that all right angles are equal, that you can draw a straight line between any two points, and that two things which are both equal to the same thing are also equal to each other.
Euclid's later books teach more advanced math, including how triangles and circles work (Euclid proved that the area of a circle had to be πr-squared), irrational numbers, and three-dimensional geometry. Euclid's interest in proofs is probably related to the interest of Jain mathematicians in India in logic and proofs about the same time, though we don't know enough to say which came first.
Euclid is famous because his books were so easy to understand that they were used as the main math book in all schools in Europe, West Asia, and America for two thousand years, until the 20th century. The geometry book Anne of Green Gables hates so much is Euclid. There are still mathematicians working today who began studying geometry from Euclid's books.
## Learn by doing: the area of a sphereMore about Central Asian mathematicians
Greek and Roman Science, by Don Nardo (1998). Nardo has written a lot of good books about the ancient world for kids; this one is no exception.
Ancient Science: 40 Time-Traveling, World-Exploring, History-Making Activities , by Jim Wiese (2003). Activities, as the title says - how to make your own sundial, and so on. The author is a science teacher.
Early Greek Science: Thales to Aristotle, by Geoffrey Lloyd (1974).
History of Greek Mathematics: From Aristarchus to Diophantus, by Thomas L. Heath (1921, reprinted 1981). A lot of Euclid, but also describes who the other major Greek mathematicians were and what they did.
Episodes from the Early History of Mathematics, by Asger Aaboe (1997).
## More about Central Asian mathematiciansHellenistic EgyptQuatr.us home
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## How do you find the invariant subspace of a matrix?
A subspace is said to be invariant under a linear operator if its elements are transformed by the linear operator into elements belonging to the subspace itself. The kernel of an operator, its range and the eigenspace associated to the eigenvalue of a matrix are prominent examples of invariant subspaces.
## How do you show invariant in subspace?
Formal description
1. An invariant subspace of a linear mapping.
2. from some vector space V to itself is a subspace W of V such that T(W) is contained in W.
3. Let v be an eigenvector of T, i.e. T v = λv.
4. An invariant vector (i.e. a fixed point of T), other than 0, spans an invariant subspace of dimension 1.
What does it mean when a matrix is invariant?
The determinant, trace, and eigenvectors and eigenvalues of a square matrix are invariant under changes of basis. In other words, the spectrum of a matrix is invariant to the change of basis. The singular values of a matrix are invariant under orthogonal transformations.
Is an operator on R2 the invariant subspaces of the operator are?
We know that {0} and R2 are automatically invariant subspaces. Thus we only need to check for 1–dimensional invariant subspaces.
### Is Eigenspace an invariant?
A subspace V of Rn is invariant if L(v) ∈ V for every v ∈ V. The simplest such situation is that in which the invariant subspace is one-dimensional, i.e., spanned by a single nonzero vector v. In this case, the subspace span{v} is called an eigenspace.
### Is the 0 vector in an eigenspace?
Eigenvectors are by definition nonzero. Eigenvalues may be equal to zero. We do not consider the zero vector to be an eigenvector: since A 0 = 0 = λ 0 for every scalar λ , the associated eigenvalue would be undefined.
Is the eigenspace the Nullspace?
Both the null space and the eigenspace are defined to be “the set of all eigenvectors and the zero vector”. They have the same definition and are thus the same.
Are Eigenspaces subspaces?
An Eigenspace Is a Subspace (In fact, this is why the word “space” appears in the term “eigenspace.”) Let A be an n × n matrix, and let λ be an eigenvalue for A, having eigenspace Eλ.
## Is the subspace W always invariant under a matrix?
A subspace W is invariant under A if and only for every w ∈ W, Aw ∈ W. The only 0 -dimensional subspace is always invariant (for any matrix), since A0 = 0 ∈ {0}. V itself (in this case, R4) is also always invariant, since Av ∈ R4 for every v ∈ R4. So, let’s deal with the in-betweens: A 1 -dimensional invariant subspace;
## Which is invariant subspace theory for linear transformations of a vector space?
An invariant subspace theory for linear transformations of a vector space of infinite dimension into itself appears in the Stieltjes treatment of integration for functions of a real variable. The factorization of a polynomial with complex coefficients as a product of linear factors is again applied.
When to use invariant subspace theory in integration theory?
An invariant subspace theory applies to the difference–quotient transformation, taking a function F(z) of zinto the function [F(z)−F(0)]/z of z, when the transformation is everywhere defined. A transformation appears in the Stieltjes integration theory which was discovered by
Which is the subspace of a linear operator?
The kernel of a linear operator is the subspace Since and all the elements of are mapped into by the operator , the kernel is invariant under . The range of a linear operator is the subspace
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# Asymptote
## Asymptote
Line in which the distance to any given point of a curve tends toward zero when this point moves away on the curve infinitely (in the positive or negative direction of the independent variable).
### Properties
A horizontal asymptote is a line that is parallel with the x-axis.
A vertical asymptote is a line that is parallel with the y-axis
### Historical Note
The terme “asymptote” comes from the Greek word “asumptotos” which means “not falling together”.
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0
# What is 34 as the sum of two primes?
Updated: 4/28/2022
Wiki User
13y ago
34 = 17+17
34 = 23 + 11
34 = 29 + 5
34 = 31 + 3
Wiki User
13y ago
Lvl 2
2y ago
17 +17
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# physics
A fisherman's scale stretches 2.8 cm when a 3.7-kg fish hangs from it. (a) What is the spring constant? (b) What will be the amplitude and frequency of vibration fi the fish is pulled down 2.5 cm more and released so that it vibrates up and down?
I have a question about finding the frequency. For some reason I keep on getting ((3.7 / 129.5)^.5) / (2 * pi) = 0.0269020955 that for my answer but the back of the book says 3.0 hz
1. 👍 0
2. 👎 0
3. 👁 106
1. I did this below:
http://www.jiskha.com/display.cgi?id=1263835390
1. 👍 0
2. 👎 0
posted by Damon
## Similar Questions
1. ### physics
A fisherman's scale stretches 2.3 cm. when a 2.1 kg. fish hangs from it. What will be the frequency of vibration if the fish is pulled down and released so that it vibrates up and down?
asked by kassy on November 20, 2016
2. ### physics
A fisherman's scale stretches 2.8 cm. when a 2.1 kg. fish hangs from it. What will be the frequency of vibration if the fish is pulled down and released so that it vibrates up and down?
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11. ### jishka
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asked by Kate on January 18, 2010
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## WHAT YOU WILL LEARN:
• Lesson 1 – Understanding Fractions as Equal Parts
• Lesson 2 – Understanding the Concept of Same Whole
• Lesson 3 – Fractions on the Number Line
• Lesson 4 – Equivalent Fractions
• Lesson 5 – Comparing Fractions
• Lesson 6 – Mixed Numbers and Improper Fractions
• Lesson 7 – Addition and Subtraction Part 1 – Like Fractions
• Lesson 8 – Addition and Subtraction Part 2 – Related Fractions
• Lesson 9 – Addition and Subtraction Part 3 – Unrelated Fractions
• Lesson 10 – Fraction of a Set
• Lesson 11 – Fraction Multiplication
• Lesson 12 – Dividing with Fractions
• Lesson 13 – Fractions and Decimals
• Lesson 14 – Fraction Multiplication as Scaling
Each Lesson is accompanied by a practice worksheet you can print out and work with your class immediately!
Scroll to Top
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# Physics 203
posted by on .
A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 5.0 m. The cylinder arrives at the bottom of the plane 2.7 s after the sphere. Determine the angle between the inclined plane and the horizontal.
• Physics 203 - ,
5m=1/2(5/7*(9.81*sin(theta))*((2.7*sqrt(1/2))^2)/((sqrt(5/7)-sqrt(1/2)))^2)
All you need to do is solve for sin(theta)
you do 5*((sqrt(5/7)-sqrt(1/2)))^2)*2*7/((5*9.81*((2.7*sqrt(1/2))^2))
sin(theta)=0.00746
then you do the inverse of sin(theta) to get your degree. degree should equal 0.428
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# What Does "AUC" Mean in Chemotherapy?
Chemotherapy is designed to kill cancerous cells. However, when it reaches a high concentration in the body it can begin to attack normal, healthy cells. It is therefore very important to monitor the level of chemotherapeutic drugs in the blood. Area Under the Curve (AUC) is a mathematical method of measuring drug concentrations.
Is This an Emergency?
If you are experiencing serious medical symptoms, seek emergency treatment immediately.
## Area Under the Curve
The "curve" referred to in AUC is the curve on a concentration-versus-time graph. The concentration of a drug in the patient's blood is plotted against the time when the sample was taken. The area beneath this curve is measured with basic calculus.
## Calvert Formula
The calvert formula is commonly used to determine the best dose of chemotherapeutic drug for a patient. It states that the dose is equal to the desired AUC multiplied by the patient's glomerular filtration rate (GFR). GFR is a measure of how quickly a patient eliminates the drug from his body.
## Clinical Uses
The calvert formula is used to determine the dose of chemotherapy drugs such as 5-fluorouracil and carboplatin. AUC is a quick measure of how much of the drug an individual has been exposed to in a given amount of time. Keeping the AUC within a defined range helps limit the dangerous side effects of chemotherapy.
The Wrap Up
GFR is a measure of how quickly a patient eliminates the drug from his body. Area Under the Curve (AUC) is a mathematical method of measuring drug concentrations. Area Under the Curve (AUC) is a mathematical method of measuring drug concentrations.
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# AQA D1 Kruskals algorithm question watch
1. I'm stuck on question 10a) doing kruskals algorithm, here is the question:
For my working out I have done
I have the wrong order and the wrong length,
the correct order is
GF, BD, EG, DC, HJ, AD, GF, CF
With total length as 31
I have no idea where I have gone wrong, the answer hasn't included I for some reason too and used GF twice so i'm confused
2. (Original post by Sayless)
I have no idea where I have gone wrong, the answer hasn't included I for some reason too and used GF twice so i'm confused
Where you have a choice of edges of the same weight, the order can vary. CD then EG, or EG then CD; both are correct.
There are 10 nodes and so a spanning tree must have 9 edges.
You cannot get a repeated edge.
It must include "I".
3. (Original post by ghostwalker)
Where you have a choice of edges of the same weight, the order can vary. CD then EG, or EG then CD; both are correct.
There are 10 nodes and so a spanning tree must have 9 edges.
You cannot get a repeated edge.
It must include "I".
Thanks a lot, just needed someone to make sure I didn't make any mistake
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Formulation to Matlab code
seems OK
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snake
This example demonstrates how to get mouse coordinates in figure
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why "MATLAB Support Package for Android Sensors" is not available any more????
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1D matrix addition weirdness
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# Choose the correc
Let x3 = cos p and y3 = cos q
cos-1x3 = p and cos-1y3 = q --- (1)
Comparing L.H.S and R.H.S we get,
Substituting value of p and q from (1)
Differentiating w.r.t x we get,
Comparing L.H.S and R.H.S we get
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https://help.xlstat.com/customer/en/portal/articles/2062237-nonlinear-multiple-regression-in-excel-tutorial?b_id=9283
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# Nonlinear multiple regression in Excel tutorial
2016-05-06
This tutorial will help you set up and interpret a nonlinear multiple regression in Excel using the XLSTAT software.
Not sure this is the modeling feature you are looking for? Check out this guide.
## Dataset to run a nonlinear multiple regression
An Excel sheet with both the data and the results can be downloaded by clicking here. Our purpose is to study the effect of the concentration of two components, C1 and C2, on the viscosity of a yogurt. The model that we want to fit writes:
F(C1, C2) = pr5 / (1+Exp(-pr1-pr2*C1-pr3*C2-pr4*C1*C2))
pr1, ..., pr5 are the parameters of the model. This logistic-like model allows to take into account both the concentrations of the components and the interaction between them.
