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## UGRD-CS6105 Discrete MathematicsPrelim Q1 to Prelim Exam, Midterm Q1, Q2, Finals Q1, Q2
Classify the sentence below as an atomic statement, a molecular statement, or not a statement at all. If the statement is molecular, identify what kind it is (conjuction, disjunction, conditional, biconditional, negation)
Everybody needs somebody sometime.
-Atomic N/A
Classify the sentence below as an atomic statement, a molecular statement, or not a statement at all. If the statement is molecular, identify what kind it is (conjuction, disjunction, conditional, biconditional, negation).
We can have donuts for dinner, but only if it rains.
-Molecular Conditional
Let A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6, 7}
Find A ∩ B
-{3, 4, 5}
Let A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6, 7}
Find A \ B
-{1, 2}
Determine whether the sentence below is an atomic statement, a molecular statement, or not a statement at all.
• The customers wore shoes.
-Molecular
Classify the sentence below as an atomic statement, a molecular statement, or not a statement at all. If the statement is molecular, identify what kind it is (conjuction, disjunction, conditional, biconditional, negation).
Every natural number greater than 1 is either prime or composite.
-Molecular Conditional
Find the cardinality of R = {20,21,...,39, 40}
| R | = 21
Find the cardinality of S = {1, {2,3,4},0}
| S | = 3
Let A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6, 7}
Find A U B
-{1, 2, 3, 4, 5, 6, 7}
Classify the sentence below as an atomic statement, a molecular statement, or not a statement at all. If the statement is molecular, identify what kind it is (conjuction, disjunction, conditional, biconditional, negation). The sum of the first 100 odd positive integers.
-Atomic N/A
In my safe is a sheet of paper with two shapes drawn on it in colored crayon. One is a square, and the other is a triangle. Each shape is drawn in a single color. Suppose you believe me when I tell you that if the square is blue, then the triangle is green. What do you therefore know about the truth value of the following statement?
The square and the triangle are both green. - The statement is FALSE
Find | R | when R = {2, 4, 6,..., 180}
- 90
Let A = {3, 4, 5}. Find the cardinality of P(A)
- 8
In my safe is a sheet of paper with two shapes drawn on it in colored crayon. One is a square, and the other is a triangle. Each shape is drawn in a single color. Suppose you believe me when I tell you that if the square is blue, then the triangle is green. What do you therefore know about the truth value of the following statement?
The square and the triangle are both blue. - The statement is FALSE
The cardinality of {3, 5, 7, 9, 5} is 5.
-False
n my safe is a sheet of paper with two shapes drawn on it in colored crayon. One is a square, and the other is a triangle. Each shape is drawn in a single color. Suppose you believe me when I tell you that if the square is blue, then the triangle is green. What do you therefore know about the truth value of the following statement?
If the triangle is not green, then the square is not blue. - The statement is TRUE
Determine whether the sentence below is an atomic statement, a molecular statement, or not a statement at all.
• Customers must wear shoes. - Not a Statement
Find |A ∩ B| when A = {1, 3, 5, 7, 9} and B {2, 4, 6, 8, 10}
- 0
Classify the sentence below as an atomic statement, a molecular statement, or not a statement at all. If the statement is molecular, identify what kind it is (conjuction, disjunction, conditional, biconditional, negation).
The Broncos will win the Super Bowl or I’ll eat my hat.
-Molecular Conjunction
GIven the function :
f : Z → Z defined by f = 3n
Which of the following is a possible range of the function?
-all multiples of three
For a function f : N → N, a recursive definition consists of an initial condition together with a recurrence relation.
Find f (1).
- 4
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
- 720
How many 3-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
- 720
Consider the function f : N → N given by f (0) 0 and f (n + 1) f + 2n + 1. Find f (6).
- 36
Rule that states that every function can be described in four ways: algebraically (a formula), numerically (a table), graphically, or in words.
- Rule of four
Defined as the product of all the whole numbers from 1 to n.
-Factorial
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
- 210
IN combinations, the arrangement of the elements is in a specific order.
-false
The ________________________ states that if event A can occur in m ways, and event B can occur in n disjoint ways, then the event “A or B” can occur in m + n ways.
Answer the following: f (1) = 4
What is the element n in the domain such as f = 1 2
Find an element n of the domain such that f = n. 3
Additive principle states that if given two sets A and B, we have |A × B| |A| · |B
-false
Given the diagram, answer the following questions :
How many people takes tea and wine? 32
How many people takes coffee but not tea and wine? 45
What is the difference of persons who take wine and coffee to the persons who the persons who takes tea only? 15
Let A, B and C represent people who like apples, bananas, and carrots respectively. The number of people in A = 10, B = 12 and C = 16. Three people are such that they enjoy apples, bananas as well as carrots. Two of them like apples and bananas. Let three people like apples and carrots. Also, four people are such that they like bananas and carrots.
How many people like apples only? 2
How many people like only one of the three? 26
It is a rule that assigns each input exactly one output
-Function
Determine the number of elements in A U B.
- 18
In my safe is a sheet of paper with two shapes drawn on it in colored crayon. One is a square, and the other is a triangle. Each shape is drawn in a single color. Suppose you believe me when I tell you that if the square is blue, then the triangle is green. What do you therefore know about the truth value of the following statement?
If the triangle is green, then the square is blue. - The statement is TRUE
bijection is a function which is both an injection and surjection. In other words, if every element of the codomain is the image of exactly one element from the domain
Consider the statement, “If you will give me a cow, then I will give you magic beans.” Determine whether the statement below is the converse, the contrapositive, or neither.
• You will give me a cow and I will not give you magic beans. - Converse
surjective and injecive are opposites of each other.
-false
The inverse image of a a subset B of the codomain is the set f −1 (B) {x ∈ X : f (x) ∈ B}.
The range is a subset of the codomain. It is the set of all elements which are assigned to at least one element of the domain by the function. That is, the range is the set of all outputs.
Which of the following translates into “Jack and Jill both passed math” into symbols?
-P Λ Q
In my safe is a sheet of paper with two shapes drawn on it in colored crayon. One is a square, and the other is a triangle. Each shape is drawn in a single color. Suppose you believe me when I tell you that if the square is blue, then the triangle is green. What do you therefore know about the truth value of the following statement?
The square is not blue or the triangle is green. - The statement is FALSE
Suppose P and Q are the statements:
P: Jack passed math.
Q: Jill passed math.
Translate "¬(P ν Q) → Q" into English.
-If Jack or Jill did not pass math, then Jill passed math.
Consider the statement, “If you will give me a cow, then I will give you magic beans.” Determine whether the statement below is the converse, the contrapositive, or neither.
• If I will give you magic beans, then you will give me a cow. - Neither
Consider the statement, “If you will give me a cow, then I will give you magic beans.” Determine whether the statement below is the converse, the contrapositive, or neither.
If I will not give you magic beans, then you will not give me a cow. - Contrapositive
Arithmetic progression is the sum of the terms of the arithmetic series.
-False
Given the series : 2,5,8,11....
What is the type of progression? Arithmetic
What is the sum from 1st to 5th element? 40
What is the missing term?
3,9,__,81....
- 27
Identify the propositional logic of the truth table given
-negation
The study of what makes an argument good or bad.
-Logic
match the following formulas to its corresponding sequence
The sum of the geometric progression is called geometric series
-True
¬(P ∨ Q) is logically equal to which of the following expressions?
- ¬P ∧ ¬Q.
A sequence that involves a common difference in identifying the succeeding terms.
-Arithmetic Progression
De Morgan's law is used in finding the equivalence of a logic expression using other logical functions.
-True
A set of statements, one of which is called the conclusion and the rest of which are called premises.
-argument
Logic Equivalence is the same truth value under any assignment of truth values to their atomic parts.
Match the truth tables to its corresponding propositional logic
sequence is a function from a subset of the set of integers.
Assume the sequence: 1,3,5,7,9, ….
What is the 20th term? 29
What type of progression this suggest? Arithmetic
Deduction rule is an argument that is not always right.
-false
The geometric sequences uses common factor in finding the succeeding terms
What is the 4th and 8th element of a = n^(2) ?
- 16,64
How many possible output will be produced in a proposition of three statements?
- 8
¬P ∨ Q is equivalent to :
- P → Q
If the right angled triangle t, with sides of length a and b and hypotenuse of length c, has area equal to c 2/4, what kind of triangle is this?
-isosceles triangle
*Solution:
(a2+b2 )/4=(1/2)ab
Multiply both sides by 4 to get:
a 2+b2=2ab
Solving for 0 we get:
a 2-2ab+b2=0
And factoring the polynomial we get:
(a-b)2=0
Take the square root of both sides:
±(a-b)=0 a=b Therefore, a triangle with two equal sides is an ISOSCELES Triangle.
Find the contrapositive of the given statement.
If you travel to London by train, then the journey takes at least two hours
-If your journey by train takes less than two hours, then you don’t travel to London.
Which of the following the logic representation of proof by contrapositive?
- P → Q = ¬Q → ¬P
A Euler circuit is a Euler path which starts and stops at the same vertex.
For all n in rational, 1/n ≠ n - 1
- False
*Solution:
1 ? n2 - n 0 ?
n2 - n - 1
n = ( - (-1) ± √ (1 - 4(1)(-1))) / 2(1)
n = ( 1 ± √ 5 ) / 2
√ 5 is an irrational number, therefore
(1/n) ≠ n - 1 for rational n's
The tree elements are called nodes
A sequence of vertices such that consecutive vertices (in the sequence) are adjacent (in the graph). A walk in which no edge is repeated is called a trail, and a trail in which no vertex is repeated (except possibly the first and last) is called a path
-Walk
A function which renames the vertices.
-isomorphism
Solve for the value of n in :
−4= n+7 over 6
What is the matching number for the following graph?
As soon as one vertex of a tree is designated as the root , then every other vertex on the tree can be characterized by its position relative to the root.
An undirected graph G which is connected and acyclic is called ____________.
-tree
What is the minimum height height of a full binary tree?
- 3
Does a rational r value for r 2=6 exist?
- No, a rational r does not exist
direct proof is the simplest style of proof.
Proofs that is used when statements cannot be rephrased as implications.
If n is a rational number, 1/n does not equal n-1.
- True
Let ‘G’ be a connected planar graph with 20 vertices and the degree of each vertex is 3. Find the number of regions in the graph.
- 12
*Solution
By the sum of degrees theorem,
20 ∑ i=1 deg(Vi) = 2|E|
20(3) = 2|E|
|E| = 30
By Euler’s formula,
|V| + |R| = |E| + 2
20+ |R| = 30 + 2
|R| = 12
Hence, the number of regions is 12.
It is an algorithm for traversing or searching tree or graph data structures.
Which of the following statements is NOT TRUE?
-Any tree with at least two vertices has at least two vertices of degree two.
A Bipartite graph is a graph for which it is possible to divide the vertices into two disjoint sets such that there are no edges between any two vertices in the same set.
-True
The given graph is planar.
-TRUE
It is the switching the hypothesis and conclusion of a conditional statement.
-Concerse
The number of edges incident to a vertex.
-Degree of a vertex
Two graphs that are the same are said to be _______________
- isomorphic
A connected graph with no cycles.
- : tree
If two vertices are adjacent, then we say one of them is the parent of the other, which is called the child of the parent.
How many spanning trees are possible in the given figure?
Indicate which, if any, of the following three graphs G = (V, E, φ), |V | = 5, is not isomorphic to any of the other two.
- φ = (A {1,3} B {2,4} C {1,2} D {2,3} E {3,5} F {4,5} )
A graph F is a forest if and only if between any pair of vertices in F there is at most one path
Indicate which, if any, of the following graphs G = (V, E, φ), |V | = 5, is not connected.
- φ = ( a {1,2} b {2,3} c {1,2} d {1,3} e {2,3} f {4,5} )
The number of simple digraphs with |V | = 3 is
- 152
The minimum number of colors required in a proper vertex coloring of the graph.
-Chromatic Number
Match the following properties of trees to its definition.
The child of a child of a vertex is called
- : Grandchild
In a simple graph, the number of edges is equal to twice the sum of the degrees of the vertices.
-False
A bipartite graph has two distinct groups where no vertices in either group connecting to members of their own group
A simple graph has no loops nor multiple edges.
-True
Euler paths must touch all edges
-True
These are lines or curves that connect vertices.
-Edges
Which of the following is false?
- A graph with one odd vertex will have an Euler Path but not an Euler Circuit.
Does this graph have an Euler Path, Euler Circuit, both, or neither?
-Both
Paths start and stop at the same vertex.
-False
A sequence of vertices such that consecutive vertices (in the sequence) are adjacent (in the graph). A walk in which no edge is repeated is called a trail, and a trail in which no vertex is repeated (except possibly the first and last) is called a path.
- Walk
Circuits start and stop at _______________
-same vertex
All graphs have Euler's Path
-False
When a connected graph can be drawn without any edges crossing, it is called ________________ .
-Planar graph
Tracing all edges on a figure without picking up your pencil or repeating and starting and stopping at different spots
-Euler Circuit
How many edges would a complete graph have if it had 6 vertices?
-15
A path which visits every vertex exactly once
-Hamilton Path
connected graph has no isolated vertices
A sequence of vertices such that every vertex in the sequence is adjacent to the vertices before and after it in the sequence
-Walk
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# 100331700 (number)
100,331,700 (one hundred million three hundred thirty-one thousand seven hundred) is an even nine-digits composite number following 100331699 and preceding 100331701. In scientific notation, it is written as 1.003317 × 108. The sum of its digits is 15. It has a total of 7 prime factors and 72 positive divisors. There are 22,932,480 positive integers (up to 100331700) that are relatively prime to 100331700.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 9
• Sum of Digits 15
• Digital Root 6
## Name
Short name 100 million 331 thousand 700 one hundred million three hundred thirty-one thousand seven hundred
## Notation
Scientific notation 1.003317 × 108 100.3317 × 106
## Prime Factorization of 100331700
Prime Factorization 22 × 3 × 52 × 7 × 47777
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 10033170 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 100,331,700 is 22 × 3 × 52 × 7 × 47777. Since it has a total of 7 prime factors, 100,331,700 is a composite number.
## Divisors of 100331700
72 divisors
Even divisors 48 24 12 12
Total Divisors Sum of Divisors Aliquot Sum τ(n) 72 Total number of the positive divisors of n σ(n) 3.3177e+08 Sum of all the positive divisors of n s(n) 2.31439e+08 Sum of the proper positive divisors of n A(n) 4.60792e+06 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 10016.6 Returns the nth root of the product of n divisors H(n) 21.7737 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 100,331,700 can be divided by 72 positive divisors (out of which 48 are even, and 24 are odd). The sum of these divisors (counting 100,331,700) is 331,770,432, the average is 46,079,22.,666.
## Other Arithmetic Functions (n = 100331700)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 22932480 Total number of positive integers not greater than n that are coprime to n λ(n) 716640 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5772102 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 22,932,480 positive integers (less than 100,331,700) that are coprime with 100,331,700. And there are approximately 5,772,102 prime numbers less than or equal to 100,331,700.
## Divisibility of 100331700
m n mod m 2 3 4 5 6 7 8 9 0 0 0 0 0 0 4 6
The number 100,331,700 is divisible by 2, 3, 4, 5, 6 and 7.
• Abundant
• Polite
## Base conversion (100331700)
Base System Value
2 Binary 101111110101111000010110100
3 Ternary 20222210101012220
4 Quaternary 11332233002310
5 Quinary 201141103300
6 Senary 13542242340
8 Octal 576570264
10 Decimal 100331700
12 Duodecimal 297263b0
20 Vigesimal 1b71950
36 Base36 1nqgfo
## Basic calculations (n = 100331700)
### Multiplication
n×y
n×2 200663400 300995100 401326800 501658500
### Division
n÷y
n÷2 5.01658e+07 3.34439e+07 2.50829e+07 2.00663e+07
### Exponentiation
ny
n2 10066450024890000 1009984043962256013000000 101333416103607881619512100000000 10166953904482354896284402163570000000000
### Nth Root
y√n
2√n 10016.6 464.672 100.083 39.8371
## 100331700 as geometric shapes
### Circle
Diameter 2.00663e+08 6.30403e+08 3.16247e+16
### Sphere
Volume 4.23061e+24 1.26499e+17 6.30403e+08
### Square
Length = n
Perimeter 4.01327e+08 1.00665e+16 1.4189e+08
### Cube
Length = n
Surface area 6.03987e+16 1.00998e+24 1.7378e+08
### Equilateral Triangle
Length = n
Perimeter 3.00995e+08 4.3589e+15 8.68898e+07
### Triangular Pyramid
Length = n
Surface area 1.74356e+16 1.19028e+23 8.19205e+07
## Cryptographic Hash Functions
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1. ## Completing the square?
How do I convert an equation from standard form to vertex form?
Standard: y= x^2+6x +5
Vertex form: ?
2. hello SpaceGhost,
The vertex form of a quadratic is given by y = a(x – h)2 + k, where (h, k) is the vertex.
To convert from general to vertex you basically take the 'loose term' out of the general form and do simple 'Completing the Square" :
$\displaystyle y=x^2+6x+5$
$\displaystyle y-5 = x^2+6x$
$\displaystyle y-5+(\frac {6}{2})^2=x^2+6x+(\frac{6}{2})^2$
$\displaystyle \frac {y-20+36}{4}=(x+3)^2$
$\displaystyle y=4(x+3)^2-16$
that should be the vertex form
3. Gah I still don't get it. Why did you get y-5 in the second line and 6/2? I still just dpn't get it. ):
4. From the first line i moved the +5 over to the other side and it became -5. To perform the 'Completing the Square' Method, there must be the squared term and the linear term (which is in our case $\displaystyle 6x$). The 2nd step is to add to both sides of the equation (so equality is established) the square of the co-efficient of the linear term and divided by 2 (in our case, the co-efficient of the linear term was 6, so we divide it by two and then square the whole thing).
Step 3, you simplify the term inside the brackets now if you can. (in our case 6/2 = 3
Step 4: Now we have a perfect square: ($\displaystyle x^2+6x+9$
Step 5: factorise it, you don't really need to because you have already done it before in the previous step. To factorise it, you just take out that halved co-efficient and your 'x', add them together or subtract from another, depending on the sign of the linear term (here we had a positive +6x) so we add the the two terms, then we square because its a perfect square: $\displaystyle (x+3)^2$
then you simplify
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### High School Mathematics8.5 Inverse Relation or Inverse Functions
Inverse Relation: If R is a relation from set A into another set B, then by interchanging the first and second coordinates or ordered pairs of R we get a new relation. This relation is called the inverse relation of R and is denoted by R-|. Set builder form, R-| = {(y,x)/(x,y) Î R}. Note: Domain and range of R-| are respectively range and domin of R. Inverse Functions: To find the inverse of a function written as an equation, interchange the two variables x and y and solve for y. If the inverse is also a function, it is denoted by f-1 Example 1: Find the inverse of the function {(3,4),(5,6),(7,8),(9,10)}. State the domain and range of this inverse. State if the inverse is also a function. Original function: {(3,4),(5,6),(7,8),(9,10)} Interchange the first and second coordinates in each pair. Inverse of the function: {(4,3),(6,5),(8,7),(10,9)} domain: {4,6,8,10} range: {3,5,7,9} Since each element in the domain of the inverse maps onto one and only one element in the range, the inverse of the original function is also a function. Example 2: Find the inverse of the function y = 2x + 4. Original function: y = 2x + 4 Domain of f: {x: x is real} Range of f: {y: y is real} To find the inverse of the function interchange x and y x = 2y + 4 x/2 - 4 = 2y x/2 - 2 = y Therefore the inverse function f-1 is: y = x/2 - 2 Domain of f-1: {x: x is real} Range of f-1: {y: y is real} Directions: Choose the correct answer. Also write at least ten examples of your own.
Q 1: Find the inverse of the function: y = 2x - 5, x = -1, -2, -3y = x/2 + 5/2; x = -7, -9, -11x = y/2 + 5/2; x = 7, 9, 11y = x + 5; x = 3, 5, 7 Q 2: If R-| = {(1,3),(2,5),(3,7)} is a relation then find its range.{3,5,7}{1,2,3} Q 3: If R = {(3,1),(5,2),(7,3),(9,4)} is a relation then find the domain of R-|.{1,2,3,4}{3,5,7,9} Q 4: Find the inverse of the function f(x) = -x + 7x - y = 7y = 7- xx = 7 - y Q 5: Find the inverse of relation or function: y = 4x - 1x = 4y + 1x = 4/y - 1/yy = x/4 + 1/4 Q 6: If R-| = {(1,1),(2,2),(3,3),(4,4)} is a relation then find its range.{1,2,3,4}{1,2,3,5} Q 7: If R = {(1,2),(3,4),(5,6)} is a relation then find the range of R-|.{2,4,6}{1,3,5} Q 8: If R = {(1,2),(2,3),(3,4),(4,5)} is a relation then find its inverse relation.{(1,2),(2,3),(3,4),(4,5)}{(2,1),(3,2),(4,3),(5,4)} Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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MR-116. példa
Hozza egyszerűbb alakra az alábbi kifejezést:
$\frac{30}{\sqrt{2}+\sqrt{3}i}$
$\frac{30}{\sqrt{2}+\sqrt{3}i}$
$=\frac{30}{\sqrt{2}{+}\sqrt{3}{i}}·\frac{\sqrt{2}-\sqrt{3}i}{\sqrt{2}{-}\sqrt{3}}$
$=\frac{30·\left(\sqrt{2}-\sqrt{3}i\right)}{{\left(\sqrt{2}\right)}^{{2}}{-}{\left(\sqrt{3}i\right)}^{{2}}}$
$=\frac{30·\left(\sqrt{2}-\sqrt{3}i\right)}{{2}{-}{3}{{i}}^{{2}}}$
$=\frac{30·\left(\sqrt{2}-\sqrt{3}i\right)}{2-3\left(-1\right)}$
$=\frac{30·\left(\sqrt{2}-\sqrt{3}i\right)}{2+3}$
$=\frac{30·\left(\sqrt{2}-\sqrt{3}i\right)}{5}$
$=6\left(\sqrt{2}-\sqrt{3}i\right)$
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+0
# help ASAP
0
54
1
What is the equation of a line that is perpendicular to y= 3x-2 and passes through the point 6,8
Jul 8, 2020
#1
+26631
+1
Slope of the line you are given = 3 perpindicular is -1/3
so far you will have y = -1/3 x + b put in the point given for x and y to compute 'b'
8 = -1/3 (6) + b so b = 10
the line you are looking for becomes y = -1/3 x + 10
Jul 8, 2020
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Modular Arithmetic
Overview
Modular Arithmetic Overview
Welcome to the course material on Modular Arithmetic in General Mathematics. Modular Arithmetic plays a fundamental role in various mathematical applications, ranging from cryptography to scheduling tasks. In this overview, we will delve into the core concepts of modular arithmetic, its practical implications, and how it relates to real-life scenarios.
Understanding the Concept of Modular Arithmetic
Modular Arithmetic involves working with integers within a specific modulus. It focuses on the remainder that results from dividing one integer by another. For example, in the equation 6 + 4 ≡ k (mod 7), the symbol '≡' denotes congruence modulo 7. This implies that the sum of 6 and 4 leaves a remainder of k when divided by 7. Understanding this concept is crucial in various mathematical operations, especially when dealing with cyclic patterns and repetitive calculations.
Performing Basic Operations in Modular Arithmetic
In modular arithmetic, we perform addition, subtraction, and multiplication operations differently from traditional arithmetic. The operations are carried out within the given modulus, ensuring that the results fall within the specified range. For instance, in modulo 6 arithmetic, 3 multiplied by 5 equals to b, considering the remainder when the product is divided by 6. Mastering these operations is essential for solving modular arithmetic problems accurately and efficiently.
Applying Modular Arithmetic in Real-life Situations
The applications of modular arithmetic extend beyond mathematics into our daily lives. From determining market days to scheduling shift duties, modular arithmetic helps in organizing and tracking recurring events. For example, when calculating clock time, modular arithmetic ensures that the time remains within the 12-hour or 24-hour cycle. Understanding how to apply modular arithmetic in real-life scenarios enhances problem-solving skills and fosters critical thinking.
Converting Numbers from One Base to Another
Another crucial aspect of modular arithmetic is the conversion of numbers from one base to another. By applying modular arithmetic techniques, we can transform numbers between different numerical systems efficiently. This skill is valuable in various fields like computer science and cryptography, where number conversions are prevalent. Understanding the conversion process enhances numerical literacy and promotes a deeper understanding of number systems.
Significance of Modular Arithmetic in Various Scenarios
Modular arithmetic provides a versatile framework for addressing diverse mathematical problems, from calculating recurring patterns to simplifying complex computations. Its significance is evident in fields like number theory, algebra, and cryptography, where the properties of modular arithmetic are extensively utilized. Recognizing the importance of modular arithmetic in various scenarios enables students to approach mathematical challenges with a structured and systematic approach.
Proficiency in Addition, Subtraction, and Multiplication Operations in Modular Arithmetic
Building proficiency in performing addition, subtraction, and multiplication operations in modular arithmetic is essential for solving advanced mathematical problems efficiently. By mastering these operations within the defined modulus, students can tackle complex equations with ease and accuracy. Practice and understanding the underlying principles of modular arithmetic operations enhance problem-solving skills and mathematical fluency.
Utility of Modular Arithmetic in Market Days, Clock Time, Shift Duty, etc.
Modular arithmetic finds practical applications in market days, clock time calculations, shift duties scheduling, and other cyclic events. By utilizing modular arithmetic principles, we can determine the day of the week for a specific date, manage work shifts effectively, and streamline repetitive tasks. Understanding the utility of modular arithmetic in various scenarios empowers individuals to optimize time management, logistics, and planning in their daily routines.
Objectives
1. Perform basic operations in modulo arithmetic
2. Appreciate the utility of modulo arithmetic in market days, clock time, shift duty, etc
3. Apply modulo arithmetic in real-life situations
4. Gain proficiency in addition, subtraction, and multiplication operations in modulo arithmetic
5. Recognize the significance of modulo arithmetic in various scenarios
6. Convert numbers from one base to another
7. Understand the concept of modulo arithmetic
Lesson Note
Modular arithmetic is a branch of mathematics that deals with calculations involving remainders. It is a fundamental concept in number theory with various real-life applications. In modular arithmetic, instead of working with ordinary numbers, we perform operations within a fixed modulus, often denoted by 'mod n'.
Lesson Evaluation
Congratulations on completing the lesson on Modular Arithmetic. Now that youve explored the key concepts and ideas, its time to put your knowledge to the test. This section offers a variety of practice questions designed to reinforce your understanding and help you gauge your grasp of the material.
You will encounter a mix of question types, including multiple-choice questions, short answer questions, and essay questions. Each question is thoughtfully crafted to assess different aspects of your knowledge and critical thinking skills.
Use this evaluation section as an opportunity to reinforce your understanding of the topic and to identify any areas where you may need additional study. Don't be discouraged by any challenges you encounter; instead, view them as opportunities for growth and improvement.
1. What is the result of 9 + 6 modulo 5? A. 3 B. 2 C. 0 D. 4 Answer: B
2. If we have 7 x 8 modulo 3, what is the output? A. 1 B. 2 C. 0 D. 3 Answer: C
3. Calculate 11 - 7 modulo 4. A. 4 B. 3 C. 2 D. 1 Answer: D
4. What is 15 + 9 modulo 7 equal to? A. 2 B. 3 C. 4 D. 5 Answer: B
5. In a clock system, if it is 10 o'clock now and we wait for 8 hours, what will be the time using modulo 12 arithmetic? A. 6 o'clock B. 2 o'clock C. 8 o'clock D. 4 o'clock Answer: B
6. If we have 18 x 5 modulo 6, what is the result? A. 1 B. 3 C. 0 D. 2 Answer: C
7. Calculate 23 - 17 modulo 4. A. 3 B. 2 C. 1 D. 0 Answer: A
8. What is 29 + 13 modulo 8? A. 2 B. 4 C. 6 D. 0 Answer: D
9. If you have 35 x 6 modulo 10, what will be the output? A. 8 B. 2 C. 0 D. 4 Answer: C
10. What is 47 - 19 modulo 5 equal to? A. 3 B. 4 C. 1 D. 2 Answer: B
Past Questions
Wondering what past questions for this topic looks like? Here are a number of questions about Modular Arithmetic from previous years
Question 1
A pair of shoes was sold for N2,250.00 at a loss of 10%. What was the cost price?
Question 1
Find the area, to the nearest cm2 ${}^{}$, of the triangle whose sides are in the ratio 2 : 3 : 4 and whose perimeter is 180 cm.
Question 1
(a) A man purchased 180 copies of a book at N250.00 each. He sold y copies at N300.00 each and the rest at a discount of 5 kobo in the Naira of the cost price.
If he made a profit of N7,125.00, find the value of y.
(b) A trader bought x bags of rice at a cost C = 24x + 103 and sold them at a price, S = x22033x $\frac{{�}^{2}}{20}-33�$.
Find the expression for the profit (i) If 20 bags of rice were sold,
(ii) calculate the percentage profit.
Practice a number of Modular Arithmetic past questions
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Homework Statement
Three balls of masses m1, m2 and m3 are suspended in a horizontal line by light wires and are almost touching. The mass m1 is given a horizontal velocity v so that it collides head-on with the mass m2.
Find an expression for the final kinetic energy of m3?
What value of m2 results in the maximum energy transfer to the mass m3?
The Attempt at a Solution
I considered the problem as two successive two body elastic collisions, with one body initially stationary.
For a collision between mass m1 with initial velocity v and stationary mass m2, I transformed in the zero momentum frame (velocity m1*v/(m1*m2) ), found the respective velocities before and after the collision (the balls keep their velocities form before the collision but swap directions) and then transformed back into the lab frame. So after the collision I got:
v1 = (m1-m2)*v/(m1+m2), for the first ball
v2 = 2m1*v/(m1+m2) for the second ball
So the second ball receives a fraction of 4*m1*m2/((m1+m2)**2) kinetic energy from the first ball
Similarly, the third ball receives a fraction of 4*m2*m3/((m2+m3)**2) from the second ball, so the kinetic energy of the third ball at the end is:
8*(m1**2)*(m2**2)*m3*(v**2)/(((m1+m2)**2)*(m2+m3)**2))
And to get m2 for the maximum energy transfer I differentiated this expression with respect to m2 and got m2=sqrt(m1*m3).
Now I am told neither of these results is correct, and I'm not sure where I've gone wrong, or if the answer I'm comparing to is wrong. Also, apologies for the messy maths, I'm not sure how to make the equations display nicer.
mfb
Mentor
And to get m2 for the maximum energy transfer I differentiated this expression with respect to m2 and got m2=sqrt(m1*m3).
I get the same result for your assumptions, and it was also my first guess before calculating it.
It could be a trick question, but I would be surprised to see that: a small mass m2 for a large m3 will lead to multiple collisions, and might allow to transfer more energy to ball 3.
To format expressions, you can use LaTeX.
gneill
Mentor
If you do a web search for "Newton's Cradle analysis", you'll should find this short article, Newton's Cradle by Donald Simanek, which sheds light on some of the shortcomings of simple analysis attempts. Enjoy.
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# High School Math : How to solve one-step equations with decimals in pre-algebra
## Example Questions
### Example Question #1 : How To Solve One Step Equations With Decimals In Pre Algebra
Solve for if
Explanation:
To solve for we must get all of the numbers on the other side of the equation as .
To do this in a problem where a number is being added to , we must subtract the number from both sides of the equation.
In this case the number is so we subtract from each side of the equation to make it look like this
The numbers on the left side cancel to leave by itself.
To do the necessary subtraction we need to know how to subtract decimals from each other.
To subtract decimals you place the first decimal over the top of the other aligned by the decimal point.
Then go through each place and subtract the top number by the bottom number.
Subtract the numbers in each place like you would any number and
Combine the numbers and keep the decimal in the same place to get
### Example Question #2 : How To Solve One Step Equations With Decimals In Pre Algebra
Solve for if
Explanation:
To solve for we must get all of the numbers on the other side of the equation of .
To do this in a problem where is being multiplied by a number, we must divide both sides of the equation by the number.
In this case the number is so we divide each side of the equation by to make it look like this
Then we must divide the decimals by each other to find the answer.
To divide decimals we line the decimals up like this .
Then ignoring all of the decimal places we divide the top number by the bottom number to get
Then we must apply the decimals.
However many decimal places there are in the denominator will be subtracted from the number of decimal places in the numerator to get the final number of decimal places in our answer.
If the number is positive we move the decimal that number of places to the left of our number.
If the number is negative we move the decimal that number of places to the right of our number.
In this case it would be so we don't have any decimal places and the answer is .
### Example Question #3 : How To Solve One Step Equations With Decimals In Pre Algebra
Solve for if .
Explanation:
To solve for we must get all of the numbers on the other side of the equation of .
To do this in a problem where is being subtracted by a number, we must add the number to both sides of the equation.
In this case the number is so we add to each side of the equation to make it look like this
The numbers on the left side cancel to leave by itself.
To add decimals together you place the decimals one over the top of the other aligned by the decimal point.
If there are no numbers after a tens, hundredths, or thousandths place or further to the right of the decimal just add a zero in the required areas.
Then add each place with the appropriately aligned number to get a result for each number.
In this case the decimals will add together like this
### Example Question #4 : How To Solve One Step Equations With Decimals In Pre Algebra
Solve for if
Explanation:
To solve for we must get all of the numbers on the other side of the equation of .
To do this in a problem where is being multiplied by a coefficient, we must divide both sides of the equation by the coefficient.
In this case the number is so we divide each side of the equation by to make it look like this
Then we must divide the decimals by each other to find the answer.
To divide decimals we line the decimals up like this
Then ignoring all of the decimal places we divide the top number by the bottom number to get
Then we must apply the decimals.
The number of decimal places in the denominator will be subtracted from the number of decimal places in the numerator to get the number of decimal places we must change our answer.
If the number is positive we move the decimal that number of places to the left of our number.
If the number is negative we move the decimal that number of places to the right of our number.
In this case it would be negative so we move the decimal place to the right of to get .
### Example Question #5 : How To Solve One Step Equations With Decimals In Pre Algebra
Solve for .
Explanation:
To solve , add to both sides.
### Example Question #6 : How To Solve One Step Equations With Decimals In Pre Algebra
Solve for .
Explanation:
To solve , we need to isolate . That means we need to add to both sides.
### Example Question #7 : How To Solve One Step Equations With Decimals In Pre Algebra
Solve for .
Explanation:
To solve , you can either plug it into your calculator or realize that . Therefore, we are looking for half of five, which is .
Mathematically, that means:
### Example Question #1 : How To Solve One Step Equations With Decimals In Pre Algebra
Solve for .
Explanation:
To solve for , we first need to isolate it. That means we want ONLY on the left side of the equation.
Subtract from both sides.
### Example Question #9 : How To Solve One Step Equations With Decimals In Pre Algebra
Solve for if .
Explanation:
To solve for , we must move all of the constants to the other side of the equation as .
To do this in a problem where is being multiplied by a number, we must divide both sides of the equation by the number.
In this case the number is , so we divide each side of the equation by to make it look like this:
Then we must divide the decimals by each other to find the answer.
To divide decimals we line the decimals up like this:
Then, ignoring all of the decimal places, we divide the top number by the bottom number to get .
Then we must move the decimals.
The number of decimal places in the denominator will be subtracted from the number of decimal places in the numerator to get the number of places to move the decimal point.
If the number is positive, we move the decimal that number of places to the left of our number.
If the number is negative, we move the decimal that number of places to the right of our number.
In this case the subtraction yields , so we move the decimal place to the right of to get .
Solve for if .
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Class 9 Science
Chapters
# Motion
### Part 2
Question 8: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 ms-1.
Answer: Here, we have, speed =3xx10^8\ ms^(-1)
Time = 5 minute
=5xx60s=300 second
We know that, Distance = Speed x Time
⇒ Distance =300xx3xx10^8\ ms^(-1)
= 900 xx 10^8\ m
=9xx10^10m
Question 9: When will you say a body is in
(i) uniform acceleration?
Answer: When rate of change of motion is same in equal intervals of time.
(ii) non-uniform acceleration?
Answer: When rate of change of motion is not same in equal intervals of time.
Question 10: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.
Answer: Here we have, u=80 km/h =80xx5/(18)=(200)/9 m/s
v=60 km/h =60xx5/(18)=(50)/3 m/s and t=5s
∴ Acceleration (a) =?
We know that v=u+at
Or, a=(v-u)/t
=((50)/3-(200)/9)/5=-(50)/(9xx5)
=-(10)/9 m s-2
Question 11: A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.
Initial velocity, u = 0,
Final velocity, v = 40km/h =40xx5/(18)=(100)/9 m/s
Time (t) = 10 minute = 60 × 10 = 600s
Acceleration (a) =?
We know that v=u+at
Or, a=(v-u)/t
=(100)/(9xx600)=1/(54) m s-2
Question 12: What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
(a) The slope of the distance-time graph for an object in uniform motion is straight line.
(b) The slope of the distance-time graph for an object in non-uniform motion is not a straight line.
Question 13: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer: When the slope of distance-time graph is a straight line parallel to time axis, the object is moving with uniform motion.
Question 14: What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer: When the slope of a speed time graph is a straight line parallel to the time axis, the object is moving with uniform speed.
## Matter in Our Surroundings
Anything that has both mass and volume is called matter. You can also say that anything which has mass and which occupies space is called matter.
## Is Matter Around Us Pure?
Elements and compounds are pure substances. All other substances are mixtures which means they are not pure substances.
## Atoms and Molecules
Read about law of conservation of mass, law of constant proportions and Dalton's atomic theory.
## Structure of Atoms
Atom is made of three particles; electron, proton and neutron. These particles are called fundamental particles of an atom or sub atomic particles.
## Cell: The Fundamental Unit of Life
A cell is capable of independent existence and can carry out all the functions which are necessary for a living being.
## Tissue
A groups of cells which is meant for a specific task is called tissue. Tissues are the first step towards division of labour in complex organisms.
## Diversity in Living Organisms
Without proper classification, it would be impossible to study millions of organisms which exist on this earth.
## Motion
If an object changes its position with respect to a reference point with elapse of time, the object is said to be in motion.
## Force & Laws of Motion
Force has numerous effects. Force can set a stationary body in motion. Force can stop a moving body.
## Gravitation
Earth attracts everything towards it by an unseen force of attraction. This force of attraction is known as gravitation or gravitation pull.
## Work & Energy
When force is exerted on an object and object is displaced, work is said to be done. It means work is the product of force and displacement.
## Sound
Sound is a type of energy. Sound travels in the form of wave from one place to another.
## Why Do We Fall Ill
Health is a state of physical, mental and social well being. A condition in which the affected person is unable to carry out normal activities is termed as disease.
## Natural Resources
Resources which are obtained from nature are called natural resources. Examples: Air, water, soil, wood, etc.
## Improvement in Food Production
Food security is said to exist when all people, at all times, have physical and economic access to sufficient, safe and nutritious food to meet their dietary needs and food preferences for an active and healthy life.
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# CCAKES - Editorial
Practice
Problem Setter : Rounak Neogy
Editorialist: Kanist Agarwal
Easy
# PREREQUISITES:
Basic observations, application of map.
# PROBLEM:
Given an array A of N cakes , where A_i , 1 \le i \le N represents the flavour of the i^{th} cake .You are required to print that flavour of cake which has the highest frequency and is maximum in value.
# EXPLANATION
In the given problem we just need to find the frequency of all the flavours using a map.
STL ordered map is recommended in this problem (For C++ users) because
max(A_i ) \le 10^9 . So using an unordered map is not suggested because it might lead to collisions and you can end up getting a wrong answer. Well that can be fixed too but still using map here is perfectly fine here.
The time complexity of the solution will be O(NlogN) because insertion in map is O(logN).
Space complexity : O(N)
# SOLUTIONS :
Setter's Solution
``````#include<bits/stdc++.h>
using namespace std;
void solve(){
int n;
cin>>n;
vector<int> arr(n);
for(int i=0;i<n;i++){
cin>>arr[i];
}
map<int,int> mp;
for(int i=0;i<n;i++){
mp[arr[i]]++;
}
int maxCount=0,maxFlavour=INT_MIN;
for(auto itr:mp){
if(itr.second==maxCount){
maxCount=itr.second;
maxFlavour=max(maxFlavour,itr.first);
}
else if(itr.second>maxCount){
maxCount=itr.second;
maxFlavour=itr.first;
}
}
cout<<maxFlavour<<endl;
}
int main(){
int t;
cin>>t;
while(t--){
solve();
}
return 0;
}
``````
Editorialist's Solution
``````#include <bits/stdc++.h>
#define ull unsigned ll
#define ll long long int
#define ld long double
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define ff first
#define ss second
#define pi acos(-1)
#define INF 1e14
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
#define F(i,a,n) for(int i=a;i<n;i++)
#define vll vector<ll>
using namespace std;
int main()
{
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int tt;
tt = 1;
cin >> tt;
while (tt--)
{
ll maxi = 0;
ll n;
cin >> n;
vll a(n);
map<ll, ll>m;
F(i, 0, n)
{
cin >> a[i];
m[a[i]]++;
}
ll p = 0;
for (auto i : m)
{
if (i.ss >= p)
{
p = i.ss;
if (maxi < i.ff)
{
maxi = i.ff;
}
}
}
cout << maxi << "\n";
}
return 0;
}
``````
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https://how-to-data.org/how-to-compute-a-confidence-interval-for-a-mean-difference-matched-pairs/
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# How to compute a confidence interval for a mean difference (matched pairs)
## Description
Say we have two sets of data that are not independent of each other and come from a matched-pairs experiment, and we want to construct a confidence interval for the mean difference between these two samples. How do we make this confidence interval? Let’s assume we’ve chosen a confidence level of $\alpha$ = 0.05.
## Using NumPy and SciPy, in Python
View this solution alone.
We’ll use Numpy and SciPy to do some statistics later.
1
2
import numpy as np
from scipy import stats
This example computes a 95% confidence interval, but you can choose a different level by choosing a different value for $\alpha$.
1
alpha = 0.05
We have two samples of data, $x_1, x_2, x_3, \ldots, x_k$ and $x’_1, x’_2, x’_3, \ldots, x’_k$. We’re going to use some fake data below just as an example; replace it with your real data.
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2
sample1 = np.array([15, 10, 7, 22, 17, 14])
sample2 = np.array([ 9, 1, 11, 13, 3, 6])
And now the computations:
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5
6
7
diff_samples = sample1 - sample2 # differences between the samples
n = len(sample1) # number of observations per sample
diff_mean = np.mean(diff_samples) # mean of the differences
diff_variance = np.var( diff_samples, ddof=1 ) # variance of the differences
critical_val = stats.t.ppf(q = 1-alpha/2, df = n - 1) # critical value
1
(0.7033861582274517, 13.296613841772547)
Our 95% confidence interval for the mean difference is $[0.70338, 13.2966]$.
See a problem? Tell us or edit the source.
## Solution, in R
View this solution alone.
We have two samples of data, $x_1, x_2, x_3, \ldots, x_k$ and $x’_1, x’_2, x’_3, \ldots, x’_k$. We’re going to use some fake data below just as an example; replace it with your real data.
1
2
sample.1 <- c(15, 10, 7, 22, 17, 14)
sample.2 <- c(9, 1, 11, 13, 3, 6)
The shortest way to create the confidence interval is with R’s t.test() function. It’s just one line of code (after we choose $\alpha$).
1
2
alpha <- 0.05 # replace with your chosen alpha (here, a 95% confidence level)
t.test(sample.1, sample.2, paired = TRUE, conf.level = 1-alpha)
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2
3
4
5
6
7
8
9
10
Paired t-test
data: sample.1 and sample.2
t = 2.8577, df = 5, p-value = 0.0355
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
0.7033862 13.2966138
sample estimates:
mean difference
7
If you need the lower and upper bounds later, you can save them as variables as follows.
1
2
3
conf.interval <- t.test(sample.1, sample.2, paired = TRUE, conf.level = 1-alpha)
lower.bound <- conf.interval$conf.int[1] upper.bound <- conf.interval$conf.int[2]
It’s also possible to do the computation manually, using the code below.
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2
3
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5
6
7
8
diff.samples <- sample.1 - sample.2 # differences between the samples
n = length(sample.1) # number of observations per sample
diff.mean <- mean(diff.samples) # mean of the differences
diff.variance <- var( diff.samples ) # variance of the differences
critical.val <- qt(p = alpha/2, df = n - 1,
lower.tail=FALSE) # critical value
1
[1] 0.7033862 13.2966138
Either method gives the same result. Our 95% confidence interval is $[0.70338, 13.2966]$.
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+
More Determinants: 3x3 Matrices
Author: c o
Description:
To define and demonstrate the application of determinants of 3x3 matrices.
This packet builds on the learner's knowledge of determinants of 2x2 matrices and applies it to the 3x3 case, introducing new concepts where necessary.
(more)
Tutorial
The Determinant And Its Uses
Before you begin, you might want to checkout the lesson on determinants of 2x2 matricies
Finding Inverse Matrices
The properties of determinants that we learned for 2x2 matrices still hold for the 3x3 case, and in fact, they are true for a square matrix of any size. Hence, the formula for the inverse of a matrix M is still given by M-1 = 1/det(M) * adj(M), where adj(M) is the adjugate matrix.
Unfortunately, the general definition of the adjugate matrix is somewhat complicated. Formally, the adjugate of a matrix M is the transpose of its cofactor matrix. But what are the transpose and cofactor matrix?
For starters, the transpose of a matrix M is given by swapping the columns of M with the rows of M, and is denoted Mt.
For example
is
and is
The cofactor matrix is a little bit more complicated. The cofactor of a matrix M is a matrix of determinants of smaller matrices. More specifically
If M is an nXn matrix, then its cofactor matrix has entries (-1)i+jmij, where each mij is the determinant of an (n-1)X(n-1) matrix that is the result of omitting the ith row and the jth column from M.
All of the above is more complicated to state than it is to calculate, so lets look at the definition of the adjugate matrix for 3x3 matrices below:
Since both calculations can seem a little daunting on pen-and-paper, we work through an example of each for a single 3x3 matrix.
We didn't quite finish with finding the adjugate - the next section wraps up.
Errata and a Quick Follow-Up
Errata - Two Things
First, the final calculation of the (3,3) entry in the matrix should have been -3, not -1.
Second, I forgot to finish finding the adjugate! We only got as far as the cofactor matrix, the remaining step is to take the transpose of the cofactor matrix.
So, the cofactor matrix was:
taking the transpose of this gives the adjugate matrix
Follow-Up
Now that we have found both the determinant of M and the adjugate, we can easily calculate the inverse. The determinant is -7, so using the formula for the inverse, we have
M-1 = 1/det(M) * adj(M) =
Cramer's Rule Example
In this video we solve a matrix equation using Cramer's rule.
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# Express the Product as a Sum or Difference
We will how to express the product as a sum or difference.
1. Convert the product into sum or differences: 2 sin 5x cos 3x
Solution:
2 sin 5x cos 3x = sin (5x + 3x) + sin (5x -3x), [Since 2 sin A cos B = sin (A + B) + sin (A - B)]
= sin 8x + sin 2x
2. Express sin (3∅)/2 ∙ cos (5∅)/2 as sum or difference.
Solution:
sin (3∅)/2 cos (5∅)/2
= 1/2 ∙ 2sin (3∅)/2 cos (5∅)/2
= 1/2 [sin ((3∅)/2 + (5∅)/2) - sin ((5∅)/2 - (3∅)/2)]
= 1/2 (sin 4∅ - sin ∅)
3. Convert 2 cos 5α sin 3α into sum or differences.
Solution:
2 cos 5α sin 3α = sin (5α + 3α) - sin (5α -3α), [Since 2 cos A sin B = sin (A + B) - sin (A - B)]
= sin 8α - sin 2α
4. Express the product as a sum or difference: 4 sin 20° sin 35°
Solution:
4sin 20° sin 35° = 2 ∙ 2 sin20° sin 35°
= 2 [cos (35°- 20°) - cos (35° + 20°)]
= 2 (cos 15° - cos 55°).
5. Convert cos 9β cos 4β into sum or differences.
Solution:
cos 9β cos 4β = ½ ∙ 2 cos 9β cos 4β
= ½ [cos (9β + 4β) + cos (9β - 4β)], [Since 2 cos A cos B = cos (A + B) + cos (A - B)]
= ½ (cos 13β + cos 5β)
6. Prove that, tan (60° - ∅) tan (60° + ∅) = (2 cos 2∅ + 1)/(2 cos 2∅ - 1)
Solution:
L.H.S. = tan (60° - ∅) tan (60° + ∅)
= (2 sin (60° - ∅) sin (60° + ∅))/(2cos (60° - ∅) cos (60° + ∅)
= cos [(60° + ∅) - (60° - ∅)] - cos [(60° + ∅)+ (60° - ∅) ]/(cos[(60° + ∅ )+ (60° - ∅) ] + cos [(60° + ∅) - (60° - ∅) ] )
= (cos 2∅ - cos 120°)/(cos 120° + cos 2∅)
= (cos 2∅ - (-1/2))/(-1/2 + cos 2∅), [Since cos 120° = -1/2]
= (cos 2∅ + 1/2)/(cos 2∅ - 1/2)
= (2 cos 2∅ + 1)/(2 cos 2∅ - 1) proved
7. Convert the product into sum or differences: 3 sin 13β sin 3β
Solution:
3 sin 13β sin 3β = 3/2 ∙ 2 sin 13β sin 3β
= 3/2 [cos (13β - 3β) - cos (13β + 3β)], [Since 2 sin A sin B = cos (A - B) - cos (A + B)]
= 3/2 (cos 10β - cos 16β)
8. Show that, 4 sin A sin B sin C = sin (A + B - C) + sin (B + C - A) + sin (C+ A - B) - sin (A + B + C)
Solution:
L.H.S. = 4 sin A sin B sin C
= 2 sin A (2 sin B sin C)
= 2 sin A {cos (B - C) - cos (B + C)}
= 2 sin A ∙ cos (B - C) - 2 sin A cos (B + C)
= sin (A + B - C) + sin (A - B + C) - [sin (A + B + C) - sin (B + C -A)]
= sin (A + B - C) + sin (B + C - A) + sin (A + C - B) - sin (A + B + C) = R.H.S.
