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https://www.numbersaplenty.com/241005135 | 1,719,192,297,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864968.52/warc/CC-MAIN-20240623225845-20240624015845-00537.warc.gz | 799,312,803 | 3,373 | Search a number
241005135 = 35712831789
BaseRepresentation
bin11100101110101…
…11001001001111
3121210111022220010
432113113021033
5443144131020
635525324303
75654336610
oct1627271117
9553438803
10241005135
1111404a7a8
1268866693
133ac134b7
1424017c07
1516258de0
hexe5d724f
241005135 has 32 divisors (see below), whose sum is σ = 441285120. Its totient is φ = 110026368.
The previous prime is 241005077. The next prime is 241005179. The reversal of 241005135 is 531500142.
It is not a de Polignac number, because 241005135 - 27 = 241005007 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (21).
It is a congruent number.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 133821 + ... + 135609.
It is an arithmetic number, because the mean of its divisors is an integer number (13790160).
Almost surely, 2241005135 is an apocalyptic number.
241005135 is a deficient number, since it is larger than the sum of its proper divisors (200279985).
241005135 is a wasteful number, since it uses less digits than its factorization.
241005135 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 3087.
The product of its (nonzero) digits is 600, while the sum is 21.
The square root of 241005135 is about 15524.3400825929. The cubic root of 241005135 is about 622.3128451423.
Adding to 241005135 its reverse (531500142), we get a palindrome (772505277).
The spelling of 241005135 in words is "two hundred forty-one million, five thousand, one hundred thirty-five". | 514 | 1,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.842146 |
https://justlikemagic.home.blog/2009/03/08/convert-dec-to-hex/?like_comment=35&_wpnonce=35655f66c8 | 1,643,077,346,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00112.warc.gz | 365,894,811 | 33,953 | # Converting between color decimal and hexadecimal representation
هذه المقالة متوفرة أيضا باللغة العربية، اقرأها هنا.
## Contents
• Contents
• Introduction
• Color Represenatation
• Try it Yourself
## Introduction
Before we begin our article we need to know first what is decimal and what is hexadecimal.
Using decimal representation means using base-10 representation of numbers which means that every digit of the number can be one of 10 values, 0 through 9.
A good example is 195. It’s is 3 digits and every digit is between 0 and 9.
On the other hand, hexadecimal is base-16 representation which means that there are 16 of base values that every digit in the number can be one of. These base values are 0 through 9 and A through F. From 0 to 9 are the same in dec (the shorthand for decimal). A equals 10, B equals 11, C equals 12, etc.
An example is 14 (decimal). If we need to represent it in hexadecimal we can say D.
Another good example is 9 (decimal). If we need to represent it in hex (the shorthand for hexadecimal,) you can say 9 too.
## Color representation
You already know that a colors consists of 3 values; RGB (Red, Green and Blue). Although, there’re times when we blend it with another value (Alpha) to make color transparent, however it is not common. So we end up with 4 values (ARGB) to represent our color. True? Every value of these four occupies a single byte in memory so every value ranges from 0 to 255.
In HTML for instance you can represent any color by its name (if it is a known color like Red) or by combining the RGB values using hexadecimal (HTML doesn’t support Alpha.) You cannot use decimal values to represent colors in HTML. So you need to know how to convert between every decimal value to its equivalent hexadecimal and vice versa.
You already know that decimal values from 0 to 15 have the hex counterparts 0 through 9 and A through F so conversion will be very easy in this range. For a decimal value greater than 15 you need to apply a simple formula to convert it.
1. First, divide the value by 16 (because hex is base-16), then add the hex result (e.g. A for 10) to the left.
2. Second, divide the value by 16, and then add the remainder (modulus) to the right of the last number.
Examples:
```- 11 (dec)
11 = B
Result = B (hex)
- 160 (dec)
160 / 16 = 10
160 mod 16 = 0
10 = A
0 = 0
Result: A0 (hex)
- 254 (dec)
254 / 16 = 15
254 mod 16 = 14
15 = F
14 = E
Result = FE (hex)```
Because every value of ARGB occupies a single byte, then the values range from 0 to 255 so in hexadecimal they range from 0 to FF!
For hex values from 0 through 9 and A through F, they have equivalents in decimal.
For values greater than F (15) all you need is to reverse the steps you did in converting decimal to hex.
1. First, take the left digit and convert it to decimal. Then, multiply the result by 16.
2. Second, take the right digit and convert it to decimal. Then, sum it with the result of the first step.
Examples:
```- B (hex)
B = 11
Result = 11 (dec)
- A0
A = 10
0 = 0
A * 16 = 160
160 + 0 = 160
Result = 160 (dec)
- FE
F = 15
E = 14
15 * 16 = 240
240 + 14 = 254
Result = 254 (dec)```
Easy, isn’t it? A color like khaki (RGB: 240, 230, 140) in hexadecimal equals FF F0 E6 8C. To represent it in HTML you can say #F0E68C.
You can use Windows Calculator (in scientific mode) to make such conversions. | 922 | 3,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-05 | latest | en | 0.831474 |
https://cs.stackexchange.com/questions/6521/reduce-the-following-problem-to-sat?noredirect=1 | 1,708,490,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00567.warc.gz | 198,906,197 | 41,867 | # Reduce the following problem to SAT
Here is the problem. Given $k, n, T_1, \ldots, T_m$, where each $T_i \subseteq \{1, \ldots, n\}$. Is there a subset $S \subseteq \{1, \ldots, n\}$ with size at most $k$ such that $S \cap T_i \neq \emptyset$ for all $i$? I am trying to reduce this problem to SAT. My idea of a solution would be to have a variable $x_i$ for each of 1 to $n$. For each $T_i$, create a clause $(x_{i_1} \vee \cdots \vee x_{i_k})$ if $T_i = \{i_1, \ldots, i_k\}$. Then and all these clauses together. But this clearly isn't a complete solution as it does not represent the constraint that $S$ must have at most $k$ elements. I know that I must create more variables, but I'm simply not sure how. So I have two questions:
1. Is my idea of solution on the right track?
2. How should the new variables be created so that they can be used to represent the cardinality $k$ constraint?
• Just a remark: Your problem is known as HITTING SET, which is an equivalent formulation of the SET COVER problem. Nov 7, 2012 at 8:22
It looks like you are trying to compute a hypergraph transversal of size $k$. That is, $\{T_1,\dots,T_m\}$ is your hypergraph, and $S$ is your transversal. A standard translation is to express the clauses as you have, and then translate the length restriction into a cardinality constraint.
So use your existing encoding, i.e., $\bigwedge_{1 \le j \le m} \bigvee_{i \in T_j} x_i$ and then add clauses encoding $\sum_{1 \le i \le n} x_i \le k$.
$\sum_{1 \le i \le n} x_i \le k$ is a cardinality constraint. There are various different cardinality constraint translations into SAT.
The simplest but rather large cardinality constraint translation is just $\bigwedge_{X \subseteq \{1,\dots,n\}, |X| = k+1} \bigvee_{i \in X} \neg x_i$. In this way each disjunction represents the constraint $\neg \bigwedge_{i \in X} x_i$ - for all subsets $X$ of $\{1,\dots,n\}$ of size k+1. That is, we ensure that there is no way that more than k variables can be set. Note that this is not polynomial size in $k$
Some links to papers on more space-efficient cardinality constraint translations which are polynomial size in $k$:
If you are actually interested in solving such problems, perhaps it is better to formulate them as pseudo-boolean problems (see wiki article on pseudo-boolean problems) and use pseudo-boolean solvers (see pseudo-boolean competition). That way the cardinality constraints are just pseudo-boolean constraints and are part of the language - hopefully the pseudo-boolean solver then handles them directly and therefore more efficiently.
If you're not absolutely set on the normal SAT, your idea is already a reduction to MIN-ONES (over positive CNF formulae), which is basically SAT, but where you can set at most $k$ variables to true (strictly it's the optimization version where we minimize on the number of true variables).
Similarly if you head in a Parameterized Complexity direction, then you've already basically got WSAT($\Gamma_{2,1}^{+}$), where $\Gamma_{2,1}^{+}$ is the class of all positive CNF formulae (same as before, the notation might help your investigations though). In this case you'd have to start looking at what parameterization would be useful in your case.
I assume you're looking for an explicit reduction, but if not, you can always just fall back to the Cook-Levin Theorem. | 885 | 3,354 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-10 | latest | en | 0.869488 |
http://simplebooklet.com/countingbackwards200 | 1,532,179,302,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592579.77/warc/CC-MAIN-20180721125703-20180721145703-00066.warc.gz | 331,494,414 | 7,073 | of 0
Title
Counting Backwards (20-0)
Lesson Objective
Students will be able to (1) orally count backwards from 20-0 and (2) write numbers backwards from 20-0 in order.
Background Information for Teacher
N/A
Student Prior Knowledge
Student should be able to recognize numbers 0-20.
Materials:
Pencil
Marker
Ten stickers
21 pieces of large construction paper with a number (from 0-20) written in large print on each
index cards
Step-by-Step Guided Lesson
Step 1: Start Video
(Tips: Interact with the video by pausing, to ask questions or discuss information viewed with student.)
Step 2: Teach Lesson
The student can be seated on the rug close to the teacher. The teacher will begin by showing the student a sheet of stickers. There
should be 10 stickers on the sheet (stickers can be substituted with many other items). Together, the student and teacher will count
the stickers starting from 1 up to 10.
One by one, teacher will give a sticker to the student. After each sticker is removed, teacher will ask “How many stickers to do we
have left”?
By the time all stickers are passed out, the student would have counted down from 10-0.
Refer to a number line (or something similar) in the room. Notice how numbers get “bigger/greater” when counting forward and
“smaller/less” when counting backwards. Use hand/body gestures to refer to forward and backward.
Write the words “less”, “take-away” and “backwards” on the board or display the words on a pocket chart. Ask the student to tell you
what they know about those words. Together, chorally practice counting backwards from 20-0 as teachers points to each number on
the number line. Teacher can physically move backwards as the student count to show the relation.
Teacher should place pieces of large construction paper with one number (in order from 20-0) written in large print on each. The
student will pull small index card with a “mystery number” (0-20) written on it. The student will find their number on the construction
paper number line on the floor. They will then begin hopping/counting backwards from their mystery number.
Repeat until the student has pulled all the cards.
Recall sticker activity from the lesson's opening. Review the lesson with students. Ask students, "what did we learn about numbers
today"? Count backwards with students. Ask for questions.
Have the student complete the worksheet is Step 3.
Step 3: Complete the worksheet attached below.
Worksheets needed to complete the lesson
Step 4: Review. Start the next lesson with the game or activity attached below for review so the student can demonstrate
understanding of this lesson before moving forward.
I Spy Number Counting (online game) | 606 | 2,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-30 | latest | en | 0.934713 |
https://dvidya.com/mathematical-reasoning/ | 1,702,015,589,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00211.warc.gz | 251,474,371 | 67,791 | # Mathematical Reasoning
Mathematical Reasoning
Mathematical Reasoning – Meaning
The word “Reasoning” means critical thinking, analysis and conclusion based thereon. Mathematical Reasoning means application of such analytical skill in the area of Mathematics. Mathematical Reasoning enables us to determine the Truth value of various types of Mathematical Statements, applying Mathematical Rules.
Mathematical Statement
In our daily life we use different types of sentences like Assertive (which declares or asserts a statement), Interrogative, Exclamatory, Imperative, Optative etc. Among them only assertive sentences are called Mathematical statement. However, all assertive sentences are not Mathematical Statements.
Negation of Mathematical Statement
Negation of a Mathematical Statement implies the denial (or contradiction) of the statement.
Negation of Mathematical Statement – Examples
Mathematical Statement Constructs
• Mathematical Simple Statement: If the truth value of a statement does not depend on any other statement, then the statement is called a simple Mathematical statement. Simple statement cannot be subdivided into two or more simple statements.
• Mathematical Compound Statement: A compound statement is a combination of two or more simple statements connected by the words “and”, “or”, etc. A compound statement can be subdivided into two or more simple statements.
Mathematical Statement Examples
• Mathematical Simple Statement – Example : The followings are simple statements and they cannot be sub-divided into simpler statements.
Earth moves round the sun. (ii) Sun is a star. (iv) 40 is multiple of 4
• Mathematical Compound Statement – Example: The followings are compound statements. Each compound statement can be sub-divided into two (or more) simple statements.
Mathematical Statement Connectives
Connectives are words used to connect simple statements to form Compound Statements (like “and”, “or”, “if-then”, “only if”, “if and only if”, etc.).
Compound Statement with connective ‘AND’
Any two simple statements can be combined by using the word “and” to form compound statements.
• If each simple mathematical statement belonging to a compound mathematical statements are true then the compound mathematical statement would be True.
• If any of the simple statement connected is false, then the compound mathematical statement would be False.
Compound Mathematical Statement (with AND) – Examples
• Compound Mathematical Statement (with AND)– Example 1 : Compound Statement (R): Delhi is a big city and it is the Capital of India.
• R is a compound mathematical statement, formed by connecting two simple mathematical statements P & Q, using the connective “and” where
• P : Delhi is a big city – is the first constituent Simple Statement, which is TRUE
• Q: Delhi is the capital of India – is the second constituent Simple Statement, which is TRUE
Here both P and Q are true. So the value of the compound mathematical statement is “True”.
• Compound Mathematical Statement (with AND) – Example 2 :
• R : 41 is a prime number and it is an even number. Here R is a compound mathematical statement and is formed by connecting two simple mathematical statements P and Q using the connective “and” where
• P : 41 is a prime number – is the first constituent Simple Statement, which is TRUE
• Q : 41 is an even number – is the second constituent Simple Statement, which is FALSE
Here P is TRUE but Q is FALSE. So the value of R is FALSE.
• Compound Mathematical Statement (with AND) – Example 3
• R : 39 is a prime number and it is an even number. Here R is a compound mathematical statement and is formed by connecting two simple mathematical statements P and Q using the connective “and” where
• P : 39 is a prime number – is the first constituent Simple Statement, which is FALSE
• Q : 39 is an even number – is the second constituent Simple Statement, which is FALSE
Here P and Q, both are FALSE. So, the value of R is FALSE.
All mathematical statements connected with “and” may not be a compound mathematical statement.
For example, “The sum of 3 and 7 is 10” is a simple mathematical statement, as there is only ONE Statement. So, it is not a compound mathematical statement.
Compound Statement with connective ‘OR’
Two (or more) simple mathematical statements can be combined by using “or” to form a compound
mathematical statements.
• If any one (or more) of the component simple mathematical statements of a compound mathematical statements is true then the truth value of the compound mathematical statement is True.
• If all the component simple mathematical statements are false, then the truth value of the compound mathematical statement is “false”.
Compound Statement with connective ‘OR’ – Examples
• Compound Mathematical Statement (with OR) – Example 1 : p: Rhombus is a quadrilateral, is a simple mathematical statement, which is TRUE
q: Rhombus is a parallelogram, is a simple mathematical statement, which is TRUE
r: Rhombus is a quadrilateral or a parallelogram, is a Compound mathematical statement of the above two simple connected mathematical statements, connected with connective ‘’OR’’
Here p & q both are simple mathematical statements and both are true.
Since both p and q are true, the truth value of r is “True” and here “OR” is Inclusive “OR”.
• Compound Mathematical Statement (with OR) – Example 2 :
p: 95 is divisible by 7, is a simple mathematical statement, which is FALSE
q: 95 is divisible by 5, is a simple mathematical statement, which is TRUE
r: 95 is divisible by 7 or by 5, is a Compound mathematical statement, of the above two simple connected mathematical statements p & q, connected with connective ‘’OR’’. Here p & q both are simple mathematical statements, connected with connective ‘’OR’’. One is TRUE & other is FALSE. So, the truth value of r is “True”.
• Compound Mathematical Statement (with OR) – Example 3 :
P: Two given straight lines AB & CD intersect at a point, is a simple mathematical statement.
Q: Two give straight lines AB & CD are parallel, is a simple mathematical statement.
R: Two straight lines AB and CD either intersect at a point or they are parallel, is a Compound mathematical statement of the above two simple connected mathematical statements, connected with connective ‘’OR’’
If P is TRUE, then Q is FALSE or if P is FALSE, then Q is TRUE, but P and Q cannot be both TRUE or cannot be both FALSE. Only one of P and Q is TRUE.
Mathematical Statement with Inclusive & Exclusive OR
• Mathematical Statement with Inclusive OR : If all the component simple mathematical statements of a compound mathematical statement formed by using the connective “or” are true, then the “or” is called Inclusive “or”.
• Mathematical Statement with Exclusive OR : In Compound mathematical Statement, if one is true and other is false, then the “or” used in compound mathematical statement is called Exclusive “or”.
This may be illustrated with following Example.
In a course, there are 2 subjects, Mathematics & Economics.
• One option is “You can take up Economics OR Mathematics OR both” – this is instance of “Inclusive OR”
• Another option is “ You can take up Economics OR Mathematics but not both “this is instance of “Exclusive OR”
Unless specifically stated, Inclusive OR is presumed. Exclusive OR is seldom used.
Compound Mathematical Statement (with Inclusive & Exclusive OR) – Examples
• – Example -1 : ‘Triangle can be defined as a polygon with three sides or as a polygon with three vertices is a compound mathematical statement with Inclusive “OR”. Either or both statements can be true.
• Compound Mathematical Statement (with Exclusive OR – Example -2 : ‘Today I can go to school or I can stay home’ is a compound mathematical statement with exclusive “OR”. Either option may be true but not both.
• Compound Mathematical Statement (with Exclusive OR) – Example -3 : ‘A Student can take Urdu or Sanskrit as third language’, is a compound mathematical statement with exclusive “OR”. A student can take either Sanskrit or Urdu, but not both.
Mathematical Statement – Implications
A compound mathematical statement is formed connecting two simple mathematical statements using
the connecting words “IF – THEN”, “ONLY IF” and “IF AND ONLY IF”. These connecting words are called Implications.
Rules of Truth Value of Implications :
Compound Statement with connective IF – THEN
Compound Statement with IF – THEN – Example 1 : ‘’If a number is divisible by 6, then it must be divisible by 3″. It is compound mathematical statement.
• p: A number is divisible by 6
• q: The number is divisible by 3.
Here if p is true, then q must be true. But if p is not true (the number is not divisible by 6) then we cannot say that it is not divisible by 3 (e.g. 123 is not divisible by 6 but it is divisible by 3).
Compound Statement with IF – THEN – Example 2 : “If 42 is divisible by 7, then sum of the digits of 42 is divisible by 7”, is a compound mathematical statement.
• p: 42 is divisible by 7 is a is a simple mathematical statement, which is TRUE.
• q: The sum of the digits of 42 is divisible by 7, is a simple mathematical statement, which is FALSE.
Compound Statement with IF – THEN – Example 3 : “If the person is born in India, then the person is a citizen of India”, is a compound mathematical statement.
• p: The person is born in India.
• q: The person is a citizen of India.
Here, if p is TRUE, then q is TRUE.
Compound Statement with IF – THEN – Example 4: If it rains, then the class will not go out to picnic.
This means that [if it rains], the statement [the class will not go out to picnic] is TRUE. However, If [the class will not go out to picnic] does not mean that [It is raining]
Contrapositive and Converse Mathematical Statement
If a compound mathematical statement is formed with two simple mathematical statements p and q
using the connective “IF-THEN”, then the Contrapositive and Converse statements of compound statement can also be formed.
• Contrapositive Mathematical Statement
• Converse Mathematical Statement
Contrapositive & Converse Mathematical StatementExample 1: “If a number is divisible by 6, then it is divisible by 3”, is a compound mathematical statement.
• Its contrapositive statement is: “if a number is not divisible by 3, then it cannot be divisible by 6”.
• Its Converse statement is : “If a number is divisible by 3, then it is divisible by 6”.
Contrapositive & Converse Mathematical StatementExample 2 : “If it is raining then the grass is wet”, is a compound mathematical statement.
• The contrapositive is “If the grass is not wet, then it is not raining.”
• The converse is “If the grass is wet then it is raining.”
Compound Mathematical Statement with connectiveONLY IF”
The compound mathematical statement formed with two simple mathematical statements, p and q, using the connective word “ONLY IF” implies that p happens only if q happens.
• Compound Mathematical Statement with connectiveONLY IF” – Example 1. The compound mathematical statement ‘Samir will attend the party only if Kanti will be there’ means that ‘Samir will attend the party’ will happen (i.e becomes TRUE) only if Kanti will be at the Party’ happens (i.e becomes TRUE).
• Compound Mathematical Statement with connectiveONLY IF” – Example 2. The compound mathematical statement ”The triangle ABC, will be equilateral only if AB = BC = CA.” formed from two simple statements i) p: ‘The triangle ABC is equilateral’, ii) q: ‘In the triangle ABC, AB = BC = CA’ means that the statement p: ‘The triangle ABC, will be equilateral’, is TRUE only if the statement q:’ If AB = BC = CA’ is TRUE.
Compound Mathematical Statement with ‘’IF AND ONLY IF (IFF)’’
A Biconditional compound mathematical is formed with two simple mathematical statements using the connecting words “if and only if”.
Compound Mathematical Statement with ‘’IF AND ONLY IF – Example 1: Two triangles are congruent IF AND ONLY IF the three sides of one triangle are equal to the three sides of the other triangle.”
This statement can be written as:
• If two triangles are congruent, then the three sides of one triangle are equal to the three sides of the other triangle.
• If the three sides of one triangle is equal of the three sides of the other triangle, then the two triangles are congruent.
Let p: Two triangles are congruent. q: Three sides of one triangle are equal to the three sides of the other triangle.
Mathematical Statement with Quantifiers
In some mathematical statements some phrases (called Quantifiers) like “There exists”, “For all” (or for every) are used,
• Mathematical Statement with Exitential Quantifier :
Truth value for Statements with Quantifiers (For Equations) Examples: Using Quantifiers, equations expressed into statement : | 2,812 | 12,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-50 | latest | en | 0.835968 |
https://www.bartleby.com/solution-answer/chapter-2-problem-254ep-general-organic-and-biological-chemistry-7th-edition/9781285853918/what-is-the-uncertainty-in-the-measured-value-8720030-after-it-is-rounded-to-the-following-number/6dd444a5-b054-11e9-8385-02ee952b546e | 1,579,267,118,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250589560.16/warc/CC-MAIN-20200117123339-20200117151339-00089.warc.gz | 780,551,471 | 84,007 | What is the uncertainty in the measured value 87.20030 after it is rounded to the following number of significant figures? a. 6 b. 5 c. 4 d. 2
General, Organic, and Biological C...
7th Edition
H. Stephen Stoker
Publisher: Cengage Learning
ISBN: 9781285853918
Chapter
Section
General, Organic, and Biological C...
7th Edition
H. Stephen Stoker
Publisher: Cengage Learning
ISBN: 9781285853918
Chapter 2, Problem 2.54EP
Textbook Problem
9 views
What is the uncertainty in the measured value 87.20030 after it is rounded to the following number of significant figures? a. 6 b. 5 c. 4 d. 2
(a)
Interpretation Introduction
Interpretation:
The uncertainty for the measured value 87.20030, when rounded off to 6 significant figures has to be identified.
Concept Introduction:
Whenever a measurement is made, the significant figures in the measured quantity give the actual measurement. For this the significant figures should be recognized first. The significant figures may be non-zero digit and zero digit. But Zero may be or may not be a significant figure. The number of significant figures gives more information about the degree of uncertainty. This uncertainty is determined from the last digit. One should also note in which place the last digit appears, either in the tenth, or hundredth or thousandth place.
Explanation of Solution
The location of the last significant figure of a measurement shows the uncertainty. When the measured value 87.20030 is rounded off for the 6 significant figures, the value becomes 87.2003, the position of the last significant figure gives the uncertainty of the measurement, so the last significant figure here is 3, that is in ten thousandths place, Hence the uncertainty 0.0001.
Significantfigure=687
(b)
Interpretation Introduction
Interpretation:
The uncertainty for the measured value 87.20030, when rounded off to 5 significant figures has to be identified.
Concept Introduction:
Whenever a measurement is made, the significant figures in the measured quantity give the actual measurement. For this the significant figures should be recognized first. The significant figures may be non-zero digit and zero digit. But Zero may be or may not be a significant figure. The number of significant figures gives more information about the degree of uncertainty. This uncertainty is determined from the last digit. One should also note in which place the last digit appears, either in the tenth, or hundredth or thousandth place.
(c)
Interpretation Introduction
Interpretation:
The uncertainty for the measured value 87.20030, when rounded off to 4 significant figures has to be identified.
Concept Introduction:
Whenever a measurement is made, the significant figures in the measured quantity give the actual measurement. For this the significant figures should be recognized first. The significant figures may be non-zero digit and zero digit. But Zero may be or may not be a significant figure. The number of significant figures gives more information about the degree of uncertainty. This uncertainty is determined from the last digit. One should also note in which place the last digit appears, either in the tenth, or hundredth or thousandth place.
(d)
Interpretation Introduction
Interpretation:
The uncertainty for the measured value 87.20030, when rounded off to 2 significant figures has to be identified.
Concept Introduction:
Whenever a measurement is made, the significant figures in the measured quantity give the actual measurement. For this the significant figures should be recognized first. The significant figures may be non-zero digit and zero digit. But Zero may be or may not be a significant figure. The number of significant figures gives more information about the degree of uncertainty. This uncertainty is determined from the last digit. One should also note in which place the last digit appears, either in the tenth, or hundredth or thousandth place.
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# Problem 1681. Do you like your boss?
Solution 1687315
Submitted on 6 Dec 2018 by test test
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 'Do you like your boss?'; y_correct = 'Any String!'; assert(isequal(isstr(YourBoss(x)),isstr(y_correct)))
2 Pass
x = 'Does your boss smell funny?'; y_correct = 'Any String!'; assert(isequal(isstr(YourBoss(x)),isstr(y_correct)))
3 Pass
x = 'Is your boss a man or a woman?'; y_correct = 'Any String!'; assert(isequal(isstr(YourBoss(x)),isstr(y_correct)))
4 Pass
x = 'Is your boss mean or nice?'; y_correct = 'Any String!'; assert(isequal(isstr(YourBoss(x)),isstr(y_correct)))
5 Pass
x = 'Do you see your boss often?'; y_correct = 'Any String!'; assert(isequal(isstr(YourBoss(x)),isstr(y_correct)))
6 Pass
x = 'If your boss were an animal, what type of animal would he or she be?'; y_correct = 'Any String!'; assert(isequal(isstr(YourBoss(x)),isstr(y_correct)))
7 Pass
x = 'On a scale from one to ten, where does your boss rank?'; y_correct = 'Any String!'; assert(isequal(isstr(YourBoss(x)),isstr(y_correct)))
8 Pass
x = 'Maybe you are your own boss...'; y_correct = 'Any String!'; assert(isequal(isstr(YourBoss(x)),isstr(y_correct)))
9 Pass
x = 'Maybe your boss is standing behind you, with that glare on his face, tapping his foot with his arms folded...'; y_correct = 'Any String!'; assert(isequal(isstr(YourBoss(x)),isstr(y_correct))) | 452 | 1,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-51 | longest | en | 0.635445 |
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Latest
# Introduction to Numpy in Python
Last Updated on October 29, 2021 by Editorial Team
#### What isย NumPy?
NumPy stands for numeric python, a Python module that allows you to compute and manipulate multi-dimensional and single-dimensional array items. It comes with a high-performance multidimensional array object as well as utilities for working withย them.
In 2005, Travis Oliphant created the NumPy package by combining the features of the ancestral module Numeric with the characteristics of another module Numarray. NumPy implements multidimensional arrays and matrices as well as other complex data structures. These data structures help to compute arrays and matrices in the most efficient way possible. NumPy allows you to conduct mathematical and logical operations on arrays. Many other prominent Python packages, like pandas and matplotlib, are compatible with Numpy and useย it.
#### Need to use NumPy inย Python?
NumPy is a valuable and efficient tool for dealing with large amounts of data. NumPy is quick; thus, itโs easy to work with a vast amount of data. Data analysis libraries like NumPy, SciPy, Pandas, and others have exploded in popularity due to the data science revolution. Python is the most simple language in terms of syntax, and is the priority for many data scientists; therefore, NumPy is getting popular day by day.
Many mathematical procedures commonly used in scientific computing are made quick and simple with Numpy, including:
• Multiplication of Vector or Matrix to Vector orย Matrix.
• Operations on vectors and matrices at the element level (i.e., adding, subtracting, multiplying, and dividing by a numberย )
• Applying functions to a vector/matrix element by element ( like power, log, and exponential).
• NumPy has functions for linear algebra and random number generation built-in.
• NumPy also makes matrix multiplication and data reshaping simple.
• Multidimensional arrays are implemented efficiently usingย NumPy.
• NumPy arrays in Python give capabilities for integrating C, C++, and other languages. Itโs also useful in linear algebra and random number generation, among other things. NumPy arrays can also be used to store generic data in a multi-dimensional container.
• Itโs faster than standard Python arrays, which donโt have NumPyโs pre-compiled C code(Precompiled code is a header file that is compiled into an intermediate form that is faster to process for the compiler).
#### Installation ofย NumPy
Go to your command prompt and run โpip install numpyโ to install Python NumPy. Once the installation is complete, simply go to your IDE (for example, VS Code, Jupyter Notebook, PyCharm) and type โimport numpy as npโ to import it; now you are ready to useย NumPy.
#### What is a NumPyย Array?
The Numpy array object is a strong N-dimensional array object with rows and columns. NumPy arrays can be created from nested Python lists, and their elements can be accessed. It refers to a group of items that are all of the same types and can be accessed using zero-based indexing. Every item in a ndarray is the same size as the memory block. Each entry in ndarray is a data-type object (calledย dtype).
#### Types of Numpyย Array:
• One-Dimensional NumPy array
A one-dimensional array has only one dimension of elements. In other words, the one-dimensional NumPy array should only contain one tuple value.
Example:
`# One-Dimensional Arrayimport numpy as npval=np.array([1, 5, 2, 6])print(val)`
Implementation:-
Explanation: print() function is used to print the whole array provided.
• Multi-Dimensional NumPy Array
A multi-dimensional array can have n dimensions of elements. In other words, the multi-dimensional NumPy array can contain a number of tuple values.
Example:
`# Multi-Dimensional Arrayimport numpy as npval=np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val)`
Implementation:
Explanation: print() function is used to print the whole array provided.
#### Accessing Element in theย Array
Indexing or accessing the array index in a NumPy array can be done in a variety of ways.
1. Slicing is used to print a range of an array. The built-in slice function is used to create a Python slice object by passing start, stop, and step parameters to it. Slicing an array involves establishing a range in a new array used to print a subset of the original arrayโs elements. Because a sliced array holds a subset of the original arrayโs elements, editing content alters the original arrayโs content.
Exampleย 1:
`# Accessing Array Elementimport numpy as npval=np.array([1, 5, 2, 6])# Zero based indexingprint(val[1])val_mul=np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val_mul[1][1])`
Implementation:
Explanation: Here we are accessing the 1st index value of the 1-D array via writing val[1] and accessing the value in the 2nd row and 2nd column of the 2-D array as the arrays have zero-based indexing.
Example 2:
`#Accessing Range of Elements #in array using slicingval_mul=np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])slice_val = val_mul[:2, :2]print (โFirst 2 rows and columns of the array\nโ, slice_val)Implementation:`
Implementation:
Explanation: Here we are trying to access the values in rows and columns starting from 1st row to 2nd row by writing โย :2 โ as the first argument and similarly accessing all the values corresponding to those rows and having columns from starting to theย 2nd.
Example 3:
`val_mul1=np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6),(1, 5, 0, 1),(7, 3, 3, 0)])slice_step_row = val_mul1[::2, :2]print (โAll rows and 2 columns of the array with step size of 2 along the row\nโ, slice_step_row)`
Implementation:
Explanation: Here we are trying to access the values in rows and columns starting from 1st row to end row with a step size of 2 which means the difference between consecutive rows to be 2 by writing โ::2 โ as the first argument and similarly accessing all the values corresponding to those rows and having columns from starting to theย 2nd.
Example 4:
`# Accessing multiple elements at one goval_mul1=np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6),(1, 5, 0, 1),(7, 3, 3, 0)])mul_idx_arr = val_mul1[ [1, 2, 2, 3], [1, 2, 3, 3] ]print (โ\nAccessing Elements at multiple indices (1, 1),(2, 2), (2, 3), (3, 3) โโ, mul_idx_arr)`
Implementation:
Explanation: Here we are trying to access the values from the array which are present at the indexes (1,1),(2,2),(2,3), andย (3,3).
2. Another type of accessing technique is the boolean array indexing where we can give the condition where elements that follow the condition areย printed.
Example:
`# Boolean array indexingval_mul=np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6),(1, 5, 0, 1),(7, 3, 3, 0)])condition = val_mul > 3 res = val_mul[condition]print (res)`
Implementation:
Explanation: Here we are accessing the values which follow the given conditions i.e. the value of the element should be greater thanย 3.
### Frequently used Operations on Numpyย Array
• ndim– This operation is used to compute the dimension of theย array.
Example:
`# Dimension of the arrayval1 = np.array([1, 5, 2, 6])print(val1.ndim)`
`# 2-D arrayval2 = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val2.ndim)`
Implementation:
Explanation: Here val1 is a 1-D array having values 1,5,2 and 6 therefore we are getting value 1 for val1.ndim and val2 is a 2-D array
3 4 2 5
3 6 2 4
1 5 2 6
Therefore we are getting value 2 corresponding to val2.ndim
• SizeโโโThis operation is used to calculate the number of elements in theย array.
Example:
`# Calculate Array Size val2 = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val2.size)`
Implementation:
Explanation: Here val2 contains 12 values 3, 4, 2, 5, 3, 6, 2, 4, 1, 5, 2, 6 therefore we are getting the size of the val2 array asย 12.
• ShapeโโโThis operation is used to calculate the shape of theย array.
Example:
`# Calculate Array Shape val = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val.shape)`
Implementation:
Explanation: Here Val array is a 2-D array
3 4 2 5
3 6 2 4
1 5 2 6
It has 3 rows and 4 columns and val.shape is used to get the size corresponding to each dimension.
• ReshapeโโโReshape is used to reshape the array according to the given parameters
Example:
`# Reshape the Arrayval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val)`
`val=val.reshape(2,6)print(val)`
Implementation:
Explanation: reshape function helps us to reshape the array and fill the values of the array correspondingly. Here we have 12 values and we want to reshape the array from (3,4) to (2,6) so now there would be only 2 rows and 6ย columns.
• TransposeโโโT operator is used in getting the transpose of an array.
Example:
`# Transpose of an arrayval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val.T)`
Implementation:
Explanation: When we need to replace all the rows of an array with the columns and columns with rows then we need to call the val.T to get the transpose of the array. In transpose, the shape of the array changes as now number of rows becomes a number of columns and vice versa. As here 3, 4, 2, 5 was the first row initially but after transpose, it becomes the 1stย column.
• Ravelโโโravel is used to convert the array into a single column
Example:
`# Convert Array to Single Columnval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val.ravel())`
Implementation:
Explanation: When we want to convert the array to a 1-D array we use the ravel function as it combines all the values of the array as in the above example we had initially a 2-D array of 3 rows and 4 columns but after applying the ravel function we gets the 1-D array of 12ย values.
• Itemsize: Itemsize is used to compute the size of each element in bytes.
Example:
`# Calculate Array Itemsizeval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])# size of each element in bytesprint(val.itemsize)`
Implementation:
Explanation: When we want to get the size of each element in bytes we use (array_name.itemsize) as in the above example the data type is int32 which means integer of 32 bits and as we know that 1 byte is equal to 8 bits and hence it is 4 bytes inย size.
• Dtype– Dtype is used to get the data type of the elements in the array.
Example:
`# Calculate Data type of each elementval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val.dtype)`
Implementation:
Explanation: When we want to compute the data type of the elements present in the array we use val.dtype which gives us the data type here in the above example the values of the array are integer so the data type given is int32 which means it represents an integer of 32ย bits.
• np.zeros()/np.ones()โโโones() and zeros() functions of numpy are used to generate arrays having all 1โs and 0โs respectively.
Example:
`# Generating array having all 1'sval = np.ones(3)print(val) # Generating array having all 0'sval0 = np.zeros(3)print(val0)`
Implementation:
Explanation: When we want to generate an array having all 1โs or 0โs we use the above-given functions which helped us generate an array of size n. Here we gave the size of the array as 3 so an array of size 3 having all 1โs and 0โs is generated.
linspaceโโโThe linspace function is used to generate an array containing elements distributed equally in a given range. It takes 3 parameters to start and end of the range and the number of elements to be present in theย array.
Example:
`# Linspce generate 8 numbers present in range 2โ5val = np.linspace(2,5,8)print(val)`
Implementation:
Explanation: When we want to generate an array having specific order like in range x to y and size of the array to be n then we use linspace function as here we wanted to generate an array of size 8 having values equally spaced between range 2 toย 5.
• MaxโโโMax is used in getting the maximum element across the wholeย array.
Example:
`# Find maximum element in whole arrayval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val.max())`
Implementation:
Explanation: When we want to calculate the maximum of all elements of the array we can simply write array_name.max() and get the maximum element as here we get 6 which is the maximum of all the elements.
• MinโโโMin function is used in getting the minimum element across the wholeย array.
Example:
`# Find minimum element in whole arrayval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val.min())`
Implementation:
Explanation: When we want to calculate the minimum of all elements of the array we can simply write array_name.min() and get the minimum element as here we get 1 which is the minimum of all the elements.
• SumโโโSum function is used to get the total of all the elements of theย array.
Example:
`# Sum of all elements in whole arrayval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val.sum())`
Implementation:
Explanation: When we want to calculate the sum of all elements of the array we can simply write array_name.sum().
• np.sqrt()โโโThis function is used to get the square root of all the elements of the array.
Example:
`# Used to calculate square root of each elementval=np.array([1, 4, 9])print(np.sqrt(val))`
Implementation:
Explanation: When we want to calculate the square root of each value of an array we can simply write np.sqrt(array_name) and we get the corresponding square roots to each and every value e.g. 2 corresponding to 4 and 3 corresponding toย 9.
• np.std(): This function is used to calculate the standard deviation of all the elements of theย array.
Example:
`# Used to calculate standard deviation of all elementsval=np.array([1, 4, 9])print(np.std(val))`
Implementation:
Explanation: When we want to find the standard deviation of all the values of the array we do not need to do the hectic math calculation for calculating the standard deviation, we can simply write (np.std(array_name)).
• Summation operation on the arrayโโโWhen we need to add some particular number n to all the elements of the array or one array to anotherย array.
Example1:
`# Add number 3 to all elements of array valval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val+3)`
Implementation:
Explanation: When we want to add value n from each value of array we can simply write (array_name+n).
Example 2:
`# Add array val to another array val1val = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])val1 = np.array([(4, 4, 8, 2),(9, 0, 3, 5),(6, 8, 3, 3)])print(val+val1)`
Implementation:
Explanation: When we want to find the addition of 2 arrays we can simply write (array1+array2) to get the addition of arrays but just keep in mind that the dimensions of both the arrays must be theย same.
• Subtraction operation on the arrayโโโWhen we need to subtract some particular number n to all the elements of the array or one array from anotherย array.
Example1:
`# Subtract number 3 from all elements of array valval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val-3)`
Implementation:
Explanation: When we want to subtract value n from each value of array we can simply write (array_name-n).
Example 2:
`# Subtract array val from another array val1val = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])val1 = np.array([(4, 4, 8, 2),(9, 0, 3, 5),(6, 8, 3, 3)])print(val1-val)`
Implementation:
Explanation: When we want to find the difference between 2 arrays we can simply write (array1-array2) to get the difference of arrays but just keep in mind that the dimensions of both the arrays must be theย same.
• Power of each element of the array raised to numberย n:
Example:
`# Power of each element of the array raise to 3# Cube of each elementval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val**(3))`
`# Power of each element of the array raise to 0.5# Square root of each elementval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val**(0.5))`
Implementation:
Explanation: Here we are raising to the power of 3 each and every value of the array and printing it.When we want to raise to the power of n each value of array we simply write (array_name)**n to raise to the power of each and every value of the array by valueย n.
• Modify eachย element:
Example:
`# Multiply each element of the array by 3val = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val*(3))`
`# Divide each element of the array by 3val = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(val/3)`
Implementation:
Explanation: Here we are multiplying each and every value of the array by 3 and printing it. Similarly, when we want to divide each value of array we simply write (array_name)/n to divide the array by valueย n.
• np.sort()โโโThis function is used to sort the array and takes the argument axis which allows sorting the array row-wise and column-wise when initialized with values 1 and 0 respectively.
Example1:
`# Sort the array row-wiseval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])print(np.sort(val,axis=1))`
Implementation:
Explanation: Here we are sorting the values of the array row-wise as we have provided the value of the axis asย 1.
Example2:
`# Sort the array column-wiseval = np.array([(3, 4, 2, 5),(3, 6, 2, 4),(1, 5, 2, 6)])np.sort(val,axis=0)`
Implementation:
Explanation: Here we are sorting the values of the array column-wise as we have provided the value of the axis asย 0.
#### What to doย next?
The NumPy library in Python is one of the most widely used libraries for numerical computation. We looked at the NumPy library in-depth with the help of various examples in this article. We also demonstrated how to utilize the NumPy library to do several linear algebra operations. I recommend that you practice the examples provided in this post. If you want to work as a data scientist, the NumPy library is one of the tools youโll need to master to be a successful and productive member of the profession. Learnย More.
Introduction to Numpy in Python was originally published in Towards AI on Medium, where people are continuing the conversation by highlighting and responding to this story.
Published via Towards AI | 5,251 | 18,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-06 | longest | en | 0.874305 |
https://www.physicsforums.com/threads/calculating-coulombs-help.16123/ | 1,542,810,209,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039748901.87/warc/CC-MAIN-20181121133036-20181121154708-00030.warc.gz | 962,340,716 | 12,386 | # Homework Help: Calculating Coulombs Help
1. Mar 11, 2004
### LTech221
Hi everyone its my first time here, read the forums greatly though and have been a help in understanding but now im completely lost on this problem: Calculate the number of coulombs of positive charge in 250 cm^3 of (neutral) water (about a glass full). So far I tried converting the volume to m^3 (2.5E-4 m^3, not sure if it matters.)then tried tried dividing by colulomb's constant and ended up with units kg/s^2 * c^2 and dont know where to go from there. Im not sure if I really approached this problem right as the section it supposedly comes from keeps mentioning q=ne, but I cannot relate that directly. Im thinking im missing something inbetween , any ideas to go about the problem would be appreciated.
2. Mar 11, 2004
### Philcorp
if you know the volume of the water then it should be easy enough for you to convert that into an equivalent mass. From there you should know the ratio of moles to grams for water ( i think its something like 18.02g/mol). This allows you to convert you volume from grams into moles. Now you can use Avagadro's number (6.02x10^23) to know the number of molecules in your sample. From there you know that each water molecule is made up of H_2O.....so thats 10 protons. And each proton have 1.6x10^-19 C of charge.
There you have it!
3. Mar 11, 2004
### LTech221
Ahhh, thats what i was missing, thanks for the help | 370 | 1,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-47 | latest | en | 0.925958 |
http://moi3d.com/forum/messages.php?webtag=MOI&msg=4930.11 | 1,638,403,358,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00576.warc.gz | 49,171,653 | 6,453 | Humble script request All 1-10 11-19
From: bemfarmer 15 Feb 2012 (11 of 19)
These are very cool springs. They make great electrical connectors in medical devices, and electronics. Of the many patents, Peter J Balsells is one of the main creators. In one patent, the springs are not just a simple helix of steel which is pushed into a groove, circular or otherwise. During the winding, different angles are created in the steel. Contrary to this, the thesis formula appears to be symetrical in the z axis... For a script, I would make the assumption that the spring runs along a straight line, not a circle, and then use flow. (?) One iteration would make one coil, of 1/w offset. To script along a circle, I think would require another equation, which would rotate each coil around a polygon, (or circle). The polygon would have the same number of sides as number of coils. Or the circle would have to have the correct circumference to accomodate an integer number of coils... Michaels method looks excellent for viewing the circular canted coil. A script would be fairly easy, I think.
From: bemfarmer 15 Feb 2012 (12 of 19)
Here is a simple menu: Attachments:
From: Michael Gibson 15 Feb 2012 (13 of 19)
4930.13 In reply to 4930.12 Hi Brian, does that one match the script above? It looks like they have different names - CantedCoilSpring.js versus CantedToroidalHelix.js ... - Michael
From: bemfarmer 15 Feb 2012 (14 of 19)
4930.14 In reply to 4930.13 Sorry, it is the same htm. I've dropped toroidal thoughts for now. Wrote a canted helix script, but of course it does not work. It sort of worked for a while, but now I've completely broken it. :-) I always get messed up with t = 1/(number of something). And number of coils per unit length.
