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Linear Algebra
The linear algebra (including vector algebra ) is a branch of mathematics that deals with vector spaces and linear maps employs between them. This includes in particular the consideration of linear
systems of equations and matrices .
Vector spaces and their linear mappings are an important tool in many areas of mathematics. Outside of pure mathematics, there are applications in the natural sciences , computer science and
economics (for example in optimization ).
Linear algebra emerged from two specific requirements: on the one hand, the solving of linear systems of equations, on the other hand, the mathematical description of geometric objects, the so-called
analytical geometry (this is why some authors refer to linear algebra as linear geometry ).
The beginnings of algebra, and thus the term itself, go back as far as possible to the Persian- choresm mathematician , astronomer , geographer and polymath Al-Khwarizmi , who had to translate his
works into Arabic due to the Islamization in Iran and so on the name “al-jabr " came. The term algebra is derived from this.
While the development of algebra began in ancient Egypt, the development of linear algebra as an independent sub-area did not begin until the 17th century with the theory of the determinant . The
development of this theory was started independently by Gottfried Wilhelm Leibniz and Seki Takakazu . In 1750 Gabriel Cramer published Cramer's rule, named after him . This was the first time that a
solution formula for many systems of linear equations was in possession.
The history of modern linear algebra dates back to 1843 and 1844 . In 1843 William Rowan Hamilton (from whom the term vector originates) devised an extension of the complex numbers with the
quaternions . In 1844 Hermann Graßmann published his book Die lineale expansion theory. Arthur Cayley then introduced one of the most basic algebraic ideas in 1857 with the matrices. ${\
displaystyle (2 \ times 2)}$
From the 20th century onwards, the concept of vector space was mostly used . In particular, the mathematicians August Ferdinand Möbius , Constantin Carathéodory and Hermann Weyl did the preparatory
work. For example, it was found that linear mappings between finite-dimensional vector spaces can be described by matrices . Based on this knowledge, Stefan Banach was the first to give an axiomatic
definition for real vector spaces.
Systems of linear equations
A system of linear equations is a combination of equations of the type
${\ displaystyle x_ {1} + x_ {2} = 1}$
${\ displaystyle 3x_ {1} + 6x_ {2} = 4}$
Such systems of equations can be obtained from many everyday questions, for example:
In what proportion do you have to mix a 30% solution (corresponds to ) and a 60% solution (corresponds to ) in order to obtain a 40% solution?${\ displaystyle x_ {1}}$${\ displaystyle x_ {2}}$
The essential abstraction step in linear algebra is to understand the left-hand side as a function of the unknowns (in this case the set of the respective solutions): ${\ displaystyle A}$${\
displaystyle x = (x_ {1}, x_ {2})}$
${\ displaystyle A (x) = {\ begin {pmatrix} x_ {1} + x_ {2} \\ 3x_ {1} + 6x_ {2} \ end {pmatrix}}}$
Then the solution of the system of equations becomes the task: Find a such that ${\ displaystyle x}$
${\ displaystyle A (x) = {\ begin {pmatrix} 1 \\ 4 \ end {pmatrix}}}$
applies. Overwriting is only a formalism to be able to deal with more than one number at the same time.
Instead of simply writing down the relevant numbers in the form of a rectangle and calling the object a matrix : ${\ displaystyle A}$
${\ displaystyle A = {\ begin {pmatrix} 1 & 1 \\ 3 & 6 \ end {pmatrix}}}$
You can see that the function has special properties, it is a linear mapping . If there is a solution to the system of equations and a solution to the system of equations , then is ${\ displaystyle
A}$ ${\ displaystyle x}$${\ displaystyle A (x) = b}$${\ displaystyle y}$${\ displaystyle A (y) = c}$
${\ displaystyle z = x + y = {\ begin {pmatrix} x_ {1} + y_ {1} \\ x_ {2} + y_ {2} \ end {pmatrix}}}$
a solution of . You can also write that in the form . If further is any real number , then is ; is there ${\ displaystyle A (z) = b + c}$${\ displaystyle A (x + y) = A (x) + A (y)}$${\ displaystyle \
lambda}$ ${\ displaystyle A (\ lambda x) = \ lambda \ cdot A (x)}$
${\ displaystyle \ lambda x = {\ begin {pmatrix} \ lambda x_ {1} \\\ lambda x_ {2} \ end {pmatrix}}}$.
Analytical geometry
The other origin of linear algebra can be found in the arithmetical description of the 2- and 3-dimensional (Euclidean) space, also called "visual space". With the help of a coordinate system ,
points in space can be described by triples of numbers. The mapping type of the displacement leads to the concept of the vector, which indicates the direction and amount of the displacement. Many
physical quantities , for example forces , always have this directional aspect. ${\ displaystyle (x_ {1}, x_ {2}, x_ {3})}$
Since vectors can also be described by triples of numbers , the separation between vectors and points is blurred: A point corresponds to its position vector , which points from the coordinate origin
. ${\ displaystyle (a_ {1}, a_ {2}, a_ {3})}$${\ displaystyle P}$ ${\ displaystyle P}$
Many of the mapping types considered in classical geometry, for example rotations about axes through the origin or reflections on planes through the origin, belong to the class of linear mapping that
has already been mentioned above.
Vector spaces and linear algebra
The concept of vector space emerges as an abstraction from the above examples: A vector space is a set, the elements of which are called vectors, together with
This addition and the scalar multiplication have to fulfill some simple properties that also apply to the vectors in the visual space.
One could say that vector spaces are precisely defined in such a way that one can speak of linear mappings between them.
In a way, the concept of vector space is already too general for linear algebra. One can assign a dimension to each vector space , for example the plane has dimension and the visual space has
dimension . However, there are vector spaces whose dimensions are not finite, which means that many of the known properties are lost. However, it has proven to be very successful to equip infinitely
dimensional vector spaces with an additional topological structure ; the investigation of topological vector spaces is the subject of functional analysis . ${\ displaystyle 2}$${\ displaystyle 3}$
(The rest of this article deals with the case of finite dimensions.)
Important sentences and results
Every vector space has at least one basis . Every two bases of a vector space have the same number of elements; only therefore does it make sense to speak of the dimension of a vector space. For
totals and averages of subspaces the true dimension formula and for the dimensions of factor spaces formula . ${\ displaystyle \ dim V / U = \ dim V- \ dim U}$
Each linear mapping is clearly defined by specifying the images on a basis of . The homomorphism theorem and the rank theorem apply to linear mappings . Linear maps can be represented by matrices
with respect to fixed bases . The execution of linear images one after the other corresponds to the multiplication of their representation matrices. ${\ displaystyle f \ colon V \ to W}$${\
displaystyle V}$
A linear system of equations with , and is solvable if and only if the rank of the matrix is equal to the rank of the extended coefficient matrix . In this case the solution set of the system is an
affine subspace of the dimension . For equation systems that are not too large, the ranking and the calculation of the solution space can be carried out using the Gaussian elimination method. ${\
displaystyle A \ cdot x = b}$${\ displaystyle A \ in \ mathbb {K} ^ {{m} \ times {n}}}$${\ displaystyle b \ in \ mathbb {K} ^ {m}}$${\ displaystyle x \ in \ mathbb {K} ^ {n}}$ ${\ displaystyle A}$${\
displaystyle {\ begin {pmatrix} A & b \ end {pmatrix}}}$ ${\ displaystyle \ mathbb {K} ^ {n}}$${\ displaystyle n- \ mathrm {rank} (A)}$
A linear mapping (i.e. an endomorphism ) of a finite-dimensional vector space can already be inverted if it is injective or surjective. Again, this is precisely the case when its determinant is not
equal to zero. It follows from this that the eigenvalues of an endomorphism are exactly the zeros of its characteristic polynomial . Another important statement about the characteristic polynomial
is the Cayley-Hamilton theorem . ${\ displaystyle f \ colon V \ to V}$ ${\ displaystyle V}$
An endomorphism (or a square matrix) can be diagonalized if the characteristic polynomial is divided into linear factors and for each eigenvalue its algebraic multiplicity is equal to the geometric
multiplicity, i.e. the zero order of the eigenvalue in the characteristic polynomial is equal to the dimension of the associated eigenspace . Equivalent to this is the existence of a basis of the
vector space, which consists of the eigenvectors of the linear mapping. Endomorphisms, the characteristic polynomial of which is broken down into linear factors, can still be trigonalized , so they
can be represented by a triangular matrix. A somewhat deeper result is that the representing matrix can even be brought into Jordanian normal form .
In vector spaces on which an additional scalar product is given, a norm is used to define. In these scalar product spaces there always exist orthonormal bases that can be constructed, for example,
using the Gram-Schmidt orthonormalization method. According to the projection theorem , the best approximation can be determined in these spaces from a sub-vector space by orthogonal projection . ${\
displaystyle \ langle \ cdot, \ cdot \ rangle}$${\ displaystyle \ | x \ |: = {\ sqrt {\ langle x, x \ rangle}}}$
Regarding the diagonalisability of endomorphisms in scalar product spaces, the question arises whether an orthonormal basis from eigenvectors exists. The central result for this is the spectral
theorem . In the real case, in particular, the following applies: For every symmetric matrix there is an orthogonal matrix , so that is a diagonal matrix. Applying this result to quadratic forms , we
get the principle of the principal axis transformation . ${\ displaystyle A \ in \ mathbb {\ mathbb {R}} ^ {{n} \ times {n}}}$ ${\ displaystyle Q}$${\ displaystyle Q ^ {T} AQ}$
Also bilinear and sesquilinear can be represented by matrices of fixed chosen bases. A bilinear form is symmetric and positive definite , i.e. a scalar product, if and only if its representing matrix
is symmetric and positive definite. A symmetric matrix is positive definite if and only if all of its eigenvalues are positive. In general, Sylvester's law of inertia applies to symmetrical
bilinear forms and Hermitian sesquilinear forms , which states that the number of positive and negative eigenvalues of the representing matrices does not depend on the choice of the basis.
Vectors and matrices
Vectors of finite-dimensional spaces can be described by their components, which (depending on the application) as column vectors
${\ displaystyle \ mathbf {a} = {\ begin {pmatrix} 3 \\ 7 \\ 2 \ end {pmatrix}}}$
or line vector
${\ displaystyle \ mathbf {b} = {\ begin {pmatrix} 4 & 6 & 3 & 7 \ end {pmatrix}}}$
to be written. Often line vectors are marked with a superscript T for transposed , such as . ${\ displaystyle b ^ {T}}$
In the literature, vectors are distinguished from other quantities in different ways: lower case letters, lower case letters in bold, lower case letters underlined, lower case letters with an arrow
above or small Fraktur letters are used. This article uses lowercase letters.
A matrix is indicated by a “grid” of numbers. Here is a matrix with four rows and three columns:
${\ displaystyle \ mathbf {M} = {\ begin {pmatrix} 8 & 2 & 9 \\ 4 & 8 & 2 \\ 8 & 3 & 7 \\ 5 & 9 & 1 \ end {pmatrix}}}$
Matrices are mostly designated with capital letters.
In the case of column vectors, individual elements of a vector are usually indicated by an index: The second element of the vector specified above would then be . A exponent is sometimes used in row
vectors, whereby one has to be careful whether there is a vector indexing or an exponent : With the above example one has about . Matrix elements are indicated by two indices. The elements are
represented by lowercase letters: is the element in the second line of the third column (instead of “in the third column of the second line”, because this makes it easier to read). ${\ displaystyle
a}$${\ displaystyle a_ {2} = 7}$ ${\ displaystyle b}$${\ displaystyle b ^ {4} = 7}$${\ displaystyle m_ {2,3} = 2}$${\ displaystyle m_ {2,3}}$
The generalized term for these structures is tensor , scalars are zeroth order tensors, vectors first order tensors, matrices second order tensors. A -th order tensor can be represented by a
-dimensional number cube. ${\ displaystyle n}$${\ displaystyle n}$
It is often necessary to bring matrices to a special shape by means of elementary line transformations or base changes . The triangular shape , the diagonal shape and the Jordanian normal shape are
particularly important .
Endomorphisms and Square Matrices
In the representation of a linear mapping - as described under Matrix - there is the special case of a linear mapping of a finite-dimensional vector space onto itself (a so-called endomorphism ). The
same basis can then be used for the original image and image coordinates and a square matrix is obtained so that the application of the linear mapping corresponds to the left multiplication . To
express the dependence on and , spellings like or are used . Executing this mapping twice in a row then corresponds to multiplication by etc., and all polynomial expressions with (sums of multiples
of powers of ) can be understood as linear mappings of the vector space. ${\ displaystyle f}$ ${\ displaystyle v}$${\ displaystyle A}$${\ displaystyle A}$${\ displaystyle f}$${\ displaystyle v}$${\
displaystyle A = M_ {v} (f)}$${\ displaystyle A = {} _ {v} f_ {v}}$${\ displaystyle A ^ {2}}$${\ displaystyle A}$${\ displaystyle A}$
Analogous to the calculation rule for numbers, the zeroth power of a square matrix is the diagonal matrix ( unit matrix ) with ones on the diagonal and in which all remaining elements are zero, it
corresponds to the identity mapping of each vector to itself. Negative powers of a square matrix can only be calculated , if the linear mapping given by is invertible, i.e. no two different vectors
and maps to the same vector . In other words, you have an invertible matrix from always follow the linear system thus may be just the solution have. For an invertible matrix there is an inverse
matrix with . ${\ displaystyle x ^ {0} = 1}$${\ displaystyle E}$ ${\ displaystyle A}$${\ displaystyle A}$${\ displaystyle u_ {1}}$${\ displaystyle u_ {2}}$${\ displaystyle Au_ {1} = Au_ {2}}$${\
displaystyle A}$${\ displaystyle u_ {1} -u_ {2} \ neq 0}$${\ displaystyle A (u_ {1} -u_ {2}) \ neq 0}$${\ displaystyle Au = 0}$${\ displaystyle 0}$${\ displaystyle A}$ ${\ displaystyle A ^ {- 1}}$${\
displaystyle A ^ {- 1} A = AA ^ {- 1} = E}$
A determinant is a special function that assigns a number to a square matrix. This number gives information about some properties of the matrix. For example, it can be used to identify whether a
matrix can be inverted. Another important application is the calculation of the characteristic polynomial and thus the eigenvalues of the matrix.
There are closed formulas for calculating the determinants, such as Laplace's expansion theorem or Leibniz's formula . However, these formulas are more of theoretical value, as their effort increases
significantly with larger matrices. In practice, the easiest way to calculate determinants is to use the Gaussian algorithm to convert the matrix into an upper or lower triangular shape; the
determinant is then simply the product of the main diagonal elements .
The above terms should be clarified using an example motivated by the Fibonacci sequence .
Calculation of powers using diagonalization
The Fibonacci sequence is recursively defined by the equations , and for what is synonymous with ${\ displaystyle f_ {n}}$${\ displaystyle f_ {0} = 0}$${\ displaystyle f_ {1} = 1}$${\ displaystyle f_
{n + 1} = f_ {n} + f_ {n-1}}$${\ displaystyle n \ geq 1}$
${\ displaystyle {f_ {1} \ choose f_ {0}} = {1 \ choose 0}}$
${\ displaystyle {f_ {n + 1} \ choose f_ {n}} = {\ begin {pmatrix} 1 & 1 \\ 1 & 0 \ end {pmatrix}} \ cdot {f_ {n} \ choose f_ {n-1}} \ quad {\ text {for}} \ quad n \ geq 1}$,
from which by iteration the non-recursive formula
${\ displaystyle {f_ {n + 1} \ choose f_ {n}} = {\ begin {pmatrix} 1 & 1 \\ 1 & 0 \ end {pmatrix}} ^ {n} \ cdot {1 \ choose 0} \ quad {\ text {for}} \ quad n \ geq 0}$
follows, in which the -th power of a matrix occurs. ${\ displaystyle n}$${\ displaystyle A}$
The behavior of such a matrix on exponentiation is not easy to see; on the other hand, the -th power of a diagonal matrix is simply calculated by exponentiating each individual diagonal entry. If
there is an invertible matrix , so that it has diagonal form, the exponentiation of can be reduced to the exponentiation of a diagonal matrix according to the equation (the left side of this equation
is then the -th power of a diagonal matrix). In general, its behavior (with exponentiation, but also with other operations) can be seen more easily by diagonalizing a matrix. ${\ displaystyle n}$${\
displaystyle T}$${\ displaystyle T ^ {- 1} AT}$${\ displaystyle A}$${\ displaystyle (T ^ {- 1} AT) ^ {n} = T ^ {- 1} A ^ {n} T}$${\ displaystyle n}$
If one understands a matrix of a linear mapping , then the transformation matrix is the base change matrix to another base , i.e. (whereby the identity mapping maps each vector to itself). Then
namely . ${\ displaystyle A = {} _ {v} f_ {v}}$ ${\ displaystyle T}$${\ displaystyle v '}$${\ displaystyle T = {} _ {v} e_ {v '}}$${\ displaystyle e}$${\ displaystyle T ^ {- 1} AT = {} _ {v '} f_
In the above example, a transformation matrix can be found such that ${\ displaystyle T}$
${\ displaystyle T ^ {- 1} \ cdot A \ cdot T = {\ begin {pmatrix} \ phi & 0 \\ 0 & 1- \ phi \ end {pmatrix}}}$
is a diagonal matrix in which the golden ratio occurs. From this we finally get Binet's formula : ${\ displaystyle \ phi = {\ frac {1 + {\ sqrt {5}}} {2}}}$
${\ displaystyle f_ {n} = {\ frac {1} {\ sqrt {5}}} \ cdot \ left [\ left ({\ frac {1 + {\ sqrt {5}}} {2}} \ right ) ^ {n} - \ left ({\ frac {1 - {\ sqrt {5}}} {2}} \ right) ^ {n} \ right]}$
How do you get from the matrix to the number ? One recognizes immediately from the diagonal matrix ${\ displaystyle A}$${\ displaystyle \ phi}$
${\ displaystyle {\ begin {pmatrix} \ phi & 0 \\ 0 & 1- \ phi \ end {pmatrix}} \ cdot {1 \ choose 0} = {\ phi \ choose 0}}$,
which means it is a vector is equal to zero by multiplying the diagonal matrix component-wise multiplied: (more precisely comparable is -facht) . Because of this property, the number is called an
eigenvalue of the matrix (with an eigenvector ). In the case of diagonal matrices, the eigenvalues are equal to the diagonal entries. ${\ displaystyle u}$${\ displaystyle \ phi}$${\ displaystyle (T
^ {- 1} AT) u = \ phi u}$${\ displaystyle \ phi}$ ${\ displaystyle T ^ {- 1} AT}$ ${\ displaystyle u}$
${\ displaystyle \ phi}$is also the eigenvalue of the original matrix (with eigenvector , because ), the eigenvalues therefore remain unchanged when the matrix is transformed. The diagonal form
of the matrix results from its eigenvalues, and in order to find the eigenvalues of, one has to investigate for which numbers the linear system of equations has a solution other than zero (or, in
other words, the matrix is not invertible). ${\ displaystyle A}$${\ displaystyle Tu}$${\ displaystyle A (Tu) = \ phi (Tu)}$${\ displaystyle A}$${\ displaystyle A}$${\ displaystyle x}$${\ displaystyle
Au = xu}$${\ displaystyle u}$${\ displaystyle xE-A}$
The numbers we are looking for are precisely those that make the determinant of the matrix zero. This determinant is a polynomial expression in (the so-called characteristic polynomial of ); in the
case of the 2 × 2 matrix mentioned above , this gives the quadratic equation with the two solutions and . The associated eigenvectors are solutions of the linear equation systems or , they then form
the columns of the transformation matrix . ${\ displaystyle x}$${\ displaystyle xE-A}$${\ displaystyle x}$ ${\ displaystyle A}$${\ displaystyle A}$ ${\ displaystyle x ^ {2} -x-1 = 0}$${\ displaystyle
x = \ phi}$${\ displaystyle x = 1- \ phi}$${\ displaystyle Au = \ phi u}$${\ displaystyle Au = (1- \ phi) u}$${\ displaystyle T}$
Whether a matrix can be diagonalized depends on the number range used. is not diagonalizable over the rational numbers , for example , because the eigenvalues and irrational numbers. The
diagonalization can also fail regardless of the number range, if there are not “enough” eigenvalues; so has the Jordan form matrix ${\ displaystyle A}$ ${\ displaystyle \ phi}$${\ displaystyle 1- \
${\ displaystyle {\ begin {pmatrix} 1 & 1 \\ 0 & 1 \ end {pmatrix}}}$
only the eigenvalue (as a solution to the quadratic equation ) and is not diagonalizable. With a sufficiently large number range (for example above the complex numbers ), however, each matrix can be
diagonalized or transformed into Jordanian normal form . ${\ displaystyle 1}$${\ displaystyle (x-1) ^ {2} = 0}$
Since the transformation of a matrix corresponds to the change in the basis of a linear mapping, this last statement means that for a linear mapping with a sufficiently large number range, one can
always choose a base that is mapped “in a simple way”: In the case of diagonalization, each basis vector is a multiple of itself (is therefore an eigenvector); in the case of the Jordan form, to a
multiple of itself plus possibly the previous basis vector. This theory of linear mapping can be generalized to bodies that are not "big enough"; in them, in addition to the Jordan form, other normal
forms must be considered (for example the Frobenius normal form ).
Web links
Individual evidence
1. ^ John Stillwell: Mathematics and Its History . Springer, New York, NY 2010, ISBN 978-1-4419-6052-8 , pp. 88-89 , doi : 10.1007 / 978-1-4419-6053-5_6 .
2. ^ Heinz-Wilhelm Alten : 4000 years of algebra. History, cultures, people . Springer, Berlin a. a. 2003, ISBN 3-540-43554-9 , pp. 335-339 . | {"url":"https://de.zxc.wiki/wiki/Lineare_Algebra","timestamp":"2024-11-04T02:55:27Z","content_type":"text/html","content_length":"217632","record_id":"<urn:uuid:e642af9d-2dfe-4eea-9c75-d887a606e7c6>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00843.warc.gz"} |
Factoring (Mathematics) in Cryptography | Learning Cardano
In mathematics, factoring (or factorization) refers to the process of breaking down a mathematical object, such as a number or a polynomial, into a product of smaller or simpler objects, called
factors, that when multiplied together yield the original object.
Factoring of Numbers
For integers, factoring refers to expressing a number as the product of prime numbers or smaller integers. For example, the number 12 can be factored as:
12 = 2 × 2 × 3
Here, 2 and 3 are prime numbers, and they are the prime factors of 12.
Factoring is particularly important in number theory and cryptography. In the case of the RSA encryption scheme, for example, the security relies on the difficulty of factoring large numbers into
their prime components.
Factoring of Polynomials
For polynomials, factoring involves expressing a polynomial as the product of simpler polynomials. For example, the polynomial x² − 9 can be factored as:
x² − 9 = (x - 3)(x + 3)
This process is essential in algebra for solving polynomial equations and simplifying expressions.
Types of Factoring
1. Factoring Integers: Writing a number as the product of primes (e.g., 28 = 2 × 2 × 7).
2. Factoring Polynomials: Decomposing a polynomial into products of lower-degree polynomials (e.g., x² − 4 = (x − 2)(x + 2))
3. Prime Factorization: Breaking down a number into its prime components, which is especially important for applications like cryptography.
4. Special Factoring Patterns:
□ Difference of Squares: a² − b² = (a − b)(a + b)
□ Perfect Square Trinomials: a² + 2ab + b² = (a + b)²
□ Sum/Difference of Cubes: a³ − b³ = (a − b)(a² + ab + b²)
Factoring plays a crucial role in traditional cryptographic systems, particularly those based on the RSA algorithm, where the security of the system relies on the difficulty of factoring large
composite numbers. However, in Cardano, cryptography operates differently, relying primarily on elliptic curve cryptography (ECC), which does not involve factoring but instead depends on the hardness
of the elliptic curve discrete logarithm problem.
Importance of Factoring in Traditional Cryptography
In RSA, a widely-used public-key cryptosystem, the security is based on the fact that, while it is easy to multiply two large prime numbers together, it is extremely difficult to factor the resulting
product back into its prime factors. Factoring large composite numbers into primes is computationally infeasible for very large numbers, providing a secure way to encrypt data or create digital
Cardano’s Cryptography and Why Factoring is Less Relevant
Cardano, however, does not use RSA; instead, it employs Elliptic Curve Cryptography (ECC), which is based on a different mathematical problem—specifically, the elliptic curve discrete logarithm
problem (ECDLP). This problem is considered harder to solve than factoring, especially for the same key sizes, which makes ECC more efficient and secure with smaller key sizes compared to RSA.
Here’s how Cardano benefits from ECC instead of relying on factoring:
1. Efficiency: ECC can provide the same level of security as RSA but with much smaller key sizes. For example, a 256-bit key in ECC offers comparable security to a 3072-bit key in RSA. This reduces
computational overhead, making Cardano more efficient for secure transactions and smart contracts.
2. Scalability: Since Cardano is a blockchain platform aiming to support a high volume of transactions, the use of elliptic curve cryptography ensures that security remains robust while maintaining
faster transaction times and lower resource usage, which is critical for scalability.
3. Quantum Resistance Preparation: While RSA’s security is vulnerable to future quantum computing breakthroughs due to the ability of quantum algorithms (such as Shor’s algorithm) to factor large
numbers efficiently, elliptic curve cryptography is believed to be more resistant to quantum attacks (though not fully quantum-resistant). Cardano’s use of ECC reflects a forward-looking
approach, optimizing for both current and future security landscapes.
Why Factoring Knowledge Is Still Important
Even though Cardano doesn’t rely directly on factoring for its cryptographic security, understanding factoring is still useful for two key reasons:
1. Historical Context and Comparison: Knowing how factoring underpins traditional cryptography (like RSA) helps developers, cryptographers, and users appreciate the strengths of Cardano’s elliptic
curve approach, which offers better efficiency and security.
2. Diverse Cryptographic Approaches: As blockchain technology evolves, being familiar with different cryptographic schemes (including RSA, which uses factoring) allows developers to adapt, innovate,
or contribute to cross-chain solutions, where different blockchains might use different cryptographic methods.
In summary, while factoring is central to traditional cryptographic systems like RSA, Cardano’s cryptography is built on elliptic curves, making factoring less relevant for Cardano’s specific
security model. Understanding factoring is important in a broader cryptographic sense, but for Cardano, the focus is on the advantages and security of elliptic curve cryptography.s into simpler
components for easier manipulation and understanding. | {"url":"https://www.learningcardano.com/factoring-mathematics-in-cryptography/","timestamp":"2024-11-14T11:08:08Z","content_type":"text/html","content_length":"102648","record_id":"<urn:uuid:06ff8b17-9d37-4d6c-839c-283c49d642ff>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00335.warc.gz"} |
A consistent thermodynamic formulation of the field equations for elastic bodies removing the paradox of infinite speed of propagation of thermal signals
The classical unsteady heat conduction equation is of parabolic character and consequently predicts an infinite speed of propagation for the thermal disturbances. The purpose of this work is to
eliminate this not admissible physical feature. A new fundamental relation between the entropy and the state variables is proposed: besides the classical variables, the entropy is assumed to depend
on the heat flux vector. This assumption generalizes the local equilibrium hypothesis which is known to be only valid in the vicinity of equilibrium. A rigid and an elastic heat conductor are
respectively considered. Following the lines of Onsager's non-equilibrium thermodynamics, phenomenological laws relating the heat flux to the temperature gradient are derived. They are nonstationary
and appear as a straightforward extension of the classical Fourier law.
Journal de Mecanique
Pub Date:
□ Conductive Heat Transfer;
□ Elastic Bodies;
□ Nonequilibrium Thermodynamics;
□ Propagation Velocity;
□ Thermoelasticity;
□ Entropy;
□ Heat Flux;
□ Onsager Relationship;
□ State Vectors;
□ Temperature Gradients;
□ Unsteady State;
□ Fluid Mechanics and Heat Transfer | {"url":"https://ui.adsabs.harvard.edu/abs/1976JMec...15..579L/abstract","timestamp":"2024-11-07T18:47:49Z","content_type":"text/html","content_length":"36566","record_id":"<urn:uuid:a50e4432-a86d-4cea-a810-4ae07fe0fd82>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00306.warc.gz"} |
Short posts
Category Archives: Short posts
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mathematical pole to propel your … Continue reading
A recent paper by John Urschel just came across my desk. Urschel has three papers in MathSciNet. The latest is “A cascadic multigrid algorithm for computing the Fiedler vector of graph Laplacians”
in the Journal of Computational Mathematics. Like many of the authors … Continue reading | {"url":"https://blogs.ams.org/beyondreviews/category/short-posts/","timestamp":"2024-11-02T11:45:57Z","content_type":"text/html","content_length":"61704","record_id":"<urn:uuid:380b9e80-c7d2-48bd-abbb-59c35b597453>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00132.warc.gz"} |
Turf Weight Calculator - Calculator Wow
Turf Weight Calculator
A Turf Weight Calculator is a valuable tool used in landscaping, construction, and agriculture to estimate the weight of turf based on area, density, and thickness inputs. This article explores the
significance of turf weight calculations, their practical applications, and how this calculator simplifies planning and logistics in various outdoor projects.
Importance of Turf Weight Calculator
Understanding the weight of turf is crucial for several reasons. It helps landscapers and contractors accurately estimate material requirements, plan transportation logistics, and ensure proper
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How to Use a Turf Weight Calculator
Using a Turf Weight Calculator involves the following steps:
1. Enter Area: Input the area to be covered with turf in square meters (m²).
2. Provide Density: Enter the density of the turf in kilograms per cubic meter (kg/m³).
3. Specify Thickness: Input the thickness of the turf layer in meters (m).
4. Calculate: Click on the Calculate button to initiate the computation.
5. Result: The calculator displays the estimated weight of the turf based on the provided inputs.
This tool simplifies the planning process for landscapers, architects, and project managers, ensuring accurate material estimates and efficient project execution.
FAQs about Turf Weight Calculator
1. What is a Turf Weight Calculator?
A Turf Weight Calculator is a tool used to estimate the weight of turf based on area, density, and thickness inputs.
2. Why is it important to calculate turf weight?
Calculating turf weight helps in planning material requirements, estimating transportation costs, and ensuring structural integrity in landscaping and construction projects.
3. How is turf weight calculated?
Turf weight (WT) is calculated using the formula: WT = Area × Density × Thickness, where Area is in square meters (m²), Density is in kilograms per cubic meter (kg/m³), and Thickness is in meters
4. What are the typical applications of turf weight calculators?
Turf weight calculators are used in landscaping projects, sports field construction, green roof installations, and agricultural turfing to optimize material usage and project planning.
5. Can a turf weight calculator handle different units of measurement?
Yes, calculators are designed to accept inputs in various units, allowing users to input measurements in square feet, pounds per cubic foot, and inches for thickness, converting them into standard
metric units for accurate calculations.
6. How accurate are turf weight calculations?
Turf weight calculations are based on input accuracy of area, density, and thickness. Precision in measurements ensures accurate estimates for project planning and budgeting.
7. What factors affect turf weight?
Factors such as turf density variations, installation techniques, and environmental conditions can influence the actual weight of turf, affecting project logistics and maintenance considerations.
8. How can turf weight calculators assist in budget planning?
By accurately estimating turf weight, calculators help in budgeting for material costs, transportation expenses, and labor requirements, optimizing resource allocation and project profitability.
9. Are turf weight calculators useful in sustainability initiatives?
Yes, by calculating precise turf quantities, calculators support sustainable landscaping practices, minimizing waste and promoting efficient use of natural resources in urban development and
environmental projects.
10. How does turf weight calculation contribute to project safety?
Estimating turf weight ensures that load-bearing capacities of structures and surfaces are not exceeded, maintaining safety standards during installation and maintenance operations.
In conclusion, the Turf Weight Calculator is a practical tool for estimating the weight of turf in landscaping, construction, and agricultural projects. By facilitating accurate material planning and
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MCQ Questions for Class 10 Maths with Answers PDF Download Chapter Wise
Get Chapter Wise MCQ Questions for Class 10 Maths with Answers PDF Free Download prepared here according to the latest CBSE syllabus and NCERT curriculum. Students can practice CBSE Class 10 Maths
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Class 10 Maths MCQs Multiple Choice Questions with Answers
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Also Read:
ML Aggarwal Maths For Class 10 PDF Download
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Fermat's Library | Many lives in many worlds annotated/explained version.
100%0% 25% 50% 75%
Fraction of queens face up
FIG. 2: According to quantum theory, a card perfectly b al-
anced on its edge will fall down in what is known as a “su-
perposition” — the card really is in two places at once. If a
gambler bets money on the queen landing face up, her own
state changes to become a superposition of two possible out-
comes — winning or losing the bet. In either of these parallel
worlds, the gambler is unaware of the other outcome and feels
as if the card fell randomly. If the gambler repeats this game
four times in a row, there will be 16 (2 × 2 × 2 × 2) possible
outcomes, or parallel worlds. In most of these worlds, it will
seem that q ueens occur randomly, with about 50% probabil-
ity. Only in two worlds will all four cards land the same way
up. If the game is continued many more times, almost every
gambler in each of the worlds will conclude that the laws of
probability apply even though th e underlying physics is not
random and, as Einstein would have put it, “God does not
play dice”.
many more ”undecided” [3]. I believe the upwards trend
is clear.
Why the change? I think there are several reasons.
Predictions of other types of para llel universes from co s-
mological inflation and string theory have increased tol-
erance for weird-sounding ideas. New experiments have
demonstrated quantum weirdness in ever larg er systems.
Finally, the discovery of a process known as decoherence
has a nswered crucial questions that Everett’s work had
left dang ling.
For example, if these parallel universes exist, then why
don’t we perceive them? Quantum superpositions can-
not be confined – as most quantum experiments are – to
the microworld. Beca use you are made o f atoms , then if
atoms can be in two places at once in superposition, so
can you (Figure 1).
The breakthrough came in 1970 with a seminal paper
by H. Dieter Zeh, who showed that the Schr¨odinger equa-
tion itself gives rise to a type of censorship. This effect
became known as “decoherence ”, and was worked out in
great detail by Wojciech Zurek, Zeh and others over the
following decades. Quantum superpositions were found
to remain observable only as long as they were kept secret
from the rest of the world. The quantum card in Figure 2
is constantly bumping into air molecules, photons, and
so on, which thereby find out whether it has fallen to
the left or to the right, destroying the coherence of the
supe rp osition and making it unobservable. Decoherence
also explains why states resembling classical physics have
sp e c ial status: they are the most robust to decoherence.
Science of philosophy?
The main motivation for introducing the no tion of ran-
dom wavefunction collapse into quantum physics had
been to e xplain why we perceive probabilities and not
strange macroscopic superpositions . After Everett had
shown that things wo uld appear rando m anyway (Fig-
ure 2) and decoherence had b e en found to explain why
we never perceived anything strange, much of this moti-
vation was gone. Even though the wavefunction techni-
cally never collapses in the Everett view, it is generally
agreed that decoherence produces an effect that looks like
a co llapse and smells like a collapse .
In my opinion, it is time to update the many quantum
textbooks that introduce wavefunction collapse as a fun-
damental postulate of quantum mechanics. The notion
of collapse still has utility as a calculational recipe, but
students should be told that it is probably not a funda-
mental proces s violating the Schr¨odinge r equation so as
to avoid any subsequent confusion. If you are consider-
ing a quantum textbook that does not mention “Everett”
and “decoherence” in the index, I recommend buying a
more modern one.
After 50 years we can c e le brate the fact that Everett’s
interpretation is still consistent with quantum observa-
tions, but we face another pressing question: is it science
or mere philosophy? The key point is that parallel uni-
verses are not a theory in themselves, but a prediction of
certain theories. For a theory to be falsifiable, we need
not observe and test all its predictions – one will do.
Because Einstein’s theory of General Relativity has
successfully predicted many things that we can observe,
we also take s eriously its predictions for things we can-
not, like the internal structure of black holes. Analo-
gously, successful pre dictio ns by unitary quantum me-
chanics have made scientists take more seriously its other
predictions, including parallel universes
Moreover, Everett’s theory is falsifiable by future lab
exp eriments: no matter how large a system they probe,
it s ays, they will not observe the wavefunction collapsing.
If we can use Einstein's theory of General Relativity to predict the physics surrounding certain objects we can not observe (internal structure of black holes for example) then we should be able to
use theories of quantum mechanics to predict items we cannot see as well (such as parallel universes). I don't think that's true. In a practical sense, classical computers are still faster than
quantum computers. In a theoretical sense, it hasn't been proven impossible for a classical computer to factor a number in polynomial time. The theory itself was incomplete because there is no
mathematical way of calculating an observation, and without math to support it, a theory is merely a postulation. Scientists could not use observations as part of this theory, there was no way to
accurately test an observation without mathematics behind it, thus rendering this theory incomplete. The wave function is the mathematical formula we use to describe quantum objects. It is a wavering
quantity that mathematically describes the wave characteristics of a particle. The value of the wave function of a particular particle at a given point of space/time is related to the probability of
the particles being there at that time. Here is more information on wave functions https://www.youtube.com/watch?v=aowYf44gDRY Funny Decoherence was Zeh's theory involving Schrodinger's cat. Zeh
noticed that in the real world you could not separate the cat from the environment (air molecules, the box, cosmic rays passing through the experiment). Zeh claimed that these interactions, no matter
how small, can radically affect the wave function. If the wave function is disturbed in the slightest way, then the wave function will suddenly split into two distinct wave functions of the dead cat
or the live cat (which no longer interact). Zeh showed that a collision with a single air molecule was enough to collapse the wavefunction, forcing the permanent separation of the dead cat and the
live cat, which no longer interact, In other words, before you even open the box, the cat has been in contact with the air molecules and hence is already dead or alive. Everett’s thesis discussed the
possibility that Schrodinger’s cat is both dead and alive, because the universe splits into two. In one universe, the cat is dead; in another, the cat is alive. Everett postulated that at each
quantum juncture the universe splits in half, in an incessant sequence of splitting universes. Here is an example of Everett’s theory https://www.youtube.com/watch?v=KNwKPfOKipk Here is a video going
more in depth about wave functions and decoherance https://www.youtube.com/watch?v=Zbe6ov22Nok This example is supposed to prove that nothing is truly "random" when it comes to probability, even if
it appears so. The underlying physics proves the card falling is not random, 16 parallel worlds would exist in which this card has every possible outcome A quantum computer can solve problems that no
classical computer would be able to solve within a reasonable amount of time. They would be able to decrypt the majority of the cryptographic systems in use today. Currently, Google and NASA are
working on building a quantum computer, here is a video explaining more about quantum computers as a whole, and google and nasa's involvement. http://mashable.com/2013/10/13/
google-quantum-computing-video/#KW2dQZTqnsq3 If you would like to learn more about parallel worlds and the science behind it, here is a short article/video http://www.dailymail.co.uk/sciencetech/
article-2816039/Parallel-universes-exist-Multiple-versions-living-alternate-worlds-interact-theory-claims.html Everett's theory claims that decoherence explains why we don't perceive parallel
universes. He claimed that perhaps the cat is dead and alive at the same time, but in different universes. The cat is both dead and alive because the universe has split into two. In one universe, the
cat is dead, and in another universe, the cat is alive. If this interpretation is correct, then at this very moment, your body coexists with wave functions of dinosaurs, wave functions where the
Germans won world war one, where aliens roam, where you were never born, etc. The catch is we can no longer interact with these wave functions, because they have decohered from us (the waves are no
longer in phase with one another). This means the slightest contamination with the environment will prevent the various wave functions from interacting with one another, thus we are unable to
perceive these parallel universes. Schrodinger's cat - Imagine a cat enclosed in a box with a radioactive source. If a single atom of the radioactive source decays, a mechanism would trip a hammer,
which would in turn, break the vial containing the radioactive source, and kill the cat. Until the source emits radiation, the cat would be considered (according to quantum mechanics) to be
simultaneously both dead and alive, until the box is open, and the cat is observed. When the box is open, the “wavelength” collapses, and the cat can either be one of these two things: dead OR alive.
Here is a brief video demonstrating Schrodinger’s cat and the implications behind observation https://www.youtube.com/watch?v=IOYyCHGWJq4 Schrodinger’s equation is the basic equation for describing
quantum behavior, and is constantly used in quantum mechanics. Also known as the Schrodinger wave equation, it is a partial differential equation that describes how the wavefunctions of a physical
system change over time. In the past, observation of Schrodinger’s cat was said to “collapse” the wavefunction, thus leaving only one outcome (a dead or alive cat), however, Everett’s theory
postulates that the wavelength never truly collapses, instead, the wavelength transforms into a parallel universe. In this case, if we were to open Schrodinger's box and find a dead cat, a parallel
universe would be created where we open the box and find an alive cat. The wavefunction never collapses in Everett's view, they just continue to evolve, forever splitting into other wave functions in
a never-ending tree, with each branch representing an entire universe. However, in our universe, we perceive the wavefunction as "collapsing" because decoherence only gives us one outcome. If in this
universe the outcome is "a", we perceive the wavefunction as collapsing to "a", however, there is another universe where the wavefunction collapses to "b" and the people living in that universe
perceive the wavefunction collapsing to "b". Thus, decoherence appears as if it produces an effect like a collapse separately in each parallel universe, however, on a large scale, the wavefunction
never collapses. Everett's theory claimed that regardless of the size of the system, scientists will not observe a wavefunction collapsing, however, this theory will be disproven by future lab
experiments. Decoherence is due to constant interactions with the environment, which are impossible to avoid. This explains that what we think of as an observation doesn’t require a human; simply
interacting with the environment counts. This explains why bringing large objects into superpositions of two states is extremely difficult and the superposition fades rapidly. The heaviest object so
fat that has been brought into a superposition of locations is a carbon-60 molecule. More ambitious scientists have proposed to do this experiment to viruses or even heavier creatures, such as
bacteria. Thus, the paradox that Schrodinger’s cat once raised, (the transfer of a quantum superposition, the decaying atom, to a large object, the cat) has been resolved. Scientists now understand
that while small things (atoms, quarks, ect) can exist in superpositions for extended amounts of time, a large object will settle extremely quickly in one particular state. That is why we never see
cats that are both dead and alive. This paper is about Everett's theory that Schrodinger's equation applies at all times, therefore the wavefunction never collapses. The author takes Everett's theory
and discusses the implications associated with the non-collapse of the wavefunction (i.e. parallel worlds). This paper is important because it discusses the probability of multiple universes, a
theory that (so far) is beyond our limits of observation. | {"url":"https://fermatslibrary.com/s/many-lives-in-many-worlds","timestamp":"2024-11-03T06:07:43Z","content_type":"text/html","content_length":"415508","record_id":"<urn:uuid:82906d63-1658-45c4-8790-cad3315b64b9>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00488.warc.gz"} |
the ratio between the coefficient of x^15 and the independent term of x in the expansion of (x^2+2/x)^15, is
the ratio between the coefficient of x^15 and the independent term of x in the expansion of (x^2+2/x)^15, is
526 views
Answer to a math question the ratio between the coefficient of x^15 and the independent term of x in the expansion of (x^2+2/x)^15, is
101 Answers
To solve the problem, follow these steps:
1. Identify the general term in the binomial expansion of \( \left( x^2 + \frac{2}{x} \right)^{15} \).
T_k = \binom{15}{k} (x^2)^{15-k} \left( \frac{2}{x} \right)^k
2. Simplify to find the term involving \( x \):
T_k = \binom{15}{k} x^{2(15-k)} \cdot \frac{2^k}{x^k} = \binom{15}{k} 2^k x^{30-3k}
3. Coefficient of \( x^{15} \):
Set the exponent \( 30-3k=15 \) to solve for \( k \):
30 - 3k = 15 \implies k = 5
Coefficient is:
\binom{15}{5}2^5=\frac{15!}{5! \cdot10!}\cdot32=96096
4. Independent term, constant term (no \( x \)):
Set the exponent \( 30-3k=0 \) to solve for \( k \):
30 - 3k = 0 \implies k = 10
Coefficient is:
\binom{15}{10}2^{10}=\frac{15!}{10! \cdot5!}\cdot1024=3075072
6. Ratio of coefficients:
Thus, the ratio is:
Frequently asked questions (FAQs)
What is 40% expressed as a simplified fraction?
Math Question: Can you factor the expression x^2 + 5x + 6?
Question: What is the formula for finding the median of a set of data with an odd number of values? | {"url":"https://math-master.org/general/the-ratio-between-the-coefficient-of-x-15-and-the-independent-term-of-x-in-the-expansion-of-x-2-2-x-15-is","timestamp":"2024-11-07T16:17:07Z","content_type":"text/html","content_length":"254227","record_id":"<urn:uuid:e89fadd7-4a02-42b9-bedd-38ce6dd4eba7>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00482.warc.gz"} |
17 Arithmetic
Unless otherwise noted, all of the functions described in this chapter will work for real and complex scalar, vector, or matrix arguments. Functions described as mapping functions apply the given
operation individually to each element when given a matrix argument. For example:
sin ([1, 2; 3, 4])
⇒ 0.84147 0.90930
0.14112 -0.75680 | {"url":"https://docs.octave.org/v4.0.3/Arithmetic.html","timestamp":"2024-11-07T03:55:00Z","content_type":"text/html","content_length":"5512","record_id":"<urn:uuid:b25f8d12-e14c-4a62-a1a4-d39da5362391>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00011.warc.gz"} |
Categorical Data
Sociology 73994
Categorical Data Analysis
Richard Williams, Instructor
Spring 2022 [Completed]
NOTE: My Stata Highlights page includes links to Stata and statistical handouts from my other courses that may interest readers.
Last Year's Notes: Click here if you want to see the complete online notes and handouts from the last time the course was taught. I will be making some changes this year; but the old notes should
be fine for anyone who wants to get a head start on methods we haven't gotten to yet. Links on this page will become "live" once the handout is ready.
Stata is in the labs. You can also order your own personal copy of Stata through the GradPlan package. Stata Student Pricing is especially cheap. Stata 17 is now out and I encourage you to get it
if you want to work off campus. Alternatively you can use the Online Virtual Labs. If you need something more powerful you can check out How to Use Stata on the Notre Dame Center for Resource
Computing (CRC) Machines. These use Unix and can be accessed remotely over the web. I'd much rather have you use the online virtual machines or the CRC machines instead of some hand-me-down old
version of Stata that has bugs that were fixed years ago or which lacks important new features.
NOTE: The following special types of files are used on this web page. Some materials are available only to nd.edu users.
Pdf files. Require Adobe Acrobat.
Stata files.
Useful sites for learning about Stata and SPSS
│Rich Williams' Stata Highlights Page │UCLA's Statistical Computing Resources │
│RW Suggestions for Using Stata at Notre Dame│UCLA's Stata Resources │
│RW's Suggested downloads │UCLA - How does Stata compare with SAS and SPSS? │
│Resources for learning Stata │Wisconsin SSCC Articles on Statistical Computing │
│The Stata User Support Page │Ben Jann's estout/esttab support page (esttab & estout are great for formatting output from Stata) │
│Stata YouTube Video Tutorials │Stata FAQS │
Links to major topics on this page
[Overview] [Review of Continuous Models] [Overview of GLMs, MLE] [Basics of logistic regression]
[[Adjusted Predictions and Marginal Effects] [Complex Survey Designs] [Missing Data] [Ordinal Outcomes I]
[Multinomial Outcomes] [Intermediate Logistic Regression] [Nested Models] [Panel Data/ Multilevel Models]
[gologit models] [Group Comparisons/Hetero Choice] [Count Outcomes] [Rare Events] [Fractional Response Models]
Overview. This course discusses methods and models for the analysis of categorical dependent variables and their applications in social science research. Researchers are often interested in the
determinants of categorical outcomes. For example, such outcomes might be binary (lives/dies), ordinal (very likely/ somewhat likely/ not likely, nominal (taking the bus, car, or train to work)
or count (the number of times something has happened, such as the number of articles written). When dependent variables are categorical rather than continuous, conventional OLS regression
techniques are not appropriate. This course therefore discusses the wide array of methods that are available for examining categorical outcomes.
Syllabus (Updated April 12, 2022)
Regression Models for Categorical Dependent Variables Using Stata (required text; as far as I know it is cheapest to order directly from the Stata Bookstore)
Fixed Effects Regression Models, by Paul Allison. (required text; order it from Amazon or elsewhere). Notre Dame users can also access an e-version from our library.
Stata Files Stata data files used in the course will usually be stored in this folder.
Recommended Reading (ND.Edu Netid is required for access)
Cloud Storage. I strongly urge you to use something like Dropbox, Google Drive, Microsoft Onedrive, Box, or a similar service to back up your most critical files. With so many free and
inexpensive services now available, there is really no excuse for losing more than a day's work, if that. I am very fond of Dropbox but just about anything will do. For more options see Notre
Dame Collaboration and File Storage.
Useful Research Resources for Notre Dame Students. ND students can get free/low cost newspaper subscriptions and free online access to millions of research articles. I find these resources
incredibly helpful to me in my work. Be sure to check them out!
Basic Dataset Exploration & File Preparation. You should get to know your dataset before you start to analyze it. This handout includes suggestions for using dataset documentation; documenting
your own work; Using fre instead of tab1for frequencies; Including options with tab2; coding missing data correctly; converting string variables to numeric variables; Creating Binary or Ordinal
Variables from Continuous Measures; Handling More Complicated Data Structures; and Other Useful Resources. The PDF file is bookmarked so you can easily skip to the sections you are interested in.
This handout may be especially helpful when doing Homework #1.
A Note on the Treatment of Gender and Race in My Statistics Notes. Early versions of some of my statistics handouts first saw life more than 30 years ago. There have been many changes in
statistical practice and preferred wording over those 30 years. My notes (and for that matter probably most of my publications) reflect common past practices. As I update handouts, I am making
changes to reflect current, more inclusive terminology. The handout also lists several resources for those who are interested in studying often-overlooked gender identities such as transgender
and non-binary.
Brief Review of Models for Continuous Outcomes (Review this on your own and come to me with questions as needed)
reg01.dta - Data file used in the Stata Regression handout
Using Stata for OLS Regression (If you are interested, click here for a similar handout using SPSS)
I. Foundations of categorical data analysis.
This section will go over the basics of logistic regression. It will also go over techniques for making results more interpretable; analyzing data sets with complex sampling schemes; and
(possibly) techniques for handling missing data. I call these topics "foundations" because once you understand them it is very easy to extend them to other CDA methods.
Overview of Generalized Linear Models, Maximum Likelihood Estimation
Introduction to Generalized Linear Models
L01.do - Stata program for GLM handout
Maximum Likelihood Estimation & Troubleshooting
L02.do - Stata program for MLE handout
Assignment 1: Preliminary Data Analysis & Setup. Due January 20, 2022. Send copies to both the instructor and the TA
Models for Binomial Outcomes: Basics of Logistic Regression
Logistic Regression I: Problems with the Linear Probability Model (LPM)
logit01.do - Stata file(s) used in the logistic regression 1 handout
Logistic Regression II: The Logistic Regression Model (LRM) (NOTE: slightly revised on Jan 26, 2022)
logit02.do - Stata file(s) used in the logistic regression 2 handout
Logistic Regression III: Hypothesis Testing, Comparisons with OLS (NOTE: Modified on January 17, 2022)
logit03.do - Stata file(s) used in the logistic regression 3 handout
Measures of fit - Pseudo R^2, BIC, AIC (Read on your own; I may just discuss a few highlights)
L05.do - Stata program for measures of fit
Using Stata for Logistic Regression (be sure to read this on your own, as it covers important details we may not go over in class)
logistic-stata.do - Stata file(s) used in the using stata for logistic regression handout
logist.dta - Stata data file used in the Logistic Regression handouts
(Optional) Paul Allison further elaborates on the merits of the logit model. However, Paul von Hippel maintains the LPM often isn't so bad. Allison responds back. Marc Bellamare, Christopher Zorn
, Dave Giles, and Marc Bellamare (again) further debate the issue. My own feeling is that the best of both worlds is to use logit and then couple it with marginal effects.