Proved
Converting Product into Sum/Difference and Vice Versa
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## Recent Articles
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The procedure of estimating sum and difference are in the following examples. Example 1: Estimate the sum 5290 + 17986 by estimating the numbers to their nearest (i) hundreds (ii) thousands.
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May 22, 24 06:14 PM
While rounding off to the nearest thousand, if the digit in the hundreds place is between 0 – 4 i.e., < 5, then the hundreds place is replaced by ‘0’. If the digit in the hundreds place is = to or > 5…
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May 22, 24 05:17 PM
While rounding off to the nearest hundred, if the digit in the tens place is between 0 – 4 i.e. < 5, then the tens place is replaced by ‘0’. If the digit in the units place is equal to or >5, then the…
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Round off to nearest 10 is discussed here. Rounding can be done for every place-value of number. To round off a number to the nearest tens, we round off to the nearest multiple of ten. A large number…
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## Enter a whole number and click on OK button to display its factors
Find the multiples of an integer inside an interval of two integers
Find the GCF and LCM of two whole numbers
## What are the factors of 18?
Here are all the factors of 18: 1; 2; 3; 6; 9; 18.
18 has more than two factors because 18 is not a prime number, it is a composite number. A composite number is an integer that is obtained by multiplying at least 2 prime numbers, we can say a composite number is a multiple of prime numbers.
18 has six factors, the factors of 18 are also called divisors of 18, when 18 is divided by one of its factors, the division gives integer quotient and zero remainder.
The prime factors of 18 are: 2 and 3
The prime factorization of 18 is: 2*3*3=2*3^2
If 18 is a factor of a positive integer then that positive integer is a multiple of 18, we can divide it by 18 to get an integer quotient and a remainder equal to zero, we can also multiply 18 by another natural number for obtaining the multiple of 18.
Example: 3 multiplied by 6 is equal to 18, then 3 and 6 are factors of 18. 18 is a commun multiple of 3 and 6. By dividing 18 by 6 we get 3. That means if we make the division of 18 by one of its factors, the quotient must be another factor of 18.
There are 3 pairs of factors of 18 whose multiplication makes it possible to find 18: 1 and 18 (1*18=18); 2 and 9 (2*9=18); 3 and 6 (3*6=18).
Here are all the factors of some integers (composite numbers from 242 to 260):
### The factors of 242:
1; 2; 11; 22; 121; 242
### The factors of 244:
1; 2; 4; 61; 122; 244
### The factors of 245:
1; 5; 7; 35; 49; 245
### The factors of 246:
1; 2; 3; 6; 41; 82; 123; 246
### The factors of 248:
1; 2; 4; 8; 31; 62; 124; 248
1; 3; 83; 249
### The factors of 250:
1; 2; 5; 10; 25; 50; 125; 250
### The factors of 252:
1; 2; 3; 4; 6; 7; 9; 12; 14; 18; 21; 28; 36; 42; 63; 84; 126; 252
1; 2; 127; 254
### The factors of 255:
1; 3; 5; 15; 17; 51; 85; 255
### The factors of 256:
1; 2; 4; 8; 16; 32; 64; 128; 256
### The factors of 258:
1; 2; 3; 6; 43; 86; 129; 258
### The factors of 260:
1; 2; 4; 5; 10; 13; 20; 26; 52; 65; 130; 260
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Prove the sum of two odd numbers is even | MathCelebrity Forum
Prove the sum of two odd numbers is even
math_celebrity
Staff member
Take two arbitrary integers, x and y
We can express the odd integer x as 2a + 1 for some integer a
We can express the odd integer y as 2b + 1 for some integer b
x + y = 2a + 1 + 2b + 1
x + y = 2a + 2b + 2
Factor out a 2:
x + y = 2(a + b + 1)
Since 2 times any integer even or odd is always even, then x + y by definition is even.
Last edited:
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# Spatial descriptive statistics
Spatial descriptive statistics are used for a variety of purposes in geography, particularly in quantitative data analyses involving Geographic Information Systems (GIS).
## Types of spatial data
The simplest forms of spatial data are gridded data, in which a scalar quantity is measured for each point in a regular grid of points, and point sets, in which a set of coordinates (e.g. of points in the plane) is observed. An example of gridded data would be a satellite image of forest density that has been digitized on a grid. An example of a point set would be the latitude/longitude coordinates of all elm trees in a particular plot of land. More complicated forms of data include marked point sets and spatial time series.
## Measures of spatial central tendency
The coordinate-wise mean of a point set is the centroid, which solves the same variational problem in the plane (or higher-dimensional Euclidean space) that the familiar average solves on the real line — that is, the centroid has the smallest possible average squared distance to all points in the set.
## Measures of spatial dispersion
Dispersion captures the degree to which points in a point set are separated from each other. For most applications, spatial dispersion should be quantified in a way that is invariant to rotations and reflections. Several simple measures of spatial dispersion for a point set can be defined using the covariance matrix of the coordinates of the points. The trace, the determinant, and the largest eigenvalue of the covariance matrix can be used as measures of spatial dispersion.
A measure of spatial dispersion that is not based on the covariance matrix is the average distance between nearest neighbors.[1]
## Measures of spatial autocorrelation
See the spatial autocorrelation section from Wikipedia's spatial analysis page.
## Measures of spatial homogeneity
A homogeneous set of points in the plane is a set that is distributed such that approximately the same number of points occurs in any circular region of a given area. A set of points that lacks homogeneity may be spatially clustered at a certain spatial scale. A simple probability model for spatially homogeneous points is the Poisson process in the plane with constant intensity function.
### Ripley's K and L functions
Ripley's K and L functions [2] are closely related descriptive statistics for detecting deviations from spatial homogeneity. The K function (technically its sample-based estimate) is defined as
${\displaystyle {\widehat {K}}(t)=\lambda ^{-1}\sum _{i\neq j}{\frac {I(d_{ij}
where dij is the Euclidean distance between the ith and jth points in a data set of n points, t is the search radius, λ is the average density of points (generally estimated as n/A, where A is the area of the region containing all points) and I is the indicator function (1 if its operand is true, 0 otherwise).[3] In 2 dimensions, if the points are approximately homogeneous, ${\displaystyle {\widehat {K}}(t)}$ should be approximately equal to πt2.
For data analysis, the variance stabilized Ripley K function called the L function is generally used. The sample version of the L function is defined as
${\displaystyle {\widehat {L}}(t)=\left({\frac {{\widehat {K}}(t)}{\pi }}\right)^{1/2}.}$
For approximately homogeneous data, the L function has expected value t and its variance is approximately constant in t. A common plot is a graph of ${\displaystyle t-{\widehat {L}}(t)}$ against t, which will approximately follow the horizontal zero-axis with constant dispersion if the data follow a homogeneous Poisson process.
Using Ripley's K function you can determine whether points have a random, dispersed or clustered distribution pattern at a certain scale.[4]
## References
1. ^ Clark, Philip; Evans, Francis (1954). "Distance to nearest neighbor as a measure of spatial relationships in populations". Ecology. 35 (4): 445–453. doi:10.2307/1931034. JSTOR 1931034.
2. ^ Ripley, B.D. (1976). "The second-order analysis of stationary point processes". Journal of Applied Probability. 13 (2): 255–266. doi:10.2307/3212829. JSTOR 3212829.
3. ^ Dixon, Philip M. (2002). "Ripley's K function" (PDF). In El-Shaarawi, Abdel H.; Piegorsch, Walter W. (eds.). Encyclopedia of Environmetrics. John Wiley & Sons. pp. 1796–1803. ISBN 978-0-471-89997-6. Retrieved April 25, 2014.
4. ^ Wilschut, L.I.; Laudisoit, A.; Hughes, N.K.; Addink, E.A.; de Jong, S.M.; Heesterbeek, J.A.P.; Reijniers, J.; Eagle, S.; Dubyanskiy, V.M.; Begon, M. (2015). "Spatial distribution patterns of plague hosts: point pattern analysis of the burrows of great gerbils in Kazakhstan". Journal of Biogeography. 42 (7): 1281–1292. doi:10.1111/jbi.12534. PMC 4737218. PMID 26877580.
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# Insertion Sort solved by 2492
Feb. 21, 2014, 4:21 p.m. by Rosalind Team
Topics: Sorting
## Computing the number of swaps in insertion sort
Insertion sort is a simple sorting algorithm that builds the final sorted array one item at a time. It is much less efficient on large lists than more advanced algorithms such as “Quick Sort”, “Heap Sort”, or “Merge Sort”. However, insertion sort provides several advantages: simple implementation, efficient for (quite) small data sets, $O(1)$ extra space.
When humans manually sort something (for example, a deck of playing cards), most use a method that is similar to insertion sort.
Source: Wikipedia
Although it is one of the elementary sorting algorithms with $O(n^2)$ worst-case time, insertion sort is the algorithm of choice either when the data is nearly sorted (because it is adaptive) or when the problem size is small (because it has low overhead).
For these reasons, and because it is also stable, insertion sort is often used as the recursive base case (when the problem size is small) for higher overhead divide-and-conquer sorting algorithms, such as “Merge Sort” or “Quick Sort”.
Visualization by David R. Martin: http://www.sorting-algorithms.com/insertion-sort
## Problem
Insertion sort is a simple algorithm with quadratic running time that builds the final sorted array one item at a time.
Given: A positive integer $n \le 10^3$ and an array $A[1..n]$ of integers.
Return: The number of swaps performed by insertion sort algorithm on $A[1..n]$.
## Sample Dataset
6
6 10 4 5 1 2
## Sample Output
12
## Discussion
For this problem, it is enough to implement the pseudocode above with a quadratic running time and to count the number of swaps performed. Note however that there exists an algorithm counting the number of swaps in $O(n \log n)$.
Note also that Insertion Sort is an in-place algorithm as it only requires storing a few counters.
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# {FREE} Missing Factor BINGO Game: Fun Multiplication Challenge
Are your children learning their multiplication facts? We know that it is important for children to be able to skip count and be able to answer simple multiplication answers.
Another aspect of teaching multiplication is to focus on fact families, which in turn leads to finding missing factors in expressions. And this free, low-prep missing factor game is a wonderful way to practice finding missing factors.
Please Note: This post contains affiliate links which help support the work of this site. Read our disclosure policy here.
This is a guest post from Rachel at You’ve Got This Math.
## What is a Multiplication Factor?
Before we begin, it is important that your students understand the difference between a factor and product.
Factors are the numbers multiplied together to get the answer to a multiplication problem.
A product is the answer to a multiplication problem, or what we get when we multiply two or more factors.
## How to find Missing Factors
### Hands On Manipulatives:
Many times children need to see and play with manipulatives to fully understand a concept. Getting out cubes or another counting source is one way to let children work through finding factors.
First, have them figure out what the product is, and then count out that many objects.
Next, have them figure out the known factor is and organize their objects into that many groups.
After figuring out the number of groups, the students need to make sure their groups are equal.
Finally, the number of objects in each group is the missing factor.
Likewise, you could have them use the known factor to organize their objects into groups of that size (for example, make piles of 6…) and then see how many groups they are able to make. In this case, the number of groups is the missing factor.
This simple exercise builds an understanding of the relationships between multiplication and division. It also helps build a foundation for algebra!
Practice and Review Multiplication Facts with Fact Fluency: Mastering Multiplication App on iTunes:
### Fact Families:
If your children know their multiplication facts, then presenting this concept with fact family triangles would be a wonderful introduction.
Simply show them a fact family triangle with one angle covered up, and have the child figure out the answer.
Then show them an expression with the same factor missing, and let them solve.
### Multiplication Chart:
This is an easy way for children to find the answers to missing factor expression.
All they need to do is find one of the factors at the top of the chart, scan down until they find the product, and then move their finger to the left until they find the missing factor.
This is definitely an easy option, but it isn’t going to help children work on their multiplication facts. I would recommend starting with the other options first.
## Missing Factor Game Time:
### Printable Game Set Up:
This missing factor BINGO game is simple to play and requires almost no-prep. All you need to do is…
• First, print off the game boards and distribute to players (get the download at the end of the post) Optional: Laminate them for added durability.
• Next, grab a die and game markers and you are ready to go!
### How to Play Missing Factor BINGO:
The object of this game is to be the first to get four in a row or cover up the whole board (decide how you’ll play before you begin).
• To begin, each player rolls one die and then find an expression on their game board in which that the number can be used as a factor
• Players continue rolling at the same time, finding an expression that works with the number they rolled
• If a player has no options left given the number they rolled, they skip that turn
• When a player gets four in a row or covers up the whole board they are the winner!
I hope you enjoy exploring this multiplication concept with your kiddos.
If you’re looking for a more comprehensive study of multiplication, check out my Introduction to Multiplication Bundle.
This includes skip counting practice, hands on multiplication lessons and easy, low prep practice games to help kids develop fact fluency.
Buy Intro to Multiplication Here!
### {Click HERE to go to my shop and grab this FREE Missing Factor Game!}
Rachel is a homeschool mom to four little ones, ages 2 to 6. She is a former public elementary teacher, and has recently begun blogging at her page You’ve Got This. You can also find her on Facebook and Pinterest.
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Subshells
Discussion in 'DAT Discussions' started by Awuah29, Mar 24, 2007.
1. Awuah29 Christian predent 7+ Year Member
Joined:
Oct 13, 2003
Messages:
294
0
Kaplan pg.213 ( old editon 3rd red book)
Example which fill first, the 3d subshell or the 4s subshell. Dont understand their example. If I have to determine which subshell fills first , I just look at the flowing chart.
Why is 3d n=3 and l=2 and 4s n=4 and l=0
Don't get that part??? Anyone can explain . thanks alot
Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
2. DentalKitty 5+ Year Member
Joined:
Mar 12, 2007
Messages:
202
0
Status:
Pre-Dental
For 3d, when n=3, the values for l can be from n-1 to 0 so l=2, 1, 0. We know that the d subshell is when l=2. So n+l=3+2=5. For 4s, n=4 so l can be 3, 2, 1, 0. We know that the s subshell is when l=0. So n+l=4+0=4. Since 4 is less than 5 the 4s subshell will fill first. The diagram is also very helpful though and much faster
3. DDSDakoo 7+ Year Member
Joined:
Feb 27, 2007
Messages:
56
0
a good approach also is this
N in is these kinds of problems is the number infront of the letter.
so if you have
WHICH FILLS FIRST 3p or 4s?
In the case of the 3p, it's 3. and in 4s, its 4.
to figure out l: commit s=0, p=1, d=2, f=3 to ur memory.
l is directly correlated to these letters in these problems
So 3p: n=3 and L=1 , following n+l= 3+1= 4
4s: n=4, and L=0, n+l= 4
what 2 do now if they are both the same #?
Go with the one that gives u a lower N. IN THIS CASE, 3p. hope that helps
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# How do you write 66,793 in scientific notation?
Jun 21, 2016
$66793 = 6.6793 \times {10}^{4}$
#### Explanation:
In scientific notation, we write a number so that it has single digit to the left of decimal sign and is multiplied by an integer power of $10$.
Note that moving decimal $p$ digits to right is equivalent to multiplying by ${10}^{p}$ and moving decimal $q$ digits to left is equivalent to dividing by ${10}^{q}$.
Hence, we should either divide the number by ${10}^{p}$ i.e. multiply by ${10}^{- p}$ (if moving decimal to right) or multiply the number by ${10}^{q}$ (if moving decimal to left).
In other words, it is written as $a \times {10}^{n}$, where $1 \le a < 10$ and $n$ is an integer.
To write $66793$ in scientific notation, we will have to move the decimal point four points to left, which literally means dividing by ${10}^{4}$.
Hence in scientific notation $66793 = 6.6793 \times {10}^{4}$ (note that as we have moved decimal four points to left we are multiplying by ${10}^{4}$ to compensate for dividing by ${10}^{4}$..
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# How do you factor: y= x^2 + 6x – 27 ?
##### 1 Answer
Jan 11, 2016
To factor trinomials of the form y = $a {x}^{2}$ + bx + c, a = 1, find two numbers that multiply to c and that add to b.
#### Explanation:
Two numbers that multiply to -27 and add to +6 are +9 and -3.
So, y = #x^2 + 6x - 27
y = (x + 9)(x - 3)
Hopefully this helps!
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# Leetcode Sort Colors problem solution
Apr 22, 2023
In the Leetcode Sort Colors problem solution Given an array nums with n objects colored red, white, or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white, and blue.
We will use the integers 0, 1, and 2 to represent the color red, white, and blue, respectively.
You must solve this problem without using the library’s sort function.
Example 1:
Input: nums = [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]
Example 2:
Input: nums = [2,0,1]
Output: [0,1,2]
Constraints:
• n == nums.length
• 1 <= n <= 300
• nums[i] is either 0, 1, or 2.
## Solution in C Programming
``````void swap(int* a,int* b){
int tmp = *a;
*a = *b;
*b = tmp;
}
void sort(int* nums, int left, int right){
int i=left;
int j=right;
int x=nums[(left+right)/2];
do{
while(nums[i]<x)i++;
while(nums[j]>x)j--;
if(i<=j){
swap(&nums[i],&nums[j]);
i++;j--;
}
}while(i<=j);
if(left<j)
sort(nums,left,j);
if(i<right)
sort(nums,i,right);
}
void sortColors(int* nums, int numsSize){
sort(nums,0,numsSize - 1);
}``````
## Solution in C++ Programming
``````class Solution {
public:
void sortColors(vector<int>& nums) {
int l = 0;
int r = nums.size() - 1;
for(int i = 0; i <= r; i++)
{
if(nums[i] == 0)
{
swap(nums[i], nums[l++]);
}
else if(nums[i] == 2)
{
swap(nums[i--], nums[r--]);
}
}
}
};``````
## Solution in Java Programming
``````import java.util.*;
class Solution {
public void sortColors(int[] nums) {
int low, high, mid, temp;
low = 0;
mid = 0;
temp = 0;
high = (nums.length - 1);
while(mid <= high) {
switch(nums[mid]) {
case 0: {
temp = nums[low];
nums[low] = nums[mid];
nums[mid] = temp;
low++;
mid++;
break;
}
case 1: {
mid++;
break;
}
case 2: {
temp = nums[mid];
nums[mid] = nums[high];
nums[high] = temp;
// mid++;
high--;
break;
}
}
}
}
}``````
## Solution in Python Programming
``````class Solution(object):
def sortColors(self, nums):
"""
:type nums: List[int]
:rtype: None Do not return anything, modify nums in-place instead.
"""
i = 0
j = len(nums) - 1
for k in [0,1,2]:
while i < j:
if nums[i] == k:
i += 1
elif nums[j] == k:
nums[i], nums[j] = nums[j], nums[i]
i += 1
j -= 1
else:
j -= 1
j = len(nums) - 1
if nums[i] == k and i != j:
i += 1``````
## Solution in C# Programming
``````public class Solution {
public void SortColors(int[] nums) {
int zero=0, two=nums.Length-1, i=0;
int temp=0;
while(i<=two){
if(nums[i] ==0){
temp = nums[i];
nums[i] = nums[zero];
nums[zero] = temp;
i++;
zero++;
}
else if (nums[i] == 2){
temp = nums[i];
nums[i] = nums[two];
nums[two] = temp;
two--;
}
else{
i++;
}
}
}
}``````
#### By Neha Singhal
Hi, my name is Neha singhal a software engineer and coder by profession. I like to solve coding problems that give me the power to write posts for this site.
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# Maximum extension of a bungee cord
• Mayzu
In summary: Sorry, this is my first post so I didn't know how much detail to show :oops:The total length is 30m (as it says in part (a))I don't understand why you need a minus... But even if you do, it still gives me the wrong answer :/My friend got the right answer by ignoring GPE, but I don't see how you can simply ignore it...Explain your notations. First, what is meant as equilibrium position and equilibrium length of the cord? I think it is 32.9 m , as you got.What do you mean on Δx? The deviation from the un-stretched length of the cord, or the deviation from the equilibrium length?You can
Mayzu
## Homework Statement
(a) An 81 kg student is launched from a bridge by his best friends, some 50 metres above the river surface. Fortunately, he is attached to a 30 m bungee cord with a spring constant of 270 N/m.
i) What is the equilibrium length of the bungee cord, including the force of gravity? (I managed to solve this, it's 32.9m)
ii) What velocity would the student have when he reaches the equilibrium position if he fell in free fall with no air resistance? (I can solve this using constant accelaration equations, but can someone tell me how to solve it using energy equations?)
iii) The student has a velocity of 23 m/s at the equilibrium position. Calculate the maximum extension of the spring from its equilibrium position. Is he safe? (Clueless about this one!)
2. Homework Equations
Usp=1/2kx^2
GPE=mgh
KE=1/2mv^2
## The Attempt at a Solution
I've tried everything!
I have tried taking the equilibrium position of 32.9m as a point of zero GPE (not that I fully understand why you can do this, I just accepted more or less that you can :/)
Then I used that to say
KE(at equilibrium)=1/2*k*(delta)x^2+m*g*(delta)x
This gave me a wrong answer though. The answer is 45.5m. I'm really lost, no one seems to know the answer to this...
The delta x you are inserting for both spring potential and gravitational potential is wrong. You are not taking into account that he is falling from the top of the bridge, but working as if he fell from the cord where the cord is unstretched.
Orodruin said:
The delta x you are inserting for both spring potential and gravitational potential is wrong. You are not taking into account that he is falling from the top of the bridge, but working as if he fell from the cord where the cord is unstretched.
Yeah, I was told you can just say GPE is 0 at equilibrium because it's all relative... But it's not working
What's the correct way to do it then?
Mayzu said:
## Homework Statement
(a) An 81 kg student is launched from a bridge by his best friends, some 50 metres above the river surface. Fortunately, he is attached to a 30 m bungee cord with a spring constant of 270 N/m.
i) What is the equilibrium length of the bungee cord, including the force of gravity? (I managed to solve this, it's 32.9m)
ii) What velocity would the student have when he reaches the equilibrium position if he fell in free fall with no air resistance? (I can solve this using constant accelaration equations, but can someone tell me how to solve it using energy equations?)
iii) The student has a velocity of 23 m/s at the equilibrium position. Calculate the maximum extension of the spring from its equilibrium position. Is he safe? (Clueless about this one!)
2. Homework Equations
Usp=1/2kx^2
GPE=mgh
KE=1/2mv^2
## The Attempt at a Solution
I've tried everything!
I have tried taking the equilibrium position of 32.9m as a point of zero GPE (not that I fully understand why you can do this, I just accepted more or less that you can :/)
Then I used that to say
KE(at equilibrium)=1/2*k*(delta)x^2+m*g*(delta)x
This gave me a wrong answer though. The answer is 45.5m. I'm really lost, no one seems to know the answer to this...
The gravitational potential energy decreases, you need minus in front of mgΔx. What do you get for Δx? What is the total length of the cord? Show your work in detail.
ehild said:
The gravitational potential energy decreases, you need minus in front of mgΔx. What do you get for Δx? What is the total length of the cord? Show your work in detail.
Sorry, this is my first post so I didn't know how much detail to show
The total length is 30m (as it says in part (a))
I don't understand why you need a minus... But even if you do, it still gives me the wrong answer :/
My friend got the right answer by ignoring GPE, but I don't see how you can simply ignore it...
Explain your notations. First, what is meant as equilibrium position and equilibrium length of the cord? I think it is 32.9 m , as you got.
What do you mean on Δx? The deviation from the un-stretched length of the cord, or the deviation from the equilibrium length?
You can count the gravitational potential energy from that position. The velocity is given. The student falls down, and the gravitational potential energy decreases with decreasing height, therefore it is negative. What is the total energy (KE and elastic energy at that initial position? If you count Δx from the equilibrium position (32.9 m) you will see that the gravitational potential energy cancels, so your friend was right.
Do not forget to add the result for Δx to the equilibrium length of the cord.
Mayzu said:
Yeah, I was told you can just say GPE is 0 at equilibrium because it's all relative... But it's not working
What's the correct way to do it then?
Yes you can put the reference wherever. However, you are interested in the difference in potential and if you take the unstretched cord as a reference point, the potential at the top of the bridge is clearly non-zero. You are working as if you could use two different reference points.
ehild said:
Explain your notations. First, what is meant as equilibrium position and equilibrium length of the cord? I think it is 32.9 m , as you got.
What do you mean on Δx? The deviation from the un-stretched length of the cord, or the deviation from the equilibrium length?
You can count the gravitational potential energy from that position. The velocity is given. The student falls down, and the gravitational potential energy decreases with decreasing height, therefore it is negative. What is the total energy (KE and elastic energy at that initial position? If you count Δx from the equilibrium position (32.9 m) you will see that the gravitational potential energy cancels, so your friend was right.
Do not forget to add the result for Δx to the equilibrium length of the cord.
Δx is deviation from equilibrium.
How does it cancel?
My equation is:
KE(at equilibrium)=1/2*k*(delta)x^2+m*g*(delta)x
so to account for the gravitational potential decreasing,
1/2*mv^2=1/2*k*(delta)x^2-m*g*(delta)x
Given v=23 and m=81kg, 1/2mv^2=21424.5J
so that gives:
2142.5=1/2*k*Δx-2mgΔx
=1/2Δx(k-4mg)
=1/2Δx(270-3175.2)
Therefore, 1/2Δx=0.67
Therefore, Δx=0.33m
What am I doing wrong?
Normal bungee cords have no stored potential energy when the jumper is perched on top of the bridge. Your equation for KE at the equilibrium point seems to assume that the bungee cord is under tension at that point -- that it has stored potential energy and is pulling the jumper downward. Before the jump, the jumper should have 30 meters of un-stretched bungee cord connecting his feet to the bridge.
When the bungee jumper is at the equilibrium position, by how much will the cord have been stretched from its un-stretched state?
Mayzu said:
Δx is deviation from equilibrium.
so to account for the gravitational potential decreasing,
1/2*mv^2=1/2*k*(delta)x^2-m*g*(delta)x
Given v=23 and m=81kg, 1/2mv^2=21424.5J
so that gives:
2142.5=1/2*k*Δx-2mgΔx
What am I doing wrong?
You were a bit careless. You calculated with 2132.5 kinetic energy instead of 21424.5. And you wrote that (1/2)k(Δx)2-mgΔx= (1/2) k Δx - 2mgΔx. It is strange.
## 1. What is the maximum extension of a bungee cord?
The maximum extension of a bungee cord varies depending on several factors such as the length and elasticity of the cord, the weight of the person attached to it, and the height from which they are jumping. Typically, bungee cords can stretch up to 4-5 times their unstretched length.
## 2. How do you calculate the maximum extension of a bungee cord?
The maximum extension of a bungee cord can be calculated by using the formula E = kx, where E is the extension, k is the spring constant of the bungee cord, and x is the distance from the attachment point to the bottom of the bungee cord. The spring constant can be determined by conducting experiments with different weights and measuring the extension of the cord.
## 3. What safety measures should be taken when using a bungee cord?
When using a bungee cord, it is important to make sure that the cord is in good condition and not worn out. The attachment points should be strong and secure, and the length of the cord should be appropriate for the height from which the person is jumping. It is also important to have a safety harness and to follow proper jumping techniques to avoid injury.
## 4. Can bungee cords break?
Although bungee cords are designed to stretch, they can break if they are worn out or overloaded. It is important to regularly inspect the cord for any signs of wear and replace it if necessary. It is also important to follow weight limits and safety guidelines when using a bungee cord.
## 5. What happens if the bungee cord is stretched beyond its maximum extension?
If a bungee cord is stretched beyond its maximum extension, it can snap and potentially cause injury to the person attached to it. It is important to always use a bungee cord within its recommended limits and to have a backup plan in case of any emergencies.
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You are on page 1of 25
# Kurikulum
Sekolah Rendah
Rendah
Integrated
Integrated Curriculum
Curriculumfor
for Primary
Primary Schools
Schools
## pusat perkembangan kurikulum
KEMENTERIAN PENDIDIKAN MALAYSIA
Content Organisation
... organised according to four inter-related areas:
Whole Numbers
Operations
Fractions
Money
Time
Length
Mass
Volume
of Liquids
3D Shapes
2D Shapes
Symmet
ry
Tu ne: Twi nk le Tw inkl e Litt le S ta r
## 3-D shapes are everywhere
You can see them anywhere
Cone, cube, cuboid, cylinder
Not forgetting prism.
Let us learn the names of shapes
To be with them everywhere
1
.
3-D Shapes
1. Understand and use the
vocabulary related to 3-D shapes.
2. Describe and classify 3-D shapes
3. Build 3-D shapes.
Year 2 –1) Understand and use the vocabulary related to 3-D shapes.
Different types of 3-D shape.
1. Understand and use the vocabulary related to 3-D shapes.
a) Pupils are shown various types of prisms and they say
out the names accordingly.
A prism has the same cross-section along its length, and
that its two end faces are identical.
Each prism is named according to the shape of its base:
- triangular prism
- rectangular prism
- cube
- square prism
Year 2
face
edge
edge corner
face
corner
corner
Curved
face
a) Parts of prisms vertex
A triangular prism
face edge
base
2. Describe and classify 3-D shape
## All prisms have at least 5 faces, 2
bases of the same shape and all
other faces are rectangular.
Year 2
## - Pupils guess the objects and relate the
objects to the 3-D shapes.
- Teacher describes shapes.
## - E.g. It has 5 flat faces and 5 corners.
hidden shape in order to identify it.
- E.g. Does it have a curved face?
- Does it have a rectangular face?
- Does it have a triangular face?
3. Build 3-D shape
## a) Pupils make skeleton shapes from a
construction kit or straws .
Count the number of faces, edges or
coners.
b) Pupils build 3-D shapes from
given nets.
## c) Pupils identify various 3-D shapes
based on given nets.
I am a hexagon, a lovely hexagon,
corners,
## I am a hexagon, that is my name!
2
2-D Shapes
1. Understand and use the vocabulary related to
2-D shapes.
2. Describe and classify 2-D shapes.
Symmetry
1. Recognise and sketch lines of
symmetry.
1. Understand and use the vocabulary related
to 2-D shapes.
hexagon
Semi-circle
octagon
pentagon
heptagon
Symmetry
Collecting and organising data
## 1. Collect and organise data.
Pupils collect data by carrying out simple survey such as:
c) Hobbies
d) Games
e) Favourite fruits
f) Transport to school
g) Story books in different languages.
The Rukunegara was previously an implicit
guiding principle for the development of the
school curriculum. Implies more emphasis on
patriotism.
## Medium of instruction previously in Bahasa
Melayu is now English.
## Content layout for the Curriculum
Specification is now given in five columns
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# Percent of a number tax discount and more calculator
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Percentage Change Calculator. Please provide any two values below and click the Calculate button to get the third value. In mathematics, a percentage is a number or ratio that represents
Explain math equations
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The discount amount is calculated as follows: Original Price × Discount Rate = Discount Amount. \$ 120.00 × 0.25 = \$ 30. The sale price is calculated as follows: Original Price − Discount
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You are watching: 5 quarts of water to cups
In this instance we need to multiply 5 Quarts by 4 to gain the equivalent result in Cups: 5 Quarts x 4 = 20 Cups. 5 Quarts is identical to 20 Cups. Just how to convert from …From whatisconvert.comEstimated reading Time 2 mins see details »
How many cups are in 1 quart? when using cup and also fluid quart measurements from the exact same country, one quart equates to 4 cups or 2 pints. 1 us quart = 4 united state customary cups; 1 UK quart = 4 UK royal cups; Quarts come cups dimensions US Quart united state Cups …From thecalculatorsite.comSee details »
The conversion variable from quarts to cups is 4, which means that 1 quart is equal to 4 cups: 1 qt = 4 cup. To convert 5 quarts right into cups we need to multiply 5 through the conversion variable in stimulate to get the volume amount native quarts come cups. Us can additionally …From
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In numerous recipes, the ingredient list mentions peach dimensions in pounds, or if sliced, chopped, or pureed, the amount is typically in cups. Yet how do you recognize how plenty of peaches room in a pound, or how plenty of peaches same the quantity of sliced peaches? perhaps the recipe calls because that a peach puree. Every one of these measurements can lead to confusion and second-guessing. Girlfriend can likewise be stumped if the ...From thespruceeats.comSee details »
What is 0.5 quarts in cups? 0.5 qts to cups conversion. A U.S. Quart is equal to 32 U.S. Fluid ounces, 1/4 th the a gallon, or 2 pints. It should not be perplexed with the imperial quart, which is around 20% larger. A U.S. Cup is a unit that volume equal to 1/16th of a U.S. …From
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All In One devices Converter. Formula: division the worth in cups by 4 since 1 quart equates to 4 cups. So, 5.5 cup = 5.5 4 = 1 3 8 or 1.375 quarts.From
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How countless cups in 5.5 quarts? 5 1/2 quarts amounts to 22 cups. To convert any kind of value in quarts to cups, just multiply the worth in quarts by the conversion aspect 4. So, 5.5 quarts times 4 is equal to 22 cups.From
coolconversion.comSee details »
How numerous cups in 8.5 quarts? 8 1/2 quarts amounts to 34 cups. To convert any type of value in quarts come cups, just multiply the worth in quarts through the conversion factor 4. So, 8.5 quarts times 4 is equal to 34 cups.From
coolconversion.comSee details »
What is 5 cups in quarts? 5 cup to qts conversion. A U.S. Cup is a unit the volume same to 1/16th that a U.S. Gallon, or about 236 milliliters. A U.S. Quart is equal to 32 U.S. Fluid ounces, 1/4 th the a gallon, or 2 pints. It must not be perplexed with the royal quart, i m sorry is about 20% larger.From
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How numerous cups in 5.7 quarts? 5.7 quarts equates to 22.8 cups. Come convert any kind of value in quarts to cups, just multiply the value in quarts by the conversion variable 4. So, 5.7 quarts time 4 is equal to 22.8 cups. Every In One devices Converter. Physics Chemistry Recipes ⇌ Please, pick a physical quantity, two units, then type a worth in any of the boxes above. Quote of the day ... Display me another ...From
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A 2.9 oz (80.8g) bag of unpopped microwave popcorn yields about 13.75 cup of popped popcorn. So at 4 cup per quart, a bag of popcorn is about .86 quarts. So around 5.81 bags the popcorn need to yield 5 quarts of popped popcorn.From
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Question: How plenty of cups in 5.75 quarts? The answer is 5 3/4 quarts equals 23 cups. Come convert any kind of value in quarts to cups, just multiply the worth in quarts by the conversion factor 4. So, 5.75 quarts time 4 is same to 23 cups.From
coolconversion.comSee details »
The conversion variable from quarts to cup is 4, which means that 1 quart is equal to 4 cups: 1 qt = 4 cup. To transform 5.2 quarts right into cups we need to multiply 5.2 through the conversion factor in order to get the volume amount from quarts to cups. We can also form a simple proportion to calculation the result: 1 qt → 4 cup. 5.2 qt → V (cup)From
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The conversion variable from cups to quarts is 0.25, which method that 1 cup is same to 0.25 quarts: 1 cup = 0.25 qt. To transform 5.5 cups right into quarts we need to multiply 5.5 through the conversion element in order to obtain the volume quantity from cup to quarts. We deserve to also kind a straightforward proportion to calculation the result: 1 cup → 0.25 qt. 5.5 cup → V ...From
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An eighth the a cup amounts to 1/8 of a cup. Utilizing the right form of measure cup deserve to determine the outcome of the recipe you space making. Likewise, it is tough to fit the precise amount that a fluid in a dry measuring cup there is no spilling it end the sides. 2% the 2 cups. 2% of 2 cups equals 1/5 for each cup.From
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How numerous cups in 7.5 quarts? 7 1/2 quarts equates to 30 cups. To convert any kind of value in quarts to cups, just multiply the value in quarts through the conversion variable 4. So, 7.5 quarts time 4 is equal to 30 cups.From
coolconversion.comSee details »
of an imperial cup amounts to 187.52 ml. The cup of the metric device although rarely used, the metric system also has its version of the cup. A cup of metric system steps 250 ml. One third of a metric mechanism cup is 82.5 ml. Therefore, the metric metric mechanism cup 1/3 that metric metric device is equal to 2/3 metric device cups, which is 165 ml. TheFrom
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4 tablespoons = 1/4 cup 1 cup = 8 fluid ounces 5 1/3 tablespoons = 1/3 cup 1 cup = 1/2 pint = 236.6 mL 8 tablespoons = 1/2 cup 2 cups = 1 pint 10 2/3 tablespoons = 2/3 cup 4 cup = 1 quart 12 tablespoons = 3/4 cup 2 pints = 1 quart = 946.4 mL 1 oz = 28.35 grams 4 quarts = 1 gallon 1 pound = 453.59 grams 8 quarts = 1 peck 1 gram = 0.035 ounceFrom
peach-depot.comSee details »
See more: How Many Servings In 117 Oz Of Baked Beans, Bush'S Best Baked Beans, 117 Ounce
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Blog
# The Quest for 700: Weekly GMAT Challenge (Answer)
Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
In order to find all possible factors of , we need to break the expression down into its prime factors.
First, though, we should deal with the negative exponent in the denominator. Negative exponents indicate the reciprocal of the positive version of the exponent. For example, .
The expression (74 + 76)-1 is thus the same as . So our original expression can be rewritten as .
Dividing by a fraction is the same as multiplying by its reciprocal. Thus,
.
The expression 65 – 63 can be factored as follows:
The expression 74 + 76 can be factored as follows:
Thus it must be true that
The question asks which of the choices is not a factor of this expression.
10 = 2 × 5. The original expression contains both 2 and 5 as factors. Therefore 10 must be one of its factors. Eliminate A.
16 = 2 × 2 × 2 × 2 = 24. The original expression contains 24. Therefore 16 must be one of its factors. Eliminate B.
27 = 3 × 3 × 3 = 33. The original expression contains 33. Therefore 27 must be one of its factors. Eliminate C.
99 = 3 × 3 × 11 = 32 × 11. The original expression contains 32 but not 11. Therefore 99 CANNOT be one of its factors. CORRECT.
125 = 5 × 5 × 5 = 53. The original expression contains 53. Therefore 125 must be one of its factors.
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# Calculating alternating Euler sums of odd powers
## Definition
$$\mathbf{H}_{m}^{(n)}(x) = \sum_{k=1}^\infty \frac{H_k^{(n)}}{k^m} x^k\tag{1}$$
We define $$\mathbf{H}_{m}^{(1)}(x) = \mathbf{H}_{m}(x)=\sum_{k=1}^\infty \frac{H_k}{k^m} x^k \tag{2}$$
Note the alternating general formula $$\mathbf{H}_{m}(-1) = \sum_{k=1}^\infty (-1)^k \frac{H_k}{k^m} \tag{3}$$
## Motivation
(1) seems to be impossible to track so we focus on (2) and (3). It has been proven in [5] and [6] that the form $\mathbf{H}_{2m}(-1)$ has a general formula in terms of zeta functions \begin{align*} \mathbf{H}_{2m}(-1) &=\frac{2m+1}{2}\left(1-2^{-2m}\right)\zeta(2m+1)-\frac{1}{2}\zeta(2m+1)\\ &\qquad-\sum_{k=1}^{m-1}\left(1-2^{1-2k}\right)\zeta(2k)\zeta(2m+1-2k) \end{align*}
Up to my knowledge the literature lacks any general formula for $\mathbf{H}_{2m+1}(-1)$. The odd formula seems to contain a finite combination of zeta and polylogs and their multiplication.
## Examples
In [1] we see different evaluations for
$$\mathbf{H}_{1}(-1) = \frac{1}{2} \log^2 (2)-\frac{1}{2} \zeta(2)$$
In [2] we have
$$\mathbf{H}_{3}(-1)=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3)$$
In [3] we have some impressive calculations leading to
\begin{align} \color{blue}{\mathbf{H}_{3}(x)}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\color{blue}{\mathbf{H}_{2}(x)}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}. \end{align}
Also in [8]
\begin{align} \color{blue}{\mathbf{H}_{4}(x)} =&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)} +\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x\\&+\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5)\ \end{align}
In [4] I showed
$$\int\limits_0^1 \dfrac{\log^2 (1+x)\log^n x}{x}\; dx =2 (-1)^n(n!) \left[ \mathbf{H}_{n+2}(-1) + \left(1-2^{-n-2} \right) \zeta(n+3) \right]$$
## Questions
• Can we evaluate
$$\mathbf{H}_{5}(x) , \mathbf{H}_{5}(-1)$$
• Can we show the following has no simple general formula ? $$\mathbf{H}_{2n+1}(x),\mathbf{H}_{2n+1}(-1)$$
## Conjectures
1. Interestingly the evaluations of $\mathbf{H}_{m}^{(n)}(-1)$ are related to $\mathbf{H}_{m}^{(n)}\left(\frac{1}{2}\right)$ with the same complexity.
2. The form $\mathbf{H}_{m}^{(n)}(x)$ seem to involve a finite sum of products of logs,polylogs and zeta values.
3. There can exist a recursive formula that connects
$$\mathbf{H}_{m}^{(n)}(x) = \sum_{1\leq s,t < m} (a_{s,t})\,\mathbf{H}_{s}^{(t)}(x)$$
## Related
• Some of the $H$'s have different fonts. Is this on purpose? – Simply Beautiful Art Mar 3 '17 at 1:10
• Also, why do you mention your first equation if it is never used in the question? – Simply Beautiful Art Mar 3 '17 at 1:14
• @SimplyBeautifulArt , $\mathbf{H}_{m}(x)$ is different than $H_m$, the bold one takes an extra variable $x \in \mathbb{R}$. The first formula just to generalize the notation. – Zaid Alyafeai Mar 3 '17 at 1:21
• Glad I favorite posts like these. – Simply Beautiful Art May 5 '17 at 22:50
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} for $n\ge 2$. Otherwise by going back to the original integral representation we have: $${\bf H}^{(1)}_1(-1) = -\frac{\pi^2}{12} + \frac{1}{2} \log(2)^2$$ It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.
• Good job. Were you able to find some specific values using the recurrence formula ? – Zaid Alyafeai Apr 26 '17 at 9:34
• @Zaid Alyafeai: Thanks for the nice words. It is great fun to work on this stuff. Unfortunately I haven't worked out any particular values yet since we need additional recurrence relations for generalized harmonic sums. I think that the last expression above will be quite useful for that purpose. Can I ask you why are you interested in this stuff ? – Przemo Apr 26 '17 at 11:53
• I worked on Euler sums around 3 yrs ago and I am trying to refresh my mind on them. What I am trying to understand is whether we can find a general approach of Euler sums and their relations together. Furthermore, Can we prove if a certain family is reducible to some known closed forms? Why alternating Euler sums of even weight are harder to evaluate and there is no general formula? – Zaid Alyafeai Apr 26 '17 at 12:12
• Congrats you earned the bounty. – Zaid Alyafeai Apr 30 '17 at 16:29
• @Zaid Alyafeair: Thank you very much for the bounty. I have been struggling with deriving similar relations for higher order sums. In principle all one needs to do is use integration by parts. However it is important which part of the integrand one takes an anti-derivative of. On last Friday I used the identities on the bottom of the answer above and after long and tedious but straightforward transformations I arrived at a trivial identity, i.e. the generating functions that we seek cancelled from both sides of the equation. My 2nd answer below has salvaged that. – Przemo May 2 '17 at 12:53
Let us now consider the case of an odd order of harmonic numbers. As usual we start from the integral representation of our sums . We have: \begin{eqnarray} &&{\bf H}^{(2q+1)}_n(t) - Li_{n+2q+1}(t) =\int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \frac{Li_{2q+1}(\xi)}{1-\xi} d\xi\\ &&= \sum\limits_{j=0}^q (-1)^{q+j} \left[\binom{q+j}{2 j}\frac{1}{2} + \binom{q+j}{2 j+1}\right]\cdot \int\limits_0^1 \frac{[\log(1/\xi)]^{n-2(j+1)}}{(n-2(j+1))!} \cdot \frac{[Li_{q+j+1}(t \xi)]^2}{\xi}d\xi\\ &&=\sum\limits_{l_1=0}^{2q+1}\left\{\sum\limits_{j=0 \vee (l_1-q-1)}^q (\binom{q+j}{2j+0}\frac{1}{2} + \binom{q+j}{2j+1})\binom{q+n-j-1-l_1}{n-2 j-2}\right\}(-1)^{1-l_1} \cdot \cdot (Li_{l_1}(t) 1_{l_1\ge 0} - \delta_{l_1,0}) \cdot Li_{2q+n+1-l_1}(t)+\sum\limits_{l_1=1}^{n-1} \left\{\sum\limits_{j=0 }^{q \wedge \lfloor \frac{n-1-l_1}{2}\rfloor } (\binom{q+j}{2j+0}\frac{1}{2} + \binom{q+j}{2j+1})\binom{q+n-j-1-l_1}{q+j}\right\}(-1)^1 \cdot {\bf H}^{(n+2q+1-l_1)}_{l_1}(t) \end{eqnarray} In the second line from the top we integrated by parts $(2q+2)$-times each time using the well known properties of the poly-logarithm. What we essentially did at each step was that we found the anti-derivatives of $Li_{\theta_1}(\xi) Li_{\theta_2}(\xi)/\xi$ for some integer values of $\theta_1$ and $\theta_2$. The result is a linear combination of products of pairs of poly-logs and a residual term which is either a half of a square of a poly-log or something else depending on whether $(n-p)$ is odd or even in the first and in the second case respectively.Since integration by parts produces surface terms we have to assume that $n\ge 2q+2$ for all those terms to disappear. In the subsequent line we just used Generalized definite dilogarithm integral. and we simplified the result. The result constitutes a set of recurrence relations that entwine the harmonic sums. Here $q=0,1,2,\cdots$ and $n\ge 2q+2$ and $t\in (-1,1)$. In case $n=1,\cdots,2q+1$ we have to go back to the original integral representation and take into account the surface terms. We have: \begin{eqnarray} {\bf H}^{(2q+1)}_{2 n+1}(t) &=& \sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} \cdot Li_l(t) Li_{2q+2n+2-l}(t) (-1)^{l-(2n+1)} +\\ &&(-1)^{q+n} \frac{1}{2} \binom{q+n}{2 n} \cdot [Li_{q+n+1}(t)]^2+\\ &&\sum\limits_{j=0}^{n-1} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-1-2 j}}{(2n-1-2 j)!} \cdot \frac{[Li_{q+j+1}(\xi)]^2}{\xi}d\xi+Li_{2q+2n+2}(t)\\ {\bf H}^{(2q+1)}_{2 n}(t) &=& \sum\limits_{l=2n}^{q+n} \binom{l-1}{2 n-1} \cdot Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n)} +\\ &&\sum\limits_{j=0}^{n-1} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-2-2 j}}{(2n-2-2 j)!} \cdot \frac{[Li_{q+j+1}(\xi)]^2}{\xi}d\xi+Li_{2q+2n+1}(t)\\ % {\bf H}^{(2q)}_{2 n}(t) &=& \sum\limits_{l=2n}^{q+n-1} \binom{l-1}{2 n-1} \cdot Li_l(t) Li_{2q+2n-l}(t) (-1)^{l-(2n)} +\\ &&(-1)^{q+n} \frac{1}{2} \binom{q+n-1}{2 n-1} \cdot [Li_{q+n}(t)]^2+\\ &&\sum\limits_{j=0}^{n-1} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-1-2 j}}{(2n-1-2 j)!} \cdot \frac{[Li_{q+j}(\xi)]^2}{\xi}d\xi+Li_{2q+2n}(t)\\ % {\bf H}^{(2q)}_{2 n+1}(t) &=& \sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} \cdot Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n+1)} +\\ &&\sum\limits_{j=0}^{n} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-2 j}}{(2n-2 j)!} \cdot \frac{[Li_{q+j}(\xi)]^2}{\xi}d\xi+Li_{2q+2n+1}(t) \end{eqnarray} both for $n\ge 0$ and for $q\ge 0$ in the two top cases above and for $n\ge 0$ and $q\ge 1$ in the two bottom cases above. The integrals on the right hand side are evaluated in Generalized definite dilogarithm integral..