From: bemfarmer 15 Feb 2012 (15 of 19)
Got the CantedHelix script to work. It was necessary to adjust the formulas. (The use of w had to be modified.) EDITED: 15 Feb 2012 by BEMFARMER Attachments: Image Attachments:
From: bemfarmer 15 Feb 2012 (16 of 19)
4930.16 In reply to 4930.15 Picture: I suppose it could be turned into a toroid, without too many hours of work? Image Attachments: | 578 | 2,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-49 | latest | en | 0.927469 |
https://getrevising.co.uk/revision-cards/similar_triangles | 1,519,198,722,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813571.24/warc/CC-MAIN-20180221063956-20180221083956-00451.warc.gz | 670,376,348 | 12,752 | Similar triangles
how to work out the lengths of sides from a ratio
HideShow resource information
• Created by: Kelly:)
• Created on: 01-06-11 09:23
How to work out a length on a similar triangle
with a question like this, you could get either:
• a large triangle with a smaller triangle inside it
• the tip of a smaller triangle balancing on the tip of the the larger one (or vice versa)
with the first option, you need to split these triangles into two separate trianlges and label the side lengths that you know of.
then you need to work out the ratio: find a side from each triangle that is the same side (but will obviously have different lengths) and divide the larger one by the smaller one to give you your ratio.
to find a length on the smaller triangle, you will need to divide the larger length from the same place by the ratio, eg., if your ratio is 1:2.5, divide the larger length by 2.5.
or if you need to find a longer length then you need to multiply your smaller length by the ratio.
1 of 4
2 of 4
continued..
we know the triangles from the last page are similar because they all have the same angles therefore we can work our their lengths.
ratio = 3:5 because that's the length of the smaller triange against the length of the larger triangle
or ratio = 1:1.6666...
therefore to find 'x' we need to 4/1.6666... = 2.4
3 of 4
Balancing Triangles
these are a little easier because you just need to work out the ratio and then do you division or multiplication to get the other lengths.
4 of 4 | 373 | 1,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-09 | longest | en | 0.924679 |
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# If a centrifuges radius and mass are both doubled without otherwise changing the design
ISBN: 9780321706690 402
## Solution for problem 83 Chapter 10
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Problem 83
If a centrifuges radius and mass are both doubled without otherwise changing the design, its rotational inertia will a. double. b. quadruple. c. increase by a factor of 8. d. increase by a factor of 16.
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##### ISBN: 9780321706690
Since the solution to 83 from 10 chapter was answered, more than 215 students have viewed the full step-by-step answer. This full solution covers the following key subjects: . This expansive textbook survival guide covers 39 chapters, and 2979 solutions. This textbook survival guide was created for the textbook: Essential University Physics: Volume 1, edition: 2. Essential University Physics: Volume 1 was written by and is associated to the ISBN: 9780321706690. The full step-by-step solution to problem: 83 from chapter: 10 was answered by , our top Physics solution expert on 03/14/18, 04:47PM. The answer to “If a centrifuges radius and mass are both doubled without otherwise changing the design, its rotational inertia will a. double. b. quadruple. c. increase by a factor of 8. d. increase by a factor of 16.” is broken down into a number of easy to follow steps, and 36 words.
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Solving time: 3 mins
# In a parallelogram , and . The bisector of meets in . and both when produced meet at . Find length of .
## Text solutionVerified
From the figure we know that is the bisector of
so we get
....(1)
We know that is a parallelogram
from the figure we know that and is a transversal and are alternate angles
....(2)
Consider equation (1) and(2)
.....(3)
We know that the sides opposite to equal angles are equal
and are vertically opposite angles
.....(4)
We know that and is the transversal
it can be written as
we know that the sides opposite to equal angles are equal
we get
by substituting the values
therefore length of is
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### Practice more questions from Secondary School Mathematics For Class 9(RS Aggarwal)
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Question Text In a parallelogram , and . The bisector of meets in . and both when produced meet at . Find length of . Topic Quadrilaterals Subject Mathematics Class Class 9 Answer Type Text solution:1 Upvotes 11 | 377 | 1,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-38 | latest | en | 0.885481 |
https://mathematica.stackexchange.com/questions/89663/funny-behavior-when-computing-dot-product-of-coefficients-with-high-order-polyno | 1,726,615,459,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00049.warc.gz | 335,393,486 | 42,751 | # Funny behavior when computing dot product of coefficients with high-order polynomials
I have a similar problem to Funny behaviour when plotting a polynomial of high degree and large coefficients. However, the thing being evaluated is not just a polynomial, but a dot product of some coefficients with a list of polynomials.
(The back story is that I want to distort a sine random variable so that it has an approximate Laplace distribution. I am doing this by approximating the nonlinear mapping with a Chebyshev approximation. This has the cool property that if I use a maxN-th order Chebyshev series then there are only maxN harmonics of the sine wave. If the exact mapping is used, there are an infinite number of harmonics.)
The dot product (coefficients times Chebyshev polynomials) begins to become inaccurate (blow up, oscillate like crazy, whatever) near $$\pm 1$$ for maxN greater than the mid-$$40$$s. I've tried the Rationalize function on the coefficients--the polynomial coefficients are all integers--but to no avail.
Here is what I've tried. I have included the exact mapping for comparison to the Chebyshev approximation. (Mathematica told me that this involves the SinIntegral so I have hard-coded that into the definition.) The mapping has odd symmetry about the origin so even-order coefficients are zero.
FLaplaceInverse[x_, β_] :=
Piecewise[{{β Log[2 x], x < 1/2}, {-β Log[2 - 2 x],
x >= 1/2}}];
FSine[x_, b_] := (π + 2 ArcSin[x/b])/(2 π);
CompositeSineLaplace[x_] := FLaplaceInverse[FSine[x, 1], 1];
maxN = 49;
chebCoeffList =
Table[If[EvenQ[n], 0,
N[(4 SinIntegral[(n π)/2])/(n π), 100]], {n, 0, maxN}];
rChebCoeffList = Rationalize[chebCoeffList, 0];
chebPolyList =
Table[If[EvenQ[n], 0, ChebyshevT[n, x]], {n, 0, maxN}];
plotBoth =
Plot[{CompositeSineLaplace[x],
rChebCoeffList.chebPolyList} , {x, -1, 1}, PlotRange -> Automatic,
ImageSize -> Medium];
plotDiff =
Plot[Evaluate[
CompositeSineLaplace[x] - chebCoeffList.chebPolyList] , {x, -1,
1}, PlotRange -> Automatic, ImageSize -> Medium];
chebCoeffsWithOrder =
Transpose[{Table[n - 1, {n, Length[chebCoeffList]}], chebCoeffList}];
plotCoeffs =
ListPlot[chebCoeffsWithOrder, PlotRange -> All, ImageSize -> Medium];
{plotBoth, plotDiff, plotCoeffs}
The Plot functions give this for maxN = 49.
• It is nice to use the Chebyshev polynomial expansion fo this purpose. However, you are in effect converting back to the ill-conditioned monomial basis here. Why not use Clenshaw? Also, you are wasting effort by not exploiting the oddness of your function; there are identities that only require you to retain the odd terms. Commented Aug 1, 2015 at 5:05
• For the time being: try cranking up WorkingPrecision. Commented Aug 1, 2015 at 5:05
• Numerical Recipes as well as GNU Scientific Library incorporate Clenshaw's method; I have been using the Numerical Recipes code for a while until I noticed something weird. The NR guys say, compute more Chebyshev coefficients than you need, say M, then just truncate to how many you want, say N < M. The assumption is that the truncated terms will cause a negligible error. However, I found that if I compute M1 then M2 terms and then truncate each time to N terms, the N terms are different for the two cases. I attribute this either to slow convergence to -+infinity at ±1 or my coding error to Ada Commented Aug 1, 2015 at 5:35
• Hmm… if I find time later I might investigate this (unless somebody beats me to it). Re: NR, they gave the identities for evaluating an odd Chebyshev series efficiently IIRC. Use them! Commented Aug 1, 2015 at 5:40
• By suggesting that I increase WorkingPrecision I gather that you mean, add it to the Plot statement. I don't think this will help—anyway, I tried with value of 100 with no change in the plots. Commented Aug 3, 2015 at 9:49
To resolve this:
Compare
Plot[chebCoeffList.chebPolyList, {x, -1, 1}]
with
Plot[chebCoeffList.Table[If[EvenQ[n], 0, ChebyshevT[n, x]], {n, 0, maxN}],
{x, -1, 1}]
The blowup seen in the first plot is, as noted, due to the fact that chebPolyList is already converted automatically to the monomial basis, which is terribly ill-conditioned for numerical evaluations of this sort. In contrast, the evaluation in the second one did not involve this conversion; that is, ChebyshevT[] was directly evaluated at the inexact numbers fed to it by Plot[]. Setting WorkingPrecision to a higher value in the first plot can stave off this ill-conditioning, but at the expense of increased effort.
As also noted in the comments, a more efficient way of proceeding would be to use the Clenshaw recurrence to evaluate your Chebyshev series. This has to be modified a bit to exploit the oddness of your function. Thus,
cheboddval[c_?VectorQ, x_] :=
Module[{n = Length[c], u, v, w, y = 4 x^2 - 2},
u = c[[n - 1]] + y (v = c[[n]]);
Do[w = v; v = u; u = c[[k]] + y v - w, {k, n - 2, 2, -1}];
x (c[[1]] + (y - 1) u - v)]
Let's generate your coefficients without the zeroes taking up space:
cc = Table[N[(4 SinIntegral[n π/2])/(n π), 50], {n, 1, maxN, 2}];
Now, the plot:
Plot[{CompositeSineLaplace[x], cheboddval[cc, x]}, {x, -1, 1}] | 1,443 | 5,122 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.838601 |
http://www.jiskha.com/display.cgi?id=1272410799 | 1,495,490,219,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607120.76/warc/CC-MAIN-20170522211031-20170522231031-00155.warc.gz | 551,082,957 | 3,965 | # calculus
posted by on .
Find the fourier series of
F(x) = 1, -L<=x<0
= 0, 0<=<L
f(x+2L) = f(x)
• calculus - ,
You can take the functions:
e_n(x) = exp(i pi n x/L)
to be your the basis functions. If you define the inner product as:
<f,g> = 1/(2 L) Integral from -L to L
of f(x)g*(x) dx
Then the e_n are orthonormal. So, you can expand a function f(x) as:
f(x) = sum over n of <f,e_n> e_n(x)
The coefficient of e_n is thus the integral from -L to 0 of
exp(-pi i n x/L) dx
If n is not equal to zero, this is:
-1/(2 pi i n) [1 - (-1)^n]
So, for nonzero n, only the coefficients for odd n are nonzero. The coefficient for n = 0 is easily found to be 1/2 . If you now combine the contribution from negative and positive n, you find that the Fourier series is given by:
1/2 - 2/pi sum from k = 0 to infinity of
2/(2k+1) Sin[(2k+1)x/L] | 292 | 845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2017-22 | latest | en | 0.839609 |
https://www.physicsforums.com/threads/carmichael-number.276246/ | 1,519,506,012,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815934.81/warc/CC-MAIN-20180224191934-20180224211934-00113.warc.gz | 938,025,849 | 13,814 | Carmichael number
1. Dec 1, 2008
peteryellow
$n=pqr$ is a Carmichael number. If $q-1|pr-1$ and $r-1|pq-1$ then show that q-1 is a divisor in
$(d+p)(p-1).$
2. Dec 1, 2008
peteryellow
ok here it is the clear version of it.
n=pqr is carmicahelnumber. if we have that r-1|pq-1 then I have shown that
pq-1=d(r-1) where d is [2;p-1]. Moreover I have shown that q-1|d(r-1)-p+1.
Now I want to show that if q-1|pr-1 is also fulfilled then q-1 is divisor in (d+p)(p-1). Do you understand it?
3. Dec 14, 2008
rscosa
Hi!
I suggest you to read more before publish your problems since this is a very well known question (Wikipedia). | 231 | 628 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-09 | longest | en | 0.92536 |
http://forums.wolfram.com/mathgroup/archive/2002/Jul/msg00341.html | 1,720,858,650,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00301.warc.gz | 13,522,779 | 7,981 | Re: Re: Factoring problem
• To: mathgroup at smc.vnet.net
• Subject: [mg35536] Re: [mg35515] Re: [mg35499] Factoring problem
• From: Rob Pratt <rpratt at email.unc.edu>
• Date: Thu, 18 Jul 2002 03:06:24 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```On Wed, 17 Jul 2002, Rob Pratt wrote:
> On Tue, 16 Jul 2002, Steven Hodgen wrote:
>
> > Hello,
> >
> > I've decided to see if any of you can factor this eq. since it's not
> > possible to have Mathematica show intermediate steps when factoring. This
> > is a problem from my precalc book, and the instructor of the class hasn't
> > been able to get to the final answer either. I've tried and tried using
> > grouping in various ways, as well as other techniques.
> >
> > I'd appreciate it if anyone can figure this out, since I just can't get over
> > this problem.
> >
> > Original problem:
> > y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2
> > (y^2 - p*q)*(y - p)*(y - q)
>
> In[1]:= Expand[(y^2 - p*q)*(y - p)*(y - q)]
>
> 2 2 2 2 3 3 4
> Out[1]= -(p q ) + p q y + p q y - p y - q y + y
>
> So the proposed factorization checks out.
>
> How could you have discovered the factorization? Pretend p and q are
> prime numbers. Then the rational roots theorem would suggest that you try
> the divisors of p^2 q^2 as possible roots. Substitute each of
>
> 1, p, q, p q, p^2 q, p q^2, and p^2 q^2
>
> for y and you will find that both p and q make the expression vanish, so
> y - p and y - q are factors. Dividing these factors out yields y^2 - p q.
>
> Rob Pratt
> Department of Operations Research
> The University of North Carolina at Chapel Hill
>
> rpratt at email.unc.edu
>
> http://www.unc.edu/~rpratt/
I inadvertently omitted p^2 and q^2 from the list of divisors of p^2 q^2
(there should be (2 + 1)*(2 + 1) = 9 total), but these two aren't roots
anyway.
Rob Pratt
Department of Operations Research
The University of North Carolina at Chapel Hill
rpratt at email.unc.edu
http://www.unc.edu/~rpratt/
```
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https://www.zbmath.org/?q=an%3A1330.35303 | 1,618,153,912,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00130.warc.gz | 1,208,098,745 | 12,296 | zbMATH — the first resource for mathematics
Anomalous dissipation for $$1/5$$-Hölder Euler flows. (English) Zbl 1330.35303
The authors consider the incompressible Euler equations, and study the celebrated Onsager conjecture. The conjecture states that (i) if the solution has a Hölder exponent $$\theta$$ satisfying $$\theta >1/3$$, then the energy integral of the solution is conserved, and (ii) if it satisfies $$\theta >1/3$$, then the energy is not necessarily conserved. The authors study the latter statement, and discuss about the Hölder exponent. They consider the case $$\theta <1/5$$ on a three-dimensional torus $$\mathbf{T}\times \mathbf{T}\times\mathbf{T}$$, which means to consider periodic soluitons. They prove that for any positive smooth function $$e(t)$$ there exists a solution of the incompressible Euler equations whose energy integral is equal to this function. This means that the Onsager conjecture for this case is true.
They use iteration for $$q=1,2,3,\cdots$$ At each step $$q$$, they construct a triple $$(v_q ,p_q ,R_q )$$ consisting of velocity $$v_q$$, pressure $$p_q$$, and the Reynolds stress $$R_q$$ which satisfy the following Euler Reynolds system: $\partial_t v_q +\roman{div}\!\;(v_q \otimes v_q )+\nabla p_q =\roman{div}\!\;R_q ,\;\roman{div}\!\;v_q =0 .$ The iteration proceeds so that $$(v_q ,p_q , R_q )\rightarrow (v,p,0)$$ in the Hölder space when $$q\rightarrow \infty$$, and $$\int|v_q |^2 dx$$ converges to a given function $$e(t)$$. Technical parts are collected in the appendix, which makes this article more accessible.
MSC:
35Q31 Euler equations 76B03 Existence, uniqueness, and regularity theory for incompressible inviscid fluids
Full Text:
References:
[1] T. Buckmaster, ”Onsager’s conjecture almost everywhere in time,” Comm. Math. Phys., vol. 333, p. 24, 2015. · Zbl 1308.35184 · doi:10.1007/s00220-014-2262-z · arxiv:1304.1049 [2] A. Cheskidov, P. Constantin, S. Friedlander, and R. Shvydkoy, ”Energy conservation and Onsager’s conjecture for the Euler equations,” Nonlinearity, vol. 21, iss. 6, pp. 1233-1252, 2008. · Zbl 1138.76020 · doi:10.1088/0951-7715/21/6/005 · arxiv:0704.0759 [3] P. Constantin, W. E, and E. S. Titi, ”Onsager’s conjecture on the energy conservation for solutions of Euler’s equation,” Comm. Math. Phys., vol. 165, iss. 1, pp. 207-209, 1994. · Zbl 0818.35085 · doi:10.1007/BF02099744 [4] S. Conti, C. De Lellis, and L. relax Székelyhidi Jr., ”$$h$$-principle and rigidity for $$C^{1,\alpha}$$ isometric embeddings,” in Nonlinear Partial Differential Equations, New York: Springer-Verlag, 2012, vol. 7, pp. 83-116. · Zbl 1255.53038 · doi:10.1007/978-3-642-25361-4_5 [5] R. Courant, K. Friedrichs, and H. Lewy, ”On the partial difference equations of mathematical physics,” IBM J. Res. Develop., vol. 11, pp. 215-234, 1967. · Zbl 0145.40402 · doi:10.1147/rd.112.0215 [6] S. Daneri, ”Cauchy problem for dissipative Hölder solutions to the incompressible Euler equations,” Comm. Math. Phys., vol. 329, iss. 2, p. 745, 2014. · Zbl 1298.35140 · doi:10.1007/s00220-014-1973-5 [7] C. De Lellis and L. Székelyhidi Jr., ”The Euler equations as a differential inclusion,” Ann. of Math., vol. 170, iss. 3, pp. 1417-1436, 2009. · Zbl 1350.35146 · doi:10.4007/annals.2009.170.1417 · annals.math.princeton.edu [8] C. De Lellis and L. Székelyhidi Jr., ”On admissibility criteria for weak solutions of the Euler equations,” Arch. Ration. Mech. Anal., vol. 195, iss. 1, pp. 225-260, 2010. · Zbl 1192.35138 · doi:10.1007/s00205-008-0201-x · arxiv:0712.3288 [9] C. De Lellis and L. Székelyhidi Jr., Dissipative Euler flows and Onsager’s conjecture. · Zbl 1307.35205 · doi:10.4171/JEMS/466 [10] C. De Lellis and L. Székelyhidi Jr., ”The $$h$$-principle and the equations of fluid dynamics,” Bull. Amer. Math. Soc., vol. 49, iss. 3, pp. 347-375, 2012. · Zbl 1254.35180 · doi:10.1090/S0273-0979-2012-01376-9 [11] C. De Lellis and L. Székelyhidi Jr., ”Dissipative continuous Euler flows,” Invent. Math., vol. 193, iss. 2, pp. 377-407, 2013. · Zbl 1280.35103 · doi:10.1007/s00222-012-0429-9 [12] J. Duchon and R. Robert, ”Inertial energy dissipation for weak solutions of incompressible Euler and Navier-Stokes equations,” Nonlinearity, vol. 13, iss. 1, pp. 249-255, 2000. · Zbl 1009.35062 · doi:10.1088/0951-7715/13/1/312 [13] G. L. Eyink, ”Energy dissipation without viscosity in ideal hydrodynamics. I. Fourier analysis and local energy transfer,” Phys. D, vol. 78, iss. 3-4, pp. 222-240, 1994. · Zbl 0817.76011 · doi:10.1016/0167-2789(94)90117-1 [14] G. L. Eyink and K. R. Sreenivasan, ”Onsager and the theory of hydrodynamic turbulence,” Rev. Modern Phys., vol. 78, iss. 1, pp. 87-135, 2006. · Zbl 1205.01032 · doi:10.1103/RevModPhys.78.87 [15] U. Frisch, Turbulence, Cambridge: Cambridge Univ. Press, 1995. · Zbl 0832.76001 [16] D. Gilbarg and N. S. Trudinger, Elliptic Partial Differential Equations of Second Order, New York: Springer-Verlag, 2001. · Zbl 1042.35002 [17] M. Gromov, Partial Differential Relations, New York: Springer-Verlag, 1986, vol. 9. · Zbl 0651.53001 · doi:10.1007/978-3-662-02267-2 [18] P. Isett, Hölder continuous Euler flows in three dimensions with compact support in time, 2012. · Zbl 1367.35001 [19] A. N. Kolmogorov, ”The local structure of turbulence in incompressible viscous fluid for very large Reynolds numbers,” Proc. Roy. Soc. London Ser. A, vol. 434, iss. 1890, pp. 9-13, 1991. · Zbl 1142.76389 · doi:10.1098/rspa.1991.0075 [20] J. Nash, ”$$C^1$$ isometric imbeddings,” Ann. of Math., vol. 60, pp. 383-396, 1954. · Zbl 0058.37703 · doi:10.2307/1969840 [21] L. Onsager, ”Statistical hydrodynamics,” Nuovo Cimento, vol. 6, iss. Supplemento, 2(Convegno Internazionale di Meccanica Statistica), pp. 279-287, 1949. [22] R. Robert, ”Statistical hydrodynamics (Onsager revisited),” in Handbook of Mathematical Fluid Dynamics, Vol. II, Amsterdam: North-Holland, 2003, pp. 1-54. · Zbl 1141.76330 · doi:10.1016/S1874-5792(03)80003-4 [23] V. Scheffer, ”An inviscid flow with compact support in space-time,” J. Geom. Anal., vol. 3, iss. 4, pp. 343-401, 1993. · Zbl 0836.76017 · doi:10.1007/BF02921318 [24] A. Shnirelman, ”Weak solutions with decreasing energy of incompressible Euler equations,” Comm. Math. Phys., vol. 210, iss. 3, pp. 541-603, 2000. · Zbl 1011.35107 · doi:10.1007/s002200050791
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2,328 | 6,663 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-17 | latest | en | 0.716898 |
https://www.unitsconverters.com/en/Tdb-To-Fdb/Utu-8774-8777 | 1,720,805,698,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00156.warc.gz | 818,870,362 | 36,627 | Formula Used
1 Decibel = 1E+15 Femtodecibel
## TdB to fdB Conversion
The abbreviation for TdB and fdB is teradecibel and femtodecibel respectively. 1 TdB is 1E+27 times bigger than a fdB. To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including TdB to fdB conversion.
Check our Teradecibel to fdB converter and click on formula to get the conversion factor. When you are converting sound from Teradecibel to fdB, you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
## TdB to Femtodecibel
The formula used to convert TdB to Femtodecibel is 1 Teradecibel = 1E+27 Femtodecibel. Measurement is one of the most fundamental concepts. Note that we have Bel as the biggest unit for length while Millidecibel is the smallest one.
## Convert TdB to fdB
How to convert TdB to fdB? Now you can do TdB to fdB conversion with the help of this tool. In the length measurement, first choose TdB from the left dropdown and fdB from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from fdB to TdB? You can check our fdB to TdB converter.
## TdB to fdB Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like sound finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like TdB to fdB through multiplicative conversion factors. When you are converting sound, you need a Teradecibel to Femtodecibel converter that is elaborate and still easy to use. Converting TdB to Femtodecibel is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Teradecibel to fdB, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in TdB to fdB conversion along with a table representing the entire conversion.
Let Others Know | 556 | 2,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.862888 |
https://mersenneforum.org/showpost.php?s=d41c483345230064a5f03f8e3b93928a&p=561260&postcount=8 | 1,619,134,425,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00419.warc.gz | 481,231,559 | 4,477 | View Single Post
2020-10-27, 17:45 #8 petrw1 1976 Toyota Corona years forever! "Wayne" Nov 2006 Saskatchewan, Canada 461110 Posts I can follow the simplification that gets one to 2 x SQRT(7!+1) = 142 Can someone explain how you would start at 2 x SQRT(7!+1) = 142 and given that the final equation must contain only 7's, get to the original equation I supplied in post 1. For example the 2 at the front could become something like (7+7)/7 OR + iff that function can be divided out in another part of the entire formula to equal 2. ....confused??? Me too. (My son has a game/puzzle app when he must find a formula using the least number of each digit from 1 to 9 to get a number) He is limited to +. -, x, /, SQRT and concatenation. ie 77 or 777 | 219 | 749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-17 | latest | en | 0.908179 |
https://phptpoint.com/convert-a-given-binary-tree-to-doubly-linked-list/ | 1,638,372,919,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.6/warc/CC-MAIN-20211201143545-20211201173545-00324.warc.gz | 537,707,113 | 15,006 | Home >>Cpp Programs >Convert a given binary Tree to Doubly Linked List
# Convert a given binary Tree to Doubly Linked List
### Convert a given binary Tree to Doubly Linked List
In this example, we will see a C++ program through which we can convert a given binary tree to a doubly linked list.
Algorithm:
• STEP 1: We start to convert the tree node to DLL from the rightmost tree node to the leftmost tree node.
• STEP 2: Every time we connect the right pointer of a node to the head of the DLL.
• STEP 3: Connect the left pointer of the DLL to that node.
• STEP 4: Make that node to the head of the linked list.
• STEP 5: Repeat the process from right to left most node of the tree.
##### Example
``````
#include <bits/stdc++.h>
using namespace std;
struct node {
int data;
node* left;
node* right;
};
//Create a new node
struct node* create_node(int x)
{
struct node* temp = new node;
temp->data = x;
temp->left = NULL;
temp->right = NULL;
return temp;
}
//convert a BST to a DLL
void BinarytoDll(node* root, node** head)
{
if (root == NULL)
return;
BinarytoDll(root->right, head);
root->right = *head;
if (*head != NULL) {
(*head)->left = root;
}
*head = root;
BinarytoDll(root->left, head);
}
//Print the list
void print(node* head)
{
struct node* temp = head;
while (temp) {
cout << temp->data << " ";
temp = temp->right;
}
}
//print the tree in inorder traversal
void print_tree(node* root)
{
if (root == NULL) {
return;
}
print_tree(root->left);
cout << root->data << " ";
print_tree(root->right);
}
int main()
{
struct node* root = create_node(5);
root->left = create_node(6);
root->right = create_node(7);
root->left->left = create_node(8);
root->left->right = create_node(1);
root->right->right = create_node(0);
cout << "Print Tree" << endl;
print_tree(root);
struct node* head = NULL;
BinarytoDll(root, &head);
cout << "\nDoubly Linked List" << endl;
print(head);
return 0;
}
```
```
Output:
Print Tree
8 6 1 5 7 0
Doubly Linked List
8 6 1 5 7 0 | 559 | 1,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-49 | latest | en | 0.537996 |
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quiz9_solutions
# quiz9_solutions - process if the resulting products are at...
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ECH 3023 Quiz 9 Fall 2006 1. (5 pts) 2 mol of solid NaOH(c) at 25 o C is mixed with 100 mol water at 25 o C. Calculate the heat removed to keep the resulting solution at 25 o C. Table B.11 is attached. ) 50 , 25 ( ˆ = = r H n Q o s = 2(-42.51) = -85.02 kJ 2. (5 pts) A mixture of 1 mol O 2 and 3 mol H 2 at 25 o C reacts at constant pressure (1 atm) to form H 2 O vapor. Calculate the amount of heat removed (or added) in this
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Unformatted text preview: process if the resulting products are at 1000 o C. Table B.8 is attached. Reaction: ½O 2 + H 2 ↔ H 2 O(v) = ∆ o r H ˆ-241.82 kJ/mol. Products: 2 mol H 2 0, 1 mol H 2 o r o H o H H C H C H Q ˆ 2 ) 1000 ( ' ˆ ) 2 ( ) 1000 ( ' ˆ ) 1 ( 20 2 ∆ +--+ = = 29.04 + 2(37.67)+2(-241.82) = - 397.26 kJ...
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# BALANCE AND MOTION UNIT
Product Description
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\$6.00
\$6.00 | 586 | 2,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-30 | latest | en | 0.919672 |
https://www.numwords.com/words-to-number/en/2773 | 1,560,725,828,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998325.55/warc/CC-MAIN-20190616222856-20190617004856-00557.warc.gz | 836,916,323 | 1,859 | NumWords.com
# How to write Two thousand seven hundred seventy-three in numbers in English?
We can write Two thousand seven hundred seventy-three equal to 2773 in numbers in English
< Two thousand seven hundred seventy-two :||: Two thousand seven hundred seventy-four >
Five thousand five hundred forty-six = 5546 = 2773 × 2
Eight thousand three hundred nineteen = 8319 = 2773 × 3
Eleven thousand ninety-two = 11092 = 2773 × 4
Thirteen thousand eight hundred sixty-five = 13865 = 2773 × 5
Sixteen thousand six hundred thirty-eight = 16638 = 2773 × 6
Nineteen thousand four hundred eleven = 19411 = 2773 × 7
Twenty-two thousand one hundred eighty-four = 22184 = 2773 × 8
Twenty-four thousand nine hundred fifty-seven = 24957 = 2773 × 9
Twenty-seven thousand seven hundred thirty = 27730 = 2773 × 10
Thirty thousand five hundred three = 30503 = 2773 × 11
Thirty-three thousand two hundred seventy-six = 33276 = 2773 × 12
Thirty-six thousand forty-nine = 36049 = 2773 × 13
Thirty-eight thousand eight hundred twenty-two = 38822 = 2773 × 14
Forty-one thousand five hundred ninety-five = 41595 = 2773 × 15
Forty-four thousand three hundred sixty-eight = 44368 = 2773 × 16
Forty-seven thousand one hundred forty-one = 47141 = 2773 × 17
Forty-nine thousand nine hundred fourteen = 49914 = 2773 × 18
Fifty-two thousand six hundred eighty-seven = 52687 = 2773 × 19
Fifty-five thousand four hundred sixty = 55460 = 2773 × 20
Sitemap | 414 | 1,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-26 | latest | en | 0.796753 |
https://mixedmediadirectory.com/126268384/ascending-order-homework/ | 1,606,413,324,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00388.warc.gz | 396,964,609 | 8,154 | # Ascending order homework
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# No homework policy executive order
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## My homework lesson 4 order numbers page 33
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#### Unit 1 algebra basics homework 4 order of operations all things algebra
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## My homework lesson 4 order numbers answers
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### Homework 10-1 order of operations
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# Ascending order homework
## No homework policy executive order
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### My homework lesson 4 order numbers page 33
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### Teacher order homework
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## Homework 4 order of operations answers
Worksheet for a math questions for the correct answer. While solving problems with answer as a graphic preview for the pre-test worksheet,. Intro: 3 gives 7.7 7and 7 12 5: evaluate numerical expressions. My dear aunt sally stands for the proper order of numbers, what is called the last problem. To use in the button titled create free algebra 1 algebra 1. May 13 4 - note: add and exponents, place. These free printable worksheets to find help your answer, place value. Learn to the previous night and the order of basic operations need to 1 algebra - see order of operations. | 2,601 | 11,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-50 | latest | en | 0.87287 |
https://www.sanfoundry.com/wireless-mobile-communications-questions-answers-diffraction/ | 1,718,711,537,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00864.warc.gz | 862,696,366 | 21,389 | # Wireless & Mobile Communications Questions & Answers – Diffraction
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers (MCQs) focuses on “Diffraction”.
1. Diffraction occurs when radio path between Tx. And Rx. Is obstructed by ____________
a) Surface having sharp irregularities
b) Smooth irregularities
c) Rough surface
d) All types of surfaces
Explanation: Diffraction occurs when radio path between transmitter and receiver is obstructed by a surface that has sharp irregularities (edges). The secondary waves resulting from the obstructing surface are present throughout the space and even behind the obstacle.
2. At high frequencies, diffraction does not depends on ___________
a) Geometry of the object
b) Distance between Tx and Rx
c) Amplitude of incident wave
d) Polarization of incident wave
Explanation: At high frequency, diffraction depends on the geometry of the object, as well as the amplitude, phase, and polarization of the incident wave at the point of diffraction. It gives rise to a bending of waves even when line of sight does not exist between transmitter and receiver.
3. Diffraction allows radio signals to propagate around ________
a) Continuous surface
b) Smooth surface
c) Curved surface of Earth
d) Does not allow propagation
Explanation: Diffraction allows radio signals to propagate around the curved surface of the Earth. Signals can propagate beyond the horizon and to propagate behind obstruction. It is the slight bending of light as it passes around the edge of an object.
4. Which principle explains the phenomenon of diffraction?
a) Principle of Simultaneity
b) Pascal’s Principle
c) Archimedes’ Principle
d) Huygen’s principle
Explanation: The phenomenon of diffraction can be explained by Huygen’s principle. It states that all points on a wavefront can be considered as point sources for the production of secondary wavelets. And these wavelets combine to produce a new wavefront in direction of propagation.
5. Diffraction is caused by propagation of secondary wavelets into _______
a) Bright region
c) Smooth region
d) Large region
Explanation: Diffraction is caused due to propagation of secondary wavelets into a shadowed region. The field strength in the shadowed region is the vector sum of the electric field components of all the secondary wavelets in the space around the obstacle.
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6. Difference between the direct path and the diffracted path is called _______
a) Average length
c) Excess path length
d) Wavelength
Explanation: Excess path length denoted by ∆, is the difference between the direct path and the diffracted path. It is calculated with the help of Fresnel zone geometry.
7. The phase difference between a direct line of sight path and diffracted path is function of _______
a) Height and position of obstruction
b) Only height
c) Operating frequency
d) Polarization
Explanation: The phase difference between a direct line of sight path and diffracted path is a function of height and position of the diffraction. It is also a function of transmitter and receiver location.
8. Which of the following explains the concept of diffraction loss?
a) Principle of Simultaneity
b) Pascal’s Principle
c) Fresnel zone
d) Archimedes’ Principle
Explanation: The concept of diffraction loss is a function of the path difference around an obstruction. It can be explained by Fresnel zones. Fresnel zones represent successive regions where secondary waves have a path length from Tx to Rx which are nλ/2 greater than total path length.
9. In mobile communication system, diffraction loss occurs due to ______
a) Dielectric medium
b) Obstruction
c) Electric field
d) Operating frequency
Explanation: Diffraction loss occurs from the blockage of secondary waves such that only a portion of the energy is diffracted around an obstacle. An obstruction causes a blockage of energy from source some of the Fresnel zones, allowing only some of the transmitted energy to reach the receiver.
10. For predicting the field strength in a given service area, it is essential to estimate ______
a) Polarization
b) Magnetic field
c) Height of transmitter
d) Signal attenuation
Explanation: Estimating the signal attenuation caused by diffraction of radio waves over hills and buildings is essential in predicting the field strength in a given service area. In practice, prediction is a process of theoretical approximation modified by necessary empirical corrections.
Sanfoundry Global Education & Learning Series – Wireless & Mobile Communications.
To practice all areas of Wireless & Mobile Communications, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 1,015 | 4,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-26 | latest | en | 0.910126 |
https://www.clutchprep.com/physics/practice-problems/138367/chameleons-catch-insects-with-their-tongues-which-they-can-rapidly-extend-to-gre | 1,610,724,688,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00472.warc.gz | 792,166,786 | 29,290 | Kinematics Equations Video Lessons
Concept
Problem: Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 280 m/s2 for 20 ms, then travels at constant speed for another 30 ms.During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?Express your answer to two significant figures and include the appropriate units.
FREE Expert Solution
Using the kinematic equation:
$\overline{){\mathbf{\Delta }}{\mathbf{x}}{\mathbf{=}}{{\mathbf{v}}}_{{\mathbf{0}}}{\mathbf{t}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{a}}{{\mathbf{t}}}^{{\mathbf{2}}}}$
In the first 20ms: t = 20/1000 = 0.02 s, and v0 = 0 m/s, a = 280 m/s2
94% (231 ratings)
Problem Details
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 280 m/s2 for 20 ms, then travels at constant speed for another 30 ms.
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Express your answer to two significant figures and include the appropriate units. | 340 | 1,194 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-04 | latest | en | 0.824465 |
https://www.khanacademy.org/math/6th-grade-illustrative-math/unit-1-area-and-surface-area/extra-practice-surface-area/a/surface-area-versus-volume | 1,713,277,250,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00205.warc.gz | 764,876,855 | 117,886 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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### Course: 6th grade (Illustrative Mathematics)>Unit 1
Lesson 11: Extra practice: Surface area
# Surface area versus volume
A 3D figure has both surface area and volume measurements, but we use them for different purposes. Learn the difference and when to use each.
## Making sense of units
We have many types of units. Some measure length in $1$ dimension. Some measure area in $2$ dimensions. Others measure volume in $3$ dimensions. Units come in larger and smaller sizes, too.
1.1
For each quantity, choose the most appropriate unit of the choices.
## Key measurement terms
Length is a $1$-dimensional measurement. It tells us the number of units between one point and another. We measure length in units like centimeters, inches, feet, meters, kilometers, and miles.
• Perimeter is a special example of length. It is the distance around a closed 2D figure.
Area is a $2$-dimensional measurement. It tells us the amount of space enclosed in a 2D figure. We measure area in square units such as square centimeters (${\text{cm}}^{2}$), square inches, and square meters.
• Surface area is a special example of area. It tells us the number of square units it would take to cover the faces of the 3D figure.
Volume is a $3$-dimensional measurement. It tells us the number of cubic units it would take to fill a 3D figure. We measure volume in units like cubic centimeters $\left({\text{cm}}^{3}$), cubic inches, and cubic meters. For liquids, we sometimes use different volume units, such as milliliters, cups, liters, and gallons.
Notice, this means that we can measure both the surface area and the volume of a 3D figure, but they tell us different things about the figure.
## Distinguishing area and volume
Let's consider the same situations from before, this time to decide whether which type of measurement makes the most sense.
2.1
Decide whether we would use area, volume, or neither in each context.
ContextMeasurement
Amount of water in the ocean
Distance around a bathroom
Space covered by a stamp
Room inside a closet
Space painted on a table
The same 3D figure can have both surface area and volume.
Let's contrast the volume and surface area of two figures.
3.1
Answer $2$ questions about the following figure.
1. What is the volume of the figure?
cubic units
1. What is the surface area of the figure?
square units
3.2
Answer $2$ questions about the following figure.
1. What is the volume of the figure?
cubic units
1. What is the surface area of the figure?
square units
So the figures have the same volume, but different surface areas!
The opposite is possible, too. Two figures could have the same surface area, but different volumes.
## Try it out!
4.1
The following figure shows a right rectangular prism.
What is the volume of the prism?
${\text{cm}}^{3}$
## Want to join the conversation?
• why the Surface area is 32
• idk but i also got 32 on the first try by adding the 4 middle parts
• What the heck is this its so hard🥲
• ikr why do we need to learn this
• I was kind of confused after reading and doing the practice, so I had to redo it three times over. Worth it though, compared to other curriculums I've done. Is there a way I can learn to understand surface area versus volume better?
• this dont make sence
• if you pay attention it does
• why is it LITTARLY so hard to do and even understand?
• You should ask your teacher for more help if you don't understand
• i dont understand how to find the surface area of the 3d shape with a hole in it
• look at the whole and notice that it has 4 faces surrounding it. Find the surface area like you normally would then add the 4 faces to the equation. The equation would look something like this: A=8+8+3+3+3+3+1+1+1+1.
• If anyone can tell me which question dose not make sense to you I am glad to help! | 976 | 4,018 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 32, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-18 | latest | en | 0.912906 |
https://cs.stackexchange.com/questions/115093/another-vertex-cover-question | 1,721,857,226,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00602.warc.gz | 159,011,568 | 40,519 | # Another vertex cover question?
I'm not sure this is equivalent to bipartite vertex cover question. The question is:
Given a BIPARTITE graph, what is the minimum number of vertex from the right side whose edges cover all vertex from the left side.
e.g. In the following graph, the answer is 1, cause vertex g has connection to all vertex on the left.
It is equivalent to the set cover problem. You can regard each vertex in the right side as a set of its neighbors in the left side. In your example, $$e,f,g$$ correspond respectively to the sets $$\{d\},\{d\},\{a,b,c,d\}$$. Now a minimum vertex cover in your problem is equivalent to a minimum set cover. | 160 | 660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.924402 |
https://au.mathworks.com/matlabcentral/cody/problems/408-back-to-basics-18-justification/solutions/387418 | 1,603,983,670,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904287.88/warc/CC-MAIN-20201029124628-20201029154628-00310.warc.gz | 219,043,025 | 16,954 | Cody
# Problem 408. Back to basics 18 - justification
Solution 387418
Submitted on 17 Jan 2014 by priyanka shrivastava
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = ' MATLAB'; y_correct = ' MATLAB '; assert(isequal(justify(x),y_correct))
y = MATLAB
2 Pass
%% x = 'MATLAB '; y_correct = ' MATLAB '; assert(isequal(justify(x),y_correct))
y = MATLAB
3 Pass
%% x = ' MATLAB '; y_correct = ' MATLAB '; assert(isequal(justify(x),y_correct))
y = MATLAB
4 Pass
%% x = ' MATLAB '; y_correct = ' MATLAB '; assert(isequal(justify(x),y_correct))
y = MATLAB
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 225 | 817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-45 | latest | en | 0.45029 |
https://www.physicsforums.com/threads/is-there-a-better-way-to-integrate-this.287560/ | 1,624,229,827,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00348.warc.gz | 848,101,117 | 14,297 | # Is there a better way to integrate this
what is the best way to integrate this functio, i integrated in parts
f(x)=x2/2x+3
i split the top of the fraction and then integrated the x2 half using integration in parts, but its just long and lots of place to make mistakes.
I would try to rid of the top bit. Here is a hint $$x^2=\frac{1}{4}((2x-3)(2x+3) + 9)$$.
Good luck
jambaugh
Gold Member
what is the best way to integrate this functio, i integrated in parts
f(x)=x2/2x+3
i split the top of the fraction and then integrated the x2 half using integration in parts, but its just long and lots of place to make mistakes.
I assume you mean:
$$\frac{x^2}{2x+3}$$
Note the degree of the numerator is higher than that of the denominator. (Improper rational function like an improper fraction.)
You should first carryout long division of the polynomials (and check by re-multiplying).
You will then get a polynomial plus a proper fraction. Since the denom. has degree 1 the numerator (remainder) will be constant and you can integrate via simple variable substitution.
Carry out the long division:
$$2x+3)\overline{x^2 + 0x + 0}$$
You should get:
$$P(x)+ \frac{C}{2x+3}$$
where P is a polynomial
You can easily integrate:
$$\int P(x)dx + C/2 \int \frac{du}{u};\quad u = 2x+3$$ | 367 | 1,283 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-25 | latest | en | 0.839906 |
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A
Levi Strauss v. Tesco and E.U. Trademark Exhaustion: What, in your opinion, would be different now if the UK had been an EU member? What would the decision be in this scenario, and how may Tesco be impacted?.
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### Levi Strauss v. Tesco and E.U. Trademark ExhaustionWhat, in your .pdf
• 1. Levi Strauss v. Tesco and E.U. Trademark Exhaustion: What, in your opinion, would be different now if the UK had been an EU member? What would the decision be in this scenario, and how may Tesco be impacted?
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Deutsche | 2,188 | 7,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-50 | latest | en | 0.586406 |
https://discourse.codecombat.com/t/decoy-drill-help/2085 | 1,722,745,095,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00693.warc.gz | 162,813,561 | 10,489 | # Decoy Drill: help
I have been trying to beat decoy drill. Code so far:
``````// We are field testing a new battle unit: the decoy.
// Build 4 decoys, then report the total to Naria.
// Each decoy costs 25 gold. Use the Quartz Sense Stone
// to know when you have more than 25 gold with
// Keep a count of decoys you built as you go along.
// Break out of the loop when you have built 4.
var decoysBuilt = 0;
var gold = 0;
loop {
var item = this.findNearestItem();
this.moveXY(item.pos.x, item.pos.y);
gold = gold + item.value;
if (gold >= 25) {
this.buildXY("decoy", this.pos.x, this.pos.y);
decoysBuilt = decoysBuilt + 1;
}
else if (decoysBuilt >= 4) {
break;
}
}
this.say("Done building decoys!");
// Go to Naria and say how many decoys you built.
this.moveXY(14, 36);
this.say("4 decoys");
``````
The problem is that once I get 25 gold, my hero builds a decoy, but then he stops doing anything. I have also tried using an if statement to collect coins, but the same thing happened with that. I assume this has something to do with my code, but I guess it could also be a bug. Please help.
Curious, does item.value still work on this level?
I used:
``````gold = this.gold;
``````
Let me comment out my code and try what you have…
Your code works after switching to:
``````gold = this.gold;
``````
Also you’ll want to move to the “X” before speaking that you are done building decoys. Remember that you will want to also let her know HOW MANY decoys you built.
Give that a shot! =D
EDIT: So this can be a bug as it accepted the `item.value` (I know this was used in a previous level…but it doesn’t seem to work here.)
If you track gold manually, you 1) aren’t decrementing your `gold` variable by 25 when you spend it and 2) can’t keep track of everything you pick up, since you only know what coins you are going for, not any other coins you pick up in the process. That’s why the quartz sense stone, with access to `this.gold`, comes in handy.
2 Likes
But how to get the decoy value Oo
Decoy value? ???
If you build a decoy you should increment a counter for decoys built by 1. You manually keep track of the decoys. A decoy is a decoy…it’s a 1:1 value.
@tjchan
Yeah i’ve already noticed that , btw ty
A solution I used:
As the level doesn’t specify you have to build 1 decoy at every 25 interval (collect all gold first)
Then build 4 decoys (probably should of looped this )
Go to marker and say 4.
On an unrelated matter game doesn’t point out if your hammer cannot produce decoys.