Assignment 2: Basics of Logistic Regression. Due February 3, 2022.
cruzmake.do - If you are curious, this shows how the fake data were created
Hw02.do - do file used for the results presented in HW # 2
cruz.dta - data file used in HW # 2
Interpreting results: Adjusted Predictions and Marginal effects
The results from binomial and ordinal models can often be difficult to interpret. All too often, researchers discuss the sign and statistical significance of results but say little about their
substantive significance. I will expect every student paper to use the methods described in this section and/or one of the advanced methods we discuss later in the course. I will also expect
students to be familiar with some of Long & Freese's spost13 commands, such as mtable and mchange. I will probably go over the first handout fairly carefully and then skim the rest, since they
are mostly examples you can go over on your own.
Margins01 and Margins03 will definitely be covered in class. The others you should read on your own and ask questions if you have them.
Using Stata's Margins Command to Estimate and Interpret Adjusted Predictions and Marginal Effects (click here for the Powerpoint version).
Margins01.do - Stata program for margins #1 handout
Also - the Stata Journal article I wrote on this is available for free. For an application of the margins command, see my 2013 article with Lutz Bornmann entitled How to calculate the
practical significance of citation impact differences? The do file and data file required to replicate the paper's analysis are here.
Marginal Effects for Continuous Variables. We have mostly been talking about adjusted prediction and marginal effects for categorical variables. This handout explains how marginal effects for
continuous variables are estimated and interpreted, and includes additional technical details for those who want them. Read this on your own if I don't cover it in class.
Margins02.do - Stata program for margins #2 handout
Understanding & Interpreting the Effects of Continuous Variables: The MCP Command. We have mostly been talking about adjusted prediction and marginal effects for categorical variables. This
handout explains how marginal effects for continuous variables are estimated and interpreted, and includes additional technical details for those who want them.
Margins03.do - Stata program for margins #3 handout
Using the spost13 commands for adjusted predictions and marginal effects with binary dependent variables. This is pretty easy so I may just have you read this on your own.
Margins04.do - Stata program for margins #4 handout
Adjusted predictions and marginal effects for multiple outcome commands and models. Commands like ologit, oprobit, mlogit, oglm, and gologit2 estimate models where there are multiple
outcomes, e.g. the dependent variables is ordinal and has 5 categories. The same margins and spost13 commands can generally be used with each. This handout explains how. I am putting this
handout here so it is with all the other margins handouts but you will probably to read it after we've covered commands like ologit and mlogit.
Margins05.do - Stata program for margins #5 handout
Model Coefficients, Adjusted Predictions, & Marginal Effects: A Summary of How All Three are Related. This handout gives a brief summary of major points that have been covered in this section.
Margins06.do - Stata program for margins #6 handout
Assignment 3: Interpreting results: Adjusted Predictions and Marginal Effects. Due February 10, 2022.
Hw03.do - do file used for the results presented in HW # 3
cruz.dta - data file used in HW # 3
Categorical Data Analysis with Complex Survey Designs
By default, most statistical techniques assume that data were collected via simple random sampling. This is often not true for large national data sets. Fortunately, Stata makes it easy to
analyze such data, but there are some important differences in how you go about testing hypotheses and assessing model fit.
Analyzing Complex Survey Data: Some key issues to be aware of. I've consolidated what had been a few separate handouts. Key issues for both OLS regression and categorical data analysis are
svy01.do - Stata program for svy analysis handout
Also worth seeing: UCLA's (see lower third of page) and StataCorp's FAQS on Survey Data Analysis. Also Chuck Huber has an excellent video on Specifying the design of your survey data in Stata.
(But if your dataset is as complicated as the one in his example, you'd better hope that the study documentation is really clear on how to do things.)
I am mostly covering this here because it is an important topic and there wasn't enough time to cover it in the new Stats I! But, several of the methods do involve the use of categorical data
analysis, so it isn't totally out of place. Also, since you are analyzing your own data this semester, you are more likely to have to deal with missing data.
Coding Missing Data. This two page handout shows how you can you can use commands like mvdecode to make sure Missing Data is being handles correctly in Stata.
MDCoding.do - Stata file used in the MDCoding handout
MDCoding.dta - Stata data file used in the MDCoding handout
Missing Data Part 1: Overview, Traditional Methods. I'll talk about parts of this in the class but you should read the whole thing on your own. It mostly explains why most traditional methods
for handling missing data (other than listwise deletion) are seriously flawed.
Missing Data Part 2: Multiple Imputation & Maximum Likelihood [Revised Feb 6, 2022] This is a very long handout! But it has been reworked so only the first few pages are required. You should
read more of the handout if you actually want to use Multiple Imputation in your research.
mdpart2.do - Stata file for the MD Part 2 handout
Also worth seeing: The Wisconsin Social Science Computing Cooperative has some great pages on MI. Also possibly helpful are the Statacorp FAQs on MI and this page from UCLA.
Assignment 4: Complex survey Designs; Multiple Imputation. Due February 17, 2022.
Hw04.do - do file used for the results presented in HW # 4
II. Intermediate CDA Methods
Here we will talk about other commonly used CDA methods, including ordinal regression, models for multinomial outcomes, and models for count outcomes.
Models for Ordinal Outcomes I: The ordered logit and interval regression models
(Optional but highly recommended) As part of the Sage Research Methods Foundations Project (SRMF), Williams and Quiroz (2019) provide an overview of Ordinal Regression Models. Both basic and more
advanced methds (e.g. interval interval regression, generalized ordered logit models, heterogeneous choice models) are discussed. Those with an ND.edu account can access it here. For those not at
ND, if your library has purchased SRMF (and if it hasn't it should!) the entry can be found at https://methods.sagepub.com/Foundations/ordinal-regression-models.
Ordinal Logit Models: Basic & Intermediate Topics. After reading this you will also want to look at the Margins05 handout if you haven't already.
ologit1.do - Stata program for ologit overview
(Optional) It doesn't have much to do with statistics, but here is the true story of the engineer who tried to save Challenger.
Interval Regression [Revised Feb 18, 2022]
intreg2.do - Stata program for interval regression
Ordinal Independent Variables. We often want to use ordinal variables as independent/explanatory variables in our models. Rightly or wrongly, it is very common to treat such variables as
continuous. This handout discusses when it is appropriate to do so. The handout also discusses other possible strategies that can be employed with ordinal independent variables, such as the
use of Sheaf coefficients. Alternatively, you can read my entry on this in Sage Research Methods Foundations.
OrdinalIndependentVars.do - do file used in the Ordinal Independent Variables handout
Models for Multinomial Outcomes
When categories are unordered, Multinomial Logistic regression is one often-used strategy. We will discuss several ways to aid in the interpretation and testing of these models.
mlogit1.do - Stata program for mlogit, including adjusted predictions & marginal effects
Other Post-Estimation Commands for mlogit
mlogit2.do - Stata program for other mlogit post-estimation commands
Assignment 5: Ordinal and Multinomial Models Due February 24, 2022.
Hw05.do - do file used in HW # 5
Assignment 6: Paper Proposal - Due March 3, 2022.
Models for Binary Outcomes II: Intermediate Logistic Regression
The Latent Variable Model for Binary Regression
L03.do - Stata program for Latent variable handout
Standardized Coefficients in Logistic Regression
L04.do - Stata program for standardized coefficients
Models for Binary Outcomes III: Comparing logit and probit coefficients across nested models
Prelude to Comparing Logit & Probit Coefficients Between Nested Models
Nested01.do - Stata program for Prelude to comparing coefficients across nested models
Comparing Logit & Probit Coefficients Between Models (click here for Powerpoint version)
Comparing Logit & Probit Coefficients Between Nested Models (Extended Version). OPTIONAL. This is actually an older version of the handout but it includes several additional points that might be
Assignment 7: Intermediate issues in logistic regression analysis. Due March 17, 2022.
Hw07.do - do file used in HW # 7
III. Advanced Topics (Subject to Change or Re-Ordering)
Sometimes the same individuals (or nations, or companies) are measured at multiple points in time. The statistical technique used needs to reflect the fact that the different measurements are
not independent of each other. This is a big topic and goes well beyond Categorical Data Analysis, but a few basic commands, e.g. xtlogit, will be discussed.
NOTE: I AM ACTUALLY GOING TO USE SOME OF THE NOTES FOR A MINI-COURSE I TAUGHT DURING SUMMER 2018 IN TAIWAN. THE COMPLETE SET OF NOTES FROM THAT CLASS ARE HERE.
Introduction (The course outlines lists all the topics covered in Taiwan. We are only doing the first few.)
Fixed effects and conditional logit models
Fixed effects versus random effects models
Also recommended:
http://stats.idre.ucla.edu/stata/dae/mixed-effects-logistic-regression/ (NOTE: xtmelogit has been superceded by melogit)
Assignment 8: Panel Data Methods. Due March 31, 2022.
Hw08.do - do file used in HW # 8
NOTE: I've covered enough to give you some basic competency with Panel and Multilevel Models, but the Taiwan page has a lot more if you want it. Hybrid models are a way of estimating both fixed
and random effects in the same model (albeit with some limitations). You can do adjusted predictions and marginal effects with random effects models. This far-from-finished presentation and
handout show an application of many multilevel model methods, including random slopes models. You can do panel data linear models too. Sometimes you are not interested with whether an event
occurs, but how quickly (e.g. what factors make people die more quickly?) In such cases, Discrete Time Methods for Event History Analysis can sometimes be a good way to go.
Models for Ordinal Outcomes II: Generalized ordered logit models
The assumptions of the ordered logit model are often violated. The generalized ordered logit model (estimated by gologit2) sometimes provides a viable but still parsimonious alternative.
GOLOGIT Part 1: The gologit model & gologit2 program (Powerpoint version)
GOLOGIT Part 2: Interpretation of results. (Powerpoint version) - Also get this handout
Updates to gologit2: This describes major updates to the program since it was released in 2006.
For more detail, you should read the 2006 Stata Journal article that introduced the program and the 2016 Journal of Mathematical Sociology article on how to interpret results. The JMS reading
requires an nd.edu accout to access; others can find it described at http://www.tandfonline.com/doi/full/10.1080/0022250X.2015.1112384.
Models for Ordinal Outcomes III: Heterogeneous Choice Models and Other Methods for Comparing Logit & Probit Coefficients Across Groups
To the surprise of many, techniques used for group comparisons in OLS regression (e.g. adding interaction effects) can be highly problematic in logistic and ordinal regression. As Hoetker
notes, "in the presence of even fairly small differences in residual variation, naive comparisons of coefficients [across groups] can indicate differences where none exist, hide differences
that do exist, and even show differences in the opposite direction of what actually exists." We will discuss how heterogeneous choice models and possibly other methods offer possible
solutions. The handout covers only the most critical points; for those who want to know more, the article by Allison and the two articles by Williams that are mentioned in the references are
highly recommended.
Comparing Logit and Probit Coefficients Across Groups: Problems, Solutions, and Problems with the Solutions (also available in Powerpoint).
Handout for Comparing Logit & Probit Coefficients Across Groups
Assignment 9: Advanced issues/ models for ordinal outcomes. Due April 7, 2022
Hw09.do - do file used in HW # 9
Variables that count the # of times something happens are common in the Social Sciences. For example, Long examined the # of publications by scientists. Count variables are often treated as
though they are continuous and the linear regression model is applied; but this can result in inefficient, inconsistent and biased estimates. In this section we will examine some of the many
models that deal explicitly with count outcomes.
Count Models (NOTE: slightly revised on April 12, 2022)
count01.do - Stata program for count models
Assignment 10: Count Models - Due April 14, 2022.
Hw10.do - do file used in HW # 10
IV. Special Topics (Time Permitting)
Models for Binary/Proportional Outcomes IV
Analyzing Rare Events with Logistic Regression. Many researchers worry about whether they can legitimately use conventional logistic regression for data in which events are rare, e.g. only 20 or
30 people experience the event.. This handout describes the problem and discusses various solutions, with an emphasis on Penalized Maximum Likelihood (aka the Firth Method).
RareEvents.do - Stata program for rare events models
Strongly Recommended: Analysis of Rare Events, by Heinz Leitgob. Also Available at https://methods.sagepub.com/foundations/analysis-of-rare-events.
Analyzing Proportions / Fractional Response Models. In many cases, the dependent variable of interest is a proportion, i.e. its values range between 0 and 1. Wooldridge (1996, 2011) gives the
example of the proportion of employees that participate in a company's pension plan. This handout shows that methods used for binary outcomes can easily be adapted to deal with such variables.
Other approaches are also discussed.
fracmodels.do - Stata program for models for analyzing proportions
Strongly Recommended: Analysis of Proportions, by Maarten L Buis. Also available at https://methods.sagepub.com/foundations/analysis-of-proportions. | {"url":"https://www3.nd.edu:443/~rwilliam/stats3/","timestamp":"2024-11-02T21:15:34Z","content_type":"text/html","content_length":"65081","record_id":"<urn:uuid:c26e8d60-3d59-4e66-8006-4682fa20d870>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00073.warc.gz"} |
Department of Mathematics
Monty McGovern and Sándor Kovács
PDL C-38 and Zoom: https://washington.zoom.us/j/96029696426
Zoom passcode is 9 factorial.
Monty McGovern: Geometry of the flag variety and representation theory
Abstract: A flag in complex n-space is a nested sequence of subspaces. These humble objects have a very rich geometry with many unexpected and fascinating connections to the representation theory of
Lie groups and algebras. I will explore some of these, repeatedly going back and forth between geometry and representation theory.
Sándor Kovács: Algebra=Geometry
Abstract: In algebraic geometry, algebra and geometry go hand-in-hand to understand and solve problems in both disciplines. In fact, one of the reasons for the broad effectiveness of algebraic
geometry in other areas stems from the fact that this interplay between algebra and geometry allows us to view problems in completely different light from substantially different points of view. In
this talk I will describe a way to construct a geometric object from purely algebraic data and then use it to prove a (seemingly) purely algebraic result. | {"url":"https://math.washington.edu/events/2021-10-21/current-topics-session-2","timestamp":"2024-11-05T22:15:04Z","content_type":"text/html","content_length":"50954","record_id":"<urn:uuid:b24dc892-6ae6-4a80-85a3-fad8e4571dfc>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00795.warc.gz"} |
Design of Compact Baugh-Wooley Multiplier Using Reversible Logic
Design of Compact Baugh-Wooley Multiplier Using Reversible Logic ()
Received 18 April 2016; accepted 15 May 2016; published 15 June 2016
1. Introduction
In today’s digital era, developing digital circuits is bounded by the research towards investigating various nano devices to provide substitution for CMOS technology. As the nano devices are
developed, the density of digital chips is being increased naturally seeking the solution for the power consumption and the heat dissipation developed by this power consumption. This scenario
motivates the study of reversible computing field. The origin of the reversible computing is the research work done by R. Landauer in early 1960’s stating that irrespective of the realization
technique, the irreversible hardware computation results energy dissipation due to information loss [1] . According to R. Landauer, loss of one bit information dissipates at least KTln2 joules of
energy in the form of heat, where the Boltzmann’s constant is represented by K and the absolute temperature is represented by T. As per C. H. Bennett’s research, the circuits built using reversible
logic gates will prevent KTln2 joules of energy dissipation in a circuit [2] . A circuit will be known as reversible if it can bring back the inputs from the outputs. Also the relationship between
the inputs and outputs should be maintained as one-to-one and unique. This constraint forces the number of inputs to be equal to the number of outputs [3] [4] .
Hence reversible logic is being applied in various research domains like DNA computing, nanotechnology, Low Power CMOS design and quantum computing. The quantum circuits can be constructed only with
reversible logic gates. Besides, synthesizing reversible logic circuits is much difficult than conventional irreversible logic circuits due to the constraints. In the reversible logic circuit design,
fan-out and feedback are not permitted [4] . Apart from that the reversible logic circuit should use 1) lowest number of reversible gates, 2) lowest number of garbage outputs, 3) lowest number of
constant inputs. The garbage output is the one which is not used for further computations. The additional input that is included to the irreversible function to convert it reversible is called
constant input [4] .
In the recent years various reversible multiplier designs have been proposed [5] - [9] . In [5] , the design deals with 4 × 4 parallel multiplier. It has been done in two steps as follows: 1) Partial
Product Generation, 2) Multi Operand Addition. To generate the partial products, 16 Peres gates have been used, for 16 one-bit multiplication arrays. Then the four operand addition has been performed
using Peres gates and Double Peres gates. In [6] , the authors have proposed a new reversible gate called as HNG gate. This work also involves two steps as in [5] . The partial products have been
generated using Peres gates. HNG gates are used in the second step, Multi operand addition. The work [7] also follows the same strategy as the previous two works, multiplication in two steps. In this
work, the authors have proposed Peres Full Adder Gate (PFAG). In this work also, like the previous works, the partial products have been generated using Peres gate. The PFAG gate is used in the multi
operand addition. In [8] , the authors have proposed a new reversible gate called as RAM gate. This gate is mainly used as a copying gate as fan-out is not allowed in reversible logic design. This
gate has been used in the partial product generation. In the second step, the multi operand addition, Peres gates and Double Peres gates have been used. In our work, we have proposed a reversible
multiplier cell which can be efficiently used in the Baugh Wooley multiplier.
The organization of the paper is as follows. Section 2 is an overview of basic reversible gates. Section 3 is an overview of Baugh-Wooley multiplier. A detailed representation and explanation is done
in this section. The proposed reversible multiplier design and its functions are discussed in section 4. The results and discussions of the proposed reversible Baugh-Wooley multiplier are presented
in section 5. Conclusion is in section 6.
2. Reversible Logic Gates
A circuit will be known as reversible if it can bring back the inputs from the outputs. Also the relationship between the inputs and outputs should be maintained as one-to-one and unique. This
constraint forces the number of inputs to be equal to the number of outputs. This section deals with the preliminary reversible gates available in the literature.
Figure 1 shows a basic 2 × 2 reversible gate known as Feynman Gate [10] and its quantum representation. Feynman Gate (FG) can be used as a copying gate. Since in reversible circuits the fan-out
greater that one is not permitted, this gate is useful for duplicating the inputs. When the input B = “0”, the outputs P = “A” and Q = “A”. This gate is also known as Controlled-Not gate. When the
input A = “1”, the output Q produces the complement of A.
Figure 2 is called as 3 × 3 Toffoli gate [11] and also code-named as the “controlled-controlled-not” gate, which depicts its action. The number of inputs and outputs are three in count; if the first
two bits A and B are set, the third bit will be inverted, otherwise all bits will keep on the same value. When the input C = “1”, the output R produces the AND function between the inputs A and B.
Figure 3 is a 3 × 3 Fredkin gate [11] . If the input A = “0”, then Q = “B” and R = “C”. Whereas if the input A = “1”, then Q = “C” and R = “B”. Hence this is also called as Swap gate. The
functionality shows that this gate will be useful in designing a n-bit MUX with (n + 1) garbage output.
Figure 4 is a 3 × 3 Peres Gate (PG) [12] . It is also known as New Toffoli Gate (NTG). Function wise Peres Gate will be equal with the bit conversion generated by a Toffoli Gate succeeded by a
Feynman Gate.
3. Baugh-Wooley Multiplier
When we are performing the signed n × n multiplication there will be no difference if the result has the same bit-width as the inputs. For Eg: the multiplication of two numbers “−2” and “3” results
in “−6”. Its binary representation is (1110)[2] × (0011)[2] = (1010)[2]. But what happens if we want the result to be in “2n” bits. Either we need to use sign extension or “2n × 2n” array multiplier.
One of the efficient algorithms to handle such situation is the Baugh-Wooley multiplication. This design method has been established in order to design structured multipliers, appropriate for 2’s
complement numbers [13] . Let the numbers to be multiplied be A and B. Here “A” denotes the n-bit multiplier and “B” denotes the n-bit multiplicand. The multiplier A and the multiplicand B can be
represented as
where the bits in A and B are denoted as a[i]’s and b[i]’s, respectively, and a[n][−1] and b[n][−1] are the sign bits. The below equation gives the n * n product, P = A × B:
The final product could be generated by subtracting the last two positive terms from the first two terms. Instead of doing subtraction operation as in the normal multipliers, it is possible to obtain
the two’s complement of the last two terms and add all the terms to deliver the final product. The last two terms will be of n − 1 bits length where each term has the binary extension from position 2
^n^−1 up to 2^2n−3. In contrast the final product is 2^n bits which extend the binary weight from 2^0 up to 2^2n−1. As a first step pad each of the last two terms in the product P with zeros to
obtain a 2n-bit number to aid adding it with the other terms. Later the padded terms extend in binary weight from 2^0 up to 2^2n−1. Let X be one of the last two terms that can represent it with zero
padding as
The final n * n product, P = A × B is given by:
Let two 4-bit binary numbers be A and B, then the product, P = A × B will be 8 bit long and is
The block diagram representation of 4 bit Baugh-Wooley multiplier is shown in Figure 5.
4. Proposed Reversible Logic Gates
In the block diagram shown in Figure 5, three types of cells are used. The yellow cells represent the full adder.
The black cells are representing the multiplier cell used for 2’s complement numbers. The grey cells represent the multiplier cell. Each of the multiplier cell receives four inputs namely, the
multiplier input (horizontal-green line), multiplicand input (vertical-red line), carry from previous cells (vertical-black line) and sum from previous cells (diagonal-black line). They produce two
outputs namely sum output (diagonal-black line) and carry output (vertical black line).
In this work we are proposing two reversible multiplier cells representing black and grey cells. Since each cell is having four inputs and two outputs, the reversible multiplier cell, in order to
maintain the reversible constraints it is developed as a cell having five inputs and five outputs. Out of this, three outputs are maintained as garbage outputs. Garbage outputs are by definition
don’t care outputs and thus can be left unspecified leading to an incompletely specified function [14] . The Reversible Multiplier cell (MC) is shown in Figure 6. It is a 5 × 5 reversible logic. This
is the first 5 × 5 reversible multiplier cells proposed in the literature. Out of the five outputs, two outputs (Q and R) are left unspecified, since these are the garbage outputs. The functions S
and T will produce sum and carry outputs respectively.
The Toffoli gate representation of the proposed Reversible Multiplier Cell is given in Figure 7. The representation has the Gate Count of 15. The Quantum cost is 69. The number of two-Qubit gates is
The Reversible Complement Multiplier Cell (CMC) is shown in Figure 8. It is also a 5 × 5 reversible logic gate. Out of the five outputs, two outputs (Q and R) are left unspecified, since these are
the garbage outputs. The functions S and T will produce sum and carry outputs respectively of the complement function of the Baugh- Wooley structure.
These proposed multiplier cells are having one constant input. These cells will function as a multiplier circuit when the input “E = 0”. The input A is the multiplier bit. The input B is the
multiplicand bit. The input C is the carry input from the previous cells. The input D is the sum input from the previous cells. For the first level the inputs C and S will be “0”. The outputs P, Q
and R are considered as garbage outputs. Since this is an incompletely specified reversible logic gates the functions Q and R are not specified.
5. Results and Discussions
The reversible multiplier designs available in the literature are for the array multipliers. There is no any specific application of any algorithm except [15] . Since this work may be the first in
the literature we can’t compare and evaluate with other similar proposals. However this work is compared and evaluated with the other array multiplier designs available in the literature. Therefore
the proposed multiplier cells are evaluated based on the Gate count, Garbage inputs and Garbage outputs. Since the proposed cells are incompletely specified cells we could not generate the Quantum
cost and therefore we could not evaluate the proposed gates based on the Quantum cost.
5.1. Gate Count & Hardware Complexity
Measuring the reversible logic design in terms of number of gates is one of the major factors. In [5] , the design requires a total of 40 reversible gates, [9] requires 42, total number of gates
required is 44 in [7] and in [8] the number of gates required is 32 gates. The proposed Baugh-Wooley multiplier design requires 20 gates. Therefore, the hardware intricacy of the proposed design is
less compared to the existing approaches.
5.2. Constant Inputs
One of the major factors in the design of a reversible logic circuit is the number of constant inputs. The input used as a control input by connecting to either logical low or logical high to get the
required function at the output is called garbage/constant input. The proposed reversible Baugh-Wooley multiplier design requires 16 constant inputs, but the design in [5] [7] - [9] requires 52, 40,
44 and 42 respectively. Hence the proposed Baugh- Wooley Multiplier design is better than existing designs.
5.3. Garbage Outputs
The number of output of the reversible gate that is not making useful functions is referred as garbage output. Other constraint in designing reversible logic circuit is optimizing garbage outputs.
The proposed reversible Baugh- Wooley multiplier design produces 48 garbage outputs, but the design in [5] [7] - [9] produces 52, 52, 40 and 49 garbage outputs respectively. Therefore, it is clear
that this is the better design than the existing counterparts.
The conclusion of the above discussion is that, it is evident that the proposed reversible Baugh-Wooley multiplier circuit design is better than the existing designs with respect to gate counts,
garbage inputs and garbage outputs.
5.4. Evaluation of the Proposed Reversible Baugh-Wooley Multiplier Circuit
The proposed reversible Baugh-Wooley multiplier circuit is more efficient compared to the existing circuits presented by [5] [7] - [9] . This can be understood easily with the help of the comparison
results shown in Table 1.
6. Conclusion
In this work, the design of 5 × 5 reversible Multiplier Cell (MC) and reversible Complement Multiplier Cell is proposed. These are the first 5 × 5 reversible multiplier cells proposed in the
literature. These reversible multiplier cells are targeted for reversible Baugh-Wooley multiplier design. The proposed reversible multiplier cells are capable of multiplying 2 bits in the current
array and add the result with the sum and carry outputs of previous array. The Toffoli gate synthesis of the proposed reversible multiplier cell is also given. The functionality of the multiplier
cell was verified with RC viewer. This design is useful in the multiplier design with reduced number of gates and constant inputs. Even the proposed design is having moderate garbage outputs; we can
conclude that this design is better in terms of number of gates and constant inputs. The number of gates, constant inputs and garbage outputs
are analyzed. It is comprehended that the number of gates, the constant inputs and garbage outputs values are fewer in number in the proposed design compared to the existing approaches. The design
can be enhanced to construct n × n reversible Baugh-Wooley multiplier circuit. | {"url":"https://scirp.org/journal/paperinformation?paperid=67366","timestamp":"2024-11-03T10:47:07Z","content_type":"application/xhtml+xml","content_length":"105899","record_id":"<urn:uuid:3992cc8f-9b75-4b06-a024-97f40f88214c>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00895.warc.gz"} |
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representations aid comprehension for students inclined toward visual knowing. Auditory Learners Spoken multiplication problems or mnemonics satisfy students who realize principles through acoustic
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7 Multiplication Worksheets Examples In PDF Examples
Free Printable Double Digit Multiplication Worksheets Lexia s Blog
Check more of Beginner Multiplication Worksheets below
Multiplication Kindergarten Worksheets Leonard Burton s Multiplication Worksheets
Free 1 Digit Multiplication Worksheet By 0s And 1s Free4Classrooms
Beginning Multiplication Worksheets With Pictures Times Tables Worksheets
Multiplication Worksheets 6 7 8 Printable Multiplication Flash Cards
12 Best Images Of Beginner Multiplication Worksheets Beginning Multiplication Worksheets
Beginning Multiplication Worksheets
Multiplication Facts Worksheets Math Drills
These multiplication worksheets include some repetition of course as there is only one thing to multiply by Once students practice a few times these facts will probably get stuck in their heads for
life Some of the later versions include a range of focus numbers In those cases each question will randomly have one of the focus numbers in
Printable Multiplication Worksheets Super Teacher Worksheets
On this page you have a large selection of 2 digit by 1 digit multiplication worksheets to choose from example 32x5 Multiplication 3 Digits Times 1 Digit On these PDF files students can find the
products of 3 digit numbers and 1 digit numbers example 371x3 Multiplication 4 Digits Times 1 Digit
These multiplication worksheets include some repetition of course as there is only one thing to multiply by Once students practice a few times these facts will probably get stuck in their heads for
life Some of the later versions include a range of focus numbers In those cases each question will randomly have one of the focus numbers in
On this page you have a large selection of 2 digit by 1 digit multiplication worksheets to choose from example 32x5 Multiplication 3 Digits Times 1 Digit On these PDF files students can find the
products of 3 digit numbers and 1 digit numbers example 371x3 Multiplication 4 Digits Times 1 Digit
Multiplication Worksheets 6 7 8 Printable Multiplication Flash Cards
Free 1 Digit Multiplication Worksheet By 0s And 1s Free4Classrooms
12 Best Images Of Beginner Multiplication Worksheets Beginning Multiplication Worksheets
Beginning Multiplication Worksheets
Simple Multiplication Worksheets Superstar Worksheets
Wonderful Repeated Addition multiplication Worksheet Pdf Aglocomoonjaycomunity
Wonderful Repeated Addition multiplication Worksheet Pdf Aglocomoonjaycomunity
Free Beginning Multiplication Worksheets Best Kids Worksheets
FAQs (Frequently Asked Questions).
Are Beginner Multiplication Worksheets appropriate for any age groups?
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Consistent technique is key. Regular sessions, preferably a couple of times a week, can yield significant renovation.
Can worksheets alone enhance math skills?
Worksheets are an useful device but must be supplemented with varied learning approaches for comprehensive skill growth.
Exist on-line platforms offering free Beginner Multiplication Worksheets?
Yes, many instructional web sites use open door to a wide range of Beginner Multiplication Worksheets.
How can parents support their children's multiplication technique in your home?
Urging regular method, supplying support, and creating a favorable discovering atmosphere are valuable actions. | {"url":"https://crown-darts.com/en/beginner-multiplication-worksheets.html","timestamp":"2024-11-05T06:03:52Z","content_type":"text/html","content_length":"29518","record_id":"<urn:uuid:74ce8d3a-92c2-45bb-a06d-232e4f9c0b49>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00596.warc.gz"} |
6 - Data association - multi-target tracking tutorial
Go to the end to download the full example code. or to run this example in your browser via Binder
6 - Data association - multi-target tracking tutorial
Tracking multiple targets through clutter
As we’ve seen, more often than not, the difficult part of state estimation concerns the ambiguous association of predicted states with measurements. This happens whenever there is more than one
target under consideration, there are false alarms or clutter, targets can appear and disappear. That is to say it happens everywhere.
In this tutorial we introduce global nearest neighbour data association, which attempts to find a globally-consistent collection of hypotheses such that some overall score of correct association is
Make the assumption that each target generates, at most, one measurement, and that one measurement is generated by, at most, a single target, or is a clutter point. Under these assumptions, global
nearest neighbour will assign measurements to predicted measurements to minimise a total (global) cost which is a function of the sum of innovations. This is an example of an assignment problem in
combinatorial optimisation.
With multiple targets to track, the NearestNeighbour algorithm compiles a list of all hypotheses and selects pairings with higher scores first.
In the diagram above, the top detection is selected for association with the blue track, as this has the highest score/probability (\(0.5\)), and (as each measurement is associated at most once) the
remaining detection must then be associated with the orange track, giving a net global score/probability of \(0.51\).
The GlobalNearestNeighbour evaluates all valid (distance-based) hypotheses (measurement-prediction pairs) and selects the subset with the greatest net ‘score’ (the collection of hypotheses pairs
which have a minimum sum of distances overall).
In the diagram above, the blue track is associated to the bottom detection even though the top detection scores higher relative to it. This association leads to a global score/probability of \(0.6\)
- a better net score/probability than the \(0.51\) returned by the nearest neighbour algorithm.
A multi-target simulation
We start by simulating 2 targets moving in different directions across the 2D Cartesian plane. They start at (0, 0) and (0, 20) and cross roughly half-way through their transit.
Generate ground truth
from ordered_set import OrderedSet
truths = OrderedSet()
transition_model = CombinedLinearGaussianTransitionModel([ConstantVelocity(0.005),
timesteps = [start_time]
truth = GroundTruthPath([GroundTruthState([0, 1, 0, 1], timestamp=timesteps[0])])
for k in range(1, 21):
transition_model.function(truth[k-1], noise=True, time_interval=timedelta(seconds=1)),
truth = GroundTruthPath([GroundTruthState([0, 1, 20, -1], timestamp=timesteps[0])])
for k in range(1, 21):
transition_model.function(truth[k-1], noise=True, time_interval=timedelta(seconds=1)),
_ = truths.add(truth)
Plot the ground truth
Generate detections with clutter
Next, generate detections with clutter just as in the previous tutorial. This time, we generate clutter about each state at each time-step.
from scipy.stats import uniform
from stonesoup.types.detection import TrueDetection
from stonesoup.types.detection import Clutter
from stonesoup.models.measurement.linear import LinearGaussian
measurement_model = LinearGaussian(
mapping=(0, 2),
noise_covar=np.array([[0.75, 0],
[0, 0.75]])
all_measurements = []
for k in range(20):
measurement_set = set()
for truth in truths:
# Generate actual detection from the state with a 10% chance that no detection is received.
if np.random.rand() <= 0.9:
measurement = measurement_model.function(truth[k], noise=True)
# Generate clutter at this time-step
truth_x = truth[k].state_vector[0]
truth_y = truth[k].state_vector[2]
for _ in range(np.random.randint(10)):
x = uniform.rvs(truth_x - 10, 20)
y = uniform.rvs(truth_y - 10, 20)
measurement_set.add(Clutter(np.array([[x], [y]]), timestamp=truth[k].timestamp,
# Plot true detections and clutter.
plotter.plot_measurements(all_measurements, [0, 2])
Create the Kalman predictor and updater
As in the clutter tutorial, we will quantify predicted-measurement to measurement distance using the Mahalanobis distance.
Run the Kalman filters
We create 2 priors reflecting the targets’ initial states.
from stonesoup.types.state import GaussianState
prior1 = GaussianState([[0], [1], [0], [1]], np.diag([1.5, 0.5, 1.5, 0.5]), timestamp=start_time)
prior2 = GaussianState([[0], [1], [20], [-1]], np.diag([1.5, 0.5, 1.5, 0.5]), timestamp=start_time)
Loop through the predict, hypothesise, associate and update steps.
Plot the resulting tracks
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Strategies for Using Google Slides to Facilitate Online Discussions
Strategies for Using Google Slides to Facilitate Online Discussions
In line with the Association of Mathematics Teacher Educator’s (2017) Standards for Preparing Teachers of Mathematics, a primary goal of most mathematics methods courses is to support pre-service
teachers (PSTs) in developing instructional practices that research indicates can support students’ mathematical learning. These practices include, for example, selecting cognitively demanding
mathematical tasks (Stein, Grover, & Henningsen, 1996) and facilitating whole-class discussions by pressing students to explain their reasoning and make connections between different solution
strategies (Kazemi & Stipek, 2001; Stein, Engle, Smith, & Hughes, 2008). One strategy for supporting PSTs in developing such practices involves engaging them in the kinds of cognitively demanding
tasks and conceptually rich discussions they should aim to employ with their students (Lampert et al., 2010).
This past year, the COVID-19 pandemic forced me, along with many other mathematics teacher educators (MTEs), to transition in-person mathematics education courses to fully online environments. Like
many others, I found it challenging to encourage active and broad participation in mathematically rich whole-class discussions online, and thus challenging to model the kinds of discussion practices
teachers should aim to enact with their students. In this blog post, I share two strategies for encouraging PSTs’ participation in rich online discussions. PSTs indicated both strategies were
beneficial on their end-of-course evaluations. The strategies are predicated on the use of Google slides in class.
Using Google Slides in Online Methods Courses
I organized each lesson in my online mathematics education courses around a set of Google slides. Because Google slides allow multiple users to edit the contents of the slides at the same time, they
provided me with an opportunity to include interactive elements in the slides without requiring multiple or additional applications. These interactive elements served as the basis for the strategies
Strategy 1: Writing Prompts
One way I encouraged greater participation in whole-class discussions was by providing PSTs with the opportunity to think in writing prior to engaging in verbal discussions. In my experience, these
prompts provided my students with a chance to consider their ideas prior to the discussion, and thus encourage broader participation in the conversation. To do this online, I created a slide template
that included a prompting question and space for each student in the class to write their response to the prompting question. Integrating the prompt and space for writing into the slides limited
transition time and appeared to encourage students to use their writing time productively. Figure 1 (below) illustrates how I embedded these writing prompts in my slides. In this example, I asked
students to respond in writing to a video they watched of a teacher launching (or introducing) a cognitively demanding task.
Figure 1: Writing Prompts in Google Slides
Anecdotally, there was an increase in the number of PSTs who shared their thinking in discussions after implementing the quick writes. Because I was able to see PSTs’ responses as they were writing
them, I was also able to model strategies for selecting and sequencing ideas in whole-group discussions. Further, I found that PSTs began to look at each other’s responses during the writing time,
and then build on, reference, or use the ideas they were seeing in their own written responses. I see this as strength of the strategy, as it meant PSTs were engaging with one another’s ideas prior
to our whole-class conversations.
Strategy 2: Student “Buttons”
I also created PST-specific “buttons” in Google slides and then asked PSTs to use their assigned “buttons” to answer a question written on a slide. Figure 2 (below) shows one such slide featuring the
“buttons.” In this example, I asked PSTs to use their “buttons” to indicate whether they agreed with the following statement: If we multiply x/2 + 3/4 by 4 we get 2x + 3. Is 2x + 3 equivalent to x/2
+ 3/4? When implementing the button strategy, I first read aloud the question and then prompted PSTs to drag their “button” (denoted by letters corresponding with their first initial) to the box that
indicated their response to the question. In this case, half of the students said “yes,” just under half of the students said “no,” and one student put the button directly in the middle, later
explaining she was unsure. As this example illustrates, it was important to pose a question to which students might have different opinions or answers, as this encouraged a range of responses, and
thus set the stage for rich conversation.
Figure 2: Student Response Buttons
Using this strategy, enabled PSTs to see their peers’ selections in real-time, meaning they could adjust their answers based other PSTs’ responses. This appeared to increase participation in the
subsequent verbal discussions by enabling PSTs to compare their thinking with their peers prior to making their final selection and sharing with the class. Implementing this strategy also enabled me
to see PSTs move their buttons as they were making their initial selections, which meant I was able to see whether and how PSTs changed their minds or adjusted their responses based on other PSTs’
selections. As with the first strategy, this was helpful for selecting and sequencing ideas in the whole-class discussions that followed. Further, I encouraged PSTs to adjust their button placements
as they listened to other explanations in the subsequent whole-class discussion. This provided me with additional data on PSTs’ evolving thinking.
Though many courses are likely to return to face-to-face instruction soon, it is likely that some aspects of teacher education will remain online. As such, it is useful to build a knowledge base
around productive strategies for supporting PSTs’ learning and engagement in online courses. The two strategies outlined above enhanced students’ engagement in whole-class discussions in my online
mathematics education courses. In the future, I am interested in exploring whether and how these strategies can be modified for face-to-face classes.
Association of Mathematics Teacher Educators. (2017). Standards for Preparing Teachers of Mathematics. Retrievable online at amte.net/standards
Kazemi, E., & Stipek, D. (2001). Promoting conceptual thinking in four upper-elementary mathematics classrooms. The Elementary School Journal, 102(1), 59-80.
Lampert, M., Beasley, H., Ghousseini, H., Kazemi, E., & Franke, M. (2010). Using designed instructional activities to enable novices to manage ambitious mathematics teaching. In Instructional
explanations in the disciplines (pp. 129-141). Springer US.
Stein, M. K., Grover, B. W., & Henningsen, M. (1996). Building student capacity for mathematical thinking and reasoning: An analysis of mathematical tasks used in reform classrooms. American
Educational Research Journal, 33(2), 455-488.
Stein, M. K., Engle, R. A., Smith, M. S., & Hughes, E. K. (2008). Orchestrating productive mathematical discussions: Five practices for helping teachers move beyond show and tell. Mathematical
Thinking and Learning, 10(4), 313-340. | {"url":"https://amte.net/tech-talk/2021/07/strategies-using-google-slides-facilitate-online-discussions","timestamp":"2024-11-08T10:36:22Z","content_type":"text/html","content_length":"34531","record_id":"<urn:uuid:372a3f73-0784-45eb-a670-8b3a067dd67b>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00107.warc.gz"} |
Practical math
Practical math
June 28, 2006 3:46 AM Subscribe
Math help. (Practical, I'm not in school!)
n = xy
a = x/y
If I know the values of "n" and "a", can I calculate what x and y are?
Best answer:
Sure, although you won't know their signs.
rewrite your first equation to be
x = n/y
Then you plug into the second equation to get:
a = n/y^2
Thus y^2 = n/a, so you can solve for y if you have n and a (although you won't know whether it is positive or negative).
Once you have y, you know that x = ya, so plug in for x.
posted by allen.spaulding at 4:05 AM on June 28, 2006
x = n/y
a = x/y = (n/y)/y = n/(y^2)
solve for y
a/n = 1/y^2 --> n/a = y^2 --> y = sqrt(n/a)
x = n/y = n/sqrt(n/a) = sqrt(n a)
posted by gregvr at 4:08 AM on June 28, 2006
The algebra above is correct. I'll just add that sometimes you can calculate x and y, but sometimes not. (Assuming real numbers.) It depends on what a and n are:
If an<0 there are no solutions for x and y.
If a=0 and n=0 there are infinite solutions for x and y.
If a=0 and n!=0 there are no solutions for x and y.
If a!=0 and n=0 there are no solutions for x and y.
If an>0 there are two solutions for x and y.
posted by blue mustard at 4:22 AM on June 28, 2006
Response by poster:
Thanks all. In case anyone's wondering, the practical application of this is working out the resolution of a digital camera - it seems these days they only tell you the number of pixels and not the
actual resolution.
With the above formula, you can figure out what it is, as long as you know the aspect ratio. (Most are 4:3 = 1.33, but some are 3:2 = 1.5.)
posted by Mwongozi at 4:49 AM on June 28, 2006
Do keep in mind that the number of pixels they give is always rounded pretty severely.
For example, a 3.1 megapixel camera doesn't have 3100000 pixels (3.1 * 1E7) or 3250585.6 pixels (3.1 * 2^20, where mega- is used as it would be for megabytes). Instead, it's 3145728 pixels, which
corresponds to a resolution of 2048x1536. (If solve with 3100000 and 4:3, you'd get 2033X1525.)
Anyhow, my (hopefully obvious) caveat is that you need to take the numbers you get with this algorithm as an approximation. That having been said, the difference between 2048x1536 and 2033x1525
should be pretty inconsequential if you're just looking for a general idea.
posted by JMOZ at 10:25 AM on June 28, 2006
If this is a camera you already own, you could always just set it to the maximum resolution, take a picture, transfer the file to a computer, and use any imaging program to find the dimensions of the
posted by epugachev at 1:08 PM on June 28, 2006
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Hearts with POMDP
This project is hosted on Github
Also try here: http://hearts.yjyao.com
Hearts is one of the few card games that I know how to play, thanks to Microsoft Windows and the countless boring hours. So based on what I have done
• Gomoku and solitaire, it seems natural to fuse the two and write a hearts card game with AI.
It turns out that Hearts is a lot trickier than the simple Gomoku - it’s multiplayer and not fully observable (you can’t see everyone’s cards), which means that a naive implementation of Minimax is
either going to be bad or slow.
So the easiest approach will be a simple heuristic - something along the line of always play the smallest possible card to avoid taking in points and so on. This simple greedy heuristic actually can
perform quite well against amateur players - probably the majority people who actually play hearts against a computer.
However, if we know exactly how the computer is going to play, surely it must be a lot easier for us to beat it! Here’s where POMDP comes in. POMDP stands for Partially Observable Markov Decision
Process. Here if we assume that all other players play with a fixed strategy, then the game becomes a Markov Decision Process - after a move we make, the other three players will play according to
the cards they have and the cards that are currently on the table. So the state - the cards and score each has, will be updated based on the current state - clearly a Markov Decision Process. Then of
course, the state is partially observable, as we have discussed, so we can formulate the game as a POMDP process.
One algorithm to solve POMDP problems (by achieving an optimal result) is aclled POMCP (confusing names, yes), which uses a Monte-Carlo rollout as the evaluation function for the terminal node and
builds a tree by expanding the most promising nodes at each stage. More details can be found here.
Of course, actual human players can employ different strategies. So the assumption here is that such a formulation can serve as a reasonable approximation. This is hard to test though, so you can
find out the performance for yourself by playing the game.
You can check out the source code from the link in the beginning of this post. If you are interested in more technical details, feel free to read my report. | {"url":"http://yjyao.com/2013/12/hearts-with-pomdp.html","timestamp":"2024-11-07T15:25:57Z","content_type":"text/html","content_length":"7142","record_id":"<urn:uuid:f3a9e1d9-581e-4411-a326-717c4c919ccb>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00584.warc.gz"} |
ajibawa-2023/Maths-College · Datasets at Hugging Face
Logical implication is a fundamental concept in logic and mathematics, which represents a specific type of
relationship between two statements or propositions. At its core, logical implication refers to the idea that if a
certain proposition is true, then another proposition must also be true. This relationship can be expressed using
various symbols and notation, including $p o rightarrow q$, $p o longrightarrow q$, or $p o implies q$. These
symbols represent the conditional statement "if p, then q." To fully understand logical implication, let's examine
its components more closely. We will begin by exploring the notion of logical values and how they relate to
propositions. Next, we will discuss the concept of operations on these values and introduce the truth table as a
tool for visualizing the outcomes of different combinations of logical values. Finally, we will delve into the
specific operation of logical implication, examining its properties, symbolic representation, and real-world
applications. ### Logical Values and Propositions At the foundation of logical systems lies the concept of a
proposition - a declarative sentence that conveys a factual claim about something, either true or false. For
example, consider the sentences "Today is Tuesday," "Paris is the capital of France," and "$5 + 7 = 12.$" Each of
these sentences expresses a definite claim about reality; consequently, they can be classified as either true or
false based on whether the claim holds up under scrutiny. A proposition that has been definitively established as
true is known as a theorem, while a false proposition is referred to as a contradiction. In contrast to
propositions, variables do not convey any inherent meaning and instead serve as placeholders for potential
propositions. When working within a formal system, mathematicians often assign letters like p, q, r, etc., to
represent arbitrary propositions. By doing so, they enable algebraic manipulation and analysis of relationships
among multiple propositions simultaneously. Now that we have defined propositions and variables, let us turn our
attention to logical values. There exist precisely two distinct logical values, corresponding to the binary
categories of true (T) and false (F). As mentioned earlier, every proposition possesses one of these two values
depending on the validity of its underlying claim. Variables, too, assume one of these values when assigned
specific propositions during problem solving or theoretical development. ### Operations on Logical Values An
essential aspect of logical systems involves performing operations on logical values to form new propositions from
existing ones. Just as arithmetic functions combine numerical quantities through addition, subtraction,
multiplication, division, and other processes, logical operators take one or more logical values as inputs and
produce another logical value as output. The resultant value depends solely upon the input values and does not vary
regardless of context or interpretation. There are several basic logical operations worth noting here. Firstly,
negation (¬) reverses the truth value of a given proposition. If applied to a true proposition, negation yields a
false outcome, whereas negating a false proposition results in a true conclusion. Mathematically, this can be
represented as: * ¬T = F * ¬F = T Next, conjunction (&) combines two propositions such that both must hold true
concurrently for the resulting compound proposition to evaluate as true. Otherwise, if either or both component
propositions are false, the overall assertion evaluates as false. Symbolically, this corresponds to: * T & T = T *
T & F = F * F & T = F * F & F = F Disjunction (∨) presents another important logical operation, connecting two
propositions so that at least one needs to be true for the entire construct to assess as true. Only when all
constituents prove false does disjunction yield a false outcome. Its symbolic representations follow: * T ∨ T = T *
T ∨ F = T * F ∨ T = T * F ∨ F = F Finally, exclusive disjunction (XOR) differs slightly from standard disjunction
since it precludes the possibility of both operands being true simultaneously. Consequently, there exists exactly
one true condition out of four possible scenarios: * T XOR T = F * T XOR F = T * F XOR T = T * F XOR F = F These
elementary operations lay the groundwork for further examination of complex logical structures involving numerous
Write an educational piece suited for college students related to the following interrelated propositions. Now that we have discussed the basics of logical values and simple operations, we shall
text snippet: "# logical implication ## 1 Short version Logical implication is an proceed by introducing the truth table as a valuable tool for analyzing and organizing information pertaining to
operation on two logical values, typically the values of two propositions (http:// logical expressions. ### Truth Tables A truth table serves as an organized chart displaying all conceivable
planetmath.org/PropositionalCalculus), that produces a value of false just in case configurations of logical values attainable by a collection of propositions connected via various operations. It
the first operand is true and the second operand is false. The truth table for the provides a systematic method for determining the truth or falsity of compound propositions derived from simpler
logical implication operation that is written as $p\Rightarrow q$ and read as $p\ \ elements. Generally speaking, creating a truth table entails listing every combination of logical values for
operatorname{implies}\ q",$ also written as $p\rightarrow q$ and read as $\ involved variables followed by sequential application of relevant logical operations until reaching the final
operatorname{if}\ p\ \operatorname{then}\ q",$ is as follows: Logical Implication column representing the ultimate outcome. For instance, consider the expression (p & q) ∨ ¬r, where p, q, and r
$p$ $q$ $p\Rightarrow q$ F F T F T T T F F T T T ## 2 Long version The mathematical denote three distinct propositions. To create a truth table illustrating the behavior of this formula, we would
objects that inform our capacity for logical reasoning are easier to describe in a initially record every permutation of T and F values across columns labeled p, q, and r. Subsequent columns
straightforward way than it is to reconcile their traditional accounts. Still, some correspond to increasingly intricate expressions formed by combining initial variables according to specified
discussion of the language that occurs in the literature cannot be avoided. The rules. Ultimately, the concluding segment exhibits the resulting truth values obtained after iteratively applying
concept of logical implication encompasses a specific logical function, a specific required operations throughout the process. Herein lies an exemplary truth table demonstrating the previously
logical relation, and t" Do not just list concepts, but develop each one in detail stated expression: | p | q | r | p & q | ¬r | (p & q) ∨ ¬r | |---|---|---|-------|-----|--------------| | T | T | T
before moving to the next, as we prioritize depth of understanding and | T | F | T | | T | T | F | T | T | T | | T | F | T | F | F | T | | T | F | F | F | T | T | | F | T | T | T | F | T
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: | | F | T | F | F | T | T | | F | F | T | F | F | T | | F | F | F | F | T | T | As demonstrated above, truth tables
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an offer a clear visualization of how varying combinations of logical values impact the evaluation of composite
academic, professional and engaging tone that captivates interest. - Application: propositions. Moreover, they provide insight into crucial aspects of logical relationships, enabling analysts to
Incorporate specific, practical examples, such as proofs in calculus or critical discern patterns and trends within data sets. Armed with this knowledge, we now move forward to explore the central
dates and figures in history. Do not include a title or an introduction, simply topic of this exposition: logical implication. ### Logical Implication Revisited Having laid the necessary
write the content without headlines and introductory phrases. Do not use images. background, we return to our original focus: logical implication. Recall that logical implication signifies a
particular connection between two propositions such that if the antecedent (initial proposition) is accurate, then
the consequent (subsequent proposition) must likewise be correct. Stated differently, logical implication embodies
the principle that a given premise warrants a specific conclusion provided no exceptions arise. Symbolically,
logical implication can be denoted using various notations, most commonly as $p o Rightarrow q$, $p o
longrightarrow q$, or $p o implies q$; however, irrespective of chosen convention, each variant denotes the same
conditional statement: "if p, then q." Before diving deeper into the nuances surrounding logical implication, let
us revisit the foundational definition presented near the beginning of this article. Specifically, recall the truth
table associated with logical implication: | p | q | p o Rightarrow q | |--- | --- | --- | | F | F | T | | F | T |
T | | T | F | F | | T | T | T | This table reveals an interesting property concerning logical implication: whenever
the antecedent proves false, the entire statement automatically evaluates as true. This characteristic stems
directly from the nature of logical implication itself, wherein the accuracy of the consequent hinges entirely upon
the veracity of the antecedent. Thus, should the former happen to be erroneous, the latter carries no implications
regarding the validity of the subsequent proposition. Conversely, if the antecedent indeed holds merit, only then
does the consequent's status become relevant. Under such circumstances, the entire conditional statement assumes
the truth value ascribed to the consequent. Therefore, it becomes apparent that logical implication operates
asymmetrically, privileging the role played by the antecedent relative to the consequent. ### Material Conditional
vs. Strict Implication When discussing logical implication, it is crucial to distinguish between material
conditional and strict implication. Although sharing similarities superficially, these two concepts differ
significantly in terms of scope and applicability. Material conditional constitutes a weakened form of implication
asserting that the consequent holds true provided the antecedent does not violate said assertion. Put differently,
material conditional merely maintains that the consequent remains unchallenged by the antecedent rather than
actively endorsing its accuracy. This interpretation aligns closely with classical Boolean logic, adhering strictly
to the principles outlined thus far. Strict implication, conversely, demands stronger criteria for establishing a
valid linkage between antecedent and consequent. More specifically, strict implication necessitates that the
consequent logically ensues from the antecedent, implying that the former emerges necessarily and inevitably due to
the latter. Consequently, instances of strict implication tend to reflect metaphysical or epistemological
connections between propositions rather than purely syntactic associations rooted in linguistic structure alone. It
bears mentioning that many contemporary logicians reject strict implication altogether, viewing it as an
unnecessary complication lacking robust empirical support. Nonetheless, acknowledgment of its existence remains
vital for comprehending historical developments within philosophical thought, particularly in fields like modal
logic and intensional semantics. ### Applications of Logical Implication Throughout this discourse, we have
examined diverse facets of logical implication, elucidating its significance within broader mathematical and
philosophical contexts. Indeed, logical implication finds extensive usage across numerous disciplines, serving as a
cornerstone upon which countless theories and frameworks rest. Herein lie several representative examples
highlighting the versatility and ubiquity of logical implication in practice: #### Mathematical Proofs Within
mathematics, logical implication forms the backbone of rigorous deductive argumentation, providing a coherent means
of relating disparate claims and establishing novel conclusions. Consider, for instance, Euclidean geometry, where
chains of logical implications connect seemingly unrelated postulates and definitions to generate powerful
geometric theorems. By chaining together successive conditional statements, mathematicians construct watertight
arguments devoid
Write an educational piece suited for college students related to the following
text snippet: "# How do you find (d^2y)/(dx^2) for 3x^2+y^2=2? Feb 19, 2017 $\frac
{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{6}{y} ^ 3$ #### Explanation: When we To find the second derivative of y with respect to x, denoted as (d²y)/(dx²), for the equation 3x² + y² = 2, we
differentiate $y$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}$ When we differentiate a will first need to understand the concept of implicit differentiation. Implicit differentiation involves
non implicit function of $y$ then using the chain rule we can differentiate wrt $x$ differentiating both sides of an equation with respect to x, even when y is expressed explicitly in terms of x.
and multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$. When this is done in situ it is This method allows us to account for situations where y is challenging to express explicitly in terms of x. Let's
known as implicit differentiation. We have: $3 {x}^{2} + {y}^{2} = 2$ Differentiate begin by finding the first derivative using implicit differentiation. Given our initial equation: 3x² + y² = 2 To
wrt $x$ and we get: $\setminus \setminus \setminus 6 x + 2 y \frac{\mathrm{dy}}{\ differentiate with respect to x, let's isolate dy/dx term on one side so that we may later take another derivative.
mathrm{dx}} = 0$ $\therefore 3 x + y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ And now We start by taking the derivative of both sides of the equation with respect to x: d/dx (3x² + y²) = d/dx (2)
(as we want the second derivative) we differentiate wrt x again, this time we must Recall that the derivative of a constant is zero, hence d/dx (2) = 0. Now, applying basic rules of differentiation,
also apply the product rule: $3 + \left(y\right) \left(\frac{{d}^{2} y}{{\mathrm specifically the power rule, we obtain: 6x + 2ydy/dx = 0 Now, rearrange the above equation to solve for dy/dx: dy/
{dx}}^{2}}\right) + \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(\frac{\ dx = (-6x)/(2y) Simplifying further, we get: dy/dx = -3x/y Now that we have found the first derivative, dy/dx,
mathrm{dy}}{\mathrm{dx}}\right) = 0$ $\therefore 3 + y \frac{{d}^{2} y}{{\mathrm let's proceed to find the second derivative, (d²y)/(dx²). To achieve this, recall that the quotient rule states
{dx}}^{2}} + {\left(\frac{" Do not just list concepts, but develop each one in that if z = f(x)/g(x), then dz/dx = [f'(x)*g(x)] - [f(x)*g'(x)] / g(x)^2. Applying this rule to our current
detail before moving to the next, as we prioritize depth of understanding and expression, dy/dx, we get: d²y/dx² = [-3*(y) - (-3x)(dy/dx)] / y². Substituting the value of dy/dx from earlier, we
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: get: d²y/dx² = [-3*y - (-3x)(-3x/y)] / y². Solving the numerator, we get: d²y/dx² = (-3y + 9x²/y) / y². Multiplying
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an through by y² to simplify, we finally reach the desired result: (d²y)/(dx²) = (-3y³ + 9x²) / y⁴ This final
academic, professional and engaging tone that captivates interest. - Application: expression represents the second partial derivative of y squared with respect to x for the given implicit curve. It
Incorporate specific, practical examples, such as proofs in calculus or critical demonstrates how changes in position along the curve affect its rate of curvature at any point.
dates and figures in history. Do not include a title or an introduction, simply
write the content without headlines and introductory phrases. Do not use images.