## Setting up a nonlinear multiple regression
After opening XLSTAT, select the XLSTAT / Modeling data / Nonlinear regression command, or click on the corresponding button of the Modeling Data toolbar (see below).
Once you've clicked on the button, the nonlinear regression dialog box appears. Select the data on the Excel sheet.
The Dependent variable (or response variable) is in our case the "Viscosity".
The quantitative explanatory variables are the concentration of the two components "C1" and "C2".
As we selected the column headers, we left the option Variable labels option activated. We left the Residuals option activated as well, because we want to analyze the predictions and the residuals.
In the Options tab we selected the values of the initial values of the five parameters.
In the Functions tab, the various functions are displayed. As the function we want to use is not listed in the Preprogrammed functions (you can notice the univariate version of the function in the list), we needed to enter the model: we first clicked on Add, then entered the function, then checked Derivatives, then selected them on the Excel sheet. In order to add this function to the user functions library, we clicked on Save. The function is then automatically added and selected.
The computations begin once you have clicked on the OK button. The results will then be displayed.
## Interpreting the results of a non linear multiple regression
The first table gives the basic statistics of the selected variables.
The second table (see below) displays the goodness of fit coefficients, including the R² (coefficient of determination), and the SSE (sum of square of errors), the later being the criterion used for the model optimization. The R² corresponds to the % of the variability of the dependant variable (the viscosity) that is explained by the two explanatory variables (the components). The closer to 1 the R² is, the better the fit.
In our case, 99% of the variability is explained by the two variables and their interaction, which is an excellent result that confirms that the selected model is appropriate.
The next table shows the results for the model parameters. As we can see, the ratios (parameter)/(std deviation) are larger for pr5 and pr4. As the same ratio is the largest for pr5 we deduce that the interaction between the two components has a greater effect on the viscosity than the concentrations themselves.
The following chart allows to visualize the quality of the fit by comparing the predicted values to the observed values.
1c26995d494fb3061dd0ae8571ffc0a4@xlstat.desk-mail.com
https://cdn.desk.com/
false
desk
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https://mathematica.stackexchange.com/questions/120723/intersection-of-volumes-defined-by-inequalities
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# Intersection of volumes defined by inequalities
I have 5 inequalities with some parameters like dh, din and thetawhich I intend to vary.
In Mathematica, I am trying to:
1. Check if the inequalities have a solution.
2. Find the common volume if they do.
I tried this:
theta = 85* Pi/180;
dh = 0.5;
din = 1.016;
dout = 1.27;
zz = z*Cos[theta] - x*Sin[theta];
xx = z*Sin[theta] + x*Cos[theta];
yy = y;
region1 = z^2 >= (dh/2 + 0.0254)^2 - (dout/2)^2*x^2/(x^2 + y^2);
region2 = (din/2)^2 < x^2 + y^2 < (dout/2)^2;
region3 = -1 <= z <= 1;
region4 = zz^2 <= (dh/2)^2 - (dout/2)^2*xx^2/(xx^2 + yy^2);
region5 = (din/2)^2 <= xx^2 + yy^2 <= (dout/2)^2;
region = region1 && region2 && region3 && region4 && region5;
AbsoluteTiming[Resolve[Exists[{x, y, z}, region], Reals]]
whose execution is not completing.
Plotting with
AbsoluteTiming[RegionPlot3D[region1 && region2 && region3 && region4 && region5,
{x , -1.27/2, 1.27/2}, {y, -1.27/2, 1.27/2}, {z, -1, 1}, PlotRange
-> {{-0.7, 0.7}, {-0.7, 0.7}, {-0.7, 0.7}}, AspectRatio -> 1, PlotPoints -> 500,
MaxRecursion -> 15]]
gives an inaccurate, broken volume after 15 minutes of running.
Numerical integration gives 0 (possibly due to some volumes being negative?)
How can I quickly and accurately find if the regions have a common volume and if they do, the magnitude of the common volume?
Show[RegionPlot3D[ region1 && region2 && region5,
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https://edurev.in/course/quiz/attempt/7813_Modem-Digital-Modulation-Detection-Techniques/24d59434-c087-404b-bcec-25f7fd0d0a44
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Courses
# Modem Digital Modulation & Detection Techniques
## 10 Questions MCQ Test GATE ECE (Electronics) 2022 Mock Test Series | Modem Digital Modulation & Detection Techniques
Description
This mock test of Modem Digital Modulation & Detection Techniques for GATE helps you for every GATE entrance exam. This contains 10 Multiple Choice Questions for GATE Modem Digital Modulation & Detection Techniques (mcq) to study with solutions a complete question bank. The solved questions answers in this Modem Digital Modulation & Detection Techniques quiz give you a good mix of easy questions and tough questions. GATE students definitely take this Modem Digital Modulation & Detection Techniques exercise for a better result in the exam. You can find other Modem Digital Modulation & Detection Techniques extra questions, long questions & short questions for GATE on EduRev as well by searching above.
QUESTION: 1
### Assertion (A): Binary phase shift keying (BPSK) is the most efficient of the three digital modulation, i.e., ASK, FSK and PSK. Reason (R): In BPSK, phase of sinusoidal carrier is changed according to the data bit to be transmitted.
Solution:
BPSK is the most efficient of the three digital modulation, i.e. ASK, FSK and PSK because BPSK is used for high bit rates. Reason is also a true statement but not the correct explanation of assertion.
QUESTION: 2
Solution:
QUESTION: 3
### Match List-I with List-ll and select the correct answer using the codes given below the lists: List-I A. BPSK B. AM C. QPSK D. ASK List-ll 1. A group of two binary bits is represented by one phase state. 2. A binary bit is represented by one phase state. 3. Carrier ON or OFF depending on whether ‘V or ‘0’ is to be transmitted. 4. Detected using envelope detector
Solution:
QUESTION: 4
The bit rate of a digital communication system is 34 Mbps. The modulation scheme is QPSK. The baud rate of the system is
Solution:
In BPSK, baud rate is half the bit rate due to which there is more effective utilization of the available bandwidth of the transmission channel. Given, bit rate = 34 Mbps
QUESTION: 5
Consider the following statements;
1. BPSK has a bandwidth which is lower than that of a BFSK signal.
2. BPSK yields the maximum value of probability of error compared to ail the three digital modulation techniques i.e. ASK, FSK and PSK.
3. Binary FSK has the highest system complexity.
4. Binary ASK is demodulated using coherent detection while binary FSK and PSK are demodulated using envelope detection
Which of the statements given above are correct?
Solution:
• Statement-1 is correct.
• BPSK has the best performance of all the three digital modulation techniques in presence of noise, it yields the minimum value of probability of error compared to ASK, FSK and PSK. Hence, statement-2 is false.
• Statement-3 is correct since binary FSK has highest system complexity while binary ASK has least system complexity.
• Binary ASK is demodulated using envelope detection while binary FSK and PSK are demodulated using coherent detection.
Hence, statement-4 is fals
QUESTION: 6
Comparison of MSKand QPSK scheme shows that
Solution:
More energy (or power) is contained in the main lobe of MSK (about 99% of the total energy or power) as compared to the QPSK (about 90% of the total energy or power).
QUESTION: 7
In a digital continuous-time communication system, the bit rate of NRZ data stream is 1 Mbps and carrier frequency of transmission is 100 MHz. What is the bandwidth requirement of the channel in BPSK and QPSK systems respectively?
Solution:
The bit period of NRZ data stream is
In BPSK, each binary bit is a symbol, therefore symbol duration Ts = Tb = 1 μs.
∴ The bandwidth required for BPSK system is
In QPSK, we group two successive bits to form one symbol, therefore symbol duration,
QUESTION: 8
The error probability of QPSK is
Solution:
The error probability of QPSK is better (or lower) than that of BPSK due to which QPSK is used for very high bit rate data transmission.
QUESTION: 9
Bandwidth of BPSK is ___ than that of BFSK.
Solution:
Minimum bandwidth required in BPSK = 2fb.
Minimum bandwidth required in BFSK = 4fb.
Hence, bandwidth of BPSK is lower than that of BFSK.
QUESTION: 10
The error probability of QPSK is
Solution:
The error probability of QPSK is better (or lower) than that of BPSK due to which QPSK is used for very high bit rate data transmission.
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https://manuals.pqstat.pl/en:statpqpl:zgodnpl:nparpl:kappalpl
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# PQStat - Baza Wiedzy
### Pasek boczny
en:statpqpl:zgodnpl:nparpl:kappalpl
### The Cohen's Kappa coefficient and the test examining its significance
The Cohen's Kappa coefficient (Cohen J. (1960)1)) defines the agreement level of two-times measurements of the same variable in different conditions. Measurement of the same variable can be performed by 2 different observers (reproducibility) or by a one observer twice (recurrence). The coefficient is calculated for categorial dependent variables and its value is included in a range from -1 to 1. A 1 value means a full agreement, 0 value means agreement on the same level which would occur for data spread in a contingency table randomly. The level between 0 and -1 is practically not used. The negative value means an agreement on the level which is lower than agreement which occurred for the randomly spread data in a contingency table. The coefficient can be calculated on the basis of raw data or a contingency table.
Unweighted Kappa (i.e., Cohen's Kappa) or weighted Kappa can be determined as needed. The assigned weights () refer to individual cells of the contingency table, on the diagonal they are 1 and off the diagonal they belong to the range .
Unweighted Kappa
It is calculated for data, the categories of which cannot be ordered, e.g. data comes from patients, who are divided according to the type of disease which was diagnosed, and these diseases cannot be ordered, e.g. pneumonia , bronchitis and other . In such a situation, one can check the concordance of the diagnoses given by the two doctors by using unweighted Kappa, or Cohen's Kappa. Discordance of pairs and will be treated equivalently, so the weights off the diagonal of the weight matrix will be zeroed.
Weighted Kappa
In situations where data categories can be sorted, e.g., data comes from patients who are divided by the lesion grade into: no lesion , benign lesion , suspected cancer , cancer , one can build the concordance of the ratings given by the two radiologists taking into account the possibility of sorting. The ratings of than may then be considered as more discordant pairs of ratings. For this to be the case, so that the order of the categories affects the compatibility score, the weighted Kappa should be determined.
The assigned weights can be in linear or quadratic form.
• Linear weights (Cicchetti, 19712)) – calculated according to the formula:
The greater the distance from the diagonal of the matrix the smaller the weight, with the weights decreasing proportionally. Example weights for matrices of size 5×5 are shown in the table:
• Square weights (Cohen, 19683)) – calculated according to the formula:
The greater the distance from the diagonal of the matrix, the smaller the weight, with weights decreasing more slowly at closer distances from the diagonal and more rapidly at farther distances. Example weights for matrices of size 5×5 are shown in the table:
Quadratic scales are of greater interest because of the practical interpretation of the Kappa coefficient, which in this case is the same as the intraclass correlation coefficient 4). To determine the Kappa coefficient compliance, the data are presented in the form of a table of observed counts , and this table is transformed into a probability contingency table .
The Kappa coefficient () is expressed by the formula:
where:
,
,
, - total sums of columns and rows of the probability contingency table.
Note
denotes the concordance coefficient in the sample, while in the population.
The standard error for Kappa is expressed by the formula:
where:
,
.
The Z test of significance for the Cohen's Kappa () (Fleiss,20035)) is used to verify the hypothesis informing us about the agreement of the results of two-times measurements and features and it is based on the coefficient calculated for the sample.
Basic assumptions:
Hypotheses:
The test statistic is defined by:
Where:
.
The statistic asymptotically (for a large sample size) has the normal distribution.
The p-value, designated on the basis of the test statistic, is compared with the significance level :
The settings window with the test of Cohen's Kappa significance can be opened in Statistics menu → NonParametric testsKappa-Cohen or in ''Wizard''.