``````var item = this.findNearestItem();
while (this.gold < 100)
{
item = this.findNearestItem();
this.moveXY(item.pos.x, item.pos.y);
}
this.buildXY("decoy", 20, 40);
this.buildXY("decoy", 25, 40);
this.buildXY("decoy", 30, 40);
this.buildXY("decoy", 35, 40);
this.say("Done building decoys!");
// Go to Naria and say how many decoys you built.
this.moveXY(14, 36);
this.say("4");
``````
what do you mean by a “1:1” value ?
We know that we could do this , but that would be a very noob code , as you can see your code has more than 15 lines and only works for 4 decoys , but our codes can be working for unlimited decoys building , you’ll have to change your way of thinking , the lvl wanted to teach you how to use “break” and how to add a value to another variables wich got another value
ex : decoy = 0
after adding 1 each time you build a decoy it would be 4
now this.say (decoy)
sorry if i’m being rude ,not trying to be looking like a pro or something just wanted you to c the difference between codes and coders
tip : use variables instead of numbers
If you build a decoy it counts as a single decoy unlike coins which depending on the type like Gold, Silver, and Copper, could be 3, 2, and 1 respectively (can’t remember true value of coins).
So collecting 3 gold and 2 copper would mean a total value of (3*3+2)=11
i can’t seem to get this level right, can someone help me? this is my cde…
``````decoysBuilt = 0
totalGold = 0
loop:
coin = self.findNearestItem()
if coin:
position = coin.pos
x = position.x
y = position.y
self.moveXY(x, y)
totalGold = self.gold + coin.value
if totalGold = 25:
self.buildXY("decoy", 36, 30)
decoysBuilt + 1
else:
if decoysBuilt = 4:
break
self.moveXY(14, 36)
self.say("Done building decoys!")
self.say("4 decoys built!")
``````
the part that I was stuck on may also be what’s hanging HAMINPANTS up - the check for number of decoys built requires a double equals sign (my non-noob big brother had to show me that - might be a useful thing to include somewhere in this lesson!)
so my code - which works - looks like:
``````#Break out of the loop when you have built 4.
decoysBuilt = 0
loop:
item = self.findNearestItem()
if self.gold < 25:
self.moveXY(item.pos.x, item.pos.y)
if self.gold >= 25:
self.buildXY("decoy", self.pos.x+1, self.pos.y)
decoysBuilt = decoysBuilt+1
self.say("thats")
self.say(decoysBuilt)
if decoysBuilt == 4:
break
self.say("Done building decoys!")
``````
etc
Refinig,
I couldn’t agree more my code is noob :), However I was trying to illustrate that it separation of gold retrieval and creation of decoys. As I believe highlighting this to OP would give him a good opportunity to tackle the problem again.
I don’t claim to be a pro coder either, but find if I’m unable to get through a hurdle, I may be able to work around.
The level specifies the requirements quite clearly unlike most RL scenarios and whilst I would normally use a variable for decoys as we already know how many we need we can just say the number hehe. Looping through a variable would add be adding additional unneeded complexity. I’m probably taking the KISS principle too seriously. Once you have a working example you can always refactor. I have noticed later levels seem to be missing the “Maximum Number of Code Statements challenges”. Hope they add them back in.
I am completely new to this. But I can’t get it to count the coin value right. At least I’m pretty sure that is where it’s hanging up at. Mainly because it doesn’t reach the amount intended and just stops since it doesn’t have enough to build a decoy. This is what I have so far. Any advice would be appreciated. Especially on how to pull the value of the coin.
``````# Declaring variables.
decoysBuilt = 0
# Setting location for decoy x and y
dx = 23
dy = 23
loop:
# Connecting variable to object.
item = self.findNearestItem()
# Collecting coins
while self.gold < 25:
itemPos = item.pos # Dot seperator
gx = itemPos.x # Assign gx to x value of coin
gy = itemPos.y # Assign gy to y value of coin
self.moveXY(gx, gy)
break
# Build a decoy
self.buildXY("decoy", dx, dy) # This is where it hangs up at!!!
dx = dx + 5
dy = dy + 5
decoysBuilt = decoysBuilt + 1
# Test if there are enough decoys
if decoysBuilt == 4:
self.say("Done building decoys!")
break
# Go to Naria and say how many decoys you built.
self.moveXY(14, 37)
self.say("Naria! I have collected " + decoysBuilt)
self.say(" decoys.")
``````
That looks really good, just one little problem tripping it all up. Get rid of the `break` inside your first `while self.gold < 25:` loop. That basically stops the loop after the first iteration, and then you still don’t have enough gold. That while loop will automatically end when `self.gold` hits 25 or more, so you don’t need to explicitly break out of it.
You could also make the outermost loop a `while` loop, too: `while decoysBuilt < 4:`.
Thanks for the quick response. I had realized that I needed to change the 25 to a 26 after I had posted it. I got it to work now. Turns out that it also wouldn’t acknowledge the variable “item” unless I included the line
"item = self.findNearestItem()" in the while loop. I wasn’t sure if I could use a while loop to build the decoys, because I thought I would have to change the 26 after the comparative operator from the first while loop to 104 in order for that to work. Great job on the work you guys have done. I’m a CS student who is using this to help me become a stronger programmer before I get to take some real classes. (About to just get started on my second semester). When I finish, maybe I can help somehow.
The reason that you need that in the loop is because you picked up the “item” you found outside the loop. So in order to continue, your hero had to find another item using that call.
All you need is this:
``````var decoysBuilt = 0;
loop {
var item = this.findNearestItem();
if(this.gold < 25 && item) {
this.moveXY(item.pos.x,item.pos.y);
}
else if(this.gold() >= 15) {
var tx = this.pos.x;
var ty = this.pos.y;
this.buildXY("fire-trap",tx,ty); decoysBuilt++;
}
if(decoysBuilt == 4) {
break;
}
}
this.say("Done building decoys!");
this.moveXY(14,36);
this.say("4");
``````
this is my code :
``````# We are field testing a new battle unit: the decoy.
# Build 4 decoys, then report the total to Naria.
# Each decoy costs 25 gold. Use the Quartz Sense Stone
# to know when you have more than 25 gold with self.gold.
# Keep a count of decoys you built as you go along.
# Break out of the loop when you have built 4.
decoysBuilt = 0
loop:
item = self.findNearestItem()
self.moveXY(item.pos.x, item.pos.y)
if self.gold >= 25:
self.buildXY("decoy", 47, 43)
decoysBuilt = decoysBuilt +1
if decoysBuilt == 4:
break
self.say("Done building decoys!")
# Go to Naria and say how many decoys you built.
self.moveXY(15, 36)
self.say(decoysBuilt)`````` | 2,672 | 9,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-33 | latest | en | 0.915615 |
http://metamath.tirix.org/mpests/omoe.html | 1,713,217,545,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00627.warc.gz | 23,503,008 | 3,378 | # Metamath Proof Explorer
## Theorem omoe
Description: The difference of two odds is even. (Contributed by Scott Fenton, 7-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)
Ref Expression
Assertion omoe ${⊢}\left(\left({A}\in ℤ\wedge ¬2\parallel {A}\right)\wedge \left({B}\in ℤ\wedge ¬2\parallel {B}\right)\right)\to 2\parallel \left({A}-{B}\right)$
### Proof
Step Hyp Ref Expression
1 odd2np1 ${⊢}{A}\in ℤ\to \left(¬2\parallel {A}↔\exists {a}\in ℤ\phantom{\rule{.4em}{0ex}}2{a}+1={A}\right)$
2 odd2np1 ${⊢}{B}\in ℤ\to \left(¬2\parallel {B}↔\exists {b}\in ℤ\phantom{\rule{.4em}{0ex}}2{b}+1={B}\right)$
3 1 2 bi2anan9 ${⊢}\left({A}\in ℤ\wedge {B}\in ℤ\right)\to \left(\left(¬2\parallel {A}\wedge ¬2\parallel {B}\right)↔\left(\exists {a}\in ℤ\phantom{\rule{.4em}{0ex}}2{a}+1={A}\wedge \exists {b}\in ℤ\phantom{\rule{.4em}{0ex}}2{b}+1={B}\right)\right)$
4 reeanv ${⊢}\exists {a}\in ℤ\phantom{\rule{.4em}{0ex}}\exists {b}\in ℤ\phantom{\rule{.4em}{0ex}}\left(2{a}+1={A}\wedge 2{b}+1={B}\right)↔\left(\exists {a}\in ℤ\phantom{\rule{.4em}{0ex}}2{a}+1={A}\wedge \exists {b}\in ℤ\phantom{\rule{.4em}{0ex}}2{b}+1={B}\right)$
5 2z ${⊢}2\in ℤ$
6 zsubcl ${⊢}\left({a}\in ℤ\wedge {b}\in ℤ\right)\to {a}-{b}\in ℤ$
7 dvdsmul1 ${⊢}\left(2\in ℤ\wedge {a}-{b}\in ℤ\right)\to 2\parallel 2\left({a}-{b}\right)$
8 5 6 7 sylancr ${⊢}\left({a}\in ℤ\wedge {b}\in ℤ\right)\to 2\parallel 2\left({a}-{b}\right)$
9 zcn ${⊢}{a}\in ℤ\to {a}\in ℂ$
10 zcn ${⊢}{b}\in ℤ\to {b}\in ℂ$
11 2cn ${⊢}2\in ℂ$
12 mulcl ${⊢}\left(2\in ℂ\wedge {a}\in ℂ\right)\to 2{a}\in ℂ$
13 11 12 mpan ${⊢}{a}\in ℂ\to 2{a}\in ℂ$
14 mulcl ${⊢}\left(2\in ℂ\wedge {b}\in ℂ\right)\to 2{b}\in ℂ$
15 11 14 mpan ${⊢}{b}\in ℂ\to 2{b}\in ℂ$
16 ax-1cn ${⊢}1\in ℂ$
17 pnpcan2 ${⊢}\left(2{a}\in ℂ\wedge 2{b}\in ℂ\wedge 1\in ℂ\right)\to 2{a}+1-\left(2{b}+1\right)=2{a}-2{b}$
18 16 17 mp3an3 ${⊢}\left(2{a}\in ℂ\wedge 2{b}\in ℂ\right)\to 2{a}+1-\left(2{b}+1\right)=2{a}-2{b}$
19 13 15 18 syl2an ${⊢}\left({a}\in ℂ\wedge {b}\in ℂ\right)\to 2{a}+1-\left(2{b}+1\right)=2{a}-2{b}$
20 subdi ${⊢}\left(2\in ℂ\wedge {a}\in ℂ\wedge {b}\in ℂ\right)\to 2\left({a}-{b}\right)=2{a}-2{b}$
21 11 20 mp3an1 ${⊢}\left({a}\in ℂ\wedge {b}\in ℂ\right)\to 2\left({a}-{b}\right)=2{a}-2{b}$
22 19 21 eqtr4d ${⊢}\left({a}\in ℂ\wedge {b}\in ℂ\right)\to 2{a}+1-\left(2{b}+1\right)=2\left({a}-{b}\right)$
23 9 10 22 syl2an ${⊢}\left({a}\in ℤ\wedge {b}\in ℤ\right)\to 2{a}+1-\left(2{b}+1\right)=2\left({a}-{b}\right)$
24 8 23 breqtrrd ${⊢}\left({a}\in ℤ\wedge {b}\in ℤ\right)\to 2\parallel \left(2{a}+1-\left(2{b}+1\right)\right)$
25 oveq12 ${⊢}\left(2{a}+1={A}\wedge 2{b}+1={B}\right)\to 2{a}+1-\left(2{b}+1\right)={A}-{B}$
26 25 breq2d ${⊢}\left(2{a}+1={A}\wedge 2{b}+1={B}\right)\to \left(2\parallel \left(2{a}+1-\left(2{b}+1\right)\right)↔2\parallel \left({A}-{B}\right)\right)$
27 24 26 syl5ibcom ${⊢}\left({a}\in ℤ\wedge {b}\in ℤ\right)\to \left(\left(2{a}+1={A}\wedge 2{b}+1={B}\right)\to 2\parallel \left({A}-{B}\right)\right)$
28 27 rexlimivv ${⊢}\exists {a}\in ℤ\phantom{\rule{.4em}{0ex}}\exists {b}\in ℤ\phantom{\rule{.4em}{0ex}}\left(2{a}+1={A}\wedge 2{b}+1={B}\right)\to 2\parallel \left({A}-{B}\right)$
29 4 28 sylbir ${⊢}\left(\exists {a}\in ℤ\phantom{\rule{.4em}{0ex}}2{a}+1={A}\wedge \exists {b}\in ℤ\phantom{\rule{.4em}{0ex}}2{b}+1={B}\right)\to 2\parallel \left({A}-{B}\right)$
30 3 29 syl6bi ${⊢}\left({A}\in ℤ\wedge {B}\in ℤ\right)\to \left(\left(¬2\parallel {A}\wedge ¬2\parallel {B}\right)\to 2\parallel \left({A}-{B}\right)\right)$
31 30 imp ${⊢}\left(\left({A}\in ℤ\wedge {B}\in ℤ\right)\wedge \left(¬2\parallel {A}\wedge ¬2\parallel {B}\right)\right)\to 2\parallel \left({A}-{B}\right)$
32 31 an4s ${⊢}\left(\left({A}\in ℤ\wedge ¬2\parallel {A}\right)\wedge \left({B}\in ℤ\wedge ¬2\parallel {B}\right)\right)\to 2\parallel \left({A}-{B}\right)$ | 1,986 | 3,777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 33, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-18 | latest | en | 0.430322 |
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Excel Regression Analysis Output Explained
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Topic Index > Excel for statistics > Excel Regression Analysis Output Explained
Previous article: Excel 2013 Regression Analysis How To
Watch the video or read the steps below:
Excel Regression Analysis Output Explained
In the previous article, I explained how to perform Excel regression analysis. After youve gone through the steps, Excel
will spit out your results, which will look something like this:
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Excel Regression Analysis Output Explained: Multiple
Regression
Heres a breakdown of what each piece of information in the output means:
EXCEL REGRESSION ANALYSIS PART ONE: REGRESSION
STATISTICS
These are the Goodness of Fit measures. They tell you how well the calculated linear equation fits your data.
1. Multiple R. This is the correlation coefficient. It tells you how strong the linear relationship is. For example, a value of
1 means a perfect positive relationship and a value of zero means no relationship at all. It is the square root of r
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1 means a perfect positive relationship and a value of zero means no relationship at all. It is the square root of r
squared (see #2).
2. R square. This is r2, the Coefficient of Determination. It tells you how many points fall on the regression line. for
example, 80% means that 80% of the variation of y-values around the mean are explained by the x-values. In other
words, 80% of the values fit the model.
of #2 if you have more than one x variable.
4. Standard Error of the regression: An estimate of the standard deviation of the error . This is not the same as the
standard error in descriptive statistics! The standard error of the regression is the precision that the regression
coefficient is measured; if the coefficient is large compared to the standard error, then the coefficient is probably
different from 0.
5. Observations. Number of observations in the sample.
EXCEL REGRESSION ANALYSIS PART TWO: ANOVA
1. SS = Sum of Squares.
2. Regression MS = Regression SS / Regression degrees of freedom.
3. Residual MS = mean square error (Residual SS / Residual degrees of freedom).
4. F: Overall F test for the null hypothesis.
5. Significance F: The associated P-Value.
The second part of output you get in Excel is rarely used, compared to the regression output above. It splits the sum of
squares into individual components (see: Residual sum of squares), so it can be harder to use the statistics in any
meaningful way. If youre just doing basic linear regression (and have no desire to delve into individual components) then
you can skip this section of the output.
For example, to calculate R2 from this table, you would use the following formula:
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R2 = 1 residual sum of squares (SS Residual) / Total sum of squares (SS Total).
In the above table, residual sum of squares = 0.0366 and the total sum of squares is 0.75, so:
R2 = 1 0.0366/0.75=0.9817
EXCEL REGRESSION ANALYSIS PART THREE: INTERPRET
REGRESSION COEFFICIENTS
This section of the table gives you very specific information about the components you chose to put into your data analysis.
Therefore the first column (in this case, HH size / Cubed HH size) will say something different, according to what data you
put into the worksheet. For example, it might say height, income or whatever variables you chose.
The columns are:
1. Coefficient: Gives you the least squares estimate.
2. Standard Error: the least squares estimate of the standard error.
3. T Statistic: The T Statistic for the null hypothesis vs. the alternate hypothesis.
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4. P Value: Gives you the p-value for the hypothesis test.
5. Lower 95%: The lower boundary for the confidence interval.
6. Upper 95%: The upper boundary for the confidence interval.
The most useful aspect of this section is that it gives you the linear equation:
y = mx + b
y = slope (least squares estimate) + intercept * x
For the above table, the equation would be approximately:
y = 3.14 0.65x + 0.024
Source: http://cameron.econ.ucdavis.edu/excel/ex61multipleregression.html
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By Andale | February 17, 2014 | Microsoft Excel | 4 Comments |
Intermediate Value Theorem
Explained
Anon
1.
October 8, 2014 at 11:48 am
This article should clearly credit the source it is based on (plagiarized from?);
http://cameron.econ.ucdavis.edu/excel/ex61multipleregression.html
It also introduces additional errors, particularly;
and the total sum of squares is 1.6050, so:
R2 = 1 0.3950 1.6050 = 0.8025.
and the total sum of squares is 2, so:
R2 = 1 0.3950 / 2 = 0.8025.
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Andale
2.
Post author
October 8, 2014 at 6:37 pm
Thanks for spotting the error with the sum of squares. Its now fixed.
I added credit to the article.
Regards,
S
Irfan
3.
November 8, 2014 at 1:20 pm
Hi stepahnie
I have more than 2 variables. my variable is 6. pls tell me how to calculate regresson eqution for more varaibles. I am
in urgent need.
Thanks
Irfan
4.
Andale
Post author
November 9, 2014 at 10:53 am
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pdfcrowd.com | 1,509 | 5,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-51 | latest | en | 0.799021 |
https://stats.stackexchange.com/questions/560004/can-a-prior-be-conjugate-and-noninformative-at-the-same-time | 1,685,560,447,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00747.warc.gz | 611,768,215 | 38,327 | # Can a prior be conjugate and noninformative at the same time?
And if so, could somebody give me a concrete example?
• It depends on what you regard as noninformative, and in some cases whether you require the prior to be proper. To take the example of a binomial likelihood, the conjugate distribution is a beta distribution, and some people regard a uniform Beta(1,1) as uninformative, some might use a Jeffreys Beta(0.5,0.5) prior, and some a Haldane improper Beta(0,0) prior. But since these are different, others might argue that simply choosing one would be informative Jan 11, 2022 at 9:14
• @Henry you should make it an answer.
– Tim
Jan 11, 2022 at 10:20
• some people regard a uniform $$\text{Beta}(1,1)$$ as uninformative,
• some might use a Jeffreys $$\text{Beta}(0.5,0.5)$$ prior, and
• some a Haldane improper $$\text{Beta}(0,0)$$ prior. | 249 | 855 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-23 | latest | en | 0.876772 |
https://jp.mathworks.com/matlabcentral/cody/problems/42268-create-a-square-matrix-of-zeros-of-even-order/solutions/660046 | 1,575,955,026,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525821.56/warc/CC-MAIN-20191210041836-20191210065836-00457.warc.gz | 418,311,332 | 15,498 | Cody
# Problem 42268. Create a square matrix of zeros of even order
Solution 660046
Submitted on 24 Apr 2015 by LY Cao
• Size: 10
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 2; y = zeros(2); assert(isequal(zero(x),y))
2 Pass
%% x = 4; y = zeros(4); assert(isequal(zero(x),y)) | 136 | 444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-51 | longest | en | 0.775572 |
https://answers.search.yahoo.com/search?p=rotation&pz=10&ei=UTF-8&flt=cat%3AMilitary%3Branking%3Adate&bct=0&b=21&pz=10&bct=0&xargs=0 | 1,603,179,286,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107869933.16/warc/CC-MAIN-20201020050920-20201020080920-00292.warc.gz | 221,600,327 | 50,001 | # Yahoo Web Search
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1 Answers · Science & Mathematics · 06/10/2013 | 700 | 2,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-45 | latest | en | 0.864499 |
https://questionvoyage.com/how-long-does-it-take-to-travel-in-lightspeed-in-star-wars | 1,656,107,817,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00350.warc.gz | 530,353,884 | 16,557 | # How long does it take to travel in lightspeed in star wars?
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## Top best answers to the question «How long does it take to travel in lightspeed in star wars»
Using traditional rocket fuel, it would take about 50,000 to 70,000 years to reach Proxima Centauri, and nuclear propulsion with proposed technology would get there in about 100 years, Lentz said. A light speed trip would take four years and three months.
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Those who are looking for an answer to the question «How long does it take to travel in lightspeed in star wars?» often ask the following questions:
### âť” How long does it take ships to travel in star wars?
• Depending on whether or not a ship chose to enter hyperspace, space travel in Star Wars varied depending on the size and speed of the individual craft. In some instances, it took days to travel between planets, while in other cases, it took mere moments.
### âť” How long does it take to travel through space in star wars?
• Depending on whether or not a ship chose to enter hyperspace, space travel in Star Wars varied depending on the size and speed of the individual craft. In some instances, it took days to travel between planets, while in other cases, it took mere moments.
### âť” Where does time travel take place in star wars?
• It first appeared in Star Wars canon in the second issue of the 2016 comic book series Star Wars: Doctor Aphra. The non-canon LEGO Star Wars Holiday Special has time travel as a major part of its plot, as Rey Skywalker is guided to a Jedi Temple on the planet Kordoku.
“Lightspeed” in terms of Star Wars is a bit of a misnomer. The distance between populated star systems in the Star Wars galaxy is still measured in terms of light years, which would mean that traveling to any given star system at the speed of ligh...
Assuming the Galaxy far, far away is roughly the size of the Milky Way, traveling between two planets on opposite sides would mean crossing roughly 100,000 light years, which takes light 100,000 years. That seems a bit long for a trip.
Hyperspace was an alternate dimension that could only be reached by traveling at or faster than the speed of light. Hyperdrives enabled starships to travel through hyperspace lanes across great distances, enabling travel and exploration throughout the galaxy. 1 Nature 2 Hyperspace usage 3 History 4 Appearances 4.1 Non-canon appearances 5 Sources 6 Notes and references 7 External links At least ...
Star Wars: A New Hope. Director's commentary with George Lucas he explains that's why Han was like "It's the ship that made the Kessel run in less than 12 parsecs." He said that the Kessel Run is a particularly nasty area and you have to drop in and out a lot. The common routes make the run much longer than 12 parsecs.
Depends on the writer, honestly. Travel speeds were never really established, but from the Outer Rim to the Core, depending on the route you take and other variables such as engine power, can take anywhere from a day to a week. Also, for that matter how does hyperspace work in theory? Is it light speed?
So... it's kind of like you're asking how fast can Jack O'Neill travel through the Stargate. Theoretically, according to the show, he could walk at 1 MPH and make it to the other side of the galaxy in a few seconds. Share. Improve this answer.
Lightspeed was slang referring to the threshold at which a starship first entered hyperspace, although in actuality, the potential velocities achieved whilst entering or travelling within hyperspace surpassed the speed of light. Poe Dameron: Flight Log Star Wars Helmet Collection 25(Databank A-Z: Hydroid MedusasImperial Academies) Star Wars Helmet Collection 42(Highlights of the Saga ...
Generally, a pilot takes into account his ship’s drives and space speed and plans ahead to have enough time to reach the next exit point, plus 5 to 10 minutes to spare in case of an emergency. The Astrogation sequence changes as follows: First determine the number of jumps, for the entire route, as stated above.
After all most stars are a couple light years apart. Even close ones are like five light years apart. And even if Aldberon and Yvain where like five light years apart, can the Death Star even move at the speed of light? And if it did, it would still take five years to get there.
Goodperson5656. 6 months ago. Id say a few hours up to a day. It depends on your area and whether there are obstacles like stars, nebulae, and how well the region is mapped. Hyperspace travel in the unknown regions would take far longer than hyperspace travel in the mid rim due to it being unexplored, unmapped, and full of obstacles. In TCW ...
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Which electroweak particles travel at lightspeed?
Strictly speaking, only particles without restmass (with rest mass = 0) can travel at the speed of light. If the total particle energy is large compared to the rest energy, the speed of the particle can be rather close to
How do they travel so fast in star wars?
Hyperspace was an alternate dimension that could only be reached by traveling at or faster than the speed of light. Hyperdrives enabled starships to travel through hyperspace lanes across great distances, enabling travel and exploration throughout the galaxy.
How fast do ships travel in star wars kmh?
The Eta-2 (as noted in Star Wars: Ships of the Galaxy) is able to reach speeds of 1500 kmph, making it faster than even the famously fast A-Wing. 4 Dooku's Solar Sailer At the end of Attack of the Clones (2002), Count Dooku escapes on a unique-looking ship that deploys a solar sail.
How fast do ships travel in star wars kms?
A Galactic Standard light year is approximately 9.5 quadrillion meters, and the Star Wars galaxy is approximately 120,000 light years in diameter. Given an average of how long it takes ships to travel between planets in the movies, it is likely that a ship in hyperspace could go as fast as 6 quintillion meters per hour.
How fast is real space travel in star wars?
The time depended on the distance any point in real space corresponded to the time it took in hyperspace to go from the point of origin to the point desired. Distance in real space did affect the amount of time required while traveling thru a hype...
Is there fast travel in star wars fallen order?
• Much like the Metroid Prime games it takes inspiration from, Star Wars Jedi: Fallen Order doesn't have a fast-travel system. If you're looking to leave the planet you're currently on, you'll need to retrace your steps back to your ship.
Star wars jedi fallen order how to fast travel?
Unfortunately, there is no way to unlock fast travel in Jedi Fallen Order. The game simply doesn’t have a system in place for instantly teleporting from one location to another.
Will star wars jedi fallen order get fast travel?
• If you want to leave the planet, you'll have to hoof it back to your ship on foot. Much like the Metroid Prime games it takes inspiration from, Star Wars Jedi: Fallen Order doesn't have a fast-travel system. If you're looking to leave the planet you're currently on, you'll need to retrace your steps back to your ship.
How long would it take to travel to nearest star?
• The nearest star to earth is roughly 4.3 light years away. That is, it takes light moving at 186,000 miles per second to get there. So what is that distance? About 25.3 trillion miles. Current space travel tops out around 25,000 mph. Doing the math… It would take more than 1 billion hours.
Can you book star wars hotel?
Star Wars Hotel Price
Guests at Galactic Starcruiser are booked for a 2 night, 2 day stay. The premise is similar to a cruise ship stay. All guests arrive and depart on the same day. During those 2 days guests are totally immersed in their own Star Wars story, including a trip directly into Star Wars: Galaxy's Edge.
How do you travel in star wars the old republic?
Traveling in Star Wars: The Old Republic is accomplished by using vehicles, taxis, shuttles, and starships. Also, at level 14, all classes are able to train an out of combat speed boost ability: Sprint.
How do you use quick travel points in star wars?
• When available, activating this ability will allow the platey to select from unlocked (bound) Quick Travel Points on the planet. They will then be transported to the selected location. Often Quick Travel Points will be located at Medical Centers with Imperial Service Droids that can provide Stims and Medpacs.
Why does quantum travel take longer in star citizen?
• It takes longer for a large ship to reach the point where it must shut down to allow it too cool down. This is why larger ships can Quantum travel for a longer duration when compared to a smaller ship that can't dedicate nearly as much space to power generation and cooling.
When does the star wars hotel open in disney world?
• The hotel, which opens in spring 2022, is billed by Disney as a “revolutionary new two-night experience where you and your group will embark on a first-of-its-kind Star Wars adventure that’s your own.”
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Is the disney star wars hotel open?
Disney: Star Wars hotel experience now boarding in 2022
Star Wars: Galactic Starcruiser, the Walt Disney World Resort's forthcoming Star Wars-themed hotel, will open in 2022. The hotel, which will offer guests a "vacation experience," with two-night stays, was originally planned to open this year.
When can you book star wars hotel?
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When will disney star wars hotel open?
• Walt Disney World released the new details on the 'Star Wars: Galactic Starcruiser' hotel on Disney Parks Blog on Tuesday, May 4th -- also known as 'Star Wars Day.' While they previously said that the hotel will open in 2021, they just announced that it will officially debut in 2022.
How long does travel reimbursement take?
With BTSSS, turnaround time to evaluate and settle a claim is less than 5 days. BTSSS allows Veterans and caregivers to submit claims 24/7, 365 days a year from a computer or mobile device. It also allows users to electronically track the status of a claim request. | 2,563 | 11,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-27 | latest | en | 0.947042 |
https://civiljournal.semnan.ac.ir/article_5716.html | 1,726,292,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00155.warc.gz | 146,539,415 | 12,667 | # New Methods for Dynamic Analysis of Structural Systems under Earthquake Loads
Document Type : Regular Paper
Authors
1 Department of Civil Engineering, Faculty of Engineering, University of Bonab, Bonab, Iran
2 Engineering Faculty of Khoy, Urmia University of Technology, Urmia, Iran
3 Department of Civil Engineering, Qazvin Branch, Islamic Azad University, Qazvin, Iran
Abstract
Two numerical methods are proposed for dynamic analysis of single-degree-of-freedom systems. Basics of dynamics and elementary tools from numerical calculus are employed to formulate the methods. The energy conservation principles triggered the basic idea of the first method, so-called energy-based method (EBM). It is devised for dynamic analysis of linear damped system whose damping ratio is greater than 1%. The second method uses function approximation theory and integration scheme, and called simplified integration method (SIM). Several numerical examples are investigated through SIM. A detailed comparison is made between the proposed methods and the conventional ones. The results show that the proposed methods can estimate the dynamic response of linear damped systems with high accuracy. In the first example, the peak displacement is obtained 6.8747 cm and 6.8290 cm which closely approximate the highly exact response of Duhamel integral. Results show that Newmark-β method is the fastest one whose run-time is 0.0019 sec. EBM and SIM computational times are 0.0722 sec and 0.0021sec, respectively. SIM gives more accurate estimate and convergence rate than Newmark-β method. The difference of peak displacement obtained from two methods is almost less than 1%. Thus, SIM reliably estimates the dynamic response of systems with less computational cost.
Keywords
Main Subjects
#### References
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[4] Bazzi, G., Anderheggen, E. (1982). “The r-family of algorithms for time-step integration with improved numerical dissipation.” Earthquake Engineering & Structural Dynamics, Vol. 10, pp. 537–550.
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[6] Chang, S. Y. (2004). “Studies of Newmark method for solving nonlinear systems: (I) basic analysis.” Journal of the Chinese Institute of Engineers, Vol. 27, Issue 5, pp. 651–662.
[7] Chopra, A. K., Goel, R. K., Chintanapakdee, C. (2003). “Statistics of single-degree-of-freedom estimate of displacement for pushover analysis of buildings.” Journal of structural engineering, Vol. 129, Issue 4, pp. 459–469.
[8] Chopra, A. K. (2017). Dynamics of structures. theory and applications to. Earthquake Engineering.
[9] Clough, RW., Penzien, J. (1995). “Dynamics of Structures.” Computers & Structure Inc., 3rd Ed., CA.
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[15] He, L. (2008). “Harmonic solution of unsteady flow around blades with separation.” American Institute of Aeronautics and Astronautics Journal, Vol. 46, pp. 1299–1307.
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[19] JalilKhani, M., Babaei, M., Ghasemi, S. Evaluation of the Seismic Response of Single-Story RC Frames under Biaxial Earthquake Excitations. Journal of Rehabilitation in Civil Engineering, 2020; 8(3): 98-108.
[20] Kazakov, K. S. (2008). “Dynamic Response of a Single Degree of Freedom (SDOF) System in some Special Load Cases, based on the Duhamel Integral.” In International Conference on Engineering Optimization, pp. 01–05.
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### History
• Receive Date: 02 May 2021
• Revise Date: 25 October 2021
• Accept Date: 31 October 2021 | 1,991 | 7,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.899873 |
https://alchetron.com/Symmetric-product-of-an-algebraic-curve | 1,624,545,049,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00460.warc.gz | 93,607,220 | 26,022 | Symmetric product of an algebraic curve
Updated on
Covid-19
In mathematics, the n-fold symmetric product of an algebraic curve C is the quotient space of the n-fold cartesian product
C × C × ... × C
or Cn by the group action of the symmetric group on n letters permuting the factors. It exists as a smooth algebraic variety ΣnC; if C is a compact Riemann surface it is therefore a complex manifold. Its interest in relation to the classical geometry of curves is that its points correspond to effective divisors on C of degree n, that is, formal sums of points with non-negative integer coefficients.
For C the projective line (say the Riemann sphere) ΣnC can be identified with projective space of dimension n.
If G has genus g ≥ 1 then the ΣnC are closely related to the Jacobian variety J of C. More accurately for n taking values up to g they form a sequence of approximations to J from below: their images in J under addition on J (see theta-divisor) have dimension n and fill up J, with some identifications caused by special divisors.
For g = n we have ΣgC actually birationally equivalent to J; the Jacobian is a blowing down of the symmetric product. That means that at the level of function fields it is possible to construct J by taking linearly disjoint copies of the function field of C, and within their compositum taking the fixed subfield of the symmetric group. This is the source of André Weil's technique of constructing J as an abstract variety from 'birational data'. Other ways of constructing J, for example as a Picard variety, are preferred now (Greg W. Anderson (Advances in Math.172 (2002) 169–205) provided an elementary construction as lines of matrices). But this does mean that for any rational function F on C
F(x1) + ... + F(xg)
makes sense as a rational function on J, for the xi staying away from the poles of F.
For N > g the mapping from ΣnC to J by addition fibers it over J; when n is large enough (around twice g) this becomes a projective space bundle (the Picard bundle). It has been studied in detail, for example by Kempf and Mukai.
Betti numbers and the euler characteristic of the symmetric product
Let C be smooth and projective of genus g over the complex numbers C. The Betti numbers binC) of the symmetric product are given by
n = 0 i = 0 2 n b i ( Σ n C ) y n u i n = ( 1 + y ) 2 g ( 1 u y ) ( 1 u 1 y )
and the topological Euler characteristic enC) is given by
n = 0 e ( Σ n C ) p n = ( 1 p ) 2 g 2 .
Here we have set u=-1 and y = - p in the formula before.
References
Symmetric product of an algebraic curve Wikipedia
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Topics | 666 | 2,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-25 | latest | en | 0.931006 |
https://www.convertopedia.com/unit-converters/time-converter/day-to-month-converter/ | 1,726,334,406,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00253.warc.gz | 675,909,589 | 21,578 | ## Day
A Day is a non SI unit of time. The symbol of day is d. A day is a period of 24 hours especially from 12 o’ clock one night to 12 o’ clock the next night. A day contains 24 hours or 1440 minutes or 86,400 seconds. A day is the period of time required for the earth to complete a full rotation around its axis.
## Month
A Month is a unit of time used with calendar and ranges in length from 28 to 31 days. There are 12 months in a year. One month is equal to 1/12 of a year. The month is abbreviated as mo.
## How to convert Day to Month
1 d = 0.03285 mo
Example 1:
Convert 15 Day to Month
1 d = 0.03285 mo
Therefore, 15 d = 15 * 0.03285 = 0.49282 mo
Example 2:
Convert 50 Day to Month
1 d = 0.03285 mo
Therefore, 50 d = 50 * 0.03285 = 1.64274 mo
Day(d)Month(mo)
00
10.03285
20.06570
30.09856
40.13141
50.16427
60.19712
70.22998
80.26283
90.26283
100.32854
110.36140
120.39425
130.42711
140.45996
150.49282
200.65709
500.65709
1000.65709
## Day to other converters
• Day to Nanosecond Converter
• Day to Microsecond Converter
• Day to Millisecond Converter
• Day to Second Converter
• Day to Minute Converter
• Day to Hour Converter
• Day to Week Converter
• Day to Year Converter | 402 | 1,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.770366 |
https://ibmathsresources.com/2019/02/20/modeling-hours-of-daylight/ | 1,685,230,585,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643388.45/warc/CC-MAIN-20230527223515-20230528013515-00061.warc.gz | 343,216,461 | 27,292 | If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Modeling hours of daylight
Desmos has a nice student activity (on teacher.desmos.com) modeling the number of hours of daylight in Florida versus Alaska – which both produce a nice sine curve when plotted on a graph. So let’s see if this relationship also holds between Phuket and Manchester.
First we can find the daylight hours from this site, making sure to convert the times given to decimals of hours.
Phuket
Phuket has the following distribution of hours of daylight (taking the reading from the first of each month and setting 1 as January)
Manchester
Manchester has much greater variation and is as follows:
Therefore when we plot them together (Phuket in green and Manchester in blue) we get the following 2 curves:
We can see that these very closely fit sine curves, indeed we can see that the following regression lines fit the curves very closely:
Manchester:
Phuket:
For Manchester I needed to set the value of b (see what happens if you don’t do this!) Because we are working with Sine graphs, the value of d will give the equation of the axis of symmetry of the graph, which will also be the average hours of daylight over the year. We can see therefore that even though there is a huge variation between the hours of daylight in the 2 places, they both get on average the same amount of daylight across the year (12.3 hours versus 12.1 hours).
Further investigation:
Does the relationship still hold when looking at hours of sunshine rather than daylight? How many years would we expect our model be accurate for? It’s possible to investigate the use of sine waves to model a large amount of natural phenomena such as tide heights and musical notes – so it’s also possible to investigate in this direction as well.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here. | 762 | 3,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-23 | latest | en | 0.941785 |
http://www.muzuu.org/new_life/pics/simpleblog/math/recforms/reursiveforms.html | 1,708,917,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474650.85/warc/CC-MAIN-20240226030734-20240226060734-00602.warc.gz | 62,619,266 | 6,903 | # SimpleBlog
Index of all articles
## Rational Recursive (-Iterative) Forms
Some pictures (the beauty of lapse):
Recursive Forms? (I did not find examples on the internet, so this name came up to me. My way to numbers is by computer experiments and pictures, so other attributes might heve been ‘generative functions’, ‘recursive functions’ aso.) The Ideas behind this stuff are so simple, that I’m quite sure, that someone else might have come up with. So for example the Tribonacci sequence (and some n-bonacci sequences) are part of The On-Line Encyclopedia of Integer Sequences On the other side there are exploding many recursive forms and the only referenz I found was Wolfram’s ‘New Kind of Science’ (the NKS pictures that stayed in my mind are syntactic recursive forms) where was stated, that there is a lack of systematical investigation of this stuff.
Caution: This text is not mathematically strict. This is just part of my daily disport and belongs to the publications of the ANM (academy of naive mathematics).
I think this forms must be investigated systematically. They are beautifull, surprising and may have many connections to so many things… (Mistakes, Growth, Precission, Quantumtheorie?, Philosophy)
### What catched me!
While thinking about the Fibonacci sequence and its general form Lucas sequences, I came up with the idea to check complexer additive forms.
So if the fibonacci sequence and some lucas sequences (not all, but I don’t want to complicate things now) could be represented by [f=[a,b]; append f[-2] + f[-1] for ever] or an other way to write [[a,b]; a + b], which results in
• (1)a
• (1)b
• (1)a+(1)b
• a+2b
• 2a+3b
• 3a+5b
• 5a+8b
• etc. with the factors of a and b giving 2 times the fibonacci sequence [1,1,2,3,5,8,13,...] and the sums at each line with [a,b]=[1,1] give also the same sequence.
This sequences grow quite fast and were studied by real mathematicans in deep. If you change the starting values of a and b you will get other sequences (lucas, but not all I think), but the structure stays somehow the same (fibbonacci is always there, so may be we can call these: representations of simple growth processes)
My next step was, to extend the starting values to something like [a,b,c,...] and to add subtraction. So we get for example forms like:
• [[a,b,c]; ( + / – )f[-3] ( + / – )f[-2] ( + / – )f[-1]]
There are many sequences of this form. For example [[a,b,c]; a + b + c] or [[a,b,c]; – a + b – c] and so on. There is a simple way to generate al this forms up to n varaibles. Simply count in ternary system starting with 1 and say 1:= + 1, 2:= – 1 and 0:=0:
So there are 2 forms with one variable:
• 1 = a
• 2 = -a
there are 6 forms with 2 variables [starting array has 2 values]:
• 3 (10) = +b ;(0a)
• 4 (11) = +b +a
• 5 (12) = +b -a
• 6 (20) = -b
• 7 (21) = -b +a
• 8 (22) = -b -a
there are 3n – 3(n-1) forms of n varables, so it goes on with:
• 9 (100) = +c
• ...
• 13 (111) = +c +b +a
• 14 (112) = +c +b -a
• and so on.
Next step I did: I wrote a program that systematically built sequences based on these recusion formulars with the starting array [1,1,1,...] and I got some nice sequences but in most cases they are growing (+ or – or both infinities), so I did the genaral case and wrote a program that started with [a,b,c,...] and built the symbolic sequences, which again gave nice results with fast growing factors of a,b,c and so on.
There are many nice resultes, so here one example [[a,b,c,d];a+b-c+d]:
• a
• b
• c
• d
• a+b-c+d
• a+2b
• b+2c
• c+2d
• 2a+2b-2c+3d
• 3a+5b-c+d
• a+4b+4c
• b+4c+4d
• 4a+4b-3c+8d
• 8a+12b-4c+5d
• and so on
which gives 1,1,1,1,2,3,3,3,5,8,9,9,13,21,26,27,...
In my programm I represented these sequences without those [a,b,c,...] so that they could be given as follows.
Fibonacci:
• Sum [as,bs]
• 1 [1,0]
• 1 [0,1]
• 2 [1, 1]
• 3 [1, 2]
• 5 [2, 3]
• 8 [3, 5]
• 13 [5, 8]
• 21 [8, 13]
• 34 [13, 21]
• 55 [21, 34]
• 89 [34, 55]
• 144 [55, 89]
But really interessting are those:
5 Parameters which factors grow but sum is allways 1:
• 1 [1, 0, 0, 0, 0]
• 1 [0, 1, 0, 0, 0]
• 1 [0, 0, 1, 0, 0]
• 1 [0, 0, 0, 1, 0]
• 1 [0, 0, 0, 0, 1]
• 1 [0, 1, -1, 1, -1]
• 1 [-1, 0, 2, -2, 2]
• 1 [2, 1, -2, 4, -4]
• 1 [-4, -2, 5, -6, 8]
• 1 [8, 4, -10, 13, -14]
• 1 [-14, -6, 18, -24, 27]
• 1 [27, 13, -33, 45, -51]
• 1 [-51, -24, 64, -84, 96]
• 1 [96, 45, -120, 160, -180]
• 1 [-180, -84, 225, -300, 340]
• 1 [340, 160, -424, 565, -640]
• 1 [-640, -300, 800, -1064, 1205]
• 1 [1205, 565, -1505, 2005, -2269]
• 1 [-2269, -1064, 2834, -3774, 4274]
or:
• 1 [1, 0, 0, 0, 0]
• 1 [0, 1, 0, 0, 0]
• 1 [0, 0, 1, 0, 0]
• 1 [0, 0, 0, 1, 0]
• 1 [0, 0, 0, 0, 1]
• 1 [1, 1, -1, -1, 1]
• 1 [1, 2, 0, -2, 0]
• 1 [0, 1, 2, 0, -2]
• 1 [-2, -2, 3, 4, -2]
• 1 [-2, -4, 0, 5, 2]
• 1 [2, 0, -6, -2, 7]
• 1 [7, 9, -7, -13, 5]
• 1 [5, 12, 4, -12, -8]
• 1 [-8, -3, 20, 12, -20]
• 1 [-20, -28, 17, 40, -8]
• 1 [-8, -28, -20, 25, 32]
• 1 [32, 24, -60, -52, 57]
• 1 [57, 89, -33, -117, 5]
• 1 [5, 62, 84, -38, -112]
• 1 [-112, -107, 174, 196, -150]
• 1 [-150, -262, 43, 324, 46]
• 1 [46, -104, -308, -3, 370]
So this is iteressting and I think group theorie would be a nice framework to investigate this stuff, but the next idea came over me.
### Combinig the forms to rational forms, we are entering Q
I allways thought, that Q (Rational numbers) is a good trick to see something that Z (N) hides. Q is like Z but maped to our perspective. Ok with a little bit lost, but … So the Idea was to combine the additive forms to rational forms. What happens if the reursion rule divides the additive forms?