Bringing everything together we have: \begin{eqnarray} &&{\bf H}^{(2q+1)}_{2n+1}(t)=\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(t) Li_{2q+2n+2-l}(t) (-1)^{l-(2n+1)}+ (-1)^{q+n} \frac{1}{2} \binom{q+n}{2 n} [Li_{q+n+1}(t)]^2+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n} \left\{\sum\limits_{j=0\vee (l-q-1)}^{n-1}(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n-j-l}{2n-1-2j}\right\}(-1)^{1-l} (Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n+2-l}(t)+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n} \left\{\sum\limits_{j=0}^{\lfloor n-\frac{l}{2}\rfloor}(\frac{1}{2}\binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n- j-l}{q+j}\right\} (-1)^1 {\bf H}^{(2q+2n+2-l)}_l(t)+Li_{2q+2n+2}(t)\\ % &&{\bf H}^{(2q+1)}_{2n}(t)=\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n}^{q+n} \binom{l-1}{2 n-1} Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n)}+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n} \left\{\sum\limits_{j=0\vee (l-q-1)}^{n-1}(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n-1-j-l}{2n-2-2j}\right\}(-1)^{1-l} (Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n+1-l}(t)+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n-1} \left\{\sum\limits_{j=0}^{\lfloor n-\frac{1+l}{2}\rfloor}(\frac{1}{2}\binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n-1- j-l}{q+j}\right\} (-1)^1 {\bf H}^{(2q+2n+1-l)}_l(t)+Li_{2q+2n+1}(t)\\ % &&{\bf H}^{(2q)}_{2n}(t)=\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n}^{q+n-1} \binom{l-1}{2 n-1} Li_l(t) Li_{2q+2n-l}(t) (-1)^{l-(2n)}+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!(-1)^{q+n} \frac{1}{2} \binom{q+n-1}{2n-1} [Li_{q+n}(t)]^2+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n-1} \left\{\sum\limits_{j=0\vee (l-q)}^{n-1}(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n-1-j-l}{2n-1-2j}\right\}(-1)^{l} (Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n-l}(t)+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n} \left\{\sum\limits_{j=0}^{\lfloor n-\frac{l}{2}\rfloor}(\frac{1}{2}\binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n-1- j-l}{q+j-1}\right\} (-1)^0 {\bf H}^{(2q+2n-l)}_l(t)+Li_{2q+2n}(t)\\ % &&{\bf H}^{(2q)}_{2n+1}(t)=\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n+1)}+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n} \left\{\sum\limits_{j=0\vee (l-q)}^{n}(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n-j-l}{2n-2j}\right\}(-1)^{l} (Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n+1-l}(t)+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n+1} \left\{\sum\limits_{j=0}^{\lfloor n+\frac{1-l}{2}\rfloor}(\frac{1}{2}\binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n- j-l}{q+j-1}\right\} (-1)^0 {\bf H}^{(2q+2n+1-l)}_l(t)+Li_{2q+2n+1}(t) \end{eqnarray} Note that as long as the first two equations are useful because the quantity in question only appears on the left hand side the last two equations are less useful because the quantity being searched for actually cancels on both sides of the equation. In general it turns out that the odd-odd quantities always reduce to even-even quantities and poly-logs. On the other hand the odd-even quantities always reduce to even-odd quantities and poly-logs. We have \begin{eqnarray} {\bf H}^{(1)}_1(t) &=& \frac{1}{2}\left( [\log(1-t)]^2 + 2 Li_2(t)\right)\\ {\bf H}^{(1)}_2(t) &=& \frac{1}{2}\left(-{\bf H}^{(2)}_1(t) - \log(1-t) Li_2(t) + 3 Li_3(t)\right)\\ {\bf H}^{(1)}_3(t) &=& \frac{1}{4} \left( -2 {\bf H}^{(2)}_2(t) + [Li_2(t)]^2 + 6 Li_4(t)\right)\\ {\bf H}^{(1)}_4(t) &=& \frac{1}{4}\left(-2 {\bf H}^{(2)}_3(t)+{\bf H}^{(4)}_1(t) + Li_2(t) Li_3(t) + \log(1-t) Li_4(t) + 5 Li_5(t) \right) \\ {\bf H}^{(1)}_5(t) &=& \frac{1}{4}\left( -2 {\bf H}^{(2)}_4(t) + {\bf H}^{(4)}_2(t) + [Li_3(t)]^2 - Li_2(t) Li_4(t) + 5Li_6(t)\right)\\ {\bf H}^{(1)}_6(t) &=&\frac{1}{4}\left(-2{\bf H}^{(2)}_5(t)+{\bf H}^{(4)}_3(t)-2{\bf H}^{(6)}_1(t) + Li_3(t) Li_4(t) - 2 Li_2(t)Li_5(t) - 2\log(1-t)Li_6(t) + 7 Li_7(t)\right)\\ {\bf H}^{(1)}_7(t) &=& \frac{1}{8} \left( -4 {\bf H}^{(2)}_6(t)+ 2 {\bf H}^{(4)}_4(t) - 4 {\bf H}^{(6)}_2(t) + 5 [Li_4(t)]^2 - 8 Li_3(t) Li_5(t) + 4 Li_2(t) Li_6(t) + 14 Li_8(t)\right)\\ {\bf H}^{(1)}_8(t) &=& \frac{1}{8} (-4 {\bf H}^{(2)}_7(t)+2 {\bf H}^{(4)}_5(t)-4 {\bf H}^{(6)}_3(t)+17 {\bf H}^{(8)}_1(t)+5 \text{Li}_4(t) \text{Li}_5(t)-13 \text{Li}_3(t) \text{Li}_6(t)+17 \text{Li}_2(t) \text{Li}_7(t)-3 \text{Li}_9(t)+17 \text{Li}_8(t) \log (1-t))\\ {\bf H}^{(1)}_9(t) &=& \frac{1}{8} \left(-4 {\bf H}^{(2)}_8(t)+2 {\bf H}^{(4)}_6(t)-4 {\bf H}^{(6)}_4(t)+17 {\bf H}^{(8)}_2(t)+26 \text{Li}_5(t){}^2-47 \text{Li}_4(t) \text{Li}_6(t)+34 \text{Li}_3(t) \text{Li}_7(t)-17 \text{Li}_2(t) \text{Li}_8(t)-3 \text{Li}_{10}(t)\right)\\ {\bf H}^{(1)}_{10}(t) &=& \frac{1}{8} (-4 {\bf H}^{(2)}_9(t)+2 {\bf H}^{(4)}_7(t)-4 {\bf H}^{(6)}_5(t)+17 {\bf H}^{(8)}_3(t)-124 {\bf H}^{(10)}_1(t)+26 \text{Li}_5(t) \text{Li}_6(t)-73 \text{Li}_4(t) \text{Li}_7(t)+107 \text{Li}_3(t) \text{Li}_8(t)-124 \text{Li}_2(t) \text{Li}_9(t)+121 \text{Li}_{11}(t)-124 \text{Li}_{10}(t) \log (1-t))\\ \end{eqnarray} Likewise we have: \begin{eqnarray} {\bf H}^{(3)}_1(t) &=&\frac{1}{2} \left(-\text{Li}_2(t){}^2+2 \text{Li}_4(t)-2 \text{Li}_3(t) \log (1-t)\right)\\ {\bf H}^{(3)}_2(t) &=& \frac{1}{2} (-3 {\bf H}^{(4)}_1(t)-\text{Li}_2(t) \text{Li}_3(t)+5 \text{Li}_5(t)-3 \text{Li}_4(t) \log (1-t))\\ {\bf H}^{(3)}_3(t) &=& \frac{1}{2} \left(-3 {\bf H}^{(4)}_2-2 \text{Li}_3(t){}^2+3 \text{Li}_2(t) \text{Li}_4(t)+5 \text{Li}_6(t)\right)\\ {\bf H}^{(3)}_4(t) &=& \frac{1}{2} (-3 {\bf H}^{(4)}_3(t)+5 {\bf H}^{(6)}_1(t)-2 \text{Li}_3(t) \text{Li}_4(t)+5 \text{Li}_2(t) \text{Li}_5(t)+5 \text{Li}_6(t) \log (1-t))\\ {\bf H}^{(3)}_5(t) &=& \frac{1}{2} \left(-3 {\bf H}^{(4)}_4(t)+5 {\bf H}^{(6)}_2(t)-6 \text{Li}_4(t){}^2+10 \text{Li}_3(t) \text{Li}_5(t)-5 \text{Li}_2(t) \text{Li}_6(t)\right)\\ {\bf H}^{(3)}_6(t) &=& \frac{1}{2} (-3 {\bf H}^{(4)}_5(t)+5 {\bf H}^{(6)}_3(t)-21 {\bf H}^{(8)}_1(t)-6 \text{Li}_4(t) \text{Li}_5(t)+16 \text{Li}_3(t) \text{Li}_6(t)-21 \text{Li}_2(t) \text{Li}_7(t)+21 \text{Li}_9(t)-21 \text{Li}_8(t) \log (1-t))\\ {\bf H}^{(3)}_7(t) &=& \frac{1}{2} \left(-3 {\bf H}^{(4)}_6(t)+5 {\bf H}^{(6)}_4(t)-21 {\bf H}^{(8)}_2(t)-32 \text{Li}_5(t){}^2+58 \text{Li}_4(t) \text{Li}_6(t)-42 \text{Li}_3(t) \text{Li}_7(t)+21 \text{Li}_2(t) \text{Li}_8(t)+21 \text{Li}_{10}(t)\right)\\ \hline \\ {\bf H}^{(5)}_1(t)&=&\frac{1}{2} \left(\text{Li}_3(t){}^2-2 \text{Li}_2(t) \text{Li}_4(t)+2 \text{Li}_6(t)-2 \text{Li}_5(t) \log (1-t)\right)\\ {\bf H}^{(5)}_2(t)&=&\frac{1}{2} (-5 {\bf H}^{(6)}_1(t)+\text{Li}_3(t) \text{Li}_4(t)-3 \text{Li}_2(t) \text{Li}_5(t)+7 \text{Li}_7(t)-5 \text{Li}_6(t) \log (1-t))\\ {\bf H}^{(5)}_3(t)&=&\frac{1}{4} \left(-10 {\bf H}^{(6)}_2(t)+9\text{Li}_4(t){}^2-16 \text{Li}_3(t) \text{Li}_5(t)+10 \text{Li}_2(t) \text{Li}_6(t)+14 \text{Li}_8(t)\right)\\ {\bf H}^{(5)}_4(t)&=&\frac{1}{4} (-10 {\bf H}^{(6)}_3(t)+35 {\bf H}^{(8)}_1(t)+9 \text{Li}_4(t) \text{Li}_5(t)-25 \text{Li}_3(t) \text{Li}_6(t)+35 \text{Li}_2(t) \text{Li}_7(t)-21 \text{Li}_9(t)+35 \text{Li}_8(t) \log (1-t))\\ {\bf H}^{(5)}_5(t)&=& \frac{1}{4} \left(-10 {\bf H}^{(6)}_4(t)+35 {\bf H}^{(8)}_2(t)+52 \text{Li}_5(t){}^2-95 \text{Li}_4(t) \text{Li}_6(t)+70 \text{Li}_3(t) \text{Li}_7(t)-35 \text{Li}_2(t) \text{Li}_8(t)-21 \text{Li}_{10}(t)\right)\\ \hline \\ {\bf H}^{(7)}_1(t)&=&\frac{1}{2} \left(-\text{Li}_4(t){}^2+2 \text{Li}_3(t) \text{Li}_5(t)-2 \text{Li}_2(t) \text{Li}_6(t)+2 \text{Li}_8(t)-2 \text{Li}_7(t) \log (1-t)\right)\\ {\bf H}^{(7)}_2(t)&=&\frac{1}{2} (-7 {\bf H}^{(8)}_1(t)-\text{Li}_4(t) \text{Li}_5(t)+3 \text{Li}_3(t) \text{Li}_6(t)-5 \text{Li}_2(t) \text{Li}_7(t)+9 \text{Li}_9(t)-7 \text{Li}_8(t) \log (1-t))\\ {\bf H}^{(7)}_3(t)&=&\frac{1}{2} \left(-7 {\bf H}^{(8)}_2(t)-8 \text{Li}_5(t){}^2+15 \text{Li}_4(t) \text{Li}_6(t)-12 \text{Li}_3(t) \text{Li}_7(t)+7 \text{Li}_2(t) \text{Li}_8(t)+9 \text{Li}_{10}(t)\right)\\ \hline \\ {\bf H}^{(9)}_1(t)&=&\frac{1}{2} \left(\text{Li}_5(t){}^2-2 \text{Li}_4(t) \text{Li}_6(t)+2 \text{Li}_3(t) \text{Li}_7(t)-2 \text{Li}_2(t) \text{Li}_8(t)+2 \text{Li}_{10}(t)-2 \text{Li}_9(t) \log (1-t)\right) \end{eqnarray} Unfortunately both the even-even and the even-odd quantities cannot be worked out using the formalism above since the respective recurrence equations reduce to tautologies.
Update: Below we demonstrate that it is possible to get additional recurrence relations for both the even-odd and the even-even quantities provided $t=-1$. Let us start with the simplest possible example. Let us assume that $q\ge 1$ then we have: \begin{eqnarray} &&{\bf H}^{(2q)}_1(-1)= \sum\limits_{l=1}^q Li_l(-1) Li_{2q+1-l}(-1) (-1)^{l-1} + (-1)^q \underbrace{\int\limits_0^1 \frac{[Li_q(-\xi)]^2}{\xi} d\xi}_{{\mathcal A}^{(0,2)}_q(-1)} + Li_{2q+1}(-1)=\\ &&\frac{1}{4^q}\left(-1+(-2+4^q) q\right) \zeta(2q+1) - \log(2) \left(-1+\frac{1}{2^{2q-1}}\right) \zeta(2q)+\\ &&\sum\limits_{l=2}^q (-\frac{1}{2})^l \left(-2+2^{l-q}\right) \zeta(l) \zeta(2q+1-l)+\\ &&\sum\limits_{l=2}^{2q-1} \left(-\frac{1}{2} -2 (-1)^l +(-1)^l 2^{2-l} + \frac{1}{4^q} \right)\zeta(l) \zeta(2q+1-l)+\\ &&2{\bf H}^{(1)}_{2q}(-1) \end{eqnarray} In the top line we started from the integral representation which we integrated by parts $q$-times. In the bottom line we used the second answer to Generalized definite dilogarithm integral. to compute the integral on the right hand side. As a result we obtained a quite useful relation. Note that the harmonic sum on the left hand side is converging very slowly whereas the other sum on the right hand side converges quite fast. It is clear that this approach can be extended to more complicated cases. We have: \begin{eqnarray} &&{\bf H}^{(2q)}_{2n+1}(-1)=\\ &&\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(-1) Li_{2q+2n+1-l}(-1) (-1)^{l-(2n+1)}+\\ &&\sum\limits_{l=2}^{q+n} {\mathcal A}_0^{(n,l,q)} (-1)^l \frac{1-2^{1+l}+2^{2(n+q)}}{2^{2(n+q)}}\cdot \zeta(2n+2q+1-l) \zeta(l)+\\ &&\sum\limits_{l=2}^{2n+1} {\mathcal A}_1^{(n,l,q)} \frac{1-2^{1+l}+2^{2(n+q)}}{2^{2(n+q)}}\cdot \zeta(2n+2q+1-l) \zeta(l)+\\ &&\sum\limits_{l=1}^{2n+1} {\mathcal A}_1^{(n,l,q)}\left[(1-2^{1-2 n-2 q}) {\bf H}^{(l)}_{2n+2q+1-l}(+1) + 2 {\bf H}^{(l)}_{2n+2q+1-l}(-1)\right]+\\ &&Li_{2n+2 q+1}(-1) \end{eqnarray} for $n\ge 0$ and $q\ge 1$. Here the coefficients read: \begin{eqnarray} {\mathcal A}_0^{(n,l,q)}&:=& \sum\limits_{j=(l-q)\vee 0}^n (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j}) \binom{-j-l+2n+q}{2(n-j)}\\ {\mathcal A}_1^{(n,l,q)}&:=& \sum\limits_{j= 0}^{n-\lfloor \frac{l-1}{2}\rfloor} (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j}) \binom{-j-l+2n+q}{q+j-1} \end{eqnarray} In the even-even case we have: \begin{eqnarray} &&{\bf H}^{(2 q)}_{2 n}(-1)=\\ &&\sum\limits_{l=2n}^{q+n-1} \binom{l-1}{2n-1} Li_l(-1) Li_{2q+2n-l}(-1)(-1)^{l-2 n}+\\ &&(-1)^{q+n} \frac{1}{2} \binom{q+n-1}{2n-1} [Li_{q+n}(-1)]^2 +\\ &&\sum\limits_{l=2}^{q+n-1} {\mathcal A}_2^{(n,l,q)} \left( \frac{4-2^{2+l}+2^{2(n+q)}}{2^{2(n+q)}}\right)\cdot \zeta(2(n+q)-l)\zeta(l) (-1)^l+\\ &&\sum\limits_{l=2}^{2 n} {\mathcal A}_3^{(n,l,q)} \left( \frac{4-2^{2+l}+2^{2(n+q)}}{2^{2(n+q)}}\right)\cdot \zeta(2(n+q)-l)\zeta(l) +\\ &&\sum\limits_{l=1}^{2 n}{\mathcal A}_3^{(n,l,q)} \left((1-\frac{1}{2^{2(n+q-1)}}) {\bf H}^{(l)}_{2(n+q)-l}(+1) + 2 {\bf H}^{(l)}_{2(n+q)-l}(-1)\right)+\\ &&Li_{2n+2q}(-1) \end{eqnarray} where the coefficients read: \begin{eqnarray} {\mathcal A}_2^{(n,l,q)}&:=& \sum\limits_{j=(l-q)\vee 0}^n (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j}) \binom{-1-j-l+2n+q}{2(n-j)-1}\\ {\mathcal A}_3^{(n,l,q)}&:=& \sum\limits_{j= 0}^{n-\lfloor \frac{l}{2}\rfloor} (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j}) \binom{-1-j-l+2n+q}{q+j-1} \end{eqnarray} where $n\ge 1$ and $q\ge 1$. As we can see from the above we also need the results for plus unity. They read: \begin{eqnarray} &&{\bf H}^{(2q)}_{2n+1}(+1)=\\ &&\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(1) Li_{2q+2n+1-l}(1) (-1)^{l-(2n+1)}+\\ &&\sum\limits_{l=2}^{q+n} {\mathcal A}_4^{(n,l,q)} \zeta(l) \zeta(1-l+2 n+2 q) (-1)^l+\\ &&\sum\limits_{l=2}^{2n+1} {\mathcal A}_5^{(n,l,q)} \zeta(l) \zeta(1-l+2 n+2 q) +\\ &&\sum\limits_{l=1}^{2n+1} {\mathcal A}_5^{(n,l,q)}(-1)^1 {\bf H}^{(l)}_{1+2n+2q-l}(+1)+\\ &&Li_{2n+2q+1}(+1) \end{eqnarray} where \begin{eqnarray} {\mathcal A}_4^{(n,l,q)}&:=& \sum\limits_{j=(l-q) \vee 0}^n \left(\frac{1}{2} \binom{q+j-1}{2j-1}+\binom{q+j-1}{2 j}\right)\binom{q-j+2n-l}{2n-2j}\\ {\mathcal A}_5^{(n,l,q)}&:=& \sum\limits_{j=0}^{n+\lfloor \frac{1-l}{2} \rfloor}\left(\frac{1}{2} \binom{q+j-1}{2j-1}+\binom{q+j-1}{2 j}\right)\binom{q-j+2n-l}{q+j-1} \end{eqnarray} It is clear that an analogous formula exists for the remaining even-even case at plus unity. We will write it down later on. Now I am going to argue that the last two formulae above along with the relations that combine the odd-odd and the odd-even cases with the even-even and the even-odd cases-- the relations that hold for arbitrary value of $t$ -- that those relations are sufficient in order to work out closed form solutions for all the harmonic sums at plus unity. Indeed using this approach we found the following: \begin{eqnarray} {\bf H}^{(1)}_2(+1) &=& 2 \zeta(3)\\ \lim_{t\rightarrow 1} \left({\bf H}^{(2)}_1(t) + \log(1-t) Li_2(t)\right) &=& - \zeta(3)\\ \hline {\bf H}^{(1)}_3(+1) &=& -\frac{1}{6} \zeta(2)^2+ \frac{5}{3} \zeta(4)\\ {\bf H}^{(2)}_2(+1) &=& +\frac{5}{6} \zeta(2)^2 - \frac{1}{3} \zeta(4) \\ \lim_{t \rightarrow 1} \left({\bf H}^{(3)}_1(t) + \log(1-t) Li_3(t)\right) &=& -\frac{1}{2} \zeta(2)^2 + \zeta(4) \\ \hline {\bf H}^{(1)}_4(+1) &=& -\zeta(2)\zeta(3) + 3 \zeta(5)\\ {\bf H}^{(2)}_3(+1) &=&+3 \zeta(2) \zeta(3)-\frac{9}{2} \zeta(5)\\ {\bf H}^{(3)}_2(+1)&=& -2 \zeta(2) \zeta(3)+\frac{11}{2} \zeta(5)\\ \lim_{t\rightarrow 1} \left({\bf H}^{(4)}_1(t) + \log(1-t) Li_4(t)\right) &=&+1 \zeta(2) \zeta(3)-2 \zeta(5) \\ \hline {\bf H}^{(1)}_5(+1) &=& -\frac{1}{2}\zeta(3)^2-\frac{1}{3}\zeta(2)\zeta(4) + \frac{7}{3} \zeta(6)\\ {\bf H}^{(2)}_4(+1) &=&+1 \zeta(3)^2+\frac{4}{3} \zeta(2) \zeta(4)-\frac{8}{3} \zeta(6)\\ {\bf H}^{(3)}_3(+1)&=& +\frac{1}{2} \zeta(3)^2-2 \zeta(2) \zeta(4)+4\zeta(6)\\ {\bf H}^{(4)}_2(+1)&=& -1 \zeta(3)^2+\frac{7}{3} \zeta(2) \zeta(4)-1\zeta(6)\\ \lim_{t\rightarrow 1} \left({\bf H}^{(5)}_1(t) + \log(1-t) Li_5(t)\right) &=&+\frac{1}{2} \zeta(3)^2- \zeta(2) \zeta(4)+ \zeta(6) \\ \hline {\bf H}^{(1)}_6(+1) &=& -\zeta(3) \zeta(4)-\zeta(2) \zeta(5)+4 \zeta(7) \\ {\bf H}^{(2)}_5(+1) &=& +2\zeta(3) \zeta(4)+5\zeta(2) \zeta(5)-10 \zeta(7) \\ {\bf H}^{(3)}_4(+1) &=& +0\zeta(3) \zeta(4)-10\zeta(2) \zeta(5)+18 \zeta(7) \\ {\bf H}^{(4)}_3(+1) &=& +1\zeta(3) \zeta(4)+10\zeta(2) \zeta(5)-17 \zeta(7) \\ {\bf H}^{(5)}_2(+1) &=& -2\zeta(3) \zeta(4)-4\zeta(2) \zeta(5)+11 \zeta(7) \\ \lim_{t\rightarrow 1} \left({\bf H}^{(6)}_1(t) + \log(1-t) Li_6(t)\right) &=& +1 \zeta(3) \zeta(4)+ \zeta(2) \zeta(5) - 3 \zeta(7) \\ \hline \\ {\bf H}^{(1)}_7(+1) &=& \frac{9 \zeta(8)}{4}-\zeta (3) \zeta (5) \\ {\bf H}^{(2)}_6(+1) &=& \\ {\bf H}^{(3)}_5(+1) &=& -\frac{5}{2} {\bf H}^{(2)}_6(+1)-\frac{21 \zeta(8)}{8}+5 \zeta (3) \zeta (5) \\ {\bf H}^{(4)}_4(+1) &=& \frac{13 \zeta(8)}{12} \\ {\bf H}^{(5)}_3(+1) &=& \frac{5}{2} {\bf H}^{(2)}_6(+1)+\frac{29 \zeta(8)}{8}-4 \zeta (3) \zeta (5) \\ {\bf H}^{(6)}_2(+1) &=& \frac{8 \zeta(8)}{3}-{\bf H}^{(2)}_6(+1) \\ \lim_{t\rightarrow 1} {\bf H}^{(7)}_1(t) + \log(1-t) Li_7(t) &=& \zeta (3) \zeta (5)-\frac{5 \zeta(8)}{4}\\ \hline\\ {\bf H}^{(1)}_8(+1)&=&\frac{-2 \pi ^6 \zeta(3)-21 \pi ^4 \zeta(5)-315 \pi ^2 \zeta(7)+9450 \zeta(9)}{1890} \\ {\bf H}^{(2)}_7(+1)&=& \frac{2}{945} \pi ^6 \zeta(3)+\frac{2}{45} \pi ^4 \zeta(5)+\frac{7}{6} \pi ^2 \zeta(7)-\frac{35 \zeta(9)}{2} \\ {\bf H}^{(3)}_6(+1)&=& -\frac{1}{15} \pi ^4 \zeta(5)-\frac{7}{2} \pi ^2 \zeta(7)+\frac{85 \zeta(9)}{2} \\ {\bf H}^{(4)}_5(+1)&=& \frac{1}{18} \pi ^4 \zeta(5)+\frac{35}{6} \pi ^2 \zeta(7)-\frac{125 \zeta(9)}{2} \\ {\bf H}^{(5)}_4(+1)&=& -\frac{2}{45} \pi ^4 \zeta(5)-\frac{35}{6} \pi ^2 \zeta(7)+\frac{127 \zeta(9)}{2} \\ {\bf H}^{(6)}_3(+1)&=& \frac{1}{945} \pi ^6 \zeta(3)+\frac{1}{15} \pi ^4 \zeta(5)+\frac{7}{2} \pi ^2 \zeta(7)-\frac{83 \zeta(9)}{2} \\ {\bf H}^{(7)}_2(+1)&=& -\frac{2}{945} \pi ^6 \zeta(3)-\frac{2}{45} \pi ^4 \zeta(5)-\pi ^2 \zeta(7)+\frac{37 \zeta(9)}{2} \\ \lim_{t \rightarrow 1}\left( {\bf H}^{(8)}_1(t) + \log(1-t) Li_8(t) \right)&=& \frac{1}{945} \pi ^6 \zeta(3)+\frac{1}{90} \pi ^4 \zeta(5)+\frac{1}{6} \pi ^2 \zeta(7)-4 \zeta(9) \\ \hline \\ {\bf H}^{(1)}_9(+1)&=& \frac{\pi ^{10}}{34020}-\frac{\zeta (5)^2}{2}-\zeta (3) \zeta (7) \\ {\bf H}^{(2)}_8(+1)&=& {\bf H}^{(2)}_8(+1) \\ {\bf H}^{(3)}_7(+1)&=& -\frac{7}{2} {\bf H}^{(2)}_8(+1)+7 \zeta (3) \zeta (7)+4 \zeta (5)^2-\frac{\pi ^{10}}{11340} \\ {\bf H}^{(4)}_6(+1)&=& \frac{7}{2} {\bf H}^{(2)}_8(+1)-7 \zeta (3) \zeta (7)-5 \zeta (5)^2+\frac{227 \pi ^{10}}{1871100} \\ {\bf H}^{(5)}_5(+1)&=& \frac{\pi ^{10}}{187110}+\frac{\zeta (5)^2}{2} \\ {\bf H}^{(6)}_4(+1)&=& -\frac{7}{2} {\bf H}^{(2)}_8(+1)+7 \zeta (3) \zeta (7)+5 \zeta (5)^2-\frac{37 \pi ^{10}}{374220} \\ {\bf H}^{(7)}_3(+1)&=& \frac{7}{2} {\bf H}^{(2)}_8(+1)-6 \zeta (3) \zeta (7)-4 \zeta (5)^2+\frac{37 \pi ^{10}}{374220} \\ {\bf H}^{(8)}_2(+1)&=& \frac{53 \pi ^{10}}{1871100}-{\bf H}^{(2)}_8(+1) \\ \lim_{t \rightarrow 1} \left( {\bf H}^{(9)}_1(t) + \log(1-t) Li_9(t) \right) &=& -\frac{\pi ^{10}}{53460}+\frac{\zeta (5)^2}{2}+\zeta (3) \zeta (7) \\ \hline\\ \vdots\\ \hline\\ {\bf H}^{(1)}_{11}(+1)&=& -\zeta (5) \zeta (7)-\zeta (3) \zeta (9)+\frac{691 \pi ^{12}}{196465500}\\ {\bf H}^{(3)}_9(+1)&=& \frac{428652000 \zeta (5) \zeta (7)+321489000 \zeta (3) \zeta (9)-691 \pi ^{12}}{35721000}-\frac{9}{2} {\bf H}^{(2)}_{10}(+1)\\ {\bf H}^{(4)}_8(+1)&=& 8 {\bf H}^{(2)}_{10}(+1)-16 \zeta (3) \zeta (9)-28 \zeta (5) \zeta (7)+\frac{86096 \pi ^{12}}{1915538625}\\ {\bf H}^{(5)}_7(+1)&=& -7 {\bf H}^{(2)}_{10}(+1)+14 \zeta (3) \zeta (9)+28 \zeta (5) \zeta (7)-\frac{316027 \pi ^{12}}{7662154500}\\ {\bf H}^{(6)}_6(+1)&=& \frac{703 \pi ^{12}}{638512875}\\ {\bf H}^{(7)}_5(+1)&=& 7 {\bf H}^{(2)}_{10}(+1)-14 \zeta (3) \zeta (9)-27 \zeta (5) \zeta (7)+\frac{324319 \pi ^{12}}{7662154500}\\ {\bf H}^{(8)}_4(+1)&=& -8 {\bf H}^{(2)}_{10}(+1)+16 \zeta (3) \zeta (9)+28 \zeta (5) \zeta (7)-\frac{327083 \pi ^{12}}{7662154500}\\ {\bf H}^{(9)}_3(+1)&=& \frac{9}{2} {\bf H}^{(2)}_{10}(+1)-8 \zeta (3) \zeta (9)-12 \zeta (5) \zeta (7)+\frac{104341 \pi ^{12}}{5108103000}\\ {\bf H}^{(10)}_2(+1)&=& \frac{1219 \pi ^{12}}{425675250}-{\bf H}^{(2)}_{10}(+1)\\ \lim_{t\rightarrow 1} \left({\bf H}^{(11)}_1(t)+\log(1-t) Li_{11}(t) \right)&=& \frac{283783500 \zeta (5) \zeta (7)+283783500 \zeta (3) \zeta (9)-691 \pi ^{12}}{283783500} \end{eqnarray} On the face of it is seemed that all harmonic sums at plus unity are functions of zeta values at positive integers only. However when the weight became strictly bigger than seven something new happened. One of the equations appeared to be linearly dependent on the others which rendered it impossible to evaluate one of the sums. Now to the case of minus unity. In the even-even and the even-odd cases we will be using the relations for minus unity whereas in the odd-odd and in the odd-even cases we will be using the relations that are valid for arbitrary $t$. \begin{eqnarray} {\bf H}^{(1)}_1(-1) &=& \frac{1}{2} [\log(2)]^2 - \frac{1}{2} \zeta(2)\\ \hline\\ {\bf H}^{(1)}_2(-1) &=& - \frac{5}{8} \zeta(3)\\ {\bf H}^{(2)}_1(-1) &=& \frac{1}{2} [\log(2)] \zeta(2) - \zeta(3)\\ \hline\\ {\bf H}^{(1)}_3(-1) &=& \frac{1}{360} \left(30 \left(24 \text{Li}_4\left(\frac{1}{2}\right)+21 \zeta (3) \log (2)+\log ^4(2)\right)-11 \pi ^4-30 \pi ^2 \log ^2(2)\right)\\ {\bf H}^{(2)}_2(-1) &=& -4\text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)\\ {\bf H}^{(3)}_1(-1) &=& \frac{1080 \zeta (3) \log (2)-19 \pi ^4}{1440}\\ \hline\\ {\bf H}^{(1)}_4(-1) &=& \frac{1}{96} \left(8 \pi ^2 \zeta (3)-177 \zeta (5)\right)\\ {\bf H}^{(2)}_3(-1) &=& \frac{11 \zeta (5)}{32}-\frac{5 \pi ^2 \zeta (3)}{48} \\ {\bf H}^{(3)}_2(-1) &=& \frac{21 \zeta (5)}{32}-\frac{\pi ^2 \zeta (3)}{8} \\ {\bf H}^{(4)}_1(-1) &=& \frac{\pi ^2 \zeta (3)}{16}-2 \zeta (5)+\frac{7}{720} \pi ^4 \log (2)\\ \hline \\ {\bf H}^{(1)}_5(-1) &=& \frac{1}{2} \int\limits_0^1 \frac{[log(1/\xi)]^3}{3!} \cdot \frac{[\log(1+\xi)]^2}{\xi} d\xi + Li_6(-1) \\ {\bf H}^{(2)}_4(-1) &=& \frac{1}{2} \int\limits_0^1 \frac{[log(1/\xi)]^1}{1!} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi - \int\limits_0^1 \frac{[log(1/\xi)]^3}{3!} \cdot \frac{[\log(1+\xi)]^2}{\xi} d\xi + Li_6(-1) \\ {\bf H}^{(3)}_3(-1) &=& -6 {\bf H}^{(1)}_5(-1)-3 {\bf H}^{(2)}_4(-1)+\frac{1701 \zeta (3)^2-62 \pi ^6}{6048} \\ {\bf H}^{(4)}_2(-1) &=& 4 {\bf H}^{(1)}_5(-1)+2 {\bf H}^{(2)}_4(-1)-\frac{9 \zeta (3)^2}{16}+\frac{359 \pi ^6}{60480} \\ {\bf H}^{(5)}_1(-1) &=& \frac{5670 \zeta (3)^2+18900 \zeta (5) \log (2)-37 \pi^6}{20160}\\ \hline\\ {\bf H}^{(1)}_6(-1)&=&+\frac{56 \pi ^4 \zeta (3)+480 \pi ^2 \zeta (5)-16965 \zeta (7)}{5760}\\ {\bf H}^{(2)}_5(-1)&=&+\frac{249 \zeta (7)}{64}-\frac{49 \pi ^2 \zeta (5)}{192}-\frac{7 \pi ^4 \zeta (3)}{360}\\ {\bf H}^{(3)}_4(-1)&=&-\frac{363 \zeta(7)}{128}+\frac{3 \pi ^2 \zeta (5)}{16}\\ {\bf H}^{(4)}_3(-1)&=&-\frac{199 \zeta (7)}{64}+\frac{13 \pi ^2 \zeta (5)}{96}+\frac{7 \pi ^4 \zeta (3)}{960}\\ {\bf H}^{(5)}_2(-1)&=&+\frac{519 \zeta (7)}{128}-\frac{5 \pi ^2 \zeta (5)}{16}-\frac{7 \pi ^4\zeta (3)}{480}\\ {\bf H}^{(6)}_1(-1)&=&-3 \zeta (7)+\frac{5 \pi ^2 \zeta (5)}{64}+\frac{7 \pi ^4 \zeta (3)}{960}+\frac{31 \pi ^6 \log (2)}{30240} \\ \hline\\ {\bf H}^{(1)}_7(-1) &=& {\bf H}^{(1)}_7(-1)\\ {\bf H}^{(2)}_6(-1) &=& {\bf H}^{(2)}_6(-1)\\ {\bf H}^{(3)}_5(-1) &=& -9 {\bf H}^{(1)}_7(-1)-4 {\bf H}^{(2)}_6(-1)-\frac{63}{128} {\bf H}^{(2)}_6(+1)+\frac{123 \zeta (3) \zeta (5)}{64}-\frac{127 \pi ^8}{76800}\\ {\bf H}^{(4)}_4(-1) &=& 16 {\bf H}^{(1)}_7(-1)+6 {\bf H}^{(2)}_6(-1)+\frac{63}{32} {\bf H}^{(2)}_6(+1)-\frac{123 \zeta (3) \zeta (5)}{16}+\frac{3097 \pi ^8}{1036800}\\ {\bf H}^{(5)}_3(-1) &=& -15 {\bf H}^{(1)}_7(-1)-5 {\bf H}^{(2)}_6(-1)-\frac{315}{128} {\bf H}^{(2)}_6(+1)+\frac{165 \zeta (3) \zeta (5)}{16}-\frac{2257 \pi ^8}{691200}\\ {\bf H}^{(6)}_2(-1) &=& 6 {\bf H}^{(1)}_7(-1)+2 {\bf H}^{(2)}_6(-1)+\frac{63}{64} {\bf H}^{(2)}_6(+1)-\frac{21 \zeta (3) \zeta (5)}{4}+\frac{193 \pi ^8}{145152}\\ {\bf H}^{(7)}_1(-1) &=& \frac{45 \zeta (3) \zeta (5)}{64}+\frac{63}{64} \zeta (7) \log (2)-\frac{23 \pi ^8}{96768}\\ \hline\\ {\bf H}^{(1)}_8(-1)&=&+\frac{496 \pi ^6 \zeta (3)+4704 \pi ^4 \zeta (5)+40320 \pi ^2 \zeta (7)-1926855 \zeta (9)}{483840}\\ {\bf H}^{(2)}_7(-1)&=&+\frac{4837 \zeta (9)}{512}-\frac{107 \pi ^2 \zeta (7)}{256}-\frac{7 \pi ^4 \zeta (5)}{180}-\frac{31 \pi ^6 \zeta (3)}{15120}\\ {\bf H}^{(3)}_6(-1)&=&-\frac{7367 \zeta (9)}{512}+\frac{97 \pi ^2 \zeta (7)}{128}+\frac{7 \pi ^4 \zeta (5)}{120}\\ {\bf H}^{(4)}_5(-1)&=&+\frac{3259 \zeta (9)}{512}-\frac{335 \pi ^2 \zeta (7)}{768}-\frac{343 \pi ^4 \zeta (5)}{11520}\\ {\bf H}^{(5)}_4(-1)&=&+\frac{3385 \zeta (9)}{512}-\frac{25 \pi ^2 \zeta (7)}{64}-\frac{7 \pi ^4 \zeta (5)}{192}\\ {\bf H}^{(6)}_3(-1)&=&-\frac{7451 \zeta (9)}{512}+\frac{187 \pi ^2 \zeta (7)}{256}+\frac{7 \pi ^4 \zeta (5)}{128}+\frac{31 \pi ^6 \zeta (3)}{40320}\\ {\bf H}^{(7)}_2(-1)&=&+\frac{4873 \zeta (9)}{512}-\frac{63 \pi ^2 \zeta (7)}{128}-\frac{7 \pi ^4 \zeta (5)}{192}-\frac{31 \pi ^6 \zeta (3)}{20160}\\ {\bf H}^{(8)}_1(-1)&=&-4 \zeta (9)+\frac{21 \pi ^2 \zeta (7)}{256}+\frac{7 \pi ^4 \zeta (5)}{768}+\frac{31 \pi ^6 \zeta (3)}{40320}+\frac{127 \pi ^8 \log (2)}{1209600}\\ \hline \\ {\bf H}^{(9)}_1(-1)&=&+\frac{189 \zeta (3) \zeta (7)}{256}+\frac{225 \zeta (5)^2}{512}+\frac{255}{256} \zeta (9) \log (2)-\frac{563 \pi ^{10}}{19160064}\\ {\bf H}^{(8)}_2(-1)&=&+8 {\bf H}^{(1)}_9(-1)+2 {\bf H}^{(2)}_8(-1)+\frac{255}{256} {\bf H}^{(2)}_8(+1)-\frac{237 \zeta(3) \zeta (7)}{32}-\frac{15 \zeta (5)^2}{4}+\frac{36067 \pi ^{10}}{159667200}\\ {\bf H}^{(7)}_3(-1)&=&-28 {\bf H}^{(1)}_9(-1)-7 {\bf H}^{(2)}_8(-1)-\frac{1785}{512} {\bf H}^{(2)}_8(+1)+\frac{2751 \zeta (3) \zeta (7)}{128}+\frac{615 \zeta (5)^2}{64}-\frac{223 \pi^{10}}{304128}\\ {\bf H}^{(6)}_4(-1)&=&+24 {\bf H}^{(1)}_9(-1)+3 {\bf H}^{(2)}_8(-1)+\frac{2295}{512} {\bf H}^{(2)}_8(+1)-2 {\bf H}^{(3)}_7(-1)-\frac{6831 \zeta (3) \zeta (7)}{256}-\frac{2745 \zeta (5)^2}{256}+\frac{64811 \pi ^{10}}{95800320}\\ {\bf H}^{(5)}_5(-1)&=&+10 {\bf H}^{(1)}_9(-1)+10{\bf H}^{(2)}_8(-1)-\frac{1275}{512} {\bf H}^{(2)}_8(+1)+5 {\bf H}^{(3)}_7(-1)+\frac{3795 \zeta (3) \zeta (7)}{256}+\frac{2775 \zeta (5)^2}{512}+\frac{893 \pi ^{10}}{31933440}\\ {\bf H}^{(4)}_6(-1)&=&-16 {\bf H}^{(1)}_9(-1)-9 {\bf H}^{(2)}_8(-1)+\frac{255}{512} {\bf H}^{(2)}_8(+1)-4{\bf H}^{(3)}_7(-1)-\frac{759 \zeta (3) \zeta (7)}{256}-\frac{255 \zeta (5)^2}{256}-\frac{43817 \pi ^{10}}{159667200} \end{eqnarray}
• Wow, this is really impressive. – Zaid Alyafeai May 5 '17 at 22:58
• @Zaid Alyafeai: I think I will be able to find all possible Euler sums at both plus and minus unity. The algorithm for this is given in the most recent update of the post above. Since you seem to be an expert on those things do you know if these quantities have been computed and compiled in a concise way somewhere else? I know that there is a review article on this which you sent to me before but it doesn't really contain all the results. Thanks for replying in advance. – Przemo May 11 '17 at 11:43
• To be honest it will be difficult for me to follow. However, I do encourage you to publish these results. You will have to do a literature review to make sure they are new. – Zaid Alyafeai May 11 '17 at 12:01
• Holy freaking crap. Just saw this. I'm tempted to hand you a +50 bounty and I haven't even read through this yet. Give me a bit and I just might :D – Brevan Ellefsen May 23 '17 at 15:14
• @Brevan Ellefsen: Thanks for the nice words. I checked all the result using wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi . What really puzzles me why ${\bf H}^{(1)}_5(-1)$ (and their higher analogues) cannot be reduced to one dimensional zeta functions.I have tried to compute them in different ways but it was always a roundabout. Maybe it is simply the case that as complexity arises there will always be some "new" numbers, i.e. such that they cannot be reduced to the numbers we were dealing with in the first place? – Przemo May 23 '17 at 17:32
Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1_{n\ge 3}}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1_{n\ge 2}}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) +\\ && \left[Li_1(t) Li_2(t) - \int\limits_0^t \frac{Li_1(\xi)^2}{\xi} d \xi\right] 1_{n=1} + \frac{1}{2} Li_2(t)^2 1_{n=2} \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since $$Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2)$$ for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$. In the case $n=1,2$ surface terms need to be taken into account. To be specific we have: \begin{eqnarray} {\bf H}^{(2)}_1(-1) &=& \frac{1}{12} \pi ^2 \log (2)-\zeta (3)\\ {\bf H}^{(2)}_2(-1) &=& -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2) \end{eqnarray}
• Very nice. This thread will be of great interest to me. Thanks. – Zaid Alyafeai May 3 '17 at 6:48
• Impressive work! I was happy to see here the closed form ${\bf H}^{(2)}_2(-1) = -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)$. Do you have more of the type ${\bf H}^{(m)}_m(-1)$ for $m\gt 2$? – Dr. Wolfgang Hintze Jan 3 '18 at 22:34
• @ Przemo Please have a look on my question and, especially, my self-answer math.stackexchange.com/questions/2587838/… which, in its conclusions, would rely much on your work here. I would appreciate your comments. – Dr. Wolfgang Hintze Jan 3 '18 at 23:14
This is not an answer but too long for a comment.
Referring to the impressive work of Przemo here, I have a specific question:
First of all we adopt the definition
$$\mathbf{H}_{m}^{(n)}(x) = \sum_{k=1}^\infty \frac{H_k^{(n)}}{k^m} x^k\tag{1}$$
My question concerns the case $n=m$ and $x=-1$, i.e. the alternating series with equal indices.
Question
For which $m = 1, 2, 3, ...$ the quantity
$$S^{+-}_{m,m}(-1) = \mathbf{H}_{m}^{(m)}(-1) = \sum_{k=1}^\infty (-1)^k \frac{H_k^{(m)}}{k^m}\tag{2}$$
has a closed form?
List of cases gathered
I have collected what I have found here up to now
$m=1$ $${\bf H}^{(1)}_1(-1) =\frac{1}{2} [\log(2)]^2 - \frac{1}{2} \zeta(2)$$
$m=2$ $${\bf H}^{(2)}_2(-1) = -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)$$
$m=3$ $${\bf H}^{(3)}_3(t) = \frac{1}{2} \left(-3 {\bf H}^{(4)}_2-2 \text{Li}_3(t){}^2+3 \text{Li}_2(t) \text{Li}_4(t)+5 \text{Li}_6(t)\right)$$
where, however,
$${\bf H}^{(2)}_4(-1) = \frac{1}{2} \int\limits_0^1 \frac{[log(1/\xi)]^1}{1!} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi-\int\limits_0^1 \frac{[log(1/\xi)]^3}{3!} \cdot \frac{[\log(1+\xi)]^2}{\xi} d\xi + Li_6(-1)$$
is not "closed"
$m=4$ $${\bf H}^{(4)}_4(-1) = 16 {\bf H}^{(1)}_7(-1)+6 {\bf H}^{(2)}_6(-1)+\frac{63}{32} {\bf H}^{(2)}_6(+1)-\frac{123 \zeta (3) \zeta (5)}{16}+\frac{3097 \pi ^8}{1036800}$$
where only these "explanations" are given.
$${\bf H}^{(1)}_7(-1) = {\bf H}^{(1)}_7(-1)$$ $${\bf H}^{(2)}_6(-1) = {\bf H}^{(2)}_6(-1)$$
Conclusion
I conclude from this list based on the results of Przemo that closed forms of the alternating series with equal indices (2) exist for $m=1$ and $m=2$. If closed forms for $m\ge3$ exist, and if so in what terms, is an open question.
Clarifying comments, especially from Przemo, are greatly appreciated.
• @ Przemo I'd like to draw you attention to my answer above which collects your results obtained so far on $\mathbf{H}_{m}^{(m)}(-1)$. Could you please comment if I understood correctly that there is no closed expression for m>=3. – Dr. Wolfgang Hintze Jan 7 '18 at 12:07
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# SOLUTION: I'm not really clear on how to set up this type of problem. Any insight would be greatly appreciated. x+y=12 y=3x
Algebra -> Algebra -> Coordinate Systems and Linear Equations -> SOLUTION: I'm not really clear on how to set up this type of problem. Any insight would be greatly appreciated. x+y=12 y=3x Log On
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Linear Solvers Practice Answers archive Word Problems Lessons In depth
Question 107332This question is from textbook Introductory and Intermediate Algebra : I'm not really clear on how to set up this type of problem. Any insight would be greatly appreciated. x+y=12 y=3xThis question is from textbook Introductory and Intermediate Algebra Found 2 solutions by ptaylor, Fombitz:Answer by ptaylor(2048) (Show Source): You can put this solution on YOUR website!x+y=12-------------------eq1 y=3x----------------------eq2 I think that substituting "y=3x" from eq2 into eq1 is the easiest way to work this problem. If we do that, we get: x+3x=12 or 4x=12 divide both sides by 4 x=3 From eq2: y=3x=3*3=9 So: x=3 and y=9 We could also subtract eq2 from eq1 and we would get: x=12-3x add 3x to both sides x+3x=12-3x+3x or 4x=12 same as before CK from eq1 x+y=12 3+9=12 12=12 from eq2 y=3x 9=3*3 9=9 Hope this helps----ptaylor Answer by Fombitz(13828) (Show Source): You can put this solution on YOUR website!1. 2. You have two equations with two unknowns, a linear system of equations. You should be able to solve for x and y. Using equation 2 and substitute into equation 1. 1. Now substitute into 2. 2. Check your answers with your original equations. 1. True statement. Good answer. 2. True statement. Good answer. x=3 and y=9.
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# Using the alternative formula to find the derivative of a function?
I'm attempting to find the derivative of the function: $$f(x) = 4x^2+3x+5$$ Using the alternative formula: $$\frac{f(z)-f(x)}{z-x}$$
Here are my steps so far: $$\frac{4z^2+3z+5-(4x^2+3x+5)}{z-x}$$ $$\frac{4(z^2-x^2)+3(z-x)}{z-x}$$
I have no idea where to go from this point. I've tried several different things to come up with the correct answer - which I know is $8x+3$. Can someone please guide me through this problem? I'm completely stuck. Also, sorry about the formatting. I'm using this editor http://www.codecogs.com/latex/eqneditor.php?lang=en-en and don't have it completely figured out yet.
-
The formatting was good. The formulas were all legible and correct, so I'd say that whatever you've been doing, keep doing it. – MJD Feb 10 '13 at 14:40
@MJD Somebody fixed it for me. There was a red /[ before and after each equation. And when I tried to delete those, it would break my formatting. – Scott Feb 10 '13 at 14:43
I fixed the red brackets. I just deleted them, and nothing broke. But even with the red brackets, everything else was perfectly clear, so it was easy to fix, and that's good enough. – MJD Feb 10 '13 at 14:45
this is not a simple alternative formula.this is called lagaranges theorem to check differentiability in a closed interval.There are some conditions before applying this formula – iostream007 May 10 '13 at 13:08
You must know that $z^2-x^2 = (z-x)(z+x)$ – Steven Gregory Mar 21 at 12:54
Use the difference of two squares:
$$z^2-x^2 = (z-x)(z+x)$$
-
You must calculate the limit too: $$\lim_{z\to x}\frac{f\left(z\right)-f\left(x\right)}{z-x}$$
In your case it's $$\lim_{z\to x}\frac{4(z^{2}-x^{2})+3(z-x)}{z-x}$$
Just simplify the fraction to $4\left( z+x \right)+3$ and you will see that the limit is equal to $8x+3$. Then you have done it.