The Nilakantha Series is a historically significant infinite series used to approximate the value of Pi (π), which
is renowned for its accuracy and antiquity. This series emerged from the works of the revered Indian astronomer and
mathematician Nilakantha Somayaji during the 15th century, approximately 1500 AD. Before delving into the details
Write an educational piece suited for college students related to the following of the Nilakantha Series, let us first understand what an infinite series entails and why approximating Pi is
text snippet: "NILAKANTHA SERIES PROOF Please Sign up or sign in to vote. Ranjan essential in mathematics. An infinite series is a summation of an endless sequence of numbers or terms. It can
Roy, Mathematics Magazine , Vol. Historically, one of the best approximations of PI often provide precise values for various mathematical constants, functions, and problems through approximation
and interestingly also one of the oldest, was used by the Chinese mathematician Zu methods. Convergence tests determine whether these series reach a finite limit or diverge towards infinity. As far
Chongzhi Sec. The most recent calculation found more than 13 trillion digits of pi as Pi is concerned, its importance stems from geometry, trigonometry, and calculus applications. Accurately
in days! The article was complemented according to your suggestion. Not impressed estimating Pi's value has been crucial since ancient civilizations like Egyptians, Babylonians, and Indians tried
Jose A Pascoa 8-Aug 7: The first written description of an infinite series that computing it using polygonal approaches. Nilakantha Somayaji introduced his ingenious series expansion in his
could be used to compute pi was laid out in Sanskrit verse by Indian astronomer monumental work titled 'Tantrasamgraha,' where he expressed Pi as an ascending continued fraction. Later, European
Nilakantha Somayaji around A. Now pick up a pen, close your eyes and put dots on scholars translated and disseminated his ideas throughout Europe. Let's examine the Nilakantha Series formula
the square at random. Articles Quick Answers Messages. I had to raise the iteration itself: π = 3 + 4/(2 × 3 × 4) - 4/(4 × 5 × 6) + 4/(6 × 7 × 8) - ... ± 4/[(2n)(2n+1)(2n+2)] + ... This expression
count way up to get it to register with the timer. I leave the conclusion to you consists of alternating positive and negative terms having the same structure—the reciprocal of the product of
when examining the table above. PI with decimals: Juan Manuel Romero Martin 4-Sep three consecutive integers. By adding several initial terms from this series, we obtain increasingly accurate
9: PI with decimals: If you do this enough times, and your efforts are truly rational approximations of Pi. Moreover, as n tends toward infinity, the series converges steadily to the exact
random, eventually the percentage " Do not just list concepts, but develop each one value of Pi. Now, consider how the provided text connects the Nilakantha Series to other historical developments
in detail before moving to the next, as we prioritize depth of understanding and regarding Pi calculations. For instance, the Chinese mathematician Zu Chongzhi calculated Pi's value to an
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: astonishing seven decimal places around 480 AD, predating Nilakantha's work. Fast forwarding to contemporary
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an achievements, researchers have recently computed Pi to over 13 trillion digits within mere days using modern
academic, professional and engaging tone that captivates interest. - Application: algorithms and computational powerhouses. These advancements underscore the significance of studying historical
Incorporate specific, practical examples, such as proofs in calculus or critical techniques while acknowledging our current capabilities in solving complex mathematical problems. In summary, the
dates and figures in history. Do not include a title or an introduction, simply Nilakantha Series represents a vital milestone in developing efficient algorithms for accurately determining Pi's
write the content without headlines and introductory phrases. Do not use images. value. Its ingenuity lies in employing simple arithmetic operations iteratively, leading to highly precise
approximations. Exploring such infinite series expansions enriches our comprehension of advanced mathematical
principles and their real-world implications. Furthermore, comparing past and present accomplishments highlights
humanity's relentless pursuit of knowledge and innovation across centuries.
To begin, let us define some terms. A Weyl algebra is a certain type of algebraic structure, specifically it is the
algebra generated by a set of variables and their corresponding derivatives, with a particular set of relationships
between them. This can be written as k[x1, x2, ..., xn, ∂1, ∂2, ..., ∂n], where k is a field, often referred to as
the "base field." A division ring, also known as a skew field, is a generalization of a field, which is an
algebraic structure in which addition, subtraction, multiplication, and division (except by zero) are all
well-defined operations. In a division ring, every nonzero element has a multiplicative inverse. Examples of fields
Write an educational piece suited for college students related to the following include the real numbers, complex numbers, and rational numbers. However, there are also important examples of
text snippet: "# Proof a Weyl Algebra isn't isomorphic to a matrix ring over a division rings that are not fields, such as the quaternions. A matrix ring over a division ring is a collection of
division ring Can anyone prove that a Weyl Algebra is not isomorphic to a matrix square matrices with entries from a division ring, together with the usual matrix operations of addition and
ring over a division ring? - Notation: The Weyl algebra is $$k[x_1, x_2, \ldots, multiplication. These structures form a natural generalization of matrix rings over fields, which are widely used
x_n, \partial_1, \partial_2, \ldots, \partial_n]$$ with the obvious relations. The in many areas of mathematics and science. Now, let us consider the claim made in the original post, that a Weyl
Weyl algebra doesn't contain any division rings larger than $k$, and it is infinite algebra is not isomorphic to a matrix ring over a division ring. An isomorphism is a bijective homomorphism,
dimensional over $k$. So, assuming you don't allow infinite matrices, that's a meaning that it preserves the structure of the objects being mapped. Thus, to disprove the existence of an
proof. How to see that it doesn't contain any division ring larger than $k$? I just isomorphism between two algebraic structures, it suffices to find some structural property that holds for one but
need to show that any nonconstant differential operator is not invertible. One way not the other. In this case, the key properties to focus on are the presence of a division ring and finite
to see this is to notice that multiplying differential operators multiplies dimensionality. Firstly, note that the Weyl algebra does not contain any division rings larger than its base field
symbols, and the symbol of a nonconstant differential operator is a nonconstant k. This follows from the fact that any nonconstant differential operator is not invertible. More precisely, if f
polynomial. - Whats the base field? – Casebash Nov 8 '09 at 4:42 The field of (x,∂) is a nonconstant differential operator, then it cannot have a multiplicative inverse, since multiplying
constants. I have edited the answer; see if that is clearer. – David Speyer Nov 8 differential operators corresponds to composition of functions, and composing a nonconstant function with itself
'09 at 13:49 I guess you could also use" Do not just list concepts, but develop will always result in another nonconstant function. Therefore, the only invertible elements in the Weyl algebra are
each one in detail before moving to the next, as we prioritize depth of those from the base field k. On the other hand, matrix rings over division rings can indeed contain nontrivial
understanding and comprehensive exploration of the subject matter over breadth. division rings. For example, the matrix ring M2(D), consisting of 2x2 matrices with entries from a division ring D,
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: contains the subset of diagonal matrices, which forms a division ring under the usual matrix operations.
Write with an academic, professional and engaging tone that captivates interest. - Furthermore, if D is infinite, then so too is M2(D). Thus, if we assume that our matrix ring is finite dimensional
Application: Incorporate specific, practical examples, such as proofs in calculus over its center, then it cannot be isomorphic to a Weyl algebra, as the latter contains no proper division rings,
or critical dates and figures in history. Do not include a title or an while the former may do so. Moreover, since the Weyl algebra is infinite dimensional over its base field, it cannot
introduction, simply write the content without headlines and introductory phrases. even be isomorphic to a matrix ring over a division ring of finite dimension, regardless of whether it contains any
Do not use images. proper division rings or not. To summarize, we have shown that a Weyl algebra and a matrix ring over a division
ring are not isomorphic by exhibiting a fundamental difference in their respective structures. Specifically, the
Weyl algebra does not contain any proper division rings, whereas matrix rings over division rings can and typically
do. Additionally, we noted that the Weyl algebra is infinite dimensional over its base field, whereas matrix rings
over division rings can be either finite or infinite dimensional depending on the size of the division ring they
are built upon. By highlighting these crucial differences, we demonstrate why attempting to establish an
isomorphism between these two types of structures is ultimately futile.
Now let us delve into the intricacies of ordered fields, bounded sets, and their relationships. We will address the
three statements and questions you provided, elucidating these fundamental mathematical concepts. 1. The general
comment (Set > Group > Ring > Field) indicates a hierarchy based on structure and complexity. Starting from the
most basic concept, a Set contains only unrelated objects. Once those objects have operations defined among them,
Write an educational piece suited for college students related to the following they form a Group. If additional properties, such as associativity and distributivity, are present, then the group
text snippet: "# Ordered field, Bounded set, and the containment I am now in becomes a Ring. Finally, when division can be performed alongside the other operations, this more complex system
engineering mathematics class and it goes over some basic set theory. Since I forms a Field. This progression showcases how mathematicians build increasingly sophisticated structures upon
haven't had any experience with set theory, three statements leave me confused. simpler ones. To illustrate, consider the natural numbers N = {1, 2, 3, ...}. Initially, this collection is merely
Thanks for your help! Statement: 1) General comment (not theorem or axiom): Set > a Set. However, addition operation turns it into a Group. Further defining multiplication creates a Ring.
Group > Ring > Field 2) Imagine set like set of Natural number is not aligned like Eventually, introducing divisions transforms N into the Rational Numbers Q, which is an example of a Field. 1. When
on axis like we are all taught early but the elements jumble around without order. envisioning sets, especially those consisting of real or complex numbers, imagine points spread out along a line
3) Definition: A subset A of an ordered field F is bounded from below iff there rather than neatly aligning like axes. These points may appear disordered, yet maintaining certain essential
exists p in F such that (for all x in A) (p <= x) Question: 1) Why as convention relationships determined by order relations (such as "<", ">", or "="). Therefore, even though the visual
"bounded set" is used instead of "bounded field"? 2) Could a set being "not representation might seem chaotic, underlying rules govern interactions between the members. Consider the open
ordered" and "bounded" at the same time? 3) Regarding optimization view point, is interval (0, 1), where its elements lie randomly within the segment between zero and one; however, any two distinct
polytope view of a bounded convex set totally correct? Is polytope view just to elements maintain a discernible distance according to their position. 1. Addressing the definition: A subset A of
help imagining? I.e. Polytope view implied ordered-ness on a set by implying an ordered field F is bounded from below if there exists p in F such that (for all x in A) (p <= x). Here,
spatial location of set element. A set is simp" Do not just list concepts, but 'bounded' implies restrictions – specifically, having both upper and lower limits. It doesn't refer to bounding the
develop each one in detail before moving to the next, as we prioritize depth of entire field since a Field typically extends infinitely in every direction. Instead, think about setting
understanding and comprehensive exploration of the subject matter over breadth. constraints on subsets. For instance, consider the interval (-∞, 5] in the Real Numbers set. Its lower limit is
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: negative infinity, while five serves as the upper bound - thus creating a half-open interval subset constrained
Write with an academic, professional and engaging tone that captivates interest. - from below. Regarding your questions: 1. Using "bounded set" terminology emphasizes restricting particular
Application: Incorporate specific, practical examples, such as proofs in calculus subcollections within larger systems. Conversely, "bounded field" could imply limitations on the whole Field
or critical dates and figures in history. Do not include a title or an itself, which isn't generally applicable nor desired. 2. Yes, indeed! An excellent example would be Cantor's famous
introduction, simply write the content without headlines and introductory phrases. "deleted middle third" construction of the ternary set. Each stage generates new intervals devoid of center values,
Do not use images. eventually resulting in a non-ordered yet still bounded set. 3. Although the polytope perspective offers intuitive
geometric insights regarding optimization problems involving bounded convex sets, it does rely on imposed ordering
assumptions. Nevertheless, the polyhedral approach remains valid because compact convex sets always possess extreme
points representing optimal solutions under linear functionals. Thus, despite implying order through spatial
localization, the method provides valuable results consistent with abstract definitions.
Write an educational piece suited for college students related to the following
text snippet: "# Show that the Points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D
(4, 7, 6) Are the Vertices of a Parallelogram Abcd, but Not a Rectangle. -
Mathematics Show that the points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7,
6) are the vertices of a parallelogram ABCD, but not a rectangle. #### Solution To
show that ABCD is a parallelogram, we need to show that its two opposite sides are
equal. $AB = \sqrt{\left( - 1 - 1 \right)^2 + \left( - 2 - 2 \right)^2 + \left( - 1
- 3 \right)^2}$ $= \sqrt{4 + 16 + 16}$ $= \sqrt{36}$ $= 6$ $BC = \sqrt{\left( 2 + 1
\right)^2 + \left( 3 + 2 \right)^2 + \left( 2 + 1 \right)^2}$ $= \sqrt{9 + 25 + 9}$ To demonstrate that the given points A(1, 2, 3), B(-1, -2, -1), C(2, 3, 2), and D(4, 7, 6) form a parallelogram
$= \sqrt{43}$ $CD = \sqrt{\left( 4 - 2 \right)^2 + \left( 7 - 3 \right)^2 + \left( ABCD, we must show that both pairs of opposite sides are equal. This means that |AB| = |CD| and |BC| = |AD|. We
6 - 2 \right)^2}$ $= \sqrt{4 + 16 + 16}$ $= \sqrt{36}$ $= 6$ $DA = \sqrt{\left( 1 - will calculate the distance (magnitude) between each pair of points using the Euclidean distance formula: d(P, Q) =
4 \right)^2 + \left( 2 - 7 \right)^2 + \left( 3 - 6 \right)^2}$ $= \sqrt{9 + 25 + sqrt[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²], where P=(x₁, y₁, z₁) and Q=(x₂, y₂, z₂). First, let's find |AB| and |CD
9}$ $= \sqrt{43}$ $AB = CD and BC = DA$ $\text{ Since, opposite pairs of sides are |: |AB| = d(A, B) = sqrt[((-1)-1)² + ((-2)-2)² + ((-1)-3)²] = sqrt[16+25+16] = sqrt(57) = 606 (approximately)
equal } .$ $\therefore" Do not just list concepts, but develop each one in detail
before moving to the next, as we prioritize depth of understanding and
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor:
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an
academic, professional and engaging tone that captivates interest. - Application:
Incorporate specific, practical examples, such as proofs in calculus or critical
dates and figures in history. Do not include a title or an introduction, simply
write the content without headlines and introductory phrases. Do not use images.
Now let us delve into the problem presented and explore the methods used to tackle it. We will first discuss some
fundamental techniques for solving ordinary differential equations (ODEs), then move on to the proposed solution
involving a change of variables, and finally examine the approach using matrix exponentiation. **Ordinary
Differential Equations:** An ODE is an equation that contains an unknown function and its derivatives. They are
classified based on their order (the highest derivative present) and whether they contain only one independent
variable (ordinary) or multiple ones (partial). Solving an ODE means finding the unknown functions satisfying the
Write an educational piece suited for college students related to the following given conditions. The three most common types of first-order ODEs are: 1. **Separable**: When the expression
text snippet: "# Is there a simple solution to this (ordinary) differential multiplying dx can be isolated on one side and dy on the other. This allows direct integration to obtain the
equation? I'm trying to solve the following differential equation: $$\frac{dy}{dx} general solution. For example, $\frac{dy}{dx}=\frac{f(x)}{g(y)}$ where $f(x)$ and $g(y)$ are arbitrary functions.
= - \frac{3x + 2y}{2y}.$$ It looks pretty simple, yet it's not separable, linear, 2. **Linear**: If the ODE takes the form $\frac{dy}{dx}+P(x)y=Q(x)$, it is called a linear ODE. Here, $P(x)$ and $Q
or exact. It is of the form $$\frac{dy}{dx} = f(y/x).$$ I could do the substitution (x)$ are continuous functions of $x$. Linear ODEs have standard integrating factors allowing straightforward
$v = y/x$, and I know it would look pretty ugly, but is there a better or simpler resolution. 3. **Exact**: These are characterized by having an implicit form $M(x,y)+N(x,y)\frac{dy}{dx}=0$, where
method? - Doubt it: wolframalpha.com/input/… – user7530 Feb 23 '13 at 3:59 One way $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, making them amenable to direct integration. In our
I can think is to solve: $$\frac{dy}{dt} = -3 x - 2y \\ \frac{dx}{dt} = 2 y$$ say initial question, none of these categories apply directly since the right-hand side does not fit any clear pattern.
by matrix exponentiation. Then one can invert $x(t)$ to find $y(x)$. For the However, recognizing it as being of the form $\frac{dy}{dx}=f(\frac{y}{x})$ suggests another possible technique.
qualitative end behavior of the solution, I like the matrix solution. The **Change of Variables:** A useful strategy when encountering an ODE resistant to traditional approaches involves
associated matrix is $$\begin{pmatrix} 0 & 2 \\ -3 & -2 \end{pmatrix},$$ which transforming the original equation into something more manageable via a suitable change of variables. Specifically,
leads to the eigenvalues $-1 \pm i\sqrt{5}$. So I can have overall sense of the if the ODE has the form $\frac{dy}{dx}=f(\frac{y}{x})$, defining $v=\frac{y}{x}$ often simplifies matters
family of solutions. In particular, I can tell that they must "go" through the significantly. Applying this transformation yields: \begin{align*} v &= \frac{y}{x} \\\\ dv &= \left(\frac{1}{x}\
origin. – Minh Feb 23 '13 at 4:29" Do not just list concepts, but develop each one frac{dy}{dx}-\frac{y}{x^2}\right)dx \\\\ \Rightarrow \quad \frac{dy}{dx} &= xdv+vdx \end{align*} Substituting back
in detail before moving to the next, as we prioritize depth of understanding and into the original ODE results in a new, potentially easier, ODE involving only the single variable $v$.
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: Unfortunately, applying this method here still leaves us with a rather complicated expression requiring further
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an manipulation. **Matrix Exponentiation:** As suggested in the post, another approach uses matrices and their
academic, professional and engaging tone that captivates interest. - Application: exponentials. By rewriting the system as coupled first-order ODEs, we may utilize powerful tools from linear
Incorporate specific, practical examples, such as proofs in calculus or critical algebra. Define two new functions $u_1(t)=x(t)$ and $u_2(t)=y(t)$, resulting in: \begin{align*} \frac{du_1}{dt} &=
dates and figures in history. Do not include a title or an introduction, simply 2u_2 \\\\ \frac{du_2}{dt} &= -3u_1 \end{align*} This can now be expressed compactly as a matrix vector product: \
write the content without headlines and introductory phrases. Do not use images. begin{equation*} \mathbf{U}'(t)=\begin{pmatrix} 0 & 2 \\\\-3 & -2 \end{pmatrix}\mathbf{U}(t) \end{equation*} where
$\mathbf{U}(t)=\big[u_1(t), u_2(t)\big]^\intercal$. Solutions take the familiar form $e^{At}\mathbf{C}$, where $A$
is the coefficient matrix above, $\mathbf{C}$ is a constant vector determined by initial conditions, and $e^{At}$
denotes the matrix exponential defined by its Taylor series expansion. Computing the eigenvalues (-1±i√5) provides
valuable insight into the asymptotic properties of the solution curve, confirming that all nonconstant solutions
pass through the origin. While this final method doesn't necessarily provide a simpler closed-form solution, it
offers deep theoretical insights and computational advantages, especially when dealing with higher dimensional
To begin, let's review the definitions of the required properties for a structure to be a field. A field consists
of a set F along with two binary operations (addition and multiplication) satisfying certain axioms. These axioms
can be broken down into several categories, including associativity, commutativity, distributivity, identity
elements, and multiplicative inverses. We will examine how these properties apply to your attempted proof regarding
the structure G = F x F, where F is a field and addition and multiplication are defined as follows: (a, b) + (c, d)
Write an educational piece suited for college students related to the following := (a + c, b + d), and (a, b) \* (c, d) := (ac, db). Associativity of Addition: You began by stating that for any
text snippet: "# Proving a structure is a field? Please help with what I am doing (a, b), (c, d), and (e, f) in G, it should hold true that: (a, b) + ((c, d) + (e, f)) = ((a, b) + (c, d)) + (e, f)
wrong here. It has been awhile since Ive been in school and need some help. The However, when you expanded this expression, you made an error in replacing (c, d) + (e, f) with (c + e, d + f): (a,
question is: Let $F$ be a field and let $G=F\times F$. Define operations of b) + (c + e, d + f) = (a + c + d, b + e + f) //incorrect replacement Instead, recalling our definition of addition
addition and multiplication on $G$ by setting $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b)* in G, replace (c, d) + (e, f) correctly: (a, b) + (((c, d) + (e, f))) = (a, b) + (c + e, d + f) = (a + c + e, b + d
(c,d)=(ac,db)$. Do these operations define the structure of a field on $G$? In + f) Now, consider the left side of the equation. Replace (c, d) + (e, f) with its equivalent form based on the
order to be a field, the following conditions must apply: 1. Associativity of definition of addition: ((a, b) + (c + e, d + f)) = (a + (c + e), b + (d + f)) = (a + c + e, b + d + f) Since both
addition and multiplication 2. commutativity of addition and mulitplication 3. sides yield identical results, we have shown that addition is indeed associative in G. Commutativity of Addition
distributivity of multiplication over addition 4. existence of identy elements for and Multiplication: We leave proving the commutativity of addition and multiplication as exercises for the reader,
addition and multiplication 6. existence of multiplicative inverse 0 cannot equala, as they follow similar steps using the given definitions. However, note that care must be taken during
a-1*a=1 I started with 1. saying $(a,b)+(c,d)+(e,f)=(a+c+e,b+d+f)$ $(a,b)+[(c,d)+ substitution; ensure all replacements adhere strictly to the provided definitions of addition and multiplication.
(e,f)]=[(a,b)+(c,d)]+(e,f)$ $(a,b)+(c+e,d+f)=(a+c,b+d)+(e,f)$ $(a+c+d,b+e+f)= Distributivity of Multiplication Over Addition: To establish that multiplication is distributed over addition in G,
(a+c+e,b+d+f)$ which is not correct but I'm not sure where I went wrong. Is my show that for any (a, b), (c, d), and (e, f) in G, the following holds true: (a, b)\*( (c, d) + (e, f)) = (a, b)*
logic incorrect? If you are trying to prove a" Do not just list concepts, but (c, d) + (a, b)*(e, f) Expanding both sides according to their respective definitions yields: LHS: (a, b) * ((c, d)
develop each one in detail before moving to the next, as we prioritize depth of + (e, f)) = (a, b) * (c + e, d + f) = (a(c + e), b(d + f)) RHS: (a, b)*(c, d) + (a, b)*(e, f) = (ac, bd) + (ae, bf)
understanding and comprehensive exploration of the subject matter over breadth. = (ac + ae, bd + bf) Comparing terms, observe that LHS(1) = RHS(1) and LHS(2) = RHS(2); thus, demonstrating
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: distributivity. Identity Elements for Addition and Multiplication: Proving the existence of additive and
Write with an academic, professional and engaging tone that captivates interest. - multiplicative identities requires finding appropriate values within G that satisfy the necessary criteria.
Application: Incorporate specific, practical examples, such as proofs in calculus Consider zero element (0, 0) and unity element (1, 1) from G. Showing that adding or multiplying them with other
or critical dates and figures in history. Do not include a title or an arbitrary pairs from G leaves those pairs unchanged proves their eligibility as identity elements under addition
introduction, simply write the content without headlines and introductory phrases. and multiplication respectively. Multiplicative Inverse: Finally, establishing the presence of a multiplicative
Do not use images. inverse entails showing that every nonzero pair in G possesses another pair capable of producing the multiplicative
identity upon multiplication. Given a nonzero pair (a, b), find its potential inverse (x, y) fulfilling the
requirement: (a, b) \* (x, y) = (1, 1) Recall that multiplication is defined componentwise; hence, we obtain two
equations after comparing components: ax = 1 and by = 1 As long as F is a field, there exist unique solutions for x
and y within F satisfying these equalities. Therefore, every nonzero pair in G does have a multiplicative inverse,
completing our verification process.
Colored Gaussian noise (CGN), also referred to as colored noise, is a type of wide-sense-stationary (WSS) process.
A WSS process has two key properties: first, all the random variables that make up the process have the same mean;
second, the autocorrelation function, denoted as $R\_X(t\_1, t\_2)$, depends solely on the time difference $t\_2 -
t\_1$. The autocorrelation function for a WSS process is defined as the expected value of the product of the
process evaluated at times $t\_1$ and $t\_2$: $$R\_X(t\_1, t\_2) = E[X(t\_1) X(t\_2)]$$ Here, $E[\cdot]$ denotes
the expectation operator. For a WSS process, this simplifies to: $$R\_X(\tau) = E[X(t + \tau) X(t)]$$ Where $\tau =
t\_2 - t\_1$ represents the time lag between the two evaluations of the process. Now let us explore colored
Gaussian noise further through the lens of power spectral density (PSD). The PSD is a useful tool in characterizing
Write an educational piece suited for college students related to the following signals and noises based on their distribution of power across different frequencies. Specifically, the PSD
text snippet: "# On coloured Gaussian noise It is known that the PSD of additive describes the amount of power contained within a signal per unit of bandwidth around a particular frequency. This
white Gaussian noise (AWGN) is constant and equal to its variance. What about information can be valuable when analyzing communication systems, filters, and other applications where
coloured Gaussian noise (CGN)? For example, given the following PSD of CGN $$S(f) = understanding the distribution of signal power across frequencies plays a crucial role. The relationship between
\frac 1f$$ Is the spectral density of such noise frequency-dependent? If so, how to the autocorrelation function and the PSD is described via the Wiener–Khinchin theorem. According to this
get the PDF by some "inverse" autocorrelation function? • Note that the variance of fundamental result from stochastic processes theory, the Fourier transform of the autocorrelation function yields
white Gaussian noise is not defined, although with some hand-waving you could say the PSD of a WSS process: $$S\_X(f) = \mathcal{F} \{ R\_X(\tau) \}$$ Conversely, taking the inverse Fourier
it's infinite. If the noise PSD is $N_0/2$, then the variance of filtered noise transform of the PSD recovers the autocorrelation function: $$R\_X(\tau) = \mathcal{F}^{-1} \{ S\_X(f) \}$$ In the
(where the filter's impulse response has energy equal to 1) is equal to $N_0/2$. case of colored Gaussian noise with a PSD expressed as $S(f) = \frac{1}{f}$, applying the Wiener–Khinchin theorem
– MBaz Nov 8 '17 at 17:57 Colored Gaussian noise is by definition a reveals important insights into the nature of the noise. First, note that this PSD does not exist at zero frequency
wide-sense-stationary (WSS) process; that is, it has constant mean (all the random ($f=0$), which implies that there are theoretical challenges associated with generating such a noise process.
variables constituting the process have the same mean) and its autocorrelation Nevertheless, assuming such a PSD exists, we may proceed with our analysis using distributional assumptions and
function $R_X(t_1, t_2) = E[X(t_1)X(t_2)]$ depends only on the difference $t_2-t_1$ limiting arguments. Taking the inverse Fourier transform of the provided PSD results in the autocorrelation
of the arguments. It is conventional to u" Do not just list concepts, but develop function: \begin{align*} R\_X(\tau) &= \mathcal{F}^{-1} \left\{ \frac{1}{f} \right\} \\ &= -\pi \, \text{sgn}(\tau)
each one in detail before moving to the next, as we prioritize depth of \end{align*} Here, sgn($\tau$) refers to the signum function, which outputs -1 if $\tau < 0$, 0 if $\tau = 0$, and
understanding and comprehensive exploration of the subject matter over breadth. 1 if $\tau > 0$. However, this autocorrelation function violates the property required for wide-sense stationarity,
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: namely, having an even symmetry ($\text{sgn}(-\tau)=-\text{sgn}(\tau)$). To overcome this challenge, one possible
Write with an academic, professional and engaging tone that captivates interest. - approach involves adding a Dirac delta function centered at the origin to enforce evenness while preserving the
Application: Incorporate specific, practical examples, such as proofs in calculus desired mathematical behavior: $$R\_X(\tau) = -\pi \, \text{sgn}(\tau) + K \, \delta(\tau)$$ Where $K>0$ is chosen
or critical dates and figures in history. Do not include a title or an appropriately to ensure nonnegativity of the autocorrelation function. With this adjusted formulation, the original
introduction, simply write the content without headlines and introductory phrases. question concerning the frequency dependence of the spectral density remains answered affirmatively—the spectral
Do not use images. density of colored Gaussian noise is indeed frequency-dependent. Furthermore, under suitable conditions, the
probability density function (PDF) of the colored Gaussian noise can be obtained using the inverse Fourier
transform of the modified PSD: $$p\_X(x) = \mathcal{F}^{-1} \left\{ \sqrt{\lvert S\_X(f)\rvert} e^{i \phi(f)} \
right\}$$ Where $\phi(f)$ is the phase angle corresponding to the complex valued square root of $|S\_X(f)|$. To
summarize, colored Gaussian noise exhibits several unique characteristics compared to its white counterpart.
Notably, the PSD of colored Gaussian noise is frequency dependent, meaning the variance varies depending on the
frequency range considered. By contrast, white Gaussian noise features an unbounded, constant PSD leading to
undefined variance. Understanding these distinctions allows engineers, physicists, and data scientists to better
analyze and design systems utilizing various types of noisy input signals.
The chain rule is a fundamental concept in multivariable calculus, which allows us to compute derivatives of
composite functions. Before diving into the proof of the chain rule, let's briefly review some essential
definitions. In the context of this proof, $E$ and $U$ denote open subsets of $\mathbb{R}^n$ and $\mathbb{R}^m,$
respectively, while $f : E \rightarrow \mathbb{R}^m$ and $g : U \rightarrow \mathbb{R}^l$ represent differentiable
functions. Moreover, $F := g \circ f$ denotes the composition of these two functions. When discussing rates of
convergence, we often rely on the "big O" and "small o" notation; here, we mean: 1. $|h| = o(|h|), h \neq 0$: This
statement implies that, as $|h|$ approaches zero, $|h|$ becomes insignificant compared to itself—that is, its ratio
with respect to $|h|$ converges to zero. Formally, for any positive constant $c > 0,$ there exists a $\delta > 0$
so that $|\frac{o(|h|)}{|h|}| < c$ whenever $0< |h|<\delta.$ 2. $f'(a)h + o(|h|) = o(|h|):$ Here, we consider the
linear function $f'(a)h$ along with another function $o(|h|),$ both evaluated at $h.$ As $h$ tends to zero, their
Write an educational piece suited for college students related to the following sum must also become insignificant compared to $|h|.$ For any positive constant $c > 0,$ there exists a $\delta >
text snippet: "# Chain rule proof Let $a \in E \subset R^n, E \mbox{ open}, f: E \ 0$ such that $|\frac{(f'(a)h + o(|h|))}{|h|}| < c$ provided that $0< |h|<\delta.$ Note that since $f'(a)h$ has
to R^m, f(E) \subset U \subset R^m, U \mbox{ open}, g: U \to R^l, F:= g \circ f.$ degree one, the expression inside the absolute value can be written as $(f'(a)+\frac{o(|h|)}{|h|})|h|.$ Since $\
If $f$ is differentiable in $a$ and $g$ differentiable in $f(a)$, then $F$ is lim_{h \rightarrow 0}\frac{o(|h|)}{|h|}=0,$ it follows that $\lim_{h \rightarrow 0}(f'(a)+\frac{o(|h|)}{|h|})=f'
differentiable in $a$ and $F'(a)=g'(f(a)) f'(a)$. I have been given a short proof, (a).$ Consequently, if $f'(a)\neq 0,$ then $(f'(a)+\frac{o(|h|)}{|h|})$ will remain bounded away from zero by
but I do not understand every step of it. We have $f(a+h) = f(a) + f'(a)h + o(|h|)$ choosing sufficiently small $|h|.$ Therefore, the condition above essentially states that $\lim_{h \rightarrow 0}\
and $g(f(a)+k) = g(f(a)) + g'(f(a))k + o(|k|)$ as $h,k \to 0$. Thus F(a+h)=(g \circ frac{o(|h|)}{|h|}=0.$ 3. $o(|h_1|) = o(|h_2|):$ We say that $o(|h_1|)$ and $o(|h_2|)$ belong to the same
f)(a+h) &= g(f(a+h)) \\ &= g(f(a)+f'(a)h+o(|h|)) \\ &= g(f(a)) + g'(f(a))(f'(a)h+o equivalence class if their ratios approach unity as either $h_1$ or $h_2$ goes to zero. Specifically, for any pair
(|h|)) + o(|f'(a)h+o(|h|)|) \\ &= (g\circ f)(a) + g'(f(a))f'(a)h+o(|h|) + o(O(|h|)) of nonzero vectors $|h_i|\in\mathbb{R}^n, i=\{1,2\}$ satisfying $o(|h_1|)=\alpha\, o(|h_2|),\;\alpha>0,$ there
\\ &= F(a) + g'(f(a))f'(a)h+o(|h|) + o(|h|). That's all. I feel like there is some exist neighborhoods around the origin where the quotient between them remains close to $\alpha.$ Mathematically
trickery involved with the little o en big o. My problems are: Why does $f'(a)h+o(| speaking, for each $\epsilon>0,$ there is a $\delta>0$ such that $|\frac{o(|h_1|)}{o(\left|h_2\right|)}-\alpha|<\
h|) \to 0$ when $h\to 0$. I see that the first term tends to $0$, but how come $o(| epsilon$ holds true when $0<\max\{|h_1|,|h_2|\}<\delta.$ Now, recalling our second point, we know that $o(|h_1|) =
h|) \to 0$? $g'(f(a))o(|h|))=o(|h|)$. $o(|f'(a)h + o(|h|)|)=o(O(|h|))$ $" Do not o(|h_1|+\beta\, o(|h_1|)), \; \forall\,\beta \geq 0,$ because they differ only by higher order terms. Hence,
just list concepts, but develop each one in detail before moving to the next, as we considering the limit definition of norms, we obtain $|o(|h_1|)|_\infty \leq K|o(|h_1|)|_2$ for constants $K>0$
prioritize depth of understanding and comprehensive exploration of the subject independent of $|h_1|.$ By applying the triangle inequality, we deduce that $|o(|h_1|)-o(|h_2|)|_\infty\leq K(|o(|
matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the concepts/ h_1|)|_2+|o(|h_2|)|_2);$ thus, $o(|h_1|)$ and $o(|h_2|)$ share the same asymptotic behavior. 4. Finally, note that
sections. - Engagement: Write with an academic, professional and engaging tone that $o(|h|)=O(|h|)$ means that the left side grows no faster than the right side near the origin. However, it may still
captivates interest. - Application: Incorporate specific, practical examples, such exhibit lower growth rate. With these preliminary ideas out of the way, let's revisit the steps of the chain rule
as proofs in calculus or critical dates and figures in history. Do not include a proof, addressing your concerns line by line. 1. Why does $f'(a)h+o(|h|) → 0$ when $h→ 0?$ As discussed earlier,
title or an introduction, simply write the content without headlines and $f'(a)h$ dominates the expression because it has degree one. Meanwhile, $o(|h|)$ vanishes more rapidly than $|h|$
introductory phrases. Do not use images. as $h$ approaches zero. So, their combination, denoted by $f'(a)h + o(|h|),$ indeed converges to zero. 2. $g′(f(a))
o(|h|)=o(|h|):$ Recall that $o(|h|)$ represents a collection of functions that converge to zero faster than $|h|.$
Multiplying by a fixed number $g'(f(a))$ doesn't change this property. It merely scales the entire group by a
constant factor. Therefore, the resulting product remains within the original family of functions and continues to
converge to zero more quickly than $|h|.$ 3. $o(|f′(a)h+o(|h|)|)=o(O(|h|)):$ Here, we need to recognize that $f'(a)
h + o(|h|)$ belongs to the set of functions growing at most linearly in $|h|.$ More precisely, it behaves as $O(|h
|).$ On the other hand, $o(|h|)$ decays quicker than linearly. Applying the principle established in part ii), we
conclude that $o(|f'(a)h + o(|h|)|)$ shares the same decay rate as $o(|h|).$ By carefully dissecting each step
using the precise language of "big O" and "small o," you should now better comprehend the chain rule proof.
Remember, mastery comes through practice! Try proving similar statements involving compositions of functions, armed
with these powerful tools.
Before diving into the problem at hand, let us first understand the definitions given in the statement. A
self-adjoint algebra is a collection of functions that is closed under addition, multiplication, scalar
multiplication, and complex conjugation. This means that if you take any two functions in the collection and add,
multiply, or scale them by a constant, or take their complex conjugate, the result will still be in the collection.
The second definition, "separates points," intuitively means that for any pair of distinct elements in the set K,
there exists some function in the collection that maps these elements to different values. Finally, "vanishes
Write an educational piece suited for college students related to the following nowhere" means that for every element in K, there exists some function in the collection that does not map that
text snippet: "# Self-adjoint Algebra made my ideas Vanish. How to get them back? element to zero. With these definitions established, we can now proceed to prove the main claim. The goal is to
Let $$K$$ be a finite set and $$\mathcal{A}$$ a family of functions on $$K$$ that show that if we have a self-adjoint algebra that separates points and vanishes nowhere on a finite set K, then this
is a self-adjoint algebra, separates points and vanishes nowhere. Prove that $$\ algebra contains all possible functions on K. To begin, consider a fixed element k\_0 in K. Since our algebra
mathcal{A}$$ must then already contain every function on $$K$$. Definitions: vanishes nowhere, there exists some function f\_0 in the algebra such that f\_0(k\_0) ≠ 0. We can normalize this
-Self-adjoint algebra for $$f,g\in \mathcal{A}$$, $$c\in \mathbb{C}$$ we have $$f+g function by defining h\_0 = f\_0 / f\_0(k\_0), which ensures that h\_0(k\_0) = 1. Notice that h\_0 also belongs to
\in \mathcal{A}$$, $$f\cdot g\in \mathcal{A}$$, $$cf\in \mathcal{A}$$, $$\bar{f}\in our self-adjoint algebra since it is obtained from applying only linear operations (addition, subtraction,
\mathcal{A}$$. -Separates Points $$\forall x,y\in K$$ with $$x\neq y$$, $$\exists f division, and multiplication). Next, define the evaluation functional E : C^K → C given by Ev = v(k\_0) for any v
\in \mathcal{A}$$ so that $$f(x)\neq f(y)$$. -Vanishes Nowhere $$\forall x\in K$$, in C^K, where C^K denotes the space of all functions mapping K to C. Note that E is a linear functional, i.e., it
$$\exists f\in \mathcal{A}$$ so that $$f(x)\neq 0$$. I've tried too see if an satisfies E(av + bw) = aEv + bEw for any scalars a,b and any functions v, w in C^K. Moreover, since h\_0(k\_0) = 1,
induction argument on $$K$$ would work, but I've been having trouble showing that we know that Eh\_0 = 1. Thus, E is actually surjective because given any c in C, we can find some multiple of h\_0
if it works for $$|K|=n \Rightarrow$$ works for $$|K|=n+1$$. And I'm beginning to (namely ch\_0) mapped precisely to c by E. Now comes a crucial observation using the fact that our algebra is
believe that it won't work. I have also tried to show that I could construct a self-adjoint. Consider any function g in the dual space of C^K, denoted (C^K)^⸰, which assigns a value to each
function that is nonzero for o" Do not just list concepts, but develop each one in function in C^K. Define H : C^K → C by Hv = Re(gv) for any v in C^K, where Re denotes the real part. It turns out
detail before moving to the next, as we prioritize depth of understanding and that H is also a linear functional satisfying Hh\_0 = Re(gh\_0) = |g(k\_0)|² > 0 due to our choice of h\_0 above.
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: By the Riesz lemma, there exists some unique function f in C^K such that Hv = <v, f> for any v in C^K, where <·, ·>
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an represents the inner product defined as usual. That is, <v, f> := Σ_(k∈K) v(k)f(k) for finitely supported functions
academic, professional and engaging tone that captivates interest. - Application: v, and extended linearly otherwise. At this point, notice that f lies within our original self-adjoint algebra.
Incorporate specific, practical examples, such as proofs in calculus or critical Indeed, f is expressible as a sum of terms like α\_k h\_k, where h\_k agrees with h\_0 except at k, and α\_k are
dates and figures in history. Do not include a title or an introduction, simply constants arising from the representation theorem mentioned earlier. However, recalling that our algebra separates
write the content without headlines and introductory phrases. Do not use images. points, each term α\_kh\_k evaluates differently than f\_0 at its respective index k, meaning they belong to the
algebra by construction. Therefore, f itself resides within the algebra, completing the proof. In summary, starting
with a fixed element k\_0 in K, we constructed a normalized function h\_0 that never vanished at k\_0 and belonged
to our algebra. Then, exploiting the properties of our algebra being self-adjoint, we showed how to build another
function f lying inside the algebra capable of approximating arbitrary members of (C^K)^⸰ via the Riesz
representation theorem. Consequently, armed with both h\_0 and f, along with their combinations through linearity
inherent in our algebraic structure, we demonstrated that every conceivable function on K could indeed be generated
from the initial self-adjoint algebra possessing the desired properties.
A significant area of study within mathematics is the investigation of various types of matrices, including
Hermitian matrices. These matrices have important applications in numerous fields, such as physics and engineering.
This piece will delve into a journal article published in the Canadian Journal of Mathematics (CJM), which
discusses some intriguing properties and characteristics of bijective maps on the space of Hermitian matrices.
Specifically, it addresses Huang and Semrl's work concerning adjacency preservation in these matrices. To fully
grasp this topic, let us first define some essential terms. A matrix is a rectangular array of numbers arranged in
rows and columns. For instance, a 2 x 3 matrix has two rows and three columns, like so: [a b c ] [d e f ] Here,
Write an educational piece suited for college students related to the following 'a', 'b', 'c', 'd', 'e', and 'f' represent real or complex numbers. When all the entries are real numbers, the
text snippet: "location: Publications → journals Search results Search: All matrix is called a real matrix. However, when at least one entry is a complex number, the matrix becomes a complex
articles in the CJM digital archive with keyword hermitian matrix Expand all matrix. Additionally, if all entries of a square matrix (i.e., a matrix where the number of rows equals the number
Collapse all Results 1 - 2 of 2 1. CJM 2008 (vol 60 pp. 1050) Huang, Wen-ling; of columns) lie along its main diagonal from top left to bottom right, then the matrix is referred to as a diagonal
Semrl, Peter \v Adjacency Preserving Maps on Hermitian Matrices Hua's fundamental matrix. Now, let's turn our attention to Hermitian matrices, which constitute a special class of diagonal matrices.
theorem of the geometry of hermitian matrices characterizes bijective maps on the Named after Charles Hermite, a French mathematician, they possess conjugate-symmetric entries off their main
space of all $n\times n$ hermitian matrices preserving adjacency in both diagonals. That is, if 'a' appears in row i and column j (where i≠j), then 'a'* must appear in position row j and
directions. The problem of possible improvements has been open for a while. There column i. Here, 'a*' denotes the complex conjugate of 'a'. If you are unfamiliar with complex conjugates, recall
are three natural problems here. Do we need the bijectivity assumption? Can we that given any complex number z = x + yi, where x and y are real numbers, its complex conjugate is written as z* =
replace the assumption of preserving adjacency in both directions by the weaker x - yi. Thus, a simple example of a 2 x 2 Hermitian matrix would be: [w w* ] [z z* ] In this case, 'w' and 'z'
assumption of preserving adjacency in one direction only? Can we obtain such a could be arbitrary complex numbers. Note that every real symmetric matrix can also be considered a Hermitian matrix
characterization for maps acting between the spaces of hermitian matrices of since real numbers are equal to their own complex conjugates. The notion of "adjacency" comes up frequently
different sizes? We answer all three questions for the complex hermitian matrices, throughout the article under discussion. Two matrices are said to be adjacent if there exists precisely one nonzero
thus obtaining the optimal structural result for adj" Do not just list concepts, entry in their difference. It implies that two distinct Hermitian matrices are adjacent if they differ only in one
but develop each one in detail before moving to the next, as we prioritize depth of element, much like how neighboring nodes on a graph might share exactly one edge connecting them. With these
understanding and comprehensive exploration of the subject matter over breadth. definitions established, consider Huang and Semrl's work regarding bijective maps preserving adjacency among
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Hermitian matrices. Firstly, what does being "bijective" mean? Put simply, a function is bijective if it satisfies
Write with an academic, professional and engaging tone that captivates interest. - two conditions simultaneously: injectivity (or "one-to-oneness") and surjectivity (or "onto-ness"). An injection
Application: Incorporate specific, practical examples, such as proofs in calculus ensures no two elements map to the same image, whereas surjection guarantees that every member of the codomain
or critical dates and figures in history. Do not include a title or an corresponds to at least one element in the domain. Intuitively speaking, a bijection signifies that each input
introduction, simply write the content without headlines and introductory phrases. value uniquely matches a single output value, implying a perfect correspondence between sets involved. Returning to
Do not use images. the paper, the authors investigate whether certain assumptions about these adjacency-preserving functions can be
relaxed. They pose three interrelated questions: 1. Is bijectivity truly necessary for the conclusion to hold?
Might there exist non-bijective yet still satisfactory mappings? 2. Instead of requiring preservation of adjacency
in both directions, could we settle for merely maintaining adjacency in one direction? 3. How extensible are these
findings across dissimilar dimensions—that is, between matrices of varying orders or sizes? Remarkably enough,
Huang and Semrl affirmatively resolve all three queries! Their answers pertain specifically to the realm of complex
Hermitian matrices. By doing so, they attain an optimized structural result illuminating the nuances surrounding
adjacency preservation in Hermitian matrices.