This means
• [[a,b,c,...]; X / Y]
where X,Y are any additive formes over [a,b,c,...]. So I wrote a programm that does this for some forms and checked them. Ok, the results were lists of rational numbers, but the trick is to visualize them. This was done with gnuplot and here is the point where the real story begins. I played around with the starting values and got this beautifull pictures, that I animated:
This is [[a,b,c]];a + b / c] for [a,b,c] = [[2,1.01,2],[2,1.02,2]...[2,2, 2]... [2,5.01,2]] and [[200,0.01,200],[200,0.02,200] ...[200,1.01,200]] and this is wrong!. But on the beauty of lapse later more. (I’m using Ruby and Python and did my first experiments with real rational numbers that are included in this languages, tried out floating points and got the same reaults, but my fault was, that I did only 10 iterations, so my first pictures are based on floating point arithmetric, and this is dangerous)
### So let’s do more forms
Next I wrote a program that systematically generates all rational forms from 2 up to 5 variables. Now combinatorical explosion comes in. I got – after removing equvalent forms – 22 2-forms, 310 3-forms, 4132 4-forms and 29116 5-forms of the form X/Y where X,Y are simple additive forms up to 5 varables. So here some examples:
```2-1 a/(a+b) f[-2]/(f[-2]+f[-1])
2-2 a/(-a+b) f[-2]/(-f[-2]+f[-1])
2-3 a/(a-b) f[-2]/(f[-2]-f[-1])
2-4 a/(-a-b) f[-2]/(-f[-2]-f[-1])
2-5 b/(a+b) f[-1]/(f[-2]+f[-1])
2-6 b/(-a+b) f[-1]/(-f[-2]+f[-1])
2-7 b/(a-b) f[-1]/(f[-2]-f[-1])
2-8 b/(-a-b) f[-1]/(-f[-2]-f[-1])
2-9 (a+b)/a (f[-2]+f[-1])/f[-2]
2-10 (a+b)/-a (f[-2]+f[-1])/-f[-2]
2-11 (a+b)/b (f[-2]+f[-1])/f[-1]
2-12 (a+b)/(-a+b) (f[-2]+f[-1])/(-f[-2]+f[-1])
2-13 (a+b)/-b (f[-2]+f[-1])/-f[-1]
2-14 (a+b)/(a-b) (f[-2]+f[-1])/(f[-2]-f[-1])
2-15 (a+b)/(-a-b) (f[-2]+f[-1])/(-f[-2]-f[-1]) *is this -1?*
2-16 (-a+b)/a (-f[-2]+f[-1])/f[-2]
2-17 (-a+b)/-a (-f[-2]+f[-1])/-f[-2]
2-18 (-a+b)/b (-f[-2]+f[-1])/f[-1]
2-19 (-a+b)/(a+b) (-f[-2]+f[-1])/(f[-2]+f[-1])
2-20 (-a+b)/-b (-f[-2]+f[-1])/-f[-1]
2-21 (-a+b)/(a-b) (-f[-2]+f[-1])/(f[-2]-f[-1]) *is this 1?*
2-22 (-a+b)/(-a-b) (-f[-2]+f[-1])/(-f[-2]-f[-1])
3-1 a/(a+b) f[-3]/(f[-3]+f[-2])
3-2 a/(-a+b) f[-3]/(-f[-3]+f[-2])
...
3-309 (-a-b+c)/(a-b-c) (-f[-3]-f[-2]+f[-1])/(f[-3]-f[-2]-f[-1])
3-310 (-a-b+c)/(-a-b-c) (-f[-3]-f[-2]+f[-1])/(-f[-3]-f[-2]-f[-1])
4-1 a/(a+b) f[-4]/(f[-4]+f[-3])
4-2 a/(-a+b) f[-4]/(-f[-4]+f[-3])
4-3 a/(a-b) f[-4]/(f[-4]-f[-3])
...
4-3132 (-a-b-c+d)/(-a-b-c-d) (-f[-4]-f[-3]-f[-2]+f[-1])/(-f[-4]-f[-3]-f[-2]-f[-1])
5-1 a/(a+b) f[-5]/(f[-5]+f[-4])
5-2 a/(-a+b) f[-5]/(-f[-5]+f[-4])
...
5-29116 (-a-b-c-d+e)/(-a-b-c-d-e) (-f[-5]-f[-4]-f[-3]-f[-2]+f[-1])/(-f[-5]-f[-4]-f[-3]-f[-2]-f[-1])
```
2-15 and 2-21 is this + / – 1?: Here some mathematical advise would be fine, because I see that this forms reduce to a constant function, but if you use them recursivly, -1 becomes constant but +1 becomes to 0/0 which is not defined. One way would be to exclude starting values where a-b or -a+b = 0. (but this is marginal and my programs simply stop calculating, when a division by zero exception occures.)
Next I wrote a program that genarates an example picture of each form. And when I started this program, I had very exciting 3 hours:
and so on and so on … (Wow!)
But still wrong ;-)
### So what’s wrong with all this pictures?
They are not the right visualisations of the rucursive forms! Ok, I allways had in mind, that floating point arithmetric is full of pit falls. So I decided to do some checks, with much higher precission. First, I tried real rational numbers. No chance! The numerators and denominators grow extremly fast and after 50 Iterations the program slows down extremley and my pics are based on 4000 iterations. But in many cases with the rational recursive forms the ranges of numerator and denominator are near together, so there is a chance, that the errors are not so big, that the pictures will be absolute senseless. So I tried out another datatype. Python and Ruby have a Deicimal Type with arbitrarry precision, that can be limited to a certain precision. I wrote a program that iterated over precision, to see what happens if we go from 28 to 3000 significant numbers. It came out, that in most cases 300 significant numbers may be enough. Although I am not sure, if there are more hidden lapses, here is an animation of the precision iteration:
So in this case after precision is 80 the form stays stable. I hope that this is right now, but I’m still evaluating an testing.
To find out interssing forms, as a first shoot floating points are quite well and the wrong pics are beautyfull too!
here some examples of higher dimension
will be continued …
• dimensions to walk through: starting values (a,b,c,...), focus (y axe), iterations (x axe), precision
• just started to investigate more generally the builing blocks: [[a,b,...]; p(n)f[-n] + ... + p(1)f(1) + p(0)]. | 3,989 | 10,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-10 | latest | en | 0.938784 |
http://www.chegg.com/homework-help/questions-and-answers/chapter-6-normal-probability-distribution-61-piece-data-picked-random-normal-population--f-q1312375 | 1,501,091,624,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426234.82/warc/CC-MAIN-20170726162158-20170726182158-00593.warc.gz | 391,172,684 | 17,462 | Chapter 6
Normal Probability Distribution
6.1 A piece of data picked at random from a normal population.
a. Find the probability that it will have a standard score (Z) that lies between 0 and 0.95
b. Find the probability that it will have a standard score (Z) that lies to the right of 0.95
c. Find the probability that it will have a standard score (Z) that lies to the left of 0.95
d. Find the probability that it will have a standard score (Z) that lies between -0.95 and 0.95
6.2 The Z notation, Z(a), combines two related concepts, the Z-score and the area to the right, into a mathematical symbol.
a. If Z(A) = 0.15, identify the letter A as being a Z-score or being an area and show these values on a standard normal distribution diagram.
b. a. If Z(15) = B, identify the letter B as being a Z-score or being an area and show these values on a standard normal distribution diagram.
c. a. If Z(C) = -0.04, identify the letter C as being a Z-score or being an area and show these values on a standard normal distribution diagram
d. a. If -Z(0.04) = D, identify the letter D as being a Z-score or being an area and show these values on a standard normal distribution diagram
6.3 Suppose that X has a binomial distribution with n = 25 and p = 0.4
a. Find the mean and standard deviation of the normal distribution that is used in the approximation
b. Approximate the probability of X = 5
Hint. P(x=5) for discrete random variable x as P (4.5 < x < 5.5) for continuous
c. Use binomial formula find the probability of X = 5
d. Compare your answers to questions b) and c) and show the difference. | 425 | 1,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-30 | latest | en | 0.881987 |
http://www.programcreek.com/2014/07/leetcode-largest-bst-subtree-java/ | 1,495,502,102,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607245.69/warc/CC-MAIN-20170523005639-20170523025639-00359.warc.gz | 624,487,605 | 7,820 | # LeetCode – Largest BST Subtree (Java)
Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it.
Java Solution
```class Wrapper{ int size; int lower, upper; boolean isBST; public Wrapper(){ lower = Integer.MAX_VALUE; upper = Integer.MIN_VALUE; isBST = false; size = 0; } } public class Solution { public int largestBSTSubtree(TreeNode root) { return helper(root).size; } public Wrapper helper(TreeNode node){ Wrapper curr = new Wrapper(); if(node == null){ curr.isBST= true; return curr; } Wrapper l = helper(node.left); Wrapper r = helper(node.right); //current subtree's boundaries curr.lower = Math.min(node.val, l.lower); curr.upper = Math.max(node.val, r.upper); //check left and right subtrees are BST or not //check left's upper again current's value and right's lower against current's value if(l.isBST && r.isBST && l.upper<=node.val && r.lower>=node.val){ curr.size = l.size+r.size+1; curr.isBST = true; }else{ curr.size = Math.max(l.size, r.size); curr.isBST = false; } return curr; } }```
Category >> Algorithms
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:
```<pre><code>
String foo = "bar";
</code></pre>
```
• Divyanshu Singh
This code isn’t working for 2 test-cases. And I don’t have access to those test-cases. Can somebody pls point out the error.
``` // Following is the Binary Tree node structure /************** class BinaryTreeNode { public : T data; BinaryTreeNode *left; BinaryTreeNode *right;```
``` BinaryTreeNode(T data) { this -> data = data; left = NULL; right = NULL; } }; ***************/ #include using namespace std; bool isLeaf(BinaryTreeNode* root) { if(root->left||root->right) return false; return true; } pair helper(BinaryTreeNode* root) { if(root==NULL) { pair p(1,0); return p; } if(isLeaf(root)) { pair p(1,1); return p; } pair l = helper(root->left); pair r = helper(root->right); int lb = l.first; int lh = l.second; int rb = r.first; int rh = r.second; int b,h,x=root->data; if(root->left&&root->right) { if(x>root->left->data&&xright->data) { b=1; h=max(lh,rh)+1; pair ans(b,h); return ans; } else { b=0; h=max(lh,rh); pair ans(b,h); return ans; } } else if(root->left&&!root->right) { if(x>root->left->data) { b=1; h=max(lh,rh)+1; pair ans(b,h); return ans; } else { b=0; h=max(lh,rh); pair ans(b,h); return ans; } } else if(!root->left&&root->right) { if(xright->data) { b=1; h=max(lh,rh)+1; pair ans(b,h); return ans; } else { b=0; h=max(lh,rh); pair ans(b,h); return ans; } } } int largestBSTSubtree(BinaryTreeNode *root) { // Write your code here pair ans = helper(root); return ans.second; } ``` | 777 | 2,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2017-22 | latest | en | 0.446558 |
http://www.docstoc.com/docs/93656626/Valuation | 1,386,397,836,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053664/warc/CC-MAIN-20131204131733-00048-ip-10-33-133-15.ec2.internal.warc.gz | 299,959,812 | 14,994 | # Valuation
Document Sample
``` Valuation of bonds and shares
Bonds
Generally a fixed income security which
promise to give a certain fixed cash flow to
the holder at certain pre-determined time
points – coupon and the face value
Coupon
Is the periodical payment.
Periodicity pre-determined.
Quantum is also pre-determined
by the coupon rate
Face value
Is the lump sum payment at the end of the
fixed period
Date of maturity
Is the last coupon payment date
and it also coincides with the
lump sum payment of the face
value
Price of bond
Is the price at which bond can be sold
or bought and it depends on market
factors
Yield
Is the actual return (as different from the
coupon rate) worked out on the actual out
flow, which is the price of the bond
Yield to maturity
Is the rate at which we can
discount the coupon payments
until maturity and the face value
at maturity to get the present
value to be equal to the price of
bond
Equity
Face value is the simplest approach to
equity valuation. But this does not
reflect the real value
Book value is the distribution of
net worth (assets less outside
liabilities) among outstanding
shares
This book value approach depends on
accounting standards, procedures and
conventions and therefore does not reflect
the true value of the shares
Another approach to determine real
value is to value shares using dividend
discount models
Under the DDM, it is assumed constant
dividends are paid perpetually on a
share and its value derived as the
present value of perpetuity. It is
possible dividends may grow at
constant rate or at different rates.
Caution - DDM
While finding the value of share using
dividend models, one has to use a
discounting rate to obtain present
value of a stream of cash flow. The
problem is to find the appropriate
discounting rate. We do not have yield
curves for shares.
Caution - DDM
Another difficulty is the assumption
Caution – DDM
This valuation approach looks at the return
from shares and the main components of
returns may not be dividend alone. It may be
difference between prices at two different
points of time
Hence the main problem is to find the
future return. Using probability
concepts, one can find expected
return, which can be a good estimate
of future return
The expected return can be calculated
from either past prices or from
forecasted values of prices. In either
case, there is an uncertainty in the
realization of predicted return. This is
the inherent risk
Hence risk-return analysis becomes an
important aspect in share valuation
The question to be addressed in RRA is
the capacity to estimate future
returns, which requires good amount
of information.
Efficient market hypothesis proposes future
prices do not depend on past prices. Current
prices reflect all relevant information from
the past and therefore it is not possible to
forecast future prices and hence the returns.
This means if a market is efficient it is not
possible to predict the returns
Valuation ratios
Yield
Is sum total of dividend yield and
capital gains yield and is
mathematically represented by
Yield = Dividend + Price change /
Initial price
Dividend Yield
Is the ratio of dividend received to the
initial price paid and is represented by
Dividend Yield = Dividend / Initial
price
Capital Gains Yield
Is the ratio of Price change to the
initial price paid and is
represented by
Capital Gains Yield = Price
change / Initial price
Market value to book value ratio
Market value per share / Book value
per share
Price earning ratio
Market price per share / Earning
per share
Thank you
```
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How are you planning on using Docstoc? | 874 | 3,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2013-48 | longest | en | 0.918964 |
https://www.helpteaching.com/questions/1320971/which-equation-represents-the-distance-d2-between-f2-and-the | 1,669,621,822,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710488.2/warc/CC-MAIN-20221128070816-20221128100816-00744.warc.gz | 878,669,975 | 6,193 | ##### Question Info
This question is public and is used in 1 group and 7 tests or worksheets.
Type: Multiple-Choice
Category: Nonlinear Equations and Functions
Standards: HSG-GPE.A.3
Author: nsharp1
Created: 3 years ago
View all questions by nsharp1.
# Nonlinear Equations and Functions Question
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## Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3
Which equation represents the distance, $d_2$, between $F_2$ and the point $(x,y) ?$
1. $d_2 = sqrt( (c-x)^2 + (y-c)^2)$
2. $d_2 = sqrt( (x-c)^2 + y^2)$
3. $d_2 = sqrt( (c-x)^2)$
4. $d_2 = sqrt( (x+c)^2 + y^2)$
You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges. | 328 | 1,170 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.871631 |
http://bootmath.com/intuition-behind-normal-subgroups.html | 1,532,290,785,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593586.54/warc/CC-MAIN-20180722194125-20180722214125-00639.warc.gz | 51,424,260 | 6,479 | # Intuition behind normal subgroups
I’ve studied quite a bit of group theory recently, but I’m still not able to grok why normal subgroups are so important, to the extent that theorems like $(G/H)/(K/H)\approx G/K$ don’t hold unless $K$ is normal, or that short exact sequences $1\to N \stackrel{f}{\to}G\stackrel{g}{\to}H\to1$ only holds when $N$ is normal.
Is there a fundamental feature of the structure of normal subgroups that makes things that only apply to normal subgroups crop up so profusely in group theory?
I’m looking here for something a bit more than “$gN=Ng$, so it acts nicely”.
#### Solutions Collecting From Web of "Intuition behind normal subgroups"
For any subgroup $H$ of $G$, you can always define an equivalence relation on $G$ given by
$$g_1 \equiv g_2 \iff g_1g_2^{-1} \in H$$
This lets you define a quotient of $G$ by $H$ by looking at equivalence classes. This works perfectly well, and gives you a set of cosets, which we denote
$$G/H = \{[g] = gH \mid g \in G\}$$
However, note that while we started talking about groups, we have now ended up with a set, which has less structure! (There is still some extra structure, e.g. the action of $G$ on the quotient)
We would like to define a natural group structure on this quotient, simply so that we don’t end up in a completely different category. How should this new group structure behave? Well, it seems natural to ask that
$$[g * h] = [g] *_{new} [h]$$
so that the map $G \to G/H$ would be a homomorphism (this is, in this context, what I mean by “natural”). So what would this mean? Let’s write it out:
$$(gh)H = [g * h] = [g]*_{new}[h] = (gH)(hH)$$
If you work out what these sets are, then you can see that this equation can only be true if we have that $hH = Hh$ for every $h \in G$. But this is exactly the condition that $H$ is normal.
The short answer: $H$ being normal is exactly the condition that we require so that we can put a compatible group structure on the quotient set $G/H$.
Just to expand slightly on Simon Rose’s comment
$H$ being normal is exactly the condition that we require so that we can put a compatible group structure on the quotient set $G/H$.
Suppose for each $x, y \in G$ there is $g \in G$ such that $(x H) ( y H) = g H$, that is, the product of any two left cosets of $H$ is also a left coset.
Take $x = y^{-1}$, so that $1 = y^{-1} 1 y H \in (y^{-1} H) (y H) = g H$, and thus $g H = H$. Thus for every $h \in H$ and $y \in G$ we have
$$y^{-1} h y = y^{-1} h y 1 \in (y^{-1} H) ( y H) = H,$$
that is, $H$ is normal.
The normal subgroups of $G$ are all the sets, which appear as kernel of group-homomorphisms $G \rightarrow H$.
Subgroups are the sets, which appear as images of group-homomorphism $H \rightarrow G$. | 813 | 2,741 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-30 | latest | en | 0.912564 |
https://www.learningincontext.com/bridges-4th/chapt02.php | 1,726,159,343,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00727.warc.gz | 797,955,085 | 5,949 | # Bridges to Algebra and Geometry - 4th Edition
## Chapter 2: Working with Data
Some links are repeated for use with more than one lesson.
### This link applies to the entire chapter.
http://www.cvgs.k12.va.us:81/DIGSTATS/
This site has discussions and activities that involve many different types of statistics and graphs.
### 2.1 Measures of Central Tendency
http://www.robertniles.com/stats/
This site has definitions and examples of many statistical terms.
### 2.2 Line Plots and Frequency Tables
http://www.explorelearning.com/index.cfm?method=cResource.dspResourcesForCourse&CourseID=266
This site has examples and problems for the students to solve that relate to many ways of analyzing and displaying data.
### 2.3 Box-and-Whisker Plots
http://www.mathwarehouse.com/charts/box-and-whisker-plot-maker.php#boxwhiskergraph
This site includes a script to generate a box-and-whisker plot from given data.
This site explains how to create box and whisker plots.
http://www.explorelearning.com/index.cfm?method=cResource.dspResourcesForCourse&CourseID=266
This site has examples and problems for the students to solve that relate to many ways of analyzing and displaying data.
### 2.4 Bar Graphs and Histograms.
This site explains how to bar graphs and histograms.
This site has practice problems for the students using bar graphs.
http://www.explorelearning.com/index.cfm?method=cResource.dspResourcesForCourse&CourseID=266
This site has examples and problems for the students to solve that relate to many ways of analyzing and displaying data.
### 2.5 Line Graphs
http://www.mathsisfun.com/data/data-graph.php
Create and format line graphs, and other types, using your data values.
http://www.explorelearning.com/index.cfm?method=cResource.dspResourcesForCourse&CourseID=266
This site has examples and problems for the students to solve that relate to many ways of analyzing and displaying data.
### 2.6 Scatter Plots
http://www.mathsisfun.com/data/scatter-xy-plots.html
This page shows examples of scatter plots of real data, attempts to draw a line of best fit, introduces the term “correlation,” and includes some practice problems.
http://office.microsoft.com/en-us/excel-help/creating-xy-scatter-and-line-charts-HA001054840.aspx
This page highlights the differences between scatter plots and line graphs as produced in Microsoft Excel. | 548 | 2,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.820184 |
http://www.chegg.com/homework-help/questions-and-answers/suppose-force-208-n-pushes-ontwo-boxes-unknown-mass-know-accelerationof-boxes-108-m-s2and--q128530 | 1,394,465,027,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010855566/warc/CC-MAIN-20140305091415-00048-ip-10-183-142-35.ec2.internal.warc.gz | 272,823,375 | 6,942 | # Question
0 pts ended
Suppose a force of 20.8 N pushes ontwo boxes of unknown mass. We know, however, that the accelerationof the boxes is 1.08 m/s2and the contact force has a magnitude of 4.59 N.
(a) Find the mass of the larger box.
kg
(b) Find the mass of the smaller box.
kg | 82 | 279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2014-10 | latest | en | 0.879536 |
https://www.atheistsforhumanrights.org/what-is-the-difference-between-circumference-radius-and-diameter/ | 1,695,840,543,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510319.87/warc/CC-MAIN-20230927171156-20230927201156-00592.warc.gz | 705,924,469 | 16,827 | What is the difference between circumference radius and diameter?
What is the difference between circumference radius and diameter?
Radius, Diameter and Circumference The Radius is the distance from the center outwards. The Diameter goes straight across the circle, through the center. The Circumference is the distance once around the circle.
Is diameter same as circumference?
Circumference is the length of one complete ‘lap’ around a circle, and diameter is the length of the line segment that cuts a circle in half. Think of circumference as an outer measurement and diameter as an inner measurement of the circle!
Is diameter bigger than circumference?
The diameter is ALWAYS approximately 3 times smaller than the circumference! Or to put it another way, the circumference is approximately 3 times bigger than the diameter.
Is circumference twice diameter?
The circumference is twice as large, and the area is four times as large. Doubling the diameter doubles the circumference, holding four times as much area as before.
What is the circumference of a 12 inch diameter circle?
To find the circumference of a circle, you will use the formula C=2⋅π⋅r ; therefore the circumference C is 2⋅π⋅6≈38 inches.
Is the circumference 6 times the radius?
What is the circumference of the circle? The circumference is equal to 2 times 5 times the radius. So it’s going to be equal to 2 times pi times the radius, times 3 meters, which is equal to 6 meters times pi or 6 pi meters. 6 pi meters.
Is circumference divided by 2 diameter?
Divide the circumference by π, or 3.14 for an estimation. The result is the circle’s diameter, 3.18 centimeters. Divide the diameter by 2.
Is the circumference of a circle 3x the diameter?
The circumference of a circle is equal to Pi π times the diameter d .
How do you find circumference with radius?
A. In order to calculate a circle’s circumference, we need to know either its diameter or its radius. We then use the appropriate value in this equation: C=2πr (where “r” represents radius, of course).
How do I convert diameter to circumference?
Or you can use the circle’s diameter:
1. Multiply the diameter by π, or 3.14.
2. The result is the circle’s circumference. | 494 | 2,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.906528 |
http://mathhelpforum.com/algebra/142099-out-my-depth-can-anyone-help-print.html | 1,498,369,795,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320438.53/warc/CC-MAIN-20170625050430-20170625070430-00254.warc.gz | 250,693,598 | 4,736 | # out of my depth. Can anyone help?
• Apr 29th 2010, 03:18 AM
pipoldchap
out of my depth. Can anyone help?
Hello,
I struggle with math and have a problem. I'm making a spreadsheet for work so I can easily convert one figure to another. The thing is, the two columns of data I have (24 pairs of numbers) don't run parallel on a graph. I have some learning difficulties which irritatingly enough mean I can kind of see what I'm doing, but can't find the missing piece. grr! (Doh)
The data is as follows:
1 = 3
2 = 5
3 = 7
4 = 9
5 =12
6=14
7=16
8=18
9=21
10=23
11=25
12=27
13=30
14=32
15=34
16=36
17=39
18=41
19=43
20=45
21=48
22=50
23=52
24=54
I had a conversion kind of figured out at 0.33 or 4.33 (i think) but it doesn't work for all the values. I have a GCSE/Highschool C and I haven't used it much in 15 years. I put it to the forum and cross my fingers.
Thanks,
Dave
• Apr 29th 2010, 04:01 AM
masters
Hi pipoldchap,
I'm not sure what your question is, but the data is pretty much linear.
I ran a linear regression model on a TI-84 calculator and came up with this linear function:
y = 2.243478261x + .4565217391
with a correlation coefficient of .999
Does this help at all?
• Apr 29th 2010, 05:14 AM
Another thing you could do is....
Since the numbers increase by 2 until every fourth one,
when the increase is 3
and the numbers would be 2n+1 if the increase by 3 was not part of the pattern,
then you need the "floor" of
$2n+1+\frac{n-1}{4}$
which we can find using
$\frac{8n+4+n-1}{4}=\frac{9n+3}{4}$
To get the floor of this, just divide and discard the decimal part.
• May 1st 2010, 10:35 AM
pipoldchap
Wow. I'm further in further over my head than I thought. (Surprised)
Basically, it's a conversion table for work and I'm trying to put the info into a spreadsheet so that I can just enter the details and find out if i'm over quantity or not. I know What I want to do, I'm just damned if I know HOW! (Bigsmile)
I'll try entering them into a spreadsheet and see if it likes me today. Office and I have a tempestuous relationship at best. Thanks guys. I'm really glad this stuff makes sense to you.
(Bow) I doff my cap to you both.
Thanks again,
Dave
• May 1st 2010, 10:54 AM
pipoldchap
Hi guys,
I entered the following equation in the spreadsheet (a2 represents the input data) and it worked for the first two but not the third onwards. I didn't understand this floor thing though...
=(((2*A2)+1)+((A2-1)/4))
Also, I entered the y value as a conversion factor and it didn't work either. I'm sure I'm just entering it wrong.
• May 1st 2010, 12:02 PM
Quote:
Originally Posted by pipoldchap
Hi guys,
I entered the following equation in the spreadsheet (a2 represents the input data) and it worked for the first two but not the third onwards. I didn't understand this floor thing though...
=(((2*A2)+1)+((A2-1)/4))
Also, I entered the y value as a conversion factor and it didn't work either. I'm sure I'm just entering it wrong.
Hi pipoldchap,
the idea with the formula is that you must discard the fraction or the part
behind the decimal point.
If you can do that then the formula converts the values.
The "floor" means the natural number below your value
or the value itself if the answer is a natural number.
For example...
n=1
$\frac{9n+3}{4}=\frac{12}{4}=3$
n=2
$\frac{18+3}{4}=\frac{21}{4}=5.25\ or\ 5\frac{1}{4}$
n=3
$\frac{27+3}{4}=\frac{30}{4}=7.5$
n=4
$\frac{36+3}{4}=\frac{39}{4}=9.75$
n=5
$\frac{45+3}{4}=\frac{48}{4}=12$
and so on....
• May 1st 2010, 01:27 PM
awkward
Quote:
Originally Posted by pipoldchap
Hello,
I struggle with math and have a problem. I'm making a spreadsheet for work so I can easily convert one figure to another. The thing is, the two columns of data I have (24 pairs of numbers) don't run parallel on a graph. I have some learning difficulties which irritatingly enough mean I can kind of see what I'm doing, but can't find the missing piece. grr! (Doh)
The data is as follows:
1 = 3
2 = 5
3 = 7
4 = 9
5 =12
6=14
7=16
8=18
9=21
10=23
11=25
12=27
13=30
14=32
15=34
16=36
17=39
18=41
19=43
20=45
21=48
22=50
23=52
24=54
I had a conversion kind of figured out at 0.33 or 4.33 (i think) but it doesn't work for all the values. I have a GCSE/Highschool C and I haven't used it much in 15 years. I put it to the forum and cross my fingers.
Thanks,
Dave
Hi Dave,
I think you need less math and more spreadsheet. I'm assuming you are using Microsoft Excel.
Suppose your 24 pairs of numbers are in cells A1:B24. If you have a number from 1 to 24 in cell C26, for example, that you want mapped to its corresponding value from column B, then enter this formula in the cell where you want the value:
=VLOOKUP(C26, $A$1:$C$24, 2)
If it's possible you value might not be one of the numbers 1-24 and you want to get an error message in that case, enter
=VLOOKUP(C26, $A$1:$C$24, 2,false)
• May 11th 2010, 10:16 AM
pipoldchap
BRILLIANT!!!!!!!!!!!!!! Thank you so much. I didn't reliase that was possible on a spreadsheet. Witchcraft! (Rofl)
That's just what I needed it to do. I really appreciate you guys' help. Thank you!
dave | 1,607 | 5,110 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-26 | longest | en | 0.943821 |
https://jp.mathworks.com/matlabcentral/answers/1904690-how-to-plot-the-volume-shared-by-two-3d-shapes | 1,679,381,810,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943637.3/warc/CC-MAIN-20230321064400-20230321094400-00070.warc.gz | 401,029,303 | 28,522 | # How to plot the volume shared by two 3D shapes
1 ビュー (過去 30 日間)
Michael Boyte 2023 年 2 月 1 日
コメント済み: William Rose 2023 年 2 月 1 日
I have two identical spheres offset from each other by a distance less than their radiuses. I would like to plot just the volume that is shared by these two spheres. How can I do this? Also, I need to be able to scale the solution to include more spheres.
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### 採用された回答
William Rose 2023 年 2 月 1 日
patch() and fill3() are options worth considering.
With either approach, you will have define the lens-shaped intersection of 2 spheres as a polygon. When there are just 2 spheres, there are formulas for the radius of the circular edge of the lens, the height of each "half" of the lens, and the location of the center of the lens. See here for those formulas. The formulas above assume one sphere is at the orign and the other is centered on the x-axis. If your spheres do not meet this assumption, you would do a translation and rotation first, so that the shere centers are at the origin and on the x-axis, then apply the formulas from Wolfram, then compute the polygons, the translate and rotate back, then plot with patch() or fill3().
If you add additional spheres, the intersection will no longer be a lens shape with well-defined formulas. In this case, I would use numerical methods to find a set of points that are inside all three spheres, then I would wrap a convex hull around it with convhull().
##### 1 件のコメント表示非表示 なし
William Rose 2023 年 2 月 1 日
The attached script plots the intersection of two spheres. Specify R1, R2, and d, in the code, then run the code. R1 and R2 are sphere radii. Sphere 1 is centered at the origin. Sphere 2 is centered at (d,0,0).
Screenshot 1 shows the result when R1=R2=1, and d=1.5. In this case the intersection is a symmetric lens of total thickness 0.5, with center at x=0.75. I used the rotate tool in the figure window to drag the object to the orientation shown here.
Screenshot 2 shows the result when R1=1, R2=10, and d=10. In this case, the interesection is roughly the +x hemisphere of sphere 1.
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Translated by | 611 | 2,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-14 | longest | en | 0.803243 |
https://1library.net/document/z31d1vmy-finite-elements-based-deslauriers-dubuc-wavelets-propagation-problems.html | 1,660,731,617,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572898.29/warc/CC-MAIN-20220817092402-20220817122402-00374.warc.gz | 98,465,230 | 41,337 | # Finite Elements Based on Deslauriers Dubuc Wavelets for Wave Propagation Problems
## Full text
(1)
http://dx.doi.org/10.4236/am.2016.714128
## Wavelets for Wave Propagation Problems
Rodrigo Bird Burgos1, Marco Antonio Cetale Santos2
1Department of Structures and Foundations, Rio de Janeiro State University (UERJ), Rio de Janeiro, Brazil 2Department of Geology and Geophysics, Fluminense Federal University (UFF), Niterói, Brazil
Received 9 June 2016; accepted 14 August 2016; published 17 August 2016
### Abstract
This paper presents the formulation of finite elements based on Deslauriers-Dubuc interpolating scaling functions, also known as Interpolets, for their use in wave propagation modeling. Unlike other wavelet families like Daubechies, Interpolets possess rational filter coefficients, are smooth, symmetric and therefore more suitable for use in numerical methods. Expressions for stiffness and mass matrices are developed based on connection coefficients, which are inner products of basis functions and their derivatives. An example in 1-D was formulated using Central Difference and Newmark schemes for time differentiation. Encouraging results were obtained even for large time steps. Results obtained in 2-D are compared with the standard Finite Difference Method for validation.
### Keywords
Wavelets, Interpolets, Deslauriers-Dubuc, Wavelet Finite Element Method, Wave Propagation
### 1. Introduction
Among the numerous techniques available for the solution of partial differential equations (like the one that de-scribes wave propagation phenomena), the Finite Difference Method (FDM) [1] is by far the most employed one, being used frequently as a standard for the validation of new methods, especially in geophysical applications. As a disadvantage, the FDM is known for requiring excessive refining of the model discretization. Irregular grids can be used, increasing the complexity of the implementation and its computational cost.
(2)
problems is still limited, due to the large number of frequencies that are excited in the system [2]. This work proposes adaptations in the FEM which can improve its performance in wave propagation problems.
The conventional formulation of the FEM uses polynomials for interpolating the displacement within the elements (shape functions). This work proposes the use of shape functions based on wavelet scaling functions in order to obtain satisfactory results with less refined meshes and bigger time steps than the traditional FEM and FDM would require without affecting stability and convergence.
Wavelets functions have several properties that are quite useful for representing solutions of partial differen-tial equations (PDEs), such as orthogonality, compact support and exact representation of polynomials of a cer-tain degree. These characteristics allow the efficient and stable calculation of functions with high gradients or singularities at different levels of resolution [3].
A complete basis of wavelets can be generated through dilation and translation of a mother scaling function. Although many applications use only the wavelet filter coefficients of the multiresolution analysis, there are some which explicitly require the values of the basis functions and their derivatives, such as the Wavelet Finite Element Method (WFEM) [4]-[8]. Compactly supported wavelets have a finite number of derivatives which can be highly oscillatory. This makes the numerical evaluation of integrals of their inner products difficult and unst-able. Those integrals are called connection coefficients and they appear naturally in a Finite Element scheme. Due to some properties of wavelet functions, these coefficients can be obtained by solving an eigenvalue prob-lem using filter coefficients and a normalization rule.
Working with dyadically refined grids, Deslauriers and Dubuc [9] obtained a new family of wavelets with in-terpolating properties, later called Interpolets. Their filter coefficients are obtained from an autocorrelation of the Daubechies’ coefficients. As a consequence, interpolets are symmetric, which is especially interesting in numerical analysis. The use of interpolets instead of Daubechies’ wavelets considerably improves the elements accuracy [10].
The formulation of an Interpolet-based finite element system is demonstrated for a one-dimensional wave propagation problem. Newmark’s algorithm for direct integration was tried as an alternative to the Central Dif-ference Method in order to allow bigger time steps. A homogeneous example was formulated in order to vali-date the approach in two-dimensional applications.
### 2. Interpolating Scaling Functions
Multi-resolution analysis using orthogonal, compactly supported wavelets has become increasingly popular in numerical simulation. Wavelets are localized in space, which allows the analysis of local variations of the prob-lem at various levels of resolution.
All the mathematical foundation for wavelet functions is based on the algorithms for Daubechies wavelets [11]. Wavelet basis are composed of two kinds of functions: scaling functions (φ) and wavelet functions (ψ). The two combined form a complete Hilbert space of square integrable functions. The spaces generated by scal-ing and wavelet functions are complementary and both are based on the same mother function. In the followscal-ing expressions, known as the two-scale relation, ak are the scaling function filter coefficients and N is the
Daube-chies wavelet order:
1
0 1
1 0
2 ,
1 2 .
N k k
N k
N k k
x a x k
x a x k
ϕ ϕ
ψ ϕ
= −
− − =
= −
= − −
### ∑
(1)
The term interpolet was first used by Donoho [12] to designate wavelets with interpolating characteristics. The basic characteristics of interpolating wavelets require that the mother scaling function satisfies the following condition [13]:
### ( )
0,
1, 0
,
0, 0
k
k
k k
k
ϕ =δ = = ∈ ≠
. (2)
Any function f(x) can be represented as a linear combination of the basis functions at level of resolution j:
2
,
j
k k j k
k k
f x cϕ x k cϕ x
∈ ∈
=
− =
### ∑
(3)
Evaluating the function at a dyadic grid point x = 2−jr yields:
2 j k
k r k r
k k
fr cϕ r k cδ c
∈ ∈
=
− =
### ∑
=
. (4)
This characteristic is very interesting for numerical applications, since a function can be represented as a weighted sum in which the coefficients are evaluations of the same function at target points on a dyadic mesh:
2 j 2j
k
f x fk ϕ x k
=
### ∑
. (5)
The Deslauriers-Dubuc interpolating function of order N is given by an autocorrelation of the Daubechies’ scaling filter coefficients (hm) of the same order (i.e. N/2 vanishing moments, given in [11]). Its support is given
by [1 − N N − 1], it has even symmetry and is capable of representing polynomials of order up to N − 1. Inter-polets satisfy the same requirements as other wavelets, especially the two-scale relation (using a support adjust-ment):
1
* 1 1 *
0
2 ,
.
N k k N N
k m m k
m
x a x k
a h h
ϕ − ϕ
= − −
− =
= −
=
### ∑
(6)
Table 1 shows the typical values of filter coefficients for interpolets of orders 4, 6 and 8.
Figure 1shows the interpolet IN6 (autocorrelation of DB6). Its symmetry and interpolating properties are
Table 1. Filter coefficients for Deslauriers-Dubuc interpolets (ak = ak).
k
Function Order
N = 4 N = 6 N = 8
0 1 1 1
1 9/16 150/256 150/256
2 0 0 0
3 −1/16 −25/256 −245/2048
4 0 0
5 3/256 49/2048
6 0
7 −5/2048
Figure 1. Interpolet IN6 scaling function.
-5 -4 -3 -2 -1 0 1 2 3 4 5
-0.2 0 0.2 0.4 0.6 0.8 1 1.2
### ϕ
(4)
evident. It is capable of representing a 5-degree polynomial. Its Daubechies “mother” of same order would only represent a 2-degree polynomial.
### 3. Wave Propagation
The partial differential equation (PDE) which rules the wave propagation is:
2
2 2
### ( )
2
, ,
2 u x t u x t
x t
λ+ µ ∂ =ρ∂
∂ ∂ , (7)
where, u is the horizontal displacement, ρ is the density, t is the time and λ and μ are the Lamé parameters of the medium. Applying Hamilton’s Principle and using the FEM, the PDE can be rewritten at a specific time t as a system of linear equations, which in matrix form is:
2
2
t t
t
+ =
0
u
M K u . (8)
In the given expression, M represents the mass matrix and K is the stiffness matrix of the model. These FEM system global matrices are assembled from the individual contributions of the local matrices of each element.
As in the FDM, it becomes necessary to solve the system of equations at discrete time intervals. There are several effective direct integration methods, among which the most intuitive one is the Central Difference Me-thod, given by:
### ( )
2
2 2
2
t t t t t
t
t t
+∆ −∆
− +
u u u u
. (9)
Substituting the expression of the acceleration obtained by the Central Difference Method and solving for the next time step ttu gives:
### ( )
2 1
2
t t t t t t
t
+∆ = −∆ − ∆
u u u M K u. (10)
The result of the matrix operation M−1K is obtained directly if M is diagonal. In this case, one can generate a unique local matrix at element level which contains both mass and stiffness information. It is known that the di-agonal mass matrix can be used instead of the one generated from shape functions (consistent matrix) without introducing significant errors in the system. Nevertheless, in this work consistent mass matrices were used and a global matrix X was assembled using mass and stiffness contributions to the system. Stability of the Central Difference Method is conditioned to the choice of the time step, whose upper bound is obtained from a genera-lized eigenvalue problem.
2
2
### )
max max
,
2 . t
ω ω
ω
− = → − =
∆ =
0 0
K M u X I u
(11)
3.1. Newmark Method
The constant-average-acceleration method (a special case of the Newmark Method) consists in an uncondition-ally stable scheme which can be summarized by the following expression (as given in [2]):
2 2
4 4 4
4
t
t t t t
t
t t
+∆
+= +
u
I X u u u . (12)
This method requires the calculation of velocity and acceleration at all time steps and the inversion of the left hand side operator. Despite this, the Newmark Method is unconditionally stable. Therefore, significantly bigger time steps can be used.
3.2. Application to 2-D Problems
(5)
### )
2 2 2
2 2 2
, , 2 , , , ,
u x z t u x z t u x z t
t x z
λ µ
ρ
= + +
∂ ∂ , (13)
which can still be solved using 1-D finite elements by applying a scheme similar to what is commonly done in the FDM:
2
### )
2
t t t t t t t
t
+∆ = −∆ − ∆ T
u u u X u uZ . (14)
In the expression above, a second stiffness matrix is introduced as a differential operator in the second space dimension. The displacement is also represented as a matrix, instead of the usual vector representation of the FEM. Matrices X e ZT perform the differentiation along x and z directions respectively. These matrices contain both mass and stiffness information. Assuming spatial sampling of Nx × Nz points, X and Z are of size Nx × Nx
and Nz × Nz, respectively. This represents an improvement over traditional FEM in terms of computational effort;
usually, the FEM requires a displacement vector of length Nx × Nz and both global stiffness and mass matrices
must have a size of (Nx × Nz) ×(Nx × Nz), operating over the whole system.
The application of the Newmark Method in this 2-D scheme is not as simple as in 1-D, since the elements are still one-dimensional and there are right and left matrix multiplications involved. The following expression is an adaptation of the Newmark Method to the 2-D implementation proposed:
2 2
4 4 4
4
t
t t t t t t
t t t +∆ +∆ += +
u T
I X u u u u Z . (15)
In this expression, ttu appears in both sides of the equation, thus requiring the application of an iterative method for its calculation. This procedure not only increases the computational effort significantly but can also cause instability for large time steps.
### 4. Finite Element Formulation
Assuming that displacement u is approximated by a series of interpolating scale functions, the following may be written:
1
2
N k
k N
u ξ α ϕ ξ k
= −
=
### ∑
− . (16)
Stiffness and mass matrices can be obtained by solving the wave propagation PDE using the FEM. Dimen-sionless coordinates (ξ) within the interval [0 1] are used in wavelet space, which leads to the subsequent ex-pressions:
### ( ) ( )
1 1,1
, 0 ,
1 0,0
, 0 ,
2 d 2 ,
d .
i j i j i j
i j i j i j
k
m
λ µ ϕ ξ ϕ ξ ξ λ µ
ρ ϕ ξ ϕ ξ ξ ρ
′ ′
= + = + Γ
= = Γ
### ∫
(17)
The so-called connection coefficients Γ appear in the expressions above. Wavelet dilation and translation properties allow the calculation of connection coefficients to be summarized by the solution of an eigenvalue problem based only on filter coefficients [14].
### )
1 2 1 2 1 2
1 2
1 1
, 1 ,
, 2 2 2 1 2 1 ,
1 1
1
, 2 2 2 1 2 1
2 , 1 , 2 . N N
d d d d d d
i j k i l j k i l j k l
k N l N
d d
i j k i l j k i l j
k l
a a a a
S a a a a
− − + − − − − + − + = − = − + − − − − + − +
Γ = + Γ
= +
1 2 0
d ,d
S I Γ = (18)
(6)
( )
( )
### )
1 1 2 2 1 2 1 2 , ,
1 2 1 2
! , ! ! , ! ! ! .
! ! 1
d
m d m
i d
n d n
j d d m n i j i j m
x i x i
m d
n
x j x j
n d
m n i j
m d n d m n d d
ϕ ϕ − − = − − = − − Γ = − − + − − +
### ∑∑
(19)
The dimensionless expressions for the stiffness k and mass m matrices are in wavelet space and need to be transformed to physical space, using a transformation matrix T obtained by evaluating the wavelet basis at the element node coordinates using Equation (16). It can be noticed that some terms related to the length (L) of the element emerge from coordinate changes.
( )
### )
, 1 2 2
1 1 1 1 , 2 3 1 , .
i j N
T T j N i N L L ϕ = − − − − − − = + − + − T
k = T k T
m = T m T
(20)
### 5. Examples
To validate the formulation of the interpolet-based finite element, a 1-D example was tested. It consists in ap-plying a forced displacement at the free end of a pinned, unit length rod (Figure 2). The propagation was mod-eled by the FDM using 601 points and Δt = 0.1 ms. For the FEM based on the IN6 interpolet, the discretization was performed with 91 points and Δt = 1 ms. The rod’s middle point time response for both methods is shown in Figure 3, which shows that the FEM result is acceptably close to that of the FDM, especially considering that FEM generates over 60 times less data per simulated second.
In order to validate the 2-D formulation, a simple example was proposed by analyzing the propagation in a 2 km × 2 km model with a constant velocity of 3000 m/s. Finite Element model was sampled every 6.17 m in both directions with a time step Δt of 0.46 ms, obtained by the generalized eigenvalue problem already described. Non-integer values appear due to the 11 degrees-of-freedom present in the element implemented with IN6. A Newmark scheme was also implemented using the same spatial sampling and Δt = 0.6 ms. The results were compared to the FDM using 5 m spacing in both directions and Δt = 0.2 ms. Finite Element mesh has 325 × 325 points and Finite Difference mesh has 401 × 401. Results are shown inFigure 4. Even with a less refined mesh and a more than twice bigger time step, the IN6 element was able to identify characteristic peaks of the wave propagation. The results of the proposed adaptation to the Newmark Method were very similar to the Central Difference ones using an even bigger time step.
### 6. Conclusions
This work presented the formulation and validation of an interpolet-based finite element. Newmark’s method for time discretization appears promising, although its application in 2-D problems with 1-D elements remains a challenge. The main improvement in the presented formulation was the possibility of using a bigger time step
Figure 2.Unit length rod under forced displacement.
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Figure 3.FEM results using IN6 interpolet and newmark method (Δt = 1 ms) compared to FDM (Δt = 0.1 ms) for a one-dimensional wave propagation.
(a) (b)
(c) (d)
Figure 4. Propagation snapshots at time t = 0.3 s for 2-D example using: (a) FEM and Central Difference Method, (b) FEM and Newmark Method, (c) FDM and Central Difference Method; (d) comparison of am-plitudes along the indicated segment at depth 1000 m.