EXAMPLE (diagnosis.pqs file)
You want to analyse the compatibility of a diagnosis made by 2 doctors. To do this, you need to draw 110 patients (children) from a population. The doctors treat patients in a neighbouring doctors' offices. Each patient is examined first by the doctor A and then by the doctor B. Both diagnoses, made by the doctors, are shown in the table below.
Hypotheses:
We could analyse the agreement of the diagnoses using just the percentage of the compatible values. In this example, the compatible diagnoses were made for 73 patients (31+39+3=73) which is 66.36% of the analysed group. The kappa coefficient introduces the correction of a chance agreement (it takes into account the agreement occurring by chance).
The agreement with a chance adjustment is smaller than the one which is not adjusted for the chances of an agreement.
The p<0.0001. Such result proves an agreement between these 2 doctors' opinions, on the significance level ,.
Radiological imaging assessed liver damage in the following categories: no changes (1), mild changes (2), suspicion of cancer , cancer . The evaluation was done by two independent radiologists based on a group of 70 patients. We want to check the concordance of the diagnosis.
Hypotheses:
Because the diagnosis is issued on an ordinal scale, an appropriate measure of concordance would be the weighted Kappa coefficient.
Because the data are mainly concentrated on the main diagonal of the matrix and in close proximity to it, the coefficient weighted by the linear weights is lower () than the coefficient determined for the quadratic weights (). In both situations, this is a statistically significant result (at the significance level), p<0.0001.
If there was a large disagreement in the ratings concerning the two extreme cases and the pair: (no change and cancer) located in the upper right corner of the table occurred far more often, e.g., 15 times, then such a large disagreement would be more apparent when using quadratic weights (the Kappa coefficient would drop dramatically) than when using linear weights.
1)
Cohen J. (1960), A coefficient of agreement for nominal scales. Educational and Psychological Measurement, 10,3746
2)
Cicchetti D. and Allison T. (1971), A new procedure for assessing reliability of scoring eeg sleep recordings. American Journal EEG Technology, 11, 101-109
3)
Cohen J. (1968), Weighted kappa: nominal scale agreement with provision for scaled disagreement or partial credit. Psychological Bulletin, 70, 213-220
4)
Fleiss J.L., Cohen J. (1973), The equivalence of weighted kappa and the intraclass correlation coeffcient as measure of reliability. Educational and Psychological Measurement, 33, 613-619
5)
Fleiss J.L., Levin B., Paik M.C. (2003), Statistical methods for rates and proportions. 3rd ed. (New York: John Wiley) 598-626
en/statpqpl/zgodnpl/nparpl/kappalpl.txt · ostatnio zmienione: 2022/02/13 21:45 przez admin
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https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-5-section-5-6-special-products-exercise-set-page-388/40
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## Algebra: A Combined Approach (4th Edition)
$b^{2}$ -9
( b + 3 ) ( b - 3) the som and difference of the same two terms $b^{2}$ - $3^{2}$ $b^{2}$ -9
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https://number.academy/102645
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# Number 102645 facts
The odd number 102,645 is spelled 🔊, and written in words: one hundred and two thousand, six hundred and forty-five. The ordinal number 102645th is said 🔊 and written as: one hundred and two thousand, six hundred and forty-fifth. Color #102645. The meaning of the number 102645 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 102645. What is 102645 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 102645.
## Interesting facts about the number 102645
### Asteroids
• (102645) 1999 VC44 is asteroid number 102645. It was discovered by CSS, Catalina Sky Survey from Catalina Station, Mount Bigelow on 11/3/1999.
## What is 102,645 in other units
The decimal (Arabic) number 102645 converted to a Roman number is (C)MMDCXLV. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
102645 seconds equals to 1 day, 4 hours, 30 minutes, 45 seconds
102645 minutes equals to 2 months, 2 weeks, 1 day, 6 hours, 45 minutes
### Codes and images of the number 102645
Number 102645 morse code: .---- ----- ..--- -.... ....- .....
Sign language for number 102645:
Number 102645 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
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#### Is Prime?
The number 102645 is not a prime number. The closest prime numbers are 102643, 102647.
#### Factorization and factors (dividers)
The prime factors of 102645 are 3 * 3 * 5 * 2281
The factors of 102645 are 1, 3, 5, 9, 15, 45, 2281, 6843, 11405, 20529, 34215, 102645.
Total factors 12.
Sum of factors 177996 (75351).
#### Powers
The second power of 1026452 is 10.535.996.025.
The third power of 1026453 is 1.081.467.311.986.125.
#### Roots
The square root √102645 is 320,382584.
The cube root of 3102645 is 46,821566.
#### Logarithms
The natural logarithm of No. ln 102645 = loge 102645 = 11,539032.
The logarithm to base 10 of No. log10 102645 = 5,011338.
The Napierian logarithm of No. log1/e 102645 = -11,539032.
### Trigonometric functions
The cosine of 102645 is -0,967215.
The sine of 102645 is 0,253958.
The tangent of 102645 is -0,262567.
## Number 102645 in Computer Science
Code typeCode value
PIN 102645 It's recommended that you use 102645 as your password or PIN.
102645 Number of bytes100.2KB
CSS Color
#102645 hexadecimal to red, green and blue (RGB) (16, 38, 69)
Unix timeUnix time 102645 is equal to Friday Jan. 2, 1970, 4:30:45 a.m. GMT
IPv4, IPv6Number 102645 internet address in dotted format v4 0.1.144.245, v6 ::1:90f5
102645 Decimal = 11001000011110101 Binary
102645 Decimal = 12012210200 Ternary
102645 Decimal = 310365 Octal
102645 Decimal = 190F5 Hexadecimal (0x190f5 hex)
102645 BASE64MTAyNjQ1
102645 SHA17dcc7528e7f1e46fd5aebb1507e9f3c77461c4e5
102645 SHA224d7fedc08c46355bc6cbae64cb3816dd87a952ae5109971d5845840da
More SHA codes related to the number 102645 ...
If you know something interesting about the 102645 number that you did not find on this page, do not hesitate to write us here.
## Numerology 102645
### Character frequency in the number 102645
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 1 2 1 6 1 4 1 5 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 102645, the numbers 1+0+2+6+4+5 = 1+8 = 9 are added and the meaning of the number 9 is sought.
## № 102,645 in other languages
How to say or write the number one hundred and two thousand, six hundred and forty-five in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 102.645) ciento dos mil seiscientos cuarenta y cinco German: 🔊 (Nummer 102.645) einhundertzweitausendsechshundertfünfundvierzig French: 🔊 (nombre 102 645) cent deux mille six cent quarante-cinq Portuguese: 🔊 (número 102 645) cento e dois mil, seiscentos e quarenta e cinco Hindi: 🔊 (संख्या 102 645) एक लाख, दो हज़ार, छः सौ, पैंतालीस Chinese: 🔊 (数 102 645) 十万二千六百四十五 Arabian: 🔊 (عدد 102,645) مائة و اثنان ألفاً و ستمائة و خمسة و أربعون Czech: 🔊 (číslo 102 645) sto dva tisíce šestset čtyřicet pět Korean: 🔊 (번호 102,645) 십만 이천육백사십오 Danish: 🔊 (nummer 102 645) ethundrede og totusindsekshundrede og femogfyrre Hebrew: (מספר 102,645) מאה ושניים אלף שש מאות ארבעים וחמש Dutch: 🔊 (nummer 102 645) honderdtweeduizendzeshonderdvijfenveertig Japanese: 🔊 (数 102,645) 十万二千六百四十五 Indonesian: 🔊 (jumlah 102.645) seratus dua ribu enam ratus empat puluh lima Italian: 🔊 (numero 102 645) centoduemilaseicentoquarantacinque Norwegian: 🔊 (nummer 102 645) en hundre og to tusen seks hundre og førtifem Polish: 🔊 (liczba 102 645) sto dwa tysiące sześćset czterdzieści pięć Russian: 🔊 (номер 102 645) сто две тысячи шестьсот сорок пять Turkish: 🔊 (numara 102,645) yüzikibinaltıyüzkırkbeş Thai: 🔊 (จำนวน 102 645) หนึ่งแสนสองพันหกร้อยสี่สิบห้า Ukrainian: 🔊 (номер 102 645) сто дві тисячі шістсот сорок п'ять Vietnamese: 🔊 (con số 102.645) một trăm lẻ hai nghìn sáu trăm bốn mươi lăm Other languages ...
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## Narrow Search
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# Family Science Night Facilitators Guide
The 9-session NASA Family Science Night program emables middle school children and their families to discover the wide variety of science, technology, engineering, and mathematics being performed at NASA and in everyday life. Family Science Night... (View More)
Audience: Informal education
# Reasons for the Seasons
This is an activity about seasons. Learners compare the seasons though identifying seasonal activities and drawing scenes in each season. Then, they compare the temperature on thermometers left under a lamp for different lengths of time to explore... (View More)
# Seasons Unraveled - The Angle of Sunlight
This is an activity about understanding how the Earth’s axial tilt changes the angle at which sunlight hits the Earth, contributing to the variations in temperature throughout the seasons. Learners will create a sun angle analyzer in order to see... (View More)
Audience: Middle school
Materials Cost: 1 cent - \$1
# Trip to the Sun
This is an activity about the size and scale of the Sun-Earth system. Learners will take an imaginary trip to the Sun by comparing images of the Sun and Earth at different points in altitude above the Earth. This is to ultimately conceptualize the... (View More)
# Tilted Earth
This is an activity about how the Earth's axial tilt causes its seasons. Learners will make a model using polystyrene spheres and a light bulb to represent the Earth-Sun system, showing why the tilt of the Earth’s spin axis causes its seasons due... (View More)
Keywords: NSTA2013
Audience: Middle school
Materials Cost: \$1 - \$5 per student
# Temperatures Around the World
This is an activity about seasons. Learners will analyze monthly temperature data from various cities around the globe in order to dispel the misconception that the distance between Earth and Sun may be responsible for the seasons. This is Activity... (View More)
Audience: Middle school
Materials Cost: Free per group of students
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http://www-users.math.umn.edu/~rogness/math1151/exam1.txt
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Here's a rough idea of what I think is important for the first midterm. You can also keep an eye out for a practice midterm, which will probably be posted on Monday. - Angles. Positive and negative angles. Conversion between radians and degrees. - Definitions of trig functions. Given a point (x,y) on a circle of radius r which corresponds to an angle t, you should be able to give me the value of the trig functions. [For example, sin(t)=y/r.] In class we usually used a circle of radius r=1, which made the definitions easier, but you may or may not run across a circle with a different radius on the test. - Exact values of the trig functions. You should be able to give me exact values (no decimals) of the trig functions for all of our "nice" angles: multiples of 30 degrees and multiples of 45 degrees. You can find all of these by drawing a 30-60-90 or 45-45-90 triangle within a cirlce and using the definitions. - properties of the trig functions: even/odd, domain/range, min/max, where the zeros are, etc. - Graphs of the trig functions. You should be able to work with the "general" equations such as y=A*sin(wx) or y=A*sin(wx-p). Same for cosine. You should be able to graph tan(x) and apply some basic transformations to it. Given a graph of (for example) cot(x) you should be able to graph something like 2*cot(x-1). - basic trig identities, such as those in section 5.4. - Inverse sine and cosine functions. (We'll continue talking about these on Monday) More details this week in class -- Jon
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NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/
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Simplicity is the ultimate sophistication. ..."da Vinci (1452-1519, Italian artist, sculptor, painter, architect, engineer and scientist) "
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Author Topic: Two mirrors at angle with light incident (Read 1300 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
samara330
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« Embed this message on: November 12, 2016, 06:19:31 pm » posted from:,,Satellite Provider
if i had two mirrors with angle (theta) between them and a light incident on one of them and reflected to the other
what is the maximum height the light ray can get?
like the lowest level between the two mirrors
h .... starts from the point the light hit the first mirror
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Graciani
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« Embed this message Reply #1 on: February 12, 2018, 01:55:31 pm » posted from:Bangkok,Krung Thep,Thailand
The source of these content is information that I am very helpful to me.