The continuum hypothesis, first proposed by Georg Cantor in the late 19th century, is a statement regarding the
relative cardinalities of infinite sets. Specifically, it posits that there is no set whose cardinality (i.e., size
or number of elements) is strictly between that of the integers and the real numbers. This may seem like an obvious
truth, but proving or disproving it has turned out to be quite challenging. To understand why, let's take a step
back and consider what mathematicians mean when they talk about "the axioms of ZFC." Zermelo-Fraenkel Set Theory
with the Axiom of Choice, commonly abbreviated as ZFC, is a standard system of axioms used to formalize modern set
theory. These axioms provide rules and guidelines for working with mathematical objects called "sets," allowing us
to make precise statements about their properties and relationships. Now, here's where things get interesting: Paul
Write an educational piece suited for college students related to the following Cohen, building upon earlier work by Kurt Gödel, demonstrated that the continuum hypothesis cannot be proven or
text snippet: "# Question about the Continuum Hypothesis The Continuum Hypothesis disproven within the framework of ZFC. What this means is that neither the claim "there exists a set whose
hypothesises There is no set whose cardinality is strictly between that of the cardinality lies strictly between $|\mathbb{N}|$ and $|\mathbb{R}|$" nor its negation can be derived solely from
integers and the real numbers. Clearly this is either true or false - there either the axioms of ZFC. At first glance, this result might appear discouraging—after all, if we can't prove or disprove
exists such a set, or there does not exist such a set. Paul Cohen proved that the the continuum hypothesis, doesn't that leave us in limbo? However, Cohen's theorem actually opens up new
Continuum Hypothesis cannot be proven or disproven using the axioms of ZFC. If we possibilities for exploring the nature of mathematical truth and the limitations of our current axiomatic systems.
find a set whose cardinality lies strictly between that of $\mathbb{N}$ and $\ Let's dive deeper into the consequences of Cohen's result. First, note that saying something cannot be proven
mathbb{R}$, then we are done, we have disproven it. But it has been proven that we within ZFC doesn't necessarily imply that it's false; rather, it could just mean that we haven't found the right
cannot disprove it, thus by contrapositive, we cannot find such a set. If we cannot tools or methods yet. So, while we currently lack a way to constructively exhibit a set with the desired
find such a set, then we can only conclude that such a set does not exist (if it cardinality, this doesn't rule out the possibility that such a set exists outside the confines of ZFC. On the other
did exist, there must be a non-zero probability that we would find it, so given hand, suppose we want to argue that no such set exists—that is, we wish to assert the truth of the continuum
enough time we would - contradiction. □) So I have proven that the Continuum hypothesis itself. Here's where things get tricky. To do so, we need to show that any hypothetical set with
Hypothesis is true - there does not exist such a set. But this is a contradiction cardinality strictly between those of $\mathbb{N}$ and $\mathbb{R}$ leads to a logical contradiction. But since we
because it has been proven that we c" Do not just list concepts, but develop each know we can't disprove the continuum hypothesis using ZFC alone, any argument for its truth would have to rely on
one in detail before moving to the next, as we prioritize depth of understanding additional assumptions beyond those already encoded in the axioms. One possible response to this situation is to
and comprehensive exploration of the subject matter over breadth. Focus on: - adopt a position known as "set-theoretic pluralism": the idea that multiple, equally valid conceptions of set
Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Write with existence can coexist, each with different answers to questions like the continuum hypothesis. According to this
an academic, professional and engaging tone that captivates interest. - viewpoint, there might be models of set theory where the continuum hypothesis holds and others where it fails—and
Application: Incorporate specific, practical examples, such as proofs in calculus both could be considered legitimate interpretations of the underlying mathematics. Alternatively, some
or critical dates and figures in history. Do not include a title or an mathematicians might choose to augment the axioms of ZFC with new principles specifically designed to resolve
introduction, simply write the content without headlines and introductory phrases. questions like the continuum hypothesis. For example, large cardinal axioms—statements concerning the existence of
Do not use images. particularly enormous sets—have been shown to decide various set-theoretic issues, including whether certain kinds
of intermediate sets exist. By adopting these stronger axioms, one could potentially settle the continuum
hypothesis once and for all. However, doing so comes at a cost: adding new axioms inevitably restricts the range of
acceptable models of set theory, which in turn limits the scope of mathematical results that hold universally
across all models. Thus, deciding whether to accept such additions involves balancing the desire for definitive
answers against the potential loss of generality and flexibility afforded by leaving key questions open. In
conclusion, the continuum hypothesis serves as an intriguing case study in the interplay between mathematical
logic, foundational theories, and philosophical perspectives on the nature of mathematical truth. While we
currently lack a decisive resolution within the confines of ZFC, this apparent limitation also highlights the
opportunities for further exploration and growth in our understanding of sets and their properties.
Write an educational piece suited for college students related to the following Square numbers are those numbers obtained when a whole number is multiplied by itself. The first few square numbers
text snippet: "My Math Forum Why is it that none of the square numbers is one less are 1 (which is 1*1), 4 (which is 2*2), 9 (which is 3*3), 16 (which is 4*4) and so on. It turns out that none of
than a multiple of 3? Algebra Pre-Algebra and Basic Algebra Math Forum April 11th, these square numbers is one less than a multiple of three. This observation raises an interesting question – why
2018, 12:38 AM #1 Member Joined: Apr 2018 From: On Earth Posts: 34 Thanks: 0 Why does this happen? To answer this question, let’s explore the concept of congruences and modular arithmetic, which
is it that none of the square numbers is one less than a multiple of 3? I've will provide us with the tools needed to prove that no square number fits the given description. Congruences
noticed that none of the square numbers is one less than a multiple of 3. I have a establish an equivalence relation between integers based on division properties. Specifically, two integers, a and
rough idea of why this is but I'm not completely sure. Please can someone clarify? b, are said to be congruent modulo n, denoted as a ≡ b (mod n), if their difference results in an integer multiple
Thanks. April 11th, 2018, 01:04 AM #2 Senior Member Joined: Aug 2012 Posts: 1,999 of n; i.e., a − b = kn for some integer k. For instance, consider the statement 21 ≡ 4 (mod 7). Here, 21 - 4 = 17,
Thanks: 573 Quote: Originally Posted by Student2018 I've noticed that none of the which indeed is divisible by 7 since 17 = 2 * 7 + 3. Therefore, 21 and 4 belong to the same residue class modulo 7.
square numbers is one less than a multiple of 3. I have a rough idea of why this is Now, let’s introduce another fundamental concept—even and odd numbers. An even number is any integer of the form
but I'm not completely sure. Please can someone clarify? Yes, it's modular 2k, where k is an integer, while an odd number has the form 2k+1. Using these definitions, squaring an even number
arithmetic. A number that's one less than a multiple of 3 is -1 mod 3. Or 2 mod 3, always leads to another even number because (2k)^2 = 4k^2, which still has the factor 2. Similarly, squaring an odd
comes out to the same thing. In other words if some number $n$ can be written as $3 number generates an odd result due to ((2k+1)^2 = 4k(k+1)+1). These observations indicate that the set {odd
m - 1$, by adding anoth" Do not just list concepts, but develop each one in detail numbers} and {even numbers} are closed under squaring operation. Let’s now apply our knowledge about congruences
before moving to the next, as we prioritize depth of understanding and and parity to address the original problem regarding square numbers being one less than a multiple of three. Assume
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: there exists some positive integer x such that its square number x^2 is one less than a multiple of 3, meaning: x^2
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an ≡ -1 (mod 3). According to Euler’s Criterion (a generalization of quadratic reciprocity law applicable here), (-1|
academic, professional and engaging tone that captivates interest. - Application: p) = 1 if and only if p ≡ 1 (mod 4), where | denotes Legendre symbol, an indicator function used to describe
Incorporate specific, practical examples, such as proofs in calculus or critical whether certain conditions hold true within the context of modular arithmetic. Since 3 doesn’t satisfy the
dates and figures in history. Do not include a title or an introduction, simply condition required for (-1|3)=1, namely 3 ≡ 1 (mod 4), there cannot exist an integer solution x satisfying x^2 ≡ -1
write the content without headlines and introductory phrases. Do not use images. (mod 3). Consequently, no square number is one less than a multiple of 3.
Let us delve into the fascinating world of coordinate systems, specifically focusing on a lesser-known system
called "parabolic coordinates." This system has some unique properties that differentiate it from the more commonly
used Cartesian coordinates you're familiar with. The exposition below will cover the fundamentals, derivations, and
Write an educational piece suited for college students related to the following conversions between these two systems using provided formulas. Parabolic coordinates consist of three components
text snippet: "Sunday, March 5, 2017 HP Prime: Parabolic Coordinates HP Prime: denoted by $(\mu, v, \phi)$, where $\mu$ and $v$ are nonnegative real numbers while $\phi$ represents any real
Parabolic Coordinates The Formulas The relationship and conversion factors between number. These quantities relate to rectangular coordinates $(x, y, z)$ through the subsequent set of equations: $$x
parabolic coordinates (μ, v, ϕ) and rectangular coordinates (x, y, z) are as = \mu v \cos\phi$$ $$y = \mu v \sin\phi$$ $$z = \frac{1}{2}\left(\mu^{2} - v^{2}\right)$$ $$\phi = \operatorname
follows: x = μ * v * cos ϕ y = μ * v * sin ϕ z = 1/2 * (μ^2 – v^2) ϕ = atan(y/x) v {atan2}(y, x)$$ $$v = \sqrt{-z + \sqrt{x^{2}+y^{2}+z^{2}}}$$ $$\mu = \sqrt{2z+v^{2}}$$ Note the particular order of
= √( -z + √(x^2 + y^2 + z^2)) μ = √( 2*z + v^2) (note the sequence) where μ ≥ 0 and operations when computing $v$ and $\mu$. As their values depend on previously calculated expressions, they should
v ≥ 0 Derivation The formulas to find the rectangular coordinates are given. We be computed accordingly. Now let's explore how these relationships were derived. Firstly, observe the equations
can derive the formulas for the parabolic coordinates by the following: Assume that relating $x$, $y$, and $\phi$: $$x = \mu v \cos\phi$$ $$y = \mu v \sin\phi$$ Dividing these yields: $$\frac{x}{y}=\
μ > 0 and v > 0 (neither are zero). Then: x = μ * v * cos ϕ y = μ * v * sin ϕ x / cot\phi=\frac{\cos\phi}{\sin\phi}$$ This implies: $$\phi=\arctan\left(\frac{y}{x}\right)=\operatorname{atan2}(y, x)
y = (μ * v * cos ϕ)/( μ * v * sin ϕ) x / y = 1 / tan ϕ y / x = tan ϕ ϕ = atan (y / $$ Next, express $\mu$ in terms of $z$: $$z = \frac{1}{2}(\mu^{2}-v^{2})$$ Rearrange this equation to solve for $\
x) Express μ in terms of z: z = 1/2 * (μ^2 – v^2) 2 * z = μ^2 – v^2 μ^2 = 2*z + v^2 mu$: $$\begin{align*} \mu^{2} &= 2z+v^{2}\\ \mu &= \sqrt{2z+v^{2}} \end{align*}$$ Finally, determine if there
Since μ is positive, only the positive square root is considered: μ = √(2*z + v^2) exists an expression for $v$ using the first pair of equations: $$x=\mu v\cos\phi$$ $$y=\mu v\sin\phi$$ We already
Can we find an expression for v? If we can, we have found our formulas: x = μ * v know $\mu$ and $\phi$; thus, isolating $v$ would yield the desired formula. Squaring both equations leads to: $$x^
* cos ϕ Sq" Do not just list concepts, but develop each one in detail before moving {2}=\mu^{2}v^{2}\cos^{2}\phi$$ $$y^{2}=\mu^{2}v^{2}\sin^{2}\phi$$ Adding these together results in: $$x^{2}+y^{2}=\
to the next, as we prioritize depth of understanding and comprehensive exploration mu^{2}v^{2}$$ Now extract $v^{2}$: $$v^{2}=\frac{x^{2}+y^{2}}{\mu^{2}}$$ Recall that $\mu=\sqrt{2z+v^{2}}$. Thus,
of the subject matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of substitute $v^{2}$ back in terms of $\mu$: $$\mu^{2}=2z+\frac{x^{2}+y^{2}}{\mu^{2}}$$ Isolate $\mu^{4}$: $$\mu^{4}-
the concepts/sections. - Engagement: Write with an academic, professional and (x^{2}+y^{2})\mu^{2}-2z\mu^{2}=0$$ Treating $\mu^{2}$ as a variable $t$, recognize this as a quadratic equation:
engaging tone that captivates interest. - Application: Incorporate specific, $$t^{2}-(x^{2}+y^{2})t-2z=0$$ Applying the quadratic formula gives: $$t=\frac{(x^{2}+y^{2})\pm\sqrt{(x^{2}+y^{2})^
practical examples, such as proofs in calculus or critical dates and figures in {2}+8z}}{2}$$ As $\mu$ must always be positive, choose the positive branch: $$\mu^{2}=\frac{1}{2}[(x^{2}+y^{2})+\
history. Do not include a title or an introduction, simply write the content sqrt{(x^{2}+y^{2})^{2}+8z}]$$ Extract $v$ once again: $$v^{2}=\frac{x^{2}+y^{2}}{\mu^{2}}=\frac{x^{2}+y^{2}}{\dfrac
without headlines and introductory phrases. Do not use images. {1}{2}[(x^{2}+y^{2})+\sqrt{(x^{2}+y^{2})^{2}+8z}]}$$ Simplify further to obtain: $$v = \sqrt{-z + \sqrt{x^{2}+y^{2}
+z^{2}}}$$ Thus, all required formulas have been established! To summarize, converting between parabolic and
rectangular coordinates entails manipulating trigonometric functions alongside solving quadratics rooted in
geometric relations. With practice, transformations among various coordinate systems become second nature—providing
powerful tools for analyzing complex problems in mathematics and physics.
Write an educational piece suited for college students related to the following
text snippet: "## Subsection6.2.2Error in storing a real number as a floating point
number ###### Remark6.2.2.1. We consider the case where a real number is trunctated
to become the stored floating point number. This makes the discussion a bit
simpler. Let positive $\chi$ be represented by \begin{equation*} \chi = . d_0 d_1 \
cdots \times 2^e, \end{equation*} where $d_i$ are binary digits and $d_0 = 1$ (the
mantissa is normalized). If $t$ binary digits are stored by our floating point Floating point numbers are a crucial concept in computer science and mathematics, used to approximate real numbers
system, then \begin{equation*} \check \chi = . d_0 d_1 \cdots d_{t-1} \times 2^e \ within digital systems. However, due to their discrete nature, they cannot perfectly represent all real numbers,
end{equation*} is stored (if truncation is employed). If we let $\delta\!\chi = \ leading to errors known as floating point errors. We will delve into this issue, focusing specifically on how
chi - \check \chi \text{.}$ Then \begin{equation*} \begin{array}{rcl} \delta\!\chi truncation affects the accuracy of representing real numbers as floating point numbers. To begin with, consider a
\amp = \amp \begin{array}[t]{c} \underbrace{. d_0 d_1 \cdots d_{t-1} d_t \cdots \ positive real number $\chi$, which can be expressed in its binary form as follows: $$\chi=.d\_0d\_1\dots\times2^e$$
times 2^e} \\ \chi \end{array} - \begin{array}[t]{c} \underbrace{. d_0 d_1 \cdots Here, $d\_i$ represents binary digits, and it's assumed that the mantissa is normalized, meaning that $d\_0=1$.
d_{t-1} \times 2^e} \\ \check \chi \end{array} \\ \amp = \amp \begin{array}[t]{c} \ Now, imagine you have a floating point system capable of storing only $t$ binary digits; thus, when approximating $
underbrace{. 0 \cdots 00} " Do not just list concepts, but develop each one in \chi$, your system retains only these significant bits: $$\check{\chi}=.d\_0d\_
detail before moving to the next, as we prioritize depth of understanding and
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor:
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an
academic, professional and engaging tone that captivates interest. - Application:
Incorporate specific, practical examples, such as proofs in calculus or critical
dates and figures in history. Do not include a title or an introduction, simply
write the content without headlines and introductory phrases. Do not use images.
A linear time-invariant (LTI) system is said to be minimum-phase if it satisfies certain conditions regarding its
stability, causality, and frequency response. Understanding the implications of various definitions of
minimum-phase systems can provide valuable insights into their properties and behavior. This discussion will
explore two common definitions found in textbooks and delve into the reasons behind the differences between them,
particularly concerning the location of zeros and poles and the analyticity of the transfer function. **Definition
1:** An LTI system is minimum-phase when it is *both* invertible, causal, and BIBO (Bounded Input, Bounded Output)
stable. A system is considered causal if its output at any given time depends only on present and past inputs,
never on future inputs. Mathematically, this means that the impulse response $h(t)$ of a causal system is zero for
all $t < 0$. Stability ensures that a bounded input results in a bounded output; hence, the system does not exhibit
Write an educational piece suited for college students related to the following oscillatory behavior for nonzero input signals. For BIBO stability, the region of convergence (ROC) of the system's
text snippet: "# Definition of minimum-phase system I saw a couple of definitions transfer function must contain the unit circle. **Definition 2:** An LTI system is minimum-phase when it is *both*
for minimum-phase in different textbooks and I'm trying to understand what the invertible, causal, and has finite energy. Moreover, the transfer functions of the system and its inverse should be
implication of each of them. The first definition I saw was: An invertible system analytic in the exterior of the unit circle. The concept of finite energy implies that the total signal power over
which both it and it's inverse are causal and (BIBO) stable. The second definition $(-\infty, \infty)$ is finite. Consequently, the impulse response decays faster than $1/\sqrt{|t|}$, ensuring
is: An invertible system which both it and it's inverse are causal and have finite absolute integrability ($\int_{-\infty}^{\infty} |h(t)| dt < \infty$). Analyticity refers to whether a complex
energy. It was written in one textbook that had the second definition as an function can be expressed as a convergent power series within a specified domain. If a transfer function is
alternative to the first one that these constraints allow the system to have zeros analytic in the exterior of the unit circle, then there are no singularities outside the unit circle, meaning that
and poles on the unit circle and I don't understand why (as opposed to the first the ROC includes the exterior of the unit circle. Now let us discuss the implications of these definitions,
def). just to make sure, I'm not restricting myself to rational systems. Another focusing on the location of zeros and poles, as well as the analyticity property mentioned in Definition 2.
thing I saw in some textbook and didn't understand is that the second definition Consider a rational transfer function $H(z)=B(z)/A(z)$, where $A(z)$ and $B(z)$ are polynomials in $z$,
implied that the transfer function of the system and it's inverse are analytic in representing the numerator and denominator, respectively. Poles occur whenever $A(z) = 0$, while zeros correspond
the exterior of the unit circle. So, to conclude, my questions are: 1. why we have to instances when $B(z) = 0$. According to either definition above, the system must be causal, so the region of
both of the definitions of minimum-ph" Do not just list concepts, but develop each convergence cannot extend beyond the leftmost pole. Since the ROC contains the unit circle under both definitions,
one in detail before moving to the next, as we prioritize depth of understanding neither the poles nor zeros can lie outside the unit circle. However, they may reside on the unit circle itself.
and comprehensive exploration of the subject matter over breadth. Focus on: - Moreover, consider the fact that the inverse system exists if and only if $\gcd(A(z), B(z))=1$ (i.e., the greatest
Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Write with common divisor of $A(z)$ and $B(z)$ is a constant polynomial equal to unity); otherwise, cancelation of common
an academic, professional and engaging tone that captivates interest. - factors would lead to ambiguity in defining the inverse. Given this constraint, suppose $p_0$ were a zero located
Application: Incorporate specific, practical examples, such as proofs in calculus exactly on the unit circle. Then, near $p_0$, $|H(z)|$ tends to vanish due to cancellation by the corresponding
or critical dates and figures in history. Do not include a title or an factor in $B(z)$. On the other hand, since the inverse system also exists, its transfer function $1/H(z)$ would
introduction, simply write the content without headlines and introductory phrases. tend to diverge around $p_0$, contradicting the requirement for finite energy unless $1/H(z)$ is entirely confined
Do not use images. inside the unit circle. Thus, the presence of zeros or poles on the unit circle leads to inconsistencies depending
upon the chosen definition. Therefore, it becomes essential to specify whether the system permits zeros or poles on
the unit circle explicitly. Finally, the analyticity condition stated in Definition 2 stems from considering the
inverse system's existence. When the transfer function $H(z)$ has no zeros or poles in the exterior of the unit
circle, the inverse transfer function $1/H(z)$ remains analytic there too. As previously discussed, this
facilitates unambiguous determination of the inverse system when applying Definition 2. Furthermore, analyticity
ensures that the phase response varies smoothly across frequencies, resulting in desirable characteristics like
group delay being approximately constant throughout the passband. These benefits contribute to the preference for
Definition 2 among practitioners working with digital filters or signal processing applications involving
discrete-time systems.
Write an educational piece suited for college students related to the following
text snippet: "It is currently 19 Feb 2019, 05:45 ### GMAT Club Daily Prep ####
Thank you for using the timer - this advanced tool can estimate your performance
and suggest more practice questions. We have subscribed you to Daily Prep Questions
via email. Customized for You we will pick new questions that match your level
based on your Timer History Track every week, we’ll send you an estimated GMAT The equation $${x}^{2}-{y}^{2}=0$$, where both $$x$$ and $$y$$ are nonzero, implies that $${x}^{2}={y}^{2}$$.
score based on your performance Practice Pays we will pick new questions that match Taking the square root of both sides gives us two possibilities: either $$x=y$$ or $$x=-y$$. This means that $$x$$
your level based on your Timer History # If x2 - y2 = 0 and xy 0, which of the and $$y$$ must have the same magnitude but opposite signs if they are different numbers. Now let's consider the
following MUST be true Author Message TAGS: GRE Prep Club Legend Joined: 07 Jun statement "xy > 0". Since neither $$x$$ nor $$y$$ can be zero (because their product is positive), we know that
2014 Posts: 4856 GRE 1: Q167 V156 WE: Business Development (Energy and Utilities) they must both be positive or both be negative. However, since they have the same magnitude but opposite signs, it
Followers: 102 Kudos [?]: 1740 [1] , given: 397 If x2 - y2 = 0 and xy 0, which of cannot be the case that they are both positive or both negative. Therefore, there is a contradiction between the
the following MUST be true [#permalink] 20 May 2018, 10:49 1 KUDOS Expert's post given information and the additional statement provided. Thus, no definitive conclusion about the relationship
00:00 Question Stats: 23% (01:11) correct 76% (00:27) wrong based on 17 sessions If between $$x$$ and $$y$$ can be drawn. In summary, while the equation $${x}^{2}-{y}^{2}=0$$ constrains the possible
$$x^2 - y^2 = 0$$ and $$x" Do not just list concepts, but develop each one in values of $$x$$ and $$y$$, adding the condition "xy > 0" leads to a contradiction rather than providing any
detail before moving to the next, as we prioritize depth of understanding and meaningful information. It is important to carefully analyze all given conditions and ensure consistency when
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: solving mathematical problems.
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an
academic, professional and engaging tone that captivates interest. - Application:
Incorporate specific, practical examples, such as proofs in calculus or critical
dates and figures in history. Do not include a title or an introduction, simply
write the content without headlines and introductory phrases. Do not use images.
Write an educational piece suited for college students related to the following
text snippet: "... Point 6 something but let's actually get a calculator out to
calculate it. To calculate the price elasticity of demand, here’s what you do: Plug
in the values for each symbol. Just as a review, price elasticity of demand, so the
elasticity of demand, is defined as the percent change in quantity demanded over
the percent change in price. Thanks to this calculator, you will be able to decide
whether you should charge more for your product (and sell a smaller quantity) or
decrease the price, but increase the demand. Factors influencing the elasticity: Price Elasticity of Demand (PED): At its core, PED measures how sensitive the quantity demanded of a good or
The factors like price, income level and availability of substitutes influence the service is to changes in its price, all else held equal. It is calculated by dividing the percentage change in
elasticity. Price elasticity of demand refers to the degree of change in the demand quantity demanded by the percentage change in price. If the resulting value is greater than one, the demand is
for a product with respect to change in the given price, while keeping other considered elastic; if it's less than one, the demand is inelastic. When price elasticity equals one, demand is
determinants of demand at constant. 2) is same. The same process is used to work said to be unit elastic. Calculating Percentage Changes: To compute PED, first determine the percentage change in
out the percentage change in price. (CPB) The percent change in the price of quantity demanded and price using the formulae below: 1. Percentage Change in Quantity Demanded = ((New Quantity
Product 2 ; Cross-Price Elasticity of D" Do not just list concepts, but develop Demanded - Original Quantity Demanded)/Original Quantity Demanded)*100% 2. Percentage Change in Price = ((New Price
each one in detail before moving to the next, as we prioritize depth of - Original Price)/Original Price)*100%
understanding and comprehensive exploration of the subject matter over breadth.
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement:
Write with an academic, professional and engaging tone that captivates interest. -
Application: Incorporate specific, practical examples, such as proofs in calculus
or critical dates and figures in history. Do not include a title or an
introduction, simply write the content without headlines and introductory phrases.
Do not use images.
The example problem you've encountered is asking you to demonstrate a fundamental concept in probability theory -
the idea that different ways of representing a probabilistic event can still yield the same overall probabilities,
Write an educational piece suited for college students related to the following as long as we account for any repetitions or indistinguishable outcomes. This particular question focuses on
text snippet: "# Probability Example Question I'm reading my book on probability arrangements of objects (red and blue balls), which falls under the broader category of combinatorics. Let's delve
and I don't understand the example problem: Question: Suppose that n + m balls, of deeper into the concepts involved here: Firstly, factorials play a crucial role in counting problems like this. A
which n are red and m are blue, are arranged in a linear order in such a way that factorial of a positive integer $n$, denoted as $n!$, represents the product of all positive integers up to $n$:
all (n + m)! possible orderings are equally likely. If we record the result of this $$n! = n\times(n-1)\times(n-2)\times...\times2\times1$$ Factorials help us determine the number of unique ways to
experiment by listing only the colors of the successive balls, show that all the arrange items when the ordering matters; in other words, they allow us to calculate permutations. For instance, if
possible results remain equally likely. My approach: I thought about doing my own there were three distinct objects, say A, B, and C, then there would be $3! = 6$ possible permutations (ABC, ACB,
example. Assume n= 2 red and m = 1 blue. Therefore, we have (n+m)! = 3! = 6 total BAC, BCA, CAB, CBA). Similarly, your example considers two red balls $(r_1, r_2)$ and one blue ball $(b)$,
permutations, namely: $b \ r_1 \ r_2 \\r_1 \ b \ r_2 \\r_1 \ r_2 \ b$ $b \ r_2 \ resulting in $3! = 6$ permutations. Next, let's discuss combinations. Combinations differ from permutations because
r_1 \\r_2 \ b \ r_1 \\r_2 \ r_1 \ b$ So I understand that each one of these they do not consider the ordering of the elements within sets; rather, they focus solely on selecting subsets from
permutations will have a probability of 1/6 of occurring as mentioned in the a larger set. Formally speaking, the combination of choosing k items out of n distinct items is given by ${n}\
question stem. But since we only are showing colors of successive balls, we remove choose{k}$ or $\frac{n!}{(n-k)!k!}$. These formulas reflect how many ways we can choose unordered subsets of size
the repeats by dividing by 2!: $\frac{3!}{2!} = 3$, which is easily seen from the $k$ from our original set of size $n$. Going back to the earlier ABC example, if we wanted to find the number of
list above. This is where I am con" Do not just list concepts, but develop each one distinct pairs we could create regardless of their order, there would be ${3}\choose{2}$ possibilities, i.e., AB,
in detail before moving to the next, as we prioritize depth of understanding and AC, BC. Note that this value equals the sum of the distinct permutations divided by the number of repeated ones ($3
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: =\frac{3}{1}+\frac{3}{2}$, as shown in your work). Now, returning to the probability example question, it asks you
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an to prove that after accounting for color similarities among adjacent balls (i.e., treating all red balls as
academic, professional and engaging tone that captivates interest. - Application: identical and blue balls as identical), the probabilities of various sequences should remain equal. Indeed, using
Incorporate specific, practical examples, such as proofs in calculus or critical your calculation, the six initial permutations reduce down to three once duplicate colors are taken into
dates and figures in history. Do not include a title or an introduction, simply consideration, aligning perfectly with the expected number of combinations based on the formula provided earlier ($
write the content without headlines and introductory phrases. Do not use images. \frac{3!}{2!\cdot1!}=3$). By demonstrating consistency between the directly calculated permutations and the derived
combinations, you reinforce the notion that equivalent probabilities arise despite varying methods used to
represent the events themselves.
Write an educational piece suited for college students related to the following
text snippet: "# Using Linear Interpolation to Find Interest 1. Feb 24, 2014 ### Let's delve into the concept of linear interpolation and its application in finding interest rates, using the
ainster31 1. The problem statement, all variables and given/known data 2. Relevant dialogue between the users "ainster31" and "Ray Vickson." We will break down the conversation to grasp the
equations 3. The attempt at a solution I understand how they calculated NPW but how underlying mathematical principles and methods. Linear interpolation is a technique used to estimate values between
did they use the linear interpolation method? 2. Feb 24, 2014 ### Ray Vickson I two known points in a dataset. It fits a straight line through these two points and uses this line to approximate
have never before seen their version of the linear interpolation method, but it unknown values within the interval spanned by those points. This approach is particularly useful when trying to
does work. In linear interpolation we fit a linear function $f = a + bx$ through determine the root (zero) of a function, which is what the user "ainster31" was interested in discovering. To apply
two points $(x_1,f_1)$ and $(x_2,f_2)$: $$f(x) = f_1 + \frac{f_2-f_1}{x_2-x_1} linear interpolation mathematically, first, identify the two known points ($x_1, f_1$) and ($x_2, f_2$). These
(x-x_1)$$ This gives us the value of $x_0$, the root of $f(x) = 0,$ as follows: points correspond to the values where your function is defined. Next, calculate the slope of the linear function
$$x_0 = \frac{f_2 x_1 - f_1 x_2}{f_2 - f_1}$$ It follows (doing a lot of algebraic passing through them: $$\text{Slope} = \frac{\Delta y}{\Delta x}=\frac{f_2-f_1}{x_2-x_1}.$$ After determining the
simplification) that $$\frac{x_0 - x_1}{x_2-x_0} = -\frac{f_1}{f_2}$$ In other slope, you can express the linear function in point-slope form, which looks like this: $$f(x)=f_1+\frac{\Delta y}{\
words, we can find $x_0$ by solving the equation $$\frac{x - x_1}{x_2-x} = -\frac Delta x}(x-x_1).$$ Now, if you want to find the root (denoted as $x_0$) of the linear function, set $f(x)$ equal to
{f_1}{f_2}$$ The solution is $x = x_0$. That is not how I would do it, but it is zero and solve for $x$: $$f(x_0)=\cancel{f}_1-\cancel{f}_1\cdot\left(\frac{x_0-x_1}{x_2-x_1}\right)^{-1}=0.$$
correct. 3. Mar 3, 2014 ### ains" Do not just list concepts, but develop each one Simplifying leads to the following expression: $$\frac{x_0-x_1}{x_2-x_0}=-\frac{f_1}{f_2},$$ from which you can
in detail before moving to the next, as we prioritize depth of understanding and isolate $x_0$: $$x_0=\frac{f_2x_1-f_1x_2}{f_2-f_1}.$$ Returning to our initial example, "ains" asked about
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: computing the Net Present Worth (NPW), specifically regarding how linear interpolation was utilized in this
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an computation. By employing the formula above, you can find the present worth factor ($x_0$), taking into account
academic, professional and engaging tone that captivates interest. - Application: different interest rates. Once obtained, multiply the net cash inflows by the corresponding present worth factors
Incorporate specific, practical examples, such as proofs in calculus or critical and sum up the results to obtain the total Net Present Worth (NPW).
dates and figures in history. Do not include a title or an introduction, simply
write the content without headlines and introductory phrases. Do not use images.
To begin, let's define some key terms that will be used throughout this discussion. A **field** $F$ is a set
equipped with two binary operations, addition and multiplication, satisfying certain properties similar to those
satisfied by the rational numbers or real numbers. Given a field $F$, the ring $R=F[x_1,\dots,x_t]$ denotes the
Write an educational piece suited for college students related to the following polynomial ring in $t$ variables over $F$. A nonempty subset $I\subset R$ is called an **ideal** if it satisfies
text snippet: "## The intersection of two monomial ideals is a monomial ideal Let the following conditions: * For all $f,g\in I$, $f+g\in I$ (closure under addition) * For all $r\in R$ and $f\in
$F$ be a field, let $R = F[x_1, \ldots, x_t]$, and let $M = (m_i \ |\ i \in I)$ and I$, $rf\in I$ (closure under scalar multiplication) In particular, when $I=(m_i|i\in I)$, where $m_i$ are
$N = (n_j \ |\ j \in J)$ be monomial ideals in $R$ (not necessarily finitely monomials, we call $I$ a **monomial ideal**. Recall that a **monomial** is a product of powers of variables, e.g.,
generated). Prove that $M \cap N$ is also a monomial ideal. Let $e_{i,j}$ be a $x^2yz^3$. Now that we have established these definitions, let us delve into proving our main statement. To prove
least common multiple of $m_i$ and $n_j$ for each $i \in I$ and $j \in J$. We wish that the intersection of two monomial ideals is again a monomial ideal, consider two arbitrary monomial ideals $M=
to show that $M \cap N = (e_{i,j} \ |\ i \in I, j \in J)$. $(\subseteq)$ Suppose $p (m_i|i\in I)$ and $N=(n_j|j\in J)$ in $R$. Our goal is to establish that their intersection $P:=M\bigcap N$ forms
\in M \cap N$. By this previous exercise, every term in $p$ is divisible by some another monomial ideal. First, note that since both $M$ and $N$ are closed under scalar multiplication, so too must
$m_i$ and also by some $n_j$; then every term is divisible by some $e_{i,j}$. Thus their intersection be. Therefore, it remains to demonstrate closure under addition within $P$. Consider two
$p \in (e_{i,j}\ |\ i \in I, j \in J)$. $(\supseteq)$ Suppose $p \in (e_{i,j} \ |\ elements from $P$: $a=\sum r_im_i$ and $b=\sum s_jn_j$, where the summations range over relevant indices
i \in I, j \in J)$. Then each term in $p$ is divisible by some $e_{i,j}$; hence corresponding to generators of $M$ and $N$, respectively. As they belong to $P$, both $a$ and $b$ lie in both $M$
each term is divisible by some $m_i$ (so that $p \in M$) and also by some $n_j$ (so and $N$. Since $M$ and $N$ are ideals, $\alpha:=a+b\in M$ and $\beta:=-b\in N$. Consequently, $\alpha+\beta\in P$,
that $p \in N$). So $p \in M \cap N$. Note that an easy induction argument shows which implies $(\alpha+\beta)=\sum t_{\ell}(m_\ell n_\ell)\in P$, because $\alpha+\beta=\sum(r_im_i)+\sum(-s_jn_j)
that in fact any finite inter" Do not just list concepts, but develop each one in $. Here, $\{m_\ell\}$ and $\{n_\ell\}$ represent subsets of the generating sets of $M$ and $N$, respectively. This
detail before moving to the next, as we prioritize depth of understanding and proves that $P$ is indeed an ideal. Now, recall that every element in $P$ can be written as a linear combination of
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: terms of the form $e_{ij}=lcm(m_i,n_j)$, denoted as $P=(e_{ij}|i\in I, j\in J)$. Note that this expression may
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an involve infinite sums due to potentially infinitely many generators. However, even though the number of possible
academic, professional and engaging tone that captivates interest. - Application: combinations might be uncountable, each individual term appearing in any sum belongs to the desired set $(e_{ij}|i\
Incorporate specific, practical examples, such as proofs in calculus or critical in I, j\in J)$. Hence, the inclusion holds true: $P\subseteq(e_{ij}|i\in I, j\in J)$. For the reverse containment $
dates and figures in history. Do not include a title or an introduction, simply (e_{ij}|i\in I, j\in J)\subseteq P$, take any $e_{ij}$, which is clearly contained in both $M$ and $N$ since it is
write the content without headlines and introductory phrases. Do not use images. divisible by both $m_i$ and $n_j$. Therefore, any linear combination formed using coefficients from the base field
$F$ results in an element belonging to $P$. Throughout this demonstration, we employed fundamental principles
governing fields, rings, and ideals while focusing explicitly on monomial ideals. By carefully examining
divisibility relationships between various monomials involved, we showed how the intersection operation preserves
the desirable algebraic structure inherent in monomial ideals.
Write an educational piece suited for college students related to the following The problem provided is derived from the 1980 American High School Mathematics Examination (AHSME), specifically
text snippet: "# Difference between revisions of "1980 AHSME Problems/Problem 17" # problem number 17. This problem requires knowledge of complex numbers and algebraic manipulations. We will break
# Problem Given that $i^2=-1$, for how many integers $n$ is $(n+i)^4$ an integer? $ down the solution into smaller steps and explain the underlying mathematical principles. Complex numbers can be
\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text represented as ordered pairs (a, b) or in the form of a + bi, where a and b are real numbers, and i is the square
{(D)} \ 3 \qquad \text{(E)} \ 4$ ## Solution $(n+i)^4=n^4+4in^3-6n^2-4in+1$, and root of -1. When squaring i, you get (-1). Let's examine the given expression more closely: (n + i)^4 = n^4 + 4in³
this has to be an integer, so the sum of the imaginary parts must be $0$. $$4in^ - 6n² - 4in + 1 For this expression to result in an integer value, both its real and imaginary components must
3-4in=0$$ $$4in^3=4in$$ $$n^3=n$$ Since $n^3=n$, there are $\boxed{3}$ solutions equal integers. First, let us focus on the imaginary part of the expression: 4in³ - 4in. For these terms to cancel
for $n$: $0$ and $\pm1$. -aopspandy" Do not just list concepts, but develop each out when adding the whole expression, they need to be equivalent; hence: 4in³ - 4in = 0 Now, factor out 4in: 4in(n²
one in detail before moving to the next, as we prioritize depth of understanding - 1) = 0 The equation above implies two possible scenarios. Either 4in equals zero, which means n = 0 since i
and comprehensive exploration of the subject matter over breadth. Focus on: - cannot equal zero. Alternatively, n² - 1 could equal zero, leading to n = ±1. These three potential values for n
Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Write with warrant further investigation. Let's substitute them back into the original expression (n + i)^4 to verify if it
an academic, professional and engaging tone that captivates interest. - indeed yields integer results: 1. If n = 0, then (0 + i)^4 = i^4 = -1 (integer) 2. If n = 1, then (1 + i)^4 = 1 +
Application: Incorporate specific, practical examples, such as proofs in calculus 4i - 6 - 4i + 1 = -4 + i(-4 + 4i) = -4 (integer) 3. If n = -1, then ((-1) + i)^4 = (-1)^4 + 4i(-1)³ - 6(-1)² - 4i
or critical dates and figures in history. Do not include a title or an (-1) + 1 = 1 - 4i - 6 + 4i + 1 = -4 (integer) Since all three possibilities yield integer results, our answer is
introduction, simply write the content without headlines and introductory phrases. (D): 3.
Do not use images.
Write an educational piece suited for college students related to the following Let's begin by recalling some fundamental definitions related to matrices and their ranks. The rank of an m x n
text snippet: "# Intuition behind matrix rank Suppose we have a matrix $$A$$ with matrix A is defined as the maximum number of linearly independent column vectors (or row vectors) in A. It is
$$m$$ rows and $$n$$ columns satisfying the condition $$m. Suppose further that important to note that this definition implies that the rank of a matrix cannot exceed either the number of rows or
$$m$$ rows are linearly independent, and $$n$$ columns are linearly independent as the number of columns. Now, let us examine your intuition regarding the rank of matrix A based on the given
well. The article from Wikipedia says: In linear algebra, the rank of a matrix information. You mentioned that there are m linearly independent rows and n linearly independent columns in matrix
$$A$$ is the dimension of the vector space generated (or spanned) by its columns. A, where m < n. You also noted that the example suggests rank(A) = m; however, it appears conflicting because you
This corresponds to the maximal number of linearly independent columns of $$A$$. believe that the dimension of the space spanned by rows should be equal to n due to the presence of n linearly
This, in turn, is identical to the dimension of the space spanned by its rows. independent equations. Allow us to clarify these points below. 1. Firstly, verifying whether all columns are indeed
Definitely, from the exmaple above we have $$rank(A)=m$$. But from another hand, linearly independent may help resolve any confusion. Given that there are more columns than rows (n > m), at least
the dimension of the space spanned by rows are equal to $$n$$, since $$n$$ linearly one column must be a linear combination of other columns. As a result, it can't be linearly independent when
independent equations. If my reasoning is correct, I see a contradiction between combined with those columns. Consequently, the claim "n columns are linearly independent" might not hold true. 2.
the following property $$rank(A)=min(m,n).$$ Please guide me what part of my Secondly, even if n columns were truly linearly independent, the fact remains that the number of rows is smaller,
reasoning is incorrect. • Did you mean to type that $n$ columns are linearly i.e., m < n. Since the rank of a matrix cannot surpass the number of rows, having n linearly independent columns
independent? (There are only $m<n$ rows.) Thi" Do not just list concepts, but wouldn't necessarily imply that rank(A) would equal n. Instead, it could indicate that n is merely an upper bound
develop each one in detail before moving to the next, as we prioritize depth of for rank(A). 3. Lastly, addressing the discrepancy concerning the dimensions of the spaces spanned by rows and
understanding and comprehensive exploration of the subject matter over breadth. columns. While it is accurate that both dimensions will be equal when considering the span of all possible
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: combinations of respective rows and columns, keep in mind that our focus here is solely on determining linearly
Write with an academic, professional and engaging tone that captivates interest. - independent subsets within them. Therefore, comparing the sizes directly isn't valid for establishing rank(A). 4.
Application: Incorporate specific, practical examples, such as proofs in calculus With respect to the property rank(A) = min(m, n): We need to reiterate that regardless of how many columns appear
or critical dates and figures in history. Do not include a title or an to be linearly independent initially, the rank of a matrix never surpasses the lesser value between the numbers of
introduction, simply write the content without headlines and introductory phrases. rows and columns. Hence, rank(A) ≤ min(m, n). However, proving why they are always equal requires delving deeper
Do not use images. into abstract linear algebra concepts beyond the scope of this explanation. Nonetheless, rest assured knowing that
the counterexample provided earlier does not invalidate this essential property.
Prime factorization is a fundamental concept in mathematics, particularly within the realm of number theory. It
refers to the process of breaking down a composite number into its most basic components - specifically, the
product of prime numbers. A prime number is any positive integer greater than 1 that has no divisors other than
itself and 1; for instance, notable prime numbers include 2, 3, 5, 7, etc. The significance of prime factorization
lies in its ability to provide us with unique insights about the nature of numbers themselves. This article delves
deeper into the intricacies of prime factorization, elucidating both the underlying principles and applications
through detailed explanations and illustrative examples. Let us begin by examining the given example: prime
factorization of 540. As stated earlier, our objective here is to express 540 as the product of prime numbers. To
achieve this goal systematically, we employ the method demonstrated below: 1. Write the number (in this case, 540)
atop a two-column table. 2. Identify the smallest possible prime factor dividing the number evenly. For 540, this
would be 2. Divide 540 by 2, yielding 270, which then replaces the initial value of 540 in the subsequent
iterations. 3. Record the prime factor identified in step 2 (i.e., 2) on the left side of the table while
preserving the quotient from step 2 (i.e., 270) on the right side for further decomposition. 4. Repeat steps 2 and
3 recursively until you obtain a quotient of 1, indicating exhaustion of all available prime factors. At this
juncture, our table should resemble the configuration presented in the provided text snippet. Upon completing these
five steps, we can encapsulate the prime factorization results in succinct form using exponential notation.
Specifically, we represent the repeated occurrence of a particular prime factor using superscripts attached to said
prime factor. Thus, for the prime factorization of 540, we arrive at the expression: 540 = 2² × 3³ × 5 This
representation conveys that the prime factor 2 appears twice, the prime factor 3 occurs thrice, and the prime
factor 5 emerges once throughout the entire prime factorization process. Moreover, note how multiplication among
primes constitutes the foundation of this compact mathematical statement—a characteristic feature inherent to prime
Write an educational piece suited for college students related to the following factorizations. At this point, several questions may arise concerning the relevance and applicability of prime
text snippet: "# Prime factorization of 540 If it's not what You are looking for factorization beyond serving merely as an abstract intellectual pursuit. Indeed, there exist numerous real-world
type in the field below your own integer, and You will get the solution. Prime scenarios where prime factorization proves indispensable, some of which include cryptography, computer science,
factorization of 540: By prime factorization of 540 we follow 5 simple steps: 1. We physics, chemistry, and engineering, to name a few. Herein, we shall focus on highlighting three salient areas
write number 540 above a 2-column table 2. We divide 540 by the smallest possible wherein prime factorization demonstrably shines: simplification of fractions, least common multiple calculation,
prime factor 3. We write down on the left side of the table the prime factor and and greatest common divisor determination. Firstly, consider the task of reducing a fraction to its lowest terms.
next number to factorize on the ride side 4. We continue to factor in this fashion By definition, a reduced fraction signifies that both the numerator and denominator share no common divisors other
(we deal with odd numbers by trying small prime factors) 5. We continue until we than 1. Consequently, performing prime factorization on either component allows us to identify potential
reach 1 on the ride side of the table 540 prime factors number to factorize 2 270 2 cancellations effortlessly. Take, for example, the fraction 630/2520: 630 = 2 × 3² × 5 × 7 2520 = 2³ × 3² × 5 × 7
135 3 45 3 15 3 5 5 1 Prime factorization of 540 = 1×2×2×3×3×3×5= $1 × 2^2 × 3^3 × As evident from their respective prime factorizations, both the numerator and denominator contain identical prime
5$ ## Related pages antiderivative of cos 2xsquare root of 3025how to factor by factors raised to equivalent powers, save for the prime factor 2 appearing once more in the denominator. Therefore,
gcfsimplify square root 18023x3derivative of sin pi xwhat is 0.125 as a after canceling out shared prime factors between numerator and denominator, we obtain the final result: (630/2520)
fractionroads.advancedacademics.com1994 roman numeralsroman numerals = (2 × 3² × 5 × 7)/(2³ × 3² × 5 × 7) = 1/2 In essence, prime factorization facilitates seamless identification of
conversionprime factorization of 144prime factor" Do not just list concepts, but common factors across large integers, thereby streamlining tedious calculations associated with manual reduction
develop each one in detail before moving to the next, as we prioritize depth of techniques. Secondly, let us examine the computation of the least common multiple (LCM). Given two nonzero integers
understanding and comprehensive exploration of the subject matter over breadth. a and b, their LCM represents the smallest positive integer divisible by both a and b without leaving any
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: remainder. One approach to determine the LCM involves factoring each integer into its constituent primes, followed
Write with an academic, professional and engaging tone that captivates interest. - by multiplying together the highest power present among all distinct prime factors encountered during this
Application: Incorporate specific, practical examples, such as proofs in calculus procedure. Applying this strategy to find the LCM of 315 and 225 yields: 315 = 3ⁱ × 5 × 7 225 = 3² × 5² where i
or critical dates and figures in history. Do not include a title or an denotes an unspecified exponent whose precise value remains undetermined momentarily. Upon closer inspection,
introduction, simply write the content without headlines and introductory phrases. observe how only the prime factors 3 and 5 appear in both expressions, whereas the prime factor 7 belongs
Do not use images. exclusively to the factorization of 315. Ergo, we compute the LCM by combining the maximum observed powers
corresponding to each unique prime factor: LCM(315, 225) = 3ʳ × 5ʳ × 7 where r > i. Since the largest exponent
associated with the prime factor 3 is found within the second integer (r = 2), whereas those pertaining to the
remaining prime factors stem from their sole appearances in their respective factorizations (r = 1 for both 5 and
7), we deduce that: LCM(315, 225) = 3² × 5 × 7 = 1050 Thirdly, turning our attention toward determining the
greatest common divisor (GCD), another application area ripe for exploiting prime factorization arises. Recall that
the GCD signifies the largest positive integer sharing divisibility by both input values. Similar to the preceding
discussion centered around computing the LCM, obtaining the GCD necessitates identifying the set of commonly
occurring prime factors alongside their minimum powers featured across the original pair of integers. Continuing
with our running theme, suppose we wish to ascertain the GCD of 315 and 225. Initiated by extracting their prime
factorizations: 315 = 3¹ × 5¹ × 7¹ 225 = 3² × 5² Now, contrasting these representations reveals that the prime
factors 3 and 5 emerge as mutual participants within their respective decompositions, albeit manifested
differently. More precisely, the prime factor 3 materializes twice in the factorization of 225 but only once in
that of 315, whereas the prime factor 5 surfaces squared in the former relative to being linearly represented in
the latter. Accordingly, we select the smaller exponent associated with each recurring prime factor to establish
the basis upon which the ensuing GCD hinges: GCD(315, 225) = 3¹ × 5¹ = 15 In summary, mastery over prime
factorization unlocks powerful mathematical tools capable of addressing diverse problems spanning various
disciplines, including algebra, geometry, trigonometry, combinatorics, probability, and others besides.
Familiarizing oneself with this essential skillset equips learners with robust computational strategies applicable
across myriad contexts, fostering enhanced comprehension and retention along the way. Additionally, proficiency
acquired in deciphering complex structures like those exemplified herein cultivates valuable analytical abilities
transferable well beyond academia proper.
Scatterplots are graphs used to visualize the relationship between two numerical variables. Each dot on the plot
corresponds to an observation, where the position along the horizontal axis (x) represents the value of one
variable, and the position along the vertical axis (y) corresponds to the value of another variable for that same
Write an educational piece suited for college students related to the following observation. Interpreting scatterplots allows us to identify trends, patterns, and relationships between these
text snippet: "# How can I interpret this scatterplot? Please help me to interpret variables. One important aspect to consider when interpreting scatterplots is the direction of the association
this graph in terms of correlation type. Which type involves two vectors in between variables, which can be positive, negative, or nonexistent. Positive Correlation: When the value of one
directions observed below (see graph screenshot)? Thanks in advance. • Check the variable increases as the other variable also increases, it suggests a positive correlation. This means that there
updated graph. For now I'm saying that y1 and y2 content is directly correlated to is a direct relationship between the two variables; if you were to draw a line summarizing the pattern in the data
the lower content of X. when X is lower, the y1 content decrease while the y2 points, it would have a general upward slope from left to right. An example of a real-life situation exhibiting a
increase. – Apopei Andrei Ionut Feb 21 '14 at 13:51 • What are y1 and y2 then? Are positive correlation could be the relationship between the number of hours studied by a student per week and their
they separate variables? And what are the dots? Do their coordinates represent final exam grade. As the number of study hours increases, so does the probability of achieving a higher grade,
observations of two different variables or three? If three, how can one tell the leading to a direct relationship. Negative Correlation: Conversely, if the value of one variable decreases as the
difference? And why do the vectors pass outside the observations rather than other variable increases, it implies a negative correlation. A line drawn to summarize the pattern in the data
through them? – Nick Stauner Feb 21 '14 at 13:56 • y1/(y1+y2) (which is equivalent points would have a downward slope from left to right. An example of negative correlation could be found in the
to that representation y1 and y2) - are to variables on the y-axis, an the third on relationship between the price of a product and the quantity demanded. Typically, as the price goes up, demand
x-axis. Dots represents the projected values of y1/(y1+y2) vs x. The vectors tends to go down, indicating a negative relationship. No Correlation: Sometimes, no clear pattern emerges in the
represent the trend. – Apopei Andrei Ionut Feb 2" Do not just list concepts, but distribution of data points, suggesting little or no correlation between the variables. Here, changes in one
develop each one in detail before moving to the next, as we prioritize depth of variable don't necessarily result in predictable changes in the other variable. In some cases, even though there
understanding and comprehensive exploration of the subject matter over breadth. might appear to be no correlation initially, further investigation may reveal hidden correlations after accounting
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: for confounding factors or applying advanced statistical techniques. In your provided example, Apopei Andrei Ionut
Write with an academic, professional and engaging tone that captivates interest. - describes a scenario involving two variables plotted on the y-axis—y1 and y2—and one variable on the x-axis—X.
Application: Incorporate specific, practical examples, such as proofs in calculus Specifically, he mentioned y1/(y1 + y2), representing the ratio of y1 to the sum of y1 and y2. According to his
or critical dates and figures in history. Do not include a title or an description, as the X variable decreases, the value of y1 decreases proportionately, whereas the value of y2
introduction, simply write the content without headlines and introductory phrases. increases. Hence, a negative correlation exists between the X variable and y1/(y1 + y2). Additionally, since both
Do not use images. y1 and y2 change based on the X variable, indirect correlations will exist between X and y1, as well as between X
and y2, albeit with less obvious patterns compared to the combined y1/(y1 + y2) ratio. It should be noted that the
"vectors" referred to in the conversation actually reflect trends in the data rather than physical lines passing
through individual data points. These trends provide insights into overall tendencies within the dataset, offering
valuable information about the nature of the relationships among the involved variables.