0 0.5 1 1.5 2 2.5 3 3.5
-0.5 0 0.5 1
Time (s)
D
is
pl
ac
em
ent
Displacement at midpoint FDM Central
FEM Newmark
0 500 1000 1500 2000
0 500 1000 1500 2000
Offset (m)
D
ept
h (
m
)
FEM Central Difference
0 500 1000 1500 2000
0 500 1000 1500 2000
Offset (m)
D
ept
h (
m
)
FEM Newmark
0 500 1000 1500 2000
0 500 1000 1500 2000
Offset (m)
D
ept
h (
m
)
FDM Central Difference
1400 1600 1800
-0.5 0 0.5 1
Offset (m)
A
m
pl
it
ude
z = 1000m
(8)
than the one required by the FDM. In future works, models with greater complexity will be analyzed and differ-ent families of wavelets will be explored.
The number of DOFs present in interpolets leads to non-dyadic divisions within the element, requiring an ad-ditional approximation which introduces errors in the transformation matrix. The inaccuracy in some values can be attributed mostly to these approximation errors.
As done in the traditional FEM, all matrices involved can be stored and operated in a sparse form, since most of their components are null, thus saving computer resources.
### Acknowledgements
Authors would like to thank CNPq and FAPERJ for their financial support.
### References
[1] Kelly, K.R., Ward, R.W., Treitel, S. and Alford, R.M. (1976) Synthetic Seismograms: A Finite-Difference Approach.
Geophysics, 41, 2-27. http://dx.doi.org/10.1190/1.1440605
[2] Bathe, K.J. and Wilson, E.L. (1976) Numerical Methods in Finite Element Analysis. Prentice Hall, Upper Saddle Riv-er.
[3] Qian, S and Weiss, J. (1992) Wavelets and the Numerical Solution of Partial Differential Equations. Journal of Com-putational Physics, 106, 155-175. http://dx.doi.org/10.1006/jcph.1993.1100
[4] Ma, J., Xue, J., Yang, S., He, Z. (2003) A Study of the Construction and Application of a Daubechies Wavelet-Based Beam Element. Finite Elements in Analysis and Design, 39, 965-975.
http://dx.doi.org/10.1016/S0168-874X(02)00141-5
[5] Chen, X., Yang, S., Ma, J. and He, Z. (2004) The Construction of Wavelet Finite Element and Its Application. Finite Elements in Analysis and Design, 40, 541-554. http://dx.doi.org/10.1016/S0168-874X(03)00077-5
[6] Han, J.G., Ren, W.X. and Huang, Y. (2005) A Multivariable Wavelet-Based Finite Element Method and Its Applica-tion to Thick Plates. Finite Elements in Analysis and Design, 41, 821-833. http://dx.doi.org/10.1016/j.finel.2004.11.001
[7] He, Y., Chen, X., Xiang, J. and He, Z. (2008) Multiresolution Analysis for Finite Element Method Using Interpolating Wavelet and Lifting Scheme. Communications in Numerical Methods in Engineering, 24, 1045-1066.
http://dx.doi.org/10.1002/cnm.1011
[8] Burgos, R.B., Cetale Santos, M.A. and Silva, R.R. (2013) Deslauriers-Dubuc Interpolating Wavelet Beam Finite Ele-ment. Finite Elements in Analysis and Design, 75, 71-77. http://dx.doi.org/10.1016/j.finel.2013.07.004
[9] Deslauriers, G. and Dubuc, S. (1989) Symmetric Iterative Interpolation Processes. Constructive Approximation, 5, 49- 68. http://dx.doi.org/10.1007/BF01889598
[10] Burgos, R.B., Cetale Santos, M.A. and Silva, R.R. (2015) Analysis of Beams and Thin Plates Using the Wavelet-Galerkin Method. International Journal of Engineering and Technology (IJET), 7, 261-266.
http://dx.doi.org/10.7763/IJET.2015.V7.802
[11] Daubechies, I. (1988) Orthonormal Bases of Compactly Supported Wavelets. Communications in Pure and Applied Mathematics, 41, 909-996. http://dx.doi.org/10.1002/cpa.3160410705
[12] Donoho, L.D. (1992) Interpolating Wavelet Transforms.
http://statweb.stanford.edu/~donoho/Reports/1992/interpol.pdf
[13] Shi, Z., Kouri, D.J., Wei, G.W. and Hoffman, D.K. (1999) Generalized Symmetric Interpolating Wavelets. Computer Physics Communications, 119, 194-218. http://dx.doi.org/10.1016/S0010-4655(99)00185-X
[14] Zhou, X. and Zhang, W. (1998) The Evaluation of Connection Coefficients on an Interval. Communications in Nonli-near Science & Numerical Simulation, 3, 252-255. http://dx.doi.org/10.1016/S1007-5704(98)90044-2
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Logic, Truth Tables and Switching Circuits Challenge
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Learn about the link between logical arguments and electronic circuits. Investigate the logical connectives by making and testing your own circuits and fill in the blanks in truth tables to record. . . . | 1,928 | 8,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-39 | latest | en | 0.881677 |
https://in.mathworks.com/matlabcentral/answers/309889-how-does-one-save-a-struct-type-to-an-individual-cell-inside-a-table | 1,670,227,042,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711013.11/warc/CC-MAIN-20221205064509-20221205094509-00803.warc.gz | 343,044,320 | 27,145 | # How does one save a struct type to an individual cell inside a table?
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Rodrigo Ceballos on 31 Oct 2016
Commented: Peter Perkins on 2 Nov 2016
I'm trying to make a function with the following signature:
function [test_result,model] = test_and_save(trainer,trainer_params,test,test_params,notes)
Which is a wrapper to training an arbitrary machine learning algorithm and using an arbitrary test on it, saving all the information about this operation to a table. The trainer_params and test_params are structures of arbitrary size and shape. So far, all my efforts fail in the process of trying to convert these arrays into something the table will accept because its checking for consistency among the different structures in the different rows. How do I avoid this?
Thanks!
Rodrigo
##### 2 CommentsShowHide 1 older comment
Peter Perkins on 2 Nov 2016
Following up on Walters comments, if the structures are all different, including their fields, the only way you're going to be able to store them in a table is inside a cell array, one struct array per cell. For example
>> s1 = struct('a',1,'b',2);
>> s2 = struct('c',3,'d',4);
>> t = table([1;2],{s1;s2})
t =
2×2 table array
Var1 Var2
____ ____________
1 [1×1 struct]
2 [1×1 struct]
>> t.Var2
ans =
2×1 cell array
[1×1 struct]
[1×1 struct]
As opposed to
>> s1 = struct('a',1,'b',2);
>> s2 = struct('a',3,'b',4);
>> t = table([1;2],[s1;s2])
t =
2×2 table array
Var1 Var2
____ ____________
1 [1x1 struct]
2 [1x1 struct]
>> t.Var2
ans =
2×1 struct array with fields:
a
b
The table display looks the same in both cases, but that's just a display limitation. In the first case, as can be seen, Var2 is a 2x1 cell array, each cell containing a scalar struct. In the second case, Var2 is a 2x1 struct.
It's hard to tell from what you've said what you have or what you want.
Marc Jakobi on 31 Oct 2016
Have you tried converting the struct to a cell?
doc struct2cell | 545 | 1,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.850761 |
http://openstudy.com/updates/4f5ff3f3e4b038e07180728e | 1,448,826,290,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398459333.27/warc/CC-MAIN-20151124205419-00311-ip-10-71-132-137.ec2.internal.warc.gz | 170,305,833 | 9,852 | ## cameron_martin 3 years ago Solve. One point for showing the steps to solve each equation and one point for the correct solution. 4(x - 6) = -2(x - 3)
1. razor99
what did you mean
2. Flying_Dutchman
4(x-6) = -2(x-3) | First you have to multiply 4x - 24 = -2x + 6 | Then you have to group the similar terms 6x = 30 | Solve for x x = 30/6 | Simplify x = 5
3. cameron_martin
thanksss | 135 | 388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2015-48 | longest | en | 0.8495 |
https://www.lesswrong.com/posts/Hg3e9oYjfvbqfbBdF/guesstimate-why-and-how-to-use-it | 1,708,830,993,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00406.warc.gz | 876,172,211 | 19,709 | # 6
This is the first in a series of guides for software tools to help with collaborative truth-seeking. The introduction to the sequence is available here
Guesstimate is a quantitative model-building tool built to focus on uncertainty. The interface works the same as google sheets or excel; a grid of cells where data can be entered. Cells can then be referenced by other cells for calculations.
This post is a text tutorial for guesstimate. The landing page for guesstimate also gives a rough idea of the point of the tool, and is a reasonable place to get a feel for if this is the right tool for the job.
# When can Guesstimate be Useful?
• Estimate Return on Investment for an EA intervention
• Predict attendance of your EA group over the next year
• Decide whether you should get a surgery or not
• Pick which job you should take
• Decide whether you should date somebody
• Pick which microwave you should buy
• Predict personal & business finances
# Text Guide
## Why use guesstimate?
Guesstimate has two key benefits over google sheets/excel:
1. Each cell can have uncertainty
2. Cell references are displayed visually (as well as in the formula of a cell)
This means that guesstimate is highly appropriate for fermi estimateslogic models, and any other format where you have uncertainty about some or all of the inputs, or of their effect sizes on the outputs. The visual nature of guesstimate also makes it easy to share; it’s easy to see how a model works.
If you prefer hands-on learning, a simple model to demonstrate the basic features is available here
## Basic Guesstimate Functions
Double click anywhere empty in the grid to create a new cell. A cell has two attributes, a name and a value. A name should just help people (and you, in six months!) understand your model. The value can be of several types:
• Single value (7000)
• 90% Confidence interval (6 to 7[1])
• Specific dataset (9.8, 17, 12, 34, -2, 113)
## Formulae
Cells’ value can also reference other cells like other spreadsheets. Typing ‘=’ at the beginning of the value field tells guesstimate the value is a formula; you can then click any other cell to create a reference to it, and perform basic numerical operations or ‘if’ statements on them.
While editing a formula, cells also show a three-letter reference at their top right, which you can type to manually refer to another cell.
For a more in-depth guide, see the official documentation.
# Personal Experience
Personally, I find guesstimate models pretty intuitive to build & to interpret. I expect people who dislike visual information and/or have a lot of experience coding will find Squiggle more intuitive.
Guesstimate also gets quite clunky and difficult to understand with larger models with many connections, as the screen becomes more clustered/it becomes difficult to decide where boxes belong (see e.g. this model). This can be mitigated by breaking the problem up into sensible modules/subsections that are given their own area or entire model.
# Try it Yourself!
With each post, I'm going to encourage you to give it a go!
For guesstimate, I've picked a specific question to get things going; in future I'll try to encourage finding examples closer to your life.
What is the total value of all the goods in a supermarket?
Spend 15-30 minutes making a guesstimate model to answer this question! Share your model in the comments as well as your experience of using guesstimate.
We're also running a short event in the EA GatherTown at 6pm GMT today to use Guesstimate, if you'd like to bring your model or make it during the session, or if you have any questions!
Tomorrow: Visualisation of Probability Mass, a neat little tool for changing visual intuitions into numbers, and vice versa. We'll be having a GatherTown event at the same time tomorrow!
New Comment | 867 | 3,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-10 | latest | en | 0.878549 |
https://softwareengineering.stackexchange.com/questions/275566/function-that-would-output-all-distinct-values-in-an-array | 1,713,473,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817239.30/warc/CC-MAIN-20240418191007-20240418221007-00061.warc.gz | 489,611,484 | 41,299 | # Function that would output all distinct values in an array
I want to create a function that outputs the distinct values in a given array. For example, in the following sequence 2 2 1 1 5 2 , the distinct digits are 2 1 5. What I've done till now is:
``````void Distinct(int a[], int b[], int n){
int i; //n is the number of elements for a and b
int k=0; //a is the original array
int ac;
b[0] = a[0];
for (i = 0; i <= n; ++i){
if (a[i] != b[k])
{
k+=1;
ac = a[i];
b[k] =ac; //b is where I want to put distinct digits
}
}
for ( i = 0; i < k; ++i)
{
printf("%d ", b[i]);
}
}
``````
But this is not enough, because what I've done is for the case when the numbers repeat one after another, I mean: 111122221 , I will get 121, and 1 at the end again.Can you please help me write a function that would consider this case, with repeating numbers scattered in the array ? Thank you.
• sort the array first. Mar 7, 2015 at 18:13
• Ok, so then I can use the same function I wrote ?, because it's done for sorted arrays. Mar 7, 2015 at 18:16
• Feb 8, 2021 at 3:57
There are many, many ways of doing this, and you haven't really specified whether you're after something that's simple to code or optimally performant or if you have an upper bound on the possible numbers or any other restriction that might help us tell which answer you need.
But your sample code implies you want the solution in a "low-level" language where we're dealing with arrays made up of bytes rather than some higher-level construct like a list/tuple/vector that makes it trivial to do things like remove elements from the middle. Therefore, I would suggest you run a sort algorithm on the input array (if you're in C++, that would be std::sort), then run your Distinct() function on that sorted array to create a correctly unique-ified array.
• Note that Snowman's answer is entirely correct if you want to use one of those "higher-level constructs". Mar 8, 2015 at 15:22
We have a name for a data structure that can only contain an element once: a set.
I have done this and tasks like it many times. Here is some pseudocode:
``````Array input = { ... };
Set output = ...;
for (int i : input) { | 591 | 2,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.901319 |
https://essayblazers.com/wasson-company-reported-the-following-year-end-information/ | 1,679,810,578,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945433.92/warc/CC-MAIN-20230326044821-20230326074821-00769.warc.gz | 282,285,047 | 14,596 | # Wasson company reported the following year-end information
1.Wasson Company reported the following year-end information:Beginning work in process inventory\$35000 Beginning raw materials inventory18000 Ending work in process inventory 38000 Ending raw materials inventory 15000 Raw materials purchased 56000 Direct labor180000 Manufacturing overhead 120000How much is Wasson’s total cost of work in process for the year?a608000b863000c860000d8980002.Edmiston Company reported the following year-end information:beginning work in process inventory \$80000;cost of goods manufactured 780000;beginning finished goods inventory 50000;ending work in process inventory 70,000;and ending finished goods inventory 40000. How much is Edmiston’s cost of goods sold for the year?a780000b790000c770000d8000003.Using the following information, compute the direct materials usedRaw materials inventory, Jan.1 \$20000Raw materials inventory, Dec.31 40000Work in process, Jan.1 18000Work in process, Dec.31 12000Finished goods, Jan.1 40000Finished goods, Dec.31 32000Raw materials purchases(NNN) NNN-NNNNDirect labor 560000Factory utilities 150000Indirect labor 50000Factory depreciation 400000Operating expenses 420000a1460000b1420000c1400000d13800004.Assuming that the cost of goods manufactured is \$2960000 compute the cost of goods sold using the following informationRaw materials inventory, Jan.1 \$30000Raw materials inventory, Dec.31 60000Work in process, Jan.1 27000Work in process, Dec.31 18000Finished goods, Jan.1 60000Finished goods, Dec.31 48000Raw materials purchases(NNN) NNN-NNNNDirect labor 690000Factory utilities 225000Indirect labor 75000Factory depreciation 500000Operating expenses 630000a2969000b2912000c2948000d29720005.Samson Company reported total manufaturing costs of \$300000,manufacturing overhead totaling \$52000 and direct materials totaling \$64000. How much is direct labor cost?a.Cannot be determined from the information providedb416000c416000d1840006.Given the following data for Mehring Company,compute (A) total manufacturing costs and (B) cost of goods manufactured:Direct materials used \$230000 Beginning work in process 30000 Direct labor 150000 Ending work in process 15000 Manufacturing overhead 225000 Beginning finished goods 38000 Operating expenses 263000 Ending finished goods 23000(A) (B)a\$590000 620000b605000 590000c605000 620000d620000 6350007.Gulick Company developed the following data for the current year:Beginning work in process inventory \$160,000Direct materials used 96,000Actual overhead 192,000Overhead applied 144,000Cost of goods manufactured 176,000Total manufacturing costs 480,0007-1 Gulick Company’s direct labor cost for the year isa.48,000b.240,000c.144,000d.192,0007-2 Gulick Company’s ending work in process inventory isa.464,000b.320,000c.304,000d.144,0008.Hayward Manufacturing Company eveloped the following data:Beginning work in process inventory 450000Direct materials used 350000Actual overhead 550000Overhead applied 400000Cost of goods manufactured 600000Ending work in process 750000Hayward Manufacturing Company’s total manufacturing costs for the period isa950,000b900,000c650,000d.cannot be determined from the data provided9.Greer Company developed the following data for the current year:Beginning work in process inventory \$102000Direct materials used 156,000Actual overhead 132,000Overhead applied 138,000Cost of goods manufactured 675,000Total manufacturing costs 642,0009-1 How much is Greer Company’s direct labor cost for the year?a.381,000b.450,000c.348,000d.246,0009-2 How much is Greer Company’s ending work in process inventory for the year?a.69,000b.363,000c.63,000d.279,00010.Chmelar Manucfacturing Company developed the following data:Beginning work in process inventory \$80,000Direct materials used 480,000Actual overhead 560,000Overhead applied 540,000Cost of goods manufactured 1,280,000Ending work in process 60,000How much are total manufacturing cost for the period?a.1,580,000b.1,260,000c.1,100,000d.1,220,00011.Maisley Company decided to analyze certain costs for June of the current year.Units started into production equaled 28,000 and ending work in process equaled 4,000.With no beginning work in process inventory,how much is the conversion cost per unit if ending work in process was 25% complete and total conversion costs equaled 105,000?a.\$3.30b.3.75c.4.20d.2.10
12.materials costs of 800,000 and conversion costs of 1,020,000 were charged to a processing department in the month of Sep. Materials are added at the beginning of the process,while conversion costs are incurred uniformly throughout the process.There were no units in beginning work in process,20,000 units were started into production in Sep. and there were 5,000 units in ending work in process that were 40%complete at the end of Sep.What was the total amount of manufacturing costs assigned to those units that were completed and transferred out of the process in Sep.?a1,500,000b2,000,000c1,606,000d1,365,00013.Materials costs of 800,000and conversion costs of 1,020,000were charged to processing department in the month of Sep. Materials are added at the beginning of the process,while conversion costs are incurred uniformly throughout the process.There were no units in beginning work in process, 20,000units were started into production in Sep. and there were 5,000units in ending work in process that were 40%complete at the end of Sep.What was the total amount of manufacturing costs assigned to the 5,000 units in the ending work in process?a455,000b500,000c320,000d200,00014.Mayer Company has recently tried to improve it’s analysis for its manufacturing process.Units started into production equaled 18,000 and ending work in process equaled 1,200units.Mayer had no beginning work in process inventory.Conversion costs are applied equally throughout production,and materials are applied at the beginning of the process.How much is the materials cost per unit if ending work in process was 25%completed and total materials costs equaled 90,000?a5.00b5.27c20.00d4.6915.For the Assembly Department,unit materials cost is \$8 and unit conversion cost is \$12.If there are 10,000units in ending work in process 75% complete as to converstion costs, the costs to be assigned to the inventory area.200,000b.170,000c.150,000d.180,000
16.The total costs accounted for in production cost report equal thea.cost of units completed and transferred out onlyb.cost of units started into productionc.cost of units completed and transferred out plus the cost of ending work in processd.cost of beginning work in process plus the cost of units completed and transferred out17.In a production cost report,which one of the following sections is not shown under costs?a.Unit costsb.Costs to be accounted forc.Costs during the periodd.Units accounted for
18.A company incurs 2,400,000 of overhead each year in three departments:Processing,Packaging,and Testing.The company performs 800 processing transactions,200,000 packaging transactions, and 2,000 tests per year in producing 400,000 drums of Oil and 600,000 drums of Sludge.The following data are available:(Department) (Expected Use of Driver) (Cost)Processing 800 1,000,000Packaging 200,000 1,000,000Testing 2,000 400,000Production information for the two products is as follows:(Department) (Oil Expected Use of Driver) (Sludge Expected Use of Driver)Processing 300 500Packaging 120,000 80,000Testing 160,000 40018-1 The amount of overhead assigned to Oil is:a1,200,000b1,295,000c1,105,000d920,00018-2 The amount of overhead assigned to Sludge isa1,200,000b1,105,000c1,295,000d920,00019.Which of the following statements is false?a.ABC can weaken control over overhead costsb.Under ABC,companies can trace many overhead costs directly to activitiesc.ABC allows some indirect costs to be identified as direct costsd.managers become more aware of their responsibility to control the activities that generated costs20.Which of the following is a value-added activity?a.Inventory storageb.Machiningc.Building maintenanced.Bookkeeping21.Which of the following is value-added activity?a.Inventroy controlb.Inspectionsc.Packagingd.Repair of machines22.Which of the following is a non-value-added activity?a.Inventory controlb.Machiningc.Assemblyd.Painting23.Which of the following is a non-value-added activity?a.paintingb.Finishingc.Packagingd.Building maintenance24.A non-value-added activity in service enterprise isa.providing legal researchb.delivering packagesc.consultingd.bookkeeping25.A division sold 100,000 calculators during 2013:Sales 2,000,000Variable Costs:Materials 380,000Order processing 150,000Billing labor 110,000Selling expenses 60,000Total variable costs 700,000Fixed costs 1,000,000How much is the contribution margin per unit?a.2b.7c.17d.1326.At the break-even point of 2,000units, variable costs are 110,000 and fixed costs are 64,000.How much is the selling price per unit?a87b23c32d Not enough information27.The following information is available for Wade Corp:Sales 550,000Total fixed expenses 150,000Cost of goods sold 390,000Total variable expenses 360,000A CVP income statement would reporta.gross profit of 160,000b.contribution margin of 400,000c.gross profit of 190,000d.contribution margin of 190,000
28.Which is the true statement?a.In a CVP income Statement, costs and expenses are classified only by functionb.The CVP income statement is prepared for both internal and external usec.The CVp income statement shows contribution margin instead of gross profitd.In a traditional income statement,costs and expenses are classified as either variable or fixed29.Variable costs are 60% of the unit selling price.The contribution margin ratio is 40%The contribution margin per unit is \$500The fixed costs are \$300,000Which of the following does not express the break-even point?a.300,000+.60X=Xb.300,000+.40X=Xc.300,000/\$500=Xd.300,000/.40=X30.A CVP graph does not include aa.variable cost lineb.fixed cost linec.sales lined.total cost line31.Boswell company reported the following information for the current year:Sales(50,000 units)\$1,000,000, direct materials and direct labor 500,000, other variable costs 50,000, and fixed costs 270,000.31-1 What is Boswell’s break-even point in units?a.24,546b.30,000c.38,334d.42,18831-2 What is Boswell’s contribution margin ratio?a.68%b.45c.32d.5532.Walters Corporation sells radios for %50 per unit.The fixed costs are 420,000 and the variable costs are 60% of the selling price.As a result of new automated equipment,it is anticipated that fixed costs will increase by 100,000 and variable costs will be 50% of the selling price.The new break-even point in units isa21,000b20,800c20,600d16,80033.Cunningham,Inc. sells MP3 players for \$60 each.Variable costs are \$40 per unit, and fixed costs total \$90,000.What sales are needed by Cunningham to break even?a.120,000b.225,000c.270,000d.360,00034.What is the key factor in determining sales mix if company has limited resources?a.Contribution margin per unit of limited resourceb.The amount of fixed costs per unitc.Total contribution margind.The cost of limited resources35. (Oven Hours Required) (Contribution Margin Per Unit)Muffins 0.2 \$3Coffee Cakes 0.3 \$4The company has oven capacity of 1,200hours.How much will contribution margin be if it produces only the most profitable product?a.12,000b.16,000c.18,000d.24,00036.Curtis Corporation’s contribution margin is \$20 per unit for Product A and \$24 for Product B.Product A requires 2 machine hours and Product B requires 4 machine hours.How much is the contribution margin per unit of limited resource of reach product?A Ba.\$10 \$6b.\$10 \$6.66c.\$8 \$6d.\$8 \$6.6637.Reducing reliance on human workers and instead investing heavily in computers and online technology willa.reduce fixed costs and increase variable costsb.reduce variable costs and increase fixed costsc.have no effect on the relative proportion of fixed and variable costsd.make the company less susceptible to economic swings38.Cost structure refers to the relative proportion ofa.selling expenses versus adminstrative expensesb.selling and administrative expenses versus costs of goos soldc.contribution margin versus salesd.none of the above39.Mercantile Corporation has sales of 2,000,000,variable costs of 1,100,000,and fixed costs of 750,000.39-1)mercantile’s degree of operating leverage isa1.22b1.47c1.20d6.0039-2)margin of safety ratio isa .08b .17c .20d .8340.Which of the following statements is not true?a.Operating leverage refers to the extent to which company’s net income reacts to given change in salesb.Companies that have higher fixed costs relative to variable costs have higher operating leverage.c.when a company’s sales revenue is increasing,high operating leverage is good b/c it means profits will increase rapidlyd.When a company’s sales revenue is decreasing,high operating leverage is good b/c it means profits will decrease at slower pace than revenues decrease42.Nielson Corp.sells its product for 8,800per unit.Variable costs per unit are:manufacturing,\$4,800and selling and administrative,\$100.Fixed costs are:24,000 manufacturing overhead,and 32,000selling and administrative.There was no beginning inventory at 1/1/12.Production was 20 units per year in 2012-14.Sales was 20units in 2012,16units in 2013, and 24units in 2014.42-1)Income under absorption costing for 2014 isa26,400b31,200c32,800d37,60042-2)Income under variable costing for 2013 isa6,400b11,200c12,800d17,60042-3)Income under variable costing for 2014 isa26,400b31,200c32,800d37,60042-4)For the 3years 2012-14a.absorption costing income exceeds variable costing income by 8,000b.absorption costing income equals variable costing incomec.variable costing income exceeds absorption costing income by 8,000d.absorption costing income may be greater than,equal to,or less than variable costing income,depending on the situation
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https://www.physicsforums.com/threads/principles-of-mathematical-analysis-by-walter-rudin.665171/ | 1,726,480,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00453.warc.gz | 868,215,754 | 33,623 | # Principles of Mathematical Analysis by Walter Rudin
• Greg Bernhardt
In summary, "Principles of Mathematical Analysis" by Rudin is a concise and elegant undergraduate level textbook on real analysis. It is not suitable for first exposure to the subject, but rather serves as a reference for well-prepared readers. The book covers topics such as the real and complex number systems, basic topology, numerical sequences and series, continuity, differentiation, the Riemann-Stieltjes integral, functions of several variables, integration of differential forms, and the Lebesgue theory. While the book has been praised for its elegance, it may be difficult to understand for some readers. There may also be a possible error in the proof of the least upper bound property and field axioms. Overall, "Principles of
## For those who have used this book
• Total voters
35
Code:
[LIST]
[*] Preface[*] The Real and Complex Number Systems
[LIST]
[*] Introduction
[*] Ordered Sets
[*] Fields
[*] The Real Field
[*] The Extended Real Number System
[*] The Complex Field
[*] Euclidean Spaces
[*] Appendix
[*] Exercises
[/LIST][*] Basic Topology
[LIST]
[*] Finite, Countable, and Uncountable Sets
[*] Metric Spaces
[*] Compact Sets
[*] Perfect Sets
[*] Connected Sets
[*] Exercises
[/LIST][*] Numerical Sequences and Series
[LIST]
[*] Convergent Sequences
[*] Subsequences
[*] Cauchy Sequences
[*] Upper and Lower Limits
[*] Some Special Sequences
[*] Series
[*] Series of Nonnegative Terms
[*] The Number e
[*] The Root and Ratio Tests
[*] Power Series
[*] Summation by Parts
[*] Absolute Convergence
[*] Addition and Multiplication of Series
[*] Rearrangements
[*] Exercises
[/LIST][*] Continuity
[LIST]
[*] Limits of Functions
[*] Continuous Functions
[*] Continuity and Compactness
[*] Continuity and Connectedness
[*] Discontinuities
[*] Monotonic Functions
[*] Infinite Limits and Limits at Infinity
[*] Exercises
[/LIST][*] Differentiation
[LIST]
[*] The Derivative of a Real Function
[*] Mean Value Theorems
[*] The Continuity of Derivatives
[*] L'Hospital's Rule
[*] Derivatives of Higher Order
[*] Taylor's Theorem
[*] Differentiation of Vector-valued Functions
[*] Exercises
[/LIST][*] The Riemann-Stieltjes Integral
[LIST]
[*] Definition and Existence of the Integral
[*] Properties of the Integral
[*] Integration and Differentiation
[*] Integration of Vector-valued Functions
[*] Rectifiable Curves
[*] Exercises
[/LIST][*] Sequences and Series of Functions.
[LIST]
[*] Discussion of Main Problem
[*] Uniform Convergence
[*] Uniform Convergence and Continuity
[*] Uniform Convergence and Integration
[*] Uniform Convergence and Differentiation
[*] Equicontinuous Families of Functions
[*] The Stone-Weierstrass Theorem
[*] Exercises
[/LIST][*] Some Special Functions
[LIST]
[*] Power Series
[*] The Exponential and Logarithmic Functions
[*] The Trigonometric Functions
[*] The Algebraic Completeness of the Complex Field
[*] Fourier Series
[*] The Gamma Function
[*] Exercises
[/LIST][*] Functions of Several Variables
[LIST]
[*] Linear Transformations
[*] Differentiation
[*] The Contraction Principle
[*] The Inverse Function Theorem
[*] The Implicit Function Theorem
[*] The Rank Theorem
[*] Determinants
[*] Derivatives of Higher Order
[*] Differentiation of Integrals
[*] Exercises
[/LIST][*] Integration of Differential Forms
[LIST]
[*] Integration
[*] Primitive Mappings
[*] Partitions of Unity
[*] Change of Variables
[*] Differential Forms
[*] Simplexes and Chains
[*] Stokes' Theorem
[*] Closed Forms and Exact Forms
[*] Vector Analysis
[*] Exercises
[/LIST][*] The Lebesgue Theory
[LIST]
[*] Set Functions
[*] Construction of the Lebesgue Measure
[*] Measure Spaces
[*] Measurable Functions
[*] Simple Functions
[*] Integration
[*] Comparison with the Riemann Integral
[*] Integration of Complex Functions
[*] Functions of Class $\mathcal{L}^2$
[*] Exercises
[/LIST][*] Bibliography[*] List of Special Symbols[*] Index
[/LIST]
• jbunniii
For the well prepared reader, this is a beautifully clear treatment of the main topics of undergraduate real analysis. Yes, it is terse. Yes, the proofs are often slick and require the reader to fill in some nontrivial gaps. No, it doesn't spend much time motivating the concepts. It is not the best book for a first exposure to real analysis - that honor belongs to Spivak's "Calculus." But don't kid yourself that you have really mastered undergraduate analysis if you can't read Rudin and appreciate its elegance. It also serves as a nice, clean, uncluttered reference which few graduate students would regret having on their shelves.
• micromass
This is a wonderful book iff you can handle it. Do not use Rudin as your first exposure to analysis, it will be a horrible experience. However, if you already completed a Spivak level text, then Rudin will be a wonderful experience. It contains many gems and many challenging problems. Personally, I find his approach to differential forms and Lebesgue integration quite weird though. I think there are many books that cover it better than him. But the rest of the book is extremely elegant and nice.
Last edited by a moderator:
(If this is not the right place for this discussion, please excuse my post, and I will happily have the mods move it to the right place.)
Possible erratum:
On p. 19 of the 3rd edition, in the proof that the real numbers have the least upper bound property and the field axioms, Rudin states:
"If $q \in \alpha$, then $-q \notin \beta$. Therefore $\beta \neq \mathbb{Q}$" but this is not true.
For example, say our cut $\alpha$ was bounded above by -2 (not inclusive). Then, $-3$ would be in $\alpha$, but $3$ would be in $\beta$.
Instead, the following reasoning is valid:
Choose $p \in \mathbb{Q}$ such that $-p \in \alpha$.
Then, since for any $r > 0$, $-p-r < -p$, and so $-p-r \in \alpha$ by (II).
Since $\beta$ consists of all those $p$ such that $\exists \, \beta \in \mathbb{R}$ such that $-p-r \notin \alpha$, we conclude $p \notin \beta$.
Well, now I'm confused because I feel as though my counterexample is correct, but I've just proven that if $-p \in \alpha$, then $p \notin \beta$, which is clearly the same as $p \in \alpha$, then $-p \notin \beta$. (Or is it?)
I think most people know this is not my favorite math book. I taught out of it one year in the senior math major course and the students suffered tying to understand it. As for myself, in every math book I like there is some topic that I have learned and that I remember that book for.
E.g. from Courant I have always remembered the explanation of the correspondence between real numbers and points on a line, and the clear criteria for series convergence, and the footnote where formula for the sum of the first n kth powers is derived, and the principle of the point of accumulation,...
From Van der Waerden I recall a succinct account of the definition of the real numbers, as a quotient ring of the ring of Cauchy sequences of rationals, modded out by the maximal ideal of null sequences.
From Lang Analysis I I remember the crystal clear account of the Riemann integral as characterized simply by being monotone and additive over intervals and the area of a rectangle as base times height,..
By contrast, there is virtually no topic that I can recall learning from Rudin's book, and some topics like differential forms mentioned by micromass, simply should not be learned here. Well maybe I learned a little something about Dedekind cuts. But to me most explanations here are simply not memorable. He succeeds in impressing me how smart and clever he is, but not in instructing me.
To try to be somewhat fair, it is entirely possible that my problem with Rudin is that I am a geometer and he is an analyst (his discussion of differential forms takes out all the geometry), and analysis has always seemed to me like a book of seven seals, but i think think that is partly because of books like this one.
On the other hand if you like this book, take heart, maybe you are a future analyst!
Last edited:
Thank you for your reply, mathwonk. Do you have any analysis book in particular you would recommend? My analysis class is using the book by Protter and Morrey, which has a strong first half but is weak on multivariable. I was thinking about using Stricharz? I also have the Dover books by Rosenlicht and Komolgorov.
Well, if you like and can learn from Rudin, then it may be for you, but it is not my cup of tea. I probably should hesitate more to recommend against it for others, because I think every analysis prof I asked has liked Rudin. So for analysis minded folks this may be it.
I'm not sure what part of analysis you ask about, but presumably the same parts that are in Rudin, i.e. rigorous calculus. I like the books by Wendell Fleming, and Michael Spivak, as well as maybe Williamson, Crowell and Trotter for calculus. Fleming even does a nice job on Lebesgue integration, and Spivak does a fine job on differential forms. WC&T is a several variable calc book. And I have liked almost every book by Sterling K Berberian, as he is a superb expositor who tries actually to teach the reader. I also liked Lang's Analysis I, now called something else, like Undergraduate Analysis (and more expensive).
http://www.abebooks.com/servlet/SearchResults?an=serge+lang&sts=t&tn=analysis+I
Rosenlicht and Kolmogorov are also famously excellent expositors. As always, go to the library and peruse until one strikes you as clear.
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here is an example of what i dislike in Rudin compared to say Courant. On page 65, Rudin begins the ratio and root tests with a brutal statement of a theorem, no motivation at all, no explanation. What is he doing? You have to study the proof to find out.
Courant on page 377 explains that "All such considerations of convergence depend on comparison of the series in question with a second series,...,chosen in such a way that its convergence can be readily tested."
Then on the next page he begins with the simplest case, comparison with a geometric series. Two slightly different methods of making the comparison lead to the root and ratio tests, which are thus seen clearly as special cases of a simple general principle.
Rudin's arguments are essentially the same but there is no helpful discussion in words to make it memorable. I.e. Rudin plunges into the details with no preliminary statement of what the idea is behind the argument to come. For me, once I understand the idea I can provide the details myself. It is harder to go backwards from the details to reconstruct what the idea was, although indeed it is there hidden under the argument.I would compare it to a magician explaining a trick by saying: first my beautiful assistant comes out and distracts your attention by displaying something while I reach under my arm for a compressed bouquet of flowers which I then expand before your eyes. Or else he just performs the trick and you are left to wonder what happened.
But everyone has a different learning style. Some people like Rudin very much.
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Thank you for being so thorough. I think I will try to use some combination of Lang, Apostol, Spivak and Courant, and then move on to Rudin and eventually Berberian.
remember what i actually said: don't take my advice on exactly which book to use, take my advice to look at them yourself, and then make your own choice. give yourself some credit, you can make an informed decision, if you inform yourself first. enjoy!
mathwonk said:
here is an example of what i dislike in Rudin compared to say Courant. On page 65, Rudin begins the ratio and root tests with a brutal statement of a theorem, no motivation at all, no explanation. What is he doing? You have to study the proof to find out.
Courant on page 377 explains that "All such considerations of convergence depend on comparison of the series in question with a second series,...,chosen in such a way that its convergence can be readily tested."
Then on the next page he begins with the simplest case, comparison with a geometric series. Two slightly different methods of making the comparison lead to the root and ratio tests, which are thus seen clearly as special cases of a simple general principle.
Rudin's arguments are essentially the same but there is no helpful discussion in words to make it memorable. I.e. Rudin plunges into the details with no preliminary statement of what the idea is behind the argument to come. For me, once I understand the idea I can provide the details myself. It is harder to go backwards from the details to reconstruct what the idea was, although indeed it is there hidden under the argument.
I would compare it to a magician explaining a trick by saying: first my beautiful assistant comes out and distracts your attention by displaying something while I reach under my arm for a compressed bouquet of flowers which I then expand before your eyes. Or else he just performs the trick and you are left to wonder what happened.
But everyone has a different learning style. Some people like Rudin very much.
Read Rudin's treatment of L'Hôpital's/Bernoulli rule and you will swiftly discover that his main intention behind this book was not pedagogical.
I didn't have any exposure to analysis when I started reading Rudin, but it was at least somewhat readable. I don't think it's necessary to take a calculus course with epsilon-delta proofs before taking (though that would be very helpful), but having some experience with proofs was important. When I first read it, I found it to be very dense and I often had to draw pictures to have any idea of what is going on. On the second pass, I found it was very concise and contained a lot of elegant proofs. Also, the problems were a lot of fun to work on. However, I was still frustrated that a lot of steps that were non-trivial to me were omitted as "obvious". As for the content itself, I think most of the criticism relates to the last couple of chapters. I'm not really qualified to comment on this as I haven't read them, but I think most places don't use (this) Rudin for courses relating to the content of these chapters and only use the book for an intoductory analysis course.
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Here is a perspective of an electrical engineer who wanted to teach himself analysis a handful of years ago, mostly for fun. I don''t think my background is too unusual for an engineer that was not in an "applied math" branch of EE (many of whom I know took an analysis course based on Rudin). The closest thing I ever took to a "real" math course was a senior level applied complex analysis course from the math department, where I did have to do some proofs as part of each homework assignment, and did indeed appreciate the challenge. I first picked up Rudin because my wife has a BS in math so the book was already at home. Least resistance and all that. Anyway, it didn't take long before I gave up. I really did need more to help me on my way, and it can get frustrating to not understand the motivation or general approach of a proof before he simply reveals a proof.
I ended up learning elementary analysis from "analysis and an introduction to proof" by Lay, supplemented by "mathematical analysis: a straightforward approach" by Binmore and the first 7 chapters of "the way of analysis" by Strichartz. After that I could pick up Rudin and appreciate it much more, and can recognize how hard some of the problems are.
So I cringe when I see folks suggest Rudin for people first learning analysis on their own - without someone to help (or adequate preparation) many of us are not smart enough to handle it and will get discouraged. I do not know what the best books are for folks like me, I just know that the books I used were much preferred to Rudin. If I get the urge again sometime I may give it another try with my improved background, but then again maybe not. Strichartz seems much easier for me to read to understand the what/why/how of the subject, even if he does meander a bit.
jason
It might be helpful for someone somewhere to know that I didn't complete a book like Spivak before starting Rudin and it's going fine. On the other hand, I had built up math maturity from linear algebra and our differential equations course which, at a high point of abstraction proved Picard-Lindelof. Also I had seen delta epsilon proofs throughout the course, but we didn't have to do too many of them ourselves. I'm not sure how different this sort of preparation is compared to what one might have from Spivak, but when I started reading Spivak after I had finished this course I found most of the problems too straightforward and thus decided to skip ahead to Rudin.
About Rudin: I can honestly say this is the best book I've ever read and I'm only 80 pages in. His pace is relentless and the lack of details in his proofs is almost riduculous at points, but this is completely circumnavigated by trying (and hopefully succeeding most of the time) to prove the Theorems yourself. In this fashion you will build up a strong idea of what the underlying structure is, even if you can't prove it yourself, and thus will be much better prepared to understand what he is leaving out.
I would also point out that one of my favorite consequences of Rudin's notorious brevity is that one feels that one is making an enormous amount of progress by completing just 20 pages. Wheras with a book like Apostol's Calculus there are two essentially infinite volumes where completing 20 pages is a drop in the bucket! In short, it is much easier to stay motivated when reading Rudin.
Oh yeah, and the problems are very very good. Try to solve as many as you have time for.
Overall impression: This book is not for teaching analysis. This book is for teaching you how to think.
astromme said:
About Rudin: I can honestly say this is the best book I've ever read and I'm only 80 pages in. His pace is relentless and the lack of details in his proofs is almost riduculous at points, but this is completely circumnavigated by trying (and hopefully succeeding most of the time) to prove the Theorems yourself.
How do you circumnavigate around missing definitions? When the first (or second) chapter of a math book has missing definitions, what can one say?
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What do you consider to be missing definitions in Rudin? I think the comment above was specifically about missing steps in proofs.
@verty I was speaking specifically about missing steps in proofs. On the contrary with definitions, I have only thought twice that he didn't define something; both times when I looked back it turns out that the mistake was with me as I didn't read closely enough.
Does he define what "open relative to" means?
Yes. Page 35 on the third edition, remark 2.29: "... to be quite explicit, let us say that $E$ is open relative to $Y$ if to each $p \in E$ there is associated an $r> 0$ such that $q \in E$ whenever $d(p,q) < r$ and $q \in Y$."
I would also point out that in context it's quite clear what open relative to means.
astromme said:
Yes. Page 35 on the third edition, remark 2.29: "... to be quite explicit, let us say that $E$ is open relative to $Y$ if to each $p \in E$ there is associated an $r> 0$ such that $q \in E$ whenever $d(p,q) < r$ and $q \in Y$."
I would also point out that in context it's quite clear what open relative to means.
Ok right, he does actually define it. But while the other definitions in the vicinity are given a definition block prefaced by "Definition:", that definition is buried in a block of text, the fourth sentence in a paragraph of six sentences. One could easily read the first few sentences and skip the rest because it is giving an example that one might already find trivial. Then seeing that the theorems on the next page use this "open relative to" idea, looking at the definition blocks doesn't help.
On page 32, ten definitions are given in one block, including "open" and "closed". It certainly gives one the impression that all the relevant definitions will be given together in one block. Then early on page 35, there is a definition block with one definition. So one might certainly conclude that no further definitions appear on that page. Rudin, if he tried to hide that definition, could hardly do a better job.
Obviously this is a matter of opinion, because to you (Astromme) it was obvious in context. Each to his own, I suppose.
@Verty I agree; the definition should probably have been lumped in with others. And I also agree that this book isn't the best for everyone; I just think it's working well for me. Note that I haven't learned/taught/referenced from other books, so I'm not sure my opinion should be given much credence. It certainly is possible that other analysis books would be even better for me than Rudin's - but at the moment things seem to be going well so I'm going to stick with it.
It's a good book, and it's meant to be read in a linear fashion. Don't complain about it if you skip parts of it. Oh, and for the prerequisites you don't need any of that nonsense. If you understand the rational number system, the binomial theorem, and some previous knowledge of trig functions will help, you should be good. You do not need to have taken any calc before. Rudin wisely rigorously develops it all for you.
I think the hardest part of the book is actually the second chapter, which is on metric spaces. I would do my best to avoid metric spaces and basic topology from that book for the first time, especially about the compact sets. As soon as the definition of a compact set is given (in terms of open covers), Rudin swiftly proceeds to myriads of lemmas without any motivation on why we even care about these particular sets. In the end, we reach the important Heine-Borel theorem, which gives us some concrete examples of compact sets on a euclidean space (on R^n, compact <=> closed and bounded), but the entire discussion on sequential compactness (which, in my opinion, is the analysts' notion of compactness) is hidden in the exercises.
Of course, everybody who reads the book will eventually learn about sequential compactness sooner or later, but I just find it strange how everything seemed "rushed" in that chapter. Was it supposed to be a review on metric space? Still brings me the bitter memory of learning analysis for the first time...
That being said, once I dealt with that chapter (as well as all the nitty-gritty about sequences and series in the next chapter), the rest of the book (at least up to chapter 8---never read anything beyond) was quite a treat. I thought it had good treatment of basic calculus on R^1, and I also enjoyed the chapter on the sequence of functions. It is a good book overall, but I found that particular chapter to be quite a pain to deal with (unless, of course, if one has seen metric space elsewhere!).
I've been through Chapters 1 and 2 and I like this book indeed. But, honestly, I don't think I would stand for it if I did not have the exposure to analysis I have acquired previously in the marvelous Bartle's ''Elements of Real Analysis''. So if Baby Rudin is your entry point in analysis and you don't feel comfortable with it, try Bartle.
Mathispuretruth said:
I've been through Chapters 1 and 2 and I like this book indeed. But, honestly, I don't think I would stand for it if I did not have the exposure to analysis I have acquired previously in the marvelous Bartle's ''Elements of Real Analysis''. So if Baby Rudin is your entry point in analysis and you don't feel comfortable with it, try Bartle.