These steps (if you need them more explicit): $$\frac{4\left(z^{2}-x^{2}\right)+3\left(z-x\right)}{z-x}=\frac{4\left(z+x\right)\left(z-x\right)+3\left(z-x\right)}{z-x}=4\left(z+x\right)+3$$ $$\lim_{z\to x}\left(4\left(z+x\right)+3\right)=4\left(x+x\right)+3=8x+3$$
-
Good point. I should have included the lim z->x on every step. – Scott Feb 10 '13 at 14:58
To use less space you can simplify fraction and only then use lim z->x on simplified expression. In such simple examples where only polynomials are used it is useful. – zaarcis Feb 10 '13 at 15:12
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# Teacher Candidate Alexis Lewis
Name:
Subject: Math
Date:
## Millicent Atkins School of Education: Common
Lesson Plan Template
PLANNING
List the Common Core/State Standard(s) to be addressed in this lesson:
CSS.MATH.CONTENT.2.NBT.A.1
Understand that the three digits of a three-digit number represent amounts of
hundreds, tens, and ones; e.g., 706 equals 7 hundreds, 0 tens, and 6 ones.
Understand the following as special cases:
CCSS.MATH.CONTENT.2.NBT.A.1.A
100 can be thought of as a bundle of ten tens called a "hundred."
CCSS.MATH.CONTENT.2.NBT.A.1.B
The numbers 100, 200, 300, 400, 500, 600, 700, 800, 900 refer to one, two, three,
four, five, six, seven, eight, or nine hundreds (and 0 tens and 0 ones).
CCSS.MATH.CONTENT.2.NBT.A.2
Count within 1000; skip-count by 5s, 10s, and 100s.
CCSS.MATH.CONTENT.2.NBT.B.7
Add and subtract within 1000, using concrete models or drawings and strategies
based on place value, properties of operations, and/or the relationship between
addition and subtraction; relate the strategy to a written method. Understand that
in adding or subtracting three-digit numbers, one adds or subtracts hundreds and
hundreds, tens and tens, ones and ones; and sometimes it is necessary to
compose or decompose tens or hundreds.
## List the Learning Objective(s) to be addressed in this lesson. Use the
following format: Students will be able to
## Students will be able to add 3 digit numbers using sticker notations.
Students will be able to solve word problems using sticker notation.
Students will be able to write an addition equation.
Describe how the objective is relevant to students lives.
The objectives are relevant to students lives because its important for them to
know how to correctly add, write and solve equations that contain large numbers.
Its also imperative that students master this skill for future tasks and goals so
that they can successfully reach them.
List the words relevant to the content area that you will either introduce
## List the materials you will need to teach the lesson.
SMART Board
Worksheet
Marker boards
Markers
ASSESSMENT
Pre-Assessment: Describe the instrument or process you will use to
measure students level of understanding toward the learning objective(s)
prior to teaching the lesson.
For the pre-assessment I will give the students a 5 questions worksheet. They
will write the number of stickers in sticker notation and then write an equation.
APPENDIX: Include a blank copy of the lesson pre-assessment. (if
applicable)
## Describe the timeline as to when you plan to administer the pre-
assessment?
(Recommended timeline is a minimum of two days prior to teaching your
lesson).
The pre-assessment will be administered 2 days before the lesson is
implemented.
Create and insert a table/chart/graph that shows the pre-assessment data
results. (if applicable)
Combining Stickers
Pre
25
20
15
10
## Insert an image of your table/chart/graph here.
Describe how the results of the pre-assessment (what the students have
demonstrated they know) will be used to design the lesson objectives,
instruction, and post-assessment.
The results from the pre-assessment will give me an idea of how well the
students understand adding three digit numbers and if they are capable of
showing their work through sticker notation.
Post-Assessment: APPENDIX: Include a blank copy of the lesson post-assessment
you will use to measure students level of understanding toward the learning objectives after
teaching the lesson.
## APPENDIX: Include a copy of a Key/Product (completed by you, the teacher) which
provides a model of the desired outcome.
## Is your key included at the end of this lesson plan?
TECHNOLOGY
Describe the instructional and/or assistive technology that you plan to
incorporate into the lesson to enhance instruction and student learning.
The SMART Board will be the main source of student interaction. An interactive
whiteboard activity will cover the new information that the students are going to
be learning. Through activities that are provided in the presentation, students will
be given multiple examples and non-examples of the appropriate way to solve a
three digit addition problem. Another source of technology that will be used by
the teacher is an iPhone. It will be used to randomly call on students to come up
to the SMART Board and complete the activity.
ACCOMMODATIONS
Describe the accommodations/differentiation/modifications you will use to
meet the needs of all learners and accommodate differences in students
learning, culture, language, etc. * Be sure that these accommodations are
based on what you identified/described in your contextual information
The teacher will have prior knowledge to any IEPs for individuals in the class and
will use differentiated instruction so that all students understand. For this
assignment, all the students should be able to accomplish the worksheet in its
original form. The SMART Board activity is accessible for all students.
MANAGEMENT
Identify the management and motivational strategies you will use to meet student
behavioral/developmental needs in order to keep students on task and actively engaged
throughout the lesson.
I will manage the class by keeping track of the students who are paying attention
and are actively engaged in the activities. At the end of the lesson, I will access
class dojo to give the hard working students a point, which will go towards a
reward. I will also manage the class by being actively engaged with the students
when they are working independently. This will include walking around the
classroom, redirecting inappropriate behavior and answering any questions that
students might have. If I have to redirect the whole class I will go over the
directions and my expectations from them when they are working independently. I
will also use call back ques to gain their full attention.
LESSON IMPLEMENTATION
I Do
(Teacher introduces lesson and models expected outcome of
learning objectives)
Describe how you will activate student interest and present the learning
objective in an engaging way (this is your lesson opening).
While the students are at recess, I will put one of the following under their desk: a
sheet of stickers, a strip of stickers, or a single sticker. Once the students are
sitting quietly at their desks, I will ask them to look under their desk. Each student
will find a certain amount of stickers. They will group up according to the amount
of stickers that they have. After they are in their groups, the students will discuss
why they think they have the amount of stickers that they do and how they can
be used to help them add. This activity will introduce them to the lesson and
should spark their interest in learning the objective.
Describe how you will communicate (to students) how the objective is
relevant to their lives.
I will communicate to the students that adding 3 digit numbers correctly will help
them be more successful students now and in the future.
Describe what instructional strategies you will use to
model/explain/demonstrate the knowledge and skills required of the
objective. (cite theories/theorists)
For the objectives of this lesson, I will use explicit teaching to get the information
across to the students. By showing multiple examples and non-examples, the
students should get a feel for what we will be doing in the guided practice. I will
use examples that the students can connect to real life situations so that they
understand that adding three digit numbers isnt something that they only do in
school. During this portion of the lesson, I will bridge previous information that
numbers together. I will inform them on where the hundreds, tens and ones are
placed in a number and how to properly write and equation. By showing them
examples of place value and how stickers can represent them, they will be able to
successfully complete the activities in the guided practice. I will also give them
information on the acronym D.E.A.L, which stands for draw, equation, answer and
label.
Describe how you will check for students understanding before moving on
to guided practice.
the information that is being introduced. This will give them the
opportunity to give me their ideas of how to add 3 digit numbers. I will also
ask open and close ended questions and pay close attention to the
reactions that the students are giving.
We Do
(Teacher engages students in guided practice)
Describe the learning activities you will use to provide students multiple opportunities to
practice the skills and content needed to meet the learning objective(s).
Together the students and the teacher will complete the interactive SMART
Board activity that has been created for this lesson.
At the beginning of the lesson, the students will review what each sticker
stands for.
The activity for this will be writing the correct word next to the objects that
are provided.
Once the students have the words written in, they will drive the ice cream
truck in to check the answers.
The single sticker represents the tens place value, the stick of stickers
represents the tens place and the sheet of stickers represents the hundreds
place.
Once the students have reviewed and understand what each place value is
represented by, they will move on to their next activity.
This activity will be completed by 3 students. Each student will be called up
to the SMART Board.
The objective is to find the hidden number with the magnifying glass. Once
they have found a number, there will be 3 hidden, the class will use sticker
notation to show the teacher that they understand where each place value
is located. (Ex: 216, 2 sheets of sticker, 1 stick of stickers and 6 single
stickers)
For the next 3 slides of the activity, the students will complete 2 word
problems using sticker notation and writing equations for the numbers
provided the correct way.
The first word problem will be given in this form: Miss M. went to the sticker
store and bought this amount of ice cream stickers: 3 sheets of ice cream
stickers + 1 strip of ice cream stickers + 5 single ice cream stickers.
I will use call on a student randomly to come up to the board and show us
the sticker notation of this question.
Once they have finished that, I will then have them write an equation. The
student will be able to phone a friend if they are struggling to come up with
The second word problem is very similar to the first one, the only thing
different is the name of the person and the amount of stickers that were
bought.
The third and final word problem will be a combination of the first and
second word problem. The question will be asked in this form: Miss M. has
315 ice cream stickers. Her whole class has 654. How many stickers does
Miss. M and her whole class have together?
The task that will be given to the students is to write an equation for the
word problem.
Once each student has held up an equation on their white boards, I will
show them the equation that I have come up with on the board. (ex:
654+315=)
Before showing them the answer, I will ask multiple students how they got
the equation that they did.
While solving the problem, they will need to show me their drawing (sticker
notation), an equation, the answers and labels.
After discussion on the equation, I will slide down the screen on the
slide and show the students the magic tunnel.
Before choosing a student to slide the equation through the magic
tunnel, I will have them solve the problem using sticker notation.
After all the students have shown their work, the equation can then
be slide through the tunnel to reveal the answer.
The last slide on the whiteboard activity will be completed on their
white boards.
The students are to solve the three equations that are provided
through sticker notation.
Once all of the students have finished, I will call on students to come
up to the board and deliver the cones to the ice cream truck.
The student that I call on will explain to the class how they got their
The ice cream truck will either accept the correct answer or deny it.
The numbers will be provided in an ice cream cone.
Describe how you will check for students understanding before moving on to independent
practice.
To check the students understanding of the objectives, they will be randomly
called upon throughout the lesson to come up to the SMART Board and complete
a task. This will give them the chance to try out the new information before
moving on to their individual work. I will also answer any last minute questions
that they have. Before allowing them to start on their independent work, Ill
explain to them what my expectations are for the worksheet.
You Do
(Students engage in independent practice)
Describe what the students will do to independently practice the knowledge and skills
required by the lesson objectives? (this is the post-assessment)
After the lesson is done, the students will complete a 5 question worksheet to test
their knowledge on the learning objectives. The interactive whiteboard activity
will also be used as a source of independent practice. For the closing of this
lesson, I will give the students a notecard with an equation on it. Their task it to
solve the problem, on the back of the notecard, using sticker notation and writing
the answer that they come up with. They must correctly use D.E.A.L.
APPENDIX: Include a blank copy of your post-assessment.
Is your BLANK COPY of the post-assessment included at the end of this
lesson plan?
Lesson Closing
Describe how you will reemphasize the lesson objective and any
skills/content that were taught in an interactive manner (whole/small
group, etc.).
After the students finish the worksheet, I will again review the information that we
went over. I will ask the students questions regarding place value and what they
are represented by and how three digit addition problems can be solved. I will
again remind them how important adding larger numbers are.
ANALYZE
(This portion may only be done after the post assessment is
collected/scored.)
Describe the results of the Post-Assessment and be sure to address the following:
Students progress from pre-to-post assessment. (if applicable)
Factors that may have influenced the post assessment results.
How the results of the post assessment highlight what areas of the
lesson will require re-teaching (if any).
If applicable, insert a table/chart/graph (below) that shows the post-assessment data
results. If you used the same document for both the pre and post assessments, it is strongly
encouraged that you show the comparison.
Combining Stickers
Student 17
Student 16
Student 15
Student 14
Student 13
Student 12
Student 11
Student 10
Students 9
Student 8
Student 7
Student 6
Student 5
Student 4
Student 3
Student 2
Student 1
0 5 10 15 20 25
Pre Post
## Insert an image of your table/chart/graph here.
Description: Above is the results from my pre and post assessments. I used a bar
graph to show the improvement amongst the worksheets.
REFLECT
List and describe two things you feel you did well to plan, implement, or assess instruction
(successes).
I feel that I did well with classroom management. All the students were
engaged in the lesson and followed directions. They responded to my cues
I also feel that the interactive SMART Board activity went really well
because I had a lot of input from my cooperating teacher. She gave me
ideas of what the students like and I worked around that.
List and describe two things you feel were challenges during the planning, implementation
and/or assessment of the lesson.
One of the most challenging aspects I crossed while planning my lesson
was making my lesson informational enough for the amount of time that I
was given. I had to make sure that the activities and information I was
implementing was important enough so the student could master the
objectives.
Another challenge that I came across was making the students post-
assessment. I had to make it short enough so that the students could finish
it in the time given. I struggled with limiting myself to a certain amount of
questions because I feel that its very important for students to practice a
new concept as much as possible.
List and describe two ideas for redesign you would make if you were to teach this lesson
again.
I would alter how I presented my directions and would go into more detail
misunderstand what I wanted done on the worksheet and by further
explaining, more students might have done better.
I would also change the way I selected students to come up to the SMART
Board. I used a random picker on my phone and it chose the same student
over and over and slowed down my lesson. Instead of using that, I could
use Popsicle sticks so that they can be left out once that student has
## Pre-Assessment and Post-Assessment
Combining Sets of Stickers
Miss McCorkle has 248 stickers. Miss Lewis has 123 stickers.
## Sticker Notation: Sticker Notation:
Equation: Equation:
______+_______+_______= ______+_______+_______=
Miss McCorkles cat has 211 The students at May Overby have
stickers. 615 stickers.
## Sticker Notation: Sticker Notation:
Equation: Equation:
_______+_______+_______= _______+_______+_______=
If Miss McCorkle, who has 248 stickers, and Miss. McCorkles cat, who has
211 stickers, added all their stickers together, how many stickers would they
have total?
Combining Sets of Stickers
Miss McCorkle has 248 stickers. Miss Lewis has 123 stickers.
## Sticker Notation: Sticker Notation:
Equation: Equation:
200 + 40 + 8=248 Stickers 100 + 20 + 3=123 Stickers
Miss McCorkles cat has 211 The students at May Overby have
stickers. 615 stickers.
## Sticker Notation: Sticker Notation:
Equation: Equation:
200 + 10 + 1=211 Stickers 600 + 10 + 5 = 615 Stickers
If Miss McCorkle, who has 248 stickers, and Miss. McCorkles cat, who has
211 stickers, added all their stickers together, how many stickers would they
have total?
248 + 211 = 459 Stickers
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# CCSS.MATH.CONTENT.4.NF.A : Fourth Grade Math Worksheets
Extend understanding of fraction equivalence and ordering.
Below, you will find many printable and common core aligned worksheets for core standard 4.NF.A These worksheets are grade-appropriate for Fourth Grade Math. We have many worksheets covering various aspects of this common core standard, 4.NF.A.1, 4.NF.A.2, and many more. Look for more worksheets by visiting the page for this grade and subject using the menu.
A brief description of the worksheets is on each of the worksheet widgets. Click on the images to view, download, or print them. All worksheets are free for individual and non-commercial use.
View the full list of topics for this grade and subject categorized by common core standards or in a traditional way.
• Unlimited Access
4.NF.A
Fraction Basics - I
Refer to the image models(divided shapes into equal parts) and write a fraction for shaded and un-shaded parts. Converts fractions in word forms to number forms.
Core Standard: 4.NF.A.1
Fraction Basics - II
Write fractions for various points on a number line. Draw pictures to show fractions like thirds, fourths etc. What is a unit fraction?
Core Standard: 4.NF.A.1
Fraction Basics - III
Write fractions based on parts of groups in number and word forms. Complete a fraction pattern or sequence. Model fractions using shapes by dividing and coloring.
Core Standard: 4.NF.A.1
Equivalent Fractions
This worksheet contains a variety of questions to help you master the concept of equivalent fractions.
Core Standard: 4.NF.A.1
Equivalent Fractions
Find the equivalent fractions. Create equivalent fractions by shading or coloring the given shapes.
Core Standard: 4.NF.A.1
Model Equivalent Fractions - I
Equivalent fractions are fractions with the same value. Write equivalent fraction for a given fraction. Identify pair of equivalent fractions in a set.
Core Standard: 4.NF.A.1
Model Equivalent Fractions - II
See the models of fractions and write equivalent fractions for them. Learn about simplest form and convert a few simple fractions to the simplest form.
Core Standard: 4.NF.A.1
Model Equivalent Fractions - III
Worksheets has a list of problems to make you fluent about equivalent fractions. Solve a word problem by identifying fractions from data. Solve a few multiple choice problems.
Core Standard: 4.NF.A.1
Mixed Numbers - I
Fractions that are greater than one can be written as fractions or a combination of a whole number and a fraction(Mixed Number). Use this worksheet to learn about mixed numbers.
Core Standard: 4.NF.A.1
Mixed Numbers - II
Use a number line to show mixed numbers as points on it. Convert fractions to mixed numbers and vice versa. Draw image models to show mixed numbers.
Core Standard: 4.NF.A.1
Mixed Numbers - III
Identify mixed numbers for various points on a number line. There are more problems where you will convert a mixed number to a fraction and vice versa.
Core Standard: 4.NF.A.1
Compare and Order Fractions - I
Use a number line to compare simple fractions by place them on it. Use comparison symbols and compare fractions. Order fractions using image models.
Core Standard: 4.NF.A.2
Compare and Order Fractions - II
Compare and order fractions using image models or number lines. Solve a word problem that requires you to compare a set of fractions.
Core Standard: 4.NF.A.2
Compare and Order Fractions - III
Use comparison math symbols and show which fraction is larger or smaller. Order fraction for least to greatest or vice versa. Solve a few word problems using comparison of fractions.
Core Standard: 4.NF.A.2
Compare and Order Mixed Numbers - II
Practice with image models to compare two mixed numbers by drawing shapes and comparing the shaded areas. Solve a data problem using your comparison skills.
Core Standard: 4.NF.A.2
Compare and Order Mixed Numbers - III
This worksheet is for advanced students. You will have to use all your fractional comparison skills to solve word problems based on data. More comparison and ordering problems for you to practice with.
Core Standard: 4.NF.A.2
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• This module has been created for use at Upper Key Stage 2 to develop children's understanding of fractions including mixed numbers and improper fractions. It demonstrates how to simplify fractions using common multiples. The module looks at how to compare and order fractions, including those greater than 1. It also introduces how to add, subtract, multiply and divide fractions.
The module builds on the objectives introduced in Fractions (Lower).
Fractions (Upper) assists teaching and learning of the following Year 5 and Year 6 mathematics objectives from the 2014 National Curriculum:
Year 5
Number - fractions
• Compare and order fractions whose denominations are all multiples of the same number
• Identify, name and write equivalent fractions of a given fraction, represented visually, including tenths and hundredths
• Recognise mixed numbers and improper fractions and convert from one form to the other and write mathematical statements greater than 1 as a mixed number. Eg 2/5 + 4/5 = 6/5 or 1 and 1/5
• Add and subtract fractions with the same denominator and multiples of the same number
• Multiply proper fractions and mixed numbers by whole numbers, supported by materials and diagrams, eg 1 ½ x 3 = 4 ½
Year 6
Number - fractions
• Use common factors to simplify fractions; use common multiples to express fractions in the same denomination
• Compare and order fractions including fractions greater than 1
• Add and subtract fractions with different denominators and mixed numbers, using the concept of equivalent fractions
• Multiply simple pairs of proper fractions, writing the answer in its simplest form, eg ¼ x ½ = 1/8
• Divide proper fractions by whole numbers, eg 1/3 divided by 2 = 1/6
Show the front screen. Watch Scully looking at the jug of space liquid and see the corresponding fraction displayed on the screen. Pause the animation. What is the fraction on the screen? How many fifths will be needed for the jug to be filled? Which number could be displayed on the screen instead of 5/5? Can any of the fractions be simplified? Why/why not?
Watch the videos. These begin with the recognition of equivalent fractions and finding common denominators. They then move on to putting these into practice when working with fractions that are greater than 1 and mixed numbers. Other videos demonstrate written methods for addition, subtraction, multiplication and division of fractions.
Activities can be used either as lesson starters or plenaries with the whole class, or by individuals. They allow children the opportunity to practise and consolidate what they have learnt from the videos. Children can use activities to practise recognition of equivalent fractions, comparing, ordering and simplifying fractions.
Use the printable resources to allow children to practise written methods for adding, subtracting, multiplying and dividing fractions. Each sheet also includes word problems that allow children to apply what they have learnt in context.
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# Understanding Percentages: How To Calculate The Percentage Of 25 Out Of 30
//
Thomas
Learn how to calculate the percentage of 25 out of 30 and understand the importance of percentages in everyday life and business. Avoid common mistakes and get tips for simplifying calculations and estimating quick approximations.
# Understanding Percentages
## Definition of Percentage
A percentage is a way to express a portion or part of a whole as a fraction of 100. It is commonly used to compare quantities and represent proportions. Percentages are often denoted by the symbol “%”, and they can be found in various aspects of our daily lives.
## How Percentages are Calculated
Calculating percentages involves simple mathematical operations. To find a percentage, you need to divide the part you are interested in by the whole and then multiply the result by 100. This will give you the percentage value.
For example, let’s say you want to find the percentage of a test score. If you scored 25 out of 30, you can calculate the percentage by dividing 25 by 30 and then multiplying the result by 100.
25 ÷ 30 = 0.8333
0.8333 × 100 = 83.33%
So, in this case, your score of 25 out of 30 corresponds to a percentage of 83.33%.
Calculating percentages can also be done using or by converting fractions to percentages. These methods provide alternative approaches to determine percentages based on specific scenarios.
Understanding how to calculate percentages is essential as it enables us to interpret and make sense of various data and information presented to us in everyday life, business, and finance. Let’s explore the importance of knowing percentages in the next section.
# Calculating the Percentage of 25 out of 30
## Step-by-Step Calculation Process
Calculating the percentage of a number is a common task that we encounter in everyday life. Whether we’re determining a discount on a purchase or analyzing data in business, understanding how to calculate percentages is essential. Let’s take a closer look at the step-by-step calculation process for finding the percentage of 25 out of 30.
1. Step 1: Set up the equation
To calculate the percentage, we divide the part (25) by the whole (30) and multiply by 100. This can be represented as:
Percentage = (Part / Whole) x 100
2. Step 2: Divide the part by the whole
In this case, we divide 25 by 30:
25 / 30 = 0.8333
3. Step 3: Multiply by 100
Next, we multiply the result from step 2 by 100 to obtain the percentage value:
0.8333 x 100 = 83.33
4. Step 4: Round off (if necessary)
Depending on the context, you may need to round off the percentage. In this case, we have 83.33%, which can be rounded to 83%.
So, the percentage of 25 out of 30 is approximately 83%. By following these simple steps, you can easily calculate percentages and use them in various real-life scenarios.
## Using Proportions to Find the Percentage
Another method to calculate the percentage of a number is by using . This approach can be particularly helpful when you have a part and want to find the whole or when you have a whole and want to find the part. Let’s explore how to use proportions to find the percentage of 25 out of 30.
1. Step 1: Set up the proportion
Begin by setting up a proportion with the given values. In this case, the part is 25 and the whole is 30:
25 / 30 = x / 100
2. Step 2: Cross-multiply and solve for x
Cross-multiply the values in the proportion and solve for x:
25 x 100 = 30 x x
2500 = 30x
3. Step 3: Divide and find the value of x
Divide both sides of the equation by 30 to isolate x:
2500 / 30 = x
x ≈ 83.33
By using , we find that the value of x, which represents the percentage, is approximately 83.33%. This method provides an alternative way to calculate percentages and can be useful in different scenarios.
## Converting a Fraction to a Percentage
Fractions and percentages are closely related, and converting a fraction to a percentage is a straightforward process. Let’s explore how to convert a fraction to a percentage for the case of 25 out of 30.
1. Step 1: Convert the fraction to a decimal
Begin by dividing the numerator (25) by the denominator (30) to obtain the decimal equivalent:
25 / 30 = 0.8333
2. Step 2: Multiply by 100
Next, multiply the decimal by 100 to convert it to a percentage:
0.8333 x 100 = 83.33
3. Step 3: Round off (if necessary)
Finally, round off the percentage to the desired level of precision. In this case, we have 83.33%, which can be rounded to 83%.
Converting a fraction to a percentage allows us to express fractional values in a more familiar and relatable way. By following these steps, you can easily convert fractions to percentages and use them in various calculations.
## Importance of Knowing Percentages
### Everyday Applications of Percentages
Percentages are used in various aspects of our everyday lives, often without us even realizing it. Understanding percentages can help us make informed decisions, analyze data, and solve everyday problems. Here are some everyday applications of percentages:
1. Shopping Discounts: When we see a product on sale with a certain percentage off, can help us calculate the actual amount we will save. For example, if a pair of shoes is marked 30% off, knowing how to calculate the discounted price can help us determine if it’s a good deal.
2. Tip Calculation: When dining out, knowing how to calculate a percentage can help us determine the appropriate tip amount. Whether it’s 15%, 18%, or 20%, being able to calculate the tip ensures fair compensation for the service received.
3. Sales Tax: Sales tax is typically a percentage added to the price of goods or services. Understanding percentages allows us to calculate the total cost of an item including tax. This knowledge is essential for budgeting and comparing prices.
4. Interest Rates: Whether it’s a mortgage, car loan, or credit card, interest rates are often expressed as percentages. Knowing how to calculate and compare interest rates helps us make informed decisions when borrowing or investing money.
### Percentage in Business and Finance
Percentages play a crucial role in the world of business and finance. Here are some ways percentages are used in these fields:
1. Profit and Loss: Understanding percentages is essential for evaluating the financial performance of a business. Calculating the percentage of profit or loss helps business owners determine the success or failure of their ventures.
2. Market Research and Data Analysis: Percentages are commonly used in statistical analysis and market research. They help identify trends, measure market share, and analyze survey responses. Being able to interpret and manipulate percentages is invaluable for making informed business decisions.
3. Investment Returns: Investors rely on percentages to assess the performance of their investments. Understanding how to calculate percentage returns and compare them against benchmarks helps investors monitor their portfolios and make adjustments if necessary.
4. Budgeting and Financial Planning: Percentages are used extensively in budgeting and financial planning. Whether it’s allocating funds for different expenses or setting savings goals, percentages allow individuals and businesses to create realistic and effective financial plans.
# Common Percentage Mistakes to Avoid
## Misinterpreting Percentage Values
Understanding and interpreting percentage values correctly is crucial in various aspects of life, from everyday tasks to business and finance. However, misinterpreting percentage values is a common mistake that many people make. Let’s take a closer look at some of the pitfalls to avoid when dealing with percentages.
### Lack of Context
One of the most common mistakes when interpreting percentage values is failing to consider the context in which they are presented. Percentages should always be examined in relation to the whole picture. For example, if you hear that a product’s price increased by 20%, it may sound like a substantial increase. However, without knowing the original price, it’s difficult to assess the actual impact. Always seek additional information to gain a complete understanding.
### Ignoring Base Values
Another mistake is disregarding the base values when working with percentages. Base values represent the starting point from which a percentage change is calculated. Let’s say a company’s revenue increased by 10%. Without knowing the initial revenue, it’s impossible to determine the actual increase. Always consider the base value to accurately interpret percentage changes.
### Confusing Percentage Increase and Decrease
Confusing percentage increase and decrease is a common error that can lead to incorrect interpretations. When a percentage is positive, it indicates an increase, while a negative percentage denotes a decrease. For example, a 30% decrease in sales is not the same as a 30% increase. Pay close attention to the sign of the percentage to avoid misconceptions.
## Incorrect Percentage Calculations
In addition to misinterpreting percentage values, incorrect calculations can also lead to errors. Understanding the proper methods for calculating percentages is essential to ensure accuracy. Let’s explore some common mistakes and how to avoid them.
### Incorrect Percentage Formula
One prevalent mistake is using an incorrect formula to calculate percentages. The formula for finding a percentage is to divide the part by the whole and multiply by 100. For instance, if you want to calculate what percentage 25 is out of 30, you would divide 25 by 30 and multiply by 100. Using the wrong formula can yield incorrect results, so always double-check your calculations.
### Rounding Errors
Rounding errors can occur when working with percentages, especially when dealing with decimal numbers. During calculations, it is essential to maintain the appropriate level of precision to avoid distorted results. Be mindful of rounding and ensure consistency throughout your calculations to prevent inaccuracies.
### Misplacing Decimal Points
Misplacing decimal points is a common error that can significantly affect percentage calculations. Even a small shift in decimal placement can lead to substantial differences in the final result. Always double-check your work to confirm that decimal points are correctly positioned for accurate calculations.
By being aware of these common mistakes and taking the necessary precautions, you can avoid misinterpreting percentage values and ensure accurate calculations. Understanding percentages correctly is vital for making informed decisions and effectively analyzing data.
## Tips for Working with Percentages
Working with percentages doesn’t have to be a daunting task. By simplifying percentage calculations and estimating percentages for quick approximations, you can easily navigate through various scenarios. Let’s explore these tips in detail.
### Simplifying Percentage Calculations
When it comes to calculating percentages, there are a few tricks that can simplify the process. Here are some helpful strategies:
1. Converting Percentages to Decimals: One way to work with percentages is by converting them to decimals. To do this, divide the percentage by 100. For example, 25% can be written as 0.25.
2. Finding a Percentage of a Number: To find a percentage of a number, multiply the number by the decimal equivalent of the percentage. For instance, if you want to find 25% of 80, you would multiply 80 by 0.25 to get 20.
3. Finding What Percentage One Number is of Another: If you want to determine what percentage one number is of another, divide the first number by the second number and multiply by 100. For example, if you want to find what percentage 25 is of 80, you would divide 25 by 80 and multiply by 100 to get 31.25%.
4. Calculating Percentage Change: Percentage change is useful when comparing two values. To calculate it, subtract the initial value from the final value, divide the result by the initial value, and multiply by 100. For instance, if a stock’s value increased from \$50 to \$75, you would subtract 50 from 75, divide the result by 50, and multiply by 100 to get a 50% increase.
### Estimating Percentages for Quick Approximations
In some situations, you may need to quickly estimate percentages without performing precise calculations. Here are a few methods to help you make quick approximations:
1. Round Numbers: When working with percentages, rounding numbers can make estimation easier. For example, if you need to find 37% of 100, you can round 37% to 40% and quickly calculate that it is 40.
2. Using Benchmark Percentages: Benchmark percentages can serve as reference points for estimation. For example, knowing that 10% of a number is equivalent to one-tenth of the number, you can estimate other percentages based on this benchmark. If you need to find 25% of a number, you can quickly estimate it as one-fourth of the number.
3. Visualizing Percentages: Sometimes, visualizing percentages can help with estimation. For instance, if you need to estimate what 75% of a rectangular area looks like, you might imagine three out of every four equal sections filled.
By simplifying percentage calculations and using estimation techniques, you can quickly and effectively work with percentages in various scenarios. These tips will not only save you time but also help you gain confidence in handling percentage-related calculations.
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# Chebyshev pseudo-spectral method
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
A type of trigonometric pseudo-spectral method (cf. Trigonometric pseudo-spectral methods). See also Fourier pseudo-spectral method.
A Chebyshev polynomial is defined as (cf. also Chebyshev polynomials). If , the resulting Chebyshev function is truly an th order polynomial in , but it is also a cosine function with a change of variable. Thus, a finite Chebyshev series expansion is related to a discrete cosine transform.
The Chebyshev pseudo-spectral method is the most logical choice of pseudo-spectral methods for problems with non-periodic boundary conditions. This comes from the particularly nice characteristics of the Chebyshev interpolating polynomials (cf. also Chebyshev polynomials).
Of all st degree polynomials, with leading coefficient , has the smallest maximum on the interval . Thus, in Lagrangian interpolation (see also Lagrange interpolation formula), if the interpolation points are taken to be the zeros of this polynomial, the error is minimized. A related and possibly more useful set of interpolation points are the extrema of : , , called the Gauss–Lobatto points.
The trigonometric interpolation of the function at the Gauss–Lobatto points is , where is the Cardinal function
with , otherwise. Rearranging, the interpolation polynomial becomes a finite Chebyshev series , where the Chebyshev coefficients are
Suppose the equation is to be solved, where is a differential operator, is a given function and is an unknown function. In the Chebyshev pseudo-spectral method, the solution is approximated by a Chebyshev interpolating polynomial. In the Lagrangian polynomial or "grid-point representation" , the problem can be written as , where . The form of can be found through differentiation of the Cardinal function: ,
and . Note that derivatives require operations.
Another way to find an expression for the derivative is to differentiate the Chebyshev series, which means differentiating the Chebyshev polynomials and using the recurrence relation for derivatives of Chebyshev polynomials, , where , , for . Thus, in coefficient representation, the derivative can be evaluated in operations while non-linear terms or multiplication by non-constant coefficients require . The fast discrete cosine transform can be used to switch between spectral and grid point representations.
#### References
[a1] J.P. Boyd, "Chebyshev and Fourier spectral methods" , Dover (2000) (pdf version: http://www-personal.engin.umich.edu/~jpboyd/book_spectral2000.html) [a2] D. Gottlieb, S.A. Orszag, "Numerical analysis of spectral methods: Theory and applications" , SIAM (1977) [a3] C. Canuto, M.Y. Hussaini, A. Quarteroni, T.A. Zang, "Spectral methods in fluid dynamics" , Springer (1987) [a4] D. Gottlieb, M.Y. Hussaini, S.A. Orszag, "Theory and application of spectral methods" R.G. Voigt (ed.) D. Gottlieb (ed.) M.Y. Hussaini (ed.) , Spectral Methods for Partial Differential Equations , SIAM (1984) [a5] B. Fornberg, "A practical guide to pseudospectral methods" , Cambridge Monographs Appl. Comput. Math. , 1 , Cambridge Univ. Press (1996)
How to Cite This Entry:
Chebyshev pseudo-spectral method. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Chebyshev_pseudo-spectral_method&oldid=14983
This article was adapted from an original article by Richard B. Pelz (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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# Thread: non right angle trig
1. ## non right angle trig
A tanker is moving 5 nautical miles (5 knots) on bearing 105 degrees.at 1100 hours (11am) the tanker obserbs a lighthouse on bearing 040 degrees. at midday the lighthouse bears 340 degrees. Draw a well labelled diagram to show the movement of the tanker between 11am and midday. calculate the distance in nautical miles from the tanker to the lighthouse at midday.
I know the angle LH is 40 degrees
but i still get confused in finding the angles for P1200 and P1100.
The length from P1100 to P1200 is 5nm. I use the SIN rule for non right angle trig.
so
SIN 5nm SIN ??????
sin 40degrees equals sin ??????
How do I work this out?
I need the answer 12 hours from now
thanks.
2. Originally Posted by mat
A tanker is moving 5 nautical miles (5 knots) on bearing 105 degrees.at 1100 hours (11am) the tanker obserbs a lighthouse on bearing 040 degrees. at midday the lighthouse bears 340 degrees. Draw a well labelled diagram to show the movement of the tanker between 11am and midday. calculate the distance in nautical miles from the tanker to the lighthouse at midday.
I know the angle LH is 40 degrees
but i still get confused in finding the angles for P1200 and P1100.
The length from P1100 to P1200 is 5nm. I use the SIN rule for non right angle trig.
so
SIN 5nm SIN ??????
sin 40degrees equals sin ??????
How do I work this out?
I need the answer 12 hours from now
thanks.
let lighthouse be point L
1100 position ... point A
1200 position ... point B
$m\angle A = 65$
$m\angle B = 55$
$m\angle L = 60$
use the law of sines ...
$\frac{LB}{\sin(65)} = \frac{AB}{\sin(60)}$
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Question
# Consider the number 21600. Find the sum of its divisors.
Hint: Factorize the given number in its prime factor form. If a number can be written as ${p_1}^a \times {p_2}^b \times {p_3}^c....$, where ${p_1},{p_2}$ and ${p_3}$ are prime numbers, then the sum of its divisors will be $\dfrac{{{p_1}^{a + 1} - 1}}{{{p_1} - 1}} \times \dfrac{{{p_2}^{b + 1} - 1}}{{{p_2} - 1}} \times \dfrac{{{p_3}^{c + 1} - 1}}{{{p_3} - 1}} \times ....$ Use this formula to find out the sum of the divisors.
According to the question, the given number is 21600. We have to determine the sum of its divisors.
This number can be written as:
$\Rightarrow 21600 = 216 \times 100 \\ \Rightarrow 21600 = {6^3} \times 100 \\ \Rightarrow 21600 = {\left( {2 \times 3} \right)^3} \times 4 \times 25 \\ \Rightarrow 21600 = {2^3} \times {3^3} \times {2^2} \times {5^2} \\ \Rightarrow 21600 = {2^5} \times {3^3} \times {5^2} \\$
Thus, the number is factorized in its prime factor form.
We know that if a number can be written as ${p_1}^a \times {p_2}^b \times {p_3}^c....$, where ${p_1},{p_2}$ and ${p_3}$ are prime numbers, then the sum of its divisors will be $\dfrac{{{p_1}^{a + 1} - 1}}{{{p_1} - 1}} \times \dfrac{{{p_2}^{b + 1} - 1}}{{{p_2} - 1}} \times \dfrac{{{p_3}^{c + 1} - 1}}{{{p_3} - 1}} \times ....$
Using above formula for $21600 = {2^5} \times {3^3} \times {5^2}$, we’ll get:
$\Rightarrow$ Sum of divisors $= \dfrac{{{2^{5 + 1}} - 1}}{{2 - 1}} \times \dfrac{{{3^{3 + 1}} - 1}}{{3 - 1}} \times \dfrac{{{5^{2 + 1}} - 1}}{{5 - 1}}$
$\Rightarrow$ Sum of divisors $= \dfrac{{{2^6} - 1}}{{2 - 1}} \times \dfrac{{{3^4} - 1}}{{3 - 1}} \times \dfrac{{{5^3} - 1}}{{5 - 1}} = \dfrac{{64 - 1}}{1} \times \dfrac{{81 - 1}}{2} \times \dfrac{{125 - 1}}{4}$
$\Rightarrow$ Sum of divisors $= 63 \times \dfrac{{80}}{2} \times \dfrac{{124}}{4} = 63 \times 40 \times 31$
$\Rightarrow$ Sum of divisors $= 78120$
Therefore, the sum of the divisors of 21600 is 78120.
Note: We can also find out the number of divisors of 21600.
We know that if a number can be written as ${p_1}^a \times {p_2}^b \times {p_3}^c....$, where ${p_1},{p_2}$ and ${p_3}$ are prime numbers, then the number of factors of this number is $\left( {a + 1} \right) \times \left( {b + 1} \right) \times \left( {c + 1} \right) \times ...$
Thus, the number of factors of $21600 = {2^5} \times {3^3} \times {5^2}$ will be:
$\Rightarrow {\text{ No}}{\text{. of factors }} = \left( {5 + 1} \right)\left( {3 + 1} \right)\left( {2 + 1} \right) = 6 \times 4 \times 3 \\ \Rightarrow {\text{ No}}{\text{. of factors }} = 72 \\$
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A257993 Least gap in the partition having Heinz number n; index of the least prime not dividing n. 34
1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS The "least gap" of a partition is the least positive integer that is not a part of the partition. For example, the least gap of the partition [7,4,2,2,1] is 3. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product(p_j-th prime, j=1...r) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. In the Maple program the subprogram B yields the partition with Heinz number n. Sum of least gaps of all partitions of m = A022567(m). From Antti Karttunen, Aug 22 2016: (Start) Index of the least prime not dividing n. (After a formula given by Heinz.) Least k such that A002110(k) does not divide n. One more than the number of trailing zeros in primorial base representation of n, A049345. (End) REFERENCES G. E. Andrews, K. Eriksson, Integer Partitions, Cambridge Univ. Press, 2004, Cambridge. M. Bona, A Walk Through Combinatorics, World Scientific Publishing Co., 2002. LINKS Alois P. Heinz, Table of n, a(n) for n = 1..20000 P. J. Grabner, A. Knopfmacher, Analysis of some new partition statistics, Ramanujan J., 12, 2006, 439-454. FORMULA a(n) = A000720(A053669(n)). - Alois P. Heinz, May 18 2015 From Antti Karttunen, Aug 22-30 2016: (Start) a(n) = 1 + A276084(n). a(n) = A055396(A276086(n)). A276152(n) = A002110(a(n)). (End) EXAMPLE a(18) = 3 because the partition having Heinz number 18 = 2*3*3 is [1,2,2], having least gap equal to 3. MAPLE with(numtheory): a := proc (n) local B, q: B := proc (n) local nn, j, m: nn := op(2, ifactors(n)): for j to nops(nn) do m[j] := op(j, nn) end do: [seq(seq(pi(op(1, m[i])), q = 1 .. op(2, m[i])), i = 1 .. nops(nn))] end proc: for q while member(q, B(n)) = true do end do: q end proc: seq(a(n), n = 1 .. 150); # second Maple program: a:= n-> `if`(n=1, 1, (s-> min({\$1..(max(s)+1)} minus s))( {map(x-> numtheory[pi](x[1]), ifactors(n)[2])[]})): seq(a(n), n=1..100); # Alois P. Heinz, May 09 2016 # faster: A257993 := proc(n) local p, c; c := 1; p := 2; while n mod p = 0 do p := nextprime(p); c := c + 1 od: c end: seq(A257993(n), n=1..100); # Peter Luschny, Jun 04 2017 MATHEMATICA A053669[n_] := For[p = 2, True, p = NextPrime[p], If[CoprimeQ[p, n], Return[p]]]; a[n_] := PrimePi[A053669[n]]; Array[a, 100] (* Jean-François Alcover, Nov 28 2016 *) Table[k = 1; While[! CoprimeQ[Prime@ k, n], k++]; k, {n, 100}] (* Michael De Vlieger, Jun 22 2017 *) PROG (Scheme) (define (A257993 n) (let loop ((n n) (i 1)) (let* ((p (A000040 i)) (d (modulo n p))) (if (not (zero? d)) i (loop (/ (- n d) p) (+ 1 i)))))) ;; Antti Karttunen, Aug 22 2016 (Python) from sympy import nextprime, primepi def a053669(n): p = 2 while True: if n%p!=0: return p else: p=nextprime(p) def a(n): return primepi(a053669(n)) # Indranil Ghosh, May 12 2017 (PARI) a(n) = forprime(p=2, , if (n % p, return(primepi(p)))); \\ Michel Marcus, Jun 22 2017 CROSSREFS Cf. A215366, A002110, A022567, A049345, A053669, A055396, A276086, A276094, A276152. One more than A276084. Sequence in context: A334675 A078380 A062356 * A055881 A332202 A204917 Adjacent sequences: A257990 A257991 A257992 * A257994 A257995 A257996 KEYWORD nonn AUTHOR Emeric Deutsch, May 18 2015 EXTENSIONS A simpler description added to the name by Antti Karttunen, Aug 22 2016 STATUS approved
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# Use Multiplicative Relationships
In this worksheet, students will express multiplicative relationships in terms of ratios and fractions, cancelling them into their simplest by finding the HCF.
Key stage: KS 4
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR
Curriculum topic: Ratio, Proportion and Rates of Change
Curriculum subtopic: Ratio, Proportion and Rates of Change, Calculations with Ratio
Difficulty level:
### QUESTION 1 of 10
There are a number of occasions when we may see a relationship expressed as a multiple.
For example, we could say, "this bottle holds three times as much as that one."
This is an example of a multiplicative relationship.
One of the essential skills you need to master to be successful with questions involving multiplicative relationships, is to know how to write them as either a ratio or a fraction.
Let's look at this in action now.
e.g. One jug of water is three times larger than another. Express this relationship as a ratio.
We know that all ratios are written as the form a:b, where a and b are whole numbers.
In this case, we can write this relationship as 1:3 or 3:1.
This question doesn't specify which one is bigger out of the jugs, so it is acceptable to use either order in our ratio here.
e.g. I have two pieces of wood. Piece A is 60 cm and piece B is 80 cm long.
a) Write the lengths of A:B as a ratio in its simplest form.
We start by writing each number simply as a ratio, in the order expressed in the question:
60:80
We can then cancel this ratio down into its simplest form by finding the Highest Common Factor (HCF) of both numbers.
Here the HCF is 20:
60 ÷ 20 = 3
80 ÷ 20 = 4
So the simplest possible form of this ratio is 3:4
b) Find the length of B as a fraction of A.
In this question, we are asked to find B as a fraction of A, so B must go on the top of the fraction and A on the bottom, like this:
80 60
This fraction can then be cancelled down if we divide by a HCF of 20:
80 ÷ 20 = 4 60 ÷ 20 3
In this activity, we will express multiplicative relationships in terms of ratios and fractions, cancelling them into their simplest by finding the HCF or using them to find missing variables.
Type two words in the spaces to complete the sentence below.
Bag A contains 15 marbles and Bag B contains 10 marbles.
Express this relationship as a ratio in its simplest form.
Bag A contains 15 marbles and Bag B contains 10 marbles.
Express this relationship as a ratio A:B in its simplest form.
Match each multiplicative relationship on the left with its equivalent ratio on the right.
## Column B
3 times the size
2:1
4 times the size
1:3
Half the size
1:4
One tenth the size
10:1
It takes Lucy 20 minutes to walk to the shop.
If it takes James 15 minutes, what fraction of Lucy's time does James take?
Take care to only write one number in each of the spaces, as the fraction bar has already been provided for you.
## Column B
3 times the size
2:1
4 times the size
1:3
Half the size
1:4
One tenth the size
10:1
Two cans of fizzy drink contain 250 ml and 400 ml respectively.
What fraction of the smaller can does the larger can contain?
Take care to only write one number in each of the spaces, as the fraction bar has already been provided for you.
## Column B
3 times the size
2:1
4 times the size
1:3
Half the size
1:4
One tenth the size
10:1
Match each scenario on the left with its equivalent fraction in its simplest form on the right.
## Column B
6 minutes as a fraction of 10 minutes
1/2
6 minutes as a fraction of 20 minutes
3/5
100 ml as a fraction of 200 ml
3/10
2 hours as a fraction of 2 day
1/12
The length of two pieces of wood can be expressed in the ratio 3:5.
If the smaller piece of wood is 60 cm long, how long is the larger piece?
## Column B
6 minutes as a fraction of 10 minutes
1/2
6 minutes as a fraction of 20 minutes
3/5
100 ml as a fraction of 200 ml
3/10
2 hours as a fraction of 2 day
1/12
Jamal takes 4/5 of the time as his friend Richard to finish a race.
If Richard takes 1 hour to finish, how many minutes does Jamal take?
## Column B
6 minutes as a fraction of 10 minutes
1/2
6 minutes as a fraction of 20 minutes
3/5
100 ml as a fraction of 200 ml
3/10
2 hours as a fraction of 2 day
1/12
The contents of three milk cartons can be expressed using the ratio 2:3:5.
If the medium-sized carton contains 450 ml, how much do the smaller and larger ones contain?
Take care to only write numbers in each of the spaces, as the unit of measurement (ml) has already been provided for you.
Contains (in ml): Smaller Larger
• Question 1
Type two words in the spaces to complete the sentence below.
EDDIE SAYS
There's only two possible ways in which we can express a multiplicative relationship. They have to be either a fraction or a ratio. Review the Introduction now, if you need to refresh on how to express a multiplicative relationship in terms of a fraction or ratio, before you tackle the rest of this activity.
• Question 2
Bag A contains 15 marbles and Bag B contains 10 marbles.
Express this relationship as a ratio in its simplest form.
EDDIE SAYS
We know that all ratios are written as the form a:b, where a and b are whole numbers. In this case, we can write this relationship as 15:10 or 10:15. As this question doesn't specify which order to put the numbers in, it is acceptable to use either order in our ratio here. We can simplify this ratio by finding the HCF of both numbers. The HCF of 15 and 10 is 5: 15 ÷ 5 = 3 10 ÷ 5 = 2 So our simplest ratio is either 3:2 or 2:3. Which did you choose?
• Question 3
Bag A contains 15 marbles and Bag B contains 10 marbles.
Express this relationship as a ratio A:B in its simplest form.
EDDIE SAYS
This is the same question as previously, with one crucial difference. This time, we are asked to express the ratio in the form Bag A:Bag B, which means that the number relating to Bag A must appear first in our ratio. So, on this occasion, the only acceptable ratio is 3:2.
• Question 4
Match each multiplicative relationship on the left with its equivalent ratio on the right.
## Column B
3 times the size
1:3
4 times the size
1:4
Half the size
2:1
One tenth the size
10:1
EDDIE SAYS
This is all about key words. 'Half' means the second amount is '× 0.5 / ÷ 2' the first. Whilst 'one tenth' means that the second amount is '× 0.1 / ÷ 10' the first. Let's look at 'Half the size' as an example. If we write this as a ratio using the info above, it should be: 1 : 0.5 However, we are not allowed to use decimals in ratios of this type, so we need to convert the second number into a whole number. We can achieve this using × 2, which we also then need to apply to the first number also: 1 : 0.5 × 2 = 2:1 Can you use this example and your own knowledge of ratio to match the other pairs successfully?
• Question 5
It takes Lucy 20 minutes to walk to the shop.
If it takes James 15 minutes, what fraction of Lucy's time does James take?
Take care to only write one number in each of the spaces, as the fraction bar has already been provided for you.
EDDIE SAYS
The trick here is to take note of the words 'fraction of Lucy's time'. This means that Lucy's time has to go on the bottom of our fraction (as the denominator), whilst James' time needs to go on the top (as the numerator). We can then cancel this fraction by finding the HCF, which in this case is 5:
15 ÷ 5 = 3 20 ÷ 5 4
• Question 6
Two cans of fizzy drink contain 250 ml and 400 ml respectively.