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Simplicity is the ultimate sophistication. ..."da Vinci (1452-1519, Italian artist, sculptor, painter, architect, engineer and scientist) "
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Related Topics Subject Started by Replies Views Last post Multiple Reflection from two plane mirrors Optics Fu-Kwun Hwang 11 93992 November 15, 2012, 12:29:19 pm by koclup1580 How many images? (Multiple Reflection from two plane mirrors/different angle) Optics Fu-Kwun Hwang 2 21677 August 07, 2012, 05:29:01 pm by rocky123 Light reflected from arbitrary number of mirrors Optics Fu-Kwun Hwang 3 17104 October 13, 2010, 12:29:15 pm by Fu-Kwun Hwang How many images (Multiple Reflection from two plane mirrors) optics ahmedelshfie 0 9567 May 20, 2010, 01:25:53 am by ahmedelshfie simultaneous combination of lenses and mirrors (with images) Request for physics Simulations sen-h 9 10850 July 24, 2018, 04:04:19 pm by sophiahustler
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# Don't get confused in 1/13
#1
var speed = 65;
// Complete the condition in the ()s on line 4
if (speed === 81) {
// Use console.log() to print "Slow down"
console.log("Slow down");
}
else {
// Use console.log() to print "Drive safe"
console.log("Drive safe");
}
#2
They might not have told you about this yet, but >= means 'greater than or equal to' so instead of 'if (speed === 81)' do if(speed >= 80).
#3
i know about that but to me it seems like its the same, thank you for that advice anyway.
#4
Well `>= 80` checks for a range of numbers anything from 80 to the maximum number of JavaScript will fall in this categorie. But === 81 checks for one specific number so in the sense of the exercise a speed of 250 would be fine and you would say "Drive safe" but someone driving exactly 81 would hear "Slow down".
#5
Try var speed = 65;
// Complete the condition in the ()s on line 4
if ("speed" === 80) {
// Use console.log() to print "Slow down"
console.log("Slow down");
}
else {
// Use console.log() to print "Drive safe"
console.log("Drive safe");
}
#6
Is this a solution or a question? Some more informations would be nice.
Also `"speed" === 80` doesn't make much as "speed" is a string and 80 number. So rather use speed without the "" to indicate that you want to use the variable which currently contains 65. And think if === is really the comparison operator of choice and it is pretty odd to tell someone with a speed of 80 to slow down but e.g. someone with 250 to "Drive safe"
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Dynamical Model with Application to the Analysis of the Sweet Spot on a Baseball Bat
American Journal of Computational Mathematics
Vol.05 No.04(2015), Article ID:62267,10 pages
10.4236/ajcm.2015.54040
Dynamical Model with Application to the Analysis of the Sweet Spot on a Baseball Bat
Li Cheng, Zaiqiang Ku*
College of Mathematics and Physics, Huanggang Normal University, Huanggang, China
Received 2 November 2015; accepted 25 December 2015; published 28 December 2015
ABSTRACT
In this paper, we analyze the effect on the ball-bat model of corking the bat, and investigate the relationship between the sweet spot and different materials. First and foremost, we develop a simple but effective theoretical model to give a rough estimate for the sweet spot of the bat. Second, we give a simplified abstract form of bat which facilitates the description and modeling study. Last but not least, according to different material caused rotary inertia and different recovery coefficient, researches on the different material hitting effect were made.
Keywords:
Baseball Bat, Sweet Spot, Dynamical Model, Finite Element
1. Introduction
Baseball, the national ball in USA and Japan, is popular and is thought as the combination of competition, wisdom, braveness and cooperation. Batters know from experience that there is a sweet spot on the baseball bat, about 17 cm from the end of the barrel, where the shock of the impact, felt by the hands, is reduced to such an extent that the batter is almost unaware of the collision (Bower [1] ). In addition to the popular book by Adair [2] , there have numerous papers which address a wide variety of issues amenable to a physics calculation. Notable experimental papers are those of Brody [3] , Cross [4] , Nicholls et al. [5] , King et al. [6] , Yee et al. [7] , and Smith [8] .
Whilst technology continues to flourish for the television audience, very few developments in batting equipment have occurred. One of the main reasons stems from the laws of cricket which limits the material composition of the cricket blade to wood. This is contrary to other sports such as tennis and squash where rackets have evolved significantly (Broe et al. [9] ; Chopp in [10] ) and numerous materials such as aluminums-based alloys, and titanium-based alloys have been utilized (Handee et al. [11] ; Bower and Sinclair [12] ). For research conducted on cricket bats, the main focus has been on modeling impacts and their vibration characteristics (Miller [13] ; Cross and Nathan [14] ). Very few studies appear to have investigated the actual impact between bat and ball (Stretch et al. [15] ), possibly due to the difficulty in accurately projecting the ball to the specified area on the bat.
Brody [3] studied the vibration spectrum of a hand-held bat during and after the collision and showed that the bat behaves as a free body on the short time scale of the collision. Cross [4] made an extensive study about the vibration spectrum of free and hand-held bats and concluded that there exists a zone of impact locations on the barrel end of the bat where the impact forces on the hands due to recoil and vibration are minimize.
After Brody [3] , Van [16] solved the problem to find the normal modes for transverse bending vibrations in the bat, applying the standard theory of beams, suitably modified for a non-uniform bat. Based on the investigations of Van [16] and Cross [4] [17] [18] , Nathan [19] proposed a model which developed for the collision between the baseball and bat, taking into account the transverse bending vibrations of the bat. Bower [1] investigated the sweet spot of atypical cricket bat in both the longitudinal and transverse direction for a low-impact velocity collision. Following the work of Nathan [19] , in this paper, we treat the sweet spot as the major study, analyze the effect on the ball-bat model of corking the bat, and investigate the relationship between the sweet spot and different materials.
This paper is organized as follows. In Section 2, we introduce some preliminary results. Section 3 makes some basic assumption for our modeling idea and gives the explanation of the symbols which will be used throughout the paper. In Section 4, we present a dynamical model to modeling the sweet spot on a baseball bat. Section 5 gives some extension models. We offer our conclusions in Section 6.
2. Preliminary
In order to indicate the origin of the baseball bats problem, the following preliminary are worth mentioning.
2.1. Engineering Concepts Relating to Baseball
Before discussing the performance of baseballs and baseball bats, a few engineering concepts are presented.
2.1.1. Coefficient of Restitution (COR)
The COR is used to quantify the elasticity or “liveliness” of the baseball as it strikes a stationary object, such as a baseball bat. It is a measure of how elastic or inelastic two bodies when they come into contact with each other and must be measured experimentally. The following is a brief derivation of the COR (William and Leroy [20] ).
As mentioned in William and Leroy’s book Engineering Mechanics: Dynamics, bodies A and B are given initial velocities, and, respectively. As shown in Figure 1.
Assuming that non-impulsive force and the friction force between the two bodies may be neglected, the total momentum for the two bodies before (i) and after (f) the collision is conserved
, (1)
Figure 1. Two bodies in motion, before (top), during (middle) and after (bottom) acollision.
where and are masses of A and B, respectively.
Considering the impulse forces acting on the individual bodies while the bodies are deforming during and after the collision, the coefficient of restitution e is defined as the ratio of the impulse during the collision and the impulse as the bodies is restored. It can be simplified as
, (2)
where the COR is the negative ratio of the relative velocities of two bodies after and before a collision.
The COR is not a value that is regarded as a material property because it not only depends on the material of both impacted bodies, but for nonlinear material systems. It also depends on the velocity at which they collide. It will also vary with respect to9different sizes, shapes and the temperature of the impacting bodies. For values of, the collision is considered to be a perfectly elastic impact, that is, there is no energy loss due to the deformation of the bodies at impact. For values of, the collision is considered to be a perfectly plastic impact. The relative velocity of the two bodies after impact is zero and the two particles move together at the same speed (Mustone [21] ).
2.1.2. Mass Moment of Inertia and Parallel Axis Theorem
The mass moment of inertia is a measure of a body to resist a rotational acceleration about an axis and is the best measure of how easily a bat can be swung. It is simply denoted as MOI, noting that it refers to the mass moment of inertia and not to be confused with an area moment of inertia. Studies described later have shown that battedball velocity increases with increasing bat swing speed. Therefore, the MOI, can provide one measure of bat performance (Mustone [21] ) because it is an indicator of swing speed.
2.1.3. The Sweet Spot
Trying to locate the exact sweet spot on a baseball or softball bat is not as simple a task as it might seem, because there are a multitude of definitions of the sweet spot (Russell [22] ):
1) The location which produces least vibration sensation (sting) in the batter’s hands;
2) The location which produces maximum batted ball speed;
3) The location where maximum energy is transferred to the ball;
4) The location where coefficient of restitution is maximum;
5) The center of percussion;
6) The node of the fundamental vibration mode;
7) The region between nodes of the first two vibration modes;
8) The region between center of percussion and node of first vibration, mode.
In the paper, we define “the sweet spot” as the region where the ball gets the maximum exit speed.
2.2. Wood and Metal
The physical differences between wood and metal baseball bats are quite obvious. A wood bat is solid but not very durable. A metal (usually aluminum) bat on the other hand, is hollow and more durable than wood. A significant difference between wood and metal bats is the energy-transfer mechanism between the bat and the baseball during the collision. The difference between the energy-transfer mechanisms is a fundamental result of the wood bat being solid and the metal bat being hollow.
3. Basic Assumption and Symbols
We make the following two assumptions, which have been used extensively in Smith [8] and Yee et al. [7] .
1) Ignore the impact of hand on the ball. Studies have shown that the impact of hand on the ball is slight at the moment of impact. So the bat can be approximated as a “free end”.
2) When the bat hits the ball, direction of the ball velocity and the axis of the bat are orthogonal. Hence, the impact of friction between the ball and bat can be ignored.
Before constructing a dynamical model for sweet spot on a baseball bat problem, let us introduce the following notations, parameters and decision variables used in this paper.
: the mass of the ball;
: the mass of the bat;
: the initial velocity of the ball;
: the velocity of the ball after collision;
: the velocity of the ball-centroid before collision;
: the velocity of the ball-centroid after collision;
: the coefficient of restitution;
: the moment of inertia;
: the angular velocity.
4. Dynamical Model
4.1. Theoretical Models for the Sweet Spot
Known from the analysis of the batter’s movements, there are two rotation systems. One is the rotation of arm and the bat to body gravity as the axis. And the other is the rotation of the bat to wrist as the axis. Specific steps of this model are described as follows:
Step 1 Take the body gravity B as the origin of coordinates, and take the axis of the bat as the X-axis, the perpendicular direction of the X-axis as the Y-axis. Then aplane rectangular coordinate system is established.
Step 2 Formulate that center-of-mass coordinate of the bat is C, coordinate of body centre-of-gravity is B, and the coordinate of hand-held point is W, the coordinate of hit point is P, the coordinate of knob end is, distance between B and W is R, the distance between W and C is H, the distance between P and C is S.
Step 3 At the moment of collision, the force between bat and ball is greater than their own weights and hand upward force. Therefore, with the ball-bat system as the research object, there is momentum conservation on the Y-axis, which can be formulated as
, (3)
where is the mass of the ball, is the mass of the bat, is the initial velocity of the ball, is the velocity of the ball after collision, andare the velocity of the ball-centroid before and after collision.