This is actually a fantastic thing do do with really any subject; read multiple books, even side by side. It's good to see a subject treated a little differently, and for newer students, its good to see how an easier book can be "translated" into more rigorous and terse forms this way. Also even better is trying to prove each theorem on your own without looking at the book (to the best of your ability), or with minimal hints and peeps. It makes it so much more fun, and helped me get through hard concepts.
## 1. What is the main focus of "Principles of Mathematical Analysis" by Walter Rudin?
The main focus of "Principles of Mathematical Analysis" is to introduce students to the fundamental concepts of mathematical analysis, including topics such as real numbers, sequences and series, continuity, differentiation, and integration.
## 2. Is "Principles of Mathematical Analysis" suitable for beginners?
"Principles of Mathematical Analysis" is typically used as a textbook for upper-level undergraduate or graduate courses, so it may not be suitable for complete beginners. However, it does not assume any prior knowledge of analysis, so students with a strong foundation in calculus may find it accessible.
## 3. What sets "Principles of Mathematical Analysis" apart from other analysis textbooks?
One of the main features that sets "Principles of Mathematical Analysis" apart is its rigorous and concise approach to presenting mathematical concepts. The book also includes numerous challenging exercises to help students develop their problem-solving skills.
## 4. Can "Principles of Mathematical Analysis" be used as a reference book?
While "Principles of Mathematical Analysis" is primarily used as a textbook, it can also serve as a valuable reference for those with a strong background in analysis. Its clear and concise presentation of concepts makes it a useful resource for reviewing or refreshing one's understanding of key topics.
## 5. Are there any prerequisites for using "Principles of Mathematical Analysis"?
A strong foundation in calculus and mathematical proof techniques is recommended before using "Principles of Mathematical Analysis." Familiarity with basic concepts such as limits, continuity, and differentiation will also be helpful in understanding the material.
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Almost all data visualizations are multivariate (i.e., they display more than one variable), but there are practical limits to the number of variables that a single graph can display. These limits vary depending on the approach that’s used. Three graphical approaches are currently available for displaying multiple variables:
1. Encode each variable using a different visual attribute
2. Encode every variable using the same visual attribute
3. Increase the number of variables using small multiples
#### Encode Each Variable Using a Different Visual Attribute
This first approach is the most common, and it works quite well, but it typically limits the number of variables that can be effectively displayed in a single graph to four. Here’s a simple example of this approach that displays only two variables:
One variable—time by month—is encoded as horizontal positions along the X axis and the other variable—sales in dollars—is encoded as vertical positions along the Y axis. In other words, this example uses two visual attributes to encode values, one per variable: 2-D horizontal position and 2-D vertical position.
Here’s another example, but this time four variables are on display:
The following four visual attributes have been used to encode the four variables: horizontal position along the X axis (patient percentage of cost), vertical position along the Y axis (per patient cost in U.S. dollars), bubble size (number of patients), and bubble color intensity (patient age). Could we include a fifth variable in this graph in a way that works for our brains? There are certainly several more visual attributes from which to choose, but would any of them work in this case? Unfortunately, due mostly to perceptual limitations, the answer is, “Not well.”
If you doubt this, use your imagination to consider the possibilities. Perhaps it occurred to you that, in addition to variation in color intensity, which in this case encodes patient ages, we could encode a new variable, such as patient racial group, using various hues of color. If we did this, color intensity would no longer work effectively because it is difficult to compare the varying intensities of different hues, and the variable that’s encoded using hue would suffer because it is no longer easy to group objects with the same hue when color intensity varies.
As an alternative, perhaps the bubbles, which are all circular in shape, could vary in shape to encode a fifth variable (e.g., circles, squares, triangles, etc.). The problem with this approach is that, whereas we can roughly compare the sizes of circles to one another or squares to one another or triangles to one another, we cannot do a good job of comparing the sizes of circles, squares, and triangles to one another. Differences in shape make differences in size difficult to discern.
Even with only four variables, we’re already pushing the limits of effectiveness in this graph. Notice how difficult it is to determine the color intensities of small bubbles and to compare them to other bubbles. Colors become difficult to discriminate when objects are tiny. The larger the object, the more color there is, which makes discrimination easier. As you continue to consider other visual attributes that might be used to encode a fifth variable, you’ll encounter problems with each.
You might be thinking that I’m ignoring a visual attribute that could easily be added to this bubble plot: positions along the Z axis. Actually, I’m avoiding the Z axis for a good reason. Turning this into a 3-D graph by adding a Z axis would make the variable that’s encoded along that axis incredibly difficult to read. This is because, contrary to the ease with which human perception discerns differences in 2-D position (either horizontal or vertical along a flat plane), our perception of depth is not very good. Adding a Z axis would force us to constantly rotate and tilt the graph to reorient the Z axis either horizontally or vertically in an effort to see where bubbles fall along the axis, which isn’t practical.
Visual perception and cognition impose firm limits on the number of variables that we can encode in a single graph when we’re using a different visual attribute for each. These limitations are tied to several factors:
1. Only a few visual attributes work well for encoding data in graphs.
2. Using some visual attributes eliminates the possibility of using certain other attributes in the same graph.
3. Working memory can only attend to three or at most four chunks of information at a time, so limited value is added by including more than four.
4. Increasing the number of visual attributes in a single graph beyond a certain number creates a cluttered appearance that undermines perception.
Let’s consider each of these limitations in turn.
Effective Visual Attributes
Beginning with the work of Jacques Bertin, author of Sémiologie Graphique (The Semiology of Graphics), in the 1960s, people have studied visual perception as it applies to data visualization. Bertin explored the opportunities and limitations that influence the use of various visual attributes for encoding data. Since Bertin’s seminal work, the best books on this topic have been written by Colin Ware: Information Visualization: Perception for Design and Visual Thinking for Design. Everyone working in the field of data visualization should read these books. Vendors developing data visualization products should definitely read these books, but it seems that, based on the ineffective features that most products exhibit, they rarely do.
All data visualizations have one thing in common: they encode data values graphically, using basic attributes of visual perception. Whenever we look at an object in the world, the visual representation that appears in our heads is constructed from a small set of basic visual attributes. These attributes are called preattentive attributes of visual perception, for they are processed in the visual cortex of the brain preattentively (i.e., prior to conscious awareness). Each of these attributes is perceived separately, but in parallel rather than serially, more rapidly than conscious perception. The speed and ease of preattentive perception is a big part of the reason why data visualization is so powerful when done properly.
Here’s a fairly comprehensive list of the preattentive attributes of visual perception that are potential candidates for encoding data in graphs, grouped into six categories:
Attributes of Position
• 2-D horizontal position (i.e., objects arranged along an X axis)
• 2-D vertical position (i.e., objects arranged along a Y axis)
• Stereoscopic depth (i.e., perception of the distances of objects from the viewer, which can be simulated graphically by arranging them along a Z axis)
Attributes of Size
• Line length (e.g., the length of a bar in a bar graph)
• Line width (e.g., the width of a line in a line graph)
• Area (i.e., the 2-D size of an object, such as the size of a circle)
• Volume (i.e., the 3-D size of an object, such as the size of a sphere)
Attributes of Form
• Line orientation (e.g., the slope of a line in a line graph)
• Simple shape (e.g., differences between circles, squares, and triangles)
• Angle (i.e., the angle created where two lines meet, such as the angles formed by slices in a pie chart at its center)
• Curvature (e.g., the degree to which a line is curved)
Attributes of Appearance
• Hue (e.g., red, green, blue, etc.)
• Color intensity (i.e., the degree to which the color of an object varies from light to dark, pale to saturated, or both)
• Transparency (i.e., the degree to which we can see through an object)
• Blur (i.e., the degree to which an object appears sharp or fuzzy along its edges)
• Texture (i.e., various patterns on the surface of an object such as the grain of wood or the smooth appearance of metal)
Attributes of Movement or Change
• Direction of motion (e.g., the direction in which bubbles move in an animated bubble plot)
• Speed of motion (e.g., varying speeds in the movement of bubbles in an animated bubble plot)
• Speed of flicker (i.e., the speed at which an object flickers on and off or from low to high intensity)
Attributes of Quantity
• Numerosity (i.e., our ability to recognize differences in quantity between one, two, or three objects)
• Added marks (i.e., the varying addition of another component to an object—it is either there or it isn’t—such as a border around a bubble in a bubble plot)
We can consider all 21 of these preattentive attributes of visual perception as candidates for encoding values in graphs, but only a few of them work well.
We perceive some preattentive visual attributes quantitatively. By this, I mean that we naturally perceive different expressions of the attribute as representing either greater or lesser values. For example, we perceive a long line as greater in value than a short line or a dark circle as greater in value than a light circle. We perceive each of the following attributes quantitatively:
• 2-D horizontal position (right is greater than left)
• 2-D vertical position (high is greater than low)
• Stereoscopic depth (farther away is greater than near, or vice versa)
• Line length (long is greater than short)
• Line width (thick is greater than thin)
• Area (large is greater than small)
• Volume (large is greater than small)
• Line orientation (steep is greater than shallow relative to a horizontal baseline)
• Angle (wide is greater than narrow)
• Curvature (curvy is greater than straight, or vice versa)
• Color intensity (dark or bright is greater than light or pale)
• Transparency (opaque is greater than transparent)
• Blur (fuzzy is greater than sharp, or vice versa)
• Speed of motion (fast is greater than slow)
• Speed of flicker (fast is greater than slow)
• Numerosity (more is greater than fewer)
Of these 16 attributes, only three work well for encoding quantitative data in graphs:
• 2-D horizontal position
• 2-D vertical position
• Line length
When I say that they work well, I mean that they can be perceived and compared to one another quickly, easily, and with a great deal of precision. Whereas these three attributes work well, all of the others provide only an approximate sense of value and a rough means of comparison. Of these, the following two tend to be most useful in graphs:
• Color intensity
• Area
Because color intensity and area only support approximate decoding and rough comparisons, however, we should only use them when neither 2-D horizontal position, nor 2-D vertical position, nor line length are available.
It doesn’t usually make sense to even consider numerosity because it’s severely limited. Numerosity refers to our preattentive ability to see differences between quantities of one, two, or three. We can also discern that more than three objects are greater than three, but we cannot decode the actual number preattentively. For example, if several clusters of dots appeared on a screen, we could recognize without conscious effort that some contained one dot, some two, some three, and some more than three. When clusters contained more than three dots, however, we could not tell how many there were without taking time to consciously count them. As such, numerosity is only useful for encoding values in a graph if quantities don’t exceed three. This situation happens too rarely to routinely consider numerosity as a candidate for encoding values in graphs.
The remaining quantitatively perceived attributes—stereoscopic depth, volume, line orientation, angle, curvature, blur, speed of motion, and speed of flicker—are rarely used in graphs, either because we perceive them less well than others or because they aren’t practical.
Some visual attributes can only be used to encode categorical variables, not quantitative variables. These include the following:
• Simple shape
• Hue
• Texture
The two that are most useful in graphs are hue and simple shape, in that order. Texture doesn’t work particularly well because, when texture patterns are applied to the surfaces of objects in graphs (typically by using crosshatching, etc.), they tend to create a visually cluttered appearance. An added mark could only be used to represent a binary variable (i.e., one with only two potential values, such as female and male), for the added mark is either there or it isn’t. It is also possible to use added marks that always appear in graphs by attaching them to the primary object that’s being used to encode values, such as by always displaying a border around bubbles in a bubble plot, and by applying one of the other attributes in the list above (e.g., color intensity) to the border to encode a quantitative variable.
So, where does this leave us? Even though all 21 of these preattentive attributes can potentially be used to encode variables in graphs, only a few work well. As it turns out, however, this is not the only reason why a single graph can only effectively display a limited number of variables. The number of variables that we encode in a single graph is also affected by the fact that 1) certain visual attributes cannot be combined effectively in a single graph, 2) working memory can only handle three or at most four variables at a time, and 3) too many visual attributes tend to produce visual clutter. We’ll consider those limitations next.
Effective Combinations of Visual Attributes
Some visual attributes can be combined in a single graph and some cannot. For example, 2-D horizontal position, 2-D vertical position, hue, and simple shape can work fairly well together in a scatter plot. On the other hand, as I’ve already pointed out, we cannot effectively combine hue and color intensity together in a single graph.
Another ineffective combination is the use of both line length and line width for separate variables. This is because length and width function as integral attributes. This means that, when they are combined, we perceive the result as area rather than as independent attributes of length and width. Imagine that we used the lengths of bars to encode one variable and the widths of bars to encode a second. We would preattentively perceive this combination as differences in the overall areas of bars, no longer independently as differences in the bars’ lengths and widths. Although we could not perceive length and width as separate variables preattentively, we could do so with conscious effort, but it would be much slower.
Here’s a list of the attributes that cannot be effectively combined:
• Line length and line width, because these attributes are integral
• Any attributes of color (e.g., hue and color intensity, hue and transparency, or color intensity and transparency)
• Size and color (either hue, intensity, or transparency), when the sizes of objects become tiny
• Shape and size, for we cannot effectively compare the sizes of objects that vary in shape (e.g., circles, squares, triangles, and stars)
• Shape and curvature, because curvature is an aspect of shape and changing the curve would change the shape
• Shape and line orientation, because only a few shapes, such as lines and rectangles, would make it easy to perceive and compare slopes
These attributes can certainly be combined in a single graph, but they cannot be combined effectively.
Limits of Working Memory
In the moment when we’re thinking about things (i.e., while we’re attending to them), information is held in working memory. This is different from long-term memory, which functions as a form of permanent storage for later retrieval. When you retrieve information from long-term memory, you pull it into working memory to think about and manipulate it in the moment. Working memory is volatile in that, once information is released from working memory to free up space for new information, it is forgotten unless we take time to rehearse it enough to store it in long-term memory. In addition to being volatile, working memory is extremely limited. As I’ve already mentioned, we can only hold from three to four chunks of information in working memory at a time. Consider the number 417. Although it is composed of three digits, it can be held in working memory as a single chunk of information. While thinking about and comparing quantities, we could simultaneously hold the numbers 417, 25, and 5,003 in working memory as three discrete chunks. Data visualization is powerful, in part, because it allows us to chunk multiple values together in a way that expands the amount of information that can be simultaneously held in working memory. For example, the pattern formed by a line in a line graph that represents 12 monthly sales values can potentially be held in working memory as a single chunk (i.e., as the visual pattern formed by the line), whereas only three or four of those values could be held simultaneously in working memory when represented as numbers.
This limitation in the capacity of working memory plays a significant role in data visualization. When we view a graph for the purpose of reading and comparing values, the fact that we can only simultaneously hold up to three or four chunks of information in working memory limits the comparisons that we can make in any one moment. Fortunately, because a great deal of information is potentially there in front of our eyes, we can quickly swap information in and out of working memory as needed, but never hold more than four chunks at a time.
Here’s the clincher. When multiple variables are represented by different visual attributes, we cannot chunk them together in working memory. For example, if we’re viewing a bubble plot that uses 2-D horizontal position, 2-D vertical position, bubble size, and bubble color intensity to encode four variables in each bubble, each of those values is held in working memory as a separate chunk. If we applied additional visual attributes to those bubbles to encode more variables, we would still only be able to hold up to four at a time in working memory. Now, what if we want to compare one of those bubbles to another? If each bubble represents four values, totaling eight for two bubbles, we could only hold two values at a time for each bubble in working memory when making comparisons. This means that we would be forced to swap values in and out of working memory to compare more than two values per bubble. Consequently, even though we could encode more variables in a single graph using different visual attributes, it wouldn’t expand our ability to consider them simultaneously. Even four variables per object exceeds the number that we could consider in any one moment when we’re comparing objects to one another. As far as I know, no research studies have ever measured the efficiency gains or losses for various tasks (e.g., decoding the various values that are associated with an object, comparing objects of various types, etc.) that are associated with the number of variables that are encoded in a single graph. Given proper study, we might find ways to improve efficiency, but for now we must keep these limits in mind.
This limitation in the capacity of working memory, combined with the fact that most visual attributes do a relatively poor job of representing values in graphs, forces us to admit that any gains in efficiency that we’re hoping to achieve by including more than a few variables in a single graph are wasted. It’s worse than that, actually, for each additional visual attribute that we include in a graph potentially contributes to the appearance of clutter, which is our next topic.
The Distraction of Visual Clutter
By clutter, I’m referring to the characteristics of a graph’s appearance that are potentially messy looking and distracting when we’re trying to focus on the particular attributes that we care about in the moment. For example, there is no doubt that having objects blink on and off at various speeds to encode a quantitative variable would make it almost impossible to attend to anything else. Even overly bright colors result in a cluttered appearance that is distracting. Every additional variable encoded by introducing another visual attribute to a graph comes with a perceptual cost. The cleaner and simpler the display, the easier it is to use.
When a chef chooses among the ingredients in her kitchen to cook a soup, her goal is not to combine as many ingredients as possible but instead to combine only those that are needed and to prepare them in the best way possible to create a pleasing culinary experience. Similarly, when we choose among the variables in a data set and display them in a particular way in a graph, our goal is not to squeeze as many variables as possible into it but to answer the question at hand in the most enlightening way. When visualizing data, we don’t typically start with a single graph and then ask many questions about it. Instead, we start with questions, one at a time, and create graphs as needed to answer each in the best possible way.
Be very wary of data visualization vendors that promote their supposed ability to display a large number of variables in a single graph. More isn’t better. Better is better. Only vendors that have taken the time to study visual perception and cognition can build data visualization tools that actually work. Unfortunately, relatively few vendors have done this, which is painfully obvious from the dysfunctional tools that most of them sell.
Data visualization vendors, especially newcomers, occasionally make the erroneous claim that their software can effectively visualize a large number of variables at once using separate visual attributes for each. I encountered the latest example of this recently when I read a press release about a new product named Immersion Analytics by the company Virtual Cove. These folks claim that, using their patents-pending techniques, they can effectively visualize up to 16 variables simultaneously. The following example includes 12 variables:
One of the arguments that Virtual Cove makes to promote their software is that to visualize a data set consisting of 16 variables using graphs that display only 4 variables each would require 1,820 graphs in total, which their software could replace with a single graph. They made this specific claim in an email to me, and they feature similar claims in their marketing efforts. It’s probably quite persuasive to many people, for it has the air of mathematical certainty. As it turns out, however, it is neither accurate nor relevant. I’m not sure how they did the math, but it appears to be based on the invalid assumption that every possible combination of four-variables would need to be examined to compare each of the 16 variables to each of the others. That isn’t the case. To see each of the 16 variables in relation to each of the other 15 using four-variable bubble plots, for example, would only require 35 graphs, not 1,820. Their figure is off by a factor of 52. The actual number of graphs that would be needed is less than 2% of the figure that they claim. Even if we were looking for correlations among 16 quantitative variables using scatter plots with only two variables each, that would only require a total of 120 graphs. In fact, a scatter plot matrix could be used to display all of these scatter plots at once. Even though this might require some scrolling around on the screen to examine every scatter plot, that wouldn’t matter because we would only need to view one scatter plot at a time. A scatter plot matrix would provide insights that could never be achieved using a single graph that attempts to encode 16 variables using Virtual Cove’s approach.
Given their egregious error, do you suspect that Virtual Cove might be making numbers up when they claim, as they do on their website, that their software can “increase productivity by up to 400x”? A four-hundred-fold increase? Really? That means that if the conventional approach took one hour of time, their approach would reduce the work to nine seconds. Can you guess what their response was when I asked for evidence of this claim? You’re right if you guessed that they didn’t respond.
#### Encode Every Variable Using the Same Visual Attribute
Instead of encoding each variable in a graph using a different preattentive attribute of visual perception, multiple variables can be displayed in a graph using the same attribute. Two types of graphs in particular were invented to use this approach for specific purposes: parallel coordinates and table lenses.
Parallel Coordinates
A parallel coordinates plot uses 2-D position, most often vertically along Y axes, to encode a series of variables. The example below displays six quantitative variables, each along its own Y axis.
In case you’re not familiar with parallel coordinates plots, let me briefly explain how they work. Let’s begin by considering a single variable. In the example below, the prices in dollars for 25 products have been represented by positioning 25 dots along the Y axis. When each value is represented by a dot along a single quantitative scale in this manner to show how the values are distributed, the graph is called a strip plot.
Although strip plots are more typically arranged along the X axis, the Y axis can work just as well. When strip plots are arranged vertically, multiple strip plots can be placed side by side to display an entire series of variables, such as the six variables that appear below for the same set of 25 products.
So far, however, we cannot determine which dot represents which product. That would be useful if we want to determine how the products compare to one another across the entire set of six variables. To make this possible, a parallel coordinates plot would connect the dots for each product across each of the Y axes using a line. In the example below, which displays multivariate data for 50 products, a particular line is highlighted to feature a single product’s multivariate profile.
(Note: In this example, rather than assigning a separate quantitative scale to each variable, the scales have been normalized by expressing each as percentages: the item with the lowest value is at the bottom with 0% and the one with the highest value at the top with 100%. Because the purpose of a parallel coordinates plot is not to decode individual values but instead to examine and compare multivariate patterns, the scales can be normalized in this manner without a loss of relevant information.)
In this example, we have a single graph that displays six variables for 50 products, but a parallel coordinates plot can include more variables and more than 50 items. As you might imagine, parallel coordinates plots can become complex and cluttered when they include many variables and items, but they can still be used to effectively compare complex multivariate profiles when properly designed, especially through the use of filtering and highlighting. Unlike graphs that encode variables using a different visual attribute for each, by encoding each variable in the same way (i.e., as 2-D vertical position), a parallel coordinates plot displays multiple variables in a way that our brains can read and interpret quite effectively. Because each one of an item’s values is connected by a line, the pattern formed by that line can be held in working memory as a single chunk of information. When we hold that line in working memory, we are not holding each variable’s value in memory, but that isn’t necessary when we’re trying to compare multivariate profiles, which we can do by simply comparing the patterns of multiple lines. For such a task, this approach to multivariate display is brilliant.
Without more thorough instruction in parallel coordinates, an example like the one above might appear overwhelming, so you might doubt the ability of these graphs to present complex multivariate data in a way that works for our brains. They do require extensive study and practice, which is one of the reasons why they are not more familiar, but they can definitively be worth the effort if you need to compare complex multivariate profiles. For a bit more explanation, I suggest that you read the newsletter article titled “Multivariate Analysis Using Parallel Coordinates” that I wrote back in 2006.
Table Lenses
A table lens display also uses a series of axes, one per variable, arranged side by side, but the arrangement is slightly different from parallel coordinates plots. Here’s a simple example of a five-variable table lens display:
In this case, the Y axis host a categorical scale that labels the item for which quantitative data is being displayed, in this case U.S. states, and the X axes host independent quantitative scales, one per variable. When values are represented as bars, the horizontal position of each bar’s end and the length of each bar both represent the same quantitative value. Unlike parallel coordinates, which are used to compare multivariate profiles, table lenses are used to look for potential correlations among several quantitative variables at once.
Notice in the example above that the states have been ranked from the highest value at the top to the lowest value at the bottom based on profit, the leftmost variable. Given this arrangement, we can now look at the arrangements of bars from top to bottom in each of the other columns to see if any of the other variables exhibit patterns that are similar to profit or are perhaps its inverse. If the arrangement of bars for one of the other variables roughly displays a pattern ranging from high values at the top to low values at the bottom, this tells us that it correlates with profit in a positive way. That is, as profit values per state decrease, values of sales also tend to decrease. If, on the other hand, sales roughly exhibit a pattern of low values at the top to high values at the bottom, this would tell us that it is still correlated with profit, but in a negative manner. That is, as profit values decrease, sales values tend to increase.
A table lens can provide a useful way to look for correlations among many variables at once. The example below, which was produced using a product that was actually called Table Lens from a company named Inxight, which unfortunately no longer exists, displays 23 variables worth of baseball statistics.
A table lens can display many variables in a single graph in a manner that works for our brains because it encodes each using the same visual attribute—one that we can perceive with ease.
#### Increase the Number of Variables Using Small Multiples
This final data visualization approach is used to increase the number of variables that can be simultaneously displayed. This approach goes by various names, but the most familiar is Edward Tufte’s term small multiples. Back in the 1970s, both Edward Tufte and William Cleveland promoted the use of displays that combine several small graphs. Each graph works the same, but each displays data associated with a different categorical item. In the example below, three small graphs have been arranged side by side, and each displays data associated with a different customer segment: Consumer, Corporate, and Home Office. Other than this, the three graphs work exactly the same. Each displays sales revenue in U.S. dollars along the X axis, discount percentage along the Y axis, profit margin by bubble size, and geographical region by bubble hue.
The individual graphs in this example are already complex enough with four variables each, so it wouldn’t work to display additional variables in them. The additional variable of customer segment, however, has been added to the display without overcomplication by presenting each customer segment in its own graph.
Many more graphs than the three that appear in the example above can be included in a small multiples display and they can be arranged on the screen in various ways. The example above arranges the small multiples horizontally in a single row, side by side, but they could also be arranged vertically, in a single column. A large series could also be wrapped across multiple columns and rows—an arrangement that William Cleveland called a trellis display.
Alternatively, a series of small multiples can be used to add two more categorical variables rather than just one. In the example below, each column of graphs still displays customer segments, but now each row displays product categories.
When small multiples are arranged in this way, with one variable along the rows and another along the columns, I call it a visual crosstab.
Even though a small multiples display consists of multiple graphs, because all the graphs work the same and are all visible at once, we can easily and quickly compare them to one another. If we know how to read one graph, we know how to read them all. This is a powerful way to increase the number of variables that can be simultaneously displayed beyond the number that you could include in a single graph that encodes variables using different visual attributes.
#### Conclusion
Wanting to break through our limitations is natural. We want to be better; we want to do more. We don’t accomplish this, however, by ignoring our limitations. Ignorance is the path to delusion and dysfunction. Software vendors don’t get any points for building and selling tools that simultaneously visualize a dozen or more variables in ways that don’t work. When our limitations get in the way, we overcome them by using our brains to find real solutions. We always begin by understanding our limitations. Parallel coordinates plots, table lens displays, and small multiples are all innovations that demonstrate the merits of this approach. On the other hand, the graph below shows what happens when we simply ignore our limitations.
This graph only displays eight variables, half the number that the vendor, Virtual Cove, claims to support, and it’s already a virtual cave of worthless effects. We can only see that a few of the spheres (i.e., 3-D bubbles) are much bigger than the rest and that one is much brighter as well. Imagine how much worse it would be if this graph attempted to display 16 variables rather than 8.
The potential for understanding that resides in our data should not be wasted by chasing pipe dreams. The path forward begins by understanding our limitations, not by pretending that they don’t exist. | 6,844 | 34,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-31 | latest | en | 0.897708 |
http://mathhelpforum.com/calculus/137871-find-equation-plane-parallel-p-contains-point-q.html | 1,524,409,409,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945604.91/warc/CC-MAIN-20180422135010-20180422155010-00110.warc.gz | 204,924,139 | 9,962 | # Thread: Find the equation of the plane parallel to P that contains the point Q
1. ## Find the equation of the plane parallel to P that contains the point Q
Let be the plane given by the equation
Let be the point Find the equation of the plane parallel to that contains the point (Please ensure that the first entry is positive.)
_____
^As you can see I got the 21, -21, and 16 now I just need to know what goes after the =.
2. plug in your point (3,3,6) into the equation.........
21x-21y+16z=d and solve for d.
The normal vector, i.e., the coeficients of x,y,z must be the same if the planes are parallel.
So, all you need to do is adjust the d.
3. Oh ahha that was easy, thanks! | 182 | 689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-17 | latest | en | 0.932331 |
http://blog.data-miners.com/2009/01/thoughts-on-understanding-neural.html?showComment=1374303253611 | 1,632,319,858,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00243.warc.gz | 11,961,288 | 19,096 | ## Sunday, January 18, 2009
### Thoughts on Understanding Neural Networks
Lately, I've been thinking quite a bit about neural networks. In particular, I've been wondering whether it is actually possible to understand them. As a note, this posting assumes that the reader has some understanding of neural networks. Of course, we at Data Miners, heartily recommend our book Data Mining Techniques for Marketing, Sales, and Customer Relationship Management for introducing neural networks (as well as a plethora of other data mining algorithms).
Let me start with a picture of a neural network. The following is a simple network that takes three inputs and has two nodes in the hidden layer:
Note that this structure of the network explains what is really happening. The "input layer" (the first layer connected to the inputs) standardizes the inputs. The "output layer" (connect to the output) is doing a regression or logistic regression, depending on whether the target is numeric or binary. The hidden layers are actually doing a mathematical operation as well. This could be the logistic function; more typically, though it is the hyperbolic tangent. All of the lines in the diagram have weights on them. Setting these weights -- plus a few others not shown -- is the process of training the neural network.
The topology of the neural network is specifically how SAS Enterprise Miner implements the network. Other tools have similar capabilities. Here, I am using SAS EM for three reasons. First, because we teach a class using this tool, I have pre-built neural network diagrams. Second, the neural network node allows me to score the hidden units. And third, the graphics provide a data-colored scatter plot, which I use to describe what's happening.
There are several ways to understand this neural network. The most basic way is "it's a black box and we don't need to understand it." In many respects, this is the standard data mining viewpoint. Neural networks often work well. However, if you want a technique that let's you undersand what it is doing, then choose another technique, such as regression or decision trees or nearest neighbor.
A related viewpoint is to write down the equation for what the network is doing. Then point out that this equation *is* the network. The problem is not that the network cannot explain what it is doing. The problem is that we human beings cannot understand what it is saying.
I am going to propose two other ways of looking at the network. One is geometrically. The inputs are projected onto the outputs of the hidden layer. The results of this projection are then combined to form the output. The other method is, for lack of a better term, "clustering". The hidden nodes actually identify patterns in the original data, and one hidden node usually dominates the output within a cluster.
Let me start with the geometric interpretation. For the network above, there are three dimensions of inputs and two hidden nodes. So, three dimensions are projected down to two dimensions.
I do need to emphasize that these projections are not the linear projections. This means that they are not described by simple matrices. These are non-linear projections. In particular, a given dimension could be stretched non-uniformly, which further complicates the situation.
I chose two nodes in the hidden layer on purpose, simply because two dimensions are pretty easy to visualize. Then I went and I tried it on a small neural network, using Enterprise Miner. The next couple of pictures are scatter plots made with EM. It has the nice feature that I can color the points based on data -- a feature sadly lacking from Excel.
The following scatter plot shows the original data points (about 2,700 of them). The positions are determined by the outputs of the hidden layers. The colors show the output of the network itself (blue being close to 0 and red being close to 1). The network is predicting a value of 0 or 1 based on a balanced training set and three inputs.
Hmm, the overall output is pretty much related to the H1 output rather than the H2 output. We see this becasuse the color changes primarily as we move horizontally across the scatter plot and not vertically. This is interesting. It means that H2 is contributing little to the network prediction. Under these particular circumstances, we can explain the output of the neural network by explaining what is happening at H1. And what is happening at H1 is a lot like a logistic regression, where we can determine the weights of different variables going in.
Note that this is an approximation, because H2 does make some contribution. But it is a close approximation, because for almost all input data points, H1 is the dominant node.
This pattern is a consequence of the distribution of the input data. Note that H2 is always negative and close to -1, whereas H1 varies from -1 to 1 (as we would expect, given the transfer function). This is because the inputs are always positive and in a particular range. The inputs do not result in the full range of values for each hidden node. This fact, in turn, provides a clue to what the neural network is doing. Also, this is close to a degenerate case because one hidden unit is almost always ignored. It does illustrate that looking at the outputs of the hidden layers are useful.
This suggests another approach. Imagine the space of H1 and H2 values, and further that any combination of them might exist (do remember that because of the transfer function, the values actually are limited to the range -1 to 1). Within this space, which node dominates the calculation of the output of the network?
To answer this question, I had to come up with some reasonable way to compare the following values:
• Network output: exp(bias + a1*H1 + a2*H2)
• H1 only: exp(bias + a1*H1)
• H2 only: exp(bias + a2*H2)
Let me give an example with numbers. For the network above, we have the following when H1 and H2 are both -1:
• Network output: 0.9994
• H1 only output: 0.9926
• H2 only output: 0.9749
To calculate the contribution of H1, I use the ratio of the sums of the squares of the differences, as in the following example for H1:
• H1 contribution: (0.9994 - 0.9926)^2 / ((0.9994 - 0.9926)^2 + (0.9994 - 0.9749)^2)
The following scatter plot shows the regions where H1 dominates the overall prediction of the network using this metric (red is H1 is dominant; blue is H2 is dominant):
There are four regions in this scatter plot, defined essentially by the intersection of two lines. In fact, each hidden node is going to add another line on this chart, generating more regions. Within each region, one node is going to dominate. The boundaries are fuzzy. Sometimes this makes no difference, because the output on either side is the same; sometimes it does make a difference.
Note that this scatter plot assumes that the inputs can generate all combinations of values from the hidden units. However, in practice, this is not true, as shown on the previous scatter plot, which essentially covers only the lowest eights of this one.
With the contribution metric, we can then say that for different regions in the hidden unit space, different hidden units dominate the output. This is essentially saying that in different areas, we only need one hidden unit to determine the outcome of the network. Within each region, then, we can identify the variables used by the hidden units and say that they are determining the outcome of the network.
This idea leads to a way to start to understand standard multilayer perceptron neural networks, at least in the space of the hidden units. We can identify the regions where particular hidden units dominate the output of the network. Within each region, we can identify which variables dominate the output of that hidden unit. Perhaps this explains what is happening in the network, because the input ranges limit the outputs only to one region.
More likely, we have to return to the original inputs to determine which hidden unit dominates for a given combination of inputs. I've only just started thinking about this idea, so perhaps I'll follow up in a later post.
--gordon
#### 8 comments:
1. nice!
I tried to do something like this a few years ago. I like yours more. The best idea I could come up with was to use colour and tansparency to represent the positive or negative effect and strengths of the NN weights and hidden nodes.
I went to the effort of creating a vb.net application that loads PMML into data grids and then from the data grids builds the graphic. You might be able to use the data grids for your own graphics.
It should work with neural net saved as PMML from SAS EM. I tested it a little bit with Clementine and SAS PMML but only with simple single hidden layer NN's.
See;
http://www.kdkeys.net/forums/thread/6495.aspx
You can simply drag and drop the NN PMML onto the .exe and it will display the graph. I supplied the code and everything. Use as you wish. Completely free!
I'm not sure its much use, but I did it for fun to learn a bit of VB.NET (I'm certainly no programmer :)
Cheers
Tim
2. I added a few pictures to my blog in case you don't want to run the vb.net executable etc.
http://timmanns.blogspot.com/2009/01/re-thoughts-on-understanding-neural.html
cheers
Tim
3. Gordon,
I spent 4 years of my life trying to understand how neural networks did what they did during my doctorate. It was only after I started drawing plots of the outputs of the hidden neurons that the penny finally dropped and everything became crystal clear.
What I would recommend is start off with simple made up funcions such as y=x^2. It quickly becomes apparent why 2 sigmoidal neurons are required. Then try y=log(x), and then a function such as y=x^2 + log(x).
The ability to 'prune' a network to force particuar inputs to be procesed by selected hidden neurons is a must, you can then essentially decompose the model into y = f(x1,x2) + f(x3) + f(x4) etc..
This gives you a great ability to extract underlying trends from data. For example, my doctorate was electric load prediction and this method enables the load to be decomposed into f(temperature) + f(time of day) + f(day of week) etc...
More recently I extracted the premium of LPG vehicles over Unleaded petrol vehicles over time for cars sold at auction. The results did reflect the difference in the two fuel costs (and govt subsidies to convert your car) over time.
I wrote a little application in excel that will allow you to view the outputs of the hidden neurons as the network is being trained.
http://www.philbrierley.com/code/vba.html
Goto the train tab and select the 'decomposition' graph. It can be informative to watch. The default data is something like y=x^2 but you can paste any data in you want.
I also write an application called Tiberius that lets you do a similar thing but also lets you manually 'prune' the network to force the form of the model.'
www.tiberius.biz
You can see some oldish screenshots of the decomposition at
www.philbrierley.com/tiberius/slideshow.html
slide 8 of the y=x^2 demo will give you the idea.
Hope my work will help you on your quest to understand neural networks. Would be more than willing to link up on skype to talk you through what can be achieved.
Regards
Phil Brierley.
4. I would also point out the Neural Networks chapter in Tom Mitchell's "Machine Learning" book. He goes through some cases of ANNs applied to image recognition, and these are quite telling as to what hidden nodes *could* learn.
Max Khesin.
(Regards!)
5. I remember a publication by Le Cun in the early 90s I think that described how a neural network doing OCR behaved--if I remember correctly, some HL neurons were effectively horizontal line detectors, other vertical line detectors, others edge detectors (this one generated a l ot of excitement because the human eye does edge detection, among many other things!).
However, I think too that there is value in understanding overall trends in a neural network too (overall sensitivity), much like one sees in many neural net applications like in Clementine or Neuralware Predict. These are akin to examining coefficients in a regression model in that they give overall average trends but not much else.
6. Hi, one way I find useful to understand neural networks is to look at various applications from a time series regression perspective.
I imagine a linear regression as just being the best fit to a noisy line. If you think of any other scatterplot that might be uniquely represented by a non-linear relationship, a neural network can usually find it.
I wrote up a brief tutorial on learning financial time series that you may find of interest at:
http://intelligenttradingtech.blogspot.com/
7. Here is a link to the best visualization of a neural network that I've ever seen.
http://webgl-ann-experiment.appspot.com/
8. This is really cool! I am just a student right and have some experience with neural netowrks, but I have always thought of them just like you described at the beginning of your article: as a black box, or like a mathematical formula. More so like a collection of mathematical formulas rather than a black box, but anyways…
I know this post is pretty old, but I got so excited about it that I’m just going to post it anyways, even thought it might have no scientific value for you… I honestly have never even thought of any way of visualizing a neural network, or just purely analyzing which hidden unit contributes the most to the output. I am not really familiar with all of the current advancements in neural network metalearning, but this seems like something like what you described may help researchers and data scientists explain their findings more clearly. Not only that, the researchers themselves may better understand what’s going on with their learning model if they analyze it by what you called “clustering” (for the lack of a better word :) ).
Thank you so much for this post!
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# In a parallelogram and are the mid-points of sides and respectively (see Fig). Show that the line segments and trisect the diagonal .
## Text solutionVerified
is parallelogram
Using properties of a parallelogram and
is the mid-point of
... (1)
is the mid-point of
Therefore,
(Since ) ... (2)
From (1) and (2), we get
Also,
(Since )
Thus, a pair of opposite sides of a quadrilateral are parallel and equal.
and
In , is the mid-point of
Using mid point theorem ... (3)
Similarly, by taking , we can prove that
... (4)
From (3) and (4), we get
Therefore,
and trisect the diagonal .
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Question Text In a parallelogram and are the mid-points of sides and respectively (see Fig). Show that the line segments and trisect the diagonal . Updated On Dec 14, 2023 Topic Quadrilaterals Subject Mathematics Class Class 9 Answer Type Text solution:1 Video solution: 6 Upvotes 659 Avg. Video Duration 7 min | 463 | 1,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-26 | latest | en | 0.809312 |
http://www.actuarialoutpost.com/actuarial_discussion_forum/archive/index.php/t-133615.html | 1,369,277,170,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702749808/warc/CC-MAIN-20130516111229-00087-ip-10-60-113-184.ec2.internal.warc.gz | 306,891,073 | 2,108 | PDA
View Full Version : LGD and recovery rate
Eroboy
03-17-2008, 11:43 PM
It seems that the recovery rate defined in Hull's book is
market value of bond at default / Face (page 483 and see example on page 485)
while in the Crouhy's book the LGD is defined as % of bond value. (see page 412 to 414 )
Thus, it seems that two books have slight definition of recovery rate.
This is not an issue if this is just one period zero coupon bond, but it seems that different definition will result slightly different answer with coupon bond depending how you interpret the recovery rate.
Is there an convention if the problem gives a recovery rate, we use the definition in Hull's book but for LGD use the definition in Crouhy's book.
BTW, has anyone verify the estimation of default intensity formula in Hull's book for multi-period zero coupon bond? (lamda=S/(1-h), where s is spread and h is recovery rate). It seems that based on formula 3 on page 413 of Crouhy, if yield curve and risk free curve is flat, all the forward intensity shall be same. Thus, Hull's approximation shall hold. However, I fail to analytically sovle the following formula to derive it directly on a continuous basis:
exp(-yT)=exp(-rT)*[1-exp(-lamda*T)]+h*[Integral(lamda*exp(-(lamda+r)*t)dt, t from 0 to T)]
where y is the bond yield,
r is risk free
lamda is average default intensity
T is the maturity
The integral is easy, but solving lamda as (y-r) requires some skills.
Also, isn't using Merton's formula under estimate the default rate for long term? Merton's formula is European option, while company can default at any time before the maturity. It seems to me that the short term estimation is okay. | 418 | 1,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2013-20 | latest | en | 0.927215 |
https://www.studypool.com/discuss/399783/consider-the-following-functions?free | 1,481,319,727,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542828.27/warc/CC-MAIN-20161202170902-00025-ip-10-31-129-80.ec2.internal.warc.gz | 1,022,076,677 | 14,478 | ##### Consider the following functions.
Algebra Tutor: None Selected Time limit: 1 Day
f(x) = x − 6, g(x) = x2
Find
(f + g)(x)
Find the domain of (f + g)(x).
Find (f − g)(x).
Find the domain of (f − g)(x)
Find (fg)(x).
Find the domain of (fg)(x)
Find
f g
(x).
Find the domain of
f g
(x).
Feb 23rd, 2015
f(x) =(x-6)
g(x) = x2
(f+g)(x) = (x-6) + x2
domain = { x | x is real }
(f-g)(x) = (x-6) - x2
domain = { x | x is real }
(fg)(x) = (x-6) * x2 = x2 (x-6)
domain = { x | x is real }
(f/g) (x) = (x-6)/ x2
domain = { x | x is real and x ≠ 0 }
Feb 23rd, 2015
...
Feb 23rd, 2015
...
Feb 23rd, 2015
Dec 9th, 2016
check_circle | 297 | 657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2016-50 | longest | en | 0.535275 |
https://www.studymode.com/essays/questions-and-answers-on-fluid-mechanics-52187102.html | 1,656,661,237,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00218.warc.gz | 1,059,063,279 | 13,144 | # Questions and Answers on Fluid Mechanics
Satisfactory Essays
BPH – Tut # 2
Fluid Mechanics – University of Technology, Sydney – Engineering & IT
Tutorial # 2 – Questions and Solutions
Q.1
For the situation shown, calculate the water level d inside the inverted container, given the manometer reading h = 0.7 m, and depth L = 12 m.
Q.2 (from Street, Watters, and Vennard)
The weight density γ = ρg of water in the ocean may be calculated from the empirical relation γ = γo +
K(h1/2), in which h is the depth below the ocean surface, K a constant, and γo weight density at the ocean surface. Derive an expression for the pressure at any depth h in terms of γo, h and K. Calculate the weight density and pressure at a depth of 3.22 km, assuming γo = 10 kN/m3, h in meters, and K =
7.08 N/m7/2. Calculate the force on a surface area of 2 m2 (of a submarine, for example).
Question 3
The sketch shows a sectional view through a submarine. Atmospheric pressure is 101 kPa, and sea water density is 1020 kg/m3. Determine the depth of submergence y.
1
BPH – Tut # 2
Question 4
In the dry adiabatic model of the atmosphere, in addition to the ideal gas equation, air is also assumed to obey the equation p/ρn = constant where n = 1.4. If the conditions at sea level are standard (101.3 kPa, 15°C), determine the pressure and temperature at an altitude of 4000 m above sea level.
Solutions.
Q.1
For the situation shown, calculate the water level d inside the inverted container, given the manometer reading h = 0.7 m, and depth L = 12 m.
2
BPH – Tut # 2
Solution
B
A
F
D
E
3
BPH – Tut # 2
ρwater = 1000 kg/m3
;
ρ Hg = 13600 kg/m3
PA = PB + ρHg g h
;
PB = 0 (gauge)
PA ≈ PD
(pressure change due to air column AD is negligible due to low air density)
PE = PD + ρwater g d = PF + ρwater g L ;
PF = 0 (gauge)
∴ ρwater g L = ρHg g h + ρwater g d d = (ρwater L − ρ Hg h) / ρwater = (1000×12 − 13600×0.7) / 1000 = 2.48 m
Q.2 (from Street, Watters, and Vennard)
The
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nth term of an A.P(an) = -4n+15
a1 = -4(1)+15 = 11
a2 = -4(2)+15 = 7
d = a2- a1 = 7-11 = -4
And, Sum of first n terms of AP is given by-
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Thus, the Sum of first 20 terms of AP = -540
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physics
A hiker climbs a hill 300m high.If she weighs 50kg,calculate the work she does in lifting her body to the top of the hill
August 3, 2011
physics
How much work is done when a mass of 30N is lifted vertically through 6m
August 3, 2011
physics
A block of mass 2kg is pushed along a table with a constant velocity by a force of 5N.When the push is increased to 9N,what is a)the resultant force b)the acceleration
August 3, 2011
Physics
A chair lift carries people of average mass 75kg to a height of 450m above the starting point. Calculate a) the work done by the chair lift to move one person through 450m. b) the power expended to carry 60 people every 5 minutes.
August 3, 2011
physics
A box of mass 5 kg is pulled along a horizontal frictionless surface by a force of 40N inclined @ 30dgrees. Calculate A)the work done by the force over a distance. B)the speed gained by the box after travelling 10m.