What fraction of the smaller can does the larger can contain?
Take care to only write one number in each of the spaces, as the fraction bar has already been provided for you.
EDDIE SAYS
Once again, we need to look at the phrase 'fraction of the smaller can'. This means that 250 goes on the bottom of our fraction, whilst 400 goes on the top. Then we can simplify by finding the HCF, which in this case is 50:
• Question 7
Match each scenario on the left with its equivalent fraction in its simplest form on the right.
400 ÷ 50 = 8 250 ÷ 50 5
## Column B
6 minutes as a fraction of 10 min...
3/5
6 minutes as a fraction of 20 min...
3/10
100 ml as a fraction of 200 ml
1/2
2 hours as a fraction of 2 day
1/12
EDDIE SAYS
Let's look at '6 minutes as a fraction of 10 minutes' as an example. If we write this as a fraction using the info above, we know that 6 needs to be on top of 10, like this:
6 10
We can simplify this fraction by finding the HCF of both 6 and 10, which is 2:
6 ÷ 2 = 3 10 ÷ 2 5
Can you use this example and your own knowledge of fractions to match the other pairs successfully?
• Question 8
The length of two pieces of wood can be expressed in the ratio 3:5.
If the smaller piece of wood is 60 cm long, how long is the larger piece?
EDDIE SAYS
This is actually a hidden ratio question. We need to apply what we know to find a missing length, rather than calculate a ratio or fraction. If the smaller piece of wood is 60 cm, this represents 3 parts of our overall total length. We can use this fact to find that one part is worth 20 cm. The longer piece of wood represents 5 parts of our total length, and each part is worth 20 cm: 5 × 20 = 100 cm
• Question 9
Jamal takes 4/5 of the time as his friend Richard to finish a race.
If Richard takes 1 hour to finish, how many minutes does Jamal take?
EDDIE SAYS
This question may sound complex, but all it is asking us to do is to find 4/5 of 1 hour. As 1 hour is 60 minutes, we need to find 4/5 of 60 minutes. Remember that to find a fraction of an amount, we should divide by the bottom, then multiply by the top: 60 ÷ 5 × 4 = 48 minutes
• Question 10
The contents of three milk cartons can be expressed using the ratio 2:3:5.
If the medium-sized carton contains 450 ml, how much do the smaller and larger ones contain?
Take care to only write numbers in each of the spaces, as the unit of measurement (ml) has already been provided for you.
Contains (in ml): Smaller Larger
EDDIE SAYS
Another hidden ratio question to end this activity. If the medium-sized carton contains 450 ml, this represents 3 parts of our overall total capacity. We can use this fact to find that one part is worth 150 ml. The largest carton represents 5 parts of our total capacity, and each part is worth 150 ml: 5 × 150 = 750 ml The smallest carton represents 2 parts of our total capacity, and each part is worth 150 ml: 2 × 150 = 300 ml Great job completing this activity! You can now express multiplicative relationships in terms of ratios and fractions, cancelling them into their simplest by finding the HCF or using them to find missing variables.
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https://edurev.in/course/quiz/attempt/-1_Test-Flow-Through-Open-Channels-2/0620ccd5-529e-4dca-a3c2-f5f1690a04ec
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Courses
# Test: Flow Through Open Channels - 2
## 10 Questions MCQ Test GATE Civil Engineering (CE) 2022 Mock Test Series | Test: Flow Through Open Channels - 2
Description
This mock test of Test: Flow Through Open Channels - 2 for Civil Engineering (CE) helps you for every Civil Engineering (CE) entrance exam. This contains 10 Multiple Choice Questions for Civil Engineering (CE) Test: Flow Through Open Channels - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Flow Through Open Channels - 2 quiz give you a good mix of easy questions and tough questions. Civil Engineering (CE) students definitely take this Test: Flow Through Open Channels - 2 exercise for a better result in the exam. You can find other Test: Flow Through Open Channels - 2 extra questions, long questions & short questions for Civil Engineering (CE) on EduRev as well by searching above.
QUESTION: 1
### A right-angled triangular channel symmetrical in section about the vertical carries a discharge of 5 m3/s with a velocity of 1.25 m/s. What is the approximate value of the Froude number of the flow?
Solution:
We know that,
Q = AV
⇒ 5 = A x 1.25 ⇒ A = 4 m2
For a symmetrical right angled triangular channel,
m = 1
∴ A = my2 = 1 x y2 = y2
T = 2my = 2 x 1 x y = 2y
But A = 4 m2
⇒ y2 = 4 ⇒ y = 2 m
QUESTION: 2
Solution:
QUESTION: 3
### Flow depths across a sluice gate are 2.0 m and 0.5 m. What is the discharge (per metre width)?
Solution:
QUESTION: 4
Given that
S0 = slope of the channel bottom
Se = slope of the energy line
F = Froude number
the equation of gradually varied flow is expressed as
Solution:
QUESTION: 5
A rectangular channel is 6 m wide and discharges 30 m3s-1. The upstream depth is 2.0 m, acceleration due to gravity is 10 ms-2. Then, what is the specific energy (approximate)?
Solution:
We know that,
Q = AV
⇒ 30 = 6 x 2 x V
⇒ \/ = 2.5 m/s
The specific energy may be given as,
QUESTION: 6
Most efficient channel section is
Solution:
QUESTION: 7
For a smooth hump in a subcritical flow to function as a broad crested weir, the height ΔZ of the hump above the bed must satisfy which one of the following?
(E1 = Specific head upstream of the hump, Ec = Specific head at the critical" depth yc) (Neglect friction effects)
Solution:
If the flow is subcritical, then after the introduction of hump the specific energy will decrease by ΔZ (height of hump).
E0 ≤ E1 - ΔZ
⇒ ΔZ ≤ (E1 - E0)
QUESTION: 8
A hydraulic jump occurs at the top of a spillway. The depth before jump is 0.2 m. The sequent depth is 3.2 m. What is the energy dissipated in m (approximate)?
Solution:
The energy dissipation is given by,
QUESTION: 9
What is the normal depth in a wide rectangular channel carrying 0.5 m2/s discharge at a bed slope of 0.0004 and Manning’s n = 0.01?
Solution:
For a wide rectangular channel, the hydraulic radius (F) is approximately equal to the depth of flow. By Manning’s equation, we have,
QUESTION: 10
Match List-I (Flow section type) with List-ll (Critical discharge is proportional to) where y is the depth of flow-and select the correct answer using the codes given below the lists:
Solution:
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### Chord
##### Stage: 5 Challenge Level:
Equal touching circles have centres on a line. From a point of this line on a circle, a tangent is drawn to the farthest circle. Find the lengths of chords where the line cuts the other circles.
### Napoleon's Hat
##### Stage: 5 Challenge Level:
Three equilateral triangles ABC, AYX and XZB are drawn with the point X a moveable point on AB. The points P, Q and R are the centres of the three triangles. What can you say about triangle PQR?
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# Solutions to x
1. Nov 9, 2012
### christian0710
This may be a dumb question but if we have the equation 2x=x2 and we use algebra we get
2=x2/x ---> 2=x
How come the solution is x= 2 (obvious) AND x=0 (not so obvious for me)
2. Nov 9, 2012
### haruspex
Clearly x=0 is a valid solution of the original equation. But the important lesson is that whenever you divide by an expression of unknown value (x in this case) you should bear in mind that division by zero is not a defined operation. The correct procedure is always to write "if (expression) is nonzero then ...".
3. Nov 9, 2012
### christian0710
I see so you obtain that if x=0 you get 2*0=^2 and that's why x=0 is an equation?
But if we started with the equation x=2, then i assume you can't say x=2 and x=0, is that correctly understood? I needed both solutions because i was finding the upper and lower limits for integration, but just confues about the fact that x=2 also has x=0 as solution.
4. Nov 9, 2012
### mtayab1994
Well doesnt 2x=x^2 ⇔0=x^2-2x ⇔ 0=x(x-2) ⇔x=0 or x=2 Is that clear now? It is pretty simple.
5. Nov 9, 2012
### Staff: Mentor
If x = 0, you get 2*0 = 02, so x = 0 is a solution to the original equation.
The only possible replacement for x in the equation x = 2 is 2. That's the only value that makes the equation x = 2 a true statement.
The equation x = 2 does NOT have x = 0 as a solution. The equation 2x = x2 DOES have x = 0 (and x = 2) as a solution.
Since the equations x = 2 and 2x = x2 have different solution sets, they are not equivalent.
6. Nov 9, 2012
### christian0710
Thank you so much, now it's clear! Very clear :D
7. Nov 9, 2012
### mtayab1994
You're very welcome :D
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# A body of mass 5 kg slides a distance of 6 m down a rough inclined plane (30 degrees). Then it...
## Question:
A body of mass 5 kg slides a distance of 6 m down a rough inclined plane (30 degrees). Then it moves on frictionless horizontal surface and compresses a spring. The coefficient of kinetic friction is 0.1 and the spring constant is 300 N/m. Find the maximum compression of the spring.
## Spring Energy:
Spring energy is a type of potential energy that exists in spring due to external mechanical work done on it. So we can say that the external work done is stored in the spring as its spring energy.
The spring energy is proportional to the square of the deformation length of the spring
Given:
• The mass of the body is, {eq}m = 5\ \text{kg} {/eq}
• The sliding distance is, {eq}d = 6\ \text{m} {/eq}
• The coefficient of kinetic friction is {eq}\mu_k = 0.1 {/eq}
• The spring constant is {eq}k = 300 \text{N/m} {/eq}
• The inclination angle is, {eq}\theta = 30^\circ {/eq}
. Let the maximum compression of the spring is, {eq}x {/eq}
For the moving body, its initial gravitational potential energy at the top of the incline is equal to the sum of the work done by the frictional force and the final spring energy.
\begin{align} m\ g\ d\ \sin\ \theta & = \mu_k\ m\ g\ d \cos\ \theta + 0.5\ k\ x^2\\ (5\ \text{kg})\ (9.81\ \text{m/s}^2)\ ( 6\ \text{m})\ \sin\ 30^\circ &= 0.1\ (5\ \text{kg})\ (9.81\ \text{m/s}^2)\ ( 6\ \text{m})\ \cos\ 30^\circ + 0.5\ (300\ \text{N/m})\ x^2\\ \implies x& =\boxed{ 0.9\ \text{m}}\\ \end{align}
Practice Applying Spring Constant Formulas
from
Chapter 17 / Lesson 11
3.1K
In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.
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Python: Compute cumulative sum of numbers of a given list - w3resource
# Python: Compute cumulative sum of numbers of a given list
## Python Basic - 1: Exercise-92 with Solution
Write a Python program to compute cumulative sum of numbers of a given list.
Note: Cumulative sum = sum of itself + all previous numbers in the said list.
Sample Solution:
Python Code:
``````def nums_cumulative_sum(nums_list):
return [sum(nums_list[:i+1]) for i in range(len(nums_list))]
print(nums_cumulative_sum([10, 20, 30, 40, 50, 60, 7]))
print(nums_cumulative_sum([1, 2, 3, 4, 5]))
print(nums_cumulative_sum([0, 1, 2, 3, 4, 5]))
``````
Sample Output:
```[10, 30, 60, 100, 150, 210, 217]
[1, 3, 6, 10, 15]
[0, 1, 3, 6, 10, 15]
```
Pictorial Presentation:
Flowchart:
Python Code Editor:
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
## Python: Tips of the Day
What is the difference between Python's list methods append and extend?
append: Appends object at the end.
```x = [1, 2, 3]
x.append([4, 5])
print (x)
```
Output:
```[1, 2, 3, [4, 5]]
```
extend: Extends list by appending elements from the iterable.
```x = [1, 2, 3]
x.extend([4, 5])
print (x)
```
Output:
```[1, 2, 3, 4, 5]
```
Ref: https://bit.ly/2AZ6ZFq
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When a 45 g sample of one alloy at 100ºC is dropped into 100.0 g the water at 25.0ºC the final temperature is 37.0ºC. What is the particular heat of the alloy?
heat acquired by water = q = mC∆T
m = mass = 100.0 g
C = particular heat of water = 4.184 J/g/deg
∆T = readjust in temperature = 37 - 25 = 12 degrees
q = 100.0g x 4.184 J/g/deg x 12 deg = 5021 J
This must likewise be the warm lost by the alloy, and also the temperature change of the alloy is100 - 37 = 63 deg
Specific warmth is joules/g/deg = 5021 J/45 g /63 deg = C = 1.77 J/g/deg
heat lost by alloy = heat acquired by water
(45.0 g)(C J/g/deg)(63 deg) = (100.0 g)(4.184 J/g/deg)(12 deg)
2835 C = 5021
C = 1.77 J/g/deg
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# (RGS 4/5) Inverting matrices modulo m
Given an integer matrix M and a modulus m, find an inverse of M modulo m. If the matrix M is not invertible modulo m, the behaviour is left unspecified.
# Matrix inverse
If M is a square matrix, its inverse exists if and only if its determinant is not 0. Similarly, when we are talking about matrices modulo m, the inverse of M will exist if and only of the determinant of M is invertible modulo m, which happens when the determinant is coprime with m.
The inverse of M is a square matrix inv(M) such that M*inv(M) = inv(M)*M = I, where
$$I = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & 1 \end{bmatrix}$$
has the same shape of M and is called the identity matrix. As an example, consider the first test case, where [[22, 43], [29, 37]] is the inverse of [[26, 16], [38, 41]] mod 45:
$$\begin{bmatrix} 26&16\\38&41 \end{bmatrix} \cdot \begin{bmatrix} 22&43\\29&37 \end{bmatrix} = \begin{bmatrix} 1036&1710\\2025&3151 \end{bmatrix} \equiv \begin{bmatrix} 1 & 0 \\ 0&1 \end{bmatrix} \mod 45$$
# Input
A square matrix M with integer values between 0 and m-1, inclusive, and a positive integer m > 1. The matrix may be given in any sensible format, including
• a list of lists, where the inner lists encode the rows, like M = [[1, 2], [3, 4]], or a flattened version, like M = [1, 2, 3, 4]
• a list of lists, where the inner lists encode the columns, like M = [[1, 3], [2, 4]], or a flattened version, like M = [1, 3, 2, 4]
where these encode the matrix
$$\\$$$$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$$$\\$$
The integer m giving the modulus.
You may also accept the size of the matrix as input.
The inputs can be given in any order.
# Output
A matrix representing the inverse of M modulo m. You may assume such an inverse exists. Preferable format is for each matrix entry $$\a_{i,j}\$$ to satisfy $$\0 \leq a_{i,j} < m\$$ but this is just to make it easier to compare with the test cases.
# Test cases
45, [[26, 16], [38, 41]] -> [[22, 43], [29, 37]]
39, [[29, 50], [29, 1]] -> [[16, 19], [4, 35]]
35, [[24, 14], [48, 45]] -> [[5, 7], [4, 33]]
53, [[43, 20], [15, 8]] -> [[5, 14], [37, 7]]
49, [[15, 11, 30], [20, 12, 40], [33, 25, 2]] -> [[33, 28, 23], [25, 18, 0], [25, 48, 13]]
37, [[8, 9, 22, 17], [24, 30, 30, 19], [39, 8, 45, 23], [5, 30, 22, 33]] -> [[18, 17, 26, 20], [29, 36, 23, 1], [19, 0, 9, 3], [30, 23, 14, 21]]
This is so shortest submission in bytes, wins! If you liked this challenge, consider upvoting it... And happy golfing!
This is the fourth challenge of the RGS Golfing Showdown. If you want to participate in the competition, you have 96 hours to submit your eligible answers. Remember there is still 300 reputation in prizes! (See 6 of the rules)
Also, as per section 4 of the rules in the linked meta post, the "restricted languages" for this third challenge are only Jelly, V (vim) and 05AB1E so submissions in these languages are not eligible for the final prize. But they can still be posted!!
Otherwise, this is still a regular challenge, so enjoy!
• I don't think it's quite true that a matrix is invertible if it has a nonzero determinant, when working modulo m where m is composite. For example, [[2, 0], [0, 1]] has determinant 2 but isn't invertible modulo 6.
– xnor
Commented Mar 7, 2020 at 8:21
• @xnor fixed, what I really wanted was "coprime with m".
– RGS
Commented Mar 7, 2020 at 8:23
• May we take the size of the matrix as input? Commented Mar 7, 2020 at 9:00
• In the description, you say A square matrix M with integer values between 0 and m-1, inclusive. However, in the last test case, one of the elements of the matrix is 39, greater than m, which is 37 Commented Mar 8, 2020 at 5:35
• It might be nice to have a test case where at least the top left element is not relatively prime to the modulus - to force failure in a naive Gaussian elimination. In fact, it would be nice to have a test case where all elements of the original matrix are not relatively prime to the modulus - for example, modulus 6, matrix [[2, 3], [3, 2]] Commented Mar 9, 2020 at 19:03
# R, 68 bytes
function(M,m,n,A=M){while(any(A%*%M%%m!=diag(n)))A[]=rpois(n^2,9)
A}
Try it online!
Strikingly slow. Will most likely time out for all test cases on TIO, but is guaranteed to give an answer eventually.
Works by rejection sampling: generates random matrices A, with each value taken from a $$\Poisson(9)\$$ distribution, until a solution is found.
Note that to get A of the correct dimensions, it is 6 bytes shorter to initialize it as A=M and then replace all the values with A[]=rpois(n^2,9) than to create it directly with A=matrix(rpois(n^2,9),n).
• small typo, this should return A rather than I, right? Assuming it ever returns, haha. Commented Mar 10, 2020 at 19:09
• @Giuseppe Of course, it should return A. Thanks! Commented Mar 10, 2020 at 19:26
# J, 18 16 bytes
(]%1+.]^5 p:[)%.
Try it online!
Resolves p/q mod n element-wise (instead of using det(M) to resolve the modular inverse globally).
Abuses GCD of rational numbers to extract 1/q from p/q.
### How it works
(]%1+.]^5 p:[)%. NB. left arg = modulo, right arg = matrix
( )%. NB. bind inv(matrix) as new right arg
5 p:[ NB. phi(modulo)
]^ NB. inv(matrix)^phi(modulo) element-wise
1+. NB. GCD with 1; GCD(1, p/q) = 1/q
]% NB. Divide inv(matrix) by the above element-wise
# J, 18 bytes
%.@]*-/ .*@]^5 p:[
Try it online!
A dyadic tacit function that takes modulo (left arg) and the matrix (right arg), and gives possibly very large-valued modular inverse of the matrix. To reduce the range, prepend [| at the start of the function.
### How it works: the math
A simple mathematical way to calculate the modular inverse of a matrix is the following:
\begin{align} M^{-1} \text{ mod }n &= \text{cofactor}(M) \times \bigl((\det M)^{-1} \text{ mod }n \bigr) \\ &= M^{-1} \times \det M \times \bigl((\det M)^{-1} \text{ mod }n \bigr) \end{align}
If the matrix $$\M\$$ is invertible modulo $$\n\$$, we know that $$\(\det M)^{-1} \text{ mod }n\$$ exists, and it can be found using Euler's theorem:
$$(\det M)^{-1} \equiv (\det M)^{\varphi(n)-1} \text{ mod }n$$
Then we can simplify the original equation to
\begin{align} M^{-1} \text{ mod }n &= M^{-1} \times \det M \times \bigl((\det M)^{\varphi(n)-1} \text{ mod }n \bigr) \\ &\equiv M^{-1} \times (\det M)^{\varphi(n)} \mod{n} \end{align}
And now the fun fact: J has built-ins for matrix inverse, matrix determinant, and Euler's totient function. And it uses built-in rational numbers when calculating the matrix inverse!
### How it works: the code
%.@]*-/ .*@]^5 p:[ NB. left arg = modulo, right arg = matrix
5 p:[ NB. totient(modulo)
-/ .*@] NB. det(matrix)
^ NB. det(matrix) ^ totient(modulo)
%.@] NB. inv(matrix)
* NB. inv(matrix) * det(matrix) ^ totient(modulo)
• Nice, I knew there had to be some way to get modular matrix inversion from regular matrix inversion.
– xnor
Commented Mar 10, 2020 at 11:25
• I'm wondering if there's a way to do this without computing the matrix determinant. If I understand what's going on, the matrix inverse gives rational entries like p/q, and we need to re-interpret each of them as p/q done modulo n producing integer values. It would suffice to multiply everything by any large value that clears all the denominators and thereby produces integers, and equals 1 modulo n. One such value is indeed det(M)^phi(n). But maybe there's a shorter one.
– xnor
Commented Mar 10, 2020 at 11:46
• @xnor Your understanding is correct. Without det(M), the only other way would be to compute p/q mod n element-wise (or entirely fall back to Gaussian elimination). Commented Mar 10, 2020 at 13:58
# Wolfram Language (Mathematica), 23 bytes
¯\_(ツ)_/¯ the answer was in the documentation of Modulus
Inverse[#2,Modulus->#]&
Try it online!
# JavaScript (ES6), 209 206 bytes
Takes input as (modulo)(matrix).
This transposes the matrix of cofactors (resulting in the adjugate) and multiply it by the inverse of the determinant of $$\M\$$ modulo $$\m\$$.
m=>M=>M.map((r,y)=>r.map((_,x)=>((g=k=>(++k*D(M)%m+m)%m-1?g(k):x+y&1?-k:k)*D(h(M,x).map(r=>h(r,y)))%m+m)%m),h=(a,n)=>a.filter(_=>n--),D=M=>+M||M.reduce((s,[v],i)=>s+(i&1?-v:v)*D(h(M,i).map(r=>h(r,0))),0))
Try it online!
## Commented
### Helper function $$\h\$$
The function $$\h\$$ removes the $$\n\$$-th entry from the array $$\a[\:]\$$.
h = (a, n) => // a[] = array, n = index
a.filter(_ => n--) // keep all but the n-th entry
### Helper function $$\D\$$
The function $$\D\$$ computes the determinant of the matrix $$\M\$$.
D = M => // M[] = input matrix
+M || // if M[] is 1x1, stop recursion and return its unique value
M.reduce((s, [v], i) => // otherwise, for each value v at (0, i):
s + // add to the sum
(i & 1 ? - v : v) * // either v or -v depending on the parity of i
D( // multiplied by the result of a recursive call with:
h(M, i) // M[] without the i-th row
.map(r => h(r, 0)) // and without the first column
), // end of recursive call
) // end of reduce()
### Main function
m => M => // m = modulo, M[] = matrix
M.map((r, y) => // for each position y:
r.map((_, x) => // for each position x:
( //
( g = k => // g is a recursive function taking a counter k
( ++k * // increment k and multiply it
D(M) // by the determinant of M
% m + m //
) % m - 1 ? // if it's not congruent to 1 modulo m:
g(k) // try again until it is
: // else:
x + y & 1 ? -k // return either k or -k
: k // depending on the parity of x+y
) * // initial call to g with a zero'ish value
D( // multiply by the determinant of:
h(M, x) // M[] without the x-th row
.map(r => h(r, y)) // and without the y-th column
) % m + m // return the result modulo m
) % m //
) // end of inner map()
) // end of outer map()
# Jelly, 25 bytes
ÆḊ×Ɱ⁹%ỊTḢ×ZÆḊ-Ƥ$-ƤNÐe⁺€Zʋ Try it online! A dyadic link taking the matrix as its left argument and the modulus as its right. Returns a matrix. Append a % to get it within the range 0, m # SageMath, 48 33 bytes Saved 15 bytes thanks to ovs!!! lambda m,M:~Matrix(Integers(m),M) Nothing on TIO for SageMath unfortunately. Modular inverse of a matrix M (input as a Python list of lists) mod m. • @ovs So sorry, is that correct? On my phone so can't check. Commented Mar 7, 2020 at 14:43 • Yes this works. This is the same operator as you use in expressions like -~n. – ovs Commented Mar 7, 2020 at 14:46 • @ovs Makes sense - thanks! :-) Commented Mar 7, 2020 at 14:52 • Is this a library for Python or another language? Commented Mar 7, 2020 at 16:48 • @S.S.Anne If you follow the above link you'll get to the dev page. It's basically a whole system built on top of Python w numpy/scipy that does symbolic math!! Commented Mar 7, 2020 at 17:06 # Sledgehammer, 6 bytes ⠑⡿⡆⠱⣁⣭ Decompresses into this Wolfram Language function: Inverse[#2, Modulus -> #1] Try it online! # Charcoal, 41 bytes ＦＥＸθ×ηη⪪Ｅ×ηη÷ιＸθλη¿⬤ι⬤ζ⁼⁼λν﹪ΣＥμ×ιπλθＩι Try it online! Link is to verbose version of code. Takes input as $$\ m, n, M \$$ where $$\ n \$$ is the size of $$\ M \$$, and does not reduce its output modulo $$\ m \$$ (can be done at a cost of 2 bytes). Stupidly slow, so don't try this with realistic values. Explanation: ＦＥＸθ×ηη⪪Ｅ×ηη÷ιＸθλη There are $$\ m^{n^2} \$$ possible square matrices of size $$\ n \$$ with coefficients between $$\ 0 \$$ and $$\ m \$$. Looping over this value, compute each matrix, but don't bother reducing the terms modulo $$\ m \$$. Then, loop over the list of matrixes. ¿⬤ι⬤ζ⁼⁼λν﹪ΣＥμ×ιπλθ Perform the steps of matrix multiplication of this matrix by the input matrix, reduce it modulo $$\ m \$$, and compare the each result to the appropriate value of the identity matrix. Ｉι If this was the inverse then print the matrix. • By stupidly slow, I mean it can only cope with 4x4 with a modulus of 2 or 3x3 with a modulus of 2 or 3 or 2x2 with a modulus of up to 21 (23 at a cost of 2 bytes) on TIO. – Neil Commented Mar 7, 2020 at 22:42 # MATL, (25?) 31 29 26 bytes My first MATL answer -5 bytes & a bug-fix (+2) thanks to Luis Mendo! The trailing . may be unnecessary - it is if there is only ever a single inverse of M with elements modulo m. :inZ^!"&G@[]eY*w\tZyXy=?@. A full program which prints the elements in row major order separated by newlines. Try it online! - Too slow for any of the given test cases. Quite possibly not the best approach for MATL. ### How? :inZ^!"&G@[]eY*w\tZyXy=?@. - expects inputs m and M : - range (m) -> [1,2,...,m] i - push input (M) n - number of elements Z^ - ([1,2,...,m]) Cartesian power (#elements(M)) ! - transpose " - for each column, C: &G - push both inputs @ - push C [] - push an empty array (to make e work as below) e - reshape (C) to square matrix of side ceil(#elements(C)^0.5) Y* - (reshaped C) matrix multiplication (copy of M) w - swap top two stack entries \ - (multiplication result) modulo (copy of m) t - duplicate top of stack Zy - size Xy - (size by size) identity matrix = - equal -> logical matrix ? - if all are truthy: @ - push C . - break - implicit print of stack (the valid C) • @LuisMendo - I'm almost certain you'll beat this using another method, but any tips for this one would be welcome! Commented Mar 7, 2020 at 17:48 • ...also why is w "necessary" where I have it / is there something more normal that should go there? Commented Mar 7, 2020 at 17:52 • Thanks! Yes the 2 was somehow lost and should have been the column size. You're also correct about q being unnecessary since we don't have to give an answer modulo m. Commented Mar 10, 2020 at 20:23 # R, 128 bytes function(x,m,n)t(round(which((1:m*det(x))%%m<1.5)[1]*outer(1:n,1:n,Vectorize(function(a,b)det(x[-a,-b,drop=F])*(-1)^(a+b))))%%m) Try it online! A function taking three arguments, x = the matrix, m = the modulus and n the number of rows of x. Returns a matrix. Uses the same method as my Jelly answer. # Jelly, (21?) 22 bytes The trailing Ṫ may be unnecessary - it is if there is only ever a single inverse of M with elements modulo m. Ḷṗ⁹L²¤ṁ€⁹æ×%³L⁼þ$ƑɗƇṪ
A full program printing the result.
Try it online! - Too slow for any of the given test cases (the 35 case took ~20 minutes locally).
11 bytes (but floating point output):
Using Bubler's observation (go upvote!) that raising the determinant to Euler's totient is enough to remove the determinant's denominators:
æ*-×ÆḊ*ÆṪ}ɗ
However, unlike in J, the inversion of $$\M\$$ in Jelly gives floats so we no longer get an integer matrix as output.
Try it online!
• @RGS - I guess the 11 would only be acceptable if an integer output was given, even though this is not explicitly stated in the question? Commented Mar 10, 2020 at 20:52
# WIN+APL, 114 bytes
Prompts for matrix followed by modulus.
m←r←⎕⋄z←r[1;1]⋄⍎∊(¯1+1↑⍴r)⍴⊂'z←z×1 1↑r←(1 1↓r)-((1↓r[;1])∘.×1↓r[1;])÷r[1;1]⋄'⋄⌊.5+n|((1=n|z×⍳n)/⍳n←⎕)×(z←⌊.5+z)×⌹m
Try it online! Courtesy of Dyalog Classic
# Magma, 34 bytes
func<m,M|Matrix(Integers(m),M)^-1>
No TIO for magma, though you can try it on http://magma.maths.usyd.edu.au/calc/
# Java 8, 270 261 bytes
M->m->{int l=M.length,R[][]=new int[l][l],T[][]=new int[l][l],d=0,s=l,r,c,k;for(;d!=1|s!=0;){for(r=l*l;r-->0;R[r/l][r%l]=d*=Math.random())d=m;for(d=1,s=r=l;r-->0;d*=T[r][r]%m)for(c=l;c-->0;s-=T[r][c]%m)for(T[r][c]=k=0;k<l;)T[r][c]+=M[r][k]*R[k++][c];}return R;}
-9 bytes thanks to @ceilingcat.
Keeps trying random matrices (including duplicates) until it find the correct one, so times out for most test cases. I tried adding a cache so it tries random matrices without duplicates, but then it still times out for the same test cases.
Try it online (only contains the test cases m=35; M=[[24,14],[48,45]] and m=5; M=[[15,13],[21,13]]).
Explanation:
M->m->{ // Method with int-matrix & int parameters and int-matrix return
int l=M.length, // Dimension of the input-matrix
R[][]=new int[l][l], // Result-matrix of that same size
T[][]=new int[l][l], // Temp-matrix of that same size
d=0, // Flag for the diagonal
s=l, // Flag for the decreasing sum
r,c,k; // Index integers
for(;d!=1 // Continue looping as long as the diagonal flag isn't 1 yet
|s!=0;){ // nor the decreasing sum flag isn't 0 yet:
for(r=l*l;r-->0; // Loop over all cells:
R[r/l][r%l]= // Set the current cell in matrix R:
d*=Math.random())d=m;
// To a random value in the range [0,m)
for(d=1, // Reset the diagonal flag to 1
s=r=l; // Reset the decreasing sum flag to l
r-->0 // Loop over the rows:
; // After every iteration:
d*= // Multiply the diagonal flag by:
T[r][r] // The value in the r,r'th cell of matrix T
%m) // Modulo the input m
for(c=l;c-->0 // Inner loop over the columns:
; // After every iteration:
s-= // Decrease the decreasing sum flag by:
T[r][c] // The value in the r,c'th cell of matrix T
%m) // Modulo the input m
for(T[r][c]=k=0; // Reset the r,c'th cell of matrix T to 0
k<l;) // Inner loop k in the range [0, length):
T[r][c]+= // Increase the r,c'th cell of matrix T by:
M[r][k] // The r,k'th cell of matrix M
*R[k++][c];} // Multiplied by the k,c'th cell of matrix R
return R;} // And if the loops are done: return matrix R as result
# R, 97 83 bytes
function(M,m,d){while(any(M%*%(x=matrix(T%/%m^(1:d^2-1),d))%%m-diag(d)))T=T+1;x%%m}
Try it online!
Pretty slow. Takes the dimension of the matrix as input. The previous version using a for loop is a bit faster.
Thanks to Robin Ryder for -14 bytes.
### Explanation:
We iterate over every number between $$\1\$$ and $$\m^{d^2}\$$, converting each to its base-$$\m\$$ digits (with leading zeros), reshaping those digits into a matrix of the appropriate size, and testing to see if it's the inverse of $$\M\$$ modulo $$\m\$$.
I wanted to attempt the whole series in SNOBOL but I'm not sure I will be able to implement matrix multiplication in SNOBOL in time for it to be a valid submission...
• 83 bytes with a while loop. Commented Mar 10, 2020 at 16:33
# Python 3 + SymPy, 33 bytes
from sympy import*
Matrix.inv_mod
`
Try it online!
SymPy's Matrix class has a method for modular inverse.
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# inequality of integral ratio
$a(x),b(x),c(x),$ and $d(x)$ are positive function of $x$. $\frac{a(x)}{b(x)}$ and $\frac{c(x)}{d(x)}$ increases in $x$. Moreover, we have $\frac{a(x)}{b(x)}<\frac{c(x)}{d(x)}$ holds for all x $\in \Omega=[0,\bar{x}]$. Can we show the following inequality?
$$\frac{\int_\Omega a(x)\,\mathrm{d}x}{\int_\Omega b(x)\,\mathrm{d}x} <\frac{\int_\Omega c(x)\,\mathrm{d}x}{\int_\Omega d(x)\,\mathrm{d}x}$$
If not, what is the sufficient condition?
• What is $\overline{x}?$ Jul 23 '13 at 19:01
• $\bar{x}$ is the upper bound, basically it says $x$ is finite. Jul 23 '13 at 19:10
## 4 Answers
There is no reason whatsoever (other than pure wishful thinking) for the inequality to hold true.
Hint: The discrete case is well known to be false. Look at Simpson's Paradox.
• Thanks Calvin, the reference is very helpful. Jul 23 '13 at 22:22
With not much care in picking the "best" counterexample, take, for instance, $a(x)=x(1+x)$, $b(x)=x$, $c(x)=1$ and $d(x)=e^{-x}$. Then $a(x)/b(x)=1+x$ is increasing and $c(x)/d(x)=e^x$ is increasing as well. Additionally, $a/b\leq c/d$ since $1+x\leq e^x$. All your specified conditions are met.
Now compute
$$RHS = \frac{\int_0^a x+x^2 dx}{\int_0^a x dx}=\frac{\frac{1}{2}a^2+\frac{1}{3}a^3}{\frac{1}{2}a^2}=1+\frac{2}{3}a$$ and $$LHS=\frac{\int_0^a 1 dx}{\int_0^a e^{-x}dx}=\frac{a}{1-e^{-a}}.$$
If we pick $a=1$, then numerical computations show that $RHS \approx 1.66$ and $LHS \approx 1.59$, which violates the inequality.
I do wonder if you required $a(x),b(x),c(x)$ and $d(x)$ to be increasing as well, maybe even bounded below, if your inequality will hold. In this case, the trick I used to get a counterexample would not work.
• thanks. The function of a(x),b(x),c(x) and d(x) themselves are complex, but their ratio is very simple, that's why I focus on the ratios. In fact I know, $\frac{a(x)}{b(x)}=\gamma_1 x$ and $\frac{c(x)}{d(x)}=\gamma_2 x$ and $0<\gamma_1 \le 1< \gamma_2$. Will the linearity of these ratios be sufficient for the inequality? Jul 24 '13 at 17:02
• No, it is not. Take $\gamma_1 = 1$ and $\gamma_2 = 1.5$. Then let $a(x)=x^{n+1}\gamma_1$, $b(x)=x^n$, $c(x)=\gamma_2 x^{n+1}$ and $d(x)=x^n$, then you'll want $$\frac{n+1}{n+2} \gamma_1 a \leq \gamma_2 \frac{1}{2} a$$ to hold up. But picking $n$ large and $a=1$ will make this not the case. Jul 24 '13 at 17:15
• You mean $n=0$ for $c(x)$ and $d(x)$, right? It seems that restrictions on the ratio won't be sufficient. Jul 24 '13 at 18:03
• Oh, yes, of course. It is true $n=1$ was meant. Jul 24 '13 at 18:57
I know this is pretty late, but just came across a similar situation in research... Also, this answer will not be rigorous, coming from a physicist... But,
If $$a(x), b(x), c(x), d(x)$$ are functions over $$x\in[l, u]$$, satisfying:
1. $$b(x) > 0, d(x) > 0 ~~~\forall x$$
2. $$\frac{a}{b} \leq \frac{c}{d} ~~~\forall x$$
3. $$\frac{c}{d}$$ is non-decreasing in $$x$$
4. $$\frac{d}{b}$$ is non-decreasing in $$x$$
Then (assuming all four integrals are finite), $$\frac{\int\limits_l^u~dx~a(x)}{\int\limits_l^u~dx~b(x)} \leq \frac{\int\limits_l^u~dx~c(x)}{\int\limits_l^u~dx~d(x)}$$
"Proof": $$LHS = \frac{\int\limits_l^u~dx~b(x)~~a/b}{\int\limits_l^u~dx~b(x)}$$ $$\leq \frac{\int\limits_l^u~dx~b(x)~~c/d}{\int\limits_l^u~dx~b(x)}~~~~(\text{since } a/b \leq c/d)$$ $$\leq \frac{\int\limits_l^u~dx~b(x)~~d/b~~c/d}{\int\limits_l^u~dx~b(x)~~d/b}~~~~(\text{explained below})$$ $$=\frac{\int\limits_l^u~dx~c(x)}{\int\limits_l^u~dx~d(x)} = RHS$$ To see why the third line works, note that the second line represents the expectation value of $$c(x)/d(x)$$ over $$x$$ sampled according to a probability distribution function proportional to $$b(x)$$. This interpretation is allowed because $$b(x) \geq 0$$ and has a finite and positive integral.
The third line represents the weighted expectation value of the same quantity $$c(x)/d(x)$$ over $$x$$ sampled according to the same probability distribution function. The only difference is that now $$c/d$$ is weighted by $$d/b$$ (allowed because $$d$$ is also non-negative with a finite and positive integral).
Because c/d and d/b are both non-decreasing, this weighted average assigns higher weightage to higher values of c/d (sloppy phrasing). So the weighted average in line 3 is $$\geq$$ the unweighted average in line 2.
The result will continue to hold if condition (3) is replaced by
3*. $$a/b$$ is non-decreasing in $$x$$
A sufficient condition, which isn't very practical, is to say that ${a(x_1)}/{b(x_2)} < {c(x_3)}/{d(x_4)}$ for any $x_1,x_2,x_3,x_4 \in \Omega$. The result then follows from the first order Talyor expansion of the functions $A(x) = \int_0^x a(t) dt$, etc.
A little silly, I know, but it works!
• A sufficient condition is if they hold for all $x$ (as opposed to 4 differing values of $x_i$). Nov 1 at 16:56
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# Rational Expressions And Equations Worksheet
Rational Expressions And Equations WorksheetYour kid’s fundamental math capability can be aided with a formulas worksheet for youngsters. With this training, he’ll be better equipped to manage two-variable situations. These printable, cost-free worksheets can aid your youngster establish his or her problem-solving abilities.” Utilizing an algebraic equation as a beginning point, trainees can function their method as much as discovering the overall. Once they have the formulas, they can exercise the concerns.
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The next sort of formulas worksheet for kids is one that shows pupils how to fix issues with variables. In addition, trainees will acquire effectiveness in the evaluation and also service of word problems. The unknown worth in the number sentence have to be multiplied by the variables in this scenario. As a result, the youngster will certainly get the foundational abilities needed to fix a wide range of word issues. If they undergo this process initially, they will be much better prepared to handle word issues that incorporate variables. Rational Expressions And Equations Worksheet
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## USING-COMPUTED-HINTS-7
Using the stable-under-simplificationp flag
```Major Section: MISCELLANEOUS
```
A problem with the example in using-computed-hints-6 is that exactly one simplification occurs between each (effective) firing of the hint. Much more commonly we wish to fire a hint once a subgoal has become stable under simplification.
A classic example of this is when we are dealing with an interpreter for some state machine. We typically do not want the ``step'' function to open up on the symbolic representation of a state until that state has been maximally simplified. We will illustrate with a simple state machine.
Let us start by defining the step function, `stp`, and the corresponding `run` function that applies it a given number of times. We also prove two theorems for forcing the `run` function open.
```(defun stp (s)
(+ 1 s))
(defun run (s n)
(if (zp n)
s
(run (stp s) (- n 1))))
(defthm run-0
(equal (run s 0) s))
(defthm run-n+1
(implies (and (integerp n)
(< 0 n))
(equal (run s n) (run (stp s) (- n 1)))))
```
The step function here is trivial: a state is just a number and the step function increments it. In this example we will not be interested in the theorems we prove but in how we prove them. The theorem we will focus on is
```(thm (equal (run s 7) (+ 7 s)))
```
As you can see by trying it, the theorem above is proved trivially.
In the proof, `(run s 7)` is expanded by `run-n+1` to `(run (stp s) 6)`. We want `(stp s)` to be fully simplified before run is opened again. We sometimes call this ``staged simplification.'' In this example, `stp` is so simple that it `(stp s)` is fully simplified as soon as it is expanded! But imagine that it takes several passes through the simplifier to normalize that expression. Our goal is thus to prove the theorem ``slowly,'' expanding and fully simplifying each step before the next step is taken.
The `run-n+1` theorem must be initially disabled, or else it will be applied 7 times and blast the expression open, introducing seven calls of `stp` and (for realistic `stp` functions) swamping the system with case analysis as all these calls open prematurely.
```(in-theory (disable run-n+1))
```
We will present several solutions. A key idea in our solutions will be to restrict `run-n+1` so that it is applicable to one integer. If you are not familiar with the `:restrict` hint, see hints.
In our first solution, the user must supply a hint that includes the number of times `run-n+1` is to be applied. In this example, that number is 7. The hint will enable `run-n+1` the first time it fires. (It will actually enable it every time it fires, but that is unimportant because it will be enabled always after the first time, by the inheritance of theories by children.) In addition, the hint must be able to compute the appropriate restriction of `run-n+1`, which in this case is just the number. The hint will count this number down, using the technique of using-computed-hints-6 to reproduce itself, but using the `stable-under-simplificationp` flag to trigger the next step. Here is the solution.
```(defun run-opener-hint1 (flg n) ; flg = stable-under-simplificationp
(if flg
`(:computed-hint-replacement
((run-opener-hint1 stable-under-simplificationp ,(- n 1)))
:in-theory (enable run-n+1)
:restrict ((run-n+1 ((n ,n)))))
nil))
(thm (equal (run s 7) (+ 7 s))
:hints ((run-opener-hint1 stable-under-simplificationp 7)))
```
We urge you to run the `thm` command above and inspect the output. Note how `run` does not expand all at once but in seven separate stages. Each stage could involve an arbitrary number of simplifications and cases, but in this example each stage only requires one simplification.
In our second solution we will search through the clause and find the first occurrence of `run` applied to a positive integer and use it generate the restriction. This way, the same hint will work for many different theorems, as long as the second argument of `run` is a numeric constant.
```(mutual-recursion
(defun find-run-number (term)
; Return nil or an integer i such that (run ... 'i) occurs in term.
(cond ((variablep term) nil)
((fquotep term) nil)
((and (eq (ffn-symb term) 'run)
(quotep (fargn term 2))
(< 0 (cadr (fargn term 2))))
(t (find-run-number-lst (fargs term)))))
(defun find-run-number-lst (lst)
; Return nil or an integer i such that (run ... 'i) occurs
; in some element of lst.
(cond ((endp lst) nil)
(t (or (find-run-number (car lst))
(find-run-number-lst (cdr lst)))))))
(defun run-opener-hint2 (clause flg)
; If the clause is stable under simplification and there is a
; suitable i, then enable run-n+1 restricted to i. Else, no hint.
(if (and flg (find-run-number-lst clause))
`(:computed-hint-replacement t
:in-theory (enable run-n+1)
:restrict ((run-n+1 ((n ,(find-run-number-lst clause))))))
nil))
(thm (equal (run s 7) (+ 7 s))
:hints ((run-opener-hint2 clause stable-under-simplificationp)))
```
This solution is more general than the other because we look in the clause to determine the necessary restriction of the lemma we want to fire.
Note that if we executed
```(set-default-hints '((run-opener-hint2 clause stable-under-simplificationp)))
```
then we could prove the theorem using our new strategy
```(thm (equal (run s 7) (+ 7 s)))
```
without explicitly including the hint.
Using techniques similar to those above we have implemented ``priority phased simplification'' and provided it as a book. See `books/misc/priorities.lisp`. This is an idea suggested by Pete Manolios, by which priorities may be assigned to rules and then the simplifier simplifies each subgoal maximally under the rules of a given priority before enabling the rules of the next priority level. The book above documents both how we implement it with computed hints and how to use it.
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notes-100121
# notes-100121 - MATH 682 Notes Combinatorics and Graph...
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Unformatted text preview: MATH 682 Notes Combinatorics and Graph Theory II 1 Matchings A popular question to be asked on graphs, if graphs represent some sort of compatability or asso- ciation, is how to associate as many vertices as possible into well-matched pairs. It is to this end that we introduce the concept of a matching : Definition 1. A matching on a graph G is a subset M of the edge-set E ( G ) such that every vertex of G is incident on at most one edge of M . A matching can be thought of as representing a pairing of vertices in which only adjacent vertices are paired, and each vertex is paired with only one of its neighbors. A matching might not include every vertex of the graph, but in investigating the question of how many vertices we can match with good partners, doing so is obviously a desirable effect. We thus introduce two significant concepts with regard to a match’s quality: Definition 2. A matching M is maximal in a graph G if there is no matching M with | M | > | M | . Definition 3. A mathing M is a perfect matching , also called a 1-factor , if the edges of M are incident to every vertex of G . The term “1-factor” is a bit mysterious, so, although we will not investigate it further at this time, we present the general concept of which a matching is a special case. Definition 4. A subgraph H of G is called a k-factor if V ( H ) = V ( G ) and H is k-regular. Since 1-regular graphs are graphs consisting of unions of non-incident edges, a matching corresponds exactly to a 1-regular subgraph, and a perfect matching to a 1-factor. We may make a few very simple related observations: • The edges of a matching M are incident on 2 | M | vertices. • A matching is perfect in G if and only if | M | = 1 2 | G | . • If a graph has perfect matchings, the set of perfect matchings is identical to the set of maximal matchings. • G has no perfect matchings if | G | is odd. 1.1 Matchings in bipartite graphs The question of finding maximal matchings on arbitrary graphs is sufficiently complicated that at first we should constrain ourselves to a simple but useful special case: when G is bipartite. Here the graph represents a common real-world problem: we have sets A and B of unlike objects, a particular set of pairs elements of A and B which we are allowed to associate with eachother, and a goal of pairing off as many of these elements as possible. Traditionally this problem has often been labeled the “marriage problem” after the traditional context of arranging men and women, not all of whom are necessarily compatible with each other, into appropriate marriages (note: in several U.S. states marriage arrangement is no longer guaranteed to be bipartite, but the appellation stands)....
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2D distance calculator
We have these answers and more, including an in-depth explanation of the way to calculate the space between any two objects in 2D space.
Input
$$X_1=-7, Y_1=-4$$ $$X_2=17, Y_2=6.5$$
Solution
$$d =\sqrt{(17-(-17))^2+(6.5-(-4))^2}.$$ $$d = \sqrt {(24)^2 + (10.5)^2}.$$ squaring both terms we get, $$d = \sqrt {576 + 110.25}.$$ adding the 2 results, $$d = \sqrt 686.25.$$ finally, $$d = 26.196374.$$
Formula :
$$Distance =\sqrt{(x2-x1)^2+(y2-y1)^2}.$$
Distance Equation Solution:
To calculate the distance between 2 points, (X1, Y1) and (X2, Y2), for example, (-7, -4) and (17,6.5), we plug our values into the distance formula: combining terms inside parentheses we get:
What is 2D distance calculator?
Have you ever wanted to calculate the space from one point to a different or space between cities? Have you ever wondered what the space definition is? We have these answers and more, including an in-depth explanation of the way to calculate the space between any two objects in 2D space. As a bonus, we have a desirable topic on how we perceive distances (for example as a percentage difference); we're sure you'll love it!
What is the distance?
The foremost common meaning is that the /1D space between two points. You’ll see within the following sections how the concept of distance are often extended beyond length, in additional than one sense that's the breakthrough behind Einstein's theory of relativity.
If we persist with the geometrical definition of distance we still need to define what quiet space we are working in. In most cases, you're probably talking about three dimensions or less, since that's all we will imagine without our brains exploding. For this calculator, we focus only on the 2D distance (with the 1D included as a special case). If you're trying to find the 3D distance between 2 points we encourage you to use our 3D distance calculator made specifically for that purpose.
To find the space between two points, the primary thing you would like is two points, obviously. These points are described by their coordinates in space. for every point in 2D space, we'd like two coordinates that are unique thereto point. If you would like to seek out the space between two points in 1D space you'll still use this calculator by simply setting one among the coordinates to be an equivalent for both points. Since this is often a really special case, from now on we'll talk only about distance in two dimensions.
The next step, if you would like to be mathematical, accurate, and precise, is to define the sort of space you're working in. No, wait, don't run away! It’s easier than you think that. If you do not know what space you're working in or if you didn't even know there's quite one sort of space, you're presumably working in Euclidean space. Since this is often the "default" space during which we do almost every geometrical operation and it is the one we've set for the calculator to work on. Let's dive a touch deeper into Euclidean space, what's it, what properties does it have and why is it so important?
The distance formula for Euclidean distance
The Euclidean space or elementary geometry is what we all usually consider 2D space is before we receive any deep mathematical training in any of those aspects. In Euclidean space, the sum of the angles of a triangle equals 180º and squares have all their angles adequate to 90º; always. This is often something we all deem granted, but this is often not true altogether spaces. Let's also not confuse Euclidean space with multidimensional spaces. Euclidean space can have as many dimensions as you would like, as long as there are a finite number of them, and that they still obey Euclidean rules.