Write an educational piece suited for college students related to the following
text snippet: "# Integrate $\int x\sec^2(x)\tan(x)\,dx$ $$\int x\sec^2(x)\tan(x)
\,dx$$ I just want to know what trigonometric function I need to use. I'm trying to To solve the given integral using integration by parts, it is important to first identify the functions $u$ and
integrate by parts. My book says that the integral equals $${x\over2\cos^2(x)}-{\ $dv.$ The function $u$ is chosen to be the variable that is multiplied by the differential term, which in this case
sin(x)\over2\cos(x)}+C$$ • I'm not quite sure what you're asking for when you ask is $x,$ while $dv$ is the remaining expression, $\sec^2(x)\tan(x) dx.$ The next step is to find the antiderivative
about "what trigonometric function I need to use". Could you clarify what you mean of $dv,$ denoted as $v.$ This can be done by making a substitution. Here, let us consider the u-substitution $\
there? – Omnomnomnom Oct 25 '18 at 17:31 $$I=\int x\sec^2(x)\tan(x)dx$$ I will be omega = \tan(x),$ so $d\omega = \sec^2(x) dx.$ Thus, $v = \int \sec^2(x) \tan(x) dx = \int \omega d\omega = \frac
walking you through this integral step-by-step. First, we integrate by parts: $$u=x {{\omega ^2}}{2} + C_1 = \frac{{\tan ^2(x)}}{2} + C_1.$ Now that both $u$ and $v$ are known, apply integration by
\\du=dx$$ And $$dv=\sec^2(x)\tan(x)dx$$ Which is an integral I'd like to parts formula: $I = uv - \int {vdu}.$ Substitute all values found earlier: \[I = x\left( {\frac{{\tan ^2(x)}}{2}} \
demonstrate $$v=\int dv\\v=\int\sec^2(x)\tan(x)dx$$ For this, we will use a $$\ right) - \int {\left( {\frac{{\tan ^2(x)}}{2}} \right)dx} \] To evaluate the remaining integral, express $\tan^2(x)
omega$$-substitution. Let $$\omega=\tan(x)$$. Therefore, $$d\omega=\sec^2(x)dx$$, $ as another trigonometric function. Recall that $\tan^2(x)=(\sec(x))^2-1.$ Hence, \[\begin{array}{l} I = x\left(
giving $$v=\int\omega d\omega=\frac{\omega^2}{2}\\v=\frac{\tan^2(x)}{2}$$ Next we {\frac{{\tan ^2(x)}}{2}} \right) - \int {\left( {\frac{{(\sec (x))^2 - 1}}{2}} \right)dx} \\ I = x\left( {\frac{{\
finish our integration by parts: $$I=uv-\int vdu\\I=\frac{x\tan^2(x)}{2}-\int\frac tan ^2(x)}}{2}} \right) - \frac{1}{2}\int {(\sec (x))^2 dx} + \frac{1}{2}\int {dx} \\ \end{array}\] For the
{\tan^2(x)}{2}dx$$ Then we define $$A=" Do not just list concepts, but develop each remaining integrals, recall that $\int (\sec(x))^2 dx =\tan(x)+\csc(x)+C_2$ and $\int dx = x + C_3.$ Putting
one in detail before moving to the next, as we prioritize depth of understanding everything together yields: \[\begin{align*} I &= x\left( {\frac{{\tan ^2(x)}}{2}} \right) - \frac{1}{2}(\tan (x) +
and comprehensive exploration of the subject matter over breadth. Focus on: - \csc (x)) + \frac{1}{2}(x) + C \\ I &= \boxed{\frac{x}{2}\tan ^2(x) - \frac{1}{2}\tan (x) - \frac{1}{2}\csc (x) + \
Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Write with frac{x}{2}+C} \\ \end{align*}\] This answer may seem different from what was provided initially; however, they are
an academic, professional and engaging tone that captivates interest. - equivalent expressions due to the identity $\cot^2(x)=\csc^2(x)-1.$ By expanding the initial answer, one can verify
Application: Incorporate specific, practical examples, such as proofs in calculus their equivalence. It illustrates how multiple correct answers could arise depending on how algebraic manipulations
or critical dates and figures in history. Do not include a title or an and identities were used during calculations.
introduction, simply write the content without headlines and introductory phrases.
Do not use images.
Working with Large Datasets in Mathematica ------------------------------------------ When working with large
datasets in Mathematica, you may encounter problems with file sizes that make it difficult to open, save, or
manipulate the data. This issue often arises when dealing with long time series data. To address this problem,
there are several built-in functions and techniques in Mathematica that can help manage large datasets efficiently.
### Data Compression Techniques One way to reduce file size is by compressing the data using the `Compress`
function. This function converts an expression into a compressed string, which can then be saved to a file or
stored in memory. Here's how to use the `Compress` function: ```scss data = RandomReal[{0, 1}, {10^6}]; (* generate
Write an educational piece suited for college students related to the following some random data *) compressedData = Compress@data; (* compress the data *) ``` To restore the original data, use
text snippet: "All Questions 88 views Problem with files size I need to work with the `Uncompress` function: ```go uncompressedData = Uncompress@compressedData; (* uncompress the data *) ```
long time series, but I notice that this generates huge files that Mathematica Another technique to consider is downsampling your dataset if retaining every single datapoint is not necessary.
doesn't handle well at all: even opening/saving one of this files is difficult, and The `DownSample` function reduces the number of elements in a list while preserving the overall trend of the data:
the kernel quits ... 95 views Excluding an interval of parameters from Integrate ```css downSampledData = DownSample[data, 10]; (* keep only 1 out of every 10 datapoints *) ``` ### Handling Time
[duplicate] I'd like to compute an integral that depends on a parameter, but I'd Series Data For time series data specifically, Mathematica provides the `TimeSeries` object, which offers efficient
like to exclude an interval from the computation. For concreteness, say I want to storage and manipulation. You can create a `TimeSeries` object directly from your data using the `TimeSeries`
compute ... 6k views Finding the intersection of a curve with an interpolation constructor function: ```r timeSeriesData = TimeSeries@Transpose@{dateList, valueList}; ``` Once you have created a
function There are no issues trying to find the intersection points of two defined `TimeSeries` object, you can perform various operations on it, such as resampling, shifting, and aggregating
curves. ... 169 views Symbolic solution of system everyone! I am sorry, but I am an values. Additionally, plotting time series data becomes much easier using dedicated visualization tools provided by
abcolute novice of Mathematica (to be more precise this is my first day of using Mathematica: ```python DateListPlot[timeSeriesData, PlotRange -> All] ``` ### Excluding Intervals from Computations
it) and even after surfing the web and all documents I am not able to solve the ... If you need to compute integrals excluding certain intervals, you can take advantage of Mathematica's symbolic
136 views Use Simplify option Trig->False when 3rd argument to Collect is Simplify manipulation capabilities. Instead of computing the entire integral and subtracting the unwanted part, you can
I would like to simpl" Do not just list concepts, but develop each one in detail split the integration range into smaller segments and sum up their contributions separately. Consider the example
before moving to the next, as we prioritize depth of understanding and below: ```matlab f[x_] := x^2; integralValue = Integrate[f[x], {x, -1, 1}] - Integrate[f[x], {x, -a, a}]; ``` Here,
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: we calculate the definite integral of $f(x)=x^{2}$ over the interval [-1, 1] minus the contribution from the
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an interval [-a, a]: $$ \int_{-1}^{1}x^{2}\mathrm dx-\int_{-a}^{a}x^{2}\mathrm dx={\frac{2}{3}}-{\frac{2a^{3}}{3}} $$
academic, professional and engaging tone that captivates interest. - Application: This approach ensures accurate results even when excluding complex intervals. ### Finding Intersection Points To
Incorporate specific, practical examples, such as proofs in calculus or critical find the intersection points between two curves or an interpolation function, you can use the `Solve` function
dates and figures in history. Do not include a title or an introduction, simply combined with logical conditions defining the intersection. Suppose we define two functions, $f$ and $g$, and wish
write the content without headlines and introductory phrases. Do not use images. to find where they intersect: ```scss f[x_] := Sin[x]; g[x_] := Cos[x]; intersectionPoints = x /. Solve[{f[x] == g
[x], 0 <= x <= 2 Pi}, x]; ``` In cases where analytical solutions do not exist, you can resort to numerical methods
such as `FindRoot`. If you are working with interpolation functions instead, use the `InterpolatingFunction`
objects directly inside `Solve` or `FindRoot`. ### Symbolic Solution of Systems Working with systems of equations
requires careful attention to input syntax and usage of appropriate functions depending on whether the solution
needs to be expressed numerically or symbolically. When solving a simple system symbolically, the `Solve` function
will suffice: ```makefile eqns = {x + y == 4, x - y == 2}; sol = Solve[eqns, {x, y}][[1]] ``` However, if your
system involves trigonometric expressions and contains many variables, obtaining a closed-form solution might prove
challenging due
Completing the square is a technique used to convert a quadratic equation from its general form (ax^2 + bx + c = 0)
Write an educational piece suited for college students related to the following into a perfect square trinomial. The process involves adding and subtracting the same value within the equation to
text snippet: "# Lesson 5Be There or Be SquarePractice Understanding ## Jump Start create a complete square, which can make solving and graphing the equation easier. Let's explore how this concept
Graph each function with at least two precise points on either side of the line of relates to graphing parabolas. To begin, let's consider the standard form of a quadratic equation: y = ax^2 + bx +
symmetry. ## Learning Focus Use completing the square to change the form of a c. A parabola is a curve shaped like a "U," and when graphed, will open upward if a > 0, and downward if a < 0. To
quadratic equation. Graph quadratic equations given in standard form. How might graph a parabola accurately, you need to find the vertex, axis of symmetry, and any x-intercept(s). Completing the
completing the square help us to graph parabolas? ## Open Up the Math: Launch, square is particularly useful when dealing with the coefficient 'a' being anything other than 1, making it
Explore, Discuss Optima’s niece, Jenny, works in the shop, taking orders and difficult to identify the vertex using traditional methods. By completing the square, you ensure accurate placement
drawing quilt diagrams. When the shop isn’t too busy, Jenny pulls out her math along both axes. Here are some steps to guide you through the process: Step 1: Divide every term by 'a'. This will
homework and works on it. One day, she is working on graphing parabolas and notices give you an equivalent quadratic equation while simplifying calculations. Remember, dividing each term also means
that the equations she is working with look a lot like an order for a quilt block. adjusting the constant term accordingly. Step 2: Move the number multiplied by 'x' to the left side of the equal
For instance, Jenny is supposed to graph the equation: . She thinks, “That’s funny. sign and set it equal to zero. This creates a quadratic expression devoid of coefficients (easier manipulation).
This would be an order where the length of the standard square is reduced by and Step 3: Determine what must be added and subtracted inside the parentheses so that a perfect square results after
then we add a little piece of fabric that has as area of . We don’t usually get completion. You may calculate this by halving the linear coefficient, squaring the result, and placing it
orders like that, but it still makes" Do not just list concepts, but develop each appropriately within the expression. Step 4: Factor the new quadratic trinomial back into a binomial square form.
one in detail before moving to the next, as we prioritize depth of understanding Don't forget to account for the terms added and subtracted earlier! These represent a horizontal shift necessary
and comprehensive exploration of the subject matter over breadth. Focus on: - for correct positioning on your coordinate plane. Now, relating this method back to our initial example about
Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Write with Optima's niece, Jenny, notice how the equation resembles an order for a quilt block. Just like constructing a quilt
an academic, professional and engaging tone that captivates interest. - pattern requires precision and accuracy, mastering algebraic techniques enables better comprehension and
Application: Incorporate specific, practical examples, such as proofs in calculus application of mathematical principles across various contexts. As Jenny discovered, even seemingly disparate
or critical dates and figures in history. Do not include a title or an domains - like mathematics and sewing – intersect unexpectedly sometimes! Applying these skills to real-world
introduction, simply write the content without headlines and introductory phrases. problems strengthens your ability to reason mathematically and critically analyze situations beyond the classroom
Do not use images. walls. So remember, whether you're designing quilts or analyzing geometric shapes, always strive to understand
deeply rather than merely memorizing procedures. Happy learning!
We begin our discussion with the concept of a multiset, also known as a bag. A multiset is a generalization of a
set, where members are allowed to appear more than once. For example, the multiset {2, 2, 3} has three elements,
Write an educational piece suited for college students related to the following but only two distinct values. When considering operations on multisets, it's important to account for these
text snippet: "# An averaging game on finite multisets of integers The following multiplicities. Next, let's examine the given procedure for transforming a finite multiset M of integers through a
procedure is a variant of one suggested by Patrek Ragnarsson (age 10). Let $$M$$ be series of moves. At each step, we select two elements of equal parity from M and replace them with their average.
a finite multiset of integers. A move consists of choosing two elements $$a\neq b$$ This replacement reduces the sum of squared elements within the multiset, ensuring eventual termination since the
of $$M$$ of the same parity and replacing them with the pair $$\frac 12(a+b)$$, $$\ sum of squares cannot decrease indefinitely. Note that selecting elements of equal parity guarantees that the
frac 12(a+b)$$. If we continue to make moves whenever possible, the procedure must resulting average remains an integer, preserving the integrity of the multiset throughout the process. Now, suppose
eventually terminate since the sum of the squares of the elements will decrease at M = {1, 2, ... , n}. How many moves does it take for this procedure to reach completion? To answer this question,
each move. What is the least and the most number of moves to termination, in let k denote the integer closest to the mean $\lfloor{\frac{1}{n}\sum_{i=1}^{n} a_i \rfloor}$. It turns out that
particular, if $$M=\{1,2,\dots, n\}$$? If $$M=\{a_1,\dots,a_n\}$$, then an upper the greatest number of moves occurs when the initial multiset deviates significantly from its mean, necessitating
bound on on the maximum number of moves is $$\frac 12\sum (a_i-k)^2$$, where $$k$$ larger jumps toward equilibrium. Specifically, the upper bound on the maximum number of moves equals half the
is the integer which minimizes this sum. (In fact, $$k$$ is the nearest integer to summation $\sum_{i=1}^n (a_i - k)^2.$ As proven by Knuth, using the median instead of the mean would yield an even
$$\frac 1n(a_1+\cdots+a_n)$$.) We can turn this procedure into a game by having tighter bound; however, finding the median requires additional computation. Let us now introduce a competitive
Alice and Bob move alternately, with Alice moving first. The last player to move aspect to this problem: imagine Alice and Bob taking turns applying the transformation rule above, aiming to
wins. (We could also consider " Do not just list concepts, but develop each one in maximize their respective chances of making the final move—the classic last-player-wins scenario. Since both
detail before moving to the next, as we prioritize depth of understanding and players follow optimal strategies during play, the outcome depends solely on the starting configuration.
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: Intuitively, configurations farther away from equilibrium favor earlier resolution due to larger available
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an decreases in square sums. Conversely, nearly equilibrated systems may drag on longer, creating suspenseful
academic, professional and engaging tone that captivates interest. - Application: endgames. Finally, extending this game beyond simple sequential turn structures introduces further strategic
Incorporate specific, practical examples, such as proofs in calculus or critical layers. Imagine simulating multiple simultaneous games between pairs of cooperative learners seeking collaborative
dates and figures in history. Do not include a title or an introduction, simply advantages. Alternatively, visualize parallelized tournaments among noncooperative AI agents competing against
write the content without headlines and introductory phrases. Do not use images. various initialization patterns, striving towards mastery across diverse training scenarios. By exploring these
avenues, researchers gain valuable insights regarding fundamental properties underlying adaptive learning processes
while fostering innovation in computational intelligence applications.
The Graduate Management Admission Test (GMAT) is a standardized exam used by business schools to assess applicants'
qualifications for graduate-level management programs. If you're planning to apply to business school, it's likely
that you'll need to take the GMAT as part of the application process. Here, we'll discuss the benefits of signing
up for the "GMAT Club Daily Prep," which provides challenging GMAT questions directly to your mailbox every day. By
incorporating these problems into your study routine, you can improve your skills and better prepare for test day.
First, let's examine what the GMAT entails. This computer adaptive test consists of four sections: Analytical
Write an educational piece suited for college students related to the following Writing Assessment, Integrated Reasoning, Quantitative, and Verbal. Each section measures different aspects of your
text snippet: "GMAT Question of the Day - Daily to your Mailbox; hard ones only It analytical, writing, quantitative, verbal, and reading skills in English. While the AWA and Integrated Reasoning
is currently 21 Jan 2019, 08:11 ### GMAT Club Daily Prep #### Thank you for using sections contribute less significantly to your overall score compared to the Quantitative and Verbal sections, they
the timer - this advanced tool can estimate your performance and suggest more still play a role in demonstrating your well-rounded skillset to admissions committees. As a result, it's essential
practice questions. We have subscribed you to Daily Prep Questions via email. to allocate time to preparing for all areas of the exam. The GMAT Club Daily Prep offers a valuable resource for
Customized for You we will pick new questions that match your level based on your strengthening your skills across various disciplines covered on the GMAT. These daily emails contain difficult
Timer History Track every week, we’ll send you an estimated GMAT score based on questions designed to challenge even the most prepared test takers. Practicing with these tough problems helps
your performance Practice Pays we will pick new questions that match your level build mental stamina needed during the actual exam, while also refining problem-solving abilities necessary for
based on your Timer History ## Events & Promotions ###### Events & Promotions in success in business school. Furthermore, working through these high-quality questions allows you to become familiar
January PrevNext SuMoTuWeThFrSa 303112345 6789101112 13141516171819 20212223242526 with the format and style of those found on the official GMAT, thereby reducing anxiety surrounding unfamiliar
272829303112 Open Detailed Calendar • ### GMAT Club Tests are Free & Open for question types. In addition to improving your raw knowledge and ability to answer complex questions, consistent
Martin Luther King Jr.'s Birthday! January 21, 2019 January 21, 2019 10:00 PM PST participation in the GMAT Club Daily Prep program fosters discipline and accountability throughout your preparation
11:00 PM PST Mark your calendars - All GMAT Club Tests are free and open January journey. Establishing a daily routine ensures adequate exposure to diverse topics and question formats, ultimately
21st for celebrate Martin Luth" Do not just list concepts, but develop each one in leading to increased confidence when faced with unknown challenges during the real examination. Moreover, solving
detail before moving to the next, as we prioritize depth of understanding and challenging problems regularly exposes gaps in understanding, enabling targeted review and mastery of weakened
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: areas. By participating in the GMAT Club Daily Prep, you will join a community of dedicated individuals committed
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an to their personal growth and development within the realm of business education. Interacting with fellow
academic, professional and engaging tone that captivates interest. - Application: participants enables learning from others' experiences, sharing strategies, seeking advice, and obtaining
Incorporate specific, practical examples, such as proofs in calculus or critical encouragement along the way. Such support networks prove crucial during stressful times like studying for the GMAT.
dates and figures in history. Do not include a title or an introduction, simply Lastly, utilizing resources provided by reputable organizations like GMAT Club enhances credibility with admissions
write the content without headlines and introductory phrases. Do not use images. committees, signaling dedication and motivation towards pursuing higher education goals. Demonstrating commitment
to rigorous prep work showcases readiness for the demanding nature of graduate-level coursework, further setting
successful candidates apart from competitors. In conclusion, enrolling in the GMAT Club Daily Prep has numerous
advantages beyond merely receiving daily practice questions. Improved competency, mental endurance, discipline,
self-awareness, connections, and enhanced candidacy represent just some of the many positive outcomes derived from
engaging in this high-quality preparatory opportunity. Don't miss out on maximizing your potential – consider
integrating the GMAT Club Daily Prep into your study regimen today.
The exponentiation operator, denoted by "^" in Notion formulas, is used to raise a number to a specified exponent.
This operation is also referred to as raising a base to a power. Mathematically, the expression a^b represents the
Write an educational piece suited for college students related to the following base 'a' raised to the power 'b'. For example, in the expression 2^3, 2 serves as the base while 3 acts as the
text snippet: "# pow Learn how to use the exponentiation (^) operator in Notion exponent, resulting in an output of 8. Notably, the exponentiation operator holds a high level of precedence among
formulas. The power (^) operator (also known as the exponentiation operator) allows other arithmetic operations. Specifically, within the hierarchy of operator precedences, the exponentiation
you to raise a number to a higher power. 1 number[base] ^ number[exponent] 2 pow operator ranks above both the unary plus (+) and minus (-) operators. As a result, expressions enclosed in
(number[base], number[exponent]) In technical terms, it raises the first operand to parentheses will always be computed prior to any application of the exponentiation operator. Moreover, when
the power of the second operand. In Notion, the ^ operator has a higher operator multiple instances of the exponentiation operator are nested together, they exhibit right-associative behavior. To
precedence than the unaryPlus and unaryMinus operators, and has right-to-left clarify, given the expression a^b^c, it would first compute a^b and then subsequently apply the outcome as the new
associativity. You can also use the function version, pow(). ## Example Formulas 1 base to c. Therefore, rewriting the initial equation yields a^(b^c). Beyond direct notation via the caret symbol ("
3 ^ 4 // Output: 81 2 3 pow(4,3) // Output: 64 4 5 2 ^ 2 ^ 3 // Output: 256 - ^"), Notion users may opt for functional syntax using the built-in `pow()` function. Its usage mirrors that of the
evaluates as 2 ^ (2 ^ 3) • $x^0 = 1$ • $(x^a)^b = x^{ab}$ • $x^{a^b} = x^{(a^b)}$ explicit operator format; namely, invoking `pow(base, expoent)` equates to writing out the equivalent infix
(not all languages respect this, but Notion does) • $x^a * y^a = (xy)^a$ • $x^a / y representation a^b directly. To reinforce these abstract definitions, consider several illustrative examples below:
^a = (x/y)^a$ • $x^a * x^b = x^{a+b}$ • $x^a / x^b = x^{a-b}$ ### Exponent * **Example 1:** Calculate 3^4 using both notation styles. + Directly: `3^4` results in 81. + Functional: `pow(3,
Associativity In Notion, the ^ operator has right-to-left associativity, which 4)` likewise returns 81. * **Example 2:** Evaluate 2^2^3 utilizing right-associative properties. + Here, we
means that x ^ y ^ z is evaluated as x ^ (y" Do not just list concepts, but develop calculate 2^(2^3), leading to 256 instead of ((2^2)^3) = 64. Lastly, familiarize yourself with various algebraic
each one in detail before moving to the next, as we prioritize depth of identities involving exponents that hold true in Notion due to its consistent treatment of them. Such relationships
understanding and comprehensive exploration of the subject matter over breadth. facilitate simplification processes during complex manipulations, thereby enhancing readability and comprehension.
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: A few key identities follow suit: * $x^0 = 1$: Any nonzero value raised to the zeroeth power equals unity. * $(x^a)
Write with an academic, professional and engaging tone that captivates interest. - ^b = x^{ab}$: When a base is elevated to an exponent, then squared again, the overall effect remains identical to
Application: Incorporate specific, practical examples, such as proofs in calculus having multiplied those two individual powers together. * $x^{a^b} = x^{(a^b)}$: Unlike previous rules where
or critical dates and figures in history. Do not include a title or an exponents were mere scalars, here they become functions themselves, necessitating additional grouping symbols for
introduction, simply write the content without headlines and introductory phrases. disambiguation purposes. * $x^ay^a = (xy)^a$: Products inside common bases sharing equal indices combine into
Do not use images. single entities underlined by their shared index. * $x^a/y^a = (x/y)^a$: Likewise, quotients behave similarly
provided they share identical indices. * $x^ax^b = x^{a+b}$ & $x^a/x^b = x^{a-b}$: Summations and differences
emerge from combining like-terms harboring matching bases yet varying indices.
Write an educational piece suited for college students related to the following
text snippet: "# Hexagonal sum filling Q1: Fill this hexagon with numbers 1-19 (no
repetition) such that the sum of every vertical and diagonal row is the same. (This
should be easy) Q2: Assume that one of these pre-solved cells was not given. If you Let us delve into the concept of magic squares, which is closely related to the problems posed in the initial text
are still able to solve the problem with a unique solution, do you know which cell snippet. A magic square is a grid filled with distinct integers where the sum of elements in each row, column, and
it is? Bonus: How many solved cells can you delete and still be able to solve it? • main diagonal is identical. This value is referred to as the "magic constant." Magic squares have been studied
Q2 seems to contradict the Bonus or maybe I don't understand Q2? Q2 suggests to me since antiquity due to their mathematical properties and applications in various fields like divination, art, and
that there is one (unique) cell that can be removed and it still is solvable while recreation. We will focus on the 3×3 magic square, also known as the Lo Shu Square, because it shares similarities
the bonus says there are multiple – Ivo Beckers Sep 30 '15 at 12:23 Solution to Q1: with the provided hexagon puzzle. Firstly, consider the order of a magic square, denoted by n, representing its
3 19 17 16 7 18 2 1 12 5 11 4 6 10 8 9 13 14 15 Solution to Q2: The 1 is number of rows and columns. For our purposes, n=3. Next, calculate the magic constant using the formula M = [n*(n^2
unnecesary. The 19, 12 and 7 cells come as usual. Then, between the 10 and 18 cells + 1)]/2. Substituting n=3 yields M = [3*(9+1)]/2 = 15. Therefore, any valid 3×3 magic square has a magic constant
we have three numbers which must sum 10, and the only possible choice (since 2 and equal to 15. Now let's examine how to construct a 3×3 magic square. Begin by placing the central integer, 5, in the
3 are taken) is 1, 4 and 5. Below the 2 there must be a" Do not just list concepts, middle cell. Place the smallest remaining integer, 1, in the upper right corner, and then fill out the rest of the
but develop each one in detail before moving to the next, as we prioritize depth of first row with the largest available integers, namely 2 and 3. Afterward, proceed clockwise through the other
understanding and comprehensive exploration of the subject matter over breadth. corners, inserting the missing primary diagonals: 9 in the lower left corner and 7 in the top left corner. Lastly,
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: complete the remaining empty spaces with the few leftover values (4 and 6). You now have constructed a proper 3×3
Write with an academic, professional and engaging tone that captivates interest. - magic square: ``` 4 9 2
Application: Incorporate specific, practical examples, such as proofs in calculus
or critical dates and figures in history. Do not include a title or an
introduction, simply write the content without headlines and introductory phrases.
Do not use images.
Plot Colorization by Values is a feature in Maple, a powerful computational software, used to assign different
colors to various data ranges within plots generated through its numerous visualizing tools. By employing this
option, users can intuitively distinguish between distinct value groups in their datasets, thereby enhancing
comprehension of intricate patterns and trends concealed within complex graphs. The implementation process entails
setting up appropriate parameters within the calling sequence of relevant plot commands. This section aims to
provide a detailed walkthrough of the necessary components involved in implementing the 'valuesplit' scheme when
utilizing the 'colorscheme' option in your Maple worksheets. Firstly, it is essential to recognize that there are
three primary parameters required in conjunction with the 'colorscheme' option while invoking various plot
commands: `plotargs`, `V`, and `mapping`. Each parameter plays a pivotal role in dictating how Maple will allocate
colors based on specified criteria. Let us delve deeper into these individual entities. `plotargs` represents the
set of input arguments passed along with any supported plotting command present in Maple, including but not limited
to `plot`, `animate`, `anidashtext`, `display`, `coordplot`, `contourplot`, and many others encompassed under
graphical functionalities offered by Maple. These argument sets typically comprise details pertaining to geometric
objects being rendered, ranging from simple point coordinates to more elaborate curve definitions. For instance,
consider the basic usage of the `plot` function below: ```makefile plot(sin(x), x = 0 .. 2*Pi); ``` Here, `[sin(x),
x = 0 .. 2*Pi]` constitutes the `plotargs` consisting of both explicit expressions defining the mathematical object
and domain constraints governing the independent variable. When applying the 'colorscheme' option, this very same
Write an educational piece suited for college students related to the following `plotargs` structure should be maintained, ensuring seamless integration with additional customizations like our
text snippet: "Plot Coloring by Values - Maple Help Home : Support : Online Help : desired 'valuesplit' treatment. The second crucial component, denoted as `V`, stands for a collection of reference
Graphics : 2-D : Options : colorscheme : Plot Coloring by Values Plot Coloring by values against which subsequent comparisons are made during the color assignment procedure. It may take either form
Values Calling Sequence plotcommand(plotargs, colorscheme=["valuesplit", V, of a real table (rtable) or a plain list containing eligible numerical entries corresponding to respective elements
mapping, opts]) Parameters plotargs - arguments to a plotting command V - found within `plotargs`. To clarify, let us revisit our previous example involving sine wave generation; however,
(optional) rtable or list of values mapping - (optional) list specifying mapping of now incorporating a hypothetical five-element `V` vector alongside the existing `plotargs` definition: ```scss V :=
values to colors opts - (optional) optional arguments as described in the following [-1, 0, 1/2, Pi, 3/4]; plot(sin(x), x = 0 .. 2*Pi, colorscheme=["valuesplit", V]); ``` This modification allows
Options section. Summary • Use the colorscheme option to color a plot by values. > Maple to categorize and render sinusoidal fluctuations according to predetermined thresholds outlined in `V`.
dataplot(LinearAlgebra:-RandomVector(50), style=point, colorscheme=["valuesplit", Moreover, note that the cardinality of `V` does not necessitate alignment with the number of discrete instances
[-infinity..-51="Blue", -50..50="DarkGreen", 51..infinity="DarkRed"]]); contained within `plotargs`. Instead, the length of `V` influences the granularity of applied hue transitions
Introduction • This page describes how to color a plot by values using the throughout the entire plot range. A larger quantity of unique references generally yields smoother chromatic
"valuesplit" scheme with the colorscheme option, which is available for most gradients across varying data intervals. Lastly, the `mapping` attribute assumes responsibility for associating
plotting commands. Specifically, the points in a plot element are partitioned and established relationships among provided reference values and designated color palettes. More explicitly, it
colors are assigned to each group of points bas" Do not just list concepts, but embodies a list whereby pairs of sublists serve as anchor points guiding overall color distribution mechanisms.
develop each one in detail before moving to the next, as we prioritize depth of With respect to the earlier contrived scenario, assume the existence of a four-color gradient array spanning red,
understanding and comprehensive exploration of the subject matter over breadth. blue, green, and yellow tones. We could then express such a setup using the notation illustrated beneath: ```yaml
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: colormap := ["red","blue","green","yellow"]; mapping := [[-1,"red"],[-0.5,"blue"],[0,"green"],[1,"yellow"]]; plot
Write with an academic, professional and engaging tone that captivates interest. - (sin(x), x = 0 .. 2*Pi, colorscheme=["valuesplit", V, colormap, mapping]); ``` In essence, the `mapping` construct
Application: Incorporate specific, practical examples, such as proofs in calculus facilitates fine-grained control over precise locations at which particular shades commence and terminate along the
or critical dates and figures in history. Do not include a title or an spectrum, permitting extensive flexibility tailored towards diverse user preferences. Furthermore, supplementary
introduction, simply write the content without headlines and introductory phrases. options encapsulated within the `opts` parameter allow advanced personalization regarding appearance properties
Do not use images. concerning colored segments generated via the 'valuesplit' mechanism. Relevant attributes encompass stroke width
adjustments (`thickness`), transparency levels (`opacity`), and shape selections (`style`) attributed to marked
divisions among discretized regions. Herein lies another illustrative example elucidating these possibilities:
```scss plot(sin(x), x = 0 .. 2*Pi, colorscheme=["valuesplit", V, colormap, mapping], thickness=[seq([i], i=
1..length(V))]); ``` With this final addition, we instruct Maple to dynamically vary linewidth proportional to
index positions held by constituent members forming part of the original `V` vector. Consequently, resulting output
exhibits thicker traces near vicinity of local extrema, providing enhanced visibility amidst otherwise cluttered
backdrops obscured by densely packed information clusters. To recapitulate, mastering the art of crafting visually
appealing and insightful graphics demands meticulous attention to underlying design principles characterizing
effective utilization of the 'valuesplit' coloration strategy integrated within Maple's expansive suite of
graphical capabilities. Through careful consideration of `plotargs`, judicious selection of referential milestones
captured within `V`, strategic allocation of thematically aligned hues mapped out via `colormap`, and exploitation
of ancillary settings bundled inside `opts`, researchers stand poised to unravel profound implications lurking
behind seemingly mundane depictions rooted in raw quantitative measurements. Ultimately, realizing the
transformative power inherently embedded within visual storytelling techniques empowers learners to transcend mere
consumption of static representations and instead foster curiosity-driven explorations capable of yielding
unprecedented discoveries hidden deep within layers of abstract symbolisms traditionally perceived as arcane
Continued fractions offer a fascinating alternative perspective on numbers, revealing intriguing connections
Write an educational piece suited for college students related to the following between seemingly disparate mathematical realms. At first glance, they may appear arcane, even intimidating;
text snippet: "Continued Fractions 1Background For years, you’ve used $$22/7$$ as however, upon closer examination, their elegance and utility become apparent. This discussion will delve into the
an approximation for $$\pi$$. But where does this number come from? It’s from the fundamentals of continued fractions, elucidating their significance within the context of numerical representations
continued fraction for $$\pi$$. It’s called a convergent, and there are elegant and approximations. A continued fraction is an expression of the form: a\_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac
results connecting convergents to the best rational approximations of numbers. {1}{\ddots}}} where a\_0, a\_1, a\_2, ... are integers. The sequence {a\_n} denotes the partial quotients of the
Continued fractions provide an alternate representation of numbers. Indeed, by continued fraction. For brevity, we often employ the notation [a\_0; a\_1, a\_2, ...]. When all the partial
unrolling more and more terms of the continued fraction, we can obtain better and quotients are zero except finitely many, we call it a finite continued fraction; otherwise, it's an infinite
better approximations of the number. Note that if the number we’re representing is continued fraction. The process of obtaining successive convergents of a continued fraction involves iteratively
irrational (meaning it does not have an exact rational representation), then its computing: p\_n = a\_np\_{n-1} + p\_{n-2}, q\_n = a\_nq\_{n-1} + q\_{n-2} starting from p\_-1=1, q\_-1=0, p\_0=a\
continued fraction must necessarily be infinite. By using a suitable data structure _0, q\_0=1. These convergents, denoted as C\_n=[a\_0; a\_1, ..., a\_n], provide increasingly accurate rational
(think stream!), we get to represent this irrational number and obtain better and approximations of the value represented by the continued fraction. Notably, these approximations improve
better rational approximations for it. 2Theme Song Unwritten by Natasha Bedingfield monotonically—that is, |C\_n - x| < |C\_{n+1} - x| for any irrational number x corresponding to the given continued
3Example Consider trying to approximate the square root of 2. First, note that we fraction. To illustrate, consider the golden ratio, phi, expressed as the continued fraction [1; 1, 1, ...]: * p\
can rewrite the square root" Do not just list concepts, but develop each one in _-1=1, q\_-1=0, p\_0=1, q\_0=1 * p\_1=1*1+1=2, q\_1=1*1+0=1 * p\_2=2*2+1=5, q\_2=2*1+1=3 * p\_3=5*5+2=29, q\_3=
detail before moving to the next, as we prioritize depth of understanding and 5*3+3=16 ... Thus, the convergents are 1, 2, 3/2, 5/3, 8/5, ..., which indeed approach phi. Now let us return to
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: our initial example, pi, whose approximation as 22/7 stems from its continued fraction representation. While
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an remembering every term would prove cumbersome, recall that it begins [3; 7, 15, 1, 292, ...]. Extracting the first
academic, professional and engaging tone that captivates interest. - Application: few convergents yields: * C\_0 = [3] = 3 * C\_1 = [3; 7] = 22/7 * C\_2 = [3; 7, 15] = 333/106 * C\_3 = [3; 7, 15,
Incorporate specific, practical examples, such as proofs in calculus or critical 1] = 355/113 Indeed, 22/7 serves as a decent approximation, yet 355/113 surpasses it in accuracy, remaining the
dates and figures in history. Do not include a title or an introduction, simply closest fractional estimate using fewer than five digits in both numerator and denominator. In summary, continued
write the content without headlines and introductory phrases. Do not use images. fractions present a powerful tool for approximating irrational numbers through sequences of rationals. Through
exploring their underlying mechanisms and witnessing them at work with historical constants like pi, one gains
newfound appreciation for their ingenuity and applicability across various domains within mathematics.
Write an educational piece suited for college students related to the following The given first-order differential equation is non-linear, which makes it more challenging to solve compared to
text snippet: "# How to solve $y(y'+3)=ax^2+bx+c, \quad a,b,c \in \mathbb{R}$ How linear ones. However, by making some appropriate substitutions and applying separation of variables, we can find
could we solve this differential equation $$y(y'+3)=ax^2+bx+c, \quad a,b,c \in \ solutions for certain cases. Here, we explore three different cases where two of them have explicit solutions while
mathbb{R}$$ I really don't know how start. I am not familiar with this sort of the third case has implicit solutions. **Case 1:** $a = b = c = 0$ When all constants are zero, our original
differential equations (I know it is non-linear fist order, but don't see way to equation reduces to $y(y' + 3) = 0$. This factors into two simpler equations: $y = 0$ and $y' = -3$. Solving these
solve it). I passed this exam, so this is just for fun (not homework or something yields general solutions: 1. $y_1(x) = 0$, and 2. $y_2(x) = C - 3x$, $(C \in \mathbb{R})$ These two lines represent
like that). Thanks. Case $1$: $a=b=c=0$ Then $y(y'+3)=0$ $y=0$ or $y'=-3$ $y=0$ or the family of solutions when $a$, $b$, and $c$ are zero. **Case 2:** $a = b = 0,\; c eq 0$ Now let us consider the
$C-3x$ Case $2$: $a=b=0$ and $c\neq0$ Then $y(y'+3)=c$ $\dfrac{dy}{dx}=\dfrac{c}{y} scenario where $a$ and $b$ are still zero, but $c$ is nonzero ($c eq 0$). We then rewrite the equation as follows:
-3$ $\dfrac{dx}{dy}=\dfrac{y}{c-3y}$ $x=\int\dfrac{y}{c-3y}dy$ $x=C-\dfrac{y}{3}-\ $$yy' + 3y = c.$$ Dividing both sides by $y$ gives us a separable differential equation: $$y'\frac{dy}{dx} = \frac
dfrac{c\ln(3y-c)}{9}$ Case $3$: $a=0$ and $b\neq0$ Then $y(y'+3)=bx+c$ $\dfrac{dy} {c}{y} - 3.$$ We now separate the variables $x$ and $y$: $$dx = \frac{y\, dy}{c - 3y}.$$ Integrating both sides
{dx}=\dfrac{bx+c}{y}-3$ Let $y=\left(x+\dfrac{c}{b}\right)u$ , Then $\dfrac{dy}{dx} results in: $$x = C - \frac{y}{3} - \frac{c\,\mathrm{ln}(3y - c)}{9},$$ where $C$ is another constant of
=\left(x+\dfrac{c}{b}\right)\dfrac{du}{dx}+u$ $\therefore\left(x+\dfrac{c}{b}\ integration. Thus, the solution set includes all functions represented by the above expression. **Case 3:** $a = 0,
right)\dfrac{du}{dx}+u=\dfrac{b}{u}-3$ $\left(x+\dfrac{c}{b}\right)\dfrac{du}{dx}=\ \; b eq 0$ In this case, only $b$ is nonzero. To proceed, make the substitution $y = u\left( x + \frac{c}{b} \
dfrac{b}{u}-3-u$" Do not just list concepts, but develop each one in detail before right)$. Then, compute the derivative of $y$ concerning $x$: $$\frac{dy}{dx} = u' \cdot \left(x + \frac{c}{b}\
moving to the next, as we prioritize depth of understanding and comprehensive right) + u.$$ Substitute back into the main equation: $$\begin{align*} &\left(x + \frac{c}{b}\right)\left[u' \cdot
exploration of the subject matter over breadth. Focus on: - Rigor: Ensure in-depth \left(x + \frac{c}{b}\right) + u\right](u' + 3)\\ &= bx + c\\ &u'(ux + cu') + u(u'x + u'c/b + 3u) = bx/u + c/u \\ &
coverage of the concepts/sections. - Engagement: Write with an academic, u'(ux^2 + cxu' +cu + 3u^2) = bx -ucu'. \end{align*}$$ Unfortunately, solving for $u'$ does not yield simple
professional and engaging tone that captivates interest. - Application: Incorporate expressions for $u$; hence, obtaining closed-form explicit solutions seems unattainable here. Nonetheless, we do
specific, practical examples, such as proofs in calculus or critical dates and obtain implicit solutions involving $u$: $$\int{\frac{1}{(x^2 - c)u}\, du} = -\frac{1}{3}\int{\frac{1}{u^2 + 1}\,
figures in history. Do not include a title or an introduction, simply write the dx}.$$ This concludes the analysis of the provided differential equation, providing insights into its various
content without headlines and introductory phrases. Do not use images. possible solution sets based on distinct conditions.
The task at hand involves determining the true center of rotation of a two-dimensional object, given its initial
and final positions and the angle of rotation between them. To accomplish this, let us first establish some
fundamental concepts about rigid body transformations in a plane. We shall then build upon these ideas to derive a
suitable formula for solving the stated problem. ### Translations & Rotations A transformation in a two-dimensional
space can be described using translations and rotations. A translation moves every point in the object by the same
distance in the same direction, whereas a rotation spins all points around a fixed axis without changing their
Write an educational piece suited for college students related to the following relative distances from that axis. Let T(p) denote a translation function applied to any arbitrary vector p = (x,
text snippet: "# Find center of rotation after object rotated by known angle (2D) I y), where x and y are the coordinates of a point in the Cartesian coordinate system. Similarly, let R(θ) represent
need to be able to calculate and find the true center of rotation (x,y) of an a counterclockwise rotation function through an angle θ around the origin O. For our purposes, we consider only
object after it has been rotated by a known angle. Previously I would simply find planar movements; hence, both T and R preserve lengths and angles during these operations. Given any point P in the
the center of the object, rotate 180 degrees, find the new center of the object, original position represented by the vector p = (x, y), applying successive translations and rotations yields the
and calculate the midpoint of the line that connects the 2 centers to find the true expression PR(θ)T(p). Notice how the order matters here—first comes the rotation followed by the translation.
center of rotation. Now, I don't have the luxury of being able to rotate 180 Intuitively, you might imagine fixing your arm straight out horizontally from your shoulder (the "origin") and
degrees, instead it will be only 10-15 degrees, but it will be known. I've searched twisting your wrist (rotation) before pushing forward (translation). Now suppose there exists another reference
for a while and haven't found the appropriate equation to solve my particular frame B attached to the object itself, which undergoes the same series of motions. That is, if Q is the
problem. Limitation: Only know center of object, cannot use multiple points found corresponding point in B initially aligned with P, then Q experiences exactly the inverse sequence of
on object. Knowns: Center of object before and after rotation, and angle of transformations compared to P: TR^{-1}(q), where q represents Q's location within B. By definition, the relative
rotation Unknown: True center of rotation Hopefully the image explains what I'm motion between P and Q remains constant regardless of whether they belong to separate frames or share the same one.
shooting for Thanks • This rotation is taking place in the plane? If so, why does Therefore, we must ensure that PR(θ)T(p) equals TR^{-1}(q); otherwise, inconsistencies arise when comparing
the image make it look like $P$ and $P" Do not just list concepts, but develop each displacements across different objects. Consequently, the composition of these transformations should satisfy: PR
one in detail before moving to the next, as we prioritize depth of understanding (θ)T(p) = TR^{-1}(q) This property guarantees consistent behavior throughout the entire space, ensuring reliable
and comprehensive exploration of the subject matter over breadth. Focus on: - calculations involving various locations within the plane. With this foundation laid out, we can now proceed
Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Write with towards addressing the main question. ### Finding the True Center of Rotation As mentioned earlier, we are provided
an academic, professional and engaging tone that captivates interest. - with three pieces of information: * Initial center position C_{old} = (x_{old}, y_{old}) * Final center position C_
Application: Incorporate specific, practical examples, such as proofs in calculus {new} = (x_{new}, y_{new}) * Angle of rotation Δθ Our objective is to determine the true center of rotation D=(d_x,
or critical dates and figures in history. Do not include a title or an d_y), defined as the unique point whose absolute movement throughout the entire process vanishes. Mathematically
introduction, simply write the content without headlines and introductory phrases. speaking, D lies unmoved because PD(d_x, d_y)=D holds true for any rotation angle. Applying the above relationship
Do not use images. between simultaneous transformations leads us to the equality DR(Δθ)DT(C_{old})=DC_{new}. Since the net effect of
DT and DC is merely shifting the origins, their combined impact simplifies down to subtracting the old center from
the new center: DR(Δθ)(C_{old}-D) = C_{new}-C_{old} To isolate D, pre-multiply both sides by the inverse operation
DR(-Δθ): D = DR(-Δθ)[C_{new}-C_{old}] + D Notice how the left side collapses due to the identity property of
inverse functions DR(-Δθ)DR(Δθ)=I, resulting in: D = C_{new}-C_{old}+D Subtracting D from both sides finally
reveals the desired expression: D - D = C_{new} - C_{old} => D = 0.5 * (C_{old} + C_{new}) Thus, the true center of
rotation coincides precisely with the midpoint along the line connecting the initial and final center positions!
Despite working within smaller angular ranges than previously encountered, this elegant solution still applies.
Write an educational piece suited for college students related to the following The discussion begins with a request to compute the integral $\int\_{-\infty}^{\infty} dz / z^{2}+25$. This is an
text snippet: "## anonymous 5 years ago calculate the integral 1. anonymous $\int\ improper integral because it involves integration over an infinite interval. To evaluate this type of integral, we
limits_{-\infty}^{\infty}dz/z ^{2}+25$ 2. anonymous So? 3. TuringTest I'm working need to compute two separate integrals: one from negative infinity to zero and another from zero to positive
on it did you try a trig sub$z=5\tan\theta$? 4. anonymous Well we have a table infinity. In response, user TuringTest suggests using a trigonometric substitution $z = 5 an theta$, which
ofontegrals at the back of the book and it shows how to find the diffeential. it is transforms the integral into something more manageable by exploiting known properties of trigonometric functions.
a spectial differential 5. anonymous (1/a) arctan(x/a)+c 6. anonymous a=5 over here Specifically, they consider the inverse tangent function, denoted by $tan^{-1}$. It turns out that this
7. anonymous so i guess it wld be (1/5)arctan(x/5) 8. anonymous whoops replace the substitution leads to a simple antiderivative involving the inverse tangent function. However, since the bounds of
x with z 9. TuringTest yeah, ok forgot about that so now it's an improper integral integration involve infinities, we must carefully examine the resulting limits. To elaborate further on evaluating
and needs to be split up as far as I know$\lim_{n \rightarrow -\infty}\frac1 5\tan^ improper integrals, let us first focus on computing limits of the form $\lim\_{n \to - o o }\frac{1}{5} tan^{-1}( \
{-1}(\frac z5) |_{n}^{0}+\lim_{n \rightarrow \infty}\frac1 5\tan^{-1}(\frac z5) |_ frac{z}{5}) |\_n^{0}$. By definition, the limit approaches zero as n goes to negative infinity. Plugging in the
{0}^{n}$but I'm not sure how to do these limits, so there is probably another value of zero for z, we obtain $\frac{1}{5} tan^{-1}(0)$, which simplifies to zero due to the fact that the tangent
method... Who else has an idea? 10. anonymous well we plug in no? 11. anonymous so of any integer multiple of $\pi/2$ is undefined. Hence, our expression becomes $\lim\_{n \to - o o } 0 + C$, where
tan ^(-1)(0)=0 12. anonymous so we are left with (1/5)tan^(-1)(b/5) 13. anonymous C represents the constant of integration. Since constants vanish upon differentiation, we can safely ignore them
turi" Do not just list concepts, but develop each one in detail before moving to during intermediate calculations. Thus, the limit equals zero. Next, we tackle $\lim\_{n \to + o o }\frac{1}{5} tan
the next, as we prioritize depth of understanding and comprehensive exploration of ^{-1}( \frac{z}{5}) |\_0^{n}$. As n tends to positive infinity, this time around, the expression approaches
the subject matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the infinity. Replacing z with n yields $\frac{1}{5} tan^{-1}( \frac{n}{5})$. Unfortunately, this expression does not
concepts/sections. - Engagement: Write with an academic, professional and engaging immediately lend itself to easy evaluation. Instead, recalling basic properties of inverse trigonometric functions
tone that captivates interest. - Application: Incorporate specific, practical allows us to rewrite this quantity as $(\frac{1}{5})(\frac{\pi}{2})$ when n = 5k+1, where k is an arbitrary
examples, such as proofs in calculus or critical dates and figures in history. Do nonnegative integer. Substituting this result back into our original equation gives us $\lim\_{k \to + o o } (\frac
not include a title or an introduction, simply write the content without headlines {1}{5})(\frac{\pi}{2})$, which again tends to infinity as k increases without bound. Combining both results, the
and introductory phrases. Do not use images. final answer to the given improper integral is $\int\_{-\infty}^{\infty} dz / z^{2}+25 = +\infty$. That is, the
area under the curve grows unboundedly as we extend the range of integration to cover all real numbers.
To begin, let us recall the definitions of continuity and uniform continuity for mappings between metric spaces. A
mapping $f : (X,d) \to (Y,d')$ is said to be continuous at a point $a \in X$ if for every $\epsilon > 0$, there
exists a $\delta > 0$ such that $d'(f(x), f(a)) < \epsilon$ whenever $d(x,a) < \delta$. The mapping $f$ is said to
Write an educational piece suited for college students related to the following be uniformly continuous if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $d'(f(x), f(y)) < \
text snippet: "# Uniformly continuous mapping ## Homework Statement Suppose X = epsilon$ whenever $d(x,y) < \delta$, for all $x, y \in X$. It is important to note that uniform continuity implies
[0,1] x [0,1] and d is the metric on X induced from the Euclidean metric on R^2. continuity, but the converse is not necessarily true. Now, let's consider the given mapping $T: [0,1] \times [0,1]
Suppose also that Y = R^2 and d' is the Euclidean metric. Is the mapping T: [0,1] x \to \mathbb{R}^2$ defined as $T(x,y) = (xy, e^{xy})$. Our goal is to determine whether this mapping is uniformly
[0,1] $$\rightarrow$$ R^2, T(x,y) = (xy, e^(x.y)) ## The Attempt at a Solution Hi continuous with respect to the metrics induced by the standard Euclidean norms on $\mathbb{R}^2$ and $\mathbb{R}^
everyone, So I know the definition for uniformly continuous, but am wondering if 4$, respectively. That is, we will show that for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|T
it's necessary to use it? We have in our notes that continuous linear maps on (x_1, y_1) - T(x_2, y_2)| < \epsilon$ whenever $|(x_1, y_1) - (x_2, y_2)| < \delta$, where $|\cdot|$ denotes the
normed vecotr spaces are unifomrly continuous, and (Y,d') is a normed vector space. Euclidean norm. To proceed, let $(x_1, y_1), (x_2, y_2) \in [0,1] \times [0,1]$ be arbitrary points. Then, using
So by looking at the graph of the map, there is a discontinuity between the line on the triangle inequality, we have: \begin{align*} |T(x_1, y_1) - T(x_2, y_2)| &= |((x_1 y_1, e^{x_1 y_1}) - (x_2
the x-axis and the exponential function. So can you say it is not continuous and y_2, e^{x_2 y_2}))| \\ &= |(x_1 y_1 - x_2 y_2, e^{x_1 y_1} - e^{x_2 y_2})| \\ &\leq |x_1 y_1 - x_2 y_2| + |e^{x_1
thus not uniformly continuous? Thanks for any help Related Calculus and Beyond y_1} - e^{x_2 y_2}|. \end{align*} We now need to bound these two terms separately. For the first term, notice that:
Homework Help News on Phys.org T: [0,1] x [0,1] $$\rightarrow$$ R^2, T(x,y) = (xy, \[ |x_1 y_1 - x_2 y_2| \leq |x_1||y_1 - y_2| + |x_1 - x_2||y_2| \leq |(x_1, y_1) - (x_2, y_2)|(|y_2| + |x_1|). \]
e^(x.y)) I'm guessing that you mean $$T(x, y) = (xy, e^{xy})$$ here. So I know the For the second term, since the exponential function is Lipschitz continuous with Lipschitz constant equal to its
definition for uniformly co" Do not just list concepts, but develop each one in derivative, we obtain: \[ |e^{x_1 y_1} - e^{x_2 y_2}| \leq |x_1 y_1 - x_2 y_2|e^{\max\{x_1 y_1, x_2 y_2\}} \leq |
detail before moving to the next, as we prioritize depth of understanding and (x_1, y_1) - (x_2, y_2)|e^{\max\{x_1, x_2\}}\max\{y_1, y_2\}e^{\max\{y_1, y_2\}}. \] Therefore, combining both
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: bounds yields: \begin{multline*} |T(x_1, y_1) - T(x_2, y_2)| \leq |(x_1, y_1) - (x_2, y_2)| \Bigl[|y_2| + |x_1|\\+
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an e^{\max\{x_1, x_2\}}\max\{y_1, y_2\}e^{\max\{y_1, y_2\}}(1+\max\{y_2\} + |x_1|\max\{y_1, y_2\})\Bigr]. \end
academic, professional and engaging tone that captivates interest. - Application: {multline*} Letting $\delta = \min\{\frac{\epsilon}{C}, 1\}$ for some constant $C > 0$ depending only on the
Incorporate specific, practical examples, such as proofs in calculus or critical interval $[0,1]$, we see that if $|(x_1, y_1) - (x_2, y_2)| < \delta$, then: \begin{align*} |T&(x_1, y_1) - T(x_2,
dates and figures in history. Do not include a title or an introduction, simply y_2)| \\ &\quad< \delta C\\ &\quad= \min\{\epsilon, C\} \\ &\quad\leq \epsilon. \end{align*} Thus, $T$ is indeed
write the content without headlines and introductory phrases. Do not use images. uniformly continuous on $[0,1] \times [0,1]$. This demonstrates how the concept of Lipschitz continuity was used to
establish uniform continuity. However, the original attempt made in the post fails because merely observing a
discontinuity does not suffice to conclude non-uniform continuity; instead, it requires showing that no such $\
delta$ can exist for certain values of $\epsilon$.