Step 4 Assuming that the angular velocity before and after collision are and, so that
(4)
As the preliminary in subsection 2.2.1 , the coefficient of restitution here is
. (5)
Step 5 Establish angular momentum conservation equation of ball-bat system to the body gravity as the axis, as follows
(6)
Step 6 From Equations (3) to (6), the expression of can be obtained, as follows
, (7)
where J is the moment of inertia and the expression of J is
. (8)
Finally the derivation of with respect to S is
. (9)
The solution of is the location of the sweet spot.
4.2. Data Verification
The general specification of wood baseball bat is shown in Table 1.
As a rotating body, the cross-section along the X-axis, where the length of, can be determined by the bat mass and center-of-gravity. The expression of the bat mass is as follows:
,
.
Based on the definition of centre-of-gravity, the moment of inertia is balance at two sides of it. So the following equation was attained.
where r is an integral function of the variable x.
The length of 0.6423 and 0.3459 are measured for and by calculating, respectively. Thus, the radius of the various parts of a long baseball can be formed as follows:
By using MATLAB, a figure reflecting the relationship of the ball exit velocity and the hitting spot is obtained (Figure 2). The conclusion drawn from Figure 2 is that the peak of the curve is the coordinates of the sweet spot. So the distance from the sweet spot to the knob end of the bat is 0.7840 m. This just explains why this spot isn’t at the end of the bat.
4.3. Analysis of “Corking” a Bat
No doubt, “corking” a bat would change a series of physical properties of the bat, such as the bat mass, the centre-of-gravity position, and the coefficient of restitution, etc. The change of these physical properties will ultimately affect the hitters’ levels of technology, which is related to the fairness of the game directly. The specific
Table 1. Physical parameter of wood bat.
Figure 2. The relationship of the ball exit velocity and the hitting spot.
steps of this model are described as follows.
According to practical experience, the cork is usually a cylinder which diameter is 1 inch (2.54 cm) and depth is 10 inch (25.4 cm). The density of the cork is. Therefore, corking the bat will bring the following changes.
1) The mass and inertia of the bat deceases, and the bat control enhances.
2) Centre-of-gravity position is closer to the knob end than before, so that the ball angular momentum decreases. Then the result is affected.
3) The moment of inertia decreases. Then, the sweet spot and the maximum velocity are affected.
The following are three aspects of quantitative analysis from the above filler effect on the ball.
1) Reduction of the mass
.
2) Variation of the centre-of-gravity position
.
In the coordinate system, original X-coordination of centre-of-gravity is. Based on the definition that the X-coordination of the corking bat’s centre-of-gravity is C, the moment of inertia is balance at two sides of centre-of-gravity. Thus, a formula to calculate the C forms
.
3) Reduction of the moment of inertia
,
where is the moment of inertia to wrist as the axis and can be expressed as follows,
Based on the dynamical model, the sweet spot and maximum velocity of the corking bat can be attained only by correcting the moment of inertia. Namely
.
After substituting the superior limit of the corking bat’s density (), the maximum exit speed of the ball is attained.
Figure 3 is a comparison of the normal bat and the corking bat. The X-axis is the distance to the body centre-of-gravity, and the Y-axis is the exit speed of the ball.
The conclusion can be drawn from Figure 3 that corking a bat can’t enhance the “sweet spot” effect. The major reason is that the moment of inertia decreases due to the decrease of the bat mass, so that the energytransferred decreases. This results in the decrease of the exit speed directly. Whereas, the superiority of the corking bat mainly reflects in the good control ability to the bat and the great acceleration, so the reaction time is lengthened. This is also the reason that major league baseball prohibits “corking”.
4.4. Bat on the Ball with Different Material
Based on the preliminary in subsection 2.2, there are two major differences between a wood bat and an aluminum bat. One is the difference of moment of inertia leaded by the difference of density and mass. The other is the coefficient of restitution leaded by the difference of metal.
Assuming that the shape of the aluminum bat is the same to the wood bat we use before, the dynamical model can be used. We define v as the exit velocity of the aluminum bat and x as the distance from coordination of the hitting spot. Analyzing from aspects of the moment of inertia and recovery coefficient, the expression of the relationship between v and x is as follows,
.
Accessing to relevant information (Greenwald, Penna, and Crisco [23] ), the moment of inertia of the aluminum bat is 3.2256. Using the dynamical model, a relationship graph is obtained (Figure 3).
The modeling results show that roughly aluminum bats have higher batted ball speed than wooden bats. The purpose is to enhance athletic baseball player’s own fitness and skill, rather than engage in competition sports equipment, so the competition is not allowed to use the metal bats.
5. Model Extension
In the dynamical model, we didn’t take the energy loss into consideration, which is not match with the actual. Here we give a model extension to analyze the energy loss.
The main factor of energy loss at the collision process is vibration of the ball. Therefore, energy loss of this part should be minimized. The region between the free vibration and the fundamental nodes of the second harmonic is where the vibration energy transfer is the smallest. In the model, the free vibration of the bat is analyzed
Figure 3. Comparison of normal wood bat and aluminum bat.
based on the beam vibration model (Mustone [21] ). And finite element analysis is used to calculate the vibration parameters of a baseball bat.
5.1. Beam Vibration Model
The following is the analysis of the process of free vibration of the bat based on the beam vibration model. We mainly analyze the transfer of the fundamental harmonic, the second harmonic and the third harmonic to find the smallest vibration area. In the process of hitting, the bat is not subject to the torque and vertical extrusion pressure, so that the longitudinal vibration and tensional vibration are neglected. So the baseball is simplified as the balanced-distributed bat, which is shown in Figure 4.
The wave equation of the transverse vibration is as follows:
,
where is the transverse vibration displacement of the beam, C is the propagation velocity of the wave in the bat, E is young’s modulus, is the density of the bat, K is the radius of gyration, S is the cross sectional area.
The boundary conditions are as follows:
.
Then the following expressions are formed,
Define the function
.
Assuming that is the solution of the function above, the relationship between A and B is obtained,
.
Then the expression of amplitude Y is formed,
.
5.2. Finite Element Analysis
The real bat is irregular. In order to well simulate the vibration energy’s transmission in a real bat, finite element analysis can be used based on the above beam vibration model. Using this method, we can refine model and decompose it into a series of beam vibration model. Substitute into the equations in above beam vibration model, we can take it for a series of beam vibration’s superposition. On the basis of the existing data, we will substitute the data into the sixth equations, then make corresponding harmonic figure in MATLAB. Meanwhile, we can find out the first three vibration mode’s frequency from it. They are classified as
.
Finally, the image of previous three waves of vibration can be made in MATLAB as follows (Figure 5). From Figure 5, we can see that the ball hitting at zero point won’t cause the corresponding mode vibration. In order to
Figure 4. Simplified model.
Figure 5. Image of previous three waves of vibration.
make the total vibration energy lost as little as possible, we make a comprehensive consideration of the first three vibration and put the optimal hit region defined as the scope between the base wave’s first node and the second harmonic first nodes, denoted as (0.7652, 0.8012). In this area, the vibration amplitude caused by hitting uniform stem is smaller. We can conclude that the sweet point determined by the dynamical model is reasonable and faithful.
6. Concluding Remarks
In this paper, we analyzed the effect on the ball-bat model of corking the bat and also studied the relationship between the sweet spot and different materials. Our substantive contributions are threefold. First and foremost, we developed a simple but effective theoretical model to give a rough estimate for the sweet spot of the bat. Secondly, a simplified abstract form of the bat is given which makes description and modeling study easy. Last but not least, because different material caused rotary inertia and different recovery coefficient, researches on the different material hitting effect were made.
There are several weaknesses of this study. Firstly, in order to make the process of solving the inertia easy, we treat the aluminum bat as a solid body. But it’s hollow in practice that may leads to the error of exit velocity of the ball. Secondly, we define that the direction of the ball velocity and the axis of the bat are orthogonal. However, it can’t be always like that. So the model can’t apply in any situation. These and other related drawbacks await future research.
Acknowledgements
This work was supported by the Key Project of Hubei Provincial Natural Science Foundation under Grant No. 2015CFA144.
Cite this paper
LiCheng,ZaiqiangKu, (2015) Dynamical Model with Application to the Analysis of the Sweet Spot on a Baseball Bat. American Journal of Computational Mathematics,05,458-467. doi: 10.4236/ajcm.2015.54040
References
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http://dx.doi.org/10.1119/1.16378
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http://dx.doi.org/10.1007/s12283-012-0087-6
4. 4. Bower, R.G. and Sinclair, P. (2007) Tennis Racket Stiffness, String Tension and Impact Velocity Effects on Post-Impact Ball Angularvelocity. Sports Engineering, 10, 111-122.
http://dx.doi.org/10.1007/BF02844208
5. 5. Broe, M., Sherwood, J. and Drane, P. (2010) Experimental Study of the Evolution of Composite Baseball Bat Performance. Procedia Engineering, 2, 2653-2658.
http://dx.doi.org/10.1016/j.proeng.2010.04.047
6. 6. Choppin, S. (2013) An Investigation into the Power Point in Tennis. Sports Engineering, 16, 173-180.
7. 7. Cross, R. (1998) The Sweet Spot of a Baseball Bat. American Journal of Physics, 66, 772-779.
http://dx.doi.org/10.1119/1.19030
8. 8. Cross, R. (2011) Physics of Baseball and Softball. Springer, New York.
9. 9. Cross, R. (2014) Impact of Sports Balls with Striking Implements. Sports Engineering, 17, 3-22.
http://dx.doi.org/10.1007/s12283-013-0132-0
10. 10. Cross, R. and Nathan, A. (2009) Performance versus Moment of Inertia of Sporting Implements. Sports Technology, 2, 7-15.
http://dx.doi.org/10.1002/jst.88
11. 11. Greenwald, R.M., Penna, L.H. and Crisco, J.J. (2001) Differences in Batted Ball Speed with Wood and Aluminum Baseball Bats: A Batting Cage Study. Journal of Applied Biomechanics, 17, 241-252.
12. 12. Handee, S.P., Greenwald, R.M. and Crisco, J.J. (1998) Static and Dynamic Properties of Various Baseballs. Journal of Applied Biomechanics, 14, 390-400.
13. 13. King, K., Hough, J., McGinnis, R. and Perkins, N.C. (2012) A New Technology for Resolving the Dynamics of a Swinging Bat. Sports Engineering, 15, 41-52.
http://dx.doi.org/10.1007/s12283-012-0084-9
14. 14. Miller, S. (2006) Modern Tennis Rackets, Balls and Surfaces. British Journal of Sports Medicine, 40, 401-405.
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15. 15. Mustone, T.J. (1996) A Method to Evaluate and Predict the Performance of Baseball Bats Using Finite Elements. University of Massachusetts, Lowell.
16. 16. Nathan, A.M. (2000) Dynamics of the Baseball-Bat Collision. American Journal of Physics, 68, 979-990.
17. 17. Nicholls, R.L., Miller, K. and Elliott, B.C. (2006) Numerical Analysis of Maximal Bat Performance in Baseball. Journal of Biomechanics, 39, 1001-1009.
http://dx.doi.org/10.1016/j.jbiomech.2005.02.020
18. 18. Russell, D.A. (2004) Physics and Acoustics of Baseball & Softball Bats. PhD Thesis, Science & Mathematics Department, Kettering University, Flint.
19. 19. Smith, L.V. (2014) Hygrothermal Effects of Baseballs and Softballs. Sports Engineering, 17, 123-130.
http://dx.doi.org/10.1007/s12283-013-0143-x
20. 20. Stretch, R., Brink, A. and Hugo, J. (2005) A Comparison of the Ball Rebound Characteristics of Wooden and Composite Cricket Bats at Three Approach Speeds. Sports Biomechanics, 4, 37-45.
http://dx.doi.org/10.1080/14763140508522850
21. 21. Van, Z.L. (1992) The Dynamical Theory of the Baseball Bat. American Journal of Physics, 60, 172-181.
http://dx.doi.org/10.1119/1.16939
22. 22. William, F. and Leroy, D. (1993) Engineering Mechanics: Dynamics. John Wiley and Sons, New York.