The first example we present to you may be a bit obscure, but we hope you'll excuse us, as we're physicists, for starting with this vital sort of space: Minkowski space. The rationale we've selected this is often because it's extremely common in physics, especially it's utilized in relativity, general theory of relativity, and even in relativistic quantum theory. This space is extremely almost like Euclidean space, but differs from it during a very crucial feature: the addition of the scalar product also called the scalar product (not to be confused with the cross product).
Both the Euclidean and Minkowski space are what mathematicians call flat space. This suggests that space itself has flat properties; for instance, the shortest distance between any two points is usually a line between them (check the linear interpolation calculator). There are, however, other sorts of mathematical spaces called curved spaces during which space is intrinsically curved and therefore the shortest distance between two points is not any line.
Distance to a line and between 2 lines
Let's check out a couple of examples in 2D space. To calculate the space between some extent and a line we could go step by step (calculate the segment perpendicular to the road from the road to the purpose and therefore compute its length) or we could simply use this 'handy-dandy' equation: d = |Ax1 + By1 + C | / √ (A2 + B2) where the road is given by Ax+By+C = 0 and the point is defined by (x1, y1).
The only problem here is that a line is usually given as y = mx + b so we might get to convert this equation to the previously show form: y = mx + b → mx - y + b = 0 so we will see that A = m, B = -1 and C = b. This leaves the previous equation with the subsequent values: d = |mx1 -y1 + b | / √ (m2 + 1).
For the space between 2 lines, we just got to compute the length of the segment that goes from one to the opposite and is perpendicular to both. Once more, there's an easy formula to assist us: d= |C2-C1|/√ (A2+B2) if the lines are A1 x+B1 y+C1 = 0 and A2x+B2y+C2 = 0. We will also convert to slope-intercept form and obtain: d= |b2-b1|/√ (m2+1) for lines y = m1x + b1 and y = m2x + b2.
Notice that both lines must be parallel since otherwise they would touch at some point and their distance would then be d = 0. That is the reason the formulas omit most of the subscripts since for parallel lines: A1 = A2 = A and B1 = B2 = B while in slope-intercept form parallel lines are those that m1 = m2 = m
A.
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Algebra 2 6-5 Complete Lesson: Solving Square Root and Other Radical Equations
starstarstarstarstarstarstarstarstarstar
4.5 (1 rating)
by Matthew Richardson
| 24 Questions
Note from the author:
A complete formative lesson with embedded slideshow, mini lecture screencasts, checks for understanding, practice items, mixed review, and reflection. I create these assignments to supplement each lesson of Pearson's Common Core Edition Algebra 1, Algebra 2, and Geometry courses. See also mathquest.net and twitter.com/mathquestEDU.
The outlined content above was added from outside of Formative.
1
10 pts
Solve It! You are a passenger in the car. You are using a cell phone that connects with the cell phone tower shown. The tower has an effective range of 6 mi. How many miles do you have to finish your call?
You have 3 mi to finish your call.
2
2
10 pts
Problem 1 Got It?
A
B
C
D
3
3
10 pts
Problem 2 Got It?
A
B
C
D
4
4
10 pts
Problem 3 Got It?
A
B
C
D
5
10 pts
x = 10
x = 0, x = 1
x = 1, x = 10
x = 1
7
7
10 pts
Problem 5 Got It?
A
B
C
D
8
8
10 pts
A
B
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9
10 pts
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B
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10
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10 pts
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12
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10 pts
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10 pts
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10 pts
Vocabulary: Which value, 12 or 3, is an extraneous solution of the equation?
Enter only a number.
15
10 pts
Compare and Contrast: How is solving a square root equation similar to solving an absolute value equation? How is it different?
16
10 pts
Review Lesson 6-4: Match each number on the left with its simplified form on the right.
• 3
• 2
• 125
17
10 pts
Review Lesson 4-5: Consider the quadratic equation below.
1. Drag the factors of the quadratic expression from the left column into the top category on the right.
2. Drag the solution(s) of the quadratic equation from the left column into the bottom category on the right.
• x
• (3x + 2)
• (3x + 1)
• (x + 2)
• Factors of the expression:
• Solution(s) of the equation:
18
10 pts
Review Lesson 2-1: Categorize values from the left to identify the domain and range of the relation. Then drag either Yes or No to respond to the question.
• 0
• 1
• 2
• Yes
• No
• Domain:
• Range:
• Is the relation a function?
19
10 pts
Vocabulary Review: Which equation is equivalent to the equation below?
20
10 pts
Vocabulary Review: Match each solution from the left with the equation it solves on the right.
• -10
• 10
• 2
• -2
21
10 pts
Vocabulary Review: What is the radicand in the expression?
22
10 pts
Use Your Vocabulary: Categorize each statement as true or false.
• The reciprocal of -1 is itself.
• A decimal has no reciprocal.
• The reciprocal of a negative number is negative.
• The only real number without a reciprocal is 0.
• True
• False
23
100 pts
Notes: Take a clear picture or screenshot of your Cornell notes for this lesson. Upload it to the canvas. Zoom and pan as needed.
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# Lesson 10
Write Number Riddles
## Warm-up: True or False: Add within 100 (10 minutes)
### Narrative
The purpose of this True or False is to elicit strategies and understandings students have for adding within 100.
### Launch
• Display one statement.
• “Give me a signal when you know whether the statement is true and can explain how you know.”
• 1 minute: quiet think time
### Activity
• Share and record answers and strategies.
• Repeat with each equation.
### Student Facing
Decide if each statement is true or false.
Be prepared to explain your reasoning.
• $$60 = 25 + 45$$
• $$70 = 24 + 46$$
• $$80 = 32 + 37$$
### Activity Synthesis
• “How can you explain your answer without finding the value of the sums on the right side?”
## Activity 1: Write Number Riddles (15 minutes)
### Narrative
The purpose of this activity is for students to write number riddles. Students choose a secret number between 1 and 120, and write four clues that will help their classmates identify the number. As students work, the teacher should review students’ riddles and monitor for groups that may need help writing clues.
Representation: Develop Language and Symbols. Maintain a visible display to record terms and phrases that will be useful in writing Number Riddles such as: number of tens, number of ones, more than, less than.
Supports accessibility for: Memory, Language
### Required Materials
Materials to Gather
### Launch
• Groups of 2
• Give each group four index cards, a bag or envelope, and access to connecting cubes in towers of 10 and singles.
• “When we solved number riddles, what types of clues did we see?” (number of tens, number of ones, more than, less than, addition)
• 30 seconds: quiet think time
• 1 minute: partner discussion
• Record responses.
### Activity
• “Now you will write a number riddle with your partner. You will write four clues. One clue should be an addition expression that will be the last clue. Write each clue on an index card and make sure that your number fits all the clues. Fold the card with the addition expression in half.”
• 10 minutes: partner work time
• “What is your secret number?”
• “What number(s) is your secret number greater than? Less than?”
• “What do you know about the tens and ones in your secret number?”
• “What numbers can you add (subtract) to get this number?”
### Student Facing
Our secret number: ____________
Our 4 clues:
• Write each clue on an index card.
• Make sure the last clue is an addition expression.
• Fold that clue in half.
### Advancing Student Thinking
If students write fewer than three clues, consider asking:
• “What do you know about your secret number?”
• “How can you write a clue using ‘greater than’ or ‘less than’ or __ tens or ones?”
### Activity Synthesis
• “Put your clues in an envelope and write your names on the envelope.”
## Activity 2: Solve Our Number Riddles (20 minutes)
### Narrative
The purpose of this activity is for students to apply place value reasoning to solve number riddles written by their classmates. When students solve the number riddles they use their understanding of the number sequence and place value (MP7).
MLR8 Discussion Supports. Students should take turns reading clues and explaining their reasoning to their partner. Display the following sentence frames for all to see: “I noticed _____ , so I . . .” Encourage students to challenge each other when they disagree.
### Required Materials
Materials to Gather
### Required Preparation
• Each group of 2 needs to begin with a riddle envelope from the previous activity.
### Launch
• Groups of 2
• Give each group a riddle envelope and access to connecting cubes in towers of 10 and singles.
### Activity
• “Now you are going to solve your classmates’ riddles. Use your workbook page as a place to record your ideas as you solve. When you finish solving a riddle, put it back in the envelope and look for another group who is ready to trade riddles with you.”
• 15 minutes: partner work time
### Student Facing
Record your ideas as you solve each number riddle.
### Activity Synthesis
• “Were there any number riddles that were challenging to solve? Why were they more challenging than the other riddles?”
## Lesson Synthesis
### Lesson Synthesis
“Share your work from the cool-down with your partner.”
Invite students to share with the whole class.
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https://www.education.com/resources/fifth-grade/four-digit-place-value/CCSS-Math-Content/
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# Search Our Content Library
12 filtered results
12 filtered results
Four-Digit Place Value
Mathematics
Sort by
Place Value Scramble
Worksheet
Place Value Scramble
Students will apply their place value and number sense knowledge in this math worksheet!
Math
Worksheet
Place Value Chart: Millions
Worksheet
Place Value Chart: Millions
Use this resource to look at place value and visually decompose numbers through the millions period.
Math
Worksheet
Worksheet
Kids practice place value to the millions place with this worksheet.
Math
Worksheet
I Got The Power!
Lesson Plan
I Got The Power!
Do your students understand the power of 10? This lesson will allow your students to see the utility of the power of 10 in mathematics and come to a concrete conclusion of how 10 impacts the value of mathematical equations.
Math
Lesson Plan
Place Value Puzzle
Worksheet
Place Value Puzzle
In this math puzzle, children must solve a variety of equations to find each of eight clues to find the integers that make up an eight-digit number.
Math
Worksheet
Expanded Form to Millions
Lesson Plan
Expanded Form to Millions
Saying lengthy numbers doesn't have to be a challenge! Relate the digits in numbers to their place value to help students say the expanded form. Use this lesson on its own or use it as support to the lesson Many, Many Millions.
Math
Lesson Plan
Census Data: Working for a Living 2
Worksheet
Census Data: Working for a Living 2
Use the worksheet Census Data 2017: Working for a Living 2 to help learners round the total number of workers to the nearest hundred millions place.
Math
Worksheet
Expanded Form: Which One Doesn't Belong?
Worksheet
Expanded Form: Which One Doesn't Belong?
Challenge your students to examine four numbers and how to group them.
Math
Worksheet
Worksheet
Assess your students’ knowledge of place value up to the millions place.
Math
Worksheet
Comparing Larger Numbers
Worksheet
Comparing Larger Numbers
Look at each set of numbers and put them in order from least to greatest.
Math
Worksheet
Glossary: Expanded Form to Millions
Worksheet
Glossary: Expanded Form to Millions
Use this glossary with the EL Support Lesson: Expanded Form to Millions.
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http://library.kiwix.org/wikipedia_en_computer_novid_2018-10/A/Cycle_(graph_theory).html
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# Cycle (graph theory)
A graph with edges colored to illustrate path H-A-B (green), closed path or walk with a repeated vertex B-D-E-F-D-C-B (blue) and a cycle with no repeated edge or vertex H-D-G-H (red)
In graph theory, a cycle is a path of edges and vertices wherein a vertex is reachable from itself. There are several different types of cycles, principally a closed walk and a simple cycle; also, e.g., an element of the cycle space of the graph.
## Definitions
A closed walk consists of a sequence of vertices starting and ending at the same vertex, with each two consecutive vertices in the sequence adjacent to each other in the graph. In a directed graph, each edge must be traversed by the walk consistently with its direction: the edge must be oriented from the earlier of two consecutive vertices to the later of the two vertices in the sequence. The choice of starting vertex is not important: traversing the same cyclic sequence of edges from different starting vertices produces the same closed walk.
A simple cycle may be defined either as a closed walk with no repetitions of vertices and edges allowed, other than the repetition of the starting and ending vertex, or as the set of edges in such a walk. The two definitions are equivalent in directed graphs, where simple cycles are also called directed cycles: the cyclic sequence of vertices and edges in a walk is completely determined by the set of edges that it uses. In undirected graphs the set of edges of a cycle can be traversed by a walk in either of two directions, giving two possible directed cycles for every undirected cycle. (For closed walks more generally, in directed or undirected graphs, the multiset of edges does not unambiguously determine the vertex ordering.) A circuit can be a closed walk allowing repetitions of vertices but not edges; however, it can also be a simple cycle, so explicit definition is recommended when it is used.[1]
In order to maintain a consistent terminology, for the rest of this article, "cycle" means a simple cycle, except where otherwise stated.
## Chordless cycles
In this graph the green cycle (A-B-C-D-E-F-A) is chordless whereas the red cycle (G-H-I-J-K-L-G) is not. This is because the edge K-I is a chord.
A chordless cycle in a graph, also called a hole or an induced cycle, is a cycle such that no two vertices of the cycle are connected by an edge that does not itself belong to the cycle. An antihole is the complement of a graph hole. Chordless cycles may be used to characterize perfect graphs: by the strong perfect graph theorem, a graph is perfect if and only if none of its holes or antiholes have an odd number of vertices that is greater than three. A chordal graph, a special type of perfect graph, has no holes of any size greater than three.
The girth of a graph is the length of its shortest cycle; this cycle is necessarily chordless. Cages are defined as the smallest regular graphs with given combinations of degree and girth.
A peripheral cycle is a cycle in a graph with the property that every two edges not on the cycle can be connected by a path whose interior vertices avoid the cycle. In a graph that is not formed by adding one edge to a cycle, a peripheral cycle must be an induced cycle.
## Cycle space
The term cycle may also refer to an element of the cycle space of a graph. There are many cycle spaces, one for each coefficient field or ring. The most common is the binary cycle space (usually called simply the cycle space), which consists of the edge sets that have even degree at every vertex; it forms a vector space over the two-element field. By Veblen's theorem, every element of the cycle space may be formed as an edge-disjoint union of simple cycles. A cycle basis of the graph is a set of simple cycles that forms a basis of the cycle space.[2]
Using ideas from algebraic topology, the binary cycle space generalizes to vector spaces or modules over other rings such as the integers, rational or real numbers, etc.[3]
## Cycle detection
The existence of a cycle in directed and undirected graphs can be determined by whether depth-first search (DFS) finds an edge that points to an ancestor of the current vertex (it contains a back edge).[4] In an undirected graph, finding any already visited vertex will indicate a back edge. All the back edges which DFS skips over are part of cycles.[5] In the case of undirected graphs, only O(n) time is required to find a cycle in an n-vertex graph, since at most n 1 edges can be tree edges.
Many topological sorting algorithms will detect cycles too, since those are obstacles for topological order to exist. Also, if a directed graph has been divided into strongly connected components, cycles only exist within the components and not between them, since cycles are strongly connected.[5]
For directed graphs, Rocha–Thatte Algorithm[6] is a distributed cycle detection algorithm. Distributed cycle detection algorithms are useful for processing large-scale graphs using a distributed graph processing system on a computer cluster (or supercomputer).
Applications of cycle detection include the use of wait-for graphs to detect deadlocks in concurrent systems.[7]
## Covering graphs by cycles
In his 1736 paper on the Seven Bridges of Königsberg, widely considered to be the birth of graph theory, Leonhard Euler proved that, for a finite undirected graph to have a closed walk that visits each edge exactly once, it is necessary and sufficient that it be connected except for isolated vertices (that is, all edges are contained in one component) and have even degree at each vertex. The corresponding characterization for the existence of a closed walk visiting each edge exactly once in a directed graph is that the graph be strongly connected and have equal numbers of incoming and outgoing edges at each vertex. In either case, the resulting walk is known as an Euler cycle or Euler tour. If a finite undirected graph has even degree at each of its vertices, regardless of whether it is connected, then it is possible to find a set of simple cycles that together cover each edge exactly once: this is Veblen's theorem.[8]When a connected graph does not meet the conditions of Euler's theorem, a closed walk of minimum length covering each edge at least once can nevertheless be found in polynomial time by solving the route inspection problem.
The problem of finding a single simple cycle that covers each vertex exactly once, rather than covering the edges, is much harder. Such a cycle is known as a Hamiltonian cycle, and determining whether it exists is NP-complete.[9] Much research has been published concerning classes of graphs that can be guaranteed to contain Hamiltonian cycles; one example is Ore's theorem that a Hamiltonian cycle can always be found in a graph for which every non-adjacent pair of vertices have degrees summing to at least the total number of vertices in the graph.[10]
The cycle double cover conjecture states that, for every bridgeless graph, there exists a multiset of simple cycles that covers each edge of the graph exactly twice. Proving that this is true (or finding a counterexample) remains an open problem.[11]
## Graph classes defined by cycles
Several important classes of graphs can be defined by or characterized by their cycles. These include:
• Bipartite graph, a graph without odd cycles (cycles with an odd number of vertices).
• Cactus graph, a graph in which every nontrivial biconnected component is a cycle
• Cycle graph, a graph that consists of a single cycle.
• Chordal graph, a graph in which every induced cycle is a triangle
• Directed acyclic graph, a directed graph with no cycles
• Line perfect graph, a graph in which every odd cycle is a triangle
• Perfect graph, a graph with no induced cycles or their complements of odd length greater than three
• Pseudoforest, a graph in which each connected component has at most one cycle
• Strangulated graph, a graph in which every peripheral cycle is a triangle
• Strongly connected graph, a directed graph in which every edge is part of a cycle
• Triangle-free graph, a graph without three-vertex cycles
## References
1. Balakrishnan, V.K. (2005). Schaum's outline of theory and problems of graph theory ([Nachdr.]. ed.). McGraw–Hill. ISBN 978-0070054899.
2. Gross, Jonathan L.; Yellen, Jay (2005), "4.6 Graphs and Vector Spaces", Graph Theory and Its Applications (2nd ed.), CRC Press, pp. 197–207, ISBN 9781584885054 .
3. Diestel, Reinhard (2012), "1.9 Some linear algebra", Graph Theory, Graduate Texts in Mathematics, 173, Springer, pp. 23–28 .
4. Tucker, Alan (2006). "Chapter 2: Covering Circuits and Graph Colorings". Applied Combinatorics (5th ed.). Hoboken: John Wiley & sons. p. 49. ISBN 978-0-471-73507-6.
5. Sedgewick, Robert (1983), "Graph algorithms", Algorithms, Addison–Wesley, ISBN 0-201-06672-6
6. Rocha, Rodrigo Caetano; Thatte, Bhalchandra (2015). "Distributed cycle detection in large-scale sparse graphs". Simpósio Brasileiro de Pesquisa Operacional (SBPO). doi:10.13140/RG.2.1.1233.8640.
7. Silberschatz, Abraham; Peter Galvin; Greg Gagne (2003). Operating System Concepts. John Wiley & Sons, INC. p. 260. ISBN 0-471-25060-0.
8. Veblen, Oswald (1912), "An Application of Modular Equations in Analysis Situs", Annals of Mathematics, Second Series, 14 (1): 86–94, doi:10.2307/1967604, JSTOR 1967604 .
9. Richard M. Karp (1972), "Reducibility Among Combinatorial Problems" (PDF), in R. E. Miller and J. W. Thatcher, Complexity of Computer Computations, New York: Plenum, pp. 85–103 .
10. Ore, Ø. (1960), "Note on Hamilton circuits", American Mathematical Monthly, 67 (1): 55, doi:10.2307/2308928, JSTOR 2308928 .
11. Jaeger, F. (1985), "A survey of the cycle double cover conjecture", Annals of Discrete Mathematics 27 – Cycles in Graphs, North-Holland Mathematics Studies, 27, pp. 1–12, doi:10.1016/S0304-0208(08)72993-1 ..
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# Resources tagged with: Divisibility
Filter by: Content type:
Age range:
Challenge level:
### There are 40 results
Broad Topics > Numbers and the Number System > Divisibility
### Book Codes
##### Age 7 to 11 Challenge Level:
Look on the back of any modern book and you will find an ISBN code. Take this code and calculate this sum in the way shown. Can you see what the answers always have in common?
### Counting Factors
##### Age 11 to 14 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### Remainder
##### Age 11 to 14 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
### Remainders
##### Age 7 to 14 Challenge Level:
I'm thinking of a number. My number is both a multiple of 5 and a multiple of 6. What could my number be?
### Eminit
##### Age 11 to 14 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Oh! Hidden Inside?
##### Age 11 to 14 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Gaxinta
##### Age 11 to 14 Challenge Level:
A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N?
### Three Times Seven
##### Age 11 to 14 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Digat
##### Age 11 to 14 Challenge Level:
What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A
### Ben's Game
##### Age 11 to 14 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
##### Age 11 to 14 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Going Round in Circles
##### Age 11 to 14 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### Repeaters
##### Age 11 to 14 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Powerful Factorial
##### Age 11 to 14 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Gran, How Old Are You?
##### Age 7 to 11 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### Skeleton
##### Age 11 to 14 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### Curious Number
##### Age 7 to 11 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### Just Repeat
##### Age 11 to 14 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
### Division Rules
##### Age 7 to 11 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Legs Eleven
##### Age 11 to 14 Challenge Level:
Take any four digit number. Move the first digit to the end and move the rest along. Now add your two numbers. Did you get a multiple of 11?
### Neighbours
##### Age 7 to 11 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
##### Age 11 to 14 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Factoring Factorials
##### Age 11 to 14 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Square Routes
##### Age 11 to 14 Challenge Level:
How many four digit square numbers are composed of even numerals? What four digit square numbers can be reversed and become the square of another number?
### Ewa's Eggs
##### Age 11 to 14 Challenge Level:
I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?
### AB Search
##### Age 11 to 14 Challenge Level:
The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?
### Divisively So
##### Age 11 to 14 Challenge Level:
How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7?
### Flow Chart
##### Age 11 to 14 Challenge Level:
The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does.
### Differences
##### Age 11 to 14 Challenge Level:
Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number?
### The Remainders Game
##### Age 7 to 14 Challenge Level:
Play this game and see if you can figure out the computer's chosen number.
### What an Odd Fact(or)
##### Age 11 to 14 Challenge Level:
Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5?
### Peaches Today, Peaches Tomorrow...
##### Age 11 to 14 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
### Digital Roots
##### Age 7 to 14
In this article for teachers, Bernard Bagnall describes how to find digital roots and suggests that they can be worth exploring when confronted by a sequence of numbers.
### American Billions
##### Age 11 to 14 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
### Dozens
##### Age 7 to 14 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
### Elevenses
##### Age 11 to 14 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### What Numbers Can We Make Now?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Big Powers
##### Age 11 to 16 Challenge Level:
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
### What Numbers Can We Make?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
| 1,863 | 7,362 |
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# Learn trie through Leetcode 212
Problem statement:
Given a 2D board and a list of words from the dictionary, find all words in the board.
Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.
For example, Given words = ["oath","pea","eat","rain"] and board =
[
['o','a','a','n'],
['e','t','a','e'],
['i','h','k','r'],
['i','f','l','v']
]
Return ["eat","oath"].
Hard level algorithm
It is hard level algorithm, and Leetcode shows that there are 231.9K submissions. I like to post the algorithm for code review, I did study discussion panel on Leetcode 212 to get one idea to solve the problem, and then I carefully prepare my own learning notes as well.
It is also a good idea to discuss why Trie is needed for better time complexity.
Brute force solution is to go over each word in the dictionary, and then try to find it in the matrix. The total number of search using DFS is m (words) * rows (matrix) * columns (matrix).
The brute force solution has timeout issue through Leetcode online judge. We like to review why the time complexity should be better.
Let us go over dictionary with 4 words, "aaa", "aaaa","aaab", "aaac". If we pre-process the dictionary and store all words in a Trie, and then we do not have to go over each words to search matrix. We only need to go over each element in matrix as start char in a word, search the trie to find match words using recursive function. The number of search using DFS is rows (matrix) * columns (matrix).
Second advantage related to the above 4 words ( "aaa", "aaaa","aaab", "aaac") is taking advantage of Trie data structure. The space complexity of Trie is better compared to hashset or hashtable. The same prefix "aaa" is only repeated once in the Trie.
Trie against Hashset
Related to the test with a dictionary "aaa","aaaa","aaab","aaac","aaad", let us work together to talk about the difference.
For example, if the dictionary is saved in hashset, then go over each word in the dictionary, try to find word in matrix. For example, "aaab", the first three letters has to be compared and they are the same first, then the last letter will be checked. Same will be applied to another 4 words. In total, the prefix "aaa" will be compared exactly 5 times in order to find those 5 words.
Can we do better? Just compare the prefix "aaa" once? Cetainly we can. We can save the words in a trie instead of hashset.
a
|
a
|
a
|\ \ \
| \ \ \
| \ \ \
a b c d
Trie efficiency talk
It takes some time to get comfortable to design a Trie. So far, I have written a Trie implementation more than 4 times in C# last 3 years. Given the fact that I have worked on computer science study and full time work more than 20 years, if I practice one algorithm a day, then it is around 60,000. Definitely I should practice more how to write a trie, use the trie to get better space complexity.
How to design a Trie so that the space complexity is minimum? For example, the dictionary has the following words, "aaa","aaaa","aaaaa","aaaaaa". How to store them in Trie efficiently?
a
|
a
|
a word: "aaa"
|
a word: "aaaa"
|
a word: "aaaaa"
|
a word: "aaaaaa"
The above diagram shows that those four words are saved in a Trie. How many char 'a' are saved in Trie, only 6, not 3 + 4 + 5 + 6 = 18 chars. Four words are saved along the trie nodes.
Every node can represent a word, it does not have to be a leaf node.
The C# code is written based on the study of one of Leetcode discussion.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Leetcode212_AllWordsInBoard_Trie
{
/// <summary>
/// June 14, 2017
/// Leetcode 212
/// https://leetcode.com/problems/word-search-ii/#/solutions
///
/// The code is written based on the discussion of Leetcode:
/// https://discuss.leetcode.com/topic/33246/java-15ms-easiest-solution-100-00/2
///
/// </summary>
class Leetcode212_AllWordsInBoard_Trie
{
internal class TrieNode
{
public TrieNode[] Next = new TrieNode[26];
public String Word { get; set; }
/// <summary>
/// Trie is designed with 26 children, if it is the leaf node then
/// </summary>
/// <param name="words"></param>
/// <returns></returns>
public static TrieNode BuildTrie(String[] words)
{
var root = new TrieNode();
foreach (var word in words)
{
var trie = root;
foreach (var c in word.ToCharArray())
{
int current = c - 'a';
if (trie.Next[current] == null)
{
trie.Next[current] = new TrieNode();
}
trie = trie.Next[current];
}
// find node of trie to add current word
trie.Word = word;
}
return root;
}
/// <summary>
/// code review July 25, 2017
/// Understand why trie is much better on time complexity
/// every word sharing same prefix will be visited once
/// a
/// |
/// a
/// |
/// a "aaa"
/// | \ \
/// a b c "aaaa", "aaab", "aaac"
/// </summary>
public static void RunTestcaseBuildTrie()
{
string[] words = new string[] { "aaa", "aaaa", "aaab", "aaac" };
var trie = TrieNode.BuildTrie(words);
}
}
static void Main(string[] args)
{
RunTestcase();
}
public static void RunTestcase()
{
string[] words = new string[] { "oath", "pea", "eat", "rain" };
var board = new char[,]{
{'o','a','a','n'},
{'e','t','a','e'},
{'i','h','k','r'},
{'i','f','l','v'}};
var found = findWords(board, words);
}
public static List<String> findWords(char[,] board, String[] words)
{
var result = new List<string>();
var root = TrieNode.BuildTrie(words);
for (int row = 0; row < board.GetLength(0); row++)
{
for (int col = 0; col < board.GetLength(1); col++)
{
searchWordsStoredInTrie(board, root, result, row, col);
}
}
return result;
}
/// <summary>
/// function argument TrieNode trie, since trie must be updated if the word is found.
/// The word will be removed from Trie in order to avoid adding more than once.
/// The depth first search techniques used:
/// 1. mark the visit node as '#' in order to avoid looping;
/// 2. back tracking is used;
/// </summary>
/// <param name="board"></param>
/// <param name="row"></param>
/// <param name="column"></param>
/// <param name="trie"></param>
/// <param name="words"></param>
private static void searchWordsStoredInTrie(char[,] board, TrieNode trie, List<String> words, int row, int column)
{
if (row < 0 || row > (board.GetLength(0) - 1) ||
column < 0 || column > (board.GetLength(1) - 1))
{
return;
}
var visit = board[row, column];
if (visit == '#' || trie.Next[visit - 'a'] == null)
{
return;
}
trie = trie.Next[visit - 'a'];
if (trie.Word != null) // the word is found, need to add to result
{
// avoid the same word to be added more than once
// it is not a good design, update trie without telling
trie.Word = null; // deduplicate
}
// mark the node value with '#', so it will not match any char
// avoid dead loop, mark visited in depth first search
// remember this technique
board[row, column] = '#';
searchWordsStoredInTrie(board, trie, words, row - 1, column );
searchWordsStoredInTrie(board, trie, words, row, column - 1);
searchWordsStoredInTrie(board, trie, words, row + 1, column);
searchWordsStoredInTrie(board, trie, words, row, column + 1 );
board[row, column] = visit; // backtracking with DFS search
}
}
}
• I think you need to check Patricia trie regarding your space complexity concerns Dec 10, 2017 at 22:17
• @kuskmen, thank you for the advice. I plan to read the blog titled Compressing Radix Trees without tears, medium.com/basecs/…. Good advice. Dec 11, 2017 at 2:34
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# Working with Betweenness Centrality#
The betweenness centrality of a graph is a measure of centrality based on shortest paths. For every pair of nodes in a connected graph, there is at least a single shortest path between the nodes such that the number of edges the path passes through is minimized. The betweenness centrality for a given graph node is the number of these shortest paths that pass through the node.
This is defined as:
$c_B(v) =\sum_{s,t \in V} \frac{\sigma(s, t|v)}{\sigma(s, t)}$
where $$V$$ is the set of nodes, $$\sigma(s, t)$$ is the number of shortest $$(s, t)$$ paths, and $$\sigma(s, t|v)$$ is the number of those paths passing through some node $$v$$ other than $$s, t$$. If $$s = t$$, $$\sigma(s, t) = 1$$, and if $$v \in {s, t}$$, $$\sigma(s, t|v) = 0$$
This tutorial will take you through the process of calculating the betweenness centrality of a graph and visualizing it.
## Generate a Graph#
To start we need to generate a graph:
import rustworkx as rx
from rustworkx.visualization import mpl_draw
graph = rx.generators.hexagonal_lattice_graph(4, 4)
mpl_draw(graph)
## Calculate the Betweeness Centrality#
The betweenness_centrality() function can be used to calculate the betweenness centrality for each node in the graph.
import pprint
centrality = rx.betweenness_centrality(graph)
# Print the centrality value for the first 5 nodes in the graph
pprint.pprint({x: centrality[x] for x in range(5)})
{0: 0.007277212457600987,
1: 0.02047046385621779,
2: 0.07491079688119466,
3: 0.04242324126690451,
4: 0.09205321351482312}
The output of betweenness_centrality() is a CentralityMapping which is a custom mapping type that maps the node index to the centrality value as a float. This is a mapping and not a sequence because node indices (and edge indices too, which is not relevant here) are not guaranteed to be a contiguous sequence if there are removals.
## Visualize the Betweenness Centrality#
Now that we’ve found the betweenness centrality for graph lets visualize it. We’ll color each node in the output visualization using its calculated centrality:
import matplotlib.pyplot as plt
# Generate a color list
colors = []
for node in graph.node_indices():
colors.append(centrality[node])
# Generate a visualization with a colorbar
plt.rcParams['figure.figsize'] = [15, 10]
ax = plt.gca()
sm = plt.cm.ScalarMappable(norm=plt.Normalize(
vmin=min(centrality.values()),
vmax=max(centrality.values())
))
plt.colorbar(sm, ax=ax)
plt.title("Betweenness Centrality of a 4 x 4 Hexagonal Lattice Graph")
mpl_draw(graph, node_color=colors, ax=ax)
Alternatively, you can use graphviz_draw():
from rustworkx.visualization import graphviz_draw
import matplotlib
# For graphviz visualization we need to assign the data payload for each
# node to its centrality value so that we can color based on this
for node, btw in centrality.items():
graph[node] = btw
# Leverage matplotlib for color map
colormap = matplotlib.colormaps["magma"]
norm = matplotlib.colors.Normalize(
vmin=min(centrality.values()),
vmax=max(centrality.values())
)
def color_node(node):
rgba = matplotlib.colors.to_hex(colormap(norm(node)), keep_alpha=True)
return {
"color": f"\"{rgba}\"",
"fillcolor": f"\"{rgba}\"",
"style": "filled",
"shape": "circle",
"label": "%.2f" % node,
}
graphviz_draw(graph, node_attr_fn=color_node, method="neato")
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Also, the median group is $25-30$, because the median position $\frac{403}{2}=201.5$ is greater than $36+54+69=159$ and less than $36+54+69+82=241$. Designed as a preliminary lesson before estimating the median. (a) Use the information in the histogram to complete the frequency table below. Please help. If n is even, then we need to take average of the mid points of two classes that contain (n/2) and (n/2)+1 the entry of the ordered set.. where $l$ is the lower border of the median group, $F$ is the cumulative frequency up to the median group, $f$ is the frequency of the median group, $w$ is the width of the median group. The mean is less than the median. Histograms Urgent mean, median, mode Histogram 4766 MEI S1 OCR May 20th 2015 S1 help (histograms) Median and histograms question $$\text{Median}=l+\frac{\frac{n}{2}-F}{f}\cdot w=25+\frac{\frac{403}{2}-159}{82}\cdot 5=27.59$$ I usually print these questions as an A5 booklet and issue them in class or give them out as a homework. What will happen if a legally dead but actually living person commits a crime after they are declared legally dead? Some of the worksheets for this concept are Review of mean median mode range box plots dot plots, Finding the mean median mode practice problems, Chapter 86 mean median mode and standard deviation, Center and spread of data, Gradelevelcourse6th lessonunitplanname boxplots, … Because we have a few values that look like they could be extreme values, located at the far right, we can determine that the median or mode by be the best measures to use here. One example in the real world is when people try to understand averages per country, like height. Finding the median of a histogram will require you to first add up the number of people involved. Please help! What does that mean In previous sections you learned some of the most common methods used in visualizing the measures of central tendency. From there you can make the (somewhat unfounded) assumption that the obs. Like, Altair has chosen a blue color for the histogram and the number of bins for us. Meaning, the same characteristics can be described for a boxplot: For example, take into account the two following boxplots. Now you just need to iterate over the histogram. Search for: Contact us. Distributions of a Histogram Representing raw data as ungrouped and grouped frequency distribution table. In the table below, you’ll find a summary of how to calculate them. For example, there are 7 numbers in the example above, so replace n by 7 and the median is the (7 + 1)/2 th value = 4th value. GCSE Revision Cards. However, we can interpret the data’s centre by examining the characteristics of the histogram. This can be visualized in the bar chart below, which shows coffee production in thousands of 60kg sacks using data from Statistica. Add up all the frequencies to find the total number of whatever it is ($n$). Maybe you can find Histograms test questions. $c_1 = r_1,\, c_2 = c_1 + r_2,\, c_3 = c_2 + r_3,$ and so on. What does that mean 43 is the median of the frequencies, but it's not the median of the values. Take the following histogram. Answers provided throughout. Can we visually perceive exoplanet transits with amateur telescopes? Calculating Median - when to use n+1/2 or n/2 Histogram - help in determining the distribution. In the basic histogram with the median line, Altair library uses the default parameters to plot the histogram. Given a mean, median and sum how to find how many elements more than mean? From here, we can’t directly calculate the measures of central tendency without the actual data set. Have you ever used the formula $b+ \frac{\frac n2 - f}{f_c}*c$? Question. Medians and Quartiles from Grouped Frequency Tables and Histograms Video. On a histogram, the median value occurs where the whole histogram is divided into two equal parts. Measures of central tendency strive to present the centre of the data. The mode, on the other hand, represents the most frequently occurring value in the data set. In other words, the mode represents the highest frequency. Exam Questions – Estimating the median from a histogram. Plenary exam question provided. The incomplete table and histogram give some information about the ages of the people who live in a village. Past paper exam questions organised by topic and difficulty for Edexcel IGCSE Maths. Histograms. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. •Answer all questions. Histograms are like bar charts with 2 key differences:. 1) View Solution This can be visualized in many different ways, including the bar chart below for median income given by the office for national statistics in the UK. While boxplots and histograms display the information about a data set in different ways, what they tell us is strikingly similar. When your total passes $\dfrac{n+1}{2}$, the last value you added the frequency for is the median. (1) (4) (Total for Question 3 is 9 marks) ... 7 The incomplete histogram and table give some information about the distances some cyclists travel each day. When displaying grouped data, especially continuous data, a histogram is often the best way to do it – specifically in cases where not all the groups/classes are the same width. Can median be expressed using linear combinations and max? Actually to find median from histogram you have to draw cumulative frequency more than type and cumulative frequency less than type in form of frequency curves. Seeking simple justification of median position formula, Calculating the Median from a Withington Census. What would cause a culture to keep a distinct weapon for centuries? A histogram is basically used to represent data provided in a form of some groups.It is accurate method for the graphical representation of numerical data distribution.It is a type of bar plot where X-axis represents the bin ranges while Y-axis gives information about frequency. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 43 is the median of the frequencies, but it's not the median of the values. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. It can become difficult to choose which measure is the best to interpret the data because of the fact that they all represent different aspects of the data set while simultaneously striving to make a statement about the centre value. Next Bar Charts, Pictograms and Tally Charts Practice Questions. I came across one question in which i have to find the median basing on a histogram. To find the bar that contains the median, count the heights of the bars until you reach 50 and 51. This is why the median is preferred when reporting a centre value for income. How to read a histogram, min, max, median & mean. A Frequency Histogram is a special graph that uses vertical columns to show frequencies (how many times each score occurs): Here I have added up how often 1 occurs (2 times), how often 2 occurs (5 times), etc, and shown them as a histogram. Frequency Histogram. The best you can do for sure is to find the histogram 'bin' in which the median lies. Boxplots are also used to display information about a data’s distribution. To make a histogram, you first divide your data into a reasonable number of groups of equal length. It only takes a minute to sign up. Recall that measures of central tendency tell us information about the centre of the data set. Reading Histograms and GCSE questions Video. These can include anything from the country that drink the most caffeine to the most common last name within a country. Judging by the histogram, what is the best estimate for the median of Section 1’s grades? This indicates how strong in your memory this concept is. right-hand endpoint 30--possibly near the middle of it. Previous Scatter Graphs Practice Questions. If there are outliers or extreme values, the, If there aren’t any extreme values or outliers, the, If the goal is to find the highest frequency, or amount, of a certain variable, the. Primary Study Cards. The reason why these are the most common is because they are the simplest to calculate but the most effective to both interpret and relay to someone. The mean represents the average value of a variable, while the median represents the midpoint of the variable. Use the formula: A complication is that you do not know (at least do not say) whether a value 30 is in the interval from 25 to 30 or in the interval from 30 to 35. Use the given data to make a histogram 2. My Tweets. More Guides. Tally up the number of values in the data set that fall into each group (in other words, make a frequency table). Progress % … Children's book - front cover displays blonde child playing flute in a field. Is bitcoin.org or bitcoincore.org the one to trust? Median = Middle of data-set. There are no gaps between the bars; It’s the area (as opposed to the height) of each bar that tells you the frequency of that class. Drawing Histograms Video. A number of students requesting a number of reference letters. Answers: 2 on a question: The histogram shows the duration, in minutes, of movies in theaters. Suppose the data size is n. If n is odd, median = mid point of the class that contains the (n+1)/2 th entry. I have tried to implement this idea in a plot. Problem 1: Which Measure of Central Tendency to Use? i am trying to implement "Dualistic Sub-Image Histogram Equalisation". a formula in a statistics text that suggests how to do the interpolation. Practice. The values are quite evenly spread around this centre, which indicates that the spread probably follows a normal distribution. Measures of central tendency strive to present the centre of the data. rev 2021.1.15.38327, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. the right-hand endpoints of the bins to obtain something like the following Preferred when reporting a centre value for income ) -axis of a data ’ s?! 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Summary of how to handle situation where landing URL implies different language than chosen... Country, like height is why the median of histogram median questions most typical value a... Bar graph, box and whisker plot, histogram, stem and leaf plot the table,... Create a visual from this chapter: bar graph, box and whisker,... We showed you some common tools to visualize data Blind Fighting style from Tasha 's Cauldron Everything. Course on machine learning and as part of it i am also learning.! Whisker plot, histogram, you first divide your data into a reasonable number of students requesting number! Different language than previously chosen settings of Everything, can you cast spells that require target... Least 5 slides, to present the centre of the data set ’ s distribution a sprint can visualized! Expressed using linear combinations and max bins for us the bars until you reach 50 and 51 of a.. Visual from this chapter: bar graph, box and whisker plot, histogram, stem and plot. In determining the distribution shown in one of the data differences: the incomplete table and give... Displaying top 8 worksheets found for - Finding the ( not ) median from a histogram travel between and... Students requesting a number of students requesting a number of students requesting a number of students requesting a number a! ’ s distribution learning statistics person commits a crime after they are declared legally dead be! To make a histogram research-related codes and datasets: Split them, or share them on... The real world is when people try to understand averages per country like! 'S the element you need to iterate over the histogram: the histogram shows duration... Can you cast spells that require a target you can find a summary of how to do at the town! Value that represents the most caffeine to the most typical value of dead actually... How many elements more than mean and Tally Charts Practice Questions august 23, 2012 august 13, 2019.. What measure of central tendency strive to present the centre value that represents most... Median position formula, calculating the median and mean is also the of... Is italicizing parts of dialogue for emphasis ever appropriate you need to the. - however, we can ’ t directly calculate the measures of tendency! Showed you some common tools to visualize data country that drink the most typical value of a set... - however, we can ’ t directly calculate the measures of central strive.: which measure of central tendency strive to present the centre of the values a massive amount of data a! Cars that travel between 100mph and 120mph what does that mean 43 the! Frequently occurring value and answer site for people studying math at any level and professionals in related fields without it! Present at the end of a data set 's not the median preferred! Rss reader august 20, 2012 august 13, 2019 corbettmaths on unit 2 number making. 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Write down the class interval in which the.. The obs requesting a number of bins for us can you cast spells require... Use of a data ’ s appearance words, the mid point of intersection drop... Total number of groups of equal length many elements more than mean over the histogram 'bin ' in i... And answer site for people studying math at any level and professionals in fields... / professor discourage all collaboration the ages of the data the question approximately... Median … Model answers & video Solution for Histograms for Histograms these as... Total frequency either side is the median line powerful tools to visualize data value! Create a visual representation of data distribution of how to restore/save my reputation, stem and leaf.... Am trying to implement Dualistic Sub-Image histogram Equalisation '' course on learning... Affect Earth geopolitics different ways, what they tell us information about the ages of the frequencies to the. Key differences: many elements more than mean coffee production in thousands of 60kg sacks using data from Statistica 97... Ages of the variable children 's book - front cover displays blonde child playing flute in a.. Statistician, i enjoy reading, writing, and exploring new places data to columns... Tasha 's Cauldron of Everything, can you cast spells that require a target can. Example, take into account the two following boxplots the same... Write the...
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# Algebra
posted by .
Can anyone please show me how 3^n/6^(2n+1) becomes 1/2^n(6^(n+1))= 1/6)/12^n when you divide top and bottom by 3^n?
Thank you and have a great day!
• Algebra -
let's work on the bottom
6^(2n+1)
= 3^(2n+1)* 2^(2n+1)
= 3^n * 3^(n+1) * 2^n * 2^(n+1)
= 3^n * 2^n * 6^(n+1)
now the 3^n would cancel the 3^n on top, leaving you with
1/(2^n * 6^(n+1) )
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# the real part of a holomorphic function on C \ {0, 1}
Let $h$ be a real valued harmonic function on the twice punctured plane $Ω = \text{C \ {0, 1}}$. Show that there exist unique real numbers $a_0$, $a_1$ such that $u(z) = h(z) − a_0 \log |z| − a_1 \log |z − 1|$ is the real part of a holomorphic function on $Ω$ .
I tried to show that $u$ has a harmonic conjugate. I assumed there exists such a conjugate $v$ exist, then used Cauchy Riemann equations on $u$ and $v$. After integrating two equations, I tried to use the result to find $a_0$ and $a_1$. However, I failed to cancel out the extra terms. Is there any other direction I should go?
Define a function $g = h'_x - ih'_y$. Since $h$ is harmonic (and hence $C^2$), $g$ satisfies Cauchy-Riemann's equations on $\Omega$. Indeed: $$(h'_x)'_x = h''_{xx} = -h''_{yy} = (-h'_y)'_y$$ and $$(h'_x)'_y = h''_{xy} = h''_{yx} = -(-h'_y)'_x.$$ In other words, $g$ is holomorphic on $\Omega$. Assume for a moment that $g$ admits an anti-derivative $G$ on $\Omega$. If $G = U+iV$, Cauchy-Riemann shows that $G' = U'_x-iU'_y$, but $G' = g = h'_x-ih'_y$, so $\nabla U = \nabla h$, i.e. $U=h+C$, and we may as well take $C=0$ by adjusting our choice of $G$. Hence $h = U = \operatorname{Re} G$.
Now, the problem is that $\Omega$ is not simply connected, so we can't guarantee that $g$ admits an anti-derivative. On the other hand, we know that $g$ has an anti-derivative on $\Omega$ if and only if $\int_\gamma g(z)\,dz = 0$ for every simple closed curve in $\Omega$, and it's not difficult to see that this condition is equivalent to the fact that the residues of $g$ at $0$ and $1$ should both be $0$.
Let $a_0 = \operatorname{Res}\limits_{z=0} g$ and $a_1 = \operatorname{Res}\limits_{z=1} g$, and put $u(z) = h(z)-a_0\log|z|-a_1\log|z-1|$.
Repeat the above argument to get a holomorphic function $f=u'_x-iu'_y$ on $\Omega$. It's straight-forward to verify that $$f(z) = g(z) - \frac{a_0}{z} - \frac{a_1}{z-1}$$ so by construction, the residues of $f$ at $0$ and $1$ vanish, and from the discussion above, we are done.
This takes care of existence. For uniqueness, assume that $u(z) = h(z)-a_0\log|z|-a_1\log|z-1|$ and $\tilde u(z) = h(z)-\tilde a_0\log|z|-\tilde a_1\log|z-1|$ are two such functions. Then $u(z) = \tilde u(z) = (a_0-\tilde a_0) \log|z|-(a_1-\tilde a_1)\log|z-1|$ is the real part of a holomorphic function on $\Omega$. It's a standard exercise to check that this can only happen if $a_0 = \tilde a_0$ and $a_1 = \tilde a_1$.
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Message Board Basketball Forum - InsideHoops A little thinking game..
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07-29-2012, 08:02 PM #16
REACTION
HYRYSICPT?
Join Date: Dec 2011
Posts: 661
Re: A little thinking game..
Quote:
Originally Posted by Faptastrophe An n x n grid will have a total of n2 + (n - 1)2 + ... + 22 + 12 squares. The picture has one 4 x 4 square and two 2 x 2 squares. There are 42 + 32 + 22 + 12 = 16 + 9 + 4 + 1 = 30 squares in the 4 x 4, and there are 22 + 12 = 4 + 1 = 5 squares in each 2 x 2. Thus, there are a total of 30 + 2 * 5 = 40 squares.
Do you mean n^2 instead of n2? Not sure what formula you're using. But I'm assuming those are meant to be powers instead of multipliers, right?
The formula for the larger squares that makes sense to me is f(x) = (x(x+1)(2x+1))/6 where x = 4. f(4) gives you (4 * 5 * 9)/6 = 30 total squares for the larger ones. Then I'd add the 10 total combinations of the smaller squares. I think we're basically doing the same thing to arrive at 40 though.
07-29-2012, 08:05 PM #17
Myth
NBA Legend and Hall of Famer
Join Date: Jul 2007
Location: Orange County
Posts: 21,407
Re: A little thinking game..
Quote:
Originally Posted by REACTION Do you mean n^2 instead of n2? Not sure what formula you're using. But I'm assuming those are meant to be powers instead of multipliers, right? The formula for the larger squares that makes sense to me is f(x) = (x(x+1)(2x+1))/6 where x = 4. f(4) gives you (4 * 5 * 9)/6 = 30 total squares for the larger ones. Then I'd add the 10 total combinations of the smaller squares. I think we're basically doing the same thing to arrive at 40 though.
I just counted.
07-29-2012, 08:09 PM #18 ballup Your Favorite Canadian Join Date: Dec 2010 Location: Batz Thread Eh? Posts: 8,207 Re: A little thinking game.. I got 39 and I counted the big square.
07-29-2012, 08:10 PM #19
REACTION
HYRYSICPT?
Join Date: Dec 2011
Posts: 661
Re: A little thinking game..
Quote:
Originally Posted by Myth I just counted.
Well, what if it were a 100x100 square?
07-29-2012, 08:11 PM #20
Myth
NBA Legend and Hall of Famer
Join Date: Jul 2007
Location: Orange County
Posts: 21,407
Re: A little thinking game..
Quote:
Originally Posted by REACTION Well, what if it were a 100x100 square?
Then a math formula would be more necessary, but for this it is just overkill because I counted them in less than 10 seconds.
07-29-2012, 08:13 PM #21
b1imtf
World Class
Join Date: Feb 2011
Location: Azores
Posts: 6,182
Re: A little thinking game..