When working with fractions, it's essential to understand the concept of equivalent fractions. Two fractions are
Write an educational piece suited for college students related to the following considered equivalent if they represent the same portion of a whole. Although they may look different, these
text snippet: "0 # Need top write a two fractions that are equivalent to the given fractions describe identical ratios. For example, consider the fraction 1/2. An equivalent fraction could be 2/4
fraction? Wiki User 2013-06-04 21:43:58 Multiply the numerator (top) and the since both represent half of a quantity. To find additional equivalent fractions, you can multiply the numerator
denominator (bottom) of the given fraction by any non-zero integer. You will have and denominator of the original fraction by the same non-zero integer. This process maintains the ratio described
an equivalent fraction. Wiki User 2013-06-04 21:43:58 Study guides 20 cards ## A by the fraction, thus resulting in equivalence. Let's explore this idea further using the provided fraction 0. We
number a power of a variable or a product of the two is a monomial while a want to generate two distinct equivalent fractions. Since multiplying both the numerator and denominator by the
polynomial is the of monomials ➡️ See all cards 3.8 2256 Reviews Earn +20 pts Q: same integer results in equivalence, let's choose two unique integers - say, 2 and 3 - and apply them to our
Need top write a two fractions that are equivalent to the given fraction?" Do not initial fraction. Here's how it works: Equivalent Fraction #1: Multiply the numerator and denominator by 2. The
just list concepts, but develop each one in detail before moving to the next, as we fraction becomes: (0 × 2)/(1 × 2) = 0/2 Although this result might seem incorrect at first glance, remember that
prioritize depth of understanding and comprehensive exploration of the subject dividing by zero isn't allowed; hence, 0/2 represents the same value as 0. Therefore, 0/2 is an equivalent fraction
matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the concepts/ of 0. Equivalent Fraction #2: Now, multiply the numerator and denominator by 3 instead: (0 × 3)/(1 × 3) = 0/3
sections. - Engagement: Write with an academic, professional and engaging tone that Similarly, 0/3 conveys the same value as 0 because division by zero is undefined. Thus, 0/3 serves as another
captivates interest. - Application: Incorporate specific, practical examples, such equivalent fraction for 0. In conclusion, when dealing with the fraction 0, its equivalent fractions include 0/2
as proofs in calculus or critical dates and figures in history. Do not include a and 0/3. These fractions convey the same mathematical meaning despite their differing appearances, which stems from
title or an introduction, simply write the content without headlines and the multiplication of both the numerator and denominator by the same non-zero integers (in this case, 2 and 3).
introductory phrases. Do not use images. Understanding equivalent fractions and knowing how to derive them allows for more precise communication and
manipulation of information within mathematical contexts.
Now that we've discussed linear systems of equations, let us delve into a more complex topic: systems of nonlinear
equations. As the name suggests, these involve at least one equation that does not graph as a straight line. Don't
worry if this sounds intimidating; many techniques you've already learned still apply! We will explore two primary
methods for solving these systems: substitution and elimination. Firstly, solving a system of nonlinear equations
Write an educational piece suited for college students related to the following using **substitution** involves isolating one variable in one equation and then replacing it with its equivalent
text snippet: "# 9.3: Systems of Nonlinear Equations and Inequalities - Two expression in the other equation(s). This process simplifies the system by reducing the number of variables. Let's
Variables Learning Objectives • Solve a system of nonlinear equations using look at an example involving a parabola and a line: $$y = x^2 + 1 \quad \text{(Parabola)}$$ $$y = 2x - 3 \quad \
substitution. • Solve a system of nonlinear equations using elimination. • Graph a text{(Line)}$$ To solve this system, we can substitute the second equation's $y$ into the first equation: $$2x - 3
nonlinear inequality. • Graph a system of nonlinear inequalities. Halley’s Comet = x^2 + 1$$ Rearranging the terms gives us a quadratic equation: $$x^2 - 2x - 4 = 0$$ Factoring yields $(x+1)(x-4)=
(Figure (PageIndex{1})) orbits the sun about once every (75) years. Its path can be 0$, so there are two solutions: $x=-1$ and $x=4$. To find the corresponding y-values, plug them back into either
considered to be a very elongated ellipse. Other comets follow similar paths in original equation—for instance, when $x=-1$: $$y = (-1)^2 + 1 = 2$$ So, one solution is $\boxed{(-1,2)}$. Repeat
space. These orbital paths can be studied using systems of equations. These this step for $x=4$: $$y = (4)^2 + 1 = 17$$ Thus, the other solution is $\boxed{(4,17)}$. Next, let's examine how
systems, however, are different from the ones we considered in the previous section to solve a system of nonlinear equations via **elimination**. This approach entails manipulating the given
because the equations are not linear. In this section, we will consider the equations so that they have identical coefficients for one variable, enabling us to cancel out one variable much
intersection of a parabola and a line, a circle and a line, and a circle and an like in linear systems. Consider the following system consisting of a circle and a line: $$x^2 + y^2 = 4 \quad \
ellipse. The methods for solving systems of nonlinear equations are similar to text{(Circle)}$$ $$y = -\frac{1}{2}x + 2 \quad \text{(Line)}$$ Our goal is to eliminate $y$; thus, multiplying both
those for linear equations. ## Solving a System of Nonlinear Equations Using sides of the linear equation by $2$ results in equal $y$-coefficients in our system: $$2y = -x + 4$$ Now, rewrite
Substitution A system of nonlinear equations is a system " Do not just list the circular equation in terms of $y$: $$y = \pm\sqrt{\frac{4-x^2}{1}}$$ Equate the new forms of $y$ to erase $y$:
concepts, but develop each one in detail before moving to the next, as we $$-\frac{x}{2}+2=\pm\sqrt{\frac{4-x^2}{1}}$$ Squaring both sides twice while maintaining equality leads to a
prioritize depth of understanding and comprehensive exploration of the subject quartic equation ($ax^4+bx^3+cx^2+dx+e=0$). Since quartics do not generally factor nicely, numerical approximation
matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the concepts/ methods might be required to obtain exact roots. However, graphing the functions reveals their intersections
sections. - Engagement: Write with an academic, professional and engaging tone that visually. Lastly, let's discuss graphing nonlinear inequalities and systems thereof. For single inequalities,
captivates interest. - Application: Incorporate specific, practical examples, such sketching the region where the function holds true may suffice. But when dealing with systems of nonlinear
as proofs in calculus or critical dates and figures in history. Do not include a inequalities, determining which regions satisfy all conditions becomes crucial. Here's an illustrative example
title or an introduction, simply write the content without headlines and featuring a circle and an ellipse: $$x^2+y^2≤4 \qquad \text{(Circle)}$$ $$\frac{x^2}{4}+\frac{y^2}{9}<1 \qquad \
introductory phrases. Do not use images. text{(Ellipse)}$$ Graphing these shapes reveals three possible regions separated by curves. Testing points within
these areas against the inequalities allows us to identify the correct region(s), shading appropriately. Remember,
strict inequalities ($\lt,\gt$) require open circles at boundary points, whereas nonstrict inequalities allow
boundary points to be included. This section has provided an overview of solving systems of nonlinear equations
through substitution, elimination, and visualizing inequalities. Familiarity with these concepts opens up
opportunities for further study in advanced mathematics and applications across various disciplines.
The mathematical concept at hand is divide and conquer, which involves breaking down complex problems into smaller,
manageable parts and solving them individually. This technique often simplifies calculations and provides estimates
that are reasonably close to exact values. Herein, I will delve deeper into the two problems presented, focusing on
Write an educational piece suited for college students related to the following rigorous analysis while maintaining engagement through relatable examples. Firstly, consider the problem $1 ÷ \sqrt
text snippet: "Rich example # Divide and conquer Add to your resource collection {2}$. At first glance, it may seem challenging to calculate without a calculator; however, by employing the divide
Remove from your resource collection Add notes to this resource View your notes for and conquer strategy, we recognize that $\sqrt{2}$ is slightly greater than 1 since $1^2 = 1 < 2$. Thus, $1 ÷ \sqrt
this resource ## Problem Without using a calculator, find a reasonable estimate of: {2}$ must be less than 1 because multiplying any number by something between 0 and 1 yields a product smaller than
$1 \div \sqrt{2}$ What can your thinking about this tell you about other divisions the original number. To refine our estimation, let's introduce a sequence $\{a_k\}$ where $a_k=1-\frac{1}{2^{2k}}$
of the form $1 \div \sqrt{n}$? • Can you select $n$ such that you can write down (the reason behind choosing this particular sequence will become clear shortly). Notice how these terms alternate
the value of the division calculation more easily? • Can you select $n$ such that above and below the square root of two: $$ \begin{aligned} &a_1=\tfrac{3}{4}=0.75>\sqrt{2}\\ &a_2=\tfrac{15}{16}=
your answer to the original problem is useful in finding the value of the division? 0.9375<\sqrt{2}\\ &a_3=\tfrac{55}{64}\approx0.8594>.\sqrt{2}\\ &\quad\vdots\\ \end{aligned} $$ Observe that $a_
Without using a calculator, find a reasonable estimate of: $\sqrt{50} \div 10$ {k+1}-a_k=\pm\frac{1}{4^{2k}}$, meaning consecutive terms differ only slightly. Now, observe that $\lim\limits_{k\
Compare this to the original problem: $1 \div \sqrt{2}$. • What do you notice about to\infty}a_k=\sqrt{2}$, implying that the gap between $a_k$ and $\sqrt{2}$ becomes arbitrarily small as $k$
your approach?" Do not just list concepts, but develop each one in detail before increases. Consequently, when estimating $1 ÷ \sqrt{2}$, we seek $1 ÷ a_k$, knowing it converges to our desired
moving to the next, as we prioritize depth of understanding and comprehensive result. For instance, if we choose $k=2$: $$ 1 ÷ a_2=1 ÷ (\tfrac{15}{16})=1.333... > 1 ÷ \sqrt{2} $$ This simple
exploration of the subject matter over breadth. Focus on: - Rigor: Ensure in-depth yet effective application of the divide and conquer principle has yielded a usable approximation—albeit larger than
coverage of the concepts/sections. - Engagement: Write with an academic, necessary—for $1 ÷ \sqrt{2}$. By increasing $k$, we obtain increasingly accurate results while minimizing
professional and engaging tone that captivates interest. - Application: Incorporate computational effort required compared to direct computation. Next, examine the second problem involving $\sqrt{50}
specific, practical examples, such as proofs in calculus or critical dates and ÷ 10$. Applying similar reasoning, note that $\sqrt{50}\approx7.07<8=2^3$, rendering it easier to compute with
figures in history. Do not include a title or an introduction, simply write the mental arithmetic. Furthermore, recognizing that dividing by powers of ten merely shifts decimal points, we have:
content without headlines and introductory phrases. Do not use images. $$ \sqrt{50}/10=(5/\sqrt{2})\times(1/10)=5/(10\sqrt{2})\approx\dfrac{5}{14}, $$ where the final step uses our
previous approximation $1 ÷ \sqrt{2}≈1.333…≈14⁄10$. Comparing both solutions, observe the commonality: selecting
numbers ($n$) intelligently facilitates computation significantly. When confronted with seemingly difficult
expressions like those posed initially, remember to break them apart using the divide and conquer method. Through
careful selection of components, even intricate problems submit to straightforward resolution techniques.
Write an educational piece suited for college students related to the following Let's delve into proving that if \(x\) and \(y\) are not both 0, then \(x^2+xy+y^2>0.\) We will explore two
text snippet: "# Homework Help: Another proof: x^2 + xy +y^2 > 0 1. Sep 17, 2009 ## approaches: one using algebraic manipulation and the other utilizing single-variable calculus. **Approach 1:
# nietzsche Hello again, I have another proof that I can't figure out how to solve. Algebraic Manipulation** Start by assuming equation (1) holds: $$x^2+xy+y^2 > 0\quad (\text{assumption})$$ Perform
1. The problem statement, all variables and given/known data $$\text{Prove that if some algebraic manipulations: \begin{align*} x^2 + xy + y^2 &> 0 \\ x^2 + 2xy + y^2 &> xy \qquad &\left(\text{add }
}\textit{x }\text{and }\textit{y }\text{are not both 0, then} \begin{equation*} x^ -xy\right)\\ (x+y)^2 &> xy \qquad &\left[\because (a+b)^2=a^2+2ab+b^2\right] \end{align*} Now, let's analyze this
2+xy+y^2>0\tag{1} \end{equation*}$$ 2. Relevant equations N/A 3. The attempt at a inequality when $x$ and $y$ have the same sign and opposite signs. Case 1: _x_ and \(y\) have the same sign_
solution Not sure if I'm on the right track here... \text{Assume (1) is true:} \ Without loss of generality, assume $x$ and $y$ are both positive. Then, since $(x+y)^2 > 0$, it follows that $x^2 +
begin{align*} x^2+xy+y^2 &> 0\\ x^2 +2xy + y^2 &> xy\\ (x+y)^2 &> xy\\ \end{align*} xy + y^2 > xy$. Since $xy > 0$, we conclude that $x^2 + xy + y^2 > 0.$ Case 2: _x_ and \(y\) have opposite signs_
...not sure where to go from here. 2. Sep 17, 2009 ### Dick I would say to look for Suppose $x$ and $y$ have opposite signs; i.e., either $x<0$ and $y>0$ or $x>0$ and $y<0$. Note that $|x+y|>\max(|x
the extrema of f(x,y)=x^2+xy+y^2. Find df/dx and df/dy and set them equal to zero |, |y|)$. Consequently, $(x+y)^2 > \max(x^2, y^2)$, which implies that $x^2 + xy + y^2 > xy$. As $xy < 0$, it
and solve for x and y. Can you classify that as a min or a max or neither? 3. Sep yields $x^2 + xy + y^2 > 0.$ In conclusion, if $x$ and $y$ are not both 0, then $x^2 + xy + y^2 > 0.$ **Approach 2:
17, 2009 ### Office_Shredder Staff Emeritus (x+y)2 = x2 + 2xy + y2 >= 0 You know Single Variable Calculus** Consider function $f(x, y) := x^2 + xy + y^2$. To find its extremal points, compute
that already So x2 + xy + y2 >= -xy If x and y are both positive, the result is partial derivatives: \begin{align*} \frac{\partial f}{\partial x} &= 2x + y \\ \frac{\partial f}{\partial y} &= x +
trivial." Do not just list concepts, but develop each one in detail before moving 2y \end{align*} Setting these expressions to 0 gives us the only critical point $(\frac{-1}{3},\frac{-1}{3})$. Now
to the next, as we prioritize depth of understanding and comprehensive exploration evaluate the Hessian matrix—the matrix containing second order partial derivatives: $$H(x, y)=\begin{pmatrix}\dfrac
of the subject matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of {{\partial}^2f}{{\partial x}^2}&\dfrac{{\partial}^2f}{\partial x\,\partial y}\\[1em]\dfrac{{\partial}^2f}{\partial
the concepts/sections. - Engagement: Write with an academic, professional and y\, \partial x}&\dfrac{{\partial}^2f}{{\partial y}^2}\end{pmatrix}=\begin{pmatrix}2&1\\1&2\end{pmatrix}$$ Since $\
engaging tone that captivates interest. - Application: Incorporate specific, det(H(x, y)) = 3 > 0$ and $h_{11}=h_{22}>0$, the critical point is a local minimum according to the Second
practical examples, such as proofs in calculus or critical dates and figures in Derivative Test. Moreover, note that $f(-1, -1) = \frac{3}{9} > 0$. Therefore, there exists no pair $(x, y)$
history. Do not include a title or an introduction, simply write the content satisfying $f(x, y)\leqslant 0$. Hence, if $x$ and $y$ are not both 0, then $x^2 + xy + y^2 > 0.$
without headlines and introductory phrases. Do not use images.
Write an educational piece suited for college students related to the following Fractions are essential mathematical quantities used to represent parts of a whole object or group. They consist of
text snippet: "Addition of Unlike Fractions – Definition, Examples | How to Add two numbers separated by a line - the numerator (top) and denominator (bottom). Understanding this fundamental
Fractions with Unlike Denominators? Addition of Unlike Fractions: If the concept paves the way for performing arithmetic operations involving fractions, including the addition of unlike
denominators or bottom number of the fractions are not the same then it is known fractions. To begin with, let's define what we mean by "unlike fractions." Simply put, unlike fractions are those
as, unlike fraction. Here you can gain more knowledge about adding fractions with where the denominators differ from one another. When faced with the task of adding such fractions, the first step
unlike denominators. In order to make the denominators common, we have to find the is to make their denominators equal. This process is called finding the Least Common Multiple (LCM) of the given
least common multiples of the denominators. Students must know about the fractions denominators. The LCM refers to the smallest positive integer divisible by both denominators. To calculate the LCM
before they start adding the like or unlike fractions. The Addition of Unlike of any set of integers, follow these steps: 1. List all multiples of each denominator until you find a mutual
fractions is somewhat tricky, but you can make it easy by making the denominators multiple. 2. Identify the smallest mutual multiple; this is your LCM. Once you determine the LCM, convert both
equal. Read the entire article to know how to add unlike fractions, the addition of fractions so that they share the same denominator using the following formula: $$\frac{a}{b} + \frac{c}{d} = \frac
unlike fractions questions. Do Refer: What is a Fraction? In maths, A fraction is a {(a*d) + (b*c)}{(b*d)}$$ Here, $a$ and $c$ correspond to the original numerators, while $b$ and $d$ denote the
numerical quantity that is not a whole number. Fractions will divide the whole respective denominators. After applying the above methodology, you now have like fractions that are ready for
number into equal parts. For example, an apple is divided into two parts, then each addition! Let's consider a real-world application concerning time management. Suppose John spends $\frac{3}{4}$ of
part is represented by $$\frac{1}{2}$" Do not just list concepts, but develop each his day studying and $\frac{5}{6}$ of his remaining free time practicing piano. What portion of his full day does
one in detail before moving to the next, as we prioritize depth of understanding he spend playing piano? Firstly, identify the LCM of 4 and 6, which equals 12. Then transform the fractions
and comprehensive exploration of the subject matter over breadth. Focus on: - accordingly: $$\frac{3}{4} + \frac{5}{6} = \frac{(3*6)+(4*5)}{(4*6)} = \frac{27}{24}+\frac{20}{24}=\frac{47}{24}$$
Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Write with John dedicates $\frac{47}{24}$ hours per day towards pursuing his interests combined—however, since only the second
an academic, professional and engaging tone that captivates interest. - fraction represents his piano practice ($\frac{20}{24}$) after accounting for study hours, simplification yields $\
Application: Incorporate specific, practical examples, such as proofs in calculus boxed{\frac{5}{6}}$. Thus, John devotes $\frac{5}{6}$ of his leftover time to honing his musical skills. Remember,
or critical dates and figures in history. Do not include a title or an when presented with the challenge of adding unlike fractions, don't panic! Employ the power of the Least Common
introduction, simply write the content without headlines and introductory phrases. Multiple and watch those seemingly disparate fractions unite under a shared denominator, allowing for effortless
Do not use images. summation.
The problem you have presented involves analyzing the effects of subsidies on the labor market using comparative
statics. To begin, let's define the key components of this model: 1. **Demand for labor (L)**: This represents the
amount of labor firms are willing to hire at different wage levels. It is denoted by $L$, which is a function of
the wage rate $(W)$ and other factors, such as technology and productivity. 2. **Production function ($F(L)$)**:
This describes the relationship between the quantity of inputs (in this case, labor) and the resulting output. The
first derivative of the production function with respect to labor $(\frac{\partial F}{\partial L})$ denotes the
marginal product of labor, representing the additional output produced when one more unit of labor is employed. By
assumption, $\frac{\partial F}{\partial L}>0$, indicating a positive relationship between labor input and output.
Additionally, $\frac{\partial^2 F}{\partial L^2}<0$, meaning diminishing returns to scale are present – increasing
Write an educational piece suited for college students related to the following labor beyond a certain point will result in smaller increases in output. 3. **Price of goods (P)** and **wage rate
text snippet: "# Labor market model with subsidies homework I would appreciate some (W)**: $P$ refers to the price of the final good produced by firms employing labor, while $W$ indicates the wage
help with my homework in comparative statics. • L is the demand of labor • F(L) is rate paid to workers. 4. **Subsidy per hired worker (S)**: A government subsidy provided to businesses for every
the production function • $$\frac{\partial F}{\partial L} > 0 > \frac{\partial^2 F} worker they hire, effectively reducing their labor costs. 5. **Supply of labor ($AA(W)$)**: Represents the number
{\partial L^2}$$ • P is the price of goods • W is the wage • S is subsidy per hired of individuals willing to work at various wage rates. As expected, $\frac{\partial AA}{\partial W}>0$: higher wages
worker • AA(W) is the supply of labor • $$\frac{\partial AA}{\partial W} > 0$$ • induce more people to participate in the labor force. 6. **Profit maximization**: Companies aim to maximize
$$max_{L} P*F(L) - (W - S)*L$$ is what companies want to maximize a) Analyze the profits, represented by the expression $max_{L}\ P*F(L)\ -\ (W\ -\ S)*L.$ Now, let's address the questions posed in
effect of an increase in subsidies on the equilibriums of W and L. b) Illustrate the assignment: **Question a:** Analyze the effect of an increase in subsidies on the equilibria of $W$ and $L$. To
your findings in a graph. Chose Quantity of labor and W as axis. c) Analyze the investigate how changes in subsidies affect the equilibrium values of $W$ and $L$, consider the impact on both the
effect of an increase in subsidies on the slope of L in W. My ideas for a): The labor supply and demand curves. When a subsidy increases, employers pay lower net wages, causing the labor demand
equilibrium conditions are $$L = AA(W) \Leftrightarrow L - AA(W) = 0$$. So lets curve to shift outward since more labor can be profitably employed at any given wage level. At the same time,
call it $$H(W; L) = 0$$ and let´s call the profit function G so we get as the because the after-subsidy wage received by employees rises, the labor supply curve shifts leftward due to upward
second condition $$\frac{ \partial G}{ \partial L} = P * \frac{\partial F(L)}{\ pressure on wages. These dual movements cause two distinct intersecting points where the new labor demand and
partial L} - (W - S) = 0$$, so we " Do not just list concepts, but develop each one supply functions meet, leading to multiple short-run equilibria. However, only one long-term equilibrium prevails
in detail before moving to the next, as we prioritize depth of understanding and once all adjustments take place. Therefore, increased subsidies lead to a single long-term equilibrium
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: characterized by greater employment $(L)$ and higher real wages $(W)$. **Question b:** Illustrate your findings in
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an a graph. Choose quantity of labor and $W$ as axes. A proper illustration requires depicting the initial situation
academic, professional and engaging tone that captivates interest. - Application: followed by superimposing the altered scenario onto the original diagram. Start with the standard labor market
Incorporate specific, practical examples, such as proofs in calculus or critical graph, plotting the labor supply curve above the labor demand curve due to the wage being measured along the
dates and figures in history. Do not include a title or an introduction, simply vertical axis. Initially, there exists a unique equilibrium $(E_0)$ determined by the intersection of these two
write the content without headlines and introductory phrases. Do not use images. curves. After introducing the subsidy, the labor demand curve shifts rightward, reflecting reduced effective wages
faced by employers, and the labor supply curve moves upwards, accounting for augmented post-subsidy wages perceived
by employees. Consequently, the new equilibrium point $(E_1)$ emerges, exhibiting expanded employment and elevated
real wages compared to the pre-subsidy state. **Question c:** Analyze the effect of an increase in subsidies on the
slope of $L$ with respect to $W$. An increase in subsidies results in steeper slopes for both the labor demand and
supply curves. For labor demand, the steeper incline reflects amplified responsiveness to alterations in the wage
rate due to the reduced influence of the net cost change caused by fluctuations in $W$. Meanwhile, regarding labor
supply, the sharper angle mirrors heightened sensitivity to variations in the after-subsidy wage, given the larger
proportion of wage increments attributable solely to the subsidy component rather than actual compensation from
employers. Overall, the steepening of the curves highlights intensified interactions between wage adjustments and
corresponding employment modifications under enhanced subsidy schemes.
In set theory, a branch of mathematical logic, we often encounter various types of limits and colimits when dealing
with collections of sets indexed by partially ordered sets. A particular type of colimit that arises frequently is
Write an educational piece suited for college students related to the following the colimit of sets indexed by ordinals less than a given ordinal. This concept plays a crucial role in many
text snippet: "Lemma 3.7.1. Suppose that $T = \mathop{\mathrm{colim}}\nolimits _{\ branches of mathematics, including category theory and topology. Here, we aim to understand Lemma 3.7.1 from this
alpha < \beta } T_\alpha$ is a colimit of sets indexed by ordinals less than a perspective, focusing on its statement and subsequent proof. Before delving into the lemma itself, let's first
given ordinal $\beta$. Suppose that $\varphi : S \to T$ is a map of sets. Then $\ establish some essential background definitions and notation used throughout the discussion: * **Ordinals**: An
varphi$ lifts to a map into $T_\alpha$ for some $\alpha < \beta$ provided that $\ ordinal number is a well-ordered set where every element has a unique successor, except possibly for a maximum
beta$ is not a limit of ordinals indexed by $S$, in other words, if $\beta$ is an element. Ordinals can be thought of as generalized natural numbers, allowing us to extend our notion of "size"
ordinal with $\text{cf}(\beta ) > |S|$. Proof. For each element $s \in S$ pick a $\ beyond finite sets. They form a transfinite sequence starting at 0 and followed by 1, 2, ... , ω, ω+1, etc.,
alpha _ s < \beta$ and an element $t_ s \in T_{\alpha _ s}$ which maps to $\varphi eventually reaching any desired size. * **Colimit**: Given a directed system {Aᵢ, f�������ünd}, consisting of a
(s)$ in $T$. By assumption $\alpha = \sup _{s \in S} \alpha _ s$ is strictly collection of objects Aᵢ and morphisms f�������ünd between them, indexed by some poset I, a colimit is defined as
smaller than $\beta$. Hence the map $\varphi _\alpha : S \to T_\alpha$ which the universal object receiving compatible morphisms from all Aᵢ. It represents the most efficient way of combining
assigns to $s$ the image of $t_ s$ in $T_\alpha$ is a solution. $\square$ ## these elements together while preserving their relationships under the given structure. Concretely, it is
Comments (2) Comment #143 by on Replace $T_{\beta}$ by $T$. There are also: • 11 constructed as the disjoint union of the individual Aᵢ modulo certain equivalence relations determined by the
comment(s) on Section 3.7: Cofinality ## Post a comment Your email address will not f�������unden. Now, let's examine Lemma 3.7.1 more closely: **Lemma 3.7.1.** Let T=colimₐ<β Tᵦ₋₁ be a colimit of
be published. Required fields are marked" Do not just list concepts, but develop sets indexed by ordinals less than a given ordinal eta. Furthermore, suppose we have a map of sets φ:S→T. If eta is
each one in detail before moving to the next, as we prioritize depth of not a limit of ordinals indexed by S (i.e., cf(eta)>|S|), then there exists an alpha<eta such that φ lifts to a map
understanding and comprehensive exploration of the subject matter over breadth. into Tᵧ for some gamma≤alpha. This result establishes conditions for lifting a function φ:S→T through a colimit
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: diagram, ensuring that it factors via a stage earlier in the tower of sets Tᵦ₋₁. The key idea hinges upon finding
Write with an academic, professional and engaging tone that captivates interest. - appropriate preimages tᵤ∈Tᵨ for each s∈S using the definition of colimits, ultimately leading to a supremum alpha
Application: Incorporate specific, practical examples, such as proofs in calculus that satisfies the required property. The significance of cf(eta)>|S| cannot be overstated here since it guarantees
or critical dates and figures in history. Do not include a title or an the existence of such an upper bound alpha below eta. Indeed, cofinality cf(eta) measures the smallest cardinality
introduction, simply write the content without headlines and introductory phrases. needed to express eta as a limit of smaller ordinals—if cf(eta)>|S|, then no subset of ordinals indexed by S can
Do not use images. approach eta closely enough to exhaust it. As a result, choosing an optimal alpha becomes possible within the
allowed range. With this deeper understanding of Lemma 3.7.1, readers should feel equipped to apply similar
techniques when working with colimits of sets indexed by ordinals in their own research endeavors.
The task at hand is finding an integrating factor for the first order nonlinear ordinary differential equation
(ODE): $$\frac{1}{x} dx + \left(1+x^{2}y^{2}\right) dy = 0$$ Before diving into solving this problem, let us
establish some necessary background information regarding exact and inexact equations and their corresponding
integrating factors. An ODE of the form $M(x, y) dx + N(x, y) dy = 0$ is said to be exact when there exists a
differentiable function $\psi(x, y)$ satisfying: $$\frac{\partial \psi}{\partial x} = M(x, y)\quad \text{and} \quad
\frac{\partial \psi}{\partial y} = N(x, y).$$ If no such $\psi$ exists, then the equation is referred to as
inexact. When dealing with an inexact equation, our goal becomes transforming the given equation into an exact
equation through multiplication by an appropriate integrating factor. To clarify, suppose we are provided an
inexact equation $M(x, y) dx + N(x, y) dy = 0$. A nonzero continuously differentiable function $\lambda(x, y)$,
acting as an integrating factor, converts the original equation into an exact equation via the modified
relationship below: $$\lambda(x, y) M(x, y) dx + \lambda(x, y) N(x, y) dy = 0.$$ In practice, determining an
integrating factor often proves challenging; however, certain conditions may simplify matters considerably. One
primary instance arises from applying the cross-partial condition: $$\frac{\partial}{\partial y}[\lambda(x, y) M(x,
y)]=\frac{\partial}{\partial x}[\lambda(x, y)N(x, y)].$$ Now, returning to our initial problem: $$\frac{1}{x} dx +
\left(1+x^{2}y^{2}\right) dy = 0,$$ we note its immediate failure to satisfy the criteria for being an exact
equation since: $$\frac{\partial}{\partial y}\left(\dfrac{1}{x}\right)=\mathbf{0},$$ while $$\frac{\partial}{\
partial x}(1+x^{2}y^{2})=2xy^{2}.$$ Consequently, the search begins for an integrating factor capable of turning
this particular expression into an exact equation. From the earlier discussion, recall the requirement dictated by
Write an educational piece suited for college students related to the following the equality involving partial derivatives: $$\frac{\partial }{\partial y}[\lambda(x, y) (\dfrac{1}{x})] =\frac{\
text snippet: "# Integration factor - First Order Nonlinear ODE I can't seem to partial }{ \partial x}[\lambda(x, y)(1+x^{2}y^{2})].$$ Expansion yields $$\frac{1}{x}\cdot\frac{\partial \lambda}{\
find the proper integrating factor for this nonlinear first order ODE. I have even partial y}=\lambda(2xy)+y^{2}\frac{\partial \lambda}{\partial x}.$$ At present, seeking separation of variables
tried pulling a bunch of substitution and equation-manipulating tricks, but I can't seems judicious—that is, assuming dependency solely upon either $x$ or $y$: \begin{align*} &\implies \frac{1}{x}\
seem to get a proper integrating factor. $$\frac{1}{x}dx + \left(1+x^2y^2\right)dy cdot c_1=\lambda(2y),\\[6pt] &\qquad\text{or}\\[6pt] &c_2 y^{2}=-\frac{\partial \lambda}{\partial x},\tag{$*$} \end
= 0$$ EDIT: Due to MSE users complaining about my lack of proof of work, intent of {align*} where $c_1$, $c_2$ represent constants resulting from integration. Let us consider both possibilities
conceptual understanding, etc, here is exactly why I am stuck. To start off, this separately. **Case 1:** Suppose $\displaystyle\frac{1}{x}\cdot c_1=\lambda(2y)$. We see immediately that no
ODE is obviously inexact: $$\frac{\partial}{\partial y}\left(\frac{1}{x}\right) \ nontrivial solution exists due to the absence of any variable dependencies within the assumed functional form.
neq \frac{\partial}{\partial x}\left(1+x^2y^2\right)$$ And so in order to make this Hence, discard this case. **Case 2:** Now examine $c_2 y^{2}=-\dfrac{\partial \lambda}{\partial x}$. By integrating
exact (if we choose to go down this route) we must (I'll stick to standard both sides with respect to $x$, notice how the constant term disappears because it would merely contribute another
convention/notation) find a function $$\mu$$ such that if we multiply the entire additive constant upon integration: $$-\int \frac{\partial \lambda}{\partial x}\,dx=-c_{3}+\lambda(x)=c_2 y^{2}$$
original ODE by it, we will be able to integrate and solve using 'exact ODE' for some new constant $c_3$. Solving for $\lambda$, we obtain $$\lambda(x,y) = -\int c_2 y^2\,dx + F(y),\label
methods. This is shown as: $$\mu \left(\frac{1}{x}\right)dx + \mu \left(1+x^2y^2\ {integral_of_lambda}$$ where $F(y)$ denotes yet another arbitrary function dependent only on $y$. Before proceeding
right)dy = 0$$ $$\frac{\partial}{" Do not just list concepts, but develop each one further, rewrite Equation $(*)$ above as follows: $$c_2 y^2=2y\,\frac{\partial \lambda}{\partial y}-\frac{\partial
in detail before moving to the next, as we prioritize depth of understanding and \lambda}{\partial x}.$$ Substitute the found integral representation of $\lambda$ back into the revised version of
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: ($*$) derived previously: \begin{align*} &c_2 y^2=2y\,\frac{\partial}{\partial y}\left[\left(-\int c_2 y^2\,dx + F
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an (y)\right)\right]-\frac{\partial}{\partial x}\left[\left(-\int c_2 y^2\,dx + F(y)\right)\right], \\ &=2yc_2 y^
academic, professional and engaging tone that captivates interest. - Application: 2+2yF'(y)-c_2 y^2. \end{align*} Observe now that all terms containing $x$ vanish after taking the derivative with
Incorporate specific, practical examples, such as proofs in calculus or critical respect to $y$; thus, focusing exclusively on those remaining results in \begin{align*} &c_2 y^2=c_2 y^2+2yF'(y),\\
dates and figures in history. Do not include a title or an introduction, simply &\Rightarrow 2yF'(y)=0,\\ &\Rightarrow F(y)=\mathrm{constant}. \end{align*} The discovered constant nature of $F(y)
write the content without headlines and introductory phrases. Do not use images. $ implies no actual dependence on $y$, implying that $\lambda$ depends solely on $x$: $$\therefore~\lambda(x) = -\
int c_2 y^2\,dx.\label{final_lambda}$$ Since our objective involves finding an integrating factor, which should
ideally convert the given inexact equation into an exact equation, substitute the determined $\lambda$ value back
into the cross-partial equation: \begin{align*} \frac{\partial }{\partial y}&[\lambda(x) (\dfrac{1}{x})]=\frac{\
partial }{\partial x}[\lambda(x)(1+x^{2}y^{2})], \\ &\Downarrow\\ \frac{\partial }{\partial y}&\left[\lambda(x) (\
dfrac{1}{x})\right]=y^{2}\,\lambda'(x)+\lambda(x)\,2xy. \end{align*} Recall the known structure of $\lambda$ from
Equation $(\ref{final_lambda})$: \begin{align*} \frac{\partial }{\partial y}&\left[\left(-\int c_2 y^2\,dx\right)
(\dfrac{1}{x})\right]=y^{2}\,\lambda'(x)+\lambda(x)\,2xy, \\ &\Downarrow\\ -\frac{c_2}{x} &=y^{2}\,\lambda'(x)+\
lambda(x)\,2xy. \end{align*} As evident, reaching a standstill occurs during attempts to separate variables or
glean additional insights leading to progression toward a concrete answer. At this juncture, alternative strategies
become essential. Recalling the definition of implicit differentiation might provide assistance: $$\frac{d}{dx}\
left(\lambda(x)-\frac{c_2}{x}\right)=0.$$ Integrating once more yields $$\lambda(x)=\frac{k}{x}+c_2,$$ with $k$
representing an arbitrary integration constant. Unfortunately, neither setting $k=0$ nor incorporating other values
leads to fruitful outcomes concerning the conversion of the initial inexact equation into an exact one via
multiplication by $\lambda(x)$. Thus, resorting to alternative techniques remains imperative. One possible approach
entails considering linear coefficients rather than attempting direct manipulation of the given equation.
Specifically, analyze whether any connection between the given equation and the generalized format $a(x)y'+b(x)y=f
(x)$ holds true. Should this prove successful, subsequent steps involve employing an integrating factor based on
the coefficient functions themselves instead of blindly attacking the original equation itself. However, pursuing
this direction lies beyond the current scope; hence, concluding the investigation herein appears warranted.
Write an educational piece suited for college students related to the following When dealing with quantitative problems on standardized tests like the GMAT, it's essential to understand how
text snippet: "It is currently 17 Nov 2017, 20:20 ### GMAT Club Daily Prep #### different mathematical operations affect variables and their properties. One common question type asks about the
Thank you for using the timer - this advanced tool can estimate your performance integer property of a variable based on certain conditions. This piece will explore the concept of integers,
and suggest more practice questions. We have subscribed you to Daily Prep Questions multiples, factors, and divisibility to help answer a problem asking if y is an integer when x is an integer. An
via email. Customized for You we will pick new questions that match your level integer is a whole number, either positive, negative, or zero, which can be expressed without decimal or fractional
based on your Timer History Track Your Progress every week, we’ll send you an parts. Examples include -3, 0, 5, and 18. Multiples are numbers obtained by multiplying an integer by another
estimated GMAT score based on your performance Practice Pays we will pick new integer; they maintain the same factor regardless of whether it's multiplied or divided within its set of
questions that match your level based on your Timer History # Events & Promotions # multiples. For instance, the first five multiples of 7 are 7, 14, 21, 28, and 35. Factors are numbers that evenly
##### Events & Promotions in June Open Detailed Calendar # If x is an integer, is y divide into another number without leaving any remainder. The factors of 12 are 1, 2, 3, 4, 6, and 12 since all
an integer? new topic post reply Question banks Downloads My Bookmarks Reviews these numbers result in no remainders upon dividing 12. Divisibility refers to whether a number divides evenly into
Important topics Author Message TAGS: ### Hide Tags Manager Joined: 10 Feb 2011 another number, meaning there is no remainder left after division. Now let's consider the original question: "If x
Posts: 112 Kudos [?]: 370 [0], given: 10 If x is an integer, is y an integer? [# is an integer, is y an integer?" To determine whether y must also be an integer under this condition, we need
permalink] ### Show Tags 15 Mar 2011, 16:46 00:00 Difficulty: 65% (hard) Question additional information linking x and y. Without further context, we cannot definitively conclude whether y is an
Stats: 37% (00:56) corre" Do not just list concepts, but develop each one in detail integer because x and y may or may not share a relationship necessary for ensuring y's integer status. Consider two
before moving to the next, as we prioritize depth of understanding and contrasting scenarios: Scenario A: Let x = 3 and y = 9. Here, both x and y are integers, and since 3 is a factor of
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: 9, y being an integer doesn't present a problem. Scenario B: However, suppose x = 3 and y = 3.14. Although x
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an remains an integer, y does not due to its decimal component. In conclusion, knowing only that x is an integer
academic, professional and engaging tone that captivates interest. - Application: provides insufficient evidence to claim that y is necessarily an integer too. Further details regarding the
Incorporate specific, practical examples, such as proofs in calculus or critical connection between x and y would be required to make a conclusive determination. Remember, testing individual cases
dates and figures in history. Do not include a title or an introduction, simply might offer some insight, but rigorous mathematical reasoning demands establishing general rules applicable across
write the content without headlines and introductory phrases. Do not use images. various situations.
Numerical differentiation is a fundamental concept in mathematical analysis and computational science, concerned
with evaluating the derivative of a function at discrete points. This process is essential when dealing with large
datasets or continuous systems represented through grids, as is often the case in engineering, physics, and
computer graphics. In your scenario, you're working with a two-dimensional grid ($$200 imes 200$$) representing a
scalar function and wish to compute its gradient (first-order derivatives along x and y axes) and Laplacian
(second-order derivatives). You've observed limitations in MATLAB's built-in functions due to their reliance on
fewer data points; thus, exploring higher-accuracy methods using finite differences seems prudent. Let us dive
deeper into these topics to enhance your comprehension. Finite Differences: At the core of numerical
differentiation lies the idea of approximating derivatives via finite differences—the discrepancy between
functional values at infinitesimally close points. There exist various techniques to calculate these differences
Write an educational piece suited for college students related to the following depending upon order, stability, and error requirements. Commonly used forms involve central, forward, and backward
text snippet: "High-accuracy numerical differentiation I have a $$200 \times 200$$ differences, primarily differing in stencil size (number of surrounding points involved in approximation) and
matrix representing the values taken by a function over an equally spaced grid. I position relative to the targeted point. For instance, the simplest central difference formula estimating the first
would like to perform derivatives on it. I am interested in its gradient (i.e. its derivative at $(i, j)$ reads: $$f\_x(i,j) approx frac{f(i+1,j)-f(i-1,j)}{2h}$$ where $h$ denotes the uniform step
derivative in direction $$x$$ and its derivative in direction $$y$$) and in its size in both directions. Likewise, the second-order central difference expression for the Laplacian becomes: $$∇^2f
Laplacian. I work on the Matlab platform, where I use the built-in functions (i,j) approx frac{f(i+1,j)+f(i-1,j)+f(i,j+1)+f(i,j-1)-4f(i,j)}{h^2}.$$ As you rightfully pointed out, increasing
gradient and del2 which work very well but their accuracy is limited by the fact stencil sizes leads to enhanced precision owing to reduced truncation errors inherent in lower-order
that they make use of a small number of points. Reading this Wikipedia page about approximations. However, larger stencils might introduce additional complications linked to boundary conditions,
Finite difference coefficients, one can understand that there are finite-difference numerical instability, and implementation complexity. Therefore, choosing optimal finite difference formulations
schemes to perform numerical differentiation in a more accurate way. The price to necessitates careful consideration balancing desired accuracy against potential challenges arising from intricate
pay to have an increased accuracy is of course to use more complex formulas, which expressions involving numerous neighboring points. Libraries & Functions: While implementing custom numerical
include more points. I would like to know if there is a library where there are differentiation routines may be insightful, reusing existing high-quality libraries provides several benefits
functions capable of doing this job which -I repeat- is: computing the first-order including efficiency, robustness, maintainability, and decreased development time. Fortunately, many powerful tools
and the second-order derivativ" Do not just list concepts, but develop each one in cater specifically to numerical computation tasks encompassing multidimensional arrays and advanced linear algebra
detail before moving to the next, as we prioritize depth of understanding and operations. Some prominent options compatible with MATLAB include: 1. Curvilinear Coordinates Toolbox: A collection
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: of utilities facilitating coordinate transformations, interpolation, visualization, and tensor calculations across
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an arbitrary orthogonal curvilinear grids. Notably, it incorporates sophisticated finite-difference algorithms
academic, professional and engaging tone that captivates interest. - Application: enabling precise evaluation of gradients, divergences, curls, and other differential operators. 2. Chebfun: An
Incorporate specific, practical examples, such as proofs in calculus or critical open-source package specializing in numerical solutions of differential equations, polynomial interpolations,
dates and figures in history. Do not include a title or an introduction, simply quadratures, and function representations based on adaptive Chebyshev expansions. It offers extensive support for
write the content without headlines and introductory phrases. Do not use images. array-based computations alongside automatic differentiation features ideal for tackling challenging multi-domain
problems requiring superior accuracy levels. 3. Diffpack: Another versatile object-oriented C++ framework developed
explicitly for solving partial differential equations utilizing diverse discretization strategies ranging from
finite differences to finite elements and spectral methods. Although predominantly focused on PDE solvers, it also
includes utility classes pertinent to generalized numerical differentiation and vector calculus. In conclusion,
pursuing high-accuracy numerical differentiation entails mastering underlying principles centered around finite
difference techniques while judiciously selecting appropriate libraries offering specialized functionality tailored
towards demanding applications. By adhering to best practices outlined above, you will significantly improve
computed results' fidelity compared to standard approaches leveraging insufficient sampling points. Moreover,
familiarity with refined procedures fosters better appreciation of subtle tradeoffs associated with algorithm
selection, thereby sharpening overall problem-solving skills germane to modern scientific research and engineering
In the study of topology, a fundamental concept is that of an "open cover." An open cover of a set within a
Write an educational piece suited for college students related to the following topological space is a collection of open sets whose union contains the entire set. The question arises as to
text snippet: "# Existence of Open Covers Do sets always have open covers exist? I whether open covers always exist for all sets within a topological space. The answer is yes, and this can be
know they are not always finite, but do infinite ones always exist? I was reading demonstrated through rigorous mathematical argumentation. To begin, let us recall the definition of a topological
baby rudin and the proofs for non-relative nature for compactness seems to require space. A topological space consists of a set X together with a collection T of subsets of X satisfying certain
that. But I couldn't find any explanations on why I can assume that open covers conditions (namely, that both the empty set and X itself belong to T, that the intersection of any finite number of
always exist. - Open covers do always exist, and in fact it will always be possible elements of T also belongs to T, and that the union of any arbitrary number of elements of T belongs to T). Each
to find a finite one. If $(X, \tau)$ is a topological space, then by definition, element of T is referred to as an open set. Now suppose we have a subset A of our topological space X. We claim
$X$ is open. So if $A \subset X$ is any subset, then $\{X\}$ is a finite open cover that {X} is always a finite open cover of A. This follows directly from the definition of a topological space -
of it. The point is, that in order to be compact, every open cover has to have a since X is an element of T, it is an open set; and clearly, the union of {X} contains A, so indeed {X} is an open
finite subcover. But that doesn't stop there being some finite cover for any cover of A. Moreover, it is a finite open cover, consisting only of the single set X. At this point, you may be
non-compact set. - One of the properties of open sets is that each point lies wondering what role the notion of compactness plays here. Compactness is a property possessed by certain
within at least one open set. If $A\subseteq X$, then for each $a\in A$ there is topological spaces which ensures that every open cover possesses a finite subcover. However, even for non-compact
some open set $U$ such that $a\in U$, and in other words $A\cap U\neq\varnothing$. sets, it remains true that open covers must exist. It is merely the existence of a finite subcover which need not
So taking $\{U\subseteq X\mid" Do not just list concepts, but develop each one in hold. Finally, let us consider another perspective on open covers. One of the defining properties of open sets is
detail before moving to the next, as we prioritize depth of understanding and that each point of the space lies within at least one open set. More formally, given any point x belonging to an
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: open set U, there exists some open set containing x as a subset. Now suppose we have a subset A of our topological
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an space X. For every element a of A, there exists some open set U_a containing a as a subset. Then the collection
academic, professional and engaging tone that captivates interest. - Application: {U_a | a ∈ A} forms an open cover of A. Again, note that this open cover is potentially infinite, depending on the
Incorporate specific, practical examples, such as proofs in calculus or critical size and structure of A. In summary, open covers always exist for subsets of topological spaces. They can be
dates and figures in history. Do not include a title or an introduction, simply constructed trivially using the whole space X, or alternatively, using the defining property of open sets
write the content without headlines and introductory phrases. Do not use images. themselves. While not all open covers necessarily contain a finite subcover, their mere existence still holds
important implications for the study of topology.
Equality in mathematics is a fundamental concept that underpins much of our ability to understand and work with
numerical and algebraic relationships. At its core, equality represents the idea that two mathematical entities
possess the same value or represent the same mathematical object. This foundational notion allows us to compare,
contrast, manipulate, and ultimately better comprehend various aspects within mathematical systems. To truly grasp
this crucial principle, let's delve into some nuances and applications associated with equality in mathematics.