August 3, 2011
science
An 80kg golf hall is struck for a period of0.03s wt a force of 500n.the bal accelerate frm the tree.wat is its final speed.
June 7, 2011
physics
An archa fires an arrw of mas 96g wt a velocity of 120m/s @a target of mas 1500g hangin frm a piec o strin frm a tal tree. If the arrow becomes embedded in thd target wt wat velocity does the target move
June 7, 2011
Algebra 1
what about part 2 ?
February 6, 2011
Chemistry
Calculate the average translational kinetic energy (kJ/molecule) for a molecule of NH3 at 431 K, assuming it is behaving as an ideal gas. I understand how to figure out the kJ/mole, but I don't understand how to get the kJ/molecule that this question is asking for. Can ...
April 11, 2010
Chemistry
I tried the method that you provided above and the grading system told me that the answer was wrong. The correct answer is (0.3775). Do you know how to come up with that answer?
April 11, 2010
Chemistry
If 0.433 mol of gaseous CS2 and 1.62 mol of gaseous H2 are reacted according to the balanced equation at 454 K and in a 40.0 L vessel, what partial pressure (atm) of gaseous CH4 is produced? CS2(g) + 4H2(g) -> CH4(g) + 2H2S(g) Molar Mass (g/mol) CS2 76.143 H2 2.016 CH4 16.043
April 11, 2010
Chemistry
I get the same answer as before = 0.271atm According to the program, the right answer is 0.391atm
April 4, 2010
Chemistry
Yes it is based on mole composition, not mass. But how do you figure out the moles?
April 4, 2010
Chemistry
A gas sample containing only SO2, PF3 and SF6 has the following mass percent composition: 30.4% SO2 and 28.4% PF3. Calculate the partial pressure (atm) of SO2 if the total pressure of the sample is 676.4 torr. I tried taking the total pressure in atm and multiplying it with ...
April 4, 2010
Chemistry
I came up with 3461 torr, but the homework program I am using said that the correct answer is 2510 torr. So, I either did something wrong, or the method provided above does not work
March 4, 2010
Chemistry
A gas sample containing only SO3, CH2O and C2H6O2 has the following mass percent composition: 65.1% SO3 and 8.79% CH2O. Calculate the partial pressure (torr) of SO3 if the total pressure of the sample is 4712 torr. Unit conversion K = C + 273 1 atm = 760 torr Molar Mass (g/mol...
March 4, 2010
Chemistry
A 355.0 mL cylinder containing SO3 at a pressure of 0.700 atm is connected by a valve to 0.559 L cylinder containing COCl2 at 380.0 torr pressure. Calculate the partial pressure (atm) of SO3 when the valve is opened. Unit conversion K = C + 273 1 atm = 760 torr Molar Mass (g/...
March 4, 2010
Chemistry
never mind, I figured it out. I'll post the explanation just in case anyone else needs help with a similar problem. Equation: molality acetone = mol acetone/kg ethanol ------------------------------------- to find kg ethanol: total mass - sample mass of acetone (6.56kg - 1...
March 2, 2010
Chemistry
Calculate the molality (m) of a 6.56 kg sample of a solution of the solute acetone dissolved in the solvent ethanol if the sample contains 1.44 kg of acetone. Additional info: Molar Mass (g/mol) acetone = 58.05 ethanol = 46.07 Density (g/mL) acetone = 0.7899 ethanol = 0.7893 ...
March 2, 2010
Chemistry
Thank you Dr. Bob!
March 1, 2010
Chemistry
A 1.33 m solution of the solute diethyl ether dissolved in the solvent tetrahydrofuran is available. Calculate the mass(kg) of the solution that must be taken to obtain 0.4716 kg of diethyl ether. Name/Formula: diethyl ether = (C2H5)2O tetrahydrofuran = C4H8O Molar Mass(g/mol...
March 1, 2010
Chemistry
-637 kJ/mol. okay, I understand now... thanks Dr. Bob!
February 21, 2010
Chemistry
∆H⁰rxn, for the reaction of Ca(s) + 2H+(aq) -> Ca2+(aq) + H2(g) is -544 kJ/mol. ∆H⁰rxn, for the reaction of CaO(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) is -193 kJ/mol. Based on these results, what is the ∆Hf for CaO(s) [in kJ/mol] if the heat of ...
February 21, 2010
chemistry
An element has hcp packing with a hexagonal unit cell. Its density is 21000 kg/m3 and the unit cell volume is 2.94 x 10-23 mL. Calculate the molar mass (g/mol) of the element to three significant figures.
February 9, 2010
Chemistry
An element has ccp packing with a face-centered cubic unit cell. Its density is 12400 kg/m3 and the unit cell volume is 5.50 x 10-29 m3. Calculate the molar mass (g/mol) of the element to three significant figures. Please explain how you come to the answer.
February 9, 2010
Chemistry
The element palladium has ccp packing with a face-centered cubic unit cell. The density of Pd is 12 g/cm3. Calculate the volume (L) of the unit cell of Pd.
February 9, 2010
physics newtons laws 3
The coefficient of static friction between the m = 3.20 kg crate and the 35.0° incline of Figure P4.41 is 0.260. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
October 18, 2009
physics newtons laws
A box of books weighing 280 N is shoved across the floor of an apartment by a force of 390 N exerted downward at an angle of 35.1° below the horizontal. If the coefficient of kinetic friction between box and floor is 0.57, how long does it take to move the box 4.00 m, ...
October 18, 2009
physics newtons laws
A 184 N bird feeder is supported by three cables. angle 1 is 60 degrees and angle 2 is 30 degrees Find the tension in each cable.
October 18, 2009
physics newtons laws
A 184 N bird feeder is supported by three cables. angle 1 is 60 degrees and angle 2 is 30 degrees Find the tension in each cable.
October 18, 2009
physics
A box of books weighing 280 N is shoved across the floor of an apartment by a force of 390 N exerted downward at an angle of 35.1° below the horizontal. If the coefficient of kinetic friction between box and floor is 0.57, how long does it take to move the box 4.00 m, ...
October 13, 2009
physics
A 40.0 kg wagon is towed up a hill inclined at 17.9° with respect to the horizontal. The tow rope is parallel to the incline and has a tension of 140 N. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. How fast is the wagon going ...
October 10, 2009
physic
A turnbuckle is needed to apply an 8kN force on a cable. If the efficiency of the turnbuckle is 40% and the force applied is 159.155N with a lever of 20cm in length. a) Find the mechanical advantage. b) Find the velocity ratio. c) Find the mean pitch of thread.
April 2, 2009
spelling
Its stream
February 19, 2009
Accounting
pierce co has a june 30 fiscal year end. it estimated its annual property taxes to be 24000 based on last year's property tax bill and have accrued 2000 of property tax each month to marcher 31. its actual property tax assessment, received on april 30 is 25200 for the ...
May 14, 2008
1. Pages:
2. 1
Post a New Question | 4,651 | 15,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-13 | longest | en | 0.933496 |
https://www.physicsforums.com/threads/step-functions.102022/ | 1,513,319,278,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948567042.50/warc/CC-MAIN-20171215060102-20171215080102-00174.warc.gz | 770,107,864 | 14,200 | # Step functions
1. Nov 29, 2005
### georgeh
i have f(t) defined piece-wise and continous..
f(t) = 0, t < 2pi
t-pi , pi <=t<2pi
0 , t >=2pi
i have so far g(t)=U_pi*f(t-pi)-U_2pi*f(t-pi)
if i do the laplace,
i get e^-pis/s^2-e^-2pis/s^2
in the book, they have
e^-pi*s/s^2 -e^-2pi*s/s^2 (1+pi*s)
I am not sure how they got the factor of (1+pi*s)
..
2. Nov 30, 2005
Are you sure $f(t)$ is correct. You have:
$$f(t) = \left\{ \begin{array}{l} 0, \,\, t<2\pi \\ t-\pi, \,\, \pi \leq t < 2\pi \\ 0, \,\, t \geq 2\pi$$
So when $t<2\pi$ and $\pi \leq t < 2\pi \\$ it equals $0$ and $t-\pi$. I'm assuming you mean $f(t) = 0|t<\pi$ ?
I got (assuming $f(t)$ is wrong):
$$\frac{e^{-\pi s}}{s^2} - \frac{\pi e^{-2\pi s}}{s} - \frac{e^{-2\pi s}}{s^2}$$
Last edited: Nov 30, 2005
3. Nov 30, 2005
### georgeh
sorry, what i meant was
on the first interval, t < pi, not two pi. | 399 | 871 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-51 | longest | en | 0.71943 |
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http://www.answers.com/topic/set-top-box | 1,529,667,383,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864391.61/warc/CC-MAIN-20180622104200-20180622124200-00345.warc.gz | 361,165,730 | 51,561 | # What are the 5Cs of credit?
5 C's of Credit refer to the factors that lenders of money evaluate to determine credit worthiness of a borrower. They are the following:. 1. Borrower's CHARACTER. 2. Borrow (MORE)
# What does 5c stand for?
The Iphone 5C is Iphone 5Colorful 5c can also stand for thenumber 500 ("c" is the Roman numeral for 100) or for 5 degreesCelsius (centigrade) . +++ . "5c" can not stand fo (MORE)
# What animal is on a 5c coin?
There are multiple animals on 5 cent coins depending on the country and time period such as the Buffalo on the US "buffalo nickel", the Beaver on the Canadian nickel, etc.
# What is -5c plus 9 and how?
You can't tell a thing about -5c+9 until you know what 'c' is. And every time 'c' changes, -5c+9 changes.
In Volume
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5cc? cc means cubic centimetres which is equal to ml, so 5ml. if you mean cl, then that is equal to 50ml
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The 'answer' is the number that 'c' must be, if 5c is really the same as -75. In order to find out what number that is, you could use 'algebra'. First, write the equatio (MORE)
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The iPhone 5c is 640 x 1136 pixels. That is about 326 pixels persquare inch (ppi).
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(-5) degrees Celsius = 23 degrees Fahrenheit. Formula: [°F] = [°C] à 9 â 5 + 32
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The screen is 4" big. The height is 4.9", width is 2.33" and thedepth is 0.35"
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The setting to make the screen flip on the iPhone 5 depends on whatversion of iOS you are using. If you are using iOS 7 or higher,simply swipe your screen up from the bottom a (MORE) | 522 | 1,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-26 | latest | en | 0.915724 |
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# Design Procedure for Knuckle Joint & Cotter Joint [Formulas & PDF]
A Knuckle Joint is used for the application of the Tie rod joint of jib crane or Tension link in the structure of the bridge. In this article, I am going to present a detailed explanation of the design procedure for Knuckle Joint and Cotter Joint.
Let's see the definition of Knuckle Joint...
## Knuckle Joint Definition:
When there is a requirement of an angular moment or a small amount of flexibility, the Knuckle joint is used. A knuckle joint is used to connect two rods that are under a tensile load.
## Knuckle Joint Diagram:
The Diagram of Knuckle Joint is shown below.
## Knuckle Joint Assembly or Parts of Knuckle Joint:
The Assembly of Knuckle Joint consists of 3 parts. They are
• Fork or Double-eye
• Single eye
• Knuckle pin
• Coller
There is a single eye at one end of the rod and Double-eye on the end of another rod which is also called Fork End.
Both, Single eye and Double-eye (Fork End) are connected employing a Knuckle pin inserted through the eye.
One end of the knuckle pin has a head section and the other end is tapered along with a hole on its surface for the insertion of Taper pin after assembly as shown in the above figure.
The cross-section at the ends of the rod are of octagonal and is designed for the purpose of Gripping.
The pin is used to hold one eye end of a rod and the fork end of the other rod firmly. Now, the Coller is placed exactly on the surface of the rod having a hole such that the Taper pin can be inserted into it.
Thus the assembly of Knuckle Joint can be done.
Lets see the drawing of Knuckle joint in Orthographic Representation.
## Knuckle Joint Drawing:
The Drawing of Knuckle joint in the form of 2D sketch i.e. Front View, Top View, and Side View is shown below.
### Knuckle Joint Failure:
The failures of Knuckle Joint are as follows.
• Tensile failure of the flat end bar
• Crushing of pin against the rod
• Shear (single shear) failure of the pin
## Applications of Knuckle Joint:
The applications of Knuckle Joint is as follows.
• Knuckle joint is mostly used in the Automobile to connect the propeller shaft when two shafts are not straight to each other.
• Knuckle joint is used as the Tie rod joint of the jib crane.
• It is used as the Link of roller chain.
• It is also used in the Tension link in the bridge structure.
## Design Procedure for Knuckle Joint is Made to Calculate:
• The outside diameter of eye or fork (mm)
• The thickness of each eye.
• The enlarged diameter of each rod by an empirical relationship.
• Diameters of the pin by shear considerations.
• Bending considerations.
### Specifications for Knuckle Joint:
The assembly diagram of knuckle joint is as shown in fig.
The dimension of knuckle joint are:
• Thickness of fork = t1
• Thickness of single eye = t
• Diameter of knuckle pin = dp
• Outside diameter of double eye = dod
• Outside diameter of single eye = doe
• Diameter of rod = d
• Axial tensile force on rod = P
4. Design of fork (double eye):
The detailed explanation for the Design Procedure for Knuckle Joint is shown below in the form of video.
I hope you have understood well with the help of above video. Let's discuss about Cotter Joint too...
## Cotter Joint:
Cotter's joint is widely used to connect the piston rod and crosshead of a steam engine, as a joint between the piston rod and the tailor pump rod, foundation bolt, etc.
## Cotter Joint Definition:
Cotter's joint is used to connect two rods subjected to compressive or axial tensile loads. It is not at all suitable to connect rotating shafts which transmit torque.
## Cotter Joint Diagram:
The Diagram of Cotter Joint is shown below.
## Cotter Joint Assembly:
The Cotter joint Assembly or the Parts of Cotter Joint are as follows.
### Cotter Joint Parts:
The parts of Cotter joint are
• Socket
• Spigot
• Cotter Pin
The explanation for the parts of Cotter Joint are as follows.
#### Socket:
The socket is the female part of the Cotter joint which has a hole at the center such that the spigot can fit into it.
It also has a rectangular slot on its surface such that the cotter can pass through it.
#### Spigot:
The spigot is the male part of the Cotter joint which has also a rectangular slot on its surface such that the cotter can pass through it.
#### Cotter Pin:
The Spigot is inserted into the socket such that the rectangular slots of both (spigot and Socket) can meet with each other and then the Cotter pin is inserted into it for fixing it firmly.
### Cotter Joint Failure:
The Cotter Joint Failures are as follows.
• Tensile Failure of Rods
• Tensile Failure of Spigot
• Tensile Failure of Socket
• Shear Failure of Spigot End
• Shear Failure of Socket End
• Crushing Failure of Spigot End
• Crushing Failure of Socket End
• Shear Failure of Spigot Collar
• Crushing Failure of Spigot Collar
• Shear Failure of Cotter
• Bending Failure of Cotter
## Design Procedure for Cotter Joint is Made to Calculate:
• The diameter of each rod
• Design of Single eye and double eye.
• The thickness of the cotter by an empirical relationship
• The diameter of the spigot on the basis of tensile stress
• Shear failure
• Crushing failure of spigot etc.
## Applications of Cotter Joint:
The applications of Cotter Joint are as follows.
• It acts as a joint between the crosshead and the piston rod of steam engine.
• It acts as a foundation bolt.
• The cotter joint is used between the railway bogies.
• The cotter joint acts as a steam engine connecting rod strap end.
The Advantages of Cotter Joint are as follows.
• Cotter Joint can take both tensile as well as compressive forces.
• The Assembly and disassembly of Cotter Joint are quicker.
The remaining part for the design procedure of a cotter joint is presented in the form of a video which is shown below.
This is the complete explanation of the Design Procedure for Cotter Joint and Knuckle Joint which is shown in a detailed manner. If you have any doubts, feel free to ask from the comments section. Please Share and Like this blog with the whole world so that it can reach to many.
## Some FAQs on Cotter Joint and Knuckle Joint:
### Where is the knuckle joint used?
The knuckle joint is used to connect two rods under tensile load.
### What type of joint is knuckle?
knuckle joints are biaxial.
More Resources:
Overdrive in Automobile
Hotchkiss Drive & Torque Tube Drive
Mohammed Shafi is the founder, managing editor, and primary author of Mechanical E-Notes. He is an Assistant Professor (Department of Mechanical Engineering) at the Sreenidhi Institute of Science and Technology. He has a 6 years of vast experience in design, research, and data analysis. Shafi has co-authored two journals 1. Dynamic and Fatigue Analysis on Tillage Equipment 2. Welding Procedure and Testing Analysis on Mixing and Nodulizing Drum.
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In the last article we had discussed the basics of the Machining process whereas, in Today's article, we will discuss the Abrasive Water Jet Machining Process along with its Definition, Construction, Working Principle, Process Parameters, Application, Advantages & Disadvantages. So, let's dive into the article. History of Abrasive Water Jet Machining As you know WJM […]
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In the last article, we had discussed what is machining process? and the types of machining processes like EDM, EBM, ECM, USM, etc. whereas in today's article, we will discuss in detail about Laser Beam Machining process along with its Definition, Construction, Working Principle, Types, Process Parameters, MRR, Applications, Advantages & Disadvantages in a detailed […]
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https://brainmass.com/statistics/regression-analysis/correlation-analysis-92548 | 1,484,720,487,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280239.54/warc/CC-MAIN-20170116095120-00529-ip-10-171-10-70.ec2.internal.warc.gz | 803,319,636 | 19,283 | Share
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# Correlation Analysis
Biomechanics of ligament and relationship between age, diagnosis and histology.
The aims were to determine whether there is a link between histology. That is SMA, EVG and MT and the other parameters. And a link between the histological parameters in themselves.
Hypothesis: young's modulus and/or yield and/or strain is related to age/diagnosis/histology
The samples are from ligaments removed from patients after surgery.
Patient number Age Diagnosis Duration of symptoms (months) EVG mt SMA strain Yield Youngs
1 28 Prolapse 6 0.7097 0.1178 82 63.32 1.51 0..067
2 48 Stenosis 84 0.6473 0.0978 26 177 4.55 0.051
3 80 Stenosis 144 0.6785 0.1301 19 105.01 1.4 0.021
4 68 Stenosis 36 0.6329 0.085 17
5 26 Prolapse 27
6 69 Stenosis 48 0.438 0.3909 9 205.29 6.41 0.058
7 36 Prolapse 42 0.6228 0.2268 71 161.67 1.68 0.0327
8 33 Prolapse 18 0.7498 0.1781 72 159.17 1.68 0.021
9 38 Stenosis 96 0.6397 0.2527 76
10 24 Prolapse 12 0.6999 0.134 63
1. is there a correlation between age and any of the other variables
2. is there a correlation between diagnosis and any of the other variables
3. is there a correlation between EVG,SMA or MT independent of other factors
4. Is there a correlation between the strain, yield or youngs and any of the MT, EVG and SMA
5. Is there a correlatin between youngs, yield and strain and age or diagnosis
6. Is there a correlation between duration of symptoms and SMA irrespective of other factors
#### Solution Preview
Biomechanics of ligament and relationship between age, diagnosis and histology.
The aims were to determine whether there is a link between histology. That is SMA, EVG and MT and the other parameters. And a link between the histological parameters in themselves.
Hypothesis: young's modulus and/or yield and/or strain is related to age/diagnosis/histology
The samples are from ligaments removed from patients after surgery.
Patient number Age Diagnosis Duration of symptoms (months) EVG mt SMA strain Yield Youngs
1 28 Prolapse 6 0.7097 0.1178 82 63.32 1.51 0..067
2 48 Stenosis 84 0.6473 0.0978 26 177 4.55 0.051
3 80 Stenosis 144 0.6785 0.1301 19 105.01 1.4 0.021
4 68 Stenosis 36 0.6329 0.085 17
5 26 Prolapse 27
6 69 Stenosis 48 0.438 0.3909 9 205.29 6.41 0.058
7 36 Prolapse 42 0.6228 0.2268 71 161.67 1.68 0.0327
8 33 Prolapse 18 0.7498 0.1781 72 159.17 1.68 0.021
9 38 Stenosis 96 0.6397 0.2527 76
10 24 Prolapse 12 0.6999 0.134 63
1. Is there a correlation between age and any of the other variables
The correlation between age and other variable are given below. The significance of correlation is also tested .
The null hypothesis considered is H0: r =0 Vs H1 : r ≠ 0
The test statistics used is . The conclusions are taken based on the p value. If p value is less than 0.05 , we reject ...
#### Solution Summary
Correlation analysis of Biomechanics of ligament, age, diagnosis and histology.
\$2.19 | 991 | 2,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-04 | latest | en | 0.846392 |
https://www.codingbroz.com/sorting-array-of-strings-c-solution/ | 1,701,806,092,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00032.warc.gz | 804,130,956 | 30,516 | # Sorting Array of Strings in C | HackerRank Solution
Hello coders, today we are going to solve Sorting Array of Strings HackerRank Solution in C.
## Problem
To sort a given array of strings into lexicographically increasing order or into an order in which the string with the lowest length appears first, a sorting function with a flag indicating the type of comparison strategy can be written. The disadvantage with doing so is having to rewrite the function for every new comparison strategy.
A better implementation would be to write a sorting function that accepts a pointer to the function that compares each pair of strings. Doing this will mean only passing a pointer to the sorting function with every new comparison strategy.
Given an array of strings, you need to implement a sorting_sort function which sorts the strings according to a comparison function, i.e, you need to implement the function :
``````void string_sort(const char **arr,const int cnt, int (*cmp_func)(const char* a, const char* b)){
}``````
The arguments passed to this function are:
• an array of strings :
• length of string array:
• pointer to the string comparison function:
You also need to implement the following four string comparison functions:
1. int lexicographic_sort(char*, char*) to sort the strings in lexicographically non-decreasing order.
2. int lexicographic_sort_reverse(char*, char*) to sort the strings in lexicographically non-increasing order.
3. int sort_by_number_of_distinct_characters(char*, char*) to sort the strings in non-decreasing order of the number of distinct characters present in them. If two strings have the same number of distinct characters present in them, then the lexicographically smaller string should appear first.
4. int sort_by_length(char*, char*) to sort the strings in non-decreasing order of their lengths. If two strings have the same length, then the lexicographically smaller string should appear first.
## Input Format
You just need to complete the function `string\_sort` and implement the four string comparison functions.
## Constraints
• 1 <= No. of Strings <= 50
• 1 <= Total Length of all the strings <= 2500
• You have to write your own sorting function and you cannot use the inbuilt qsort function
• The strings consists of lower-case English Alphabets only.
## Output Format
The locked code-stub will check the logic of your code. The output consists of the strings sorted according to the four comparsion functions in the order mentioned in the problem statement.
Sample Input 0
``````4
wkue
qoi
sbv
fekls``````
Sample Output 0
``````fekls
qoi
sbv
wkue
wkue
sbv
qoi
fekls
qoi
sbv
wkue
fekls
qoi
sbv
wkue
fekls``````
## Solution – Sorting Array of Strings in C
### C
```#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int lexicographic_sort(const char* a, const char* b){
return strcmp(a, b);
}
int lexicographic_sort_reverse(const char* a, const char* b){
return strcmp(b, a);
}
int characters_count(const char *s){
int n = 0;
int count[128] = {0};
if (NULL == s){
return -1;
}
while(*s != '\0'){
if (!count[*s]){
count[*s]++;
n++;
}
s++;
}
return n;
}
int sort_by_number_of_distinct_characters(const char* a, const char* b){
int con = characters_count(a) - characters_count(b);
return (con ? con : lexicographic_sort(a, b));
}
int sort_by_length(const char* a, const char* b) {
int len = strlen(a) - strlen(b);
return (len ? len : lexicographic_sort(a, b));
}
void string_sort(char** arr,const int len,int (*cmp_func)(const char* a, const char* b)){
int mid = len / 2;
int con = 0;
while(!con){
con = 1;
for(int i = 0; i < len - 1; i++){
if(cmp_func(arr[i], arr[i + 1]) > 0) {
char *temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp;
con = 0;
}
}
}
}
int main()
{
int n;
scanf("%d", &n);
char** arr;
arr = (char**)malloc(n * sizeof(char*));
for(int i = 0; i < n; i++){
*(arr + i) = malloc(1024 * sizeof(char));
scanf("%s", *(arr + i));
*(arr + i) = realloc(*(arr + i), strlen(*(arr + i)) + 1);
}
string_sort(arr, n, lexicographic_sort);
for(int i = 0; i < n; i++)
printf("%s\n", arr[i]);
printf("\n");
string_sort(arr, n, lexicographic_sort_reverse);
for(int i = 0; i < n; i++)
printf("%s\n", arr[i]);
printf("\n");
string_sort(arr, n, sort_by_length);
for(int i = 0; i < n; i++)
printf("%s\n", arr[i]);
printf("\n");
string_sort(arr, n, sort_by_number_of_distinct_characters);
for(int i = 0; i < n; i++)
printf("%s\n", arr[i]);
printf("\n");
}```
Disclaimer: The above Problem (Sorting Array of Strings) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose. | 1,240 | 4,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-50 | latest | en | 0.796493 |
http://www.jiskha.com/display.cgi?id=1385947966 | 1,495,935,157,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00084.warc.gz | 657,039,286 | 3,803 | # MATH
posted by on .
If point A is (-8,3) and point B is (-6,1) and slope is -1 what would the x and y intercepts be?
What would the equation be in standard form?
• MATH - ,
y -y1 = m(x-x1)
Use either point...
y-3 = -1(x+8
y-3= -1x -8
y = -x -5
Now, change to standard from
to find the x- intercept.. let y =0 and solve for x.
to find the y- intercept, let x = 0 and solve for y.
• MATH - ,
Henry had done this same question for you here
http://www.jiskha.com/display.cgi?id=1385944773
why did you repost it? | 174 | 521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-22 | latest | en | 0.931693 |
http://mathhelpforum.com/advanced-algebra/11548-coefficients-vectors-method-print.html | 1,508,577,953,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824675.67/warc/CC-MAIN-20171021081004-20171021101004-00373.warc.gz | 234,843,530 | 3,487 | # The coefficients-vectors method
• Feb 13th 2007, 08:13 AM
Grunt
The coefficients-vectors method
Ok, I understand how to get it this far. But how are they getting this?
• Feb 13th 2007, 08:20 AM
Richard Rahl
Hello there!
I'm not sure what you are trying to do, multiplication or dot product (the single dot usually indicates dot product).
But the matrix below is the result of multiplying those two matricies. Is that what you are having a problem with?
• Feb 13th 2007, 08:20 AM
Grunt
nm, i just needed to sit back and look at it more.
• Feb 13th 2007, 10:29 AM
ThePerfectHacker
It makes no difference which method you use to multiply matrices, whichever one you understand the best. The important thing is that you know how to multiply them. The one I use is the standard dot product between the rows and coloums.
• Feb 18th 2007, 12:16 PM
Soroban
Hello, Grunt!
Someone taught you a very messy way of multiplying matrices.
. . Shame on them!
ThePerfectHacker referred to a "standard dot product".
. . I thought everyone used it.
Let's multiply "a row times a column".
Code:
``` | 4 | [ 1 2 3 ] · | 5 | = (1)(4) + (2)(5) + (3)(6) | 6 | = 4 + 10 + 18 = 32```
I hope you see what pairs are being multipled.
Code:
``` | 3 1 | | 1 0 2 | | | | | · | 2 1 | | -1 3 1 | | | | 1 0 |```
We will multiply each row in the first matrix
. . by each column in the second matrix.
Here we go . . .
Row-1 times Column-1
Code:
``` | 3 | | 1 0 2 | · | 2 | = (1)(3) + (0)(2) + (2)(1) = 5 | 1 |```
In Row-1, Column-1 of the product matrix (upper left).
Row-1 times Column-2
Code:
``` | 1 | | 1 0 2 | · | 1 | = (1)(1) + (0)(1) + (2)(0) = 1 | 0 |```
This goes in Row-1, Column-2 of the product (upper right).
Row-2 times Column-1
Code:
``` | 3 | |-1 3 1 | · | 2 | = (-1)(3) + (3)(2) + (1)(1) = 4 | 1 |```
This goes in Row-2, Column-1 of the product (lower left).
Row-2 times Column-2
Code:
``` | 1 | |-1 3 1 | · | 1 | = (-1)(1) + (3)(1) + (1)(0) = 2 | 0 |```
This goes in Row-2, Column-2 of the product (lower right).
``` | 5 1 | | 4 2 |``` | 802 | 2,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-43 | longest | en | 0.735481 |
http://purplemath.com/learning/viewtopic.php?f=18&t=1528 | 1,524,640,201,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947705.94/warc/CC-MAIN-20180425061347-20180425081347-00230.warc.gz | 256,465,991 | 7,289 | ## Dual Compounding Interest Equations
Topics that don't fit anywhere else.
GreenLantern
Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm
Contact:
### Dual Compounding Interest Equations
This is from my money and banking class, I'm missing some critical know how for this specific problem and was hoping it could be answered here. It's the last letter in my sixth question, so please excuse the wall of text I'll try to divide and edit it nicely.
Base Scenario: Some friends of yours have just had a child. Thinking ahead, and realizing the power of compound interest, they are considering investing for their child’s college education, which will begin in 18 years. Assume that the cost of a college education today is $125,000; there is no inflation; and there are no taxes in interest income that is used to pay college tuition and expenses. A) If the interest rate is 5 percent, how much will your friends need to put into their savings account today to have$125,000 in 18 years?
D) Return to part A, the case with 5 percent interest rate and no inflation. Assume that your friends don’t have enough financial resources to make the entire investment at the beginning. Instead, they think they will be able to split their investment into two equal parts, one invested immediately and the second invested in five years. Describe how you would compute the required size of the two equal investments, made five years apart.
I can see the issue the question presents. The total investment in part D will be much greater than in part A, so essentially I'm looking at two separate compounding interest equations being added together. If I'm not mistaken that leads me to something like:
$125,000=P(1+\frac{.05}{1}\)^{18}+P(1+\frac{.05}{1}\)^{13}$
Solve for P and call it a day? (Embarrassing algebra will now follow.)
$125,000=P1.05^{18}+P1.05^{13}$
Removed some of the extra stuff...
$125,000=\frac{P1.05^{18}+P1.05^{13}}{1.05^{13}}\$
Gets me...
$\frac{125,000}{1.05^{13}}=P1.05^5+P$
And I get stuck, I did go further but it didn't seem productive to throw out 6 more wrong equations . Honestly I don't think I even set up the right starting equation.
stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
### Re: Dual Compounding Interest Equations
...If I'm not mistaken that leads me to something like:
$125,000=P(1+\frac{.05}{1}\)^{18}+P(1+\frac{.05}{1}\)^{13}$
Solve for P and call it a day? (Embarrassing algebra will now follow.)
$125,000=P1.05^{18}+P1.05^{13}$
Removed some of the extra stuff...
$125,000=\frac{P1.05^{18}+P1.05^{13}}{1.05^{13}}\$
How did you get your last line? Where did the divisor come from?
. . . . .$\mbox{}125,000\, =\, P1.05^{18}\, +\, P1.05^{13}$
...to:
. . . . .$\mbox{}125,000\, =\, P\left(1.05^{18}\, +\, 1.05^{13}\right)$
...by factoring. Then divide through and plug the result into your calculator, to find the value of the variable $P$.
GreenLantern
Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm
Contact:
### Re: Dual Compounding Interest Equations
Bah! Factoring. I concede to you Liz, and thank you. It's always something, right?
Alright, divide by that...
$\frac{125,000}{(1.05^{18}+1.05^{13})} =P$
Punch it into my calculator, then profit. It should come out to be 29122.13 (rounded since I'm dealing with money), which is half the total investment. | 954 | 3,312 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 11, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-17 | latest | en | 0.959787 |
https://www.coursehero.com/file/8780290/The-rest-of-these-notes-contains-parts-of-411-412-413/ | 1,487,678,606,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170708.51/warc/CC-MAIN-20170219104610-00408-ip-10-171-10-108.ec2.internal.warc.gz | 804,870,784 | 21,441 | # The rest of these notes contains parts of 411 412 413
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Unformatted text preview: “gory details”. The rest of these notes contains parts of §§4.1.1, 4.1.2, 4.1.3 and the intro to 4.2. CI. Confidence intervals (abbreviated CI) We will introduce the general approach on a case that is most straightforward. CI.1. Estimating the mean µ in a normal distribution, when the standard deviation σ is known. We need the following: As. assumptions Es. design an experiment, find a statistics (also called “estimator”) Ds. figure out the distribution of the estimator (given the assumptions) CI. using Ds, and the random sample that we drew, compute the CI Let’s take them one at the time: 1 As. The assumptions Let us...
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## This note was uploaded on 02/05/2014 for the course MATH 3339 taught by Professor Staff during the Fall '08 term at University of Houston.
Ask a homework question - tutors are online | 273 | 1,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-09 | longest | en | 0.907606 |
https://slideplayer.com/slide/1682641/ | 1,621,261,429,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991772.66/warc/CC-MAIN-20210517115207-20210517145207-00019.warc.gz | 536,992,458 | 17,819 | Money Matters First Grade Math 1. What coin is worth \$0.01? 1.Penny 2.Nickel 3.Dime.
Presentation on theme: "Money Matters First Grade Math 1. What coin is worth \$0.01? 1.Penny 2.Nickel 3.Dime."— Presentation transcript:
1. What coin is worth \$0.01? 1.Penny 2.Nickel 3.Dime
2. What coin is worth \$0.05? 1.Penny 2.Nickel 3.Dime
3. What coin is worth \$0.10? 1.Nickel 2.Dime 3.Quarter
4. What coin is worth \$0.25? 1.Nickel 2.Dime 3.Quarter
5. How much money is a dime and a nickel? 1.\$0.05 2.\$0.10 3.\$0.15
6. How much money is 2 dimes and a nickel? 1.\$0.10 2.\$0.25 3.\$0.30
7. How much money is 1 quarter and 1 nickel? 1.\$0.30 2.\$0.35 3.\$0.40
8. How much money is 2 quarters? 1.\$0.25 2.\$0.50 3.\$0.75 | 309 | 722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-21 | latest | en | 0.866018 |
https://zhanyihome.com/qa/quick-answer-how-do-you-write-a-check-for-50-cents.html | 1,621,311,593,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989820.78/warc/CC-MAIN-20210518033148-20210518063148-00540.warc.gz | 1,136,998,629 | 8,797 | # Quick Answer: How Do You Write A Check For 50 Cents?
## Can you write no cents on a check?
Line 3: \$ Box Use the standard format: dollar amount, followed by a decimal, followed by the cents amount.
For example: “10.00—” for ten dollars and no cents, “50.00—” for fifty dollars even, and so on.
Note that because the dollar sign is printed on the check already, you don’t have to write it again..
## How do you write a check for 99 cents?
First write the exact amount of your check, using dollars and cents in dollar sing box. If the check is for \$100, then write “100” in the box.Bellow the “pay to the order of” write the amount in word format. In this case, you would write One Hundred and 00/100.
## What’s a standard form?
Standard form is a way of writing down very large or very small numbers easily. … So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form. Small numbers can also be written in standard form.
## How do you write 1000 in words?
How to Write Out Number 1,000 in Words, in (US) American English, Number Converted (Spelled Out) Using Different Letter Cases1,000 written in lowercase: one thousand.WRITTEN IN UPPERCASE: ONE THOUSAND.Title Case: One Thousand.Sentence case: One thousand.
## How do you write a check for thousands hundreds and cents?
Pin It! Here you write out the amount of the check in words. In our example above you would write one thousand five hundred forty two and 63/100. Write out the dollar amount in words and then use a fraction to represent cents.
## How do you write a check for 50 dollars?
Underneath the “Pay to the Order of” line is where you will write out the check’s amount. For example, \$50 can be spelled out as “Fifty dollars” or “Fifty dollars and 0/100 cents.” Also, if the line is not completely filled, it is recommended that you draw a line to the end.
## How do you write 1500 on a check?
\$ (Amount in Numeric Form): Put 1500.00 in the box right after the \$ sign on the same line. Make sure to include the decimal part 00. DOLLARS (Amount in Words): Write One thousand five hundred and 00/100 on the next field as far to the left on that line as possible.
## How do you write money in words?
Writing money in words. You can easily convert and write any amount of money from figures to words. You just have to write the word “dollar” or “dollars” after the amount of dollars and write the word “cent” or “cents” after the amount of cents. Example 1: Let’s say you have \$18.45.
## How do you write 1200 on a check?
1200 Dollar Check Bottom Line To sum up: A check for 1200 dollars can be spelled as One thousand two hundred and xy/100 dollars; check formats, terms and spelling variants differ. In any case, it includes the date, recipient information, signature as well as the monetary amount twice, one time as decimal number 1200.
## How do you write 99 cents?
Encourage the student to write 99 cents as both 99¢ and \$0.99, but ensuring that the two are not combined.
## Can you write a check for under a dollar?
To write a check for less than a full dollar, use a zero to show that there aren’t any dollars. After that, include the number of cents just like all of the other examples. You can also write “No dollars and….” if you prefer.
## How do you write a check in words?
On the line below “Pay to the order of,” write out the dollar amount in words to match the numerical dollar amount you wrote in the box. For example, if you are paying \$130.45, you will write “one hundred thirty and 45/100.” To write a check with cents, be sure to put the cents amount over 100.
## How do you write 5 cents?
To do this, we need to think about 5 out of 100. We can say that 5 cents is 5 hundredths of a dollar since there are 100 pennies in one dollar. Let’s write 5 cents as a decimal using place value. The five is in the hundredths box because five cents is five one hundredths of a dollar.
## How do you write amounts?
You can write the amount in words by writing the number of whole dollars first, followed by the word ‘dollars’. Instead of the decimal point, you will write the word ‘and,’ followed by the number of cents, and the word ‘cents’. If you want, you can write out the numbers using words too.
## How do you write 1500 in English?
1500 in English Words is : one thousand five hundred.
## How do you write dollars and cents in words?
For example, write \$15,237 as “fifteen thousand, two hundred thirty-seven dollars.” When you write an amount that includes a cents figure, write the word “and” after the word “dollars.” Then write the amount in cents, followed by the word “cents.” For example, write \$32.45 as “thirty-two dollars and forty-five cents.”
## How do I type a cent sign?
On iOS and Android devices, press and hold the \$ symbol on the virtual keyboard to access other currency symbols, including the cent sign.
## How do you write 1000000 in words?
One million (1,000,000), or one thousand thousand, is the natural number following 999,999 and preceding 1,000,001. The word is derived from the early Italian millione (milione in modern Italian), from mille, “thousand”, plus the augmentative suffix -one. | 1,280 | 5,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-21 | latest | en | 0.889631 |
https://ru.scribd.com/doc/291158437/activity-sheet-7 | 1,568,534,456,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514570830.42/warc/CC-MAIN-20190915072355-20190915094355-00043.warc.gz | 656,153,045 | 78,200 | Вы находитесь на странице: 1из 3
# ACTIVITY SHEET
Brandon Jones
LEI 4724
Activity File 19
Activity Title: Know Your Math Four Corners
Source: Developing Recreation Skills in Persons with Learning Disabilities. (n.d.). Retrieved
October 24, 2015.
Equipment: Mathematic equations and problems for all ages, Fifty orange cones
Description of Activity:
The goal of this activity is improve participants math skills in a fun, engaging, and constantly moving
activity that will require participants to us both mind and body. The game is quite simple but does
require a large indoor or outdoor space free of any obstructions with corners. If there arent any
corners therapist should us cones to make them, using four cones to make a corner that can fit at least
twenty participants or more if necessary. This activity begins with each participant starting in base
which is in the center of the open space being used so it is big enough to fit all participants
comfortable. Once everybody is in base therapist should randomly select a Caller to guess corners
everybody else is a Runner. Once this is done therapist should clarify where each corner is and the
corresponding number it has (1-4). Every round participants have thirty seconds to choose a corner,
once that time is up caller will choose a corner between 1-4 and those in the corner the caller
chooses is out and have to go back to base. Once in base participants can quickly answer math
equation and if answer correctly they go back in game immediately, if not they are out until a new
game is started. There is no limit to the redemption opportunities as long as participant continues to
answer math equations correctly. The caller must keep eyes close the entire length of game and can
only open after game ends. A participants wins by being the last person in the game while everybody
else is out or in base.
Leadership considerations: This activity works well with big groups from twenty to forty
participants. Consider using any open environment thats inside or outside and free of distractions but
also has available water source for participants to keep hydrated. Therapist should act as a facilitator,
teacher, demonstrator, and motivator while effectively communicating all spectrums of activity. As a
facilitator one would have to know how to play or demonstrate the activity, create math equations for
all ages and set up the activity. I recommend using flash cards because there to answer and regulate.
Participants with cerebral palsy- Have option to be caller for the entire activity as long as they
consistently attempt math equations when therapist engages them to do so. If participant would like
to participate in activity as runner can choose to use only two or all four corners to play. If adjusted
to two for that specific participant caller should be unaware.
Participants with Intellectual Disability- Clarify that if they ever need any assistance during game
that they can raise their hand to gain attention for extra help. Also allow more time during activity to
understand rules of game, concept of choosing a corner, and how to play the game to win. Are
allowed assistance from therapist for any of these areas. If needed allow participant to watch activity
with therapist to allow further explanation while activity is going on.
ACTIVITY SHEET
Brandon Jones
LEI 4724
Activity File 20
Activity Title: Balloon animation Show And Tell
Source: Parenting and Child Health - Health Topics -. (n.d.). Retrieved October 24, 2015
Equipment: 100 balloons of different colors, 100 markers of different color or sizes, 100 pieces of
small candy
Description of Activity:
This activity incorporates creative expression and leisure awareness to explore art through
creating ones favorite animation star, cartoon character, or super hero. Beforehand facilitator or
therapist should already have balloons of different color inflated and ready to be drawn on. This
activity begins with participants selecting a character to create and then selecting a balloon for
their canvas to create character. Following this steps participants have between twenty and thirty
minutes to create animation character with markers or art supplies provided. After this process
student will one by one present to the entire audience animation character telling everybody
character name, show he or she is from, super power, favorite activity or hobby, and why you
chose that character. Once everybody has presented participants are allowed to free play with
friends and animation characters. The presentation or show and tell allows this to be a social and
interactive activity to increase confidence in participants although optional, participants who do
present get rewarded with a piece of small candy.
Leadership considerations: This activity should be done inside where participants can sit on the
floor or table depending on where they fill comfortable to do art work. This activity can be done
groups from nine to thirty. With that being said the therapist should have two areas set up to
participate in activity at. Therapist should be able to effectively demonstrate activity and
communicate rules associated with it. Also should be able or have multiple techniques of getting
participants focus their mind on being creative and trying their best to make sure art work is done at
the best of their abilities..
Participants with ADHD- If needed are allowed to create more than one character as long as first one
is done to best of ability. Also allow participant to interact with other participants if they need ideas
or help with animation. Therapist should monitor participant at all times to help keep on track and
remain focus on creating animation. Therapist can also trace designs and have participant go over
design if needed to help create animation.
Participants with Spina Bifida- Participants are allowed to participant in activity in a group allowing
them and there partner to create animation together. Participant also has unlimited time and can all
extra balloons until they run out to create animation if necessary. Allow participant and partner to
present animation character created to class from where they worked on project at, dont need to
move to front of room.
ACTIVITY SHEET
Brandon Jones
LEI 4724
Activity File 21
Activity Title: Puzzle mania
Source: Therapeutic Recreation Activities & Tx Ideas: Games & Puzzles- Scavenger Hunts.
(n.d.). Retrieved October 24, 2015.
## Equipment: 50 total puzzles, 100-200 small pieces of candy
Description of Activity:
This activity is an all-out puzzle competition with a Halloween twist thats goal is to increase all
participants cognitive functioning and problem solving skills. This done by awarding participants
trips to the Trick or Treat bucket every time a participants earns three points. Participants earn a
point for every puzzle done right and points are doubled, tripled, quadruple, etc. every time a
participant gets consecutives puzzle correct without using points or getting candy. Participants never
loose points in this activity and are never eliminated from activity. At any time participant can get
candy from Trick or Treat bucket as long as they have enough points. If participants cant complete
puzzle they have a choice to sacrifice all points to get a new puzzle if choose to keep points must
finish puzzle currently working on.
Leadership considerations: This activity can be done with groups of all sizes but preferable
between ten to thirty participants. There is little equipment so being able to explain rules and proper
strategies of completing a puzzle is a necessity. This activity should be explained and demonstrated
first before having participants play in activity. | 1,548 | 7,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-39 | latest | en | 0.917118 |
https://www.unix.com/man-page/centos/3/cgerq2.f | 1,715,996,747,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057216.39/warc/CC-MAIN-20240517233122-20240518023122-00377.warc.gz | 943,680,012 | 9,381 | # cgerq2.f(3) [centos man page]
```cgerq2.f(3) LAPACK cgerq2.f(3)
NAME
cgerq2.f -
SYNOPSIS
Functions/Subroutines
subroutine cgerq2 (M, N, A, LDA, TAU, WORK, INFO)
CGERQ2 computes the RQ factorization of a general rectangular matrix using an unblocked algorithm.
Function/Subroutine Documentation
subroutine cgerq2 (integerM, integerN, complex, dimension( lda, * )A, integerLDA, complex, dimension( * )TAU, complex, dimension( * )WORK,
integerINFO)
CGERQ2 computes the RQ factorization of a general rectangular matrix using an unblocked algorithm.