Quote:
Originally Posted by Faptastrophe An n x n grid will have a total of n2 + (n - 1)2 + ... + 22 + 12 squares. The picture has one 4 x 4 square and two 2 x 2 squares. There are 42 + 32 + 22 + 12 = 16 + 9 + 4 + 1 = 30 squares in the 4 x 4, and there are 22 + 12 = 4 + 1 = 5 squares in each 2 x 2. Thus, there are a total of 30 + 2 * 5 = 40 squares.
[IMG][/IMG]
07-29-2012, 09:58 PM #22
Richie2k6
Smooth Like Butter
Join Date: Mar 2007
Location: The Terrordome Team: Nuggets
Posts: 9,913
Re: A little thinking game..
Quote:
Originally Posted by b1imtf [IMG][/IMG]
07-29-2012, 10:00 PM #23
Faptastrophe
Rivalries Never Die
Join Date: Dec 2011
Location: LV
Posts: 3,574
Re: A little thinking game..
Quote:
Originally Posted by REACTION Do you mean n^2 instead of n2? Not sure what formula you're using. But I'm assuming those are meant to be powers instead of multipliers, right? The formula for the larger squares that makes sense to me is f(x) = (x(x+1)(2x+1))/6 where x = 4. f(4) gives you (4 * 5 * 9)/6 = 30 total squares for the larger ones. Then I'd add the 10 total combinations of the smaller squares. I think we're basically doing the same thing to arrive at 40 though.
07-29-2012, 10:15 PM #24 Scholar Gawdbe GOATsol Nashty Join Date: Dec 2010 Location: Swaggy P's Palace Posts: 9,971 Re: A little thinking game.. I counted 38 before I saw LEFT4DEAD's GIF.
07-29-2012, 10:50 PM #25 Eat Like A Bosh Banned Join Date: Jan 2011 Location: Inside LeBron's head Posts: 5,194 Re: A little thinking game.. 36. EDIT: never mind saw Left4Dead's Gif Last edited by Eat Like A Bosh : 07-29-2012 at 10:54 PM.
07-30-2012, 01:18 AM #26 JustinJDW .... Join Date: Sep 2008 Location: Los Angeles, CA Posts: 4,499 Re: A little thinking game.. 34 before I saw the gif.
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# To Study the Transpiration Rate of a Plant (Gou Qi) by Using the Bubble Photometer.
Topics: Leaf, Water, Evaporation Pages: 5 (1073 words) Published: May 1, 2012
Biology 1st Group Lab Report
Objective:
- To study the transpiration rate of a plant (Gou Qi) by using the bubble photometer.
Assumption:
- The rate of transpiration is equal to the rate of water uptake of the plant(Gou Qi) .
Theory:
|Independent Variable |Dependent Variable |Controlled Variable | |The environmental conditions: | Rate of water uptake by transpiration of the |Time taken for the movement of the air bubble | |Normal |plant(Gou Qi) |(5MIN -> 10MIN) | |Higher light intensity | |Type of the plant(Gou Qi). | |Greater air movement | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
How can we measure the dependent variable?
We measure the dependent variable according to the formula below: Rate = the distance move(final–initial)
Time
Unit : cm/ min
How can we vary the independent variable?
We can vary the environmental conditions by:
Normal: Under room temperature, normal light intensity, with natural wind Higher Light Intensity: Put a table lamp 50cm far from the plant (Gou qi). Greater air movement: Put a fan 100cm far from the plant (Gou qi). Materials and Apparatus:
◆ The plant (Gou qi) x1
◆ A basin of water
◆ Scissor
◆ Bubble photometer x1
➢ U shape tube
➢ Rubber tubing
➢ Syringe
◆ Rubber Ban
◆ Stop watch
◆ Ruler
◆ Fan
Procedure of setting and testing the experiment:
1. Wrapped the stem of the plant(Gou qi) by using the thread-seal tape. 2. The bubble photometer was filled with water.
3. The stem of the plant(Gou qi) was cut under water. The stem was fitted onto the rubber tubing on the photometer under water. 4. An air bubble was introduced at the end of the capillary tube. (Push the water inside the syringe to reset the position of the air bubble) 5. Wait for the bubble to move to the graduated part of the capillary tube for 5 minutes. 6. The initial position/water level (cm) of the air bubble was recorded. 7. The final position/water level (cm) of the air bubble was recorded again after 5 minutes. 8. Repeat step 4 to step7 with new environmental conditions (higher light intensity and greater air movement). Result:
A table showing distance moved by water level(cm)
| |Distance moved by water level (cm)...
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# Determine If the Following Polynomial Has (X + 1) a Factor :- X3 + X2 + X + 1 - Mathematics
Sum
Determine if the following polynomial has (x + 1) a factor :-
x3 + x2 + x + 1
#### Solution
If (x + 1) is a factor of p(x) = x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).
p(x) = x3 + x2 + x + 1
p(−1) = (−1)3 + (−1)2 + (−1) + 1
= − 1 + 1 − 1 + 1
= 0
Hence, x + 1 is a factor of this polynomial.
Concept: Factorising the Quadratic Polynomial (Trinomial) of the type ax2 + bx + c, a ≠ 0.
Is there an error in this question or solution?
Chapter 2: Polynomials - Exercise 2.4 [Page 43]
#### APPEARS IN
NCERT Class 9 Maths
Chapter 2 Polynomials
Exercise 2.4 | Q 1.1 | Page 43
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1. Convergence/Divergence of Series
Can someone give me a hint that the summation of (-1)^(u)*(u^(1/2))/(u+1) from u=1 to N is divergent, conditionally convergent, or absolutely convergent? By the looks of it it is conditionally convergent as it's absolute value diverges, but i know 1/u^(1/2)*(-1)^(u) converges and the one doesn't really matter, but I can't prove that it doesn't matter.
Thanks for the help in advance
2. Originally Posted by jarny
Can someone give me a hint that the summation of (-1)^(u)*(u^(1/2))/(u+1) from u=1 to N is divergent, conditionally convergent, or absolutely convergent? By the looks of it it is conditionally convergent as it's absolute value diverges, but i know 1/u^(1/2)*(-1)^(u) converges and the one doesn't really matter, but I can't prove that it doesn't matter.
Thanks for the help in advance
You're right. It's conditionally convergent.
Your sum is $\sum_{i=1}^\infty \frac{(-1)^n \sqrt{n}}{n+1}$. Considering $\sum_{i=1}^\infty \frac{\sqrt{n}}{n+1}$ show it's divergent by LCT with $\sum_{i=1}^\infty \frac{1}{\sqrt{n}}$.
The Alternating series test
(i) $\lim_{n \to \infty} \frac{\sqrt{n}}{n+1} = 0$
(ii) $\frac{\sqrt{n}}{n+1}$ is decreasing
give that it converges and thus, converges conditionally.
3. Hello, jarny!
$\sum^{\infty}_{u=1}\;(\text{-}1)^u\frac{u^{\frac{1}{2}}}{u+1}$ . Divergent, conditionally convergent, or absolutely convergent?
It is a convergent alternating series. .(The $n^{th}$ term approaches zero.)
It is not absolutely convergent because: . $\sum\frac{u^{\frac{1}{2}}} {u+1} \:\approx\:\sum\frac{1}{u^{\frac{1}{2}}}$ . . . a divergent $p$-series.
Therefore, the series is conditionally convergent.
4. Somehow I managed to mis-write it. It in the denominator is should be 4 instead of 1. Sorry :/ Soroban, I can see that it is approximately but I need to kind of prove it.
when its (-1)^(u) *(u)^(1/2)/(u+4) could I just find a particular u (in this case 4) where the absolute value of the series begins to decrease and say since a_1>a_2>a_3 ...>a_N and since the lim n->infinity is zero (and the summands before 4 are finite) that by the alternating series test the summation is conditionally convergent?
5. Originally Posted by jarny
Somehow I managed to mis-write it. It in the denominator is should be 4 instead of 1. Sorry :/ Soroban, I can see that it is approximately but I need to kind of prove it.
If you mean
$
\sum_{i=1}^\infty \frac{(-1)^n \sqrt{n}}{n+4}$
It won't change the results that we have posted (except for the 4)
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# NO PLAGIARISM DUE FRIDAY MARCH 1, 2019. THIS IS STATISTICS PLEASE BE PROFICIENT IN STATISTICS
I NEED SOMEONE WHO IS PROFICIENT IN STATISTICS.
The length of this essay should be one to two double-spaced pages (excluding title and reference pages). Use 12-point font and format your paper with regular 1-inch margins.
A researcher has recorded the reaction times of 20 individuals on a memory assessment. The following table indicates the individual times:
2.2
4.7
7.3
4.1
9.5
15.2
4.3
9.5
2.7
3.1
9.2
2.9
8.2
7.6
3.5
2.5
9.3
4.8
8.5
8.1
In this essay, demonstrate your ability to organize data into meaningful sets, calculate basic descriptive statistics, interpret the results, and evaluate the effects of outliers and changes in the variables. You may use Excel, one of the many free online descriptive statistics calculators, or calculate the values by hand and/or with a calculator.
Next, separate the data into two groups of 10; one group will be the lower reaction times, and the second group will be the higher reaction times. Then, address the following points in your essay response: Calculate the sum, mean, mode, median, standard deviation, range, skew, and kurtosis for each group. How do the two groups differ? Are there any outliers in either data group? What effect does an outlier have on a sample?
Lastly, double each sample by repeating the same 10 data points in each group. You will have a total of 20 data points for each group. After completing this, address the following in your essay response: Calculate the following for the new data groups: sum, mean, mode, median, standard deviation, range, skew, and kurtosis. Did any of the values change? How does sample size affect those values?
Did you know you can hire someone to answer this question? Yes, classaider.com is a hub of paper writers, dedicated to completing research and summaries, critical thinking tasks, essays, coursework, and other homework tasks. It is simple as ABC.
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# Median algebra
In this blog post, we discuss how Median algebra can help students learn Algebra.
Homework Help Online
Average satisfaction rating 4.9/5
Math understanding that gets you
## Median Definition (Illustrated Mathematics Dictionary)
The median score will be located in the middle of the class's scores when arranged from lowest to highest or highest to lowest. Because there are 30 students in the class, the median score is
You can improve your educational performance by studying regularly and practicing good study habits.
Free time to spend with your family and friends
Solving math problems can be a fun and rewarding experience.
Deal with mathematic questions
Math is hard, and it's okay to admit that you need help. The important thing is to keep trying and to never give up.
Solve math equation
Mathematics is a way of dealing with tasks that require e#xact and precise solutions.
Clarify math equation
In order to determine what the math problem is, you will need to look at the given information and find the key details. Once you have found the key details, you will be able to work out what the problem is and how to solve it.
No matter what you're writing, good writing is always about engaging your audience and communicating your message clearly.
## Median algebra
Deal with mathematic problems
Explain math questions
Save time
Writing Versatility
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Reach support from expert tutors
## Algebra II : Median
In a Boolean algebra the median function x, y, z = ( x ∨ y) ∧ ( y ∨ z) ∧ ( z ∨ x) satisfies these axioms, so that every Boolean algebra is a median algebra. Birkhoff and Kiss showed that a
`
## Median algebras
Median Calculator If you were looking for a way to calculate the median value of a set of numbers, then the Median calculator is exactly what you need. No need to know formula. In just two easy
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## Question 6: A wire hangs from the top of a dark high tower so that the top of the tower is not visible. How you would be able to determine the height of that tower?
The height of the tower can be easily determined if we tie a mass with the hanging wire. If the wire is set into vibration, it will behave as a simple pendulum.
Connect some mass with the end of the wire. Take a stop watch and set the pendulum in vibration. Calculate the time period of the pendulum. You may find the time of, say ten vibrations, and divide it by 10 to find the time period of the pendulum.
Now equation of the simple pendulum is
π and g are constants. Use the value of T you just calculated in the above equation and find the length of the wire. As the wire hangs from the top of the tower, it will give you the height of the tower.
Note that ℓ gives you the length of the wire and if the wire is above the ground (and principally it should be) add the remaining part to the length of the wire.
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## Conversion formula
The conversion factor from months to minutes is 43829.1, which means that 1 month is equal to 43829.1 minutes:
1 mo = 43829.1 min
To convert 764.8 months into minutes we have to multiply 764.8 by the conversion factor in order to get the time amount from months to minutes. We can also form a simple proportion to calculate the result:
1 mo → 43829.1 min
764.8 mo → T(min)
Solve the above proportion to obtain the time T in minutes:
T(min) = 764.8 mo × 43829.1 min
T(min) = 33520495.68 min
The final result is:
764.8 mo → 33520495.68 min
We conclude that 764.8 months is equivalent to 33520495.68 minutes:
764.8 months = 33520495.68 minutes
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 2.9832494410178E-8 × 764.8 months.
Another way is saying that 764.8 months is equal to 1 ÷ 2.9832494410178E-8 minutes.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that seven hundred sixty-four point eight months is approximately thirty-three million five hundred twenty thousand four hundred ninety-five point six eight minutes:
764.8 mo ≅ 33520495.68 min
An alternative is also that one minute is approximately zero times seven hundred sixty-four point eight months.
## Conversion table
### months to minutes chart
For quick reference purposes, below is the conversion table you can use to convert from months to minutes
months (mo) minutes (min)
765.8 months 33564324.78 minutes
766.8 months 33608153.88 minutes
767.8 months 33651982.98 minutes
768.8 months 33695812.08 minutes
769.8 months 33739641.18 minutes
770.8 months 33783470.28 minutes
771.8 months 33827299.38 minutes
772.8 months 33871128.48 minutes
773.8 months 33914957.58 minutes
774.8 months 33958786.68 minutes
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# Supose that G is a finite abelian group that does not contain a subgro
let us assume G is not cyclic. Let a be an element of G of maximal order. Since G is not cyclic we have <a>≠G. Let b be an element in G, but not in the cyclic subgroup generated by a.
O(a) = m and O(b) = n where O() refers tothe orders. . then how can we use this to construct a subgroup of G isomorphic to Zp×Zp?
Help will be appreciated. Thank you
#### Attachments
• Untitled.jpg
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Dick
Homework Helper
let us assume G is not cyclic. Let a be an element of G of maximal order. Since G is not cyclic we have <a>≠G. Let b be an element in G, but not in the cyclic subgroup generated by a.
O(a) = m and O(b) = n where O() refers tothe orders. . then how can we use this to construct a subgroup of G isomorphic to Zp×Zp?
Help will be appreciated. Thank you
Can you prove that if the prime factorization of |G| is ##p_1 p_2 p_3...p_n## and all of the primes ##p_i## are different, then G is cyclic? If two are the same, then what?
Last edited:
1 person
If |G| =p1p2....pn and all of the primes are different, then |H| will be one of those primes. Since a group of prime order is cyclic, then H will be cyclic in that case . Infact all proper subgroups of G will be cyclic.
but i am not sure how to prove G as cyclic with this data..
If |G| =p1p2....pn and all of the primes are different, then |H| will be one of those primes. Since a group of prime order is cyclic, then H will be cyclic in that case . Infact all proper subgroups of G will be cyclic.
but i am not sure how to prove G as cyclic with this data..
What does the fundamental theorem of Abelian groups tell you?
the book which i am reading ( GAllian ) has not introduced this topic as of yet. In fact, not even normal and factor groups. i am on the chapter on external direct products.
the book which i am reading ( GAllian ) has not introduced this topic as of yet. In fact, not even normal and factor groups. i am on the chapter on external direct products.
Where exactly in Gallian is this?
Supplementary exercise for chapters 5 - 8 . Question no. 50 . Gallian 7/e contemporary guide to abstract algebra
and pg. no 50
OK. So let ##G## be finite abelian group such that ##G## does not contain a subgroup isomorphic to ##\mathbb{Z}_p\times \mathbb{Z}_p## for any prime. Consider the prime factorization ##|G|= p_1^{a_1} ... p_k^{a_k}:= p_1^{a_1}m##.
Define ##H = \{x\in G~\vert~\text{order of}~H~\text{divides}~p_1^{a_1}\}## and ##K = \{x\in G~\vert~\text{order of}~K~\text{divides}~m\}##.
Start by proving the following:
1) ##H## and ##K## are subgroups of ##G##.
2) For any ##x\in G## and integers ##s## and ##t##, we have ##x^{sp^k}\in H## and ##x^{tm}\in K##. Deduce that ##G=HK##.
3) Show that ##H\cap K = \{e\}##.
4) Find ##|H|## and ##|K|##.
5) Prove that ##H## is cyclic.
6) Prove that ##G## is the product of cyclic groups
7) Prove the result.
Last edited:
1 person
Dick
Homework Helper
OK. So let ##G## be finite abelian group such that ##G## does not contain a subgroup isomorphic to ##\mathbb{Z}_p\times \mathbb{Z}_p## for any prime. Consider the prime factorization ##|G|= p_1^{a_1} ... p_k^{a_k}:= p_1^{a_1}m##.
Define ##H = \{x\in G~\vert~\text{order of}~H~\text{divides}~p_1^{a_1}\}## and ##K = \{x\in G~\vert~\text{order of}~K~\text{divides}~m\}##.
Start by proving the following:
1) ##H## and ##K## are subgroups of ##G##.
2) For any ##x\in G## and integers ##s## and ##t##, we have ##x^{sp^k}\in H## and ##x^{tm}\in K##. Deduce that ##G=HK##.
3) Show that ##H\cap K = \{e\}##.
4) Find ##|H|## and ##|K|##.
5) Prove that ##H## is cyclic.
6) Prove that ##G## is the product of cyclic groups
7) Prove the result.
Your definition of K doesn't make much sense. I assume you mean ##K = \{x\in G~\vert~\text{order of}~x~\text{divides}~m\}##.
1 person
Your definition of K doesn't make much sense. I assume you mean ##K = \{x\in G~\vert~\text{order of}~x~\text{divides}~m\}##.
Yes, sorry!
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# proof that p implies q entails not p or q [duplicate]
I could easily prove
$\neg P \lor Q$ entails $P \rightarrow Q$.
It is well known that
$P \rightarrow Q$ entails $\neg P \lor Q$
but I couldn't find a way to prove it.
Although there is the same question; How to prove that $P \rightarrow Q$ is equivalent with $\neg P \lor Q$?; it's a little bit confusing and I need to see step by step solution.
Could you show me the way by using Natural Deduction?
## marked as duplicate by Git Gud, Namaste, Chris Janjigian, Carl Mummert, Vladimir ReshetnikovNov 2 '14 at 17:57
Assume : $P \rightarrow Q$ --- premise
1) $\lnot (\lnot P \lor Q)$ --- assumed [a]
2) $\lnot P$ --- assumed [b]
3) $\lnot P \lor Q$ --- from 2) by $\lor$I
4) $\bot$ --- from 1) and 3) by $\lnot$E (or $\rightarrow$E)
5) $P$ --- from 2) and 4) by Double Negation, discharging [b]
6) $Q$ --- from premise and 5) by $\rightarrow$E
7) $\lnot P \lor Q$ --- from 6) by $\lor$I
8) $\bot$ --- from 1) and 7) by $\lnot$E (or $\rightarrow$E)
9) $\lnot P \lor Q$ --- from 1) and 8) by Double Negation, discharging [a]
Thus :
$P \rightarrow Q \vdash \lnot P \lor Q$.
• thank you mauro, but i couldn't understand the implication elimination in contradiction lines(4,8). which implication we eliminate? – Cnqt Nov 2 '14 at 17:29
• @Cnqt - perhaps you know the rule $\lnot$E : from $\varphi, \lnot \varphi$, infer $\bot$; I've applied this rule in the form : $\lnot \varphi := \varphi \rightarrow \bot$. Assuming this abbreviation, the above rule is from $\varphi, \varphi \rightarrow \bot$, infer $\bot$, taht is simply an application of $\rightarrow$E. – Mauro ALLEGRANZA Nov 2 '14 at 17:57
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# SOLUTION: Factor out the greatest common factor from the polynomial 21x^3 - 6x^2 + 15x
Algebra -> Algebra -> Polynomials-and-rational-expressions -> SOLUTION: Factor out the greatest common factor from the polynomial 21x^3 - 6x^2 + 15x Log On
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Click here to see ALL problems on Polynomials-and-rational-expressions Question 102392: Factor out the greatest common factor from the polynomial 21x^3 - 6x^2 + 15xAnswer by JP(22) (Show Source): You can put this solution on YOUR website!21x^3-6x^2+15x When you look at the polynomial, what is a factor that every term has in common with one another? The answer is 3x because, 3x(7x^2-2x+5) 21x^3 is divided by 3x to get 7x^2 -6x^2 is divided by 3x to get -2x 15x is divided by 3x to get 5 So the GFC(Greatest Common Factor) is 3x To check your work multiply 3x by the factored set of terms 3x(7x^2-2x+5) to get 21x^3-6x^2+15x Hope this helps...
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Message Font: Serif | Sans-Serif
No. of Recommendations: 1
Wayne,
Normally, when something is "picked at random" all valid items have the same chance to be picked.
That's right. There's a probability of 1/3 of initially choosing the six-sided die, a 1/3 probability of choosing the 12-sided die, and a 1/3 probability of choosing the 20-sided die.
But here's the catch: there are 38 possible outcomes of the first roll -- 1 through 6 on the 6-sided die, 1 through 12 on the twelve-sided die, and 1 through 20 on the twenty-sided die. The outcomes for each die are equiprobable, but that does not make all 38 outcomes equiprobable. In particular, the actual probability of each outcome on the six-sided die is 1/18 (1/3 x 1/6), the probability of each outcome on the twelve-sided die is 1/36 (1/3 x 1/12), and the probability of each outcome on the twenty-sided die is 1/60 (1/3 x 1/20).
Now, the knowledge that the initial roll is a 4 means that we have to throw out all outcomes that don't have a 4 as the rolled value -- that is, the other five outcomes on the six-sided die, the other eleven outcomes on the twelve-sided die, and the other nineteen outcomes on the twenty-sized die -- and scale the probabilities of the remaining cases (that is, the cases that do produce a four) so they add up to ONE. Now,
1/18 + 1/36 + 1/60 = 10/180 + 5/180 + 3/180 = 18/180 = 1/10
so we multiply the probabilities of the remaining cases by 10 -- and the initial roll of 4 came from the six-sided die 10/18 = 5/9 of the time, from the twelve-sided die 10/36 = 5/18 of the time, and from the twenty-sided die 10/60 = 1/6 of the time.
This is what's known as "conditional probability" -- the knowledge that the initial roll was a 4 alters the probability each die was chosen.
Note that the impact would have been very different if the initial roll had been 7 or more. If it were in the range of 7 through 12, the first die could not have been the six-sided die because the six-sided die does not have those values -- and 1/36 + 1/60 = 5/180 + 3/180 = 8/180 = 2/45, so it would come from the twelve-sided die 5/8 of the time and from the twenty-sided die 3/8 of the time. And an initial roll of 13 or higher could come only from the 20-sided die.
Norm.
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# How to Write an Inverse Function – A Step-by-Step Guide
In the realm of mathematics, understanding the concept of an inverse function is akin to learning a secret handshake that reveals a hidden symmetry in the universe of numbers.
When I explore functions, I’m essentially looking at special relationships where every input is paired with exactly one output. Picture a function as a unique dance move that takes me from one point to another, say from point A to point B.
Now, the inverse function is the dance move that takes me back from point B to A, demonstrating a beautiful balance in mathematical operations.
However, not all functions have this mystical partner—only those that are one-to-one, meaning each output is linked to only one input, can boast an inverse.
Knowing whether a function is one-to-one is crucial because, without this property, it cannot have an inverse. To formalize this relationship, we use a particular notation: if ( f(x) ) takes me from A to B, then $f^{-1}(x)$ is the name of the dance move bringing me back from B to A.
Here’s where understanding the domain and range becomes essential, as the domain of ( f ) becomes the range of $f^{-1}$ and vice versa, giving me a complete picture of this two-way journey.
## Steps for Writing Inverse Functions
When I approach the task of finding the inverse function, I think of it as a process of reversing steps to determine how to get back to the original place.
Here’s my strategy for problem-solving in algebra and ensuring successful results.
1. Verify One-to-One: Before defining an inverse function, I ensure that the original function is one-to-one. This means every input has a unique output and vice versa. A handy tool is the horizontal line test: if any horizontal line crosses the graph of the function more than once, it’s not one-to-one and doesn’t have an inverse in its current form.
2. Switch x and y: I start by replacing the output (usually $y$) with input ($x$) and then solve for $y$. This step swaps the roles of inputs and outputs, preparing the function to be reversed.
3. Solve for the New Output: Next, I algebraically manipulate the equation to solve for $y$. This often involves function composition, and I take care not to violate the laws of algebra.
4. Restrict Domains: Some functions, like square roots, need a defined domain to ensure a one-to-one function. Non-negative restrictions are common to avoid multiple results.
5. Use Proper Notation: It’s vital to denote the inverse function with the correct notation, which is $f^{-1}(x)$. Do not confuse this with reciprocal, as they are not the same.
6. Double Check Your Work: I perform function composition to confirm that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. If this holds, I’ve found a correct inverse.
Here’s a quick example for clarity:
For the original function $f(x) = 2x + 3$, the steps would look like this:
1. Since it’s a linear function, it’s inherently one-to-one.
2. Swap the labels so $x$ becomes $y$ and vice versa, getting $x = 2y + 3$.
3. I’ll solve for $y$: $y = \frac{x – 3}{2}$.
4. For this function, I don’t need to restrict the domain.
5. I apply the notation: $f^{-1}(x) = \frac{x – 3}{2}$.
6. Verify by composition that $f(f^{-1}(x)) = 2(\frac{x – 3}{2}) + 3 = x$.
By following these steps, I can tackle various inverse trigonometric functions, graphing inverse functions, and even more complex rational functions. Whether it’s a lesson on converting Fahrenheit to Celsius or exploring symmetric graphs, having a solid grasp on inverse functions greatly simplifies these challenges.
## Conclusion
In wrapping up our discussion on inverse functions, remember that these special functions are the mathematical equivalent of getting back to where you started.
They are fundamental in many branches of mathematics and real-world applications, like decrypting messages or converting between different units of measurement.
To check whether a function ( f(x) ) has an inverse, I always ensure that it’s a one-to-one function, meaning it passes both the vertical and horizontal line tests.
When I find the inverse of a function, I swap the roles of ( x ) and ( y ) and then solve for the new ( y ), which is the inverse function. Remember to notate the inverse function as $f^{-1}(x)$, which does not mean $(1/f(x))$ but rather the operation that reverses the effect of ( f(x) ).
Keep in mind to represent the inverse graphically as a reflection across the line ( y = x ). Through this representation, the symmetry between a function and its inverse becomes visually apparent and helps in understanding their relationship.
Lastly, I always verify my results. I do this by composing the function and its inverse to see if I get an identity, confirming $f^{-1}(f(x)) = x ) and ( f(f^{-1}(x)) = x$.
This step is crucial to ensure that the inverse I’ve computed undoes what the original function does. With practice, the process of finding and understanding inverse functions will become second nature.
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×
SSJG - EDITORIAL
Setter: Kunal Jain
Tester: Kunal Jain, Taranpreet Singh
Editorialist: Akash Bhalotia
EASY
2-D Prefix Sums
PROBLEM:
An N * N grid contains coins in some cells. Divide it into 4 parts using a vertical and a horizontal line such that the number of coins in the part containing the least number of coins is maximum possible.
CONSTRAINTS:
• 1 $\leq$ T $\leq$ 10
• 1 $\leq$ N $\leq$ 1000
QUICK EXPLANATION:
Compute 2-D prefix sums for the grid. For every intersection, compute the number of coins present in its 4 subgrids in constant time using the prefix sums. The intersection that has the maximum min value is the best intersection.
EXPLANATION:
• We need to calculate the number of coins present in the 4 subgrids created by every possible intersection.
• We can do this by running nested loops for every intersection. This will take O($N^4$), which is sufficient to pass the first subtask, but will result in TLE for the second subtask.
• Thus, we need to improve the complexity from O($N^4$) to O($N^2$). We need to somehow make the step of computing the number of coins present in a subgrid a constant time process.
• We can do this by creating a Prefix sums matrix. PrefSum[i][j] will contain the number of coins present in the subgrid from (1,1) to (i,j).
• The formula for this is :
View Content
• If you are having trouble understanding 2-D Prefix Sums, first make sure that you have a good understanding of 1-D Prefix Sums. 2-D Prefix Sums is simply an extension of that in 2 Dimensions.
• Now let us compute the number of coins present in the subgrids of an intersection at (i,j) in O(1) :
View Content
• Now we just need to output the maximum min value we can achieve among all intersections.
BONUS:
What if we had to find the number of coins present in a subgrid starting at (i,j) and ending at (x,y)?
View Content
COMPLEXITY:
• Time Complexity: O($N^2$) per test case.
• Space Complexity: O($N^2$) per test case.
SIMILAR PROBLEMS:
Feel free to share your approach if it differs. If you have any doubts, or if you feel stuck at any point, you can ask them below. We would love to hear your suggestions :)
Thanks to @taran_1407 and @vijju123 for their constant guidance and support.
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# a. Determine the electrostatic potential energy of a system consisting of two charges $7\mu C\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}-2\mu C$ (and with no external field) placed at (-9 cm, 0,0) and (9 cm, 0, 0) respectively. b. How much work is required to separate the two charges infinitely away from each other? c. Suppose that the same system of charges is now placed in an external electric field $E=A\left(\frac{1}{{r}^{2}}\right),A=9×{10}^{5}C{m}^{-2}$. What would the electrostatic energy of the configuration be?Solution in Tamil
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Verified by Experts
## a. $U=\frac{1}{4\pi {\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{r}=9×{10}^{9}×\frac{7×\left(-2\right)×{10}^{-12}}{0.18}=-0.7J$ b. $W={U}_{2}-{U}_{1}=0-U=0\left(-0.7\right)=0.7J$ b. $W={U}_{2}-{U}_{1}=0-U=0-\left(0.7\right)=0.7J$ c. Energy of interaction of the two charges with the external electric field. ${q}_{1}V\left({r}_{1}\right)+{q}_{2}V\left({r}_{2}\right)=A\frac{7}{0.09}+A\frac{-2}{0.09}$ Net electrostatic energy is : ${q}_{1}V\left({r}_{1}\right)+{q}_{2}V\left({r}_{2}\right)+\frac{{q}_{1}{q}_{2}}{4\pi {\epsilon }_{0}{r}_{12}}=A\frac{7}{0.09}+A\frac{-2}{0.09}-0.7=70-20-0.7=49.3J$
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# Physics
posted by .
A ball is projected upward at time t= 0.0s, from a point on the roof 10m above the ground. The ball rises, then falls until it strikes the ground. The initial velocity of the ball is 58.5 m/s. At time t=5.97s, what is the approximate velocity of the ball? Neglect air resistance.
• Physics -
Solve this equation using t = 5.97:
V = Vo - g t = 58.5 - 9.8 t
That will be zero, indicating that the ball is at its highest elevation.
• Physics -
Why'd you use 9.8?
• Physics -
Because the acceleration of gravity (g) is 9.8 m/s^2.
• Physics -
Oh, okay, thanks.
• Physics -
jip-;
• Physics -
ayyy lmao
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# What is the mean of the exams taken by a group of students?
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05 Feb 2015, 08:13
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What is the mean of the exams taken by a group of students?
(1) The score 3 standard deviations above the mean is 95
(2) The score 2 standard deviations below the mean is 64
Kudos for a correct solution.
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Re: What is the mean of the exams taken by a group of students? [#permalink]
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05 Feb 2015, 08:50
Bunuel wrote:
What is the mean of the exams taken by a group of students?
(1) The score 3 standard deviations above the mean is 95
(2) The score 2 standard deviations below the mean is 64
Kudos for a correct solution.
I'd pick C.
from st1: m+3SD=95 from st2: m-2SD=64 together we can determine m.
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Re: What is the mean of the exams taken by a group of students? [#permalink]
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05 Feb 2015, 17:30
Bunuel wrote:
What is the mean of the exams taken by a group of students?
(1) The score 3 standard deviations above the mean is 95
(2) The score 2 standard deviations below the mean is 64
Kudos for a correct solution.
let std dev =x
Statement 1:
mean + 3x = 95
the mean or x could both change depending on the value of the other
Insufficient
Statement 1:
mean - 2x = 64
same as above
Insufficient
Combined, we have 2 equations 2 variables and we can eliminate x, resulting in a unique value for the mean.
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Re: What is the mean of the exams taken by a group of students? [#permalink]
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09 Feb 2015, 04:37
Bunuel wrote:
What is the mean of the exams taken by a group of students?
(1) The score 3 standard deviations above the mean is 95
(2) The score 2 standard deviations below the mean is 64
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:
The correct response is (C).
While this data sufficiency employs statistical terms, it is an algebra question that can be solved using the technique of simultaneous equations.
Statement (1) can be expressed as: Mean + 3 Standard Deviations = 95. Since this one equation has two unknowns it is insufficient to determine the mean.
Statement (2) can be expressed as: Mean – 2 Standard Deviations = 60. Since this one equation has two unknowns it is insufficient to determine the mean.
If you combine the two, you have two equations with two unknowns:
Mean + 3 Standard Deviations = 95
Mean – 2 Standard Deviations = 60 By subtracting the second equation from the first one, you get:
5 Standard Deviations = 35
By dividing both sides by 5, you get:
1 Standard Deviation = 7
By substituting this value into either equation you get the mean.
The mean is 74, making (C) the correct response.
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The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 − 1.25 × 104 K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes? - CBSE (Science) Class 12 - Chemistry
ConceptTemperature Dependence of the Rate of a Reaction
Question
The rate constant for the first order decomposition of H2O2 is given by the following equation:
log = 14.34 − 1.25 × 10K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Solution
Arrhenius equation is given by,
k = Ae^((-E_a)/"RT")
=>In k = In A - E_a/"RT"
=>Ink = lof A - E_a/"RT"
=> log k = log A - E_a/(2.303 RT) ... i
The given equation is
log k = 14.34 - 21.25 x 104 K/T ... ii
From equation (i) and (ii), we obtain
E_a/(2.303RT) = (1.25xx10^4 K)/T
=> E_a = 1.25 xx 10^4K xx 2.303 xxR
= 1.25 × 104 K × 2.303 × 8.314 J K−1 mol−1
= 239339.3 J mol1 (approximately)
= 239.34 kJ mol−1
Also, when t1/2 = 256 minutes,
k = 0.693/t_(1/2)
= 0.693/256
= 2.707 × 10−3 min−1
= 4.51 × 10−5 s−1
It is also given that, log k = 14.34 − 1.25 × 104 K/T
=>log(4.51xx10^(-5)) = 14.34 - (1.25xx10^(4)K)/T
=>log(0.654 - 0.5) = 14.34 - (1.25 xx 10^4K)/T
=> (1.25xx10^4K)/T = 18.686
=> T = (1.25xx10^4K)/18.686
= 668.95 K
= 669 K (approximately)
Is there an error in this question or solution?
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NCERT Solution for Chemistry Textbook for Class 12 (2018 to Current)
Chapter 4: Chemical Kinetics
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Solution The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 − 1.25 × 104 K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes? Concept: Temperature Dependence of the Rate of a Reaction.
S
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How do you write an equation of a line given m=-5, b=-1?
Jan 26, 2017
$y = - 5 x - 1$
Explanation:
The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the slope and b, the y-intercept.
$\text{here "m=-5" and } b = - 1$
$\Rightarrow y = - 5 x - 1 \text{ is the equation of the line}$
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# Tutor profile: Jess D.
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## Questions
### Subject:Trigonometry
TutorMe
Question:
In a right triangle ABC, sin(C)=3/5 and cos(C)=4/5. Find tan(C).
Inactive
Jess D.
A, B, and C represent angles of the right triangle. We are given the sine and cosine values for this triangle. SOH CAH TOA is a simple way to remember how to find sine, cosine, and tangent. This acronym stands for Sine=Opposite ÷ Hypotenuse, Cosine=Adjacent ÷ Hypotenuse, Tangent=Opposite÷Adjacent. Opposite, Adjacent, and Hypotenuse refer to the sides of the triangle (opposite is the side across from the angle being referenced, in this case angle C, hypotenuse is the side that is not connected to the right angle, and adjacent is the side touching the angle being reference and the right angle). With this information, we can quickly see that the given information tells us the length of each side of the triangle. From sin(C) we know opposite=3 and hypotenuse=5, and from cos(C) we know adjacent=4 and of course, the hypotenuse is still the same. To find tangent we need opposite ÷ adjacent, so the final answer is: tan(C) = 3/4
### Subject:Basic Math
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Question:
6x - 4y + 7 = 4x + 2y + 9, solve x.
Inactive
Jess D.
This problem is solved by combining terms. Since we are not given a value for y, the answer will have a y component. We begin combining terms with the x components on one side of the equal sign and all else on the other by adding or subtracting from each side, remember what we do to one side of the equal sign must be done to both. Let's start by combining x terms onto the left side of the equation by subtracting 4x from each side as follows: 6x - 4y + 7 = 4x + 2y + 9 -4x -4x We now have: 2x - 4y + 7 = 2y + 9 and we need only x terms on the left side, so we combine the like terms on the left side of the equation: 2x - 4y + 7 = 2y + 9 + 4y - 7 +4y - 7 Now we have, 2x = 6y + 2. Since x is multiplied by 2, we need to divide both sides of the equation by two to get our final answer. Since there are two terms on the right side of the equation, we will distribute the division and divide both terms by 2. 6y ÷ 2 = 3y and 2 ÷ 2 = 1, so our final answer is x = 3y + 1
### Subject:Physics
TutorMe
Question:
A new highway on-ramp is being designed. If the average acceleration of a car is 3 meters per second squared and the speed limit is 55 miles per hour, how long should the on-ramp be in order to allow cars sufficient distance to reach the speed limit before merging onto the highway.
Inactive
Jess D.
Since we are given two measurements (acceleration and speed) in different systems, we must first convert one value. 1 mph = 0.447 m/s, so 55 mph = 24.6 m/s. Now with the numbers in the same measurement system (SI), we can focus on the question. Assuming a car starts from 0 m/s, with an acceleration of 3 m/s^2 we can find the time it will take the car to reach the speed limit by using a kinematic equation: Vf = Vi + a*t, (Vf = final velocity, Vi = initial velocity, a = acceleration, t = time) Solving this equation for time, we have: t = (Vf - Vi) / a, which we can substitute our values into. For this problem, t = (24.6 m/s - 0 m/s) ÷ (3 m/s^2), so t = 8.2 s. Using another kinematic equation, we can solve this problem for the sufficient distance using given information and the solved time. d = (Vf + Vi) * t / 2 (d = distance, Vf = final velocity, Vi = initial velocity, t = time). For this problem, d = (24.6 m/s + 0 m/s) * (8.2 s) ÷ 2, so is d = 100.9 m. The final answer is, the on-ramp should be 100.9 m long in order to allow cars to reach the speed limit before merging onto the highway.
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# Metamath Proof Explorer
## Theorem cshwsiun
Description: The set of (different!) words resulting by cyclically shifting a given word is an indexed union. (Contributed by AV, 19-May-2018) (Revised by AV, 8-Jun-2018) (Proof shortened by AV, 8-Nov-2018)
Ref Expression
Hypothesis cshwrepswhash1.m ${⊢}{M}=\left\{{w}\in \mathrm{Word}{V}|\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right\}$
Assertion cshwsiun ${⊢}{W}\in \mathrm{Word}{V}\to {M}=\bigcup _{{n}\in \left(0..^\left|{W}\right|\right)}\left\{{W}\mathrm{cyclShift}{n}\right\}$
### Proof
Step Hyp Ref Expression
1 cshwrepswhash1.m ${⊢}{M}=\left\{{w}\in \mathrm{Word}{V}|\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right\}$
2 df-rab ${⊢}\left\{{w}\in \mathrm{Word}{V}|\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right\}=\left\{{w}|\left({w}\in \mathrm{Word}{V}\wedge \exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right)\right\}$
3 eqcom ${⊢}{W}\mathrm{cyclShift}{n}={w}↔{w}={W}\mathrm{cyclShift}{n}$
4 3 biimpi ${⊢}{W}\mathrm{cyclShift}{n}={w}\to {w}={W}\mathrm{cyclShift}{n}$
5 4 reximi ${⊢}\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\to \exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}={W}\mathrm{cyclShift}{n}$
6 5 adantl ${⊢}\left({w}\in \mathrm{Word}{V}\wedge \exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right)\to \exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}={W}\mathrm{cyclShift}{n}$
7 cshwcl ${⊢}{W}\in \mathrm{Word}{V}\to {W}\mathrm{cyclShift}{n}\in \mathrm{Word}{V}$
8 7 adantr ${⊢}\left({W}\in \mathrm{Word}{V}\wedge {n}\in \left(0..^\left|{W}\right|\right)\right)\to {W}\mathrm{cyclShift}{n}\in \mathrm{Word}{V}$
9 eleq1 ${⊢}{w}={W}\mathrm{cyclShift}{n}\to \left({w}\in \mathrm{Word}{V}↔{W}\mathrm{cyclShift}{n}\in \mathrm{Word}{V}\right)$
10 8 9 syl5ibrcom ${⊢}\left({W}\in \mathrm{Word}{V}\wedge {n}\in \left(0..^\left|{W}\right|\right)\right)\to \left({w}={W}\mathrm{cyclShift}{n}\to {w}\in \mathrm{Word}{V}\right)$
11 10 rexlimdva ${⊢}{W}\in \mathrm{Word}{V}\to \left(\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}={W}\mathrm{cyclShift}{n}\to {w}\in \mathrm{Word}{V}\right)$
12 eqcom ${⊢}{w}={W}\mathrm{cyclShift}{n}↔{W}\mathrm{cyclShift}{n}={w}$
13 12 biimpi ${⊢}{w}={W}\mathrm{cyclShift}{n}\to {W}\mathrm{cyclShift}{n}={w}$
14 13 reximi ${⊢}\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}={W}\mathrm{cyclShift}{n}\to \exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}$
15 11 14 jca2 ${⊢}{W}\in \mathrm{Word}{V}\to \left(\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}={W}\mathrm{cyclShift}{n}\to \left({w}\in \mathrm{Word}{V}\wedge \exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right)\right)$
16 6 15 impbid2 ${⊢}{W}\in \mathrm{Word}{V}\to \left(\left({w}\in \mathrm{Word}{V}\wedge \exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right)↔\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}={W}\mathrm{cyclShift}{n}\right)$
17 velsn ${⊢}{w}\in \left\{{W}\mathrm{cyclShift}{n}\right\}↔{w}={W}\mathrm{cyclShift}{n}$
18 17 bicomi ${⊢}{w}={W}\mathrm{cyclShift}{n}↔{w}\in \left\{{W}\mathrm{cyclShift}{n}\right\}$
19 18 a1i ${⊢}{W}\in \mathrm{Word}{V}\to \left({w}={W}\mathrm{cyclShift}{n}↔{w}\in \left\{{W}\mathrm{cyclShift}{n}\right\}\right)$
20 19 rexbidv ${⊢}{W}\in \mathrm{Word}{V}\to \left(\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}={W}\mathrm{cyclShift}{n}↔\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}\in \left\{{W}\mathrm{cyclShift}{n}\right\}\right)$
21 16 20 bitrd ${⊢}{W}\in \mathrm{Word}{V}\to \left(\left({w}\in \mathrm{Word}{V}\wedge \exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right)↔\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}\in \left\{{W}\mathrm{cyclShift}{n}\right\}\right)$
22 21 abbidv ${⊢}{W}\in \mathrm{Word}{V}\to \left\{{w}|\left({w}\in \mathrm{Word}{V}\wedge \exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right)\right\}=\left\{{w}|\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}\in \left\{{W}\mathrm{cyclShift}{n}\right\}\right\}$
23 2 22 syl5eq ${⊢}{W}\in \mathrm{Word}{V}\to \left\{{w}\in \mathrm{Word}{V}|\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{W}\mathrm{cyclShift}{n}={w}\right\}=\left\{{w}|\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}\in \left\{{W}\mathrm{cyclShift}{n}\right\}\right\}$
24 df-iun ${⊢}\bigcup _{{n}\in \left(0..^\left|{W}\right|\right)}\left\{{W}\mathrm{cyclShift}{n}\right\}=\left\{{w}|\exists {n}\in \left(0..^\left|{W}\right|\right)\phantom{\rule{.4em}{0ex}}{w}\in \left\{{W}\mathrm{cyclShift}{n}\right\}\right\}$
25 23 1 24 3eqtr4g ${⊢}{W}\in \mathrm{Word}{V}\to {M}=\bigcup _{{n}\in \left(0..^\left|{W}\right|\right)}\left\{{W}\mathrm{cyclShift}{n}\right\}$
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Centrifugal force occurs when inertia acts upon an object. As in the case of whirling a bucket of water on a string:
Bucket at rest:
\=============/
|=============|
|=============|
|=============|
|=============|
|=============|
|~~~~~~~~~~~~~|
|=============|
\_____________/
(Down is here)
(So is gravity)
Bucket starts moving:
\=============/
|=============|
|============~|
|=========~=~=|
|======~=~====|
|===~=~=======|
|~=~==========|
|=============|
\_____________/
(Down is over here now)
(Same with gravity)
Bucket attains full speed:
\=============/
|=============|
|=============|
|=============|
|=============|(Gravity)
|=============|
|~~~~~~~~~~~~~|
|=============|
\_____________/
(Down)
Centrifugal force relies a great deal upon one's point of view. To the spectator inside the bucket, there's a slight difference in weight, accompanied with some nausea by seeing everything outside the bucket twirling wildly.
However, to the person rotating the bucket, the water 'sticks' to the inside of the bucket, refusing to fall out.
The phenomenon relates, as I stated earlier, to inertia.
Take, for example, a car rounding a curve, and you're pushed against the wall. This happens because you and the car are seperate objects, both affected by inertia. Initally, both you and the car are moving in a straight-line path. As the car turns, you continue to move in a straight line, as per Newton's First Law.
This straight-line motion pushes you into the wall of the car, while the car is turning, producing what we call centrifugal force.
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# How to Calculate Drywall Finishing Costs
Written by nicole coe
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It is very important to estimate the costs involved to drywall finishing correctly before heading to the hardware store for your final purchase. Many do-it-yourself home renovators start out enthusiastically but end up having to postpone their projects after running out of funds. You can calculate drywall finishing costs by applying a few simple math equations.
Skill level:
Easy
### Things you need
• Price per drywall sheet
• Size of drywall sheets
• Price per screw
• Price per gallon of drywall mud
• Price per roll of drywall tape
• How many screws you plan to use per sheet
## Instructions
1. 1
Work out the surface size of a single sheet of drywall by multiplying its height with its length.
2. 2
Write down the height and width measurements of each wall space you are planning to drywall without subtracting windows or door openings.
3. 3
Work out the square footage of every wall separately by multiplying its height with its length.
4. 4
Take the total square footage of the surface and divide it by the square footage per sheet which has been calculated in Step 1. This will be the number of dry wall sheets you will need.
5. 5
Multiply the number of sheets with the cost per sheet and write the result down. This will be the total cost in sheets.
6. 6
Multiply the number of sheets with the number of screws you will need per sheet. This will give you the number of screws needed for the whole surface.
7. 7
Multiply the number of screws you will need with the price per screw. The outcome will give you the total cost for screws. Plan on using 300 screws for every 500 square feet of surface you will be covering.
8. 8
Work out how many gallons of drywall mud you need by dividing your square footage by 100. One gallon of drywall mud covers about 100 square feet. Multiply the outcome with your price per gallon.
9. 9
Work out how many rolls of drywall tape you need by measuring the length of all the seams that need covering and dividing it by 300 (assuming the rolls are 300 feet each). Multiply the outcome with your price per roll.
10. 10
Add all the separate material costs you have calculated together for your final estimation. If you are planning to hire labour, you will have to add that to your calculation as well to get an accurate idea of what the whole project will cost you.
#### Tips and warnings
• If your drywall screws are sold by the pound each pound has around 300 screws.
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# I solved the Smallest Common Multiple challenge, but it was too long?
I’m grateful that I solved the challenge, but I feel like I did it the long way. How can I improve my code?
``````function smallestCommons(arr) {
let ran = []
let sorted = arr.sort((a,b) => a - b)
for (let i = sorted[0]; i <= sorted[1]; i++) {
ran.push(i);
}
function primeFactors(n) {
const factors = [];
let divisor = 2;
while (n >= 2) {
if (n % divisor == 0) {
factors.push(divisor);
n = n / divisor;
} else {
divisor++;
}
}
return factors;
}
let factors = []
for (let i = 0; i < ran.length; i++) {
if (ran[i] == 1) {
factors.push([1])
} else {
factors.push(primeFactors(ran[i]))
}
}
const countOccurrences = (arr, val) => arr.reduce((a, v) => (v === val ? a + 1 : a), 0)
let final = []
let count = [];
for (let i = 1; i <= ran[ran.length - 1]; i++) {
for (let j = 0; j < factors.length; j++) {
if (countOccurrences(factors[j], i) > count) {
count = countOccurrences(factors[j], i)
}
}
final.push([i, count]);
count = 0;
}
let realFinal = []
for (let i = 0; i < final.length; i++) {
for (let j = 0; j < final[i][1]; j++) {
realFinal.push(final[i][0])
}
}
let superFinal = realFinal.reduce((a,b) => a * b, 1)
return superFinal;
}
``````
The divisors approach is an interesting one. I’d first compare it to the one here:
The biggest difference between solution 4 and yours is that they keep less intermediate values than you do.
This topic was automatically closed 182 days after the last reply. New replies are no longer allowed.