23. 23. Yee, J., Sherwood, J.A. and Fitzgerald, S. (2014) Batted-Ball Performance of a Composite Softball Bat as a Function of Ball Type. Procedia Engineering, 72, 465-470.
http://dx.doi.org/10.1016/j.proeng.2014.06.081
NOTES
*Corresponding author.
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# Simple Machines - PowerPoint PPT Presentation
1 / 7
Simple Machines. Levers, Pulleys, and Wheel and Axles. Don’t forget BrainPop. The Lever.
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Simple Machines
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
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## Simple Machines
Levers, Pulleys, and
Wheel and Axles
Don’t forget BrainPop
### The Lever
• A lever is a simple machine that has a bar that pivots at a fixed point, called a fulcrum. Levers are used to apply a force to a load. There are three classes of levers, which are based on the locations of the fulcrum, the load, and the input force.
• First Class Levers – the fulcrum is between the input force and the load
• See Figure 1, page 564
• First class levers always change the direction of the input force and depending on the location of the fulcrum, first class levers can be used to increase force or to increase distance.
### The lever
• Second Class Lever
• The load of a second class lever is between the fulcrum and the input force.
• See Figure 2, page 565
• Second class levers do not change the direction of the input force. But they all you to apply less force than the force exerted by the load. Because the output force is greater than the input force, you must exert the input force over a greater distance.
### The lever
• Third class lever
• The input force in a third class lever is between the fulcrum and the load.
• See Figure 3, page 565
• Third class levers do not change the direction of the input force. In addition, they do not increase the input force. Therefore, the output force is always less than the input force.
### The pulley
• When you open window blinds by pulling on a cord, you’re using a pulley. A pulley is a simple machine that has a grooved wheel that holds a rope or a cable. A load is attached to one end of the rope, and an input force is applied to the other end.
• Fixed Pulley – See Figure 4, page 566
• Movable Pulley – See Figure 4, page 566
• Block and Tackles – See Figure 4, page 566
### The wheel and axle
• Did you know that a faucet is a machine? The faucet shown in Figure 5, page 567 is an example of a wheel and axle, a simple machine consisting of two circular objects of different sizes. Doorknobs, wrenches, and steering wheels all use a wheel and axle.
• Mechanical advantage of a wheel and axle – page 567
### Compound machines
• You are surrounded by machines. You even have machines in your body! But most of the machines in your world are compound machines, machines that are made of two or more simple machines.
• The block and tackle is a compound machine.
• A can opener is three machines combined…page 570
• Mechanical efficiency of compound machines is low. The efficiency is low because compound machines have more moving parts than simple machines do, thur there is more friction to overcome.
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## The Math Property of Equality
When you first learned about the math property of equality in elementary school, the symbol was used to signify this is the answer.
Here is an example of how the equal sign is used in elementary school Old Use: But things are about to change...
In the figure below, you see a set of hands each containing \$0.25. The left hand has a quarter, and the right hand has 2 dimes and a nickel. Both sides have different coins, but equal money
New Use: \$0.25 = \$0.25!
Equality says that I can perform an operation (add, subtract, multiply divide) on my equation as long as I do the same thing to both sides. Watch how I add \$0.01 to each hand...
I now added one penny to each side of the equation. Because I did the same thing to both hands, they are still equal. \$0.26 = \$0.26!
Using the Math Property of Equality for Equations
In the illustration above, we used the property of equality by adding a penny to each side of our equation.
Let's take a look at a few examples of how we can use the property to solve equations.... The properties demonstrated above are known as the addition property, the subtraction property, the division property, and the multiplication property. They all are possible because of the equal sign.
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# zgels (l) - Linux Manuals
## NAME
ZGELS - solves overdetermined or underdetermined complex linear systems involving an M-by-N matrix A, or its conjugate-transpose, using a QR or LQ factorization of A
## SYNOPSIS
SUBROUTINE ZGELS(
TRANS, M, N, NRHS, A, LDA, B, LDB, WORK, LWORK, INFO )
CHARACTER TRANS
INTEGER INFO, LDA, LDB, LWORK, M, N, NRHS
COMPLEX*16 A( LDA, * ), B( LDB, * ), WORK( * )
## PURPOSE
ZGELS solves overdetermined or underdetermined complex linear systems involving an M-by-N matrix A, or its conjugate-transpose, using a QR or LQ factorization of A. It is assumed that A has full rank. The following options are provided:
1. If TRANS = aqNaq and m >= n: find the least squares solution of
an overdetermined system, i.e., solve the least squares problem
minimize || B - A*X ||.
2. If TRANS = aqNaq and m < n: find the minimum norm solution of
an underdetermined system A B.
3. If TRANS = aqCaq and m >= n: find the minimum norm solution of
an undetermined system A**H B.
4. If TRANS = aqCaq and m < n: find the least squares solution of
an overdetermined system, i.e., solve the least squares problem
minimize || B - A**H X ||.
Several right hand side vectors b and solution vectors x can be handled in a single call; they are stored as the columns of the M-by-NRHS right hand side matrix B and the N-by-NRHS solution matrix X.
## ARGUMENTS
TRANS (input) CHARACTER*1
= aqNaq: the linear system involves A;
= aqCaq: the linear system involves A**H.
M (input) INTEGER
The number of rows of the matrix A. M >= 0.
N (input) INTEGER
The number of columns of the matrix A. N >= 0.
NRHS (input) INTEGER
The number of right hand sides, i.e., the number of columns of the matrices B and X. NRHS >= 0.
A (input/output) COMPLEX*16 array, dimension (LDA,N)
On entry, the M-by-N matrix A. if M >= N, A is overwritten by details of its QR factorization as returned by ZGEQRF; if M < N, A is overwritten by details of its LQ factorization as returned by ZGELQF.
LDA (input) INTEGER
The leading dimension of the array A. LDA >= max(1,M).
B (input/output) COMPLEX*16 array, dimension (LDB,NRHS)
On entry, the matrix B of right hand side vectors, stored columnwise; B is M-by-NRHS if TRANS = aqNaq, or N-by-NRHS if TRANS = aqCaq. On exit, if INFO = 0, B is overwritten by the solution vectors, stored columnwise: if TRANS = aqNaq and m >= n, rows 1 to n of B contain the least squares solution vectors; the residual sum of squares for the solution in each column is given by the sum of squares of the modulus of elements N+1 to M in that column; if TRANS = aqNaq and m < n, rows 1 to N of B contain the minimum norm solution vectors; if TRANS = aqCaq and m >= n, rows 1 to M of B contain the minimum norm solution vectors; if TRANS = aqCaq and m < n, rows 1 to M of B contain the least squares solution vectors; the residual sum of squares for the solution in each column is given by the sum of squares of the modulus of elements M+1 to N in that column.
LDB (input) INTEGER
The leading dimension of the array B. LDB >= MAX(1,M,N).
WORK (workspace/output) COMPLEX*16 array, dimension (MAX(1,LWORK))
On exit, if INFO = 0, WORK(1) returns the optimal LWORK.
LWORK (input) INTEGER
The dimension of the array WORK. LWORK >= max( 1, MN + max( MN, NRHS ) ). For optimal performance, LWORK >= max( 1, MN + max( MN, NRHS )*NB ). where MN = min(M,N) and NB is the optimum block size. If LWORK = -1, then a workspace query is assumed; the routine only calculates the optimal size of the WORK array, returns this value as the first entry of the WORK array, and no error message related to LWORK is issued by XERBLA.
INFO (output) INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
> 0: if INFO = i, the i-th diagonal element of the triangular factor of A is zero, so that A does not have full rank; the least squares solution could not be computed.
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# Proving tanx/2 = cscx - cotx
1. Feb 28, 2009
### xonicole
1. The problem statement, all variables and given/known data
tanx/2 = (1-cosx)/sinx
2. Relevant equations
3. The attempt at a solution
This is where i got on the right side
i dont know where to finish...(1-cosx)/sinx
2. Feb 28, 2009
### Gib Z
Well, are you familiar with Weierstrass substitution? They are also known as t-formula:
http://pear.math.pitt.edu/Calculus2/week3/3_2li5.html
Another method would be to use the Double angle identities:
$$\cos (2\theta) = 1- 2\sin^2 \theta$$ and $$\sin (2\theta) = 2 \sin \theta \cos \theta$$.
Into those, let $\theta = x/2$, and put those back into what you have and simplify.
Last edited by a moderator: Apr 24, 2017
3. Mar 1, 2009
### xonicole
I found that a lot of people have been telling me that but i dont understand how you use that to make the sides equal is there a way you can show me step by step
4. Mar 2, 2009
### Gib Z
No really I can't ! You didn't even tell me which of the two methods I posted you want help with. Show us what you have tried to do yourself, so we can help you get over what you're stuck with.
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+0
# help
0
52
1
If \(x\%\) of four-digit numbers have a repeated digit (the repeated digits do not need to be adjacent), then what is \(x\)? Express your answer as a decimal to the nearest tenth.
Nov 29, 2019
#1
0
a=1;b=0;c=0;d=0;p=0; cycle:if(a==b or a==c or a==d or b==c or b==d or c==d, goto loop, goto next); loop:print a*1000+b*100+c*10+d," ",;p=p+1; next:d++;if(d<10, goto cycle, 0);d=0;c++;if(c<10, goto cycle, 0);d=0;c=0;b++;if(b<10, goto cycle,0);b=0;c=0;d=0;a++;if(a<10, goto cycle,0);print"Total = ",p
OUTPUT =4,464 integers - have at least 2 repeated digits out of total of 9,000 integers.
So x =4,464 / 9,000 = 62 / 125 =49.6%
Nov 30, 2019
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http://www.talkstats.com/threads/probability-given-i-already-know-some-event-has-occurred.53302/?p=149490
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# Probability GIVEN I already know some event has occurred
#### jnscollier
##### New Member
I ask 14 people to buy something, 2 say yes, 12 say no.
Given this happened, what is the probability that the next person I ask will buy something? So that would make it 3 yes, 12 no, 15 total.
What is the probability that the next person I ask will buy given 0/14 have bought currently?
This isn't a homework problem, I work at a marketing agency and am trying to predict online conversions given the fact I already know some data (clicks, conversions) ex: 14 clicked, 2 people converted, 12 didn't.
I'm not sure how to approach this problem because I'm trying to find some calculation given I already know something.
Thanks
#### staassis
##### Member
"I ask 14 people to buy something, 2 say yes, 12 say no.
Given this happened, what is the probability that the next person I ask will buy something? So that would make it 3 yes, 12 no, 15 total."
The estimate of the probability is 2 / 14. The estimate will have high standard error, since 14 is a small sample size.
What is the probability that the next person I ask will buy given 0/14 have bought currently?"
If 0 people buy out of 14, then we are dealing with estimating the probability of an extremely rate event. In this case, a sample size of the order of 500-1000 would be necessary to say anything meaningful about the probability.
#### jnscollier
##### New Member
"I ask 14 people to buy something, 2 say yes, 12 say no.
Given this happened, what is the probability that the next person I ask will buy something? So that would make it 3 yes, 12 no, 15 total."
The estimate of the probability is 2 / 14. The estimate will have high standard error, since 14 is a small sample size.
What is the probability that the next person I ask will buy given 0/14 have bought currently?"
If 0 people buy out of 14, then we are dealing with estimating the probability of an extremely rate event. In this case, a sample size of the order of 500-1000 would be necessary to say anything meaningful about the probability.
No, you are getting too technical in your thinking. I'm using these numbers as a 'for example'. The underlying fundamental question I have is how to calculate the conditional probability using these numbers. The ACTUAL numbers are much much larger. This is a simplification of the real-life problem.