Firstly, it is essential to distinguish between extensional equality and intensional equality. Extensional equality
Write an educational piece suited for college students related to the following refers to comparing objects by examining all of their components; if these components match exactly, then the
text snippet: "TheInfoList OR: In mathematics Mathematics is an area of knowledge objects themselves are considered equal. For example, consider sets A = {1, 2} and B = {2, 1}. Although arranged
that includes the topics of numbers, formulas and related structures, shapes and differently, both sets contain identical elements, making them extensionally equivalent. Intensional equality,
the spaces in which they are contained, and quantities and their changes. These however, focuses on whether two objects share the exact definition or properties, regardless of their constituent
topics are represented in modern mathematics ... , equality is a relationship parts. An illustration lies in functions f(x)=x^2+1 and g(x)=(x+i)(x-i). Though expressed differently, they both
between two quantities or, more generally two mathematical expression In describe the complex parabola opening upward and intercepting the real axis at y=1—hence being identically defined
mathematics, an expression or mathematical expression is a finite combination of despite distinct forms. Thus, f(x) and g(x) exhibit intensional equality. Understanding these distinctions enhances
symbols that is well-formed according to rules that depend on the context. clarity when working through intricate problems involving multiple mathematical constructs. Another vital aspect
Mathematical symbols can designate numbers ( constants), variables, operations, ... pertains to reflexivity, symmetry, and transitivity inherently linked to equality relations. Reflexivity denotes
s, asserting that the quantities have the same value, or that the expressions self-equivalence, where any entity e is always equivalent to itself (e = e); symmetry implies equivalence remains
represent the same mathematical object A mathematical object is an abstract concept unchanged upon switching places (if a = b, then b = a); finally, transitivity signifies preserved equivalence
arising in mathematics. In the usual language of mathematics, an ''object'' is across successive replacements (should a = b and b = c, then a equals c too). Mastery of these qualities solidifies
anything that has been (or could be) formally defined, and with which one may do comprehension of how equality operates within diverse mathematical settings. Equipped with rudimentary familiarity
deductive reasoning and mathematical ... . The equality b" Do not just list of what constitutes mathematical equality, let us now explore several salient instances showcasing application
concepts, but develop each one in detail before moving to the next, as we thereof: solving equations in calculus and verifying trigonometric identities. Solving equations often necessitates
prioritize depth of understanding and comprehensive exploration of the subject applying inverse operations while preserving equality throughout the process. Take, for instance, finding roots of
matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the concepts/ polynomial quartic x^4 - 3x^3 + 5x^2 - 7x + 9 = 0. By employing substitution, factorization, and root isolation
sections. - Engagement: Write with an academic, professional and engaging tone that techniques sequentially, we arrive at solutions: x = {1, 2, (-1 ± i sqrt(3)) / 2}. Here, maintaining rigorous
captivates interest. - Application: Incorporate specific, practical examples, such adherence to upholding equality ensures accuracy during computations and permits validation post-resolution.
as proofs in calculus or critical dates and figures in history. Do not include a Verifying trigonometric identities exemplifies another avenue whereupon demonstrating equivalency plays a pivotal
title or an introduction, simply write the content without headlines and role. Consider validating sin²(θ)+cos²(θ) ≡ 1: sin²(θ)+cos²(θ) = (1-cos²(θ)) + cos²(θ) (Pythagorean identity
introductory phrases. Do not use images. applied) = 1*(1-cos²(θ)+cos²(θ)) (Associative property utilized here) = 1*1 (Simplification ensues) ∴ sin²(θ)+cos²
(θ) ≡ 1 Throughout this exercise, meticulous attention was paid towards retaining equivalencies via appropriate
manipulations until arriving at the desired conclusion. To recapitulate, equality serves as a cornerstone pillar
sustaining structural integrity amidst multitudinous facets pervading mathematical discourse. Whether juxtaposing
extensional vs. intensional comparisons, acknowledging underlying reflexivity, symmetry, and transitivity
attributes, or implementing problem-solving strategies rooted in equation resolution or identity verification,
mastery of this topic propels learners further along their journey towards becoming proficient mathematicians
capable of dissecting increasingly sophisticated dilemmas.
Let us begin by discussing the concept of an arithmetic progression, which is central to this topic. An arithmetic
progression is a sequence of numbers in which the difference between any two successive members is constant. In
Write an educational piece suited for college students related to the following this case, our sequence ${a\_m}$ is defined recursively as ${a\_{m+1}} = {a\_m} + b{m}$, where $b$ is some fixed
text snippet: "# Help on an application of Dirichlet's theorem for primes in integer. Thus, every term in our sequence is obtained by adding $b$ times its position to the previous term. Note
progression Suppose that I have an infinite sequence of positive integers $$a_1,\ that when $b=1,$ our sequence becomes ${a\_m} = \frac{{m(m+1)}}{2},$ which starts as ${1,3,6,10,...}.$ Now, let us
ldots,a_m,\ldots$$ with the following recursion $$a_{m+1} -a_m =b(m+1)$$ So that examine the condition $(b,{a\_m}) = 1.$ Here, $(\cdot,\cdot)$ denotes the greatest common divisor function, so this
$$a_{m+1} =b(m+1) +a_m$$ Suppose further that I know that $$(b,a_m)=1$$ Would condition states that $b$ and ${a\_m}$ share no common factors other than 1. It ensures that the terms in our
Dirichlet's Theorem tell us that there are infinitely many primes of the form $a_ sequence do not all share a factor because, otherwise, they cannot be prime. Indeed, if $c > 1$ divides both $b$
{m+1}$ • Is there a typo? The two displayed equations are not equivalent. – André and ${a\_m},$ then $c$ would divide every ${a\_n}$ with $n \geqslant m,$ making them composite. By requiring $(b,{a
Nicolas Jul 15 '16 at 2:15 Take $b=1$. Then, (with $a_1=1$), $a_{m+1}=1+2+\cdots + \_m}) = 1,$ we avoid this issue. Next, we must understand what Dirichlet's Theorem for Primes in Arithmetic
(m+1)$. This is non-prime for $m\gt 1$, since it is $\frac{(m+1)(m+2)}{2}$. • For Progressions entails. Stated informally, this theorem says that given any relatively prime positive integers $a$
general $b$ and $a_1$, we will have that $a_{m+1}$ is a quadratic in $m$. With such and $q,$ the arithmetic progression $\{a, a+q, a+2q, ...\}$ contains infinitely many prime numbers. Relatively
quadratics, if there is no obvious reason for the quadratic not to represent prime means that $a$ and $q$ share no common factor greater than 1, denoted as $(a, q) = 1.$ So, returning to our
infinitely many primes, it probably does. There are even conjectured estimates for original problem, if we could rewrite ${a_{m+1}}$ as $a +mq$ for constants $a$ and $q,$ and show that $(b,{a\_m}) =
the number of such primes $\le n$, which for a number of polynomials that have been 1,$ implies $(b,{a_{m+1}}) = 1,$ then Dirichlet's Theorem guarantees infinitely many primes of the desired form.
examined fit the experimental fa" Do not just list concepts, but develop each one However, observe that André Nicolas points out a discrepancy in your initial question regarding whether ${a_{m+1}}$
in detail before moving to the next, as we prioritize depth of understanding and should equal $bm + a$ or $bm + {a_m}.$ These expressions yield different sequences; thus, clarifying this
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: distinction is crucial. Moreover, whichever version you choose impacts how we proceed with applying Dirichlet's
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an Theorem. Assuming you meant ${a_{m+1}}=bm+{a_m},$ note that ${a_{m+1}}$ can be written as $a+mq,$ where $a={a_1}$
academic, professional and engaging tone that captivates interest. - Application: and $q=b.$ However, showing $(b,{a_m}) = 1$ implying $(b,{a_{m+1}}) = 1$ requires proving that changing $m$ does
Incorporate specific, practical examples, such as proofs in calculus or critical not introduce new factors shared by $b$ and ${a_{m+1}}.$ Unfortunately, this may not hold depending on the choice
dates and figures in history. Do not include a title or an introduction, simply of ${a_1}$ and $b.$ Therefore, while Dirichlet's Theorem indeed promises infinitely many primes in certain
write the content without headlines and introductory phrases. Do not use images. arithmetic progressions, extra care must be taken when selecting appropriate values for ${a_m}$ and $b.$ In
summary, we discussed arithmetic progressions, gcd conditions, and Dirichlet's Theorem, revealing complexities in
directly applying his theorem here. While fascinating patterns might emerge among polynomial sequences like ours,
rigorous mathematical arguments demand meticulous attention to details concerning their construction.
Compound interest is a powerful mathematical concept that describes how an initial principal amount grows over time
when interest is added regularly. The formula for calculating the present value (PV) of a future sum using compound
Write an educational piece suited for college students related to the following interest takes into account four key variables: the future value (FV), the annual interest rate (r), the number of
text snippet: "# Compound Interest (PV) Calculator ## Calculates the present value years (n), and the frequency at which the interest is compounded per year (k). Understanding these components and
using the compound interest method. Annual interest rate % (r) nominal effective their relationship is crucial for solving complex financial problems. The first variable, FV, represents the total
Future value (FV) Number of years (n) Compounded (k) anuually semiannually amount of money you expect to receive in the future. For example, imagine you plan to save $10,000 by the end of
quarterly monthly daily five years. This lump sum is your future value. Next, r denotes the annual interest rate expressed as a decimal.
6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit When comparing different investment opportunities, it's essential to understand whether the given interest rates
Present value (PV) $\normal Compound\ interest\ method\\[10](1)\ PV={\large\frac are nominal or effective. A nominal interest rate does not consider the impact of compounding within the same year,
{FV}{(1+{\large\frac{r}{k}})^{nk}}}\hspace{35}r:\ nominal\ rate\\(2)\ PV={\large\ whereas the effective interest rate incorporates this effect. To convert from a nominal interest rate to its
frac{FV}{(1+R)^{\normal{n}}}}\hspace{30}R:\ effective\ rate\\[10]\hspace{75} k:\ equivalent effective rate, use the equation \(\text{{Effective Rate}} = {\left({1 + \dfrac{{\text{{Nominal Rate}}}}
compounded\ times\ a\ year\\$ Compound Interest (PV) [1-10] /12 {k}}\right)^k - 1}\). For instance, let's say you have two savings accounts offering 5% annual interest; however,
Disp-Num5103050100200 [1] 2018/09/24 05:58 Female / 20 years old level / Account A compounds interest quarterly while Account B compounds only once annually. Although both offer the same
High-school/ University/ Grad student / Very / Purpose of use I needed to find the nominal interest rate, Account A's effective interest rate will be higher due to more frequent compounding. Here, k
solution to a homework problem. Comment/Request It would be wonderful if it showed refers to the number of times the interest is compounded per year. Common values of k include 1 (annual), 2
what numbers and keys are entered into the financial calculator. It is necessary (semiannual), 4 (quarterly), 12 (monthly), and even 365 (daily). Now, n signifies the length of time—typically
for me to use the formulas, just t" Do not just list concepts, but develop each one measured in whole numbers representing years—over which the growth occurs. Using our previous example, if you want
in detail before moving to the next, as we prioritize depth of understanding and to accumulate the targeted $10,000 in three years instead of five, then n equals 3. Finally, armed with knowledge
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: about FV, r, and n, we can calculate the present value (PV) of an asset today based on its expected future worth
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an using either formula provided in the original text snippet: * \(PV=\dfrac{FV}{\left(1+\dfrac{r}{k}\right)^{nk}}\):
academic, professional and engaging tone that captivates interest. - Application: Utilize this version when dealing with a stated nominal interest rate and knowing how many times it gets compounded
Incorporate specific, practical examples, such as proofs in calculus or critical per year. * \(PV=\dfrac{FV}{\left(1+R\right)^n}\): Opt for this variant when working with an effective interest
dates and figures in history. Do not include a title or an introduction, simply rate already adjusted for compounding frequency. By inputting appropriate numerical values into these equations,
write the content without headlines and introductory phrases. Do not use images. you can determine precisely how much you need to invest now to achieve a desired future return. Keep practicing
these calculations until they become second nature because mastery of compound interest principles leads to
informed decision-making regarding loans, investments, pensions, annuities, and other vital aspects of personal
finance management.
Let's begin by unpacking the problem presented. We have a box containing three coins - one is two-headed (coin A),
another is a fair coin (coin B), and the last one is a biased coin that shows heads 75% of the time (coin C). One
coin was randomly chosen from this assortment and flipped, resulting in heads. Our objective is to calculate the
probability that the selected coin was the two-headed coin based on this information. To do so, let's delve into
the Total Probability Theorem (TPT) concept. The TPT enables us to compute the overall likelihood of an event by
Write an educational piece suited for college students related to the following summing up the products of individual probabilities and their associated conditional probabilities. Mathematically
text snippet: "# Tagged: total probability theorem ## Problem 739 There are three speaking, if there exists a set of mutually exclusive events E1, E2, ... En covering the sample space S, then given
coins in a box. The first coin is two-headed. The second one is a fair coin. The any event A within S, the TPT states: P(A) = P(E1) * P(A|E1) + P(E2) * P(A|E2) + ... + P(En) * P(An|En) In our
third one is a biased coin that comes up heads $75\%$ of the time. When one of the case, we can apply the TPT using three possible events corresponding to selecting each coin type: Event D:
three coins was picked at random from the box and tossed, it landed heads. What is Selecting the two-headed coin (coin A) Event F: Selecting the fair coin (coin B) Event J: Selecting the biased coin
the probability that the selected coin was the two-headed coin?" Do not just list (coin C) By definition, these events form a partition of the sample space since only one coin will be drawn; hence
concepts, but develop each one in detail before moving to the next, as we they are pairwise disjoint and exhaustive. Now, applying the TPT yields: P(Heads) = P(D)*P(Heads|D) + P(F)*P(Heads|
prioritize depth of understanding and comprehensive exploration of the subject F) + P(J)*P(Heads|J) Let's denote PD,PF, and PJ as the respective selection probabilities for coins A, B, and C.
matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the concepts/ Also, recall that the flipped coin showed heads. Since both coins A and C always land on heads when used, while
sections. - Engagement: Write with an academic, professional and engaging tone that coin B does so half the time, we get: P(Heads | D) = 1 (since coin A is two-headed) P(Heads | F) = 0.5 (since coin
captivates interest. - Application: Incorporate specific, practical examples, such B is fair) P(Heads | J) = 0.75 (since coin C lands on heads 75% of the time) Now, since all three coins have equal
as proofs in calculus or critical dates and figures in history. Do not include a chances of being selected, we obtain: PD = PF = PJ = 1/3 Substituting these values back into our TPT equation
title or an introduction, simply write the content without headlines and gives: P(Heads) = (1/3)*1 + (1/3)*0.5 + (1/3)*0.75 = 0.333... + 0.166... + 0.25 ≈ 0.7498 However, what we actually
introductory phrases. Do not use images. want is P(D | Heads): the probability that the selected coin was indeed coin A, given that it landed on heads. By
Bayes' rule, which extends the TPT to incorporate new evidence, we express P(D | Heads) as follows: P(D | Heads) =
P(Heads | D) * P(D) / P(Heads) = (1/3)/0.7498 ≈ 0.4013 Thus, despite having seen a single heads outcome, there
remains roughly a 40.13% chance that the chosen coin was indeed the two-headed one! This underscores how even
seemingly conclusive evidence might still leave room for significant uncertainty, particularly when accounting for
underlying probabilities.
Write an educational piece suited for college students related to the following To begin, let's examine the original equation: k(k+1)(k+2)/3 + (k+1)(k+2) = 1/3(k+1)(k+2)(k+3) Our goal is to
text snippet: "# Need help simplifying one expression into another: $\frac{k(k+1) transform the left side of this equation so that it matches the right side. The key insight here is to recognize
(k+2)}{3} + (k+1)(k+2) = \frac{1}{3}(k+1)(k+2)(k+3)$ I am looking through the mark that the expression (k+1)(k+2) appears in both terms on the left side of the equation. This observation suggests
scheme of a past A level paper and I cannot work out how they have simplified an that there may be an opportunity to factor out this term. To proceed, let's first find a common denominator for the
expression: $$\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{1}{3}(k+1)(k+2)(k+3)$$ Can two terms on the left side. Since one term has a denominator of 3 and the other term does not have any explicit
anyone walk me through it? Thanks. • Consider getting a common denominator then denominator, we can take 3 as our common denominator. Thus, we rewrite the left side of the equation as follows: [k
factoring out like terms. – kmeis Jan 26 '17 at 18:05 We have: $$\frac{k(k+1)(k+2)} (k+1)(k+2)] / 3 + [3 × (k+1)(k+2)] / 3 Now, we can see that the fraction on the left side shares the same
{3}+(k+1)(k+2)$$ Then: $$\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3}$$ We can now denominator as the number 3 multiplying the second term. Therefore, we can add these two terms together by adding
combine the two fractions: $$\frac{\color{blue}{k}\color{#cc0000}{(k+1)(k+2)}+\ their numerators while keeping the same denominator: [(k(k+1)(k+2)) + (3 × (k+1)(k+2))] / 3 Next, notice that the
color{blue}{3}\color{#cc0000}{(k+1)(k+2)}}{3}$$ Note that $(k+1)(k+2)$ are alike on quantity (k+1)(k+2) occurs twice in the numerator of the resulting fraction. As mentioned earlier, this repetition
both. Hence, we can factorise: $$\frac{\color{blue}{(k+3)}\color{#cc0000}{(k+1) suggests that we might be able to factor something out from the numerator. Indeed, if we do so, we obtain: [(k+3)
(k+2)}}{3}$$ Which is the answer given. Take $(k+1)(k+2)$ common, we have - $(k+1) ((k+1)(k+2))] / 3 At this point, we can compare the result above to the desired form on the right side of the
(k+2) \left[\frac k3 + 1\right]$ = $(k+1)(k+2) \left[\frac{k+3}{3}\right]$ = $\ original equation. Observe that the numerator of our current result differs only by a multiplicative constant from
frac13(k+1)(k+2)(k+3)$ $\begin{a" Do not just list concepts, but develop each one the numerator on the right side. Specifically, dividing the numerator in the previous line by (k+1)(k+2) yields
in detail before moving to the next, as we prioritize depth of understanding and k+3, which is exactly what appears in the numerator of the desired form. Thus, all that remains is to divide the
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: entire numerator and denominator on the left side by (k+1)(k+2), yielding: [1/((k+1)(k+2))][(k+3)((k+1)(k+2))]/3
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an Canceling the common factors between the numerator and denominator in the first bracketed term, we get: (k+3)/3
academic, professional and engaging tone that captivates interest. - Application: Finally, multiplying this last expression by (k+1)(k+2), as required on the right side of the original equation,
Incorporate specific, practical examples, such as proofs in calculus or critical gives us: (k+3)(k+1)(k+2)/3 This confirms that the left and right sides of the original equation are indeed
dates and figures in history. Do not include a title or an introduction, simply equivalent forms of the same algebraic expression.
write the content without headlines and introductory phrases. Do not use images.
The slew rate of an operational amplifier (op-amp) refers to the maximum rate of change of its output voltage per
unit of time. This specification indicates how quickly an op-amp can respond to changes in its input voltage. For
instance, if an op-amp has a slew rate of 0.5 V/µs, it means its output voltage can increase or decrease by 0.5
Write an educational piece suited for college students related to the following volts every microsecond. The relationship between the slew rate and frequency response of an op-amp is crucial when
text snippet: "# The slew rate for a 741 is 0.5 V / μ second. The combination of dealing with high-frequency signals. As the amplitude of a sinusoidal waveform increases, so does its slope, which
maximum frequencies for an undistorted sine-wave output of 10 V peak and 1 V peak requires a higher slew rate for accurate reproduction. If the slew rate is insufficient, the output waveform will
is approximately:(1) 8 kHz and 80 kHz(2) 48 kHz and 4.8 kHz(3) 20 kHz and 2 kHz(4) become distorted due to the slow response of the op-amp. Consequently, determining the minimum slew rate required
2 kHz and 20 kHz This question was previously asked in UGC NET Paper 2: Electronic for faithful replication of a desired waveform is essential. To calculate the slew rate necessary for a particular
Science Nov 2020 Official Paper View all UGC NET Papers > 1. 1 2. 2 3. 3 4. 4 application, consider a sinusoidal output voltage with a peak value of Vo_peak and a frequency of f: Vo\_peak =
Option 1 : 1 Free Official Paper 1: Held on 24 Sep 2020 Shift 1 9740 50 Questions Vmax - Vmin where Vmax and Vmin are the positive and negative peaks of the sinusoidal waveform, respectively. For a
100 Marks 60 Mins ## Detailed Solution Slew Rate: • The output of an operational full cycle (T), the angular velocity (ω) is defined as: ω = 2πf During half of the period T/2, the voltage swings
amplifier can only change by a certain amount in a given time this limit is called from zero to Vo\_peak or vice versa, covering half of the total phase angle (π). Therefore, dV/dt during this
the slew rate of the op-amp. • Operational amplifier slew rate can limit the transition represents the slew rate needed for accurately rendering the sinusoidal waveform. Applying basic
performance of a circuit if the slew rate requirement is exceeded. • It can distort trigonometry, we get: Vo\_peak = Vmax × sin(ωt) At t=T/2, sin(ωt)=1; thus, Vo\_peak = Vmax. During this interval,
the waveform and prevent the input signal from being faithfully represented at the dv/dt becomes: dv/dt = d(Vo\_peak)/dt = d(Vmax×sin(ωt))/dt Since Vmax is constant, d(Vmax)/dt = 0, leaving us with:
output if the slew rate is exceeded. Mathematically, Slew Rate (SR) is calculated dv/dt = Vmax × cos(ωt) × ω Substituting t=T/2 and solving further: dv/dt = Vmax × cos(π/2) × ω Since cos(π/2) = 0,
as: $$SR = \frac{{dV}}{{dt}}$" Do not just list concepts, but develop each one in theoretically, there should be no need for slewing at this point. However, real-world scenarios introduce
detail before moving to the next, as we prioritize depth of understanding and non-idealities like finite rise times requiring some slew rate. To address these issues, let's assume a small but
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: nonzero error term (ε): cos(π/2 + ε) ≈ ε Now our expression looks like: dv/dt = Vmax × ε × ω Given that slew rates
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an are usually expressed in units of volts per second or volts per microsecond, we must convert angular velocity into
academic, professional and engaging tone that captivates interest. - Application: seconds^-1 using the relation: ω = 2πf Thus, dv/dt = Vmax × ε × 2πf Rearranging terms to isolate f yields: f <= (dv
Incorporate specific, practical examples, such as proofs in calculus or critical /dt) / (Vmax × ε × 2π) Typical values for ε range around 0.01 to 0.001 radians. Thus, assuming ε=0.001 radians and
dates and figures in history. Do not include a title or an introduction, simply applying the provided slew rate of 0.5 V/µs along with Vmax=10 V and Vmax=1 V results in two possible sets of
write the content without headlines and introductory phrases. Do not use images. frequencies meeting the criteria: Scenario A: Vmax=10 V Frequency F1 = (0.5 V/µs) / (10 V × 0.001 rad × 2π) ≈ 8 kHz
Frequency F2 = (0.5 V/µs) / (1 V × 0.001 rad × 2π) ≈ 80 kHz Scenario B: Vmax=1 V Frequency F1 = (0.5 V/µs) / (1 V ×
0.001 rad × 2π) ≈ 80 kHz Frequency F2 = (0.5 V/µs) / (1 V × 0.001 rad × 2π) ≈ 80 kHz However, Scenario B violates
the condition where F1 must be less than F2, making it invalid. Hence, based on the provided information, the
correct answer would be option (A): approximately 8 kHz and 80 kHz.
The Fast Fourier Transform (FFT) is a powerful tool for analyzing data in various fields such as physics,
Write an educational piece suited for college students related to the following engineering, and computer science. It allows us to efficiently compute the discrete Fourier transform (DFT), which
text snippet: "# fftlog #### 01 October 2016 If you are looking for a fast Fourier decomposes a signal into its frequency components. However, traditional FFT algorithms may encounter difficulties
transform (FFT) that works flawlessly over many orders of magnitudes you might want when dealing with signals that span many orders of magnitude, especially when the dynamic range is large. This is
to search for a logarithmic FFT. And if you do search for a logarithmic FFT, you where logarithmic FFT comes into play. Instead of using a uniform grid in the linear domain, logarithmic
most likely will end up at Andrew Hamilton’s website for FFTLog (published in FFTransformLog uses a logarithmically spaced grid, making it more suitable for handling data with exponential
Appendix B of Hamilton, 2000). Because, as it states on the website, “FFTLog can be growth or decay. One popular implementation of logarithmic FFT is FFTLog by Andrew Hamilton, available on his
regarded as a natural analogue to the standard Fast Fourier Transform … [for] a personal website. As stated in the quote, "FFTLog can be regarded as a natural analogue to the standard Fast
logarithmically spaced periodic sequence“. FFTLog is written in Fortran. And if you Fourier Transform ... [for] a logarithmically spaced periodic sequence." With its roots in Appendix B of Hamilton's
followed this blog than you might be aware that I do most of my scripting in 2000 publication, FFTLog has become a widely adopted solution among researchers working with multi-scale phenomena.
Python. I used FFTLog a while ago during my Ph.D. Last week I needed for a project Although FFTLog is implemented in Fortran, there are ways to integrate it seamlessly with other programming
again a logarithmically spaced FFT routine, so I dug out my old files. I improved languages like Python—a language commonly employed by scientists nowadays due to its flexibility and extensive
the wrapping slightly over my old version and moved them from Python 2 to Python 3. library support. To make the connection between Fortran and Python, the `f2py` module was introduced; fortunately,
For the wrapping I use the great f2py-module, baked right into numpy these days. In it is already bundled within NumPy packages, easing the integration process significantly. Using `f2py`, users can
order to be faster up and r" Do not just list concepts, but develop each one in create wrapper functions around existing Fortran routines, effectively extending their functionality to Python
detail before moving to the next, as we prioritize depth of understanding and applications. The author of the quoted post had previously utilized FFTLog in their PhD studies and recently
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: required a similar logarithmically spaced FFT function for another project. After revisiting their archived
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an codebase, they decided to enhance the original Python interface and port it from Python 2 to Python 3. By doing so,
academic, professional and engaging tone that captivates interest. - Application: the researcher ensured compatibility with modern scientific computing environments while retaining the core
Incorporate specific, practical examples, such as proofs in calculus or critical capabilities provided by FFTLog. To summarize, logarithmic FFT plays a crucial role in processing signals
dates and figures in history. Do not include a title or an introduction, simply encompassing wide ranges of scales, particularly those exhibiting exponential behavior. Implementations like FFTLog
write the content without headlines and introductory phrases. Do not use images. offer reliable solutions backed by solid mathematical foundations and demonstrated performance benefits.
Furthermore, integrating such tools with high-level programming languages like Python enables broader communities
to leverage these techniques in diverse domains, enhancing interdisciplinary collaborative efforts.
Exponential functions are mathematical expressions defined by the formula f(x) = ab^x, where 'a' is a constant, 'b'
is the base (positive number greater than zero but not equal to one), and x is any real number. These functions
have unique properties compared to linear or polynomial functions, especially because their rates of change vary
depending on the value of x. This tutorial will guide you through completing tables for two given exponential
functions, determining the multipliers, and writing equations based on the completed tables. Let us begin! First,
Write an educational piece suited for college students related to the following let's examine the first example provided: | x | f(x) | | --- | --- | | -1 | 3 | | 0 | 15 | | 1 | 75 | | 2 | 375 | |
text snippet: "Home > INT1 > Chapter 8 > Lesson 8.2.2 > Problem8-112 8-112. Each 3 | 1875 | To fill out this table, start by finding the initial value (when x=0). Here, it is clear from the table
table below represents an exponential function in $f(x) = ab^x$ form. Copy and that f(0)=15. Next, identify the common ratio between consecutive terms; this is called the "multiplier." To do so,
complete each table on your paper and then write an equation for the function. 1. divide any term by its preceding term. For instance, f(1)/f(0) = 75 / 15 = 5. Now, using this multiplier, calculate
$x$ $f(x)$ $-1$ $3$ $0$ $1$ $75$ $2$ $3$ What is the multiplier? $y = 15 · 5^{x}$ subsequent values: * When x=-1, f(-1) = f(0)*(1/5) = 15\*(1/5) = 3 * Similarly, when x>1, multiply the previous
$x$ $f(x)$ $-1$ $3$ $0$ $\mathbf{15}$ $1$ $75$ $2$ $\mathbf{375}$ $3$ $\mathbf result by the multiplier to find f(x): f(2) = f(1)\*5 = 75\*5 = 375, and so forth. Once the table has been filled
{1875}$ 1. $x$ $f(x)$ $0$ $1$ $2$ $96.64$ $3$ $77.312$ $4$ $x$ $f(x)$ $0$ $\mathbf completely, determine the general pattern governing these numbers. Notice how the values relate directly to the
{151}$ $1$ $\mathbf{120.8}$ $2$ $96.64$ $3$ $77.312$ $4$ $\mathbf{\approx 61.85}$" expression 15 \* 5^x – here, our initial value is 15 (the coefficient), and the base is 5. Thus, the equation
Do not just list concepts, but develop each one in detail before moving to the representing this particular exponential relationship is y = 15 * 5^x. Now, consider the second example presented:
next, as we prioritize depth of understanding and comprehensive exploration of the | x | f(x) | |---|---| | 0 | 151 | | 1 | 120.8 | | 2 | 96.64 | | 3 | ~77.312 | | 4 | ~61.85 | As done previously,
subject matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the locate the starting point at x=0; here, f(0)=151. However, unlike the first example, there isn't an integer power
concepts/sections. - Engagement: Write with an academic, professional and engaging of a single base that yields these exact results. Instead, observe how closely the sequence resembles geometric
tone that captivates interest. - Application: Incorporate specific, practical decay towards zero. We can approximate the underlying exponential relationship by iteratively adjusting our
examples, such as proofs in calculus or critical dates and figures in history. Do candidate base until the calculated values align well with those given in the table. After several trials, we
not include a title or an introduction, simply write the content without headlines discover that using approximately b≈0.88 provides reasonably accurate estimates. However, due to rounding errors
and introductory phrases. Do not use images. introduced during calculations, the resulting decimal approximations may differ slightly from the actual tabulated
values. Nevertheless, within acceptable error bounds, this approximation holds true: f(x) ≈ 151 \* (0.88)^x In
summary, constructing and interpreting exponential functions involve filling out tables using identified bases and
initial conditions, deriving patterns, and forming compact representations encapsulating these relationships. While
some problems might require direct application of standard forms like y = ab^x, others necessitate creative problem
solving involving estimation techniques. Regardless, mastering the fundamentals discussed above equips you with
powerful tools necessary for tackling various quantitative disciplines.
Ring Homomorphisms and Kernels In abstract algebra, a homomorphism is a mapping between two algebraic structures
that preserves their operations. A ring homomorphism specifically refers to a map between two rings that respects
addition, multiplication, and the additive identity (0). Given a ring homomorphism φ : R -> S, there are several
Write an educational piece suited for college students related to the following important properties we can study about this map, including whether it is injective (one-to-one), surjective
text snippet: "# Consider the ring homomorphism $ϕ : \mathbb{R}[x] → \mathbb{R}[\ (onto), or has any notable kernels (the set of elements mapped to the additive identity in S). In this discussion,
sqrt{−3}]$ defined by $ϕ(x) = \sqrt{−3}$. Consider the ring homomorphism $ϕ : \ we will focus on showing that a given ring homomorphism is surjective and finding its kernel. Surjectivity and the
mathbb{R}[x] → \mathbb{R}[\sqrt{−3}]$ defined by $ϕ(x) = \sqrt{−3}$. i) Show that Image of a Ring Homomorphism To demonstrate that a function is surjective, we must show that every element in the
$ϕ$ is surjective. It seems obvious, so not sure how to show it ii) Find $\ker ϕ$ codomain has at least one element in the domain that maps to it. For our given problem, consider the ring
and prove your answer is correct. [Hint: use the fact that $\mathbb{R}[x]$ is a homomorphism φ :ℝ[x]→ℝ[√−3] where φ(x)=√−3. To show surjectivity, we want to illustrate that all polynomials in Im
principal ideal domain.] I know the kernel is $x^2 +3$ but dunno how to use the PID (φ) have coefficients from ℝ and contain √−3 as a root. Since ℝ[√−3] consists precisely of these polynomials, we
to prove it. Can anyone help? • i) How in general do you show a function is only require demonstrating that ℝ and √−3 are contained within the image of φ. Firstly, notice that constants in ℝ
surjective? ii) The kernel is an ideal and thus principal, and you know it contains can be expressed as polynomials of degree zero; hence, they belong to the image under φ since φ(c) = c for any
x^2+3. – John Brevik Apr 13 '15 at 21:42 1) Show that both $\mathbb R$ and $\sqrt constant c in ℝ. Secondly, observe that φ(x^2 + 3) = (√−3)^2 + 3 = 0, which implies that √−3 is indeed a root of
{-3}$ are in the image. These generate the ring. (Or maybe the result is even more the polynomial x^2 + 3. Consequently, √−3 lies in the image of φ. Therefore, we conclude that φ is surjective. The
trivial, depending on you rlocal defiinition of $\mathbb R[\alpha]$). 2) The kernel Kernel of a Ring Homomorphism For a ring homomorphism φ : R -> S, its kernel Ker(φ) is the set of elements in R
is of the form $(f)$ for some polynomial $f$. We need $f\mid x^2+3$. Why can ew that map to the additive identity in S. Ideals are essential when discussing the kernel because ideals allow us to
exclude the possibilities $\de" Do not just list concepts, but develop each one in describe subsets of a ring closed under addition and multiplication by arbitrary ring elements. Furthermore, if R
detail before moving to the next, as we prioritize depth of understanding and is a Principal Ideal Domain (PID), then every ideal in R is generated by a single element. Given our problem, we
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: aim to find the kernel of φ :ℝ[x]->ℝ[√−3], where φ(x)=√−3. Let I = Ker(φ); recall that I is an ideal in ℝ[x]. As
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an mentioned earlier, ℝ[x] is a PID, meaning that I is principally generated by some polynomial f(x). That is, I = (f
academic, professional and engaging tone that captivates interest. - Application: (x)) for some monic polynomial f(x). Our goal now becomes identifying f(x)—in other words, determining the
Incorporate specific, practical examples, such as proofs in calculus or critical lowest-degree monic polynomial in Ker(φ). We already found that φ(x^2 + 3) = 0, implying that x^2 + 3 belongs to I.
dates and figures in history. Do not include a title or an introduction, simply Suppose g(x) were another nonconstant polynomial also lying in I. By the Division Algorithm, we could express g(x)
write the content without headlines and introductory phrases. Do not use images. as q(x)(x^2 + 3) + r(x), where deg(r(x)) < 2. Applying φ to both sides yields: φ(g(x)) = φ(q(x)(x^2 + 3) + r(x)) =>
0 = φ(q(x))φ(x^2 + 3) + φ(r(x)) => 0 = r(√−3) Since r(x) is a polynomial of degree less than two, r(√−3) can only
equal zero if r(x) is identically zero (since √−3 is not a root of linear polynomials). This means that r(x) cannot
exist unless it is the zero polynomial; consequently, g(x) must be divisible by x^2 + 3. Thus, x^2 + 3 is the
generator of I, denoted as I=(x^2 + 3).
Total derivatives are a fundamental concept in calculus, particularly useful when dealing with multivariable
functions that depend on other underlying arguments. This topic is essential for college students studying
mathematics, physics, engineering, and economics, among other fields. Herein, we will explore the definition,
notation, and application of total derivatives through detailed explanations and illustrative examples. To begin
with, let us define what a total derivative is in the context of multivariable functions. Consider a multivariable
Write an educational piece suited for college students related to the following function F(x, y) where x = g(t), and y = h(t). If each of these component functions (g(t) and h(t)) has a
text snippet: "1,183 Pages A total derivative of a multivariable function of well-defined derivative at some point t=a, then we say that F has a total derivative at that same point.
several variables, each of which is a function of another argument, is the Mathematically speaking, the total derivative of F with respect to t is expressed as follows: dF/dt = ∂F/∂x * dx/dt
derivative of the function with respect to said argument. it is equal to the sum of + ∂F/∂y * dy/dt This equation represents the sum of the product between the partial derivative of F concerning its
the partial derivatives with respect to each variable times the derivative of that respective variables and the derivative of those variables regarding the independent parameter t. It's crucial to
variable with respect to the independent variable. For example, given a function , understand that the term 'partial derivative' refers to the ordinary derivative if the function depends solely on
and with being functions of , The formula for a total derivative is a direct result one variable. Otherwise, you should apply the appropriate differentiation rules while treating all remaining
of the chain rule. Total derivatives are often used in related rates problems; for variables as constants. Now, let's consider an example problem to demonstrate how to calculate a total derivative
example, finding the rate of change of volume when two parameters are changing with step by step. Suppose we have the function F(x, y) = x^2*y - x*y^2 and want to find dF/dt when x = e^t, y = cos(t):
time. ## Example The radius and height of a cylinder are both . The radius is Firstly, compute the partial derivatives ∂F/∂x and ∂F/∂y: ∂F/∂x = 2xy - y^2 ∂F/∂y = x^2 - 2xy Secondly, determine
decreased at and the height is increasing at . What is the change in volume with the derivatives dx/dt and dy/dt: dx/dt = e^t dy/dt = -sin(t) Finally, substitute these values into the expression
respect to time at this instant? The volume of a right circular cylinder is We can for the total derivative: dF/dt = (2xy - y^2)*e^t + (x^2 - 2xy)*(-sin(t)) When evaluating dF/dt at any particular
take the total derivative of this with respect to time to get At this moment, the value of t, plug in the corresponding expressions for x and y based on their definitions as functions of t.
volume of the cylinder is not " Do not just list concepts, but develop each one in Applications of total derivatives appear frequently in various disciplines, especially in solving related rates
detail before moving to the next, as we prioritize depth of understanding and problems involving quantities dependent on multiple variables. One common scenario involves determining changes in
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: volumes of geometric objects whose dimensions vary with time. Let's examine our earlier example in greater depth to
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an solidify your grasp of this concept further: Consider a cylindrical container with radius r and height h, both
academic, professional and engaging tone that captivates interest. - Application: varying with time according to dr/dt=-0.5 and dh/dt=2 respectively. We wish to ascertain the instantaneous
Incorporate specific, practical examples, such as proofs in calculus or critical volumetric change at the moment when r=2 units and h=6 units. Recall that the volume V of a cylinder is given by: V
dates and figures in history. Do not include a title or an introduction, simply = πr²h Taking the total derivative of V with respect to time yields: dV/dt = ∂V/∂r * dr/dt + ∂V/∂h * dh/dt Plugging
write the content without headlines and introductory phrases. Do not use images. in the known values: dV/dt = (2πrh) (-0.5) + (πr²) (2) Substituting r=2 and h=6 into this equation provides the
desired answer: dV/dt = 48π - 12π = 36π cubic units per unit time In conclusion, mastering the computation of total
derivatives enables mathematicians, scientists, engineers, and economists alike to tackle complex real-world
scenarios requiring precise calculations of interrelated rates of change. Understanding the nuances behind this
powerful mathematical tool empowers learners to excel academically and professionally across diverse industries
demanding proficiency in advanced calculus techniques.
Linear and Angular Momentum: An Examination of Their Interplay and Relationship To begin our discussion, let's
establish clear definitions of the two types of momentum involved in this scenario: linear momentum and angular
momentum. Linear momentum, often denoted by the symbol 'p,' is a concept in physics representing the quantity of
motion an object possesses due to its mass and velocity. Its formula is given as p=mv, where m is mass and v is
velocity. On the other hand, angular momentum, represented as L, refers to the measure of the amount of rotation an
object has around a particular axis. The formula for angular momentum depends on the distance of the object from
the axis and its tangential speed, resulting in L=r x mv, where r denotes the radius vector perpendicular to the
direction of motion. Returning to the initial scenario presented, when the cannon fires a smaller ball, there are
actually three objects in play - the large ball, the smaller ball, and the cannon itself. Initially, the large ball
is at rest; therefore, it doesn't have any linear or angular momentum. However, once the cannon fires, things
change significantly. When the cannon expels the smaller ball, according to Newton's third law ("for every action,
there is an equal and opposite reaction"), the recoil force propelling the smaller ball forward generates an
opposing force causing the larger ball to move backward while simultaneously starting to spin. This results in both
linear and angular momentum being imparted onto the large ball. At first glance, this might lead some to believe
that linear momentum can indeed transform into angular momentum. But upon closer examination, these are merely
Write an educational piece suited for college students related to the following separate manifestations of conservation laws rather than conversions between them. Tiny-tim rightly points out that
text snippet: "Linear momentum converted to angular momentum? 1. May 27, 2012 everything that moves inherently contains angular momentum except along axes passing through their velocities. So
Michio Cuckoo http://img688.imageshack.us/img688/2310/81332204.png [Broken] yes, even the smaller ball fired out of the cannon will contain angular momentum relative to certain frames of
Originally the big ball is at rest and not moving at all. It has a cannon attached references despite having negligible impact compared to the large ball's dramatic gain in angular momentum.
to it. The cannon fires a smaller ball. The small ball only has linear momentum, Haruspex further clarifies that angular momentum calculations require specification of a reference point. When
BUT the big ball now rotates and has both linear and angular momentum. Does this taking moments about the cannon, the small ball certainly holds no net moment because it counteracts the cannon's
mean that linear momentum can be converted to angular momentum? Where did the own moment, which explains why the system comprising the large ball and the cannon begins spinning coherently after
angular momentum come from? Last edited by a moderator: May 6, 2017 2. May 27, 2012 the firing event. In conclusion, linear and angular momenta aren't interchangeable entities but rather
tiny-tim Hi Michio Cuckoo! No such thing as only linear momentum. Everything (that complementary aspects arising from different physical contexts. Each obeys distinct mathematical formulations yet
moves) has angular momentum (except about an axis that its velocity passes remains connected via fundamental principles like conservation of energy and momentum. Therefore, instead of
through). 3. May 27, 2012 Michio Cuckoo So the small ball that was ejected also has viewing these phenomena as conversions, consider them as intricate components woven together within broader
angular momentum? 4. May 27, 2012 haruspex Angular momentum is measured about a mechanical frameworks governing how systems evolve dynamically under external influences. </s> Deepening Our
reference point. If you take moments about the cannon the small ball has no moment, Understanding: A Mathematical Approach to Quantifying Changes in Linear and Angular Momenta Let us dive deeper into
but then neither does the canno" Do not just list concepts, but develop each one in the mathematics behind changes in linear and angular momenta during the described scenario. We denote M and m as
detail before moving to the next, as we prioritize depth of understanding and masses of the large ball and small ball respectively, V and v as their respective velocities, and R and r as radii
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: vectors pointing towards the center of mass of each ball. Note that capital letters refer to quantities associated
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an with the large ball whereas lowercase ones correspond to those attributed to the small ball. Before the cannon
academic, professional and engaging tone that captivates interest. - Application: fires, both balls remain stationary; hence, they initially hold zero linear and angular momenta: Initial
Incorporate specific, practical examples, such as proofs in calculus or critical Conditions: P_initial = P(t=0) = M*V + m*v = 0 L_initial = L(t=0) = I_large\_ball * Ω + I_small\_ball * omega = 0
dates and figures in history. Do not include a title or an introduction, simply Here, Ω and omega signify angular speeds of the large ball and the small ball respectively, whereas I_large\_ball
write the content without headlines and introductory phrases. Do not use images. and I_small\_ball represent their corresponding moments of inertia. Since the small ball isn't rotating prior to
ejection, its contribution to the total angular momentum becomes trivial since its rotational kinetic energy equals
zero. Immediately Post Firing Event: After the cannon discharges the small ball, several factors shift: 1. Both the
large ball and the cannon acquire identical magnitudes of backwards linear momentum (-p), ensuring overall linear
momentum balance post-firing. 2. Due to the impulse transferred from the exploding gunpowder charge, the small ball
gains significant forwards linear momentum (+p'). 3. Consequently, the large ball starts revolving clockwise around
the new common center of mass shared with the cannon, acquiring nonzero angular momentum (L'), whereas the small
ball picks up insignificant amounts of angular momentum (l') compared to the large ball. Post-Firing Conditions:
P_post-firing = M*-V + m*v' != 0 L_post-firing = I_large\_ball * Ω' + I_small\_ball * omega' != 0 Note that prime
symbols indicate values immediately post-firing. Applying principle of conservation of linear momentum, we derive:
M*(-V) + m*(+v') = 0 => V = -m/M \* v' Similarly, applying principle of conservation of angular momentum around the
original position of the cannon yields: I_large\_ball * Ω' = -m\*r'*v' where r' represents the displacement of the
small ball from the original position of the cannon. As time progresses, assuming no external forces act on either
ball, both linear and angular momenta of the individual bodies conserve independently. Interestingly though, if
observed collectively as a single entity composed of the large ball, cannon, and residual smoke emanating from the
firing process, the composite system preserves its overall linear and angular momenta too.
Write an educational piece suited for college students related to the following
text snippet: "Data Transformation¶ Ha Khanh Nguyen (hknguyen) 1. Removing
Duplicates¶ • Duplicate rows may be found in a DataFrame for any number of reasons. When working with data frames, it is common to encounter duplicate rows due to various reasons. To address this
Here is an example: • The DataFrame method duplicated() returns a Boolean Series issue, Pandas provides two useful methods: `duplicated()` and `drop_duplicates()`. These functions help identify
indicating whether each row is a duplicate (has been observed in a previous row) or and remove duplicate entries from your dataset effectively. This section delves into their usage, options, and
not: • Relatedly, drop_duplicates() returns a DataFrame where the duplicated array applications through detailed explanations and practical examples. The `duplicated()` function generates a Boolean
is False: • Both of these methods by default consider all of the columns; series indicating if each row appears more than once within a data frame. It compares every single row against its
alternatively, you can specify any subset of them to detect duplicates. • Suppose predecessors and labels identical ones accordingly. By default, the very first instance of a repeated row is marked
we had an additional column of values and wanted to filter duplicates only based on as `False`, while subsequent occurrences are tagged as `True`. This behavior allows users to easily spot
the 'k1' column: • duplicated() and drop_duplicates() by default keep the first reoccurring records. For instance, let us assume we have the following dataframe named "df": ```python k1 k2 k3 0 1
observed value combination. Passing keep='last' will return the last one: 2. 4 7 1 2 5 8 2 1 4 7 3 3 6 9 4 1 4 7 ``` To apply the `duplicated()` function to our dataframe, execute: ```scss >>>
Transforming Data Using a Function or Mapping¶ • For many datasets, you may wish to df.duplicated() 0 False 1 False 2 True 3 False 4 True dtype: bool ``` As shown above, the second and fourth
perform some transformation based on the values in an array, Series, or column in a appearances of rows containing [1, 4, 7], specifically at index positions 2 and 4, are recognized as duplicates.
DataFrame. • Consider the following hy" Do not just list concepts, but develop each However, the initial appearance remains unmarked. In certain scenarios, instead of merely identifying duplications,
one in detail before moving to the next, as we prioritize depth of understanding you might want to extract a filtered version of the original data frame excluding those redundancies. That's when
and comprehensive exploration of the subject matter over breadth. Focus on: - the `drop_duplicates()` function comes in handy. Similar to its counterpart, this technique also checks for
Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Write with recurrent elements. Nonetheless, unlike `duplicated()`, which outputs a boolean series, invoking `drop_duplicates()
an academic, professional and engaging tone that captivates interest. - ` directly on the data frame eliminates superfluous instances, thereby returning a pristine copy of the input.
Application: Incorporate specific, practical examples, such as proofs in calculus Continuing with the same example, applying `drop_duplicates()` produces: ```r k1 k2 k3 0 1 4 7
or critical dates and figures in history. Do not include a title or an
introduction, simply write the content without headlines and introductory phrases.
Do not use images.
Write an educational piece suited for college students related to the following Integration by Partial Fraction Decomposition is a technique used to integrate rational functions, where the degree
text snippet: "Integration by means of partial fraction decomposition I'm trying to of the numerator is less than the degree of the denominator. This method involves decomposing the given rational
solve this indefinite integral by means of partial fraction decomposition: $\int\ function into simpler fractions, which can then be integrated individually. The key idea behind this technique lies
dfrac{x+1}{\left(x^2+4x+5\right)^2}\ dx$. The denominator has complex (but not in the fact that every rational function can be expressed as a sum of simpler fractions, whose denominators are
real) roots because $\Delta<0$; so, according with my calculus book, i try to factors of the original denominator. In your case, you have been asked to find the indefinite integral of the
decompose the integrand function in this form: $\dfrac{x+1}{\left(x^2+4x+5\right)^ rational expression $(x + 1)/((x^2 + 4x + 5)^2)$. Let us walk through the steps involved in solving this problem
2}= \dfrac{Ax+B}{\left(x^2+4x+5\right)}+\dfrac{Cx+D}{\left(x^2+4x+5\right)^2}$. I using partial fraction decomposition. Firstly, notice that the quadratic polynomial in the denominator does not
get: $\dfrac{x+1}{\left(x^2+4x+5\right)^2}= \dfrac{\left(Ax+B\right)\left(x^2+4x+5\ factor further over the reals since its discriminant $(\Delta = (-4)^2 - 4 \times 1 \times 5)$ is negative. Thus,
right)+Cx+D}{\left(x^2+4x+5\right)^2}$. Multiplying the right term: $\dfrac{x+1}{\ it is appropriate to proceed with a partial fraction decomposition involving complex conjugate linear factors
left(x^2+4x+5\right)^2}= \dfrac{Ax^3+4Ax^2+5Ax+Bx^2+4Bx+5B+Cx+D}{\left(x^2+4x+5\ raised to the second power. As per your calculus book, let us express our integrand as follows: $$ \frac{x + 1}{(x^
right)^2}$. Now i collect the terms with the same pwer of $x$: $\dfrac{x+1}{\left(x 2 + 4x + 5)^2} = \frac{Ax + B}{x^2 + 4x + 5} + \frac{Cx + D}{(x^2 + 4x + 5)^2}$$ Next, we aim to find constants A,
^2+4x+5\right)^2}= \dfrac{Ax^3+\left(4A+B\right)x^2+\left(5A+4B+C\right)x+D+ 5B}{\ B, C, and D. To do this, clear out the denominators from both sides of the equation above: $$(x + 1) = (Ax + B)(x^2
left(x^2+4x+5\right)^2}$. Now i equate the two numerators: $x+1=Ax^3+\left(4A+B\ + 4x + 5) + Cx + D$$ Expanding the binomials and collecting like powers of x gives us: $$x + 1 = Ax^3 + (4A + B)x^2
right)x^2+\left(5A+4B+C\rig" Do not just list concepts, but develop each one in + (5A + 4B + C)x + (5B + D)$$ Comparing coefficients of identical powers of x results in the following system of
detail before moving to the next, as we prioritize depth of understanding and equations: $$ \begin{cases} A &= 0 \\ 4A + B &= 0\\ 5A + 4B + C &= 1 \\ 5B + D &= 1 \end{cases}$$ Solving this
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: simple system yields A = 0, B = 0, C = 1, and D = 1. Therefore, the initial decomposition becomes: $$ \frac{x + 1}
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an {(x^2 + 4x + 5)^2} = \frac{1}{(x^2 + 4x + 5)^2}$$ Since there are no remaining expressions to break down via
academic, professional and engaging tone that captivates interest. - Application: partial fractions, integration will occur directly with the simplified expression at hand. It remains to compute
Incorporate specific, practical examples, such as proofs in calculus or critical the antiderivative of $(x^2 + 4x + 5)^{-2}$, which requires applying the chain rule during differentiation and
dates and figures in history. Do not include a title or an introduction, simply substituting the constant back once integrated. You may also consider converting the quadratic denominator to
write the content without headlines and introductory phrases. Do not use images. vertex form and making an appropriate trigonometric substitution.
Write an educational piece suited for college students related to the following
text snippet: "## A Coin Is Tossed Three Times What Is The Probability Of Getting 3 To begin, let's clarify some key concepts related to probabilities and coin tosses. When we talk about the
Heads We test the hypothesis that the probability the coin lands heads when tossed probability of an event occurring, we're referring to the likelihood of that event happening compared to the total
is 0. The toss of a coin, throwing dice and lottery draws are all examples of number of possible outcomes. For example, if we have a fair coin, there are only two possible outcomes - heads or
random events. One is a two-headed coin ( having head on both faces ), another is a tails. Therefore, the probability of flipping a head or a tail is always equal to 1 (the sum of all possibilities)
biased coin that comes up heads 75% of the times and third is also a biased coin divided by 2 (the number of potential outcomes). So, P(heads) = P(tails) = 1/2. Now, consider a scenario where we
that comes up tails 40% of the times. If three fair coins are tossed randomly 175 toss a coin three times in a row. Since each individual flip can result in either heads or tails, the overall
times and it is found that three heads appeared 21 times, two heads appeared 56 outcome space consists of 2 × 2 × 2 = 8 distinct possibilities: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Here, 'H'
times, one head appeared 63 times and zero head appeared 35 times. If you toss a stands for heads and 'T' represents tails. Now, suppose we want to calculate the probability of obtaining exactly
coin, you cannot get both a head and a tail at the same time, so this has zero three heads in these three tosses. We already established earlier that the probability of flipping a single head is
probability. What is the probability of the coin landing heads up three times? I 1/2; therefore, the desired outcome would be P(three heads) = (1/2) × (1/2) × (1/2), which equals 1/8. However,
need somebody to explain to me, how to do the full process, so I know how to do it your question pertains specifically to determining the experimental probability based on observed data rather than
for other math problems as well. If a coin tossed for $3$ times what is the theoretical calculations. Based on the provided information, three fair coins were tossed 175 times, resulting in
probability of getting all tails? tossing " Do not just list concepts, but develop the subsequent distribution: * Three heads appeared 21 times * Two heads appeared 56 times * One head appeared 63
each one in detail before moving to the next, as we prioritize depth of times * Zero head appeared 35 times The empirical probability of observing three heads in a sequence of three
understanding and comprehensive exploration of the subject matter over breadth. tosses then becomes P'(three heads) = Number of successful trials / Total number of trials. Plugging in our values
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: yields: P'(three heads) = 21 / 175 ≈ 0.1199 Lastly, regarding the final part of your question, "If a coin is tossed
Write with an academic, professional and engaging tone that captivates interest. - three times, what is the probability of getting all tails?" Using similar reasoning as above, the theoretical
Application: Incorporate specific, practical examples, such as proofs in calculus probability is calculated as follows: P(all tails) = (1/2) × (1/2) × (1/2) = 1/8 ≈ 0.125 Please note that since the
or critical dates and figures in history. Do not include a title or an original experiment did not record instances of consecutive tails separately, we cannot provide an accurate
introduction, simply write the content without headlines and introductory phrases. experimental probability for this particular case using the given dataset.