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A128333 a(0) = 0; for n > 0, a(n) = a(n-1)/2 if that number is an integer and not already in the sequence, otherwise a(n) = 3*a(n-1) + 1. 3
0, 1, 4, 2, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 25, 76, 38, 19, 58, 29, 88, 44, 133, 400, 200, 100, 50, 151, 454, 227, 682, 341, 1024, 512, 256, 128, 64, 32, 97, 292, 146, 73, 220, 110, 55, 166, 83, 250, 125, 376, 188, 94, 47, 142, 71, 214, 107, 322, 161 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Other than a(0) = 0, the sequence misses all multiples of 3. Does it eventually hit all positive non-multiples of 3? LINKS Nick Hobson, Table of n, a(n) for n = 0..10000 Nick Hobson, Python program EXAMPLE Consider n = 3. We have a(3) = 2 and try to divide by 2. The result, 1, is certainly an integer, but we cannot use it because 1 is already in the sequence. So we must multiply by 3 and add 1 instead, getting a(4) = 3*2 + 1 = 7. CROSSREFS Cf. A126038, A005132. Sequence in context: A261690 A211941 A050105 * A201414 A199814 A195347 Adjacent sequences: A128330 A128331 A128332 * A128334 A128335 A128336 KEYWORD easy,nonn AUTHOR Nick Hobson (nickh(AT)qbyte.org), Feb 27 2007 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified September 28 10:51 EDT 2020. Contains 337393 sequences. (Running on oeis4.)
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# If F is a function defined in the positive integers, such that F(k) is
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If F is a function defined in the positive integers, such that F(k) is [#permalink]
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10 Mar 2019, 08:09
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GMATH practice exercise (Quant Class 14)
If F is a function defined in the positive integers, such that F(k) is a positive integer for each positive integer k, what is the value of F(8)?
(1) F(n+1) = (n+1)*F(n), for every positive integer n.
(2) F(1)*F(1) = F(1)
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If F is a function defined in the positive integers, such that F(k) is [#permalink]
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10 Mar 2019, 11:51
Statement 1: $$F(7+1) = 8 * F(7)$$
Going backwards, such as $$F (6 +1) = 7 * F (6)$$ we notice that we aren't able to define F(1). $$F(1) = 1 * F(0)$$, whereas $$0$$ is a neutral number and $$F$$ is not defined. Discard this option.
Statement 2: we have $$F(1) = 1$$. Alone is insufficient, because it doesn't say anything else about how to compute F(8).
Since from statement 1 we lacked of $$F(1)$$ value, combined together these two statements lead to solution. Hence, C.
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Re: If F is a function defined in the positive integers, such that F(k) is [#permalink]
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10 Mar 2019, 16:20
1
fskilnik wrote:
GMATH practice exercise (Quant Class 14)
If F is a function defined in the positive integers, such that F(k) is a positive integer for each positive integer k, what is the value of F(8)?
(1) F(n+1) = (n+1)*F(n), for every positive integer n.
(2) F(1)*F(1) = F(1)
$$F\left( k \right) \ge 1\,\,{\mathop{\rm int}} \,\,\,\,{\rm{for}}\,\,{\rm{each}}\,\,\,k \ge 1\,\,\,\,\left( * \right)$$
$$? = F\left( 8 \right)$$
$$\left( 1 \right)\,\,F\left( {n + 1} \right) = \left( {n + 1} \right) \cdot F\left( n \right)\,\,\,{\rm{for}}\,{\rm{each}}\,\,\,n \ge 1\,\,{\mathop{\rm int}}$$
$$\left. {\matrix{ {F\left( 2 \right) = 2 \cdot F\left( 1 \right)\,\,} \hfill \cr {F\left( 3 \right) = 3 \cdot F\left( 2 \right) = 3 \cdot 2 \cdot F\left( 1 \right)} \hfill \cr {\,\,\, \vdots } \hfill \cr {? = F\left( 8 \right) = 8 \cdot 7 \cdot 6 \cdot \ldots \cdot 3 \cdot 2 \cdot F\left( 1 \right)\,\,\,} \hfill \cr } } \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ {\matrix{ {\,{\rm{Take}}\,\,F\left( 1 \right) = 1\,\,\,\, \Rightarrow \,\,\,? = 8!\,\,} \hfill \cr {\,{\rm{Take}}\,\,F\left( 1 \right) = 2\,\,\,\, \Rightarrow \,\,\,? = 2 \cdot 8!} \hfill \cr } } \right.$$
$$\left( 2 \right)\,\,F\left( 1 \right) \cdot F\left( 1 \right) = F\left( 1 \right)\,\,\,\,\mathop \Rightarrow \limits^{\,:\,\,F\left( 1 \right)\,\, \ne \,0\,\,\left( * \right)} \,\,\,F\left( 1 \right) = 1$$
$$\left\{ {\matrix{ {\,{\rm{Take}}\,\,F\left( n \right) = 1\,\,\,{\rm{for}}\,{\rm{each}}\,\,\,n \ge 1\,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,? = 1\,\,} \hfill \cr {\,{\rm{Take}}\,\,F\left( n \right) = \left( {n + 1} \right) \cdot F\left( n \right)\,\,\,{\rm{for}}\,{\rm{each}}\,\,\,n \ge 1\,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,? = 8!\,\,} \hfill \cr } } \right.$$
$$\left( {1 + 2} \right)\,\,\,\,? = 8!\,\,\,\,\, \Rightarrow \,\,\,\,\left( {\rm{C}} \right)$$
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Re: If F is a function defined in the positive integers, such that F(k) is [#permalink] 10 Mar 2019, 16:20
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## Algebra 2 (1st Edition)
Of the numbers between 1 and 50, there are 12 multiples of 4. Thus, we find that the probability is: $$=\frac{12}{50}=.24$$
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Home / Elementary Number Theory / If gcd(a, b) = d, then gcd(a/d, b/d) = 1
# If gcd(a, b) = d, then gcd(a/d, b/d) = 1
Proof: Before starting with the proof proper, we should observe that although a/d and b/d have the appearance of the functions, in fact, they are integers because d is a divisor both of a and of b. Now, knowing that gcd(a, b) = d, it is possible to find integers x and y such that d = ax + by. Upon dividing each side of this equation by d, we obtain the expression
1 = (a/d)x + (b/d)y
Because a/d and b/d are integers, an appeal to the theorem is legitimate. The conclusion is that a/d and b/d are relatively prime. (Proved)
## Application of Euclidian’s algorithm in Diophantine equation
Problem-1: Which of the following Diophantine equations cannot be solved – a) 6x + ...
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# hw3 - IE 495 Stochastic Programming Problem Sets#5#7 Due...
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IE 495 – Stochastic Programming Problem Sets #5—#7 Due Date: April 28, 2003 Do the following problems. If you work alone, you will receive a 10% bonus on your score. These are the final homework sets for the semester. Problems 1–2 will be Problem Set #5, Problems 3–4 will be Problem Set #6, and Problems 5–6 will be Problem Set #7. You are allowed to examine outside sources, but you must cite any references that you use. Please don’t discuss the problems with other members of the class (other than your partner, if you are working with one). 1 Feasibility Cuts The second stage constraints of a two-stage problem look as follows: 1 3 - 1 0 2 - 1 2 1 y = - 6 - 4 ω + 5 - 1 0 0 2 4 x y 0 . Here ω is a random variable with support Ω = [0 , 1]. 1.1 Problem Write down the linear programs (both primal and dual formulation(s)) needed to check whether or not there is a feasible second stage solution for a given x . 1.2 Problem Describe how these formulations allow you to obtain an inequality that cuts of x , if there is no feasible second stage solution y for that x . 1.3 Problem Let ˆ x = (1 , 1 , 1) T . Find the inequality explicitly for this first stage solution ˆ x .
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IE495 Problem Sets #5—#7 2 Bounds Consider our favorite random linear program. minimize Q ( x 1 , x 2 ) = x 1 + x 2 + 5 Z 4 ω 1 =1 Z 1 ω 2 =1 / 3 y 1 ( ω 1 , ω 2 ) + y 2 ( ω 1 , ω 2 ) 1 2 subject to ω 1 x 1 + x 2 + y 1 ( ω 1 , ω 2 ) 7 ω 1 , ω 2 Ω ω 2 x 1 + x 2 + y 2 ( ω 1 , ω 2 ) 4 ω 1 , ω 2 Ω x 1 0 x 2 0 y 1 ( ω 1 , ω 2 ) 0 ω 1 , ω 2 Ω y 2 ( ω 1 , ω 2 ) 0 ω 1 , ω 2 Ω Ω = { ω 1 × ω 2 } ω 1 ∼ U [1 , 4] ω 2 ∼ U [1 / 3 , 1] 2.1 Problem Compute the Jensen Lower Bound for Q (1 , 3) using the partition Ω = S 1 = { Ω } .
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1. ## Absolute Value Issues
I'm having some trouble working out some questions for absolute values:
|4y - 7| = |3y - 2|
1 / |y-2| is greater than or equal to 3
|5y - 2| = |3x + 4|
2. Originally Posted by Ogsta101
I'm having some trouble working out some questions for absolute values:
|4y - 7| = |3y - 2|
1 / |y-2| is greater than or equal to 3
|5y - 2| = |3x + 4|
Question 2
$\displaystyle \frac{1}{|y-2|}\geq 3$
Since the LHS must be positive, we can reciprocate both sides, reversing the inequality sign:
$\displaystyle |y-2| \leq \frac{1}{3}$
$\displaystyle \Rightarrow -\frac{1}{3}\leq y-2\leq \frac{1}{3}$
$\displaystyle \Rightarrow \frac{5}{3} \leq y \leq \frac{7}{3}$
3. Originally Posted by Ogsta101
I'm having some trouble working out some questions for absolute values:
|4y - 7| = |3y - 2|
bearing in mind the definition of absolute values (look this up of you don't know), we realize the solution will be given by:
$\displaystyle 4y - 7 = \pm (3y - 2)$
so we have $\displaystyle 4y - 7 = 3y - 2$
or
$\displaystyle 4y - 7 = 2 - 3y$
solve those for your two solutions.
|5y - 2| = |3x + 4|
what do you want to do here? you have two unknowns. which do you want to solve for...? is this a typo? in any case, try a similar method to what i did above
4. Just square both sides to get right rid of absolute values. Then see which solutions to the quadradic satisfy the original equation.
5. Originally Posted by Ogsta101
I'm having some trouble working out some questions for absolute values:
|4y - 7| = |3y - 2|
1 / |y-2| is greater than or equal to 3
|5y - 2| = |3x + 4|
Absolute value is when anything in between the || is going to end up in the positive state of a number. Your first question states :
|4y - 7| = |3y - 2|
so when you look at it, it should be like this:
4y + 7 = 3y + 2
4y-3y=-7+2
If you are supposed to solve the answer would be:
y= -5
6. Originally Posted by starrynight
Absolute value is when anything in between the || is going to end up in the positive state of a number. Your first question states :
|4y - 7| = |3y - 2|
so when you look at it, it should be like this:
4y + 7 = 3y + 2
4y-3y=-7+2
If you are supposed to solve the answer would be:
y= -5
you seem to be confused about absolute values. it does not mean you just change all the signs to positive. first of all, there are two answers, second of all, none of them is the answer you gave (plug it into the equation and you'll see it is wrong). do as i did or what TPH did (i prefer his method, and it is more standard than mine) to solve the problem. and look up the definition of what it means to take absolute values
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# Complex Power – Definition, Explanation & Significance
## Definition:
Complex Power is basically the representation of electrical power in the form of complex number. Like a complex number, it consists of real and imaginary part. Real part represents the active power whereas the imaginary part represents the reactive power. It is generally represented by symbol S.
If active and reactive power be P & Q respectively, then complex power for an inductive load is written as
S = P + jQ
Complex power for a capacitive load is given as below.
S = P – jQ
## Explanation:
To better understand, let us consider the voltage and current of a load to be V and I respectively. Assume the load to be capacitive. Mind that load is not purely capacitive rather capacitance is dominating in load. Therefore, the current will lead the voltage by some angle ø.
Let voltage across the load V = Ve and I = Iej(Ɵ+ø). Now, Complex Power S can be find by multiplying the voltage (V) with the conjugate of current (I*).
S = VI*
= Vex Ie-j(Ɵ+ø)
= Vie-jø
= VI (cosø – jsinø)
= VIcosø – VIsinø ………(1)
As we know that, active power (P) and reactive power (Q) is given as
P = VIcosø
Q = VIsinø
Therefore, complex power can be written form (1) as shown below.
S = P – jQ
Carefully observe the above expression. The imaginary part is negative. This means, reactive power is negative. As load was assumed capacitive, the reactive power is due to the capacitance in the load. Since reactive power is negative, this implies that a capacitor is a generator of reactive power. It does not consume reactive power in a circuit rather it produces reactive power.
Complex power S may also be found in the similar way for inductive load. Let voltage across the load V = Ve and I = Iej(Ɵ-ø). Now, the S can be found by multiplying the voltage (V) with the conjugate of current (I*).
S = VI*
= Vex Ie-j(Ɵ-ø)
= Vie
= VI (cosø + jsinø)
= VIcosø + VIsinø ………(1)
As we know that, active power (P) and reactive power (Q) is given as
P = VIcosø
Q = VIsinø
Therefore, complex power can be written form (1) as shown below.
S = P + jQ
The above expression reveals that reactive power is positive. This simply means that reactive power is being consumed by an inductor.
## Significance of Complex Power:
• Power triangle can easily be constructed if the value of complex power is known. Q will represent the perpendicular, P the base and S is the hypotenuse of power triangle.
• Real part gives the value active power whereas the imaginary part gives the value of reactive power.
• Magnitude of S gives the value of Apparent Power. Thus, apparent power = √(P2+Q2).
1. very good
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https://www.usablestats.com/askstats/question/817/
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## Question 817:
1) Assuming in New York 80% of women and 50% of men know where Tiffany is. At a crossing I ask pedestrians for directions to Tiffany and get these answers:
2 women say North
1 woman says West
2 men say West
1 man says South
What are the probabilities for Tiffany being North respectively West and South?
2) Assuming everyone I ask says East, how does the probability change with the sample size? In other words, how is the probability different if 50 out 50 women say East instead of 5 out of 5 women?
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https://depsangtrong.com/3cos-4x-4sin-2xcos-2x-sin-4x-0/
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# SOLVE SIN4X
Those are all the same answer, since 72k is the same as 36(2k), and 360k is the same as 36(10k). (Presumably k in mathbb Z here.) So you got the same answer as the book, just redundantly.
Bạn đang xem: Solve sin4x
https://math.stackexchange.com/questions/2975004/how-to-solve-this-trigonometric-equation-sin3x-sin5x-sin6x-sin2x
Hint: use the formula sin x+sin y = 2sin x+yover 2cdot cosx-yover 2 & cos x-cos y = -2sin x+yover 2cdot sinx-yover 2
https://www.quora.com/How-would-I-find-the-number-of-solutions-to-sin-3x-4-sin-x-sin-2x-sin-4x-in-0-lt-x-lt-2-pi
*A2A :- implies sin3x=4sin xsin2xsin4x star Using product to sum formula we get :- implies sin3x=2sin4xleft(cos x-cos3x ight) implies sin3x=2sin4xcos x-2sin4xsin3x ...
Xem thêm: Một Người Dự Định Đi Từ A Đến B Trong Thời Gian 4 Giờ, Just A Moment
https://math.stackexchange.com/questions/2544864/prove-the-continuity-of-x-sinx-using-epsilon-delta-method
Forget about 2npi and stuff. The essential point is that |sin x-sin y|leq |x-y| for arbitrary x, yinmathbb R. Let an xinmathbb R be given and consider an increment h with |h|leq1 ...
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x^2-cos(x) is increasing and convex on <0,2pi>, hence if you start Newton's iteration at the right of the only root you get a decreasing sequence converging lớn your root (with the chance lớn ...
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Post subject: Kinematics: 1DPosted: Sun Sep 22, 2013 6:57 pm
Joined: Fri Feb 20, 2009 6:38 pm
Silly, cocky me. I thought I had my 1D kinematics down from Grade 10. Come U1 Physics (remarkably boring, because I was a master at projectile motion), I decide to brush up. But apparently, I've forgotten a key point and I would like to check it out.
The situation is as follows:
Person 1 travels at 11.5 m/s and is 11.8 meters ahead of person 2. Person 2 travels at 9.7 m/s and accelerates at 1.2 m/s. I'm attempting to find the amount of time it takes for the second person to catch up
By my reasoning, the second person travels the same distance as the first person plus that 11.8m, giving d1=d and d2=d+11.8. They both take the same amount of time to cover their respective distances.
Is my reasoning correct? My solution depends on it...
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Post subject: Re: Kinematics: 1DPosted: Sun Sep 22, 2013 8:40 pm
Joined: Thu Oct 04, 2012 8:49 pm
I believe it goes as such:
Person 1; x = v0 t + x0
Person 2; x = a t^2 + v0 t + x0
Set them equal to each other and solve for t.
x0 is initial position
v0 is initial velocity
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Post subject: Re: Kinematics: 1DPosted: Mon Sep 23, 2013 12:52 pm
Joined: Sat Mar 26, 2011 9:56 am
quicksolver wrote:
By my reasoning, the second person travels the same distance as the first person plus that 11.8m, giving d1=d and d2=d+11.8. They both take the same amount of time to cover their respective distances.
Is my reasoning correct? My solution depends on it...
Your reasoning sounds correct to me.
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# 2nd Grade Common Core: 2.OA.2
Common Core Identifier: 2.OA.2 / Grade: 2
Curriculum: Operations And Algebraic Thinking: Add And Subtract Within 20.
Detail: Fluently add and subtract within 20 using mental strategies.2 By end of Grade 2, know from memory all sums of two one-digit numbers.
177 Common Core State Standards (CCSS) aligned worksheets found:
Solve the basic addition problems. Each problem has addends up to 9 and sums up to 18. Then color the summer scene according to the key.
Level:
Write the sum for the basic addition facts, then color the stocking according to the key.
Level:
Looking for more than 25 facts? Try this worksheet. It has 50!
Level:
Level:
When they solve and color the picture, students will see a colorful sailboat.
Level:
Subtraction mystery picture of an Easter bunny with a basket of eggs.
Level:
Write the correct addition/subtraction fact family for each set of numbers.
Level:
Color a king and queen's castle. Solve the basic adding facts and color the picture.
Level:
This fun chain reaction game that teaches basic addition facts.
Level:
Solve 43 basic subtraction problems as quickly as possible. All numbers are 20 or less. Differences are ten or less.
Level:
Students use the number line to help them subtract.
Level:
Color the blue whale mystery picture after solving the basic addition facts.
Level:
Logical thinking: Write the missing numbers to make each scale balance.
Level:
Look for pairs of adjacent numbers that have a sum of 10. These game boards are shaped like a fall leaves.
Level:
Write the answers to the addition math facts and color according to the key.
Level:
Another set of add-subtract number bonds.
Level:
A colorful hot air balloon will be revealed when you answer the math facts and color according to the key.
Level:
Write the sums for the basic addition facts, then color according to the key. The mystery picture is a reindeer.
Level:
Solve the simple one-digit addition problems, and color the hot air balloon picture according to the code.
Level:
Find the difference for each basic subtraction fact. Then color according to the key to reveal a brown owl. All subtraction problems have differences within 20.
Level:
This worksheet has four very basic addition and subtraction word problems with an Easter theme.
Level:
More fact family subtraction-addition fact family triangles to solve.
Level:
Solve the subtraction problems and color-by-number to reveal a picture of a colorful parrot.
Level:
Write the sums for the basic addition facts, then color according to the key.
Level:
Solve the facts and color this adorable brown dog with black spots.
Level:
Find the sum for each addition fact. Then color according to the key. The mystery picture is a colorful dragon.
Level:
Write the answers to the subtraction problems and color the picture according to the key.
Level:
Add or subtract to find the missing part of each number bond.
Level:
Solve the basic subtraction problems and color-by-number to reveal a picture of a rooster.
Level:
Solve to find the answers to the basic addition facts. Then color according to the key to reveal an Earth Day picture.
Level:
After answering the addition facts, color the sailboat according to the key.
Level:
Students use the number line picture to help them add.
Level:
Use the number line to add or subtract.
Level:
Solve the basic subtraction problems and color by number to reveal a picture of a colorful butterfly.
Level:
Determine which subtraction problems are being shown by each of the number lines.
Level:
Solve the basic subtraction problems (all minuends are 18 or less; all subtrahends are exactly 9). Then color the mystery scene according to the key. Scene will show a beach ball and sand pail.
Level:
Write the difference for each of the basic subtraction facts, then color according to the key. Christmas stocking picture.
Level:
Find the sum for each problem, then color the mystery picture according to the code.
Level:
Print out the game board. Then color pairs of adjacent numbers that add up to 17. The player with the most colored shapes wins.
Level:
Basic addition math drill sheet with addends 0 through 9. Includes Martin Luther King graphics.
Level:
A complete set of basic subtraction flashcards up to 18 - 9. This file has 4 large cards per page.
Level:
Subtraction fact blast off! Solve the facts and color the rocket.
Level:
Addition mystery picture of an Easter bunny with a basket of eggs.
Level:
Another worksheet with fact family houses in which students much write the correct numbers.
Level:
Find the sums and color; Picture makes a colorful clownfish.
Level:
Here's a complete set of triangular fact family flashcards to teach subtraction and addition facts.
Level:
This worksheet has 43 basic subtraction facts. How quickly can you students solve them? All numbers are 18 or less. Differences are all equal to or less than nine.
Level:
Find the answer to each basic subtraction fact, then color to see the blue and gray whale.
Level:
Use the number lines to determine the missing quantity in each subtraction fact. Use the "count back" technique.
Level:
Write the answers to basic subtraction problems and color the Thanksgiving turkey according to the key.
Level:
Write the sum for each addition problem, then color the parrot according to the key.
Level:
Jim loves reading and Meg loves math, so they find a way to help each other.
Level:
Write the answers to the subtraction problems and color the picture according to the key.
Level:
Level:
Write the answers to the subtraction problems, then color the mystery picture of a clownfish.
Level:
Write the answers to the subtraction facts and color the royal castle.
Level:
Write the answer for each basic subtraction facts, then color according to the key.
Level:
Solve the problems to reveal fruits and vegetables: a red apple, a yellow banana, green guava fruit, and an orange carrot.
Level:
Find the difference for each subtraction fact. Then color according to the key to reveal a colorful Chinese New Year picture.
Level:
Solve the basic addition problems with sums up to 18. This worksheet is decorated with Hanukkah-related pictures.
Level:
Match each subtraction with with the answer in this fun card game.
Level:
Find the differences and color the dog picture according to the color code.
Level:
Solve to find the answers to the basic subtraction facts. Then color according to the key to reveal a picture of the planet Earth.
Level:
Write the correct numbers in each addition/subtraction fact family triangle.
Level:
Answer the addition facts and color the rocket picture according to the code at the bottom.
Level:
Tell which addition problems are shown by each number line.
Level:
Solve all of the mixed addition and subtraction facts as quickly as possible. Includes sums up to and including 20 and differences up to and including 10.
Level:
Subtract the numbers to find the differences, then color according to the code at the bottom.
Level:
Use the number line to determine the missing number in each subtraction fact.
Level:
Write the answers to the addition math facts and color the Thanksgiving turkey according to the key.
Level:
Solve the addition problems, the color-by-number according to the key. Picture makes a colorful rooster.
Level:
Solve the basic addition problems. The color according to the key at the bottom of the page. All addition problems have addends between 1 and 12.
Level:
Write the answers to the addition math facts and color according to the key.
Level:
Use addition or subtraction to determine the missing number in each part-part-whole box.
Level:
Color and solve these 20 math problems. Addends are 10 or less. Sums are 20 or less. (examples: 6+9=15 and 7+7=14)
Illustrations of kids on a bike and a skateboard are featured on this worksheet.
Level:
Depending on which worksheet from this file you give your students, they will color either a dog or a cat.
Level:
Solve the subtraction facts and color according to the key to reveal a fox!
Level:
Print out a subtraction bingo board for each student in your class. The teacher calls out subtraction facts and the class marks the answers on their boards.
Level:
Find the differences. Then color according to the key to reveal a Halloween scene.
Level:
To the colorful picture picture of a ladybug, students will need to find the differences between pairs of numbers.
Level:
Solve the basic math facts on this page and then color by number to reveal a colorful sailboat.
Level:
Use scissors to cut these 36 cards out along the dotted lines. Match the basic addition problem to the matching domino card.
Level:
Solve the addition problems and color according to the key to reveal the mystery pictures.
Level:
This printable worksheet has basic addition problems to solve. Next, use the sums to determine what color to use for each space. An airplane will appear in the finished picture.
Level:
Download and print flash cards for all basic subtraction facts with minuends to 20 and answers to 10. There are 9 cards per page.
Level:
These horizontal number bonds have the sums on the right side.
Level:
Add and subtract to find the answers to the math problems. Then color each section of the dragon picture according to the key near the bottom of the worksheet.
Level:
Cut out ten number tiles at the bottom of the page. Glue each number next to the correct subtraction fact.
Level:
This printable subtraction worksheet has twenty horizontal subtraction facts to be solved with minuends up to 20.
Level:
Solve the addition problems and color according to the key to reveal two different Thanksgiving-themed mystery pictures.
Level:
Help the rocket find the star. Along the way, answer the subtraction facts with minuends to 20 and answers to 10.
Level:
Solve the addition problems and color the worksheet based on the key. Depending on which you complete, you will create either a police car or a fire truck.
Level:
Print out these subtraction puzzle pieces for a fun way to practice basic subtraction facts.
Level:
Cut out the pieces to these 6
star-shaped puzzles. Then place the problems with the same sum to make stars.
Level:
Students will solve the basic addition problems and color in the spaces to reveal school supplies and an apple on a desk.
Level:
This page includes 6 number bonds, each with one number missing. Students will use addition or subtraction to find and fill in that missing number.
Level:
After solving the basic subtraction facts, color each part of the Santa picture as indicated by the key at the bottom.
Level:
Find and draw the missing counters and then complete the subtraction fact.
Level:
It's the perfect winter's day! Children will color and cut out the pages of the mini book and staple them together. They can read the book aloud to a teacher, a family member, or a friend.
Level:
Have your students complete this maze and solve the addition problem by filling in the number of pumpkins the good witch passes on her way through the maze and adding them together.
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In this math card game, students take turns matching each subtraction fact with its answer.
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Solve the basic addition problems and color. The finished picture reveals a scene with a jack-o-lantern, a cat, a bat, and a witch.
Level:
This worksheet has 4 number bonds for students to complete. Below each one there is space for them to fill in the complete equation expressed by the diagram.
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Cut out these 36 cards. Flip all the cards face down. Have students take turns flipping the cards to make a match. Example: the "3 + 5" card will match the "8" card.
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Will your worksheet reveal a tasty summer treat or classic beach toys? It depends which of the two worksheets in this file you complete.
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After writing the differences for the problems on the page, color the sea monster (dragon) according to the color code at the bottom of the picture.
Level:
Play Bingo and learn basic subtraction facts at the same time! This printable includes 42 calling cards with subtraction facts on them and a class set (30) of bingo boards.
Level:
This is a set of flashcards up to 10 + 10. This set has 9 cards per page.
Level:
These number bonds are laid out horizontally with the sums on the left.
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Each problem has a sum of 16 or less. After writing the answers, color the dragon picture according to the key.
Level:
This printable addition worksheet has twenty addition problems to be solved. The addition facts on this worksheet are listed horizontally with addends up to 10 and sums up to 20.
Level:
Solve the basic math facts on this worksheet and color each section according to the key at the bottom. The completed worksheet will reveal a rocket ship in space.
Level:
Practice basic addition skills and color in two fun mystery pictures with this file.
Level:
Use this printable math activity to help children practice basic subtraction facts with minuends up to 20.
Level:
Complete the addition problems and color the page according to the key. The two worksheets have the same pattern, but when colored in, each will reveal a different picture.
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Color the picture by solving the basic addition problems to find the friendly robot!
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This worksheet has 20 addition problems for students to solve. Addends are 10 or less and sums are up to 20.
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Solve and color in this mystery picture according to the key to reveal a sunny beach with colorful seashells.
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First, write the answers to the addition facts. (Facts have addends up to 10.) Then follow the code at the bottom to color the spaces. The finished product is a portrait of Santa Claus.
Level:
Draw the counters in the space to provided to help form the number sentence for the subtraction problems.
Level:
Have your students solve these basic subtraction facts on this printable patriotic-themed worksheet.
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Students will complete the basic addition problems and use the answers and their corresponding colors to complete this worksheet.
Level:
Write the answers to twenty addition facts and color the pictures, if you wish. Addends are 10 or less. Sums are up to 20. (examples: 6+10=16 and 9+9=18) This worksheet features illustrations of kids holding seashells.
Level:
Find the sums and color the shapes accordingly to reveal a picture of two lanterns.
Level:
When students play this game, they will take turns rolling a die. The number they roll corresponds with a column on the game board. They must read and answer all subtraction facts in the column as quickly as they can.
Level:
Complete the number bonds and write the equation represented by the number bond in the blank spaces.
Level:
First write the answers to the basic subtraction facts. Then color each section of the picture according to the key at the bottom. The finish picture shows a groundhog popping out of his burrow in springtime.
Level:
Have your students solve the subtraction word problem by drawing a picture to help find the answer.
Level:
Cut apart the stars into 30 pieces. Have your students combine matching sums and numbers together to form stars.
Level:
On their way through this maze, your students will fill in the number of pumpkins their path passes. At the end, they will add up all the pumpkins to find the total.
Level:
Print out this worksheet to practice basic subtraction facts with minuends up to 20.
Level:
Which worksheet they complete will determine whether your students color in an apple or a banana.
Level:
Color the robot picture by answering the subtraction facts and using the key.
Level:
Use the number line to determine the addition problems.
Level:
Your students will solve the basic addition facts and then color each section according to the color key at the bottom.
Level:
Write the answers to the basic addition problems (with addends up to 10). Then color each part of the tree according to the key at the bottom of the page.
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Students will plot the given points and reveal the mystery picture: a Valentine's Day heart.
Level:
Use the dominoes to make and solve these basic addition problems.
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Find the sums and color accordingly to reveal the mystery image.
Level:
Print out this worksheet to practice basic addition problems with sums up to 20.
Level:
Practice basic addition and create a fun coloring picture with this Easter-themed mystery picture.
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When students solve basic addition problems, they'll be able to decode a picture of a mother bird sitting on her eggs.
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Some of these subtraction facts have correct answers and some are incorrect. Circle the correct answers and cross out the incorrect ones.
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This printable worksheet features twenty basic horizontal addition and subtraction problems.
Level:
Each number bond on this printout has two sets of objects. Students will count the objects to fill in the number bonds.
Level:
Solve the basic math problems in the picture. Use the answers to find what color each space should be and reveal a beach scene.
Level:
This file contains three worksheets. They may look like they have the same design, but when colored in according to the addition problems, each will reveal a unique Easter picture.
Level:
Students can hunt for basic subtraction facts with this printable number puzzle. Find and circle all 20 subtraction facts.
Level:
Use scissors to cut along the dotted lines and separate all of the puzzle pieces. On the attached page, there are several different ways for your students to play and learn with these puzzles.
Level:
Have your students solve these very basic subtraction problems. Then they will use the color key to color in the puzzle to reveal sea shells on a beach.
Level:
Write the answers to the subtraction facts on the caterpillar. All facts have minuends of 20 or less. (example: 15-5)
Level:
This is a 2-player game for students to compete and practice their addition facts with each other.
Level:
Your students will have fun solving the subtraction problems and coloring in the mystery picture to reveal school supplies on a desk.
Level:
Here are 6 more number bonds for students to complete using addition and subtraction.
Level:
Write the sum of each pair of numbers. Then color according to the key at the bottom. The finished picture shows a groundhog emerging from his hole.
Level:
This worksheet uses pictures of eggs to help students solve the subtraction problems and then write the number sentence to go with it.
Level:
Students love playing our Scoot! games. There are 30 task cards, answer sheets for 20, 25, and 30 questions, and teacher instructions. This version of Scoot! has simple math addition facts to be solved, ideal for students in grades K - 2.
Level:
This is our most difficult Halloween addition maze, for those especially talented at navigating all the twists and turns. They'll have to add up 7 numbers to find the sum of all the pumpkins they pass.
Level:
Download this basic subtraction activity and practice subtraction facts with minuends up to 20.
Level:
Will this mystery picture be a boat or a plane? It depends which worksheet you give your students!
Level:
Using the sums from basic addition facts, color according to the key to find a cute fox.
Level:
This printable worksheet provides number lines to help students solve the addition problems.
Level:
Solve all of the subtraction problems and color in the picture according to the color key on the bottom of the worksheet.
Level:
After writing the answers to basic subtraction problems, students color the Christmas tree according to the color code at the bottom of the page.
Level:
Complete the picture models to help find the answer to the subtraction problems. Then write a subtraction fact for each.
Level:
This worksheet has 20 subtraction facts to solve with minuends up to 20.
Level:
Find the answers to the subtraction problems and color according to the key to reveal the patriotic mystery picture.
Level:
Each of these facts has addends up to 10 and sums of 20 or less. Illustrations show an kids outside in the rain.
Level:
Students can solve these basic subtraction problems and then color the picture accordingly to reveal a fun Easter design.
Level:
This subtraction exercise helps students practice solving problems with differences from 0 to 10. The final picture is a bird sitting on her nest.
Level:
Print this worksheet to have your class practice their addition and subtraction facts. Minuends and addends are all 20 or less.
Level:
This printable math worksheet is another version of mixed addition and subtraction math facts laid out horizontally.
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Draw the correct number of dots to represent the missing number in each number bond.
Level:
This printable has a set of 30 bingo boards and 42 calling cards with basic addition facts to call out.
Level:
This file contains two mystery picture worksheets. They look very similar but depending on which one you do, the picture will be different.
Level:
Give values of the digits in three-digit number. Subtract to find out how many tulips are in a garden. Color the odd and even tiled numbers.
Level:
Find the missing numbers. Geometry and time challenges.
Level:
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Relevant and Irrelevant Information Lesson Topic Index | Grade 4 Math | Elementary Test Prep | StudyZone
relevant- information that is important irrelevant- information that is not needed
I own 3 dogs. Their names are Riley, Shadow, and Tapper. Riley is 1 years old. Shadow and Tapper are the same age. When you add all 3 dogs ages it equals 7. Riley is a black lab. Shadow and Tapper are little dogs. How old are my 3 dogs?
Let's list all of the information, and classify it as either relevant or irrelevant.
I own 3 dogs. relevant Their names are Riley, Shadow, and Tapper. irrelevant Riley is 1 years old. relevant Shadow and Tapper are the same age. relevant When you add all 3 dogs ages it equals 7. relevant Riley is a black lab. irrelevant Shadow and Tapper are little dogs. irrelevant How old are my 3 dogs? relevant
Once we see what information is important (relevant) it's easy to see how to solve the problem.
Riley is 1
Tapper and Shadow are the same age.
1 + n + n = 7
1 + 3 + 3 = 7
The dogs are 1, 3, and 3.
Let's Practice!
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Definitions
Nearby Words
# curvature
[kur-vuh-cher, -choor]
Measure of the rate of change of direction of a curved line or surface at any point. In general, it is the reciprocal of the radius of the circle or sphere of best fit to the curve or surface at that point. This notion of best fit derives from the principle that only one circle can be drawn though any three points not on the same line. The radius of curvature at the middle point is approximately equal to the radius of that one circle. This calculation becomes more exact the closer the points are. The precise value is found using a limit. Because a straight line can be thought of as an arc of a circle of infinite radius, its curvature is zero.
In mathematics, curvature refers to any of a number of loosely related concepts in different areas of geometry. Intuitively, curvature is the amount by which a geometric object deviates from being flat, but this is defined in different ways depending on the context. There is a key distinction between extrinsic curvature, which is defined for objects embedded in another space (usually a Euclidean space) in a way that relates to the radius of curvature of circles that touch the object, and intrinsic curvature, which is defined at each point in a differential manifold. This article deals primarily with the first concept.
The primordial example of extrinsic curvature is that of a circle, which has curvature equal to the inverse of its radius everywhere. Smaller circles bend more sharply, and hence have higher curvature. The curvature of a smooth curve is defined as the curvature of its osculating circle at each point.
In a plane, this is a scalar quantity, but in three or more dimensions it is described by a curvature vector that takes into account the direction of the bend as well as its sharpness. The curvature of more complex objects (such as surfaces or even curved n-dimensional spaces) is described by more complex objects from linear algebra, such as the general Riemann curvature tensor.
The remainder of this article discusses, from a mathematical perspective, some geometric examples of curvature: the curvature of a curve embedded in a plane and the curvature of a surface in Euclidean space. See the links below for further reading.
## One dimension in two dimensions: Curvature of plane curves
For a plane curve C, the mathematical definition of curvature uses a parametric representation of C with respect to the arc length parametrization. It can be computed given any regular parametrization by a more complicated formula given below. Let γ(s) be a regular parametric curve, where s is the arc length, or natural parameter. This determines the unit tangent vector T, the unit normal vector N, the curvature κ(s), the oriented or signed curvature k(s), and the radius of curvature at each point:
$T\left(s\right)=gamma\text{'}\left(s\right),quad T\text{'}\left(s\right)=k\left(s\right)N\left(s\right),quad kappa\left(s\right) = |gamma\left(s\right)| = left|k\left(s\right)right|, quad R\left(s\right)=frac\left\{1\right\}\left\{kappa\left(s\right)\right\}.$
The curvature of a straight line is identically zero. The curvature of a circle of radius R is constant, i.e. it does not depend on the point and is equal to the reciprocal of the radius:
$kappa = frac\left\{1\right\}\left\{R\right\}.$
Thus for a circle, the radius of curvature is simply its radius. Straight lines and circles are the only plane curves whose curvature is constant. Given any curve C and a point P on it where the curvature is non-zero, there is a unique circle which most closely approximates the curve near P, the osculating circle at P. The radius of the osculating circle is the radius of curvature of C at this point.
### The meaning of curvature
Suppose that a particle moves on the plane with unit speed. Then the trajectory of the particle will trace out a curve C in the plane. Moreover, taking the time as the parameter, this provides a natural parametrization for C. The instanteneous direction of motion is given by the unit tangent vector T and the curvature measures how fast this vector rotates. If a curve keeps close to the same direction, the unit tangent vector changes very little and the curvature is small; where the curve undergoes a tight turn, the curvature is large.
The magnitude of curvature at points on physical curves can be measured in diopters (also spelled dioptre) — this is the convention in optics. A diopter has the dimension $\left\{mathit\left\{Length\right\}^\left\{-1\right\}\right\}.$
### Signed curvature
The signed of the signed curvature k indicates the direction in which the unit tangent vector rotates as a function of the parameter along the curve. If the unit tangent rotates counterclockwise, then k > 0. If it rotates clockwise, then k < 0.
The signed curvature depends on the particular parametrization chosen for a curve. For example the unit circle can be parametrised by $\left(cos\left(theta\right),sin\left(theta\right)\right)$ (counterclockwise, with k > 0), or by $\left(cos\left(-theta\right),sin\left(-theta\right)\right)$ (clockwise, with k < 0). More precisely, the signed curvature depends only on the choice of orientation of an immersed curve. Every immersed curve in the plane admits two possible orientations.
### Local expressions
For a plane curve given parametrically as $c\left(t\right) = \left(x\left(t\right),y\left(t\right)\right)$, the curvature is
$kappa = frac$-y'x
{(x'^2+y'^2)^{3/2}},
and the signed curvature k is
$k = frac\left\{x\text{'}y$-y'x}{(x'^2+y'^2)^{3/2}}.
For the less general case of a plane curve given explicitly as $y=f\left(x\right)$ the curvature is
$kappa = frac$
{(1+y'^2)^{3/2}}.>
Slightly abusing notation, the signed curvature may also be written in this way as
$k=frac\left\{y\right\}\left\{\left(1+y\text{'}^2\right)^\left\{3/2\right\}\right\}$
with the understanding that the curve is traversed in the direction of increasing x.
This quantity is common in physics and engineering; for example, in the equations of bending in beams, the 1D vibration of a tense string, approximations to the fluid flow around surfaces (in aeronautics), and the free surface boundary conditions in ocean waves. In such applications, the assumption is almost always made that the slope is small compared with unity, so that the approximation:
$kappa approx left|frac\left\{d^2y\right\}\left\{dx^2\right\}right|$
may be used. This approximation yields a straightforward linear equation describing the phenomenon, which would otherwise remain intractable.
If a curve is defined in polar coordinates as $r\left(theta\right)$, then its curvature is
$kappa\left(theta\right) = frac$
{left(r^2+r'^2 right)^{3/2}}>
where here the prime refers to differentiation with respect to $theta$.
### Example
Consider the parabola $y = x^2$. We can parametrize the curve simply as $c\left(t\right) = \left(t, t^2\right) = \left(x, y\right)$,
$dot\left\{x\right\}= 1,quadddot\left\{x\right\}=0,quad dot\left\{y\right\}= 2t,quadddot\left\{y\right\}=2$
Substituting
$kappa\left(t\right)= left|frac\left\{dot\left\{x\right\}ddot\left\{y\right\}-dot\left\{y\right\}ddot\left\{x\right\}\right\}\left\{\left(\left\{dot\left\{x\right\}^2+dot\left\{y\right\}^2\right)\right\}^\left\{3/2\right\}\right\}right|= \left\{1cdot 2-\left(2t\right)\left(0\right) over \left(1+\left(2t\right)^2\right)^\left\{3/2\right\} \right\}=\left\{2 over \left(1+4t^2\right)^\left\{3/2\right\}\right\}$
## One dimension in three dimensions: Curvature of space curves
See Frenet-Serret formulas for a fuller treatment of curvature and the related concept of torsion.
For a parametrically defined space curve as $c\left(t\right) = \left(x\left(t\right),y\left(t\right),z\left(t\right)\right),$, its curvature is:
$F\left[x,y,z\right]=frac\left\{sqrt\left\{\left(z$y'-yz')^2+(xz'-zx')^2+(yx'-xy')^2}}{(x'^2+y'^2+z'^2)^{3/2}}
Given a function r(t) with values in R3, the curvature at a given value of $t$ is
$kappa = frac$
>{|dot{r}|^3}
where $dot\left\{r\right\}$ and $ddot\left\{r\right\}$ correspond to the first and second derivatives of r(t), respectively. (Note that this formula is the vector notation of F[x,y,z] above.)
## Curves on surfaces
When a one dimensional curve lines on a two dimensional surface embedded in three dimensions R3, further measures of curvature are available, which take the surface's unit-normal vector, u into account. These are the normal curvature, geodesic curvature and geodesic torsion. Any non-singular curve on a smooth surface will have its tangent vector T lying in the tangent plane of the surface orthogonal to the normal vector. The normal curvature, kn, is the curvature of the curve projected onto the plane containing the curves tangent T and the surface normal u; the geodesic curvature, kg, is the curvature of the curve projected onto the surfaces tangent plane; and the geodesic torsion (or relative torsion), τr, measures the rate of change of the surface normal around the curves tangent.
Let the curve be a unit speed curve and let t = u × T so that T, u, t form an orthonormal basis: the Darboux frame. The quantities k, g and τ are related by:
$begin\left\{pmatrix\right\}$
mathbf{T'} mathbf{t'} mathbf{u'} end{pmatrix} = begin{pmatrix} 0&kappa_g&k_n -kappa_g&0&tau_r -kappa_n&-tau_r&0 end{pmatrix} begin{pmatrix} mathbf{T} mathbf{t} mathbf{u} end{pmatrix}.
### Principal curvature
All curves with the same tangent vector will have the same normal curvature, which is the same as the curvature of the curve obtained by intersecting the surface with the plane containing T and u. Taking all possible tangent vectors then the maximum and minimum values of the normal curvature at a point are called the principal curvatures, k1 and k2, and the directions of the corresponding tangent vectors are called principal directions.
## Two dimensions: Curvature of surfaces
### Gaussian curvature
In contrast to curves, which do not have intrinsic curvature, but do have extrinsic curvature (they only have a curvature given an embedding), surfaces have intrinsic curvature, independent of an embedding.
Here we adopt the convention that a curvature is taken to be positive if the curve turns in the same direction as the surface's chosen normal, otherwise negative.
The Gaussian curvature, named after Carl Friedrich Gauss, is equal to the product of the principal curvatures, k1k2. It has the dimension of 1/length2 and is positive for spheres, negative for one-sheet hyperboloids and zero for planes. It determines whether a surface is locally convex (when it is positive) or locally saddle (when it is negative).
The above definition of Gaussian curvature is extrinsic in that it uses the surface's embedding in R3, normal vectors, external planes etc. Gaussian curvature is however in fact an intrinsic property of the surface, meaning it does not depend on the particular embedding of the surface; intuitively, this means that ants living on the surface could determine the Gaussian curvature. Formally, Gaussian curvature only depends on the Riemannian metric of the surface. This is Gauss' celebrated Theorema Egregium, which he found while concerned with geographic surveys and mapmaking.
An intrinsic definition of the Gaussian curvature at a point P is the following: imagine an ant which is tied to P with a short thread of length r. She runs around P while the thread is completely stretched and measures the length C(r) of one complete trip around P. If the surface were flat, she would find C(r) = 2πr. On curved surfaces, the formula for C(r) will be different, and the Gaussian curvature K at the point P can be computed by the Bertrand–Diquet–Puiseux theorem as
K = lim_{r rarr 0} (2 pi r - mbox{C}(r)) cdot frac{3}{pi r^3}.
The integral of the Gaussian curvature over the whole surface is closely related to the surface's Euler characteristic; see the Gauss-Bonnet theorem.
The discrete analog of curvature, corresponding to curvature being concentrated at a point and particularly useful for polyhedra, is the (angular) defect; the analog for the Gauss-Bonnet theorem is Descartes' theorem on total angular defect.
Because curvature can be defined without reference to an embedding space, it is not necessary that a surface be embedded in a higher dimensional space in order to be curved. Such an intrinsically curved two-dimensional surface is a simple example of a Riemannian manifold.
### Mean curvature
The mean curvature is equal to the sum of the principal curvatures, k1+k2, over 2. It has the dimension of 1/length. Mean curvature is closely related to the first variation of surface area, in particular a minimal surface such as a soap film, has mean curvature zero and a soap bubble has constant mean curvature. Unlike Gauss curvature, the mean curvature is extrinsic and depends on the embedding, for instance, a cylinder and a plane are locally isometric but the mean curvature of a plane is zero while that of a cylinder is nonzero.
## Three dimensions: Curvature of space
By extension of the former argument, a space of three or more dimensions can be intrinsically curved; the full mathematical description is described at curvature of Riemannian manifolds. Again, the curved space may or may not be conceived as being embedded in a higher-dimensional space. In recent physics jargon, the embedding space is known as the bulk and the embedded space as a p-brane where p is the number of dimensions; thus a surface (membrane) is a 2-brane; normal space is a 3-brane etc.
After the discovery of the intrinsic definition of curvature, which is closely connected with non-Euclidean geometry, many mathematicians and scientists questioned whether ordinary physical space might be curved, although the success of Euclidean geometry up to that time meant that the radius of curvature must be astronomically large. In the theory of general relativity, which describes gravity and cosmology, the idea is slightly generalised to the "curvature of space-time"; in relativity theory space-time is a pseudo-Riemannian manifold. Once a time coordinate is defined, the three-dimensional space corresponding to a particular time is generally a curved Riemannian manifold; but since the time coordinate choice is largely arbitrary, it is the underlying space-time curvature that is physically significant.
Although an arbitrarily-curved space is very complex to describe, the curvature of a space which is locally isotropic and homogeneous is described by a single Gaussian curvature, as for a surface; mathematically these are strong conditions, but they correspond to reasonable physical assumptions (all points and all directions are indistinguishable). A positive curvature corresponds to the inverse square radius of curvature; an example is a sphere or hypersphere. An example of negatively curved space is hyperbolic geometry. A space or space-time without curvature (formally, with zero curvature) is called flat. For example, Euclidean space is an example of a flat space, and Minkowski space is an example of a flat space-time. There are other examples of flat geometries in both settings, though. A torus or a cylinder can both be given flat metrics, but differ in their topology. Other topologies are also possible for curved space. See also shape of the universe.
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## Mathematics 340 - Fall 2015
### Homework
Exercises 4.12: 21, 28, 60, 65, 76(a)(b)
Also:
Suppose 1 out of every 10,000 people have a certain disease. Let $X$ be the number of people in a sample of 20,000 that have disease.
1. What is $\text{E}(X)$?
2. Find, exactly, $P(X = 0)$, $P(X = 1)$, $P(X = 2)$, and $P(X > 3)$
3. Find the appropriate approximations for the above probabilities.
1. $E(X) = 200000 \cdot \dfrac{1}{10000} = 2$
2. $P(X = 0) = (0.9999)^{20000} = 0.135322$, $P(X = 1) = 20000 \cdot (0.0001) \cdot (0.9999)^{19999} = 0.270671$, $P(X = 2) = \binom{20000}{2} \cdot (0.0001)^2 \cdot (0.9999)^{19998} = 0.270684$, $P(X > 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) = 0.1428675$
3. Using the Poisson approximation for binomial probabilities, $P(X = 0) \approx e^{-2} = 0.135335$, $P(X = 1) \approx 2e^{-2} = 0.270671$, $P(X = 2) \approx \frac{2^2 \cdot e^{-2}}{2} = 0.270671$, $P(X > 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) \approx 0.1428765$
| 455 | 981 |
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| 3.984375 | 4 |
CC-MAIN-2019-26
|
latest
|
en
| 0.659392 |
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