So, to modify the question, forget about sample size for a moment. I'm more interested in the calculations.
Thanks anyways.
#### staassis
##### Member
I am not getting too technical because this is simple, 7-th grade, arithmetics. There is no simpler mathematics than my calculations... Again, you cannot say anything accurately if your sample size is small. Does not matter whether you phrase it as an example, advice or something else. What you want to use this for is irrelevant. 2 + 2 does not equal 5, no matter how easy going you are about the whole thing.
#### jnscollier
##### New Member
I am not getting too technical because this is simple, 7-th grade, arithmetics. There is no simpler mathematics than my calculations... Again, you cannot say anything accurately if your sample size is small. Does not matter whether you phrase it as an example, advice or something else. What you want to use this for is irrelevant. 2 + 2 does not equal 5, no matter how easy going you are about the whole thing.
I still don't think you comprehend staassis. You're getting hung up on the low sample size.
Anyone else know how to use conditional probability mathematics to compute this in an example using the SAMPLE DEMO NUMBERS that staassis seems to be hung up on?
#### staassis
##### Member
Check out my home page to see who are talking to. Good luck.
#### Dason
Without prior information I would probably use a bayesian approach to updating this probability. You can find some details here: http://en.wikipedia.org/wiki/Rule_of_succession
It basically amounts to using: (s+1)/(n+2) as your estimate where s is the total number of successes so far and n is the total number of observations so far. Now you can use different 'prior' distributions to arrive at slightly different estimates but if you have a large sample size then it won't make much of a difference anyways.
Check out my home page to see who are talking to. Good luck.
An unnamed PhD. Oh well. Throwing out your credentials isn't much of a help when somebody is just asking for help. You didn't actually explain anything.
#### staassis
##### Member
Without prior information I would probably use a bayesian approach to updating this probability. You can find some details here: http://en.wikipedia.org/wiki/Rule_of_succession
It basically amounts to using: (s+1)/(n+2) as your estimate where s is the total number of successes so far and n is the total number of observations so far. Now you can use different 'prior' distributions to arrive at slightly different estimates but if you have a large sample size then it won't make much of a difference anyways.
An unnamed PhD. Oh well. Throwing out your credentials isn't much of a help when somebody is just asking for help. You didn't actually explain anything.
Dason, your estimate (s+1) / (n+2) is extremely biased in the case of rare events. It means that you are implicitely using a very rough prior, for no apparent reason. You should have mentioned that to the user (in plain English)... Also, you do realize that he does not even know college math that well. So you just confused him with the word "prior". You did not actually explain anything. Also, formula (s+1) / (n+2) is BIASED and is especially bad in HIS case of 0 out of 14...... It is your duty to draw his attention to the fact that running any well-justified formula on a handful of points will only lead to inaccurate estimates, false confidence and wrong decision making. Seriously, would YOU bet \$100 on applying your fomula to his 14 data points?
I did provide the user with most statistically efficient (accurate) formula, which is s/n. But every statistical method has its limitations, which must be clearly stated, especially for not quantitatively fit users...
#### jnscollier
##### New Member
Thanks Dason and Staassis. Either way, it sounds like there is no advanced statistical formula to predict try and predict the probability of the next outcome (they purchase or don't). It sounds like both of you are saying essentially something similar which is it's simply s / n or (s + 1)/ (n + 2) (which doesn't really get impacted when a large sample is present).
I was hoping that there was some more advanced formula that I didn't know about for predicting the next outcome given pre-existing data.
Anyways, thanks to both of you for your help.
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15,121,414 members
Articles / Multimedia / GDI+
Article
Posted 6 Mar 2007
536.5K views
196 bookmarked
# Extensions to DrawTools
Rate me:
DrawTools library extended to include Layers, Zoom, Pan, Rotation
## Introduction
Alex Fr provided an excellent set of drawing tools in his DrawTools article and these tools serve as a basis for this article, which expands on the original toolset in the following ways:
1. In addition to the basic Rectangle, Ellipse, Line and Scribble tools, this version adds PolyLine, Filled Ellipse, Filled Rectangle, Text and Image tools
2. Multiple drawing Layers
3. Zooming
4. Panning
5. Rotation
In this article, I will describe how Layers were implemented, as well as the Text and Image tools.
## Background
See the original DrawTools article for details on how the basic application is built, class structure, etc.
It is also assumed that the reader has a working understanding of GDI+ fundamentals, including Matrices. For an excellent introduction to GDI+, see www.bobpowell.net.
## Implementing Layers
Adding Layers to the application involved adding two classes, `Layer` and `Layers`, where `Layer` defines a single `Layer `and `Layers` defines the collection of `Layers `in an `ArrayList`.
Each `Layer `exposes the following properties:
C#
```private string _name;
private bool _isDirty;
private bool _visible;
private bool _active;
private GraphicsList _graphicsList;
```
Note that the `Layer` contains the `GraphicsList` - this is the key to the whole thing - each `Layer` contains its own list of drawing objects instead of `DrawArea`. `DrawArea` is modified to declare a `Layers` collection instead of a `GraphicsList` collection:
C#
```// Define the Layers collection
private Layers _layers;
```
When `DrawArea `is initialized, the `Layers` are initialized by creating the first `Layer` and setting it `Active `and `Visible`:
C#
```public DrawArea()
{
// create list of Layers, with one default active visible layer
_layers = new Layers();
_layers.CreateNewLayer("Default");
_panning = false;
_panX = 0;
_panY = 0;
// This call is required by the Windows.Forms Form Designer.
InitializeComponent();
}
```
In the `Layers `class, the `CreateNewLayer()` method actually creates the new `Layer`:
C#
```/// <summary>
/// Create a new layer at the head of the layers list and set it
/// to Active and Visible.
/// </summary>
/// <param name="theName">The name to assign to the new layer</param>
public void CreateNewLayer(string theName)
{
// Deactivate the currently active Layer
if(layerList.Count > 0)
((Layer)layerList[ActiveLayerIndex]).IsActive = false;
// Create new Layer, set it visible and active
Layer l = new Layer();
l.IsVisible = true;
l.IsActive = true;
l.LayerName = theName;
// Initialize empty GraphicsList for future objects
l.Graphics = new GraphicsList();
}
```
Note that any one or all `Layers `can be visible at the same time, but only one `Layer `may be active at any time.
You can control the `Layers `in the sample application by clicking on the Current `Layer`: name at the bottom of the application window - Click on the name ("`Default`") to open the `Layers `dialog:
From this dialog, you can Add new `Layers`, change the names of the `Layer`(s), and change the `Layer`(s) visibility and which `Layer `is `Active`. The "New Layer" column is checked whenever you click the "Add Layer" button. To delete Layer(s), simply check the "Deleted" column and close the dialog with the "Close" button. Remember only one Layer may be active at any one time. You will be reminded of this if you attempt to have more than one `Layer `active. Also note the Active `Layer `must be `Visible`.
When the application runs, each object that is drawn is added to the `GraphicsList` maintained by the active `Layer`. Note this relationship is preserved through saving and re-opening a drawing file.
Layers come in very handy when you want to draw "on top of" another image. For example, the image at the top of this article contains two layers. The following image shows the same picture with the Background Layer turned off:
Here is the same drawing with the Drawing Layer invisible and the Background Layer visible:
Objects on Layers which are visible but not active cannot be selected, moved, deleted, etc.
Each drawing object is added to the correct `Layer `by the `AddNewObject()` method in the `ToolObject` class:
C#
```protected void AddNewObject(DrawArea drawArea, DrawObject o)
{
int al = drawArea.TheLayers.ActiveLayerIndex;
drawArea.TheLayers[al].Graphics.UnselectAll();
o.Selected = true;
o.Dirty = true;
drawArea.Capture = true;
drawArea.Refresh();
} ```
## Implementing Zooming, Panning, and Rotation
Zooming, Panning, and Rotation are implemented by adding a few variables and some code to the `MainForm` and `DrawArea` classes.
Zooming is controlled by buttons on the form, and also by the mouse wheel when Ctrl is held down.
Pan is controlled by the Hand button on the form, and can be cancelled by a right-click.
Rotation is controlled by buttons on the form - note Rotation affects the entire drawing.
Here is an example of all three in use:
The heart of this code is the `BackTrackMouse()` method, which takes the "apparent" mouse position and converts it to a valid point based on the current Zoom level, Pan position, and Rotation:
C#
```/// <summary>
/// Back Track the Mouse to return accurate coordinates regardless of
/// zoom or pan effects.
/// Courtesy of BobPowell.net <seealso cref="http://www.bobpowell.net/backtrack.htm"/>
/// </summary>
/// <param name="p">Point to backtrack</param>
/// <returns>Backtracked point</returns>
public Point BackTrackMouse(Point p)
{
// Backtrack the mouse...
Point[] pts = new Point[] { p };
Matrix mx = new Matrix();
mx.Translate(-this.ClientSize.Width / 2, -this.ClientSize.Height / 2,
MatrixOrder.Append);
mx.Rotate(_rotation, MatrixOrder.Append);
mx.Translate(this.ClientSize.Width / 2 + _panX, this.ClientSize.Height / 2 +
_panY, MatrixOrder.Append);
mx.Scale(_zoom, _zoom, MatrixOrder.Append);
mx.Invert();
mx.TransformPoints(pts);
return pts[0];
}```
This routine comes from Bob Powell's excellent website. Through the use of the GDI+ Matrix class, the mouse point passed to this method is moved (Translate), Rotated, and Scaled based on the current PanX, PanY, Zoom, and Rotation values. The important thing to remember is that anytime you need to determine where the mouse pointer actually is in your drawing, you must call this method. You will see this method used throughout the program in the `DrawArea` class as well as others. An example of its usage is shown here:
C#
```private void DrawArea_MouseDown(object sender, System.Windows.Forms.MouseEventArgs e)
{
lastPoint = BackTrackMouse(e.Location);
if (e.Button == MouseButtons.Left)
tools[(int)activeTool].OnMouseDown(this, e);
else if (e.Button == MouseButtons.Right)
{
if (_panning == true)
_panning = false;
ActiveTool = DrawArea.DrawToolType.Pointer;
}
}```
The current zoom level is controlled by the following simple routine:
C#
```private void AdjustZoom(float _amount)
{
drawArea.Zoom += _amount;
if (drawArea.Zoom < .1f)
drawArea.Zoom = .1f;
if (drawArea.Zoom > 10)
drawArea.Zoom = 10f;
drawArea.Invalidate();
SetStateOfControls();
}```
Then in the `DrawArea.Paint()` method, the zoom, pan, and rotation values are used to alter the way the canvas is painted:
C#
```private void DrawArea_Paint(object sender, System.Windows.Forms.PaintEventArgs e)
{
Matrix mx = new Matrix();
mx.Translate(-this.ClientSize.Width / 2, -this.ClientSize.Height / 2,
MatrixOrder.Append);
mx.Rotate(_rotation, MatrixOrder.Append);
mx.Translate(this.ClientSize.Width / 2 + _panX, this.ClientSize.Height / 2 +
_panY, MatrixOrder.Append);
mx.Scale(_zoom, _zoom, MatrixOrder.Append);
e.Graphics.Transform = mx;
SolidBrush brush = new SolidBrush(Color.FromArgb(255, 255, 255));
e.Graphics.FillRectangle(brush,
this.ClientRectangle);
// Draw objects on each layer, in succession so we get the correct layering.
// Only draw layers that are visible
if (_layers != null)
{
int lc = _layers.Count;
for (int i = 0; i < lc; i++)
{
if(_layers[i].IsVisible == true)
if(_layers[i].Graphics != null)
_layers[i].Graphics.Draw(e.Graphics);
}
}
DrawNetSelection(e.Graphics);
brush.Dispose();
}```
## Update - 8/25/2007 - Individual Object Rotation & Bug Fixes
The primary advancement in this update is the ability to rotate individual objects - when one or more objects are selected, clicking the Rotate tools will rotate those objects instead of the entire drawing surface.