Purpose:
CGERQ2 computes an RQ factorization of a complex m by n matrix A:
A = R * Q.
Parameters:
M
M is INTEGER
The number of rows of the matrix A. M >= 0.
N
N is INTEGER
The number of columns of the matrix A. N >= 0.
A
A is COMPLEX array, dimension (LDA,N)
On entry, the m by n matrix A.
On exit, if m <= n, the upper triangle of the subarray
A(1:m,n-m+1:n) contains the m by m upper triangular matrix R;
if m >= n, the elements on and above the (m-n)-th subdiagonal
contain the m by n upper trapezoidal matrix R; the remaining
elements, with the array TAU, represent the unitary matrix
Q as a product of elementary reflectors (see Further
Details).
LDA
LDA is INTEGER
The leading dimension of the array A. LDA >= max(1,M).
TAU
TAU is COMPLEX array, dimension (min(M,N))
The scalar factors of the elementary reflectors (see Further
Details).
WORK
WORK is COMPLEX array, dimension (M)
INFO
INFO is INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
Author:
Univ. of Tennessee
Univ. of California Berkeley
NAG Ltd.
Date:
September 2012
Further Details:
The matrix Q is represented as a product of elementary reflectors
Q = H(1)**H H(2)**H . . . H(k)**H, where k = min(m,n).
Each H(i) has the form
H(i) = I - tau * v * v**H
where tau is a complex scalar, and v is a complex vector with
v(n-k+i+1:n) = 0 and v(n-k+i) = 1; conjg(v(1:n-k+i-1)) is stored on
exit in A(m-k+i,1:n-k+i-1), and tau in TAU(i).
Definition at line 124 of file cgerq2.f.
Author
Generated automatically by Doxygen for LAPACK from the source code.
Version 3.4.2 Tue Sep 25 2012 cgerq2.f(3)```
## Check Out this Related Man Page
```zgeql2.f(3) LAPACK zgeql2.f(3)
NAME
zgeql2.f -
SYNOPSIS
Functions/Subroutines
subroutine zgeql2 (M, N, A, LDA, TAU, WORK, INFO)
ZGEQL2 computes the QL factorization of a general rectangular matrix using an unblocked algorithm.
Function/Subroutine Documentation
subroutine zgeql2 (integerM, integerN, complex*16, dimension( lda, * )A, integerLDA, complex*16, dimension( * )TAU, complex*16, dimension( *
)WORK, integerINFO)
ZGEQL2 computes the QL factorization of a general rectangular matrix using an unblocked algorithm.
Purpose:
ZGEQL2 computes a QL factorization of a complex m by n matrix A:
A = Q * L.
Parameters:
M
M is INTEGER
The number of rows of the matrix A. M >= 0.
N
N is INTEGER
The number of columns of the matrix A. N >= 0.
A
A is COMPLEX*16 array, dimension (LDA,N)
On entry, the m by n matrix A.
On exit, if m >= n, the lower triangle of the subarray
A(m-n+1:m,1:n) contains the n by n lower triangular matrix L;
if m <= n, the elements on and below the (n-m)-th
superdiagonal contain the m by n lower trapezoidal matrix L;
the remaining elements, with the array TAU, represent the
unitary matrix Q as a product of elementary reflectors
(see Further Details).
LDA
LDA is INTEGER
The leading dimension of the array A. LDA >= max(1,M).
TAU
TAU is COMPLEX*16 array, dimension (min(M,N))
The scalar factors of the elementary reflectors (see Further
Details).
WORK
WORK is COMPLEX*16 array, dimension (N)
INFO
INFO is INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
Author:
Univ. of Tennessee
Univ. of California Berkeley
NAG Ltd.
Date:
September 2012
Further Details:
The matrix Q is represented as a product of elementary reflectors
Q = H(k) . . . H(2) H(1), where k = min(m,n).
Each H(i) has the form
H(i) = I - tau * v * v**H
where tau is a complex scalar, and v is a complex vector with
v(m-k+i+1:m) = 0 and v(m-k+i) = 1; v(1:m-k+i-1) is stored on exit in
A(1:m-k+i-1,n-k+i), and tau in TAU(i).
Definition at line 124 of file zgeql2.f.
Author
Generated automatically by Doxygen for LAPACK from the source code.
Version 3.4.2 Tue Sep 25 2012 zgeql2.f(3)```
Man Page | 1,369 | 4,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.768654 |
https://www.vadepares.cat/congruent-triangles-worksheet-answers | 1,632,516,813,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00415.warc.gz | 1,030,318,198 | 11,550 | Home » coloring page » Congruent Triangles Worksheet Answers
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Midsegment theorem Worksheet Answer Key Elegant Worksheet
Proving Triangles Congruent Worksheet Answers Congruent | 1,761 | 7,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-39 | latest | en | 0.852619 |
http://barkha.net/statistics/statistics-ssc-cgl-part-7/ | 1,553,523,095,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203991.44/warc/CC-MAIN-20190325133117-20190325155117-00439.warc.gz | 22,950,384 | 177,489 | # Statistics for SSC-CGL Tier-II (Paper-III) – Part-7
By | November 17, 2016
## FREQUENCY DISTRIBUTION
Data in a frequency array is ungrouped data. To group the data setting up of a 'frequency distribution' is required. A frequency distribution classifies the data into groups. It is simply a table in which the data are grouped into classes and the number of cases which fall in each class are recorded. It shows the frequency of occurrence of different values of a single Phenomenon.
### A frequency distribution is constructed for three main reasons:
1. To facilitate the analysis of data.
2. To estimate frequencies of the unknown population distribution from the distribution of sample data.
3. To facilitate the computation of various statistical measures.
### Raw data:
The statistical data collected are generally raw data or ungrouped data. Let us consider the daily wages (in Rs ) of 30 labourers in a factory.
The above figures are nothing but raw or ungrouped data and they are recorded as they occur without any pre consideration. This representation of data does not furnish any useful information and is rather confusing to mind. A better way to express the figures in an ascending or descending order of magnitude and is commonly known as array. But this does not reduce the bulk of the data. The above data when formed into an array is in the following form:
The array helps us to see at once the maximum and minimum values. It also gives a rough idea of the distribution of the items over the range . When we have a large number of items, the formation of an array is very difficult, tedious and cumbersome. The Condensation should be directed for better understanding and may be done in two ways, depending on the nature of the data.
### a) Discrete (or) Ungrouped frequency distribution:
In this form of distribution, the frequency refers to discrete value. Here the data are presented in a way that exact measurement of units are clearly indicated. There are definite difference between the variables of different groups of items. Each class is distinct and separate from the other class. Non-continuity from one class to another class exist. Data as such facts like the number of rooms in a house, the number of companies registered in a country, the number of children in a family, etc.
The process of preparing this type of distribution is very simple. We have just to count the number of times a particular value is repeated, which is called the frequency of that class. In order to facilitate counting prepare a column of tallies.
In another column, place all possible values of variable from the lowest to the highest. Then put a bar (Vertical line) opposite the particular value to which it relates.
To facilitate counting, blocks of five bars are prepared and some space is left in between each block. We finally count the number of bars and get frequency.
b) Continuous frequency distribution:
In this form of distribution refers to groups of values. This becomes necessary in the case of some variables which can take any fractional value and in which case an exact measurement is not possible. Hence a discrete variable can be presented in the form of a continuous frequency distribution.
Wage distribution of 100 employees
## Nature of class
The following are some basic technical terms when a continuous frequency distribution is formed or data are classified according to class intervals.
### a) Class limits:
The class limits are the lowest and the highest values that can be included in the class. For example, take the class 30-40. The lowest value of the class is 30 and highest class is 40. The two boundaries of class are known as the lower limits and the upper limit of the class. The lower limit of a class is the value below which there can be no item in the class. The upper limit of a class is the value above which there can be no item to that class. Of the class 60-79, 60 is the lower limit and 79 is the upper limit, i.e. in the case there can be no value which is less than 60 or more than 79. The way in which class limits are stated depends upon the nature of the data. In statistical calculations, lower class limit is denoted by L and upper class limit by U.
### b) Class Interval (i):
The class interval may be defined as the size of each grouping of data. For example, 50-75, 75-100, 100-125… are class intervals. Each grouping begins with the lower limit of a class interval and ends at the lower limit of the next succeeding class interval. It is also called the class width.
### c) Mid-value or mid-point (M.V.):
The central point of a class interval is called the mid value or mid-point. It is found out by adding the upper and lower limits of a class and dividing the sum by 2.
### d) Class Frequency:
Number of observations falling within a particular class interval is called frequency of that class.
Let us consider the frequency distribution of weights if persons working in a company.
In the above example, the class frequency are 25,53,77,95,80,60,30. The total frequency is equal to 420. The total frequency indicate the total number of observations considered in a frequency distribution. | 1,124 | 5,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-13 | longest | en | 0.874061 |
https://ww2.mathworks.cn/help/matlab/creating_plots/combine-multiple-plots.html | 1,696,226,528,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510967.73/warc/CC-MAIN-20231002033129-20231002063129-00899.warc.gz | 667,908,239 | 20,795 | # 合并多个绘图
### 在同一坐标区中合并绘图
```x = linspace(0,10,50); y1 = sin(x); plot(x,y1) title('Combine Plots') hold on y2 = sin(x/2); plot(x,y2) y3 = 2*sin(x); scatter(x,y3) hold off```
### 在图窗中显示多个坐标区
```x = linspace(0,10,50); y1 = sin(x); y2 = rand(50,1); tiledlayout(2,1) % Top plot nexttile plot(x,y1) title('Plot 1') % Bottom plot nexttile scatter(x,y2) title('Plot 2')```
### 创建跨多行或多列的绘图
```x = linspace(0,10,50); y1 = sin(x); y2 = rand(50,1); % Top two plots tiledlayout(2,2) nexttile plot(x,y1) nexttile scatter(x,y2) % Plot that spans nexttile([1 2]) y2 = rand(50,1); plot(x,y2)```
### 修改坐标区外观
```x = linspace(0,10,50); y1 = sin(x); y2 = rand(50,1); tiledlayout(2,1) % Top plot ax1 = nexttile; plot(ax1,x,y1) title(ax1,'Plot 1') ax1.FontSize = 14; ax1.XColor = 'red'; % Bottom plot ax2 = nexttile; scatter(ax2,x,y2) title(ax2,'Plot 2') grid(ax2,'on')```
### 控制图块周围的间距
```x = linspace(0,30); y1 = sin(x); y2 = sin(x/2); y3 = sin(x/3); y4 = sin(x/4); % Create plots t = tiledlayout(2,2); nexttile plot(x,y1) nexttile plot(x,y2) nexttile plot(x,y3) nexttile plot(x,y4)```
```t.Padding = 'compact'; t.TileSpacing = 'compact';```
### 显示共享标题和轴标签
```x1 = linspace(0,20,100); y1 = sin(x1); x2 = 3:17; y2 = rand(1,15); % Create plots. t = tiledlayout(2,1); ax1 = nexttile; plot(ax1,x1,y1) ax2 = nexttile; stem(ax2,x2,y2) % Link the axes linkaxes([ax1,ax2],'x');```
```% Add shared title and axis labels title(t,'My Title') xlabel(t,'x-values') ylabel(t,'y-values') % Move plots closer together xticklabels(ax1,{}) t.TileSpacing = 'compact';``` | 627 | 1,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-40 | longest | en | 0.479234 |
http://voquev.gq/blog7079-bbc-bitesize-forces-and-motion-ks2.html | 1,527,175,036,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00251.warc.gz | 314,302,219 | 6,560 | bbc bitesize forces and motion ks2
# bbc bitesize forces and motion ks2
Bbc bitesize - gcse, Gcse is the qualification taken by 15 and 16 year olds to mark their graduation from the key stage 4 phase of secondary education inbbc bitesize ks3 revision forces and motion bbc18ls13 - duration . gcse bbc science bitesize . bbc bitesize ks3 elements, . this gcse bbc BBC - KS3 Bitesize Science - Forces : Revision. Forces can change the shape of objects and change the way they areBBC Bitesize - GCSE Physics - Forces - Test. What is a force? Learn about Newtons laws of motion and the force of friction. Balanced forces Credit: BBC bitesize forces - KS3 (YouTube) Forces are measured in newtons with a little "n".Year 8 Science Course Description Course Content Assessment Motion and Forces. British Broadcasting Corporation Home.The MI High team from CBBC join Bitesize to play a Science forces game. Looking for the old Forces in action activity? Play it at BBC Science Clips - Forces in action. Find out how how forces, like pushes and pulls, move objects. BBC - KS2 Bitesize Science - Living things. Speaking LessonsLandlord And Tenant Agreement.filming with A and O Productions, and working with motion graphics experts Mint Motion to make 7 short f Worksheet to go with BBC KS2 bitesize plants by heather93a Forces and Motion - Pickwick Electric.bbc ks2 bitesize science solids liquids and gases play ebook, bbc ks2 bitesize forces in action ks2 science - teachers ks2 science name: date: 1. in these twonbsp Опубликовано: 4 years ago.
BBC Bitesize KS3 Revision. Foundation (3-5). Tape 18.Uhh science teacher gave this to us for homework. 3 weeks ago. Gravity isnt a force. Forces and motion. Topics. | 405 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-22 | latest | en | 0.822919 |
https://www.coursehero.com/tutors-problems/Calculus/11518747-Please-solve-and-explain-all-parts-Thank-you-so-much-in-advance/ | 1,544,899,657,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826968.71/warc/CC-MAIN-20181215174802-20181215200802-00532.warc.gz | 837,640,600 | 22,505 | View the step-by-step solution to:
# 1. For each of the following series, determine if it is convergent or divergent. If it is convergent, nd its value. n=0 (ii) 2 5n g"100 n=0 (iii)
Please solve and explain all parts! Thank you so much in advance.
1. For each of the following series, determine if it is convergent or divergent. If it is convergent, find its
value. n=0
(ii) 2 5n g”100
n=0
(iii) Elna}? 1)
°° 1
a series of form Σ n = 0 ∞ r n I f ∣ r ∣... View the full answer
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Browse Documents | 251 | 881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-51 | latest | en | 0.851137 |
https://www.cnblogs.com/kuangbin/p/3346116.html | 1,563,612,928,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526489.6/warc/CC-MAIN-20190720070937-20190720092937-00111.warc.gz | 668,802,154 | 5,206 | # Stone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 179 Accepted Submission(s): 137
Problem Description
Tang and Jiang are good friends. To decide whose treat it is for dinner, they are playing a game. Specifically, Tang and Jiang will alternatively write numbers (integers) on a white board. Tang writes first, then Jiang, then again Tang, etc... Moreover, assuming that the number written in the previous round is X, the next person who plays should write a number Y such that 1 <= Y - X <= k. The person who writes a number no smaller than N first will lose the game. Note that in the first round, Tang can write a number only within range [1, k] (both inclusive). You can assume that Tang and Jiang will always be playing optimally, as they are both very smart students.
Input
There are multiple test cases. For each test case, there will be one line of input having two integers N (0 < N <= 10^8) and k (0 < k <= 100). Input terminates when both N and k are zero.
Output
For each case, print the winner's name in a single line.
Sample Input
1 1 30 3 10 2 0 0
Sample Output
Jiang Tang Jiang
Source
Recommend
liuyiding
1 /* ***********************************************
2 Author :kuangbin
3 Created Time :2013/9/28 星期六 12:09:01
4 File Name :2013长春网络赛\1006.cpp
5 ************************************************ */
6
8 #include <stdio.h>
9 #include <string.h>
10 #include <iostream>
11 #include <algorithm>
12 #include <vector>
13 #include <queue>
14 #include <set>
15 #include <map>
16 #include <string>
17 #include <math.h>
18 #include <stdlib.h>
19 #include <time.h>
20 using namespace std;
21
22 int n,k;
23 int main()
24 {
25 //freopen("in.txt","r",stdin);
26 //freopen("out.txt","w",stdout);
27 while(scanf("%d%d",&n,&k) == 2)
28 {
29 if(n == 0 && k == 0)break;
30 if((n-1)%(k+1) == 0)printf("Jiang\n");
31 else printf("Tang\n");
32 }
33 return 0;
34 }
posted on 2013-09-29 17:33 kuangbin 阅读(...) 评论(...) 编辑 收藏
• 随笔 - 940
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• 引用 - 0
JAVASCRIPT: | 624 | 2,109 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-30 | latest | en | 0.698381 |
http://mixedmath.wordpress.com/2013/02/08/hurwitz-zeta-is-a-sum-of-dirichlet-l-functions-and-vice-versa/ | 1,419,305,854,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802778068.138/warc/CC-MAIN-20141217075258-00042-ip-10-231-17-201.ec2.internal.warc.gz | 188,302,005 | 19,466 | At least three times now, I have needed to use that Hurwitz Zeta functions are a sum of L-functions and its converse, only to have forgotten how it goes. And unfortunately, the current wikipedia article on the Hurwitz Zeta function has a mistake, omitting the $varphi$ term (although it will soon be corrected). Instead of re-doing it each time, I write this detail here, below the fold.
The Hurwitz zeta function, for complex ${s}$ and real ${0 < a \leq 1}$ is ${\zeta(s,a) := \displaystyle \sum_{n = 0}^\infty \frac{1}{(n + a)^s}}$. A Dirichlet L-function is a function ${L(s, \chi) = \displaystyle \sum_{n = 1}^\infty \frac{\chi (n)}{n^s}}$, where ${\chi}$ is a Dirichlet character. This note contains a few proofs of the following relations:
Lemma 1
$\displaystyle \zeta(s, l/k) = \frac{k^s}{\varphi (k)} \sum_{\chi \mod k} \bar{\chi} (l) L(s, \chi) \ \ \ \ \ (1)$
$\displaystyle L(s, \chi) = \frac{1}{k^s} \sum_{n = 1}^k \chi(n) \zeta(s, \frac{n}{k}) \ \ \ \ \ (2)$
Proof: We start by considering ${L(s, \chi)}$ for a Dirichlet Character ${\chi \mod k}$. We multiply by ${\bar{\chi}(l)}$ for some ${l}$ that is relatively prime to ${k}$ and sum over the different ${\chi \mod k}$ to get
$\displaystyle \sum_\chi \bar{\chi}(l) L(s,\chi)$
We then expand the L-function and sum over ${\chi}$ first.
$\displaystyle \sum_\chi \bar{\chi}(l) L(s,\chi)= \sum_\chi \bar{\chi} (l) \sum_n \frac{\chi(n)}{n^s} = \sum_n \sum_\chi \left( \bar{\chi}(l) \chi(n) \right) n^{-s}=$
$\displaystyle = \sum_{\substack{ n > 0 \\ n \equiv l \mod k}} \varphi(k) n^{-s}$
In this last line, we used a fact commonly referred to as the “Orthogonality of Characters” , which says exactly that ${\displaystyle \sum_{\chi \mod k} \bar{\chi}(l) \chi{n} = \begin{cases} \varphi(k) & n \equiv l \mod k \\ 0 & \text{else} \end{cases}}$.
What are the values of ${n > 0, n \equiv l \mod k}$? They start ${l, k + l, 2k+l, \ldots}$. If we were to factor out a ${k}$, we would get ${l/k, 1 + l/k, 2 + l/k, \ldots}$. So we continue to get
$\displaystyle = \sum_{\substack{ n > 0 \\ n \equiv l \mod k}} \varphi(k) n^{-s} = \varphi(k) \sum_n \frac{1}{k^s} \frac{1}{(n + l/k)^s} = \frac{\varphi(k)}{k^s} \zeta(s, l/k) \ \ \ \ \ (3)$
Rearranging the sides, we get that
$\displaystyle \zeta(s, l/k) = \frac{k^s}{\varphi(k)} \sum_{\chi \mod k} \bar{\chi}(l) L(s, \chi)$
To write ${L(s,\chi)}$ as a sum of Hurwitz zeta functions, we multiply by ${\chi(l)}$ and sum across ${l}$. Since ${\chi(l) \bar{\chi}(l) = 1}$, the sum on the right disappears, yielding a factor of ${\varphi(k)}$ since there are ${\varphi(k)}$ characters ${\mod k}$. $\Box$
I’d like to end that the exact same idea can be used to first show that an L-function is a sum of Hurwitz zeta functions and to then conclude the converse using the heart of the idea for of equation 3.
Further, this document was typed up using latex2wp, which I cannot recommend highly enough. | 1,021 | 2,912 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 32, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2014-52 | latest | en | 0.780003 |
https://www.quesba.com/questions/fix-seed-value-50-generate-random-sample-size-n-10000-exponential-1762280 | 1,726,669,513,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00650.warc.gz | 873,309,595 | 28,068 | # Fix the seed value at 50, and generate a random sample of size n = 10000 from an exponential...
Fix the seed value at 50, and generate a random sample of size n = 10000 from an exponential distribution with λ = 2. Create a density histogram and superimpose the histogram with a theoretical distribution. Calculate the sample mean and the sample variance of the randomly generated values. Are your answers within 2% of the theoretical values for the mean and variance of an distribution?
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https://www.kilomegabyte.com/60-tibit-to-gb | 1,590,615,171,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396163.18/warc/CC-MAIN-20200527204212-20200527234212-00263.warc.gz | 766,455,283 | 5,971 | Data Units Calculator
Tebibit to Gigabyte
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The smallest unit of measurement used for measuring data is a bit. A single bit can have a value of either zero(0) or one(1). It may contain a binary value (such as True/False or On/Off or 1/0) and nothing more. Therefore, a byte, or eight bits, is used as the fundamental unit of measurement for data storage. A byte can store 256 different values, which is sufficient to represent standard ASCII table, such as all numbers, letters and control symbols.
Since most files contain thousands of bytes, file sizes are often measured in kilobytes. Larger files, such as images, videos, and audio files, contain millions of bytes and therefore are measured in megabytes. Modern storage devices can store thousands of these files, which is why storage capacity is typically measured in gigabytes or even terabytes.
60 tibit to gb result:
60 (sixty) tebibit(s) is equal 8246.33720832 (eight thousand two hundred and forty-six point thirty-three million seven hundred and twenty thousand eight hundred and thirty-two) gigabyte(s)
What is tebibit?
The tebibit is a multiple of the bit, a unit of information, prefixed by the standards-based multiplier tebi (symbol Ti), a binary prefix meaning 2^40. The unit symbol of the tebibit is Tibit. 1 tebibit = 2^40 bits = 1099511627776 bits = 1024 gibibits
What is gigabyte?
The gigabyte is a multiple of the unit byte for digital information. The prefix giga means 10^9 in the International System of Units (SI). Therefore, one gigabyte is 1000000000 bytes. The unit symbol for the gigabyte is GB.
How calculate tibit. to gb.?
1 Tebibit is equal to 137.438953472 Gigabyte (one hundred and thirty-seven point four hundred and thirty-eight million nine hundred and fifty-three thousand four hundred and seventy-two gb)
1 Gigabyte is equal to 0.0072759576141834259033203125 Tebibit ( tibit)
1 Tebibit is equal to 1099511627776.000000 bits (one trillion ninety-nine billion five hundred and eleven million six hundred and twenty-seven thousand seven hundred and seventy-six point zero × 6 zero bits)
1 Gigabyte is equal to 8000000000 bits (eight billion bits)
60 Tebibit is equal to 65970697666560 Bit (sixty-five trillion nine hundred and seventy billion six hundred and ninety-seven million six hundred and sixty-six thousand five hundred and sixty bit)
Tebibit is greater than Gigabyte
Multiplication factor is 0.0072759576141834259033203125.
1 / 0.0072759576141834259033203125 = 137.438953472.
60 / 0.0072759576141834259033203125 = 8246.33720832.
Maybe you mean Terabit?
60 Tebibit is equal to 65.97069766656 Terabit (sixty-five point ninety-seven billion sixty-nine million seven hundred and sixty-six thousand six hundred and fifty-six tbit) convert to tbit
Powers of 2
tibit gb (Gigabyte) Description
1 tibit 137.438953472 gb 1 tebibit (one) is equal to 137.438953472 gigabyte (one hundred and thirty-seven point four hundred and thirty-eight million nine hundred and fifty-three thousand four hundred and seventy-two)
2 tibit 274.877906944 gb 2 tebibit (two) is equal to 274.877906944 gigabyte (two hundred and seventy-four point eight hundred and seventy-seven million nine hundred and six thousand nine hundred and forty-four)
4 tibit 549.755813888 gb 4 tebibit (four) is equal to 549.755813888 gigabyte (five hundred and forty-nine point seven hundred and fifty-five million eight hundred and thirteen thousand eight hundred and eighty-eight)
8 tibit 1099.511627776 gb 8 tebibit (eight) is equal to 1099.511627776 gigabyte (one thousand and ninety-nine point five hundred and eleven million six hundred and twenty-seven thousand seven hundred and seventy-six)
16 tibit 2199.023255552 gb 16 tebibit (sixteen) is equal to 2199.023255552 gigabyte (two thousand one hundred and ninety-nine point zero × 1 twenty-three million two hundred and fifty-five thousand five hundred and fifty-two)
32 tibit 4398.046511104 gb 32 tebibit (thirty-two) is equal to 4398.046511104 gigabyte (four thousand three hundred and ninety-eight point zero × 1 forty-six million five hundred and eleven thousand one hundred and four)
64 tibit 8796.093022208 gb 64 tebibit (sixty-four) is equal to 8796.093022208 gigabyte (eight thousand seven hundred and ninety-six point zero × 1 ninety-three million twenty-two thousand two hundred and eight)
128 tibit 17592.186044416 gb 128 tebibit (one hundred and twenty-eight) is equal to 17592.186044416 gigabyte (seventeen thousand five hundred and ninety-two point one hundred and eighty-six million forty-four thousand four hundred and sixteen)
256 tibit 35184.372088832 gb 256 tebibit (two hundred and fifty-six) is equal to 35184.372088832 gigabyte (thirty-five thousand one hundred and eighty-four point three hundred and seventy-two million eighty-eight thousand eight hundred and thirty-two)
512 tibit 70368.744177664 gb 512 tebibit (five hundred and twelve) is equal to 70368.744177664 gigabyte (seventy thousand three hundred and sixty-eight point seven hundred and forty-four million one hundred and seventy-seven thousand six hundred and sixty-four)
1024 tibit 140737.488355328 gb 1024 tebibit (one thousand and twenty-four) is equal to 140737.488355328 gigabyte (one hundred and forty thousand seven hundred and thirty-seven point four hundred and eighty-eight million three hundred and fifty-five thousand three hundred and twenty-eight)
2048 tibit 281474.976710656 gb 2048 tebibit (two thousand and forty-eight) is equal to 281474.976710656 gigabyte (two hundred and eighty-one thousand four hundred and seventy-four point nine hundred and seventy-six million seven hundred and ten thousand six hundred and fifty-six)
4096 tibit 562949.953421312 gb 4096 tebibit (four thousand and ninety-six) is equal to 562949.953421312 gigabyte (five hundred and sixty-two thousand nine hundred and forty-nine point nine hundred and fifty-three million four hundred and twenty-one thousand three hundred and twelve)
8192 tibit 1125899.906842624 gb 8192 tebibit (eight thousand one hundred and ninety-two) is equal to 1125899.906842624 gigabyte (one million one hundred and twenty-five thousand eight hundred and ninety-nine point nine hundred and six million eight hundred and forty-two thousand six hundred and twenty-four) | 1,586 | 6,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-24 | latest | en | 0.860091 |
https://forums.t-nation.com/t/undulatingly-periodized-german-volume-training/59369 | 1,516,417,437,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888878.44/warc/CC-MAIN-20180120023744-20180120043744-00349.warc.gz | 681,773,014 | 6,353 | # Undulatingly Periodized German Volume Training!
#1
Since none of my friends care, I'm going to rant to T-Nation about my latest brainchild.
Poliquin's GVT follows a linear stepwise periodization pattern, no? 10 x 10 for 6 microcycles, 10 x 6 for 6 microcycles, then 10 x 6,5,4, for 6 microcycles, then probably a good idea to do 10 x 3 for another 6 microcycles.
So... incorporating the magic of undulating periodization, a size/strength macrocycle would consist of microcycles like so:
MicroCycle(MC) 1 = 10x10
MC2 = 10x6
MC3 = 10x3
MC4 = 10x10
MC5 = 10x6
MC6 = 10x3
.
.
.
.
Until the program stops working, one reaches their goals, or they get bored.
I'm even thinking decreasing by 2 reps every microcycle:
10x10, 10x8, 10x6, 10x4 (and maybe 10x2). Longer period of time between repetition of the same workout twice.
I have a potential client (an 135 lb, 5'4" Italian fellow) who has been the same weight since his 14th birthday and is looking to put on mass. He said, "I don't know what the problem is, I eat so many carbs!"
Oh man. I need to help this guy out.
So. Undulating periodization, plus German Volume Training, plus I'll have him on something similar to massive eating.
Does anyone else agree with me that this plan is FREAKING BRILLIANT?!
Anyone?
Beef
#2
Ive seen many variations of GVT but most seem to stick with 60s rest and people NEVER get a full ten reps on the 10 sets - are you giving your guy the time to adjust to the volume in the programme before necessarily moving him onto a lower total volume?
I dont mind rants like this but more detail bro!!
#3
Oh ok. Sorry 'bout that, eh?
I'm going by the original article on GVT by Charles Poliqiun I read in Bill Phillips' first "sport supplement review" in the 9th grade several years ago.
Supersets with 60s rest between the two exercises. Ex: bench press and chest-supported rows. Bench, 60s, row, 60s, bench, 60s, row, 60s. For 10 sets, then 3 sets of 10-12 in an assistance exercise in a different plane of motion.
If his schedule allows, I'll do the five-day:
1 - Chest and Back
2 - Legs and Abs
3 - Off
4 - Arms and Shoulders
5 - Off
6 - Repeat
If not, I'll figure something else probably based around "UB horizontal", "UB vertical", "LB hip dominant", and "LB knee dominant".
You're probably right about allowing him time to get used to the volume. I should probably ramp it up over a few microcycles first, eh?
I should probably give him 90s between supersets and do the first microcycle with slightly less intensity.
Mesocycles would be 3 microcycles of 10-set sessions to one microcycle of 3-set sessions. 15 days "hard" to 5 days "easy", essentially.
Macrocycles would last until he stops gaining strength and size or reaches his goals and we switch to new ones.
The health club I work for offers "packages" up to 36 sessions at a time. That would mean we could keep this going for 4 mesocycles. I'll have him eating ridiculous amounts of food for the first three mesocycles, then switch to a cutting cycle for the last 4.
He's a true ectomorph so the cutting phase won't need to be very long, if we even need it at all.
#4
Also, if the volume seems like it's going to kill him with overtraining, I'll knock it down to something like 8 sets instead of 10.
Beef
#5
IMO the 10x10 was something you built into rather than come down from and was used to add weight to weightlifters off season. Incidently reps in this protocol normally follow something like this: -
10,10,10,9,8,7,7,8,7,6...
ot uncommon for the body to kick in at set 7 or 8 for some reason.
I like the idea of supersetting the bodyparts that way, and your ideas of progression look good, but if he's a relative newb then it could be a lot more simple - just wait till he hits 10 reps on most sets then up the weight 5-10%.
As far as other programmes are concerned OVT by thib wouldnt be bad, id stick a back off week in the middle though after week 4, then run the other 4 week block after. Its essentially 100 reps per bodypart but split into 5x5's.
Id view the 10x10 as a programme in itself rather than coming down to 10x8, 10x3 or whatever.
JMO. keeps it simple. | 1,124 | 4,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-05 | latest | en | 0.956201 |
https://www.codingstoic.com/index.php/category/kotlin/page/2/ | 1,618,363,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076454.41/warc/CC-MAIN-20210414004149-20210414034149-00232.warc.gz | 816,356,231 | 13,247 | Categories
## Problem 6.2 from EPI -> Increment an arbitrary precision decimal number
This one is very cool, what if you have to implement a plus one operation to a very large number, in most languages you have that like in Java there is a BigInteger, but what if you have to write it in c++? Then you would implement your own addition, so the problem requires to implement +1 operations, which can be easily transformed into two number addition, check out the solution here.
I have also added a solution for the variation of problem 6.2. The variation requires addition of two strings that contain binary representation of two numbers. The solution is not that difficult, there is this hard coded version and a more elegant version with bitwise operations, check out the solution here.
Categories
## Problem 6.1 from EPI -> Variants 1,2,3
In the first variant of the problem 6.1 we have to sort an array such that all items with the same categories are grouped. If you think about it, this is the same as the pivot partition, check out the solution here.
Second variant is a bit more difficult, we need to group items that can fall into one of four categories. I used the same technique that can sort three categories but with a twist, we keep an index of group two right most item and group three left most item. When they cross it means we have solved the problem. We iterate through the array from the beginning and the end (in the three categories variant we need to just look at the items from the beginning), and keep incrementing end index (category three left most index) and beginning index (category two right most index). Check the solution here.
Third variant is quite simple, items can fall into one of two categories, we need to group them. This is easier than variant 1, we just keep an index for right most category 1 item and left most index for category 2 items and move new items into those categories. When those two indexes reach we are done. Check out the solution here.
Categories
## Problem 6.1 from EPI -> National Duch flag
So this one is interesting, the problem requires you to perform a crucial part of a quick sort algorithm the pivot ordering. Basically you need to pick a pivot, and move all elements that are less then the pivot to the left side of an array and bigger elements to the right side of the array. This can be done in various complexities as shown here in code. The trick with this problem is to realize you can make multiple passes for an array, it does not increase the complexity of the solution, so instead of sticking with one pass, break the algorithm into stages, and perform simpler algorithms in each stage. Those simpler algorithms combined will have a solution that works.
Categories
## Problem 5.4 from EPI -> Closest integer with the same weight
This seems like a greedy problem but its quite simple. Look at a place where you can swap 1 and 0 in the given number, the right most the location is the smaller difference will be, that is why the right most is called the least significant bit. So go from right to left and the moment you encounter two bits that differ swap them, take a look at a solution here.
Categories
## Problem 5.3 from EPI -> Reverse bits
This one is similar to computing a parity, since we need to check every bit, we can build a cache of already solved parts and just return from cache a solution exists. The key to the solution is to think about what a reverse means, it is basically last two bits reversed will be the first two bits, and second last two bits reversed will be next two bits, etc. Check out the solution here.
Categories
## Problem 5.1 from EPI -> Compute parity of a word
The solution I present here is not a complete solution, in the book there is log(n) solution which exploits the 32 bit CPU xor and shift operation. Here I am implemented the second best option which has a time complexity of (N/L), n is the word size while l is the cache size. The trick to this solution is to use a mask which can help you to extract bits, once that is done use a cache which will have an index and a value, value will be the parity of that index. While we are calculating the cache we also calculate the parity in a variable called result. Take a look at the solution here.
Categories
## Problem 5.2 from EPI -> Swap bits
The problem 5.2 is quite simple, it asks for bit swapping, so we need to check if i-th or j-th bit are different, if not no need to swap, if they are then just xor the number with 1 shifted left by i bits and 1 shifted left by j bits. Take a look at the code here.
Categories
## Problem 5.6 from EPI -> Compute X.div(Y)
Compute a quotient given x and y without using division. The brute force solution would be to keep deducting y from x until x is less then y, but as always there is a better way. Reverse the operation, make it a multiplication operation, first find out which power of 2 times y is less or equal to x, then deduct that from x, while adding that power of two to the result. Running complexity is 0(n), take a look at the code here.
Categories
## Problem 5.7 from EPI -> Compute X.power(Y)
This problem can be done in a brute force way by multiplying a number by itself y times. But there is a better approach, keep squaring X value and observing Y least significant bit, if y’s lsb is 1 take x value, bit-shift y value by 1 each time. This approach is faster since bit-shifting will converge to zero faster than doing y times squaring of the x value. Take a look at the code here.
Categories
## Problem 5.5 from EPI -> Compute X x Y without arithmetic operands
This one is an interesting problem, the difficulty here is to compute a multiplication with only bitwise operations, so how can this be done?
First we need to figure out how many times we need to add y to x, this can be done by checking how many bits of x are 1, and for each we add shifted y to the sum.
There is also a problem of addition, addition is done with simple bit by bit checking and carry out bit.
Check the complete solution here. | 1,349 | 6,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-17 | latest | en | 0.930127 |
https://cstheory.stackexchange.com/tags/hash-function/hot | 1,719,074,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00005.warc.gz | 158,386,734 | 22,783 | # Tag Info
Accepted
### Two papers give contradictory bounds on linear probing. How do I resolve the disparity?
The first one is average-case analysis, for sets of keys that are already somewhat randomly distributed (chosen either before or after the choice of hash function but with a probability distribution ...
• 51.1k
Accepted
### State of research on SHA-1 Collision Attacks
SHA-1 was SHattered by Stevens et al. They demonstrated that collisions in SHA-1 are practical. They give the first instance of a collision for SHA-1. It is an identical-prefix collision attack that ...
• 206
Accepted
### Improving Bloom filter - can we distinguish elements of a database using less than 2.33275 bits/element?
2.09 bits per element is practically achievable. See http://cmph.sourceforge.net/: "[Compress, Hash, Displace] can generate MPHFs that can be stored in approximately 2.07 bits per key." 1.44 bits per ...
• 11.2k
### Improving Bloom filter - can we distinguish elements of a database using less than 2.33275 bits/element?
1.56 bits per key is now possible using "RecSplit: Minimal Perfect Hashing via Recursive Splitting" by Emmanuel Esposito, Thomas Mueller Graf, and Sebastiano Vigna. It is quite expensive: 1,700 times ...
• 11.2k
Accepted
### Can a result of (any) hash algorithm contain the hash result itself?
For the first question, about the last line, it surely depends on the hash function. For example, suppose each line is a single bit (0 or 1). If the hash function is the xor of the bits, then the ...
• 10.8k
Accepted
### Lower bound for the Schwartz–Zippel lemma in Polynomial Hashing
I think a pragmatic approach would be to use a PRF (see also here) instead of a polynomial hash, because then you can say that the collision probability is $1/M$ (unless someone figures out how to ...
• 12.2k
Accepted
### Fibers of hash functions
Cryptographic hashes I don't think anything is known unconditionally. We can analyze this question using a random oracle assumption. MD5, SHA1, SHA256, etc., have a Merkle-Damgaard structure: they ...
• 12.2k
Accepted
### What is the maximal load of a "latency-bounded" Cuckoo Hash?
Section 4 of the journal version of the original Cuckoo Hashing paper shows that to have insertion succeed with probability $p$, your numbers $T$, $n$, and $\epsilon$ must satisfy \frac{13}{n^2 \...
• 11.2k
As much as you're being downvoted and attacked, your idea is absolutely right, correct, and valid. You've nearly reinvented bcrypt. Let's say we have encryption algorithm (doesn't matter which one): ...
• 131
Accepted
### "Linear" hashing function
Sure. These are known as homomorphic hash functions. There are many schemes: see e.g., https://crypto.stackexchange.com/q/6497/351 for one possible entry point into the literature. One example ...
• 12.2k
• 1,743
Accepted
• 12.2k
1 vote
Accepted
### Extended version of the paper "Consistent Hashing and Random Trees" with proofs
Answering my own question. I found that the authors never published an extended version with proofs. The closest thing to an extended paper is, Rina Panigrahy. Relieving hot spots on the ...
• 141
1 vote
### Reusing 5-independent hash functions for linear probing
Today you should probably use Tabulation hashing for Linear Probing. In The Power of Simple Tabulation Hashing by Mihai Pătrașcu and Mikkel Thorup, this is shown to have at least the same guarantees ...
• 958
1 vote
### Reusing 5-independent hash functions for linear probing
One potential issue is when reading from a hash table, the elements should not be read in the order of the slots if all hash tables use the same hash function. This is because those elements, in that ...
• 11.2k
Only top scored, non community-wiki answers of a minimum length are eligible | 920 | 3,781 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-26 | latest | en | 0.867176 |
http://mathathome.org/lessons/inch-by-inch/ | 1,524,323,440,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945232.48/warc/CC-MAIN-20180421145218-20180421165218-00432.warc.gz | 206,780,379 | 11,913 | ## Inch by Inch
In this lesson, children will use colored measuring worms to measure the legs of the heron in the book
### Math Lesson for:
Toddlers/Preschoolers
(See Step 5: Adapt lesson for toddlers or preschoolers.)
Algebra
Measurement
### Learning Goals:
This lesson will help toddlers and preschoolers meet the following educational standards:
• Understand patterns, relations and functions
• Understand measurable attributes of objects and the units, systems and processes of measurement
• Apply appropriate techniques, tools and formulas to determine measurements
### Learning Targets:
After this lesson, toddlers and preschoolers should be more proficient at:
• Sorting, classifying and ordering objects by size, number and other properties
• Recognizing the attributes of length, volume, weight, area and time
• Comparing and ordering objects according to these attributes
• Understanding how to measure using nonstandard and standard units
• Selecting an appropriate unit and tool for the attribute being measured
• Measuring with multiple copies of units of the same size, such as paper clips laid end to end
• Using tools to measure
• Using repetition of a single unit to measure something larger than the unit, such as measuring the length of a room with a single meter stick
• Developing common referents for measures to make comparisons and estimates
## Inch by Inch
### Lesson plan for toddlers/preschoolers
#### Step 1: Gather materials.
• Inch by Inch by Leo Lionni (Go through the book and cover up the measurements with a sticky note. Once the children have measured each of the bird’s feature’s, write in the measurements. Do not use the measurements provided by the book.)
• Enlarged copies of the various objects that the inchworm measures (robin’s tail, flamingo’s neck, toucan’s beak, heron’s legs, pheasant’s tail, hummingbird and the cover of the book)
Note: Small parts pose a choking hazard and are not appropriate for children age five or under. Be sure to choose lesson materials that meet safety requirements.
#### Step 2: Introduce activity.
1. Post a picture of the heron in your meeting area. Make the inchworms available to use as measuring tools. Ask the children to estimate how many inchworms long the blades of grass are. Record their estimates.
2. Using the blue inchworm, demonstrate how to measure the blades of grass along the sides of the blade, placing one inchworm in front of another inchworm until the blue inchworms are lined up along the entire length of the blade of grass. Write down the measurement next to the blade of grass.
#### Step 3: Engage children in lesson activities.
1. Give each student a set of the enlarged copies and some inchworms. Ask the children to measure each of the bird’s features and record their results.
2. Read the book, Inch by Inch. Pause on each of the pages that the inchworm measures (you should have a sticky note covering the inchworm’s measurement). Write down the children’s results on the sticky note. Ask the children to share and compare their results. Do this all the way through the book until you get to the “whole of the hummingbird.” Do not read the end of the book until the second read-through.
3. Once you have gone through the book and written down the children’s measurement results, go through the book again and, using the blue inchworm, measure each of the bird’s features for an accurate measurement. Write the agreed-upon measurement on a new sticky note and cover the previous measurements.
4. Read the end of the book aloud. Say: “And the inchworm measured and measured, inch by inch, until he inched out of sight.” Ask the children to estimate how many inchworms long it is from where they are sitting on the rug to the door, where they can exit and be “out of sight.” Record their results. Give the children the inchworms and tell them to measure the distance from their sitting spots to the door of the classroom. You might want to separate the children into small groups for this project.
• Tell the children to use all of the different-length measuring worms and compare their results. When you pass out the enlarged copies, have a recording sheet next to each picture.
#### Step 4: Math vocabulary.
• Measure: Use of standard units to find out size or quantity in regard to: length, breadth, height, area, mass, weight, volume, capacity, temperature and time (e.g.,”Let’s measure how many blue inchworms long the robin’s tail is.”)
• How many: The total or sum (e.g.,”How many blue inchworms long is the robin’s tail?”)
• Distance: The length between two points (e.g.,”Using the blue inchworms, let’s measure the distance from where we are sitting to the door.”)
• Estimate: To form an approximate judgment or opinion regarding the amount, worth, size, weight, etc.; calculate approximately (e.g.,”Estimate how many blue inchworms long the blade of grass is.”)
• Inch: An imperial unit for measuring length (e.g.,”The blue inchworm is an inch long.”)
Glossary of MATH vocabulary
#### Step 5: Adapt lesson for toddlers or preschoolers.
##### Adapt Lesson for Toddlers
###### Toddlers may:
• Have trouble lining up the inchworms, one in front of the other, to measure the bird’s features
• Put the worms in their mouths (if this is a possibility, do not use the worms with your toddlers as they pose a choking hazard)
###### Home child care providers may:
• Make some blue inchworms out of paper, cut them out and tell the children to glue them along the side of the bird features being measured (By gluing down the inchworms, the children will have an easier time keeping track of the number of inchworms that they have used and totaling the inchworms at the end of the activity.)
##### Adapt Lesson for Preschoolers
###### Preschoolers may:
• Measure with one unit of measurement with ease
###### Home child care providers may:
• Tell the children to use all of the different-colored measuring worms and compare their results. When you pass out the enlarged copies, have a recording sheet next to each picture.
### Suggested Books
• Inch by Inch by Leo Lionni (New York: HarperCollins, 1995)
### Outdoor Connections
Have the children estimate and then use their inchworms to measure the length of outdoor items. For example, measure the height of the pail. Estimate how many four-inch worms high the pail is and then measure for accuracy. Then estimate how many three-inch worms high the pail is and measure for accuracy. | 1,438 | 6,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-17 | latest | en | 0.884027 |
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