There is one caveat, however - the selection rectangle for the rotated object is not rotated - if someone can help with this, I would greatly appreciate it!
This update also includes several small bug fixes reported by users - thanks to all for reporting!
## History
• 3/6/2007
• Original article uploaded to The Code Project
• 3/6/2007
• 8/25/2007
• Updated Individual Object Rotation
• 9/27/2007
• 12/23/2009
• Added Tooltip control which appears when mouse is over an object. Tooltip displays the Center coordinates of the object for `Rectangle`, `Ellipse `and `Image `objects. For other objects, Tooltip displays starting and ending coordinates. `Text `objects do not display Tooltip.
This was implemented adding the Tooltip control to the `ToolPointer `class. Each `Draw `Object fills the `TipText `property and the `MouseMove `event in `ToolPointer `controls when the Tooltip is displayed and removed from the canvas. This implementation is not perfect, as the Tooltip flashes when displayed so is meant as an example of one way information about the object can be displayed.
Perhaps a better way to do this would be to display information about the object in a separate "Information Window" and then only when the object is selected.
• See the new source code for details.
• 6/23/2010
• Updated project to include object ordering fix that corrects the way objects are stacked when a file is opened
• Updated project to Visual Studio 2010
• See the new source code for details
• 10/4/2011
• Corrected several issues with Layers and layering
## Share
Software Developer (Senior) United States
I develop software for a leading healthcare system in Northern Illinois.
Zoom problem using Mouse Wheel and Zoom To Selected Area Manuel Salvatori11-May-10 6:19 Manuel Salvatori 11-May-10 6:19
Re: Zoom problem using Mouse Wheel and Zoom To Selected Area Mark Miller11-May-10 11:38 Mark Miller 11-May-10 11:38
Re: Zoom problem using Mouse Wheel and Zoom To Selected Area k3rmit18-Dec-12 23:27 k3rmit 18-Dec-12 23:27
Draw lines with opacity/transparency property? kippow10-May-10 4:13 kippow 10-May-10 4:13
Re: Draw lines with opacity/transparency property? Mark Miller10-May-10 8:26 Mark Miller 10-May-10 8:26
Transparent Layer Demaker3-Feb-10 4:19 Demaker 3-Feb-10 4:19
Re: Transparent Layer Mark Miller3-Feb-10 7:17 Mark Miller 3-Feb-10 7:17
Re: Transparent Layer Demaker2-Mar-10 20:31 Demaker 2-Mar-10 20:31
Now i think how better implement watermaking between layers... Have you any ideas?
Re: Transparent Layer Mark Miller10-May-10 8:29 Mark Miller 10-May-10 8:29
Draw object while keeping the current taishoo24-Jan-10 18:41 taishoo 24-Jan-10 18:41
Re: Draw object while keeping the current Mark Miller27-Jan-10 19:27 Mark Miller 27-Jan-10 19:27
Re: Draw object while keeping the current taishoo27-Jan-10 21:40 taishoo 27-Jan-10 21:40
Re: Draw object while keeping the current Mark Miller2-Feb-10 7:43 Mark Miller 2-Feb-10 7:43
Re: Draw object while keeping the current taishoo2-Feb-10 17:11 taishoo 2-Feb-10 17:11
Re: Draw object while keeping the current Mark Miller3-Feb-10 6:50 Mark Miller 3-Feb-10 6:50
Image Disappraers while panning James Shelke20-Dec-09 21:47 James Shelke 20-Dec-09 21:47
Re: Image Disappraers while panning Mark Miller22-Dec-09 5:17 Mark Miller 22-Dec-09 5:17
Re: Image Disappraers while panning James Shelke22-Dec-09 18:41 James Shelke 22-Dec-09 18:41
Re: Image Disappraers while panning James Shelke22-Dec-09 21:22 James Shelke 22-Dec-09 21:22
Re: Image Disappraers while panning Mark Miller23-Dec-09 5:31 Mark Miller 23-Dec-09 5:31
Re: Image Disappraers while panning Mark Miller23-Dec-09 5:30 Mark Miller 23-Dec-09 5:30
Add tooltip on mouse over of any shape jamesyla8-Nov-09 21:51 jamesyla 8-Nov-09 21:51
Re: Add tooltip on mouse over of any shape Mark Miller10-Nov-09 5:31 Mark Miller 10-Nov-09 5:31
Re: Add tooltip on mouse over of any shape Mark Miller11-Nov-09 6:16 Mark Miller 11-Nov-09 6:16
Re: Add tooltip on mouse over of any shape jamesyla14-Nov-09 2:18 jamesyla 14-Nov-09 2:18
Last Visit: 31-Dec-99 19:00 Last Update: 4-Dec-21 14:14 Refresh ᐊ Prev1234567 Next ᐅ
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# TCC Jim Bolen Math Competition Rules and Facts. Rules:
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Transcription
2 2014 Jim Bolen Mathematics Competition Rules 1. Candidates must be currently enrolled at TCC 2. Candidates must not have achieved a two year degree, a degree that is considered higher than a two year degree or have junior standing (completed 70+ credit hours) upon completion of the fall semester to be eligible to earn prizes 3. Candidates must be enrolled in a minimum of 6 hours at TCC at the time of the exam 4. Candidates may be enrolled in high school and be taking dual credit courses as long as they meet the other requirements set forth 5. Candidates may be enrolled in a four year institution, but cannot be taking more credit hours there than they are currently enrolled in at TCC 6. Students with fewer than 30 credit hours upon completion of the fall semester may be considered for 1st year scholarships to return to TCC 7. Students with 30 or more credit hours upon completion of the fall semester may be considered for transfer scholarships 8. Results of all eligible Jim Bolen Math Competition participants will automatically be submitted to the national competition (AMATYC, Student Math League -
3 Test #1 AMATYC Student Mathematics League October/November If the standard order of operations is reversed (that is, additions and subtractions are done first and exponentiation is done last), what is the value of 2 3 ^ 2 + 3? A. 21 B. 24 C. 39 D. 486 E The price of a stock rose 20% on Monday, fell 10% on Tuesday, and increased by 1/6 on Wednesday. By what percent did the price rise from Monday to Wednesday? A. 24 B. 26 C. 28 D. 30 E The system of equations ax - by = 8 and ax + by = 20 has the solution (x, y) = (2, 3). Find a + b. A. 6 B. 7 C. 8 D. 9 E The positive integers a, b, and c satisfy a 6 + b 2 + c 2 = Find a + b + c. A. 50 B. 51 C. 52 D. 53 E Different shades of pink, red, and white can be made by mixing whole numbers of quarts of red and white paint. Shades are different if the ratio of red to white paint is different. Find the number of different possible shades that can be made from at most 4 quarts of red and 5 quarts of white paint. A. 15 B. 16 C. 17 D. 18 E The function y = f(x) has zeros -2 and 6. Find the zeros of y = -3f(2-2x). A. 2, -2 B. 5, 1 C. 4, -1 D. -1, -5 E. 1, One population P 1 (t) grows exponentially at the same rate that another population P 2 (t) decays exponentially. If the populations were both equal to P on Jan. 1, 2009, how will the populations be related on Jan. 1, 2012? A. P 1 (t)p 2 (t) P B. P 1 (t)p 2 (t) P 2 C. P 1 (t)/p 2 (t) P D. P 1 (t) P 2 (t) P E. P 1 (t) P 2 (t) 2P 8. For b > c > 0, both x 2 + bx + 8 and x 2 + cx + 8 factor over the integers. Find b - c. A. 1 B. 2 C. 3 D. 4 E Ed drives from San Mateo to Atascadero, a distance of mi. He starts driving at a constant speed and reduces his speed by 5 mph after each half hour of driving. If the trip takes 3 hr 20 min, how far did he travel in the first 2 hours? A. 127 B. 132 C. 137 D. 142 E Sun fills her 10 liter radiator with 20% antifreeze and 80% water. She removes some of the mixture and replaces it with antifreeze. If the radiator is now one quarter antifreeze, how many liters of the original mixture did she remove? A B C. 0.5 D E How many numbers with no more than six digits can be formed using only the digits 1 through 7, with no digit used more than once in a given number? A. 879 B C D E. 8659
4 October/November 2011 AMATYC Student Mathematics League Page The lines with equations 2x + 3y = 24 and 3x + 2y = 6 are symmetric with respect to a line with equation y = mx + b with m > 0. Find m + b. A. 5 B. 12 C. 17 D. 19 E A square of area 45 is inscribed in circle C. Find the area of a square inscribed in a semicircle of circle C. (Inscribed means having all 4 vertices on the given figure). A. 5 5 B. 18 C. 9 5 D. 20 E The left edge of a dollar bill is folded against the bottom edge to form an isosceles right triangle at the left end. The new left edge is again folded against the bottom edge. A vertex of the new triangle is the upper right corner of the bill. If a dollar bill is 157 mm long, find its width to the nearest millimeter. A. 63 B. 64 C. 65 D. 66 E Five boxes are placed inside an empty box. Each of the 5 new boxes is either left empty or has 5 new boxes placed inside it. This process is repeated until there are 18 boxes containing other boxes. Find the number of empty boxes. A. 73 B. 75 C. 77 D. 79 E Al, Bo, Cy, and Di are to receive math, physics, chem, and bio awards. Al thinks Di will win bio, Bo thinks Cy will win chem, Cy thinks Al won t win math, and Di thinks Bo will win physics. The math and bio winners are both right, and the other winners are both wrong. Who wins the math award? A. Al B. Bo C. Cy D. Di E. not enough information given 17. The digits 1 through 9 are separated into 3 groups of three digits, and the product of each group is found. Let P be the largest of the 3 products. Find the smallest possible value of P. A. 70 B. 71 C. 72 D. 73 E Out of 10 red chips and 15 green chips, 6 are placed into a bag, 10 into a box, and 9 into a bowl. In how many ways can the chips be distributed, if only the number of red and green chips in each container matters? A. 45 B. 49 C. 50 D. 55 E Square ABCD has side length 72. Let E be the midpoint of side AB, and let BD and CE intersect at G. Find the length of the altitude to BE in GEB. A. 12 B. 18 C. 21 D. 24 E Let r be the positive real zero of P(x) = 9x 5 + 7x 2-9. The sum r 4 2r 9 kr 5k 1 can be represented as the rational number a/b in lowest terms. Find a + b. A. 110 B. 115 C. 120 D. 125 E. 130
5 2014 TCC Jim Bolen Math Competition Registration Form (Please print legibly) Name Student ID# Address Math Instructor (if known) _ Cell Phone Home Phone 1. Do you have an associate s degree or higher from any institution? Yes No 2. Are you currently enrolled in a 4-year institution? Yes No If yes, how many credit hours are you enrolled in at the four year institution? (To be eligible for prizes and scholarships you may be required to submit an unofficial transcript from all all 4-year institutions you are attending) How many credit hours are you enrolled in at TCC? 3. Estimate the number of credit hours you will have upon completion of the fall semester (To be eligible for prizes and scholarships the students must be below 70 credit hours at the end of the fall semester) 4. Are there guests you would like to invite to the awards ceremony in the spring? Yes No If yes, Name(s) and Addresses: 5. T-shirt size (T-shirts will be distributed during the awards ceremony or can be picked up from the math department after the ceremony in spring) I give permission to Tarrant County College, the TCC Foundation, and/or campus coordinators to take, distribute and publish my photographs, personal narratives, and all other Bolen Mathematics competition data for news, advertising, donor relations and/or promotional purposes, in print and electronic media. I understand that I will not be compensated for any photograph, narrative or other application data that may be used in this capacity. Student Signature Date
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