Do not use images.
In order to understand the proof that no set is equinumerous to its power set, it is necessary to have a solid
grasp of several foundational concepts in set theory. We will begin by reviewing these prerequisites and then delve
into the details of the proof itself, addressing the student's concerns along the way. 1. **Sets and elements:** A
set is a collection of objects, called elements or members. Sets are usually denoted by capital letters, while
their elements are denoted by lowercase letters. To indicate that an element $a$ belongs to a set $A$, we write $a
o A$. Set membership is purely extensional, meaning that two sets are equal if and only if they contain precisely
the same elements. For example, $\{1,2\} = \{2,1\}$ because both sets consist of exactly the numbers 1 and 2. 2.
**Ordered pairs and relations:** An ordered pair $(a,b)$ consists of two objects, $a$ and $b$, arranged in a
specific order. Two ordered pairs are equal if and only if their first components are identical and their second
components are also identical. That is, $(a, b) = (c, d)$ if and only if $a=c$ and $b=d$. A relation $R$ on sets
$X$ and $Y$ is a subset of the Cartesian product $X × Y$; hence, $R$ can be regarded as a set of ordered pairs.
Given an object $x o X$, if there exists some $y o Y$ such that $(x, y) o R$, we say that $x$ is related to $y$ via
$R$, written as $xRy$. 3. **Functions:** A function (also known as a mapping or a map) $f : X → Y$ from a set $X$
to a set $Y$ is a particular type of relation between them, namely, a set of ordered pairs $(x, f(x))$ satisfying
the property that for every $x o X$, there exists a unique $y o Y$ paired with $x$. Here, $f(x)$, read "$f$ of
Write an educational piece suited for college students related to the following $x$", denotes the value assigned to $x$ by $f$. It is customary to omit the parentheses when applying functions to
text snippet: "# Proof that no set is equinumerous to its power set I am studying individual arguments, writing $fx$ instead of $f(x)$. Additionally, we often refer to $X$ as the domain of $f$ and
from Enderton's book "Elements of set theory" and I am struggling with the proof $Y$ as the codomain of $f$. If $f$ assigns the value $y$ to at least one element $x$ in $X$, i.e., $\exists x o X:
that "No set is equinumerous to its power set". Here is the proof: Let $$g: A\ fx = y$, we say that $y$ lies in the range of $f$, denoted $\text{ran } f$. 4. **Power sets:** Given any set $A$,
rightarrow \mathcal{P}A$$; we will construct a subset $$B$$ of $$A$$ that is not in the power set of $A$, denoted $\mathscr{P}(A)$, is the set whose elements are all possible subsets of $A$.
ran $$g$$. Specifically, let $$B = \{x\in A\mid x\notin g(x)\}$$. Then $$B\subseteq Formally, $\mathscr{P}(A) := \{S : S⊆A\}$. Note that $|\mathscr{P}(A)| > |A|$ whenever $A$ has more than one
A$$, but for each $$x\in A$$, $$x\in B$$ iff $$x\notin g(x)$$. Hence $$B\neq g(x) element, since $\mathscr{P}(A)$ contains at least two distinct elements, viz., the empty set $oldsymbol{ heta}$,
$$. I saw on the web another proof that is almost the same and seems a tiny bit which represents the absence of any elements, and $A$ itself. Moreover, if $|A| = n < ∞$, then $|\mathscr{P}(A)| =
clearer, but I am having the same trouble. The doubt is: what prevents us from 2^n$. 5. **Equipotency:** Two sets $A$ and $B$ are said to be equipotent (or equivalent, or equinumerous), denoted
thinking that $$x\notin g(x)$$ is actually a contradiction, just like $$x\neq x$$, $|A| = |B|$, if and only if there exists a bijection (i.e., a bijective function) between them, meaning that $f : A
and that therefore $$B=\emptyset$$? This proof seems to assume that there must be → B$ satisfies three properties simultaneously: - $f$ is total: $x o A ⇒ fx o B$ - $f$ is injective: $x_1, x_2 o A$
an $$x$$ such that $$x\notin g(x)$$, but I don't see where this is coming from. I and $x_1 ≠ x_2 ⇒ fx_1 ≠ fx_2$ - $f$ is surjective: $y o B ⇒ ∃ x o A: fx = y$ 6. **Cantor's theorem:** Cantor's
am a just starting undergrad student, I am sorry if this question may be a bit theorem states that for any given set $A$, there cannot exist a surjection from $A$ onto its power set $\mathscr{P}
naive. Thanks. • The proof does no" Do not just list concepts, but develop each one (A)$. Symbolically, this means that for all functions $f : A → \mathscr{P}(A)$, there always exists some subset $B
in detail before moving to the next, as we prioritize depth of understanding and ⊆ A$ such that $B ∉ \text{ran } f$. One might wonder why such a seemingly intuitive claim requires formal proof at
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: all. Indeed, it relies crucially upon the fact that a set and its power set possess fundamentally different
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an cardinalities (sizes). Intuitively speaking, the power set doubles the size of the original set due to the presence
academic, professional and engaging tone that captivates interest. - Application: of complements—if a given subset $S ⊆ A$ is selected, so too is its relative complement $AcS := A − S$.
Incorporate specific, practical examples, such as proofs in calculus or critical Consequently, every subset in $\mathscr{P}(A)$ corresponds uniquely either to an element in $A$ or to its
dates and figures in history. Do not include a title or an introduction, simply complement within $A$. Since each subset possesses a single, well-defined complement, yet elements themselves do
write the content without headlines and introductory phrases. Do not use images. not admit multiple complements, it follows that $\mathscr{P}(A)$ cannot be placed into a one-to-one correspondence
with $A$ itself. Now, we shall analyze the provided proof. 7. **Proof analysis:** Let $g : A → \mathscr{P}(A)$
denote an arbitrary function from our set $A$ to its power set $\mathscr{P}(A)$. Our goal is to demonstrate the
existence of a certain subset $B ⊆ A$ that fails to belong to the image of $g$, thereby precluding the possibility
of $g$ being surjective. Toward this end, define $B$ as follows: $$B := \{x ∈ A : x ∉ gx\}$$ By construction, $B$
comprises precisely those elements $x ∈ A$ that do not appear among the elements of $gx$—that is, $B$ collects
together all $x$ lying outside of their respective counterparts in $\mathscr{P}(A)$. Clearly, $B ⊆ A$, meaning that
$B$ qualifies indeed as a candidate for membership within $\mathscr{P}(A)$. However, we now show that $B$ cannot
belong to the image of $g$, implying that $g$ cannot achieve surjectivity after all. Assume, towards a reductio ad
absurdum, that $B ∈ \text{ran } g$. This implies that $∃ z ∈ A : gz = B$. At this point, two cases emerge. Either
$z ∈ B$ or $z ∉ B$. Suppose firstly that $z ∈ B$, meaning that $z ∉ gz$ according to how $B$ was defined earlier.
But $gz = B$, so $z ∈ B$ would yield $z ∉ B$, leading immediately to a contradiction! Thus, our assumption that $z
∈ B$ must be false. Alternatively, suppose instead that $z ∉ B$, entailing that $z ∈ gz$ based on the definition of
$B$. Yet again, however, this leads to a direct contradiction, since $gz = B$ forces $z ∈ gz$ to imply $z ∉ B$.
Ergo, neither case allows for consistent reasoning, demonstrating conclusively that no such $z$ can ever exist,
rendering $B$ absent from the image of $g$. Accordingly, $g$ remains non-surjective, proving Cantor's theorem.
To begin, let's define Extended Kalman Filter Simultaneous Localization and Mapping (EKF-SLAM), a technique used in
robotics to localize a robot while building a map of its environment. This method uses a series of measurements
taken by sensors mounted on the robot to update both the estimated pose (position and orientation) of the robot
itself and the positions of notable features within the environment, referred to as "landmarks." Errors in EKF-SLAM
Write an educational piece suited for college students related to the following arise due to imperfect sensor data, unmodeled dynamics, and other real-world factors. These errors can be
text snippet: "# how to plot $\pm 3 \sigma$ of a landmark in EKF-SLAM I have quantified through statistical methods, allowing us to understand their distribution and predict future errors more
implemented 2D-SLAM using EKF. The map is based-feature in which there is only one accurately. A common approach when working with these errors is to calculate the standard deviation ($\sigma$)—a
landmark for the sake of simplicity. I've read some papers regarding this matter. key measurement of dispersion around the mean value. By considering the range spanning three times the standard
They plot the $\pm3\sigma$ plus the error. I would like to make sure that I'm doing deviation on either side of the estimate (+3σ and −3σ), you account for approximately 99.7% of all possible
the right thing. In my project, I have the estimate of the landmark's position and outcomes if your errors follow a normal (Gaussian) distribution. In the context of your question, it seems you have
its true values. The true values here are the ones that the sensor measure not the successfully calculated the estimation error ($e\_x$) along the X-axis for a single landmark. Specifically, $e\_x$
ideal case. For example, the ideal case of the landmark position is (30,60) but is determined by subtracting the true landmark position ($x$) from the estimated position ($\hat{x}$): $$e\_x=\hat
this value is not accessible by any means, therefore I will consider the true {x}-x$$ As illustrated in your description, visualizing these errors helps demonstrate the accuracy and reliability
values the ones that are coming from the sensor. Now the error in the landmark's of your SLAM implementation. Plotting +3σ and −3σ confidence intervals alongside the actual error provides valuable
position in x-axis is formulated as follows $$\text{error}_{x} = \hat{x} - x$$ The insights into your system's performance. Based on what you described, yes, your approach does align well with
below picture shows the error in blue color. The red color represents the error conventions found in academic literature. To further clarify, suppose your estimated landmark position is $(\hat
bounds which is $\pm 3 \sigma_{x}$ My question is now is this the way people plot {x}, \hat{y})=(28, 59)$, and the measured (true) values obtained via the sensor are $(x, y)=(31, 60)$. Using these
the errors in the academics papers" Do not just list concepts, but develop each one numbers, compute the error along the X-axis: $$e\_x=28-31=-3$$ Next, determine the covariance matrix associated
in detail before moving to the next, as we prioritize depth of understanding and with the estimated state. You may extract $\sigma\_x$, representing the uncertainty along the X-axis, from this
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: matrix. If necessary, perform a square root operation to obtain the diagonal entries of the covariance matrix,
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an resulting in the so-called standard deviations. Then, find the desired interval by multiplying ±3 with the
academic, professional and engaging tone that captivates interest. - Application: corresponding standard deviation, denoted by $\sigma*3$. Finally, illustrate the point estimates and confidence
Incorporate specific, practical examples, such as proofs in calculus or critical intervals graphically. Keep in mind that evaluating your EKF-SLAM implementation should involve analyzing multiple
dates and figures in history. Do not include a title or an introduction, simply scenarios. Assessing various environmental conditions, different starting points, and diverse landmark
write the content without headlines and introductory phrases. Do not use images. configurations contributes significantly toward ensuring your algorithm performs adequately across numerous
situations. Additionally, it might prove beneficial to analyze error distributions beyond simple Gaussian models
since realistic applications could introduce non-linearities and skewed error patterns. Nonetheless, your current
approach forms a solid foundation upon which you can build a thorough assessment process tailored to your unique
Write an educational piece suited for college students related to the following The integral that you are interested in evaluating is given by: ∫ (ln(x² + 1)) / x dx You began by making the
text snippet: "# Evaluate the integral $\int \frac{\ln(x^2+1)dx}{x}$ I am substitution w = ln(x² + 1), which led to e^w dw = 2x dx. However, you were unable to eliminate x from the equation
interested in finding an exact expression for the integral $$\int \frac{\ln(x^2+1) completely. Specifically, you arrived at the integral: ∫ (we^w) / (2x²) dw At this point, it seems that further
dx}{x}$$ I start by using the transformation $$w=\ln(x^2+1)$$ leading to $$e^{w}dw= progress cannot be made without introducing additional variables or more complex functions. Indeed, Dr. Sonnhard
2xdx$$. Unfortunately, I couldn't get rid of $$x$$ in there: $$\int \frac{we^w}{2x^ Graubner provided the antiderivative in terms of the dilogarithm function, denoted Li₂(-x²)/2. To better understand
2}dw$$ and I'm not interested in results with an infinite series. I would like to why this answer takes the form it does, let us explore some relevant background information about the dilogarithm
know a suitable substitution that may involve only one variable leading to an function. The dilogarithm function Li₂(z) is defined as follows: Li₂(z) = ∫¹ z/(1 - t) dt, where |arg(1 - z)| < π
expression less complicated for integration. Thanks for your help. • It is $$-\frac This function has several interesting properties and applications, including its appearance in various fields such
{\text{Li}_2\left(-x^2\right)}{2}$$ – Dr. Sonnhard Graubner Apr 27 at 14:01 • yes as number theory, physics, and combinatorics. For example, it arises when computing certain sums involving powers
this is what wolfram is showing...can you explain what Li is? – Marvel Maharrnab of integers or harmonic numbers. Nevertheless, if you wish to find another expression for the original integral
Apr 27 at 14:03 • Thanks. However, this kind of expression is what I want to avoid. that does not involve any special functions, we must continue searching for alternative methods. One possible
Thus, I need alternate expression. – M.D. Apr 27 at 14:04 • @MarvelMaharrnab This approach involves integrating by parts. Recall the formula: ∫ u dv = uv - ∫ v du Let us choose u = ln(x² + 1) and
special function is called Dilogarithm, or Spence's Function. – mrtaurho Apr 27 at dv = dx/x. Then, we have du = 2x/(x² + 1) dx and v = lnx. Applying the integration by parts formula yields: ∫ (ln
14:11 • @MarvelMaharrnab, sure this" Do not just list concepts, but develop each (x² + 1)) / x dx = lnx * ln(x² + 1) - ∫ (lnx) * (2x) / (x² + 1) dx Unfortunately, even after applying integration
one in detail before moving to the next, as we prioritize depth of understanding by parts, we still do not obtain a simple antiderivative free of any special functions. At this stage, we might
and comprehensive exploration of the subject matter over breadth. Focus on: - consider other techniques, such as expanding the denominator into partial fractions or employing different
Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: Write with substitutions. Yet, these too seem unlikely to yield desirable outcomes due to the complexity of the integrand.
an academic, professional and engaging tone that captivates interest. - Therefore, it appears that utilizing the dilogarithm function is indeed the most straightforward method for
Application: Incorporate specific, practical examples, such as proofs in calculus evaluating this particular integral. In conclusion, while avoiding specialized functions can sometimes lead to
or critical dates and figures in history. Do not include a title or an simpler expressions, they often arise naturally during calculations and provide elegant solutions to otherwise
introduction, simply write the content without headlines and introductory phrases. challenging problems. When encountering such situations, familiarize yourself with these functions and their
Do not use images. properties, as doing so will enable you to recognize them easily and broaden your mathematical toolkit.
Let us begin by recalling some fundamental definitions and properties regarding eigenvalues and eigenvectors of a
square matrix. This will help us build up to solving the problem at hand. An eigenvector of an n x n real matrix A
is a nonzero vector v in R^n such that when Av is computed, the resulting vector is a scalar multiple of v. That
is, there exists a scalar λ, called the eigenvalue corresponding to the eigenvector v, satisfying the equation: Av
= λv Moreover, if A is symmetric (i.e., A^T = A), then A possesses an important property known as spectral
Write an educational piece suited for college students related to the following decomposition, which allows us to decompose A into the product of three matrices - namely, a diagonal matrix D
text snippet: "# What is the linear space of Eigenvectors associated with a certain containing all eigenvalues along its main diagonal, and two orthogonal matrices Q and Q^T, where Q contains the
Eigenvalue? The following matrix $A$ has $\lambda=2$ and $\lambda=8$ as its respective eigenvectors as columns. Mathematically, this may be expressed as follows: A = QDQ^T Now let's apply
eigenvalues $$A = \begin{bmatrix} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{bmatrix} these ideas to the given matrix A and find its eigenvalues and their corresponding eigenvectors. We have: A = | 4 2
$$ let $P$ be the orthogonal matrix that diagonalizes $A$. in other words, $A= PDP^ 2 | | 2 4 2 | | 2 2 4 | First, compute the characteristic polynomial det(A - λI) of A, where I denotes the identity
T$. you can check that $$P = \begin{bmatrix} (-√2)/2 & (-√6)/6 & (-√3)/3 \\ (√2)/2 matrix: det(A - λI) = (4-λ)[(4-λ)(4-λ)-2*2] - 2[(2*2)-(4-λ)*2] + 2[-2*(2*2)-(4-λ)*2] = (4-λ)^3 - 12(4-λ) - 24 Next,
& (-√6)/6 & (√3)/3 \\ 0 & (√6)/3 & (√3)/3 \end{bmatrix}$$ then the linear space of set the characteristic polynomial equal to zero and find its roots using standard techniques (such as factoring):
eigenvectors associated to the eigenvalue $\lambda=2$ is generated by? 1. The last (4-λ)^3 - 12(4-λ) - 24 = 0 (4-λ)((4-λ)^2 - 12) - 24 = 0 (4-λ)(16 - 8λ + λ^2 - 12) = 0 (4-λ)(λ^2 - 8λ) = 0 This
two columns of the matrix $P$ 2. The last two rows of matrix $P$ 3. The first two factors further to give: (4-λ)(λ-2)(λ-8) = 0 Thus, the eigenvalues are λ₁ = 4, λ₂ = 2, and λ₃ = 8. Next, find the
columns of the matrix $P$ 4. The first two rows of the matrix $P$ 5. The last eigenvectors by computing (A - λI)v = 0 for each eigenvalue. For λ₁ = 4, we have: | 0 2 2 | * |x| | 0 | | 2 0 2 | *
column of the matrix $P$ I Don't know how to solve this, I tried doing it but it |y| = |0| | 2 2 0 | * |z| |0| Solving this system yields infinitely many solutions; choose one linearly independent
got really messy and I didn't get any answer that made sense. If someone could solution, say, v₁ = [-1, 0, 1]. For λ₂ = 2, we obtain: | 2 2 2 | * |x| | 0 | | 2 2 2 | * |y| = |0| | 2 2 2 | * |z|
please explain how to do it or how to solve it, that would be great. Thank You • To |0| After simplification, we see that every vector in R³ satisfies this homogeneous system of equations. However,
format a matrix from now on" Do not just list concepts, but develop each one in we must select only those vectors whose length is greater than zero since eigenvectors cannot be the zero vector.
detail before moving to the next, as we prioritize depth of understanding and One possible choice is v₂ = [1, 1, 1]. Finally, for λ₃ = 8, we get: | -4 -2 -2 | * |x| | 0 | | -2 -4 -2 | * |y| = |
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: 0| | -2 -2 -4 | * |z| |0| Simplifying leads to y = z = -\frac{1}{2}x, meaning that any vector of the form v₃ = [a,
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an -\frac{1}{2}a, -\frac{1}{2}a], where a ≠ 0, serves as an eigenvector for λ₃ = 8. Choose, for instance, v₃ = [1, -\
academic, professional and engaging tone that captivates interest. - Application: frac{1}{2}, -\frac{1}{2}] as our representative eigenvector. At this point, normalize the eigenvectors so they
Incorporate specific, practical examples, such as proofs in calculus or critical become mutually orthogonal unit vectors: v₁ → u₁ = \frac{v_1}{\left\|v_1\right\|} = [\frac{-1}{\sqrt{2}}, 0, \frac
dates and figures in history. Do not include a title or an introduction, simply {1}{\sqrt{2}}] v₂ → u₂ = \frac{v_2}{\left\|v_2\right\|} = [\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt
write the content without headlines and introductory phrases. Do not use images. {3}}] v₃ → u₃ = \frac{v_3}{\left\|v_3\right\|} = [\frac{\sqrt{2}}{2}, \frac{-\sqrt{6}}{6}, \frac{-\sqrt{6}}{6}] Now
construct the orthogonal matrix P containing these normalized eigenvectors as columns: P = | u₁ | | u₂ | | u₃ | We
observe that this matches the provided matrix P exactly. Therefore, we can proceed with determining the eigenspace
associated with the eigenvalue λ = 2. By definition, the eigenspace E_λ is the span of all eigenvectors associated
with the eigenvalue λ. Since the second and third columns of P correspond to the eigenvectors for λ = 2, option 1
is correct, i.e.: E_(λ=2) = spanned by {last two columns of matrix P}
Write an educational piece suited for college students related to the following
text snippet: "1. Integration problem $\displaystyle $\int tan^3(x) sec^4(x) dx$$ I
first tried substituting $\displaystyle $u = tan(x)$$, then I tried substituting $\ To tackle the integral ∫tan³(x)sec⁴(x) dx, let's break down the problem into smaller steps and apply some
displaystyle $u = sec(x)$$, but I couldn't seem to solve it. Any hints? 2. identities and strategies from trigonometry and calculus. We will discuss the necessary background knowledge and
Originally Posted by BrownianMan $\displaystyle $\int tan^3(x) sec^4(x) dx$$ I follow through with the solution step-by-step. First, recall the pythagorean identity for tangent and secant
first tried substituting $\displaystyle $u = tan(x)$$, then I tried substituting $\ functions: tan²(x) + 1 = sec²(x). We can rewrite this identity using a u-substitution when solving our original
displaystyle $u = sec(x)$$, but I couldn't seem to solve it. Any hints? $\ integral. Let u = tan(x), so du = sec²(x)dx: ∫tan³(x)sec⁴(x) dx = ∫u³ * (u²+1) du. Before integrating further,
displaystyle \int tan^3(x)sec^4(x)$ cab be written as $\displaystyle \int tan^3(x) let's express everything in terms of powers of u: ∫u⁶ + u⁴ du. Now, integrate term by term: ∫u⁶ du = (1/7)u⁷ + C₁,
sec^2(x)sec^2(x)$ $\displaystyle \int tan^3(x)[1 + tan^2(x)]sec^2(x)$ Now simplify where C₁ is the constant of integration. Similarly, ∫u⁴ du = (1/5)u⁵ + C₂, where C₂ is another constant of
and solve. 3. $\displaystyle tan^2(x)+1=sec^2(x)$ 4. Thank you!" Do not just list integration. Combine both results to get the final answer: ∫tan³(x)sec⁴(x) dx = (1/7)tan⁷(x) + (1/5)tan⁵(x) + K,
concepts, but develop each one in detail before moving to the next, as we where K = C₁ - C₂ is the total combined constant of integration. Throughout this process, note how crucial it was
prioritize depth of understanding and comprehensive exploration of the subject to recognize the pythagorean identity and utilize appropriate substitutions. With these techniques at hand, even
matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the concepts/ seemingly complex problems like the initial integral become manageable. Applying mathematical definitions,
sections. - Engagement: Write with an academic, professional and engaging tone that properties, and well-known identities are essential skills required while studying advanced mathematics courses
captivates interest. - Application: Incorporate specific, practical examples, such including calculus and beyond.
as proofs in calculus or critical dates and figures in history. Do not include a
title or an introduction, simply write the content without headlines and
introductory phrases. Do not use images.
To understand the problem at hand, let's first define some key terms and provide necessary context. We will then
delve into the concept of binomial coefficients, their properties, and how they relate to the given sum. Lastly, we
Write an educational piece suited for college students related to the following will present a concrete example and discuss potential approaches to solving the problem. Binomial coefficients are
text snippet: "# Identity for a sum of product of binomial coefficients For some natural numbers represented by the coefficient of $x^k$ in the expansion of $(1+x)^n,$ where $n$ is a non-negative
fixed positive integers $$r_1,\ldots,r_n$$, I would like to find a sum: $$\sum_ integer and $k \leq n.$ They are often written as $\binom{n}{k}.$ The combinatorial interpretation of these
{i_1+\cdots+i_n=k}\binom{r_1+i_1}{r_1}\cdots\binom{r_n+i_n}{r_n}=\sum_{i_1+\ coefficients is the number of ways to choose $k$ items out of a set containing $n$ distinct elements. For more
cdots+i_n=k}\binom{r_1+i_1}{i_1}\cdots\binom{r_n+i_n}{i_n},$$ where $$k=0,\ information about binomial coefficients and their properties, refer to section 5.1 in Concrete Mathematics by
ldots,r_1+\cdots+r_n$$ ($$i_j$$ ranges from $$0$$ to $$r_j$$, for $$j=1,\ Graham, Knuth, and Patashnik. The problem deals specifically with the summation of products of binomial
ldots,n$$). If reformulate the problem. Multiply $$n$$ finite sums: $$\sum_{i_1=0}^ coefficients, satisfying certain conditions involving subscripts and constraints. More precisely, we aim to
{r_1}\binom{r_1+i_1}{r_1}\cdots\sum_{i_n=0}^{r_n}\binom{r_n+i_n}{r_n}$$ collect and determine: $$\sum_{\substack{i\_1 + i\_2 + ... + i\_n = k \\ 0 \le i\_j \le r\_j}} \binom{r\_1 + i\_1}{r\_1} \binom
sum parts such that $$i_1+\cdots+i_n=k$$. What is the result of every such sum. I {r\_2 + i\_2}{r\_2}...\binom{r\_n + i\_n}{r\_n}$$ for $k = 0,...,r\_1 + r\_2 + ... + r\_n,$ where $r\_1, r\_2, ...,
have found similar question here, but I can not connect it to this problem. Also r\_n$ are fixed positive integers. This expression can also be rewritten as: $$\left(\sum_{i\_1 = 0}^{r\_1} \binom
found a paper which uses probabilistic method to establish several generalisations {r\_1 + i\_1}{r\_1}\right) \cdot \left(\sum_{i\_2 = 0}^{r\_2} \binom{r\_2 + i\_2}{r\_2}\right)\cdot...\cdot \left(\
of Vandermonde identity (which to my dilettante view is somewhat similar to my sum_{i\_n = 0}^{r\_n} \binom{r\_n + i\_n}{r\_n}\right)$$ subject to the constraint $i\_1 + i\_2 + ... + i\_n = k.$
problem). Here is a small example just to be clear what I want to achieve. Let $$n= As mentioned in the initial statement, there exist connections between this problem and the Vandermonde identity—a
3$$ and $$r_1=1$$, $$r_2=2$$, $$r_3=3$$. Now tak" Do not just list concepts, but well-known combinatorial identity relating three binomial coefficients. However, establishing these connections
develop each one in detail before moving to the next, as we prioritize depth of requires advanced knowledge of generating functions and manipulation techniques beyond basic algebraic operations.
understanding and comprehensive exploration of the subject matter over breadth. To further illustrate the complexity involved in solving this problem, consider the case when $n=3, r\_1 = 1, r\_2
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: = 2,$ and $r\_3 = 3$: $$\sum_{\substack{i\_1 + i\_2 + i\_3 = k \\ 0 \le i\_j \le r\_j}} \binom{1 + i\_1}{1} \binom
Write with an academic, professional and engaging tone that captivates interest. - {2 + i\_2}{2} \binom{3 + i\_3}{3}$$ Expanding the expressions yields: $$\sum_{\substack{i\_1 + i\_2 + i\_3 = k \\ 0
Application: Incorporate specific, practical examples, such as proofs in calculus \le i\_j \le r\_j}} \frac{(1+i\_1)!}{(1!)(i\_1!)} \frac{(2+i\_2)(1+i\_2)}{(2!)(i\_2!)}\frac{(3+i\_3)(2+i\_3)(1+i\
or critical dates and figures in history. Do not include a title or an _3)}{(3!)(i\_3!)}$$ It becomes apparent that simplifying this expression relies heavily upon recognizing patterns
introduction, simply write the content without headlines and introductory phrases. within factorials and applying appropriate identities. While direct computation might yield a solution for small
Do not use images. values of $n, r\_1, ..., r\_n,$ and $k,$ finding a closed form or proving a pattern generally demands sophisticated
methods. One possible approach involves using multinomial coefficients and generating functions, although
discussing those topics goes beyond the scope of this explanation.
The problem presented involves finding the number of combinations of a sequence of streetlights, where a certain
number of consecutive lights must be functional. This type of problem falls under the category of combinatorics,
which deals with counting the number of ways certain patterns or structures can occur within a finite set. To
approach this problem, let's first review some fundamental concepts in combinatorics. **Combinations:** A
combination refers to selecting a subset of items from a larger set without regard to order. Given a set S
Write an educational piece suited for college students related to the following containing n distinct objects, the number of possible combinations of r items chosen from S is given by the
text snippet: "# Combination with a minimum number of elements in a fixed length binomial coefficient C(n,r), which can be calculated using the following formula: C(n,r)=n!rn!(n−r)!where ! denotes
subset I have been searching for long but unable to find a solution for this. My factorial, defined as the product of all positive integers up to that value. For example, 5!=5⋅4⋅3⋅2⋅1=120. In our
question is "Suppose you have n street lights(cannot be moved) and if you get any m context, we need to modify these standard combinations since there is a requirement for the selected subsequence to
from them then it should have atleast k working.Now in how many ways can this be consist of consecutive numbers. We will refer to this concept as *consecutive combinations.* Before diving into the
done" This seems to be a combination problem, but the problem here is "m" must be calculation of consecutive combinations, it is important to understand another essential idea in
sequential. Eg: 1 2 3 4 5 6 7 (Street lamps) Let m=3 Then the valid sets are, 1 2 3 combinatorics—*pigeonhole principle:* **Pigeonhole Principle:** If n pigeons occupy m holes, and n > m, then at
2 3 4 3 4 5 4 5 6 5 6 7 Whereas, 1 2 4 and so are invalid selections. So every set least one hole contains more than one pigeon. This simple yet powerful principle has far-reaching implications in
must have atleast 2 working lights. I have figured how to find the minimum lamps various mathematical problems, including those involving combinations and sequences. Applying the pigeonhole
required to satisfy the condition but how can I find the number of ways in it can principle, we know that when considering n streetlights and requiring at least k functioning lights out of any m
be done ? There should certainly some formula to do this but I am unable to find taken, the minimum number of operational lights needed would always be m - k + 1. Indeed, if fewer than m - k + 1
it.. :( Eg: Let n=7,m=4,k=3. The minimum number of lights that must be working to were operational, two groups of size m could be formed without having k functioning lights in either group.
satisfy the condition that "atleast 3 of any 4 lights must be working" is 5. It is Returning to the original example, we see that indeed five operating lights are necessary when choosing four
if 23467 are working. B" Do not just list concepts, but develop each one in detail adjacent lights with three needing to work. To compute the number of consecutive combinations satisfying the
before moving to the next, as we prioritize depth of understanding and conditions outlined above, consider arranging k − 1 dividers among the remaining n - m + k positions after fixing
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: the initial position of the first functional light. There will be n - m + k available slots, while only n - m needs
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an to be filled because the last k - 1 positions are already accounted for due to the contiguity constraint.
academic, professional and engaging tone that captivates interest. - Application: Therefore, the number of arrangements becomes equivalent to placing k - 1 indistinguishable dividers among n - m +
Incorporate specific, practical examples, such as proofs in calculus or critical k - 1 total locations. By applying the combinations formula mentioned earlier, substituting r = k - 1 and n = n - m
dates and figures in history. Do not include a title or an introduction, simply + k - 1, we obtain: C(n−m+k−1,k−1)=\frac{(n−m+k−1)!}{(k−1)!\left[(n−m)+k−1\right]!}Consecutive combinations now
write the content without headlines and introductory phrases. Do not use images. become accessible through this adjusted formulation based upon traditional combination theory adapted to fit the
constraints provided. Let us revisit the illustrative case introduced previously: If n = 7, m = 4, and k = 3, the
minimal count of illuminated bulbs equals 5. Hence, we apply the revised formula stated earlier: C(7−4+3−1,3−1)=\
frac{9!}{2!\cdot8!}=\frac{9×8}{2}=36Since there are 36 different ways to select three consecutive working
streetlights leaving at most one nonfunctional light between pairs of functional ones, we successfully solved the
The relationship between the sine of an angle and the measure of that angle is a fundamental concept in
trigonometry. It may initially seem plausible that if the sine of angle A is greater than the sine of angle B
within a triangle, then angle A must necessarily be greater than angle B. However, this intuitive notion is only
correct under certain conditions. This article will delve into the reasons behind this observation, providing a
deeper understanding of the properties of trigonometric functions and their applications in geometry. To begin
with, let us clarify some terminology. When discussing angles in this context, we refer to measurements made in
relation to a circle, where an angle's measure indicates the amount of rotation required to move from the initial
side of the angle to its terminal side. Angles are usually measured in degrees, with one full circle corresponding
to 360 degrees, or in radians, with one complete circle equal to 2π radians. For the purpose of our discussion, we
shall consider angles that fall within the range of 0 to 180 degrees or, equivalently, 0 to π radians, which
encompasses acute, right, and obtuse angles. Now, let us examine the behavior of the sine function across this
domain. The sine function, often denoted as sin(x) or sometimes abbreviated as just 'sin x', maps each input
value—namely, an angle expressed either in degrees or radians—to an output value representing the ratio of the
length of the opposite side of a right triangle with respect to the hypotenuse when the given angle serves as the
angle of reference. Within the interval of 0 to 180 degrees, the sine function exhibits several notable features:
1. It attains its maximum value at 90 degrees (or π/2 radians), where sin(90°)=sin(π/2)=1, reflecting the fact that
the height of a right triangle reaches its peak when the angle of consideration equals a right angle. 2. As the
angle decreases from 90 degrees towards zero, the sine value descends monotonically, eventually reaching zero as
the angle approaches zero. Specifically, sin(0°)=sin(0)=0. 3. Upon surpassing 90 degrees and continuing up to 180
degrees (or π radians), the sine function resumes increasing after momentarily dipping to its minimum value of -1
at 180 degrees (or π radians). 4. Throughout the entire interval spanning 0 to 180 degrees (or 0 to π radians), the
sine function remains continuous; that is, there are no abrupt jumps or gaps in its values. With these observations
in mind, let us revisit the original proposition: whether the inequality sin(A)>sin(B) implies that angle A exceeds
angle B. To analyze this claim more closely, consider Figure 1 below, depicting two arbitrary angles A and B
situated within a single unit circle. Here, point P represents the terminal side of angle A, while point Q marks
the endpoint associated with angle B. Furthermore, let O denote the center of the circle, thereby forming three
right triangles OPX, OQY, and POQ, respectively. By invoking the definition of the sine function mentioned earlier,
we can express sin(A) and sin(B) as follows: sin(A) = \frac{PY}{OQ} sin(B) = \frac{PX}{OP} Since the radius of the
Write an educational piece suited for college students related to the following unit circle measures unity, we obtain the relationships OP=1 and OQ=1. Consequently, the above expressions simplify
text snippet: "# If sinA > sinB, then A>B? 1. Sep 19, 2011 ### Michael_Light I am to: sin(A) = PY sin(B) = PX At first glance, comparing these quantities directly might suggest that sin(A) indeed
in pre-u level, i apologize if i have asked something dumb but it bothers me a lot. outranks sin(B) whenever angle A eclipses angle B. Nevertheless, recall that the position of points P and Y along
Here it goes my question: If A and B are angles inside a triangle, and sinA > sinB, the vertical axis depends crucially upon the location of the terminal side of angle A—a factor that has been
can i conclude that A>B? If yes, how can we prove it? Can i use this argument in deliberately omitted thus far. Thus, additional information concerning the relative positions of angles A and B is
proving question? 2. Sep 19, 2011 ### AlephZero Your restriction to "angles inside indispensable for arriving at a definitive conclusion regarding the validity of the proposed statement. As alluded
a triangle" means angles between 0 and 180 degrees (or 0 and pi radians). Look at a to by previous respondents, the truthfulness of the assertion hinges critically on constraining the allowable range
graph of the sin function between 0 and 180 degrees, and you should be able to see of angles A and B further. More precisely, the contention holds unassailably provided that angles A and B remain
why this is false. If you restrict the range to between 0 and 90 degrees not 180, strictly confined to the interval spanning from 0 to 90 degrees (or 0 to π/2 radians). Under such restrictions, the
then it is true (that is also clear from looking at the graph). Of course looking sine function becomes strictly increasing throughout this subinterval, implying that any increment in the angle
at a graph isn't really a "proof", but proving this rigorously needs math beyond leads unequivocally to a proportional uptick in the corresponding sine value. Mathematically speaking, this
what you have probably studied at pre-university level. 3. Sep 19, 2011 ### dalcde property translates to the existence of a well-defined inverse function, commonly designated as arcsin or sin^-1,
Can't we just find the derivative of sin A and see that it is always positive in allowing one to recover the unique angle associated with any specified sine value lying within the permissible
(0,90), henc" Do not just list concepts, but develop each one in detail before bounds. In contrast, once the upper limit of 90 degrees (or π/2 radians) is transgressed, complications inevitably
moving to the next, as we prioritize depth of understanding and comprehensive arise due to the periodic nature of the sine function. Indeed, beyond this threshold, multiple distinct angles
exploration of the subject matter over breadth. Focus on: - Rigor: Ensure in-depth correspond to identical sine values, rendering the sought-after ordering ambiguous and necessitating additional
coverage of the concepts/sections. - Engagement: Write with an academic, constraints to ensure uniqueness. One possible resolution entails mandating that angles A and B adhere to the
professional and engaging tone that captivates interest. - Application: Incorporate stipulation outlined previously, namely that they both assume values exclusively within the interval from 0 to 90
specific, practical examples, such as proofs in calculus or critical dates and degrees (or 0 to π/2 radians). Alternatively, one could impose alternative requirements guaranteeing unambiguous
figures in history. Do not include a title or an introduction, simply write the comparisons among sine values belonging to larger angles—for instance, demanding that neither angle exceeds half of
content without headlines and introductory phrases. Do not use images. a full circle (i.e., 180 degrees or π radians). Regardless of the particular choice adopted, however, it becomes
manifestly clear that global statements linking absolute magnitudes of sine values to angular hierarchies cannot
dispense entirely with auxiliary hypotheses delimiting the admissible set of configurations. Before concluding this
exposition, allow us to address a tangential yet germane issue raised during the course of our investigation:
specifically, the suggestion proffered by user "dalcde" pertaining to the recourse of examining the derivative of
the sine function as a potential avenue for establishing the desired result. While the idea itself is commendable
owing to its grounding in the formalism of advanced calculus, employing differentiation techniques presupposes
familiarity with sophisticated mathematical machinery generally unfamiliar to most high school or even early
university curricula. Nonetheless, since the topic warrants brief elaboration, we provide herewith a concise sketch
elucidating the underlying rationale. Recall that the derivative of a function quantifies its instantaneous rate of
change at each point within its domain. Moreover, the sign of the derivative conveys valuable information about the
local behavior of the function, distinguishing between regions characterized by growth, decay, or stability. For
the sine function defined over real numbers—which, incidentally, extends beyond the narrower scope considered
heretofore—one can demonstrate via elementary geometric arguments that its derivative coincides with the cosine
function, denoted as cos(x) or 'cos x'. Symbolically, this correspondence manifests as: \frac{d}{dx} [\sin(x)] = \
cos(x) Concentrating now on the restricted interval from 0 to 90 degrees (or 0 to π/2 radians), observe that the
cosine function assumes solely nonnegative values throughout this segment, achieving its maximum value of 1 exactly
at the midpoint (i.e., 45 degrees or π/4 radians). Accordingly, the derivative of the sine function stays
consistently positive across this entire subdomain, underscoring the reality that the sine function increases
monotonically within this region—as already substantiated through alternate means earlier. Therefore, although
appealing to derivatives constitutes an elegant approach rooted firmly within the framework of mathematical
analysis, deploying such tools lies outside the purview of standard pre-college mathematics education. In summary,
we have explored in depth the intricate connection between the sine values of two angles and their relative angular
magnitude. Initially posited as a seemingly straightforward query, the problem ultimately revealed a rich tapestry
of interrelated concepts drawn from various branches of mathematics, including geometry, trigonometry, and
calculus. Through careful examination, we established that the conjecture holds true under stringent conditions
limiting the allowable range of angles to intervals wherein the sine function retains its strict monotonicity. By
traversing this intellectual journey together, we hope to have instilled a profound appreciation for the beauty and
complexity inherent in seemingly innocuous questions, inspiring curiosity and fostering enthusiasm for continued
learning and discovery.
The binomial distribution is a fundamental concept in statistics and probability theory. It models the number of
"successes" in a fixed number of independent experiments, also known as Bernoulli trials. Each experiment has two
possible outcomes, traditionally referred to as "success" and "failure", with associated probabilities denoted by
Write an educational piece suited for college students related to the following $p$ and $1-p$, respectively. The binomial distribution describes the sum of these binary outcomes across multiple
text snippet: "Binomial {stats} R Documentation ## The Binomial Distribution ### trials. In R programming language, there are four functions available to work with the binomial distribution:
Description Density, distribution function, quantile function and random generation `dbinom()`, `pbinom()`, `qbinom()`, and `rbinom()`. These functions allow users to compute densities, cumulative
for the binomial distribution with parameters size and prob. This is conventionally distributions, quantiles, and generate random samples from the binomial distribution, respectively. We will discuss
interpreted as the number of ‘successes’ in size trials. ### Usage dbinom(x, size, each of these functions along with their arguments and applications. `dbinom(x, size, prob, log = FALSE)` computes
prob, log = FALSE) pbinom(q, size, prob, lower.tail = TRUE, log.p = FALSE) qbinom the probability mass function (PMF) of the binomial distribution at point $x$. This function returns the
(p, size, prob, lower.tail = TRUE, log.p = FALSE) rbinom(n, size, prob) ### probability of observing exactly $x$ successes in $size$ independent Bernoulli trials, where each trial has a
Arguments x, q vector of quantiles. p vector of probabilities. n number of success probability of $prob$. By default, this function returns the PMF values directly; however, setting `log=
observations. If length(n) > 1, the length is taken to be the number required. size TRUE` provides the logarithmic values instead. For example, let's calculate the probability of obtaining 3 heads
number of trials (zero or more). prob probability of success on each trial. log, when flipping a fair coin (i.e., success probability $p = 0.5$) five times ($size = 5$). ```R > dbinom(3, 5, 0.5)
log.p logical; if TRUE, probabilities p are given as log(p). lower.tail logical; if [1] 0.2734375 ``` `pbinom(q, size, prob, lower.tail = TRUE, log.p = FALSE)` calculates the cumulative distribution
TRUE (default), probabilities are P[X \le x], otherwise, P[X > x]. ### Details The function (CDF) of the binomial distribution up to point $q$. When $lower.tail = TRUE$ (the default value), it gives
binomial distribution with size = n and prob = p has density p(x) = {n \choose x} the probability of having no more than $q$ successes in $size$ trials. Conversely, if $lower.tail = FALSE$, it
{p}^{x} {(1-p)}^{n-x} f" Do not just list concepts, but develop each one in detail computes the probability of having strictly more than $q$ successes. Setting `log.p=TRUE` yields the logarithmic
before moving to the next, as we prioritize depth of understanding and CDF values rather than direct probabilities. Let us find the probability of getting at most three heads in ten coin
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: tosses. ```R > pbinom(3, 10, 0.5) [1] 0.171875 ``` `qbinom(p, size, prob, lower.tail = TRUE, log.p = FALSE)` finds
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an the smallest integer $k$ satisfying the condition $P[X <= k] >= p$, i.e., the $(1 - p)$ percentile of the binomial
academic, professional and engaging tone that captivates interest. - Application: distribution. Here, $size$ denotes the total number of trials and $prob$ indicates the success probability per
Incorporate specific, practical examples, such as proofs in calculus or critical trial. Again, you can set `log.p=TRUE` to obtain the logarithmic values. Suppose we want to determine the median
dates and figures in history. Do not include a title or an introduction, simply number of heads in twenty coin tosses. ```R > qbinom(0.5, 20, 0.5) [1] 10 ``` Finally, `rbinom(n, size, prob)`
write the content without headlines and introductory phrases. Do not use images. generates a sample of $n$ random numbers drawn from the specified binomial distribution with parameters $size$ and
$prob$. For instance, generating five random samples representing the number of heads obtained after flipping a
fair coin ten times. ```R > rbinom(5, 10, 0.5) [1] 6 9 6 4 5 ``` Understanding and applying these R functions
allows researchers and analysts to perform essential statistical analyses involving discrete random variables and
apply them confidently in various fields like genetics, finance, engineering, social sciences, among others.
Combinatorial mathematics, also known as combinatorics, deals with counting, arranging, and selecting objects
according to certain rules. Problems in this field often involve discrete structures rather than continuous ones.
The AMC problem #1326 presented above is an example of a classic combinatorial problem that requires us to count
the number of distinct ways to paint the faces of a regular tetrahedron under equivalence relations. To solve it
rigorously, let's first clarify some essential concepts from group theory and Burnside's Lemma. These tools will
allow us to tackle similar combinatorial problems systematically. Group Theory Basics: In abstract algebra, a group
$G$ consists of a set of elements together with an operation satisfying four fundamental properties: closure,
associativity, identity element existence, and inverse element existence. For instance, consider the set of all
permutations of three letters $\{a,b,c\}$; there are six such arrangements: $(a\, b \, c)$, $(a\, c\, b)$, $(b\, a
\, c)$, $(b\, c\, a)$, $(c\, a\, b)$, and $(c\, b\, a)$. We denote these permutations as $\{(1)(2)(3), (1\,2)(3),
(1\,3)(2), (1)(2\,3), (1\,2\,3), (1\,3\,2)\}$. Here, the operation is function composition, denoted
multiplicatively. Clearly, when combining any pair of these permutations, say $(\tau_1,\tau_2)$, we always obtain
another permutation within our original set since no new symbols appear after applying both transformations. Hence,
closure holds. Associativity follows directly from the definition of function composition, while the identity
Write an educational piece suited for college students related to the following element and inverse element exist due to the nature of permutations. Stabilizers & Orbits: Let $X$ be a finite set
text snippet: "Problem #1326 1326 Each face of a regular tetrahedron is painted acted upon by a group $G$. Specifically, suppose we have a map $\varphi : G \times X \rightarrow X$, where $\varphi
either red, white, or blue. Two colorings are considered indistinguishable if two (\sigma, x)$ denotes the action of $\sigma \in G$ on $x \in X$. Then, for fixed $x \in X$, define its orbit as the
congruent tetrahedra with those colorings can be rotated so that their appearances subset of $X$ given by: $$Orb(x)=\{\varphi(\sigma,x):\,\sigma\in G\}$$ Intuitively, the orbit of $x$ comprises all
are identical. How many distinguishable colorings are possible? $\mathrm {(A)} 15\ elements reachable via actions induced by $G$. Now, fix $\sigma \in G$; then, the stabilizer subgroup of $\sigma$
qquad \mathrm {(B)} 18\qquad \mathrm {(C)} 27\qquad \mathrm {(D)} 54\qquad \mathrm at point $x$, denoted $Stab_{\sigma}(x)$, is defined as: $$Stab_{\sigma}(x)=\{g\in G:\,\varphi(g,x)=x\}$$ That is,
{(E)} 81$ This problem is copyrighted by the American Mathematics Competitions. the stabilizer subgroup contains precisely those elements leaving $x$ unchanged when acting through them. It turns
Note: you aren't logged in. If you log in, we'll keep a record of which problems out that every orbit has the same cardinality as its stabilizer subgroups up to conjugacy - meaning they share the
you've solved. • Reduce fractions to lowest terms and enter in the form 7/9. • same order. Thus, computing the size of an arbitrary stabilizer suffices for determining the number of elements in
Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square an orbit. Burnside's Lemma: Now, we introduce Burnside's lemma, named after William Burnside, a prominent British
roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as mathematician who contributed significantly to group theory and representation theory. Let $X$ again represent a
appropriate. • Exponents should be entered in the form 10^10. • If the problem is finite set equipped with a group action $\varphi : G \times X \rightarrow X$. Define the cycle index polynomial
multiple choice, enter the appropriate (capital) letter. • Enter points with $P_G(z_1, z_2, ..., z_n)$ as: $$P_G(z_1, z_2, ..., z_n)=\frac{1}{|G|}\sum_{\sigma\in G}z^{c_1(\sigma)}_{1}\,z^{c_2
parentheses, like so: (4,5) • Complex numbers sho" Do not just list concepts, but (\sigma)}_{2}\,\cdots z^{c_n(\sigma)}_{n}$$ where $|G|$ counts the total number of elements in $G$, and $c_i(\
develop each one in detail before moving to the next, as we prioritize depth of sigma)$ gives the number of cycles of length $i$ present in the unique disjoint cycle decomposition associated with
understanding and comprehensive exploration of the subject matter over breadth. $\sigma$. With these definitions established, Burnside's lemma states that the number of orbits, denoted $N$,
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: equals: $$N=\frac{1}{|G|}\sum_{\sigma\in G}|Fix(\sigma)|$$ Here, $Fix(\sigma)$ represents the set of fixed points
Write with an academic, professional and engaging tone that captivates interest. - under the transformation $\sigma$: $$Fix(\sigma)=\{x\in X:\,\varphi(\sigma,x)=x\}$$ Solution to Problem #1326 using
Application: Incorporate specific, practical examples, such as proofs in calculus Burnside's Lemma: Returning now to the initial question regarding painting the faces of a regular tetrahedron, we
or critical dates and figures in history. Do not include a title or an may apply Burnside's lemma to find the answer. Firstly, observe that the symmetry group of a regular tetrahedron,
introduction, simply write the content without headlines and introductory phrases. denoted $Td$, consists of 24 rigid motions preserving shape and orientation—these include 12 proper rotations
Do not use images. ($T$), 6 improper rotations ($Th$), and 6 reflections along planes ($Td$). Since only colors differ here, we focus
solely on the rotation group $T$. Label the vertices of our tetrahedron $v_1, v_2, v_3,$ and $v_4$. Under the
action of $T$, any vertex permutes among itself and others, yielding potential orbits $\{v_1\}, \{v_1, v_2\}, \
{v_1, v_2, v_3\}$, and $\{v_1, v_2, v_3, v_4\}$. As mentioned earlier, equivalent stabilizer groups share the same
cardinalities. Therefore, we need only compute the sizes of stabilizer subgroups corresponding to each orbit
representative: * Identity ($\epsilon$): All vertices remain fixed; thus, $|Stab_\epsilon(v_i)|=4!$. * Reflection
across plane containing edge ${v_i, v_j}$: Fixes two adjacent vertices and swaps other pairs; hence, $|Stab_{ij}
(v_k)|=2$. * Rotation around axis passing through midpoints of edges ${v_i, v_j}$ and ${v_k, v_l}$: Interchanges
opposite vertices while fixing remaining ones; thus, $|Stab_{ijkl}(v_m)|=2$. Next, calculate the contribution of
each case towards the cycle index polynomial: $$P_T(z_1, z_2, z_3, z_4)=\frac{1}{24}[z^4_1+6z_1z^2_2+8z_3^1+3z_1]
$$. Finally, recalling Burnside's lemma, plug in values for each variable representing colors used: $$N=\frac{1}
{24}[(R+W+B)^4 + 6(R+W+B)^2(R^2+W^2+B^2)+8(R+W+B)+3]$$ Expanding and simplifying expressions yields: $$N = \frac{1}
{24}[15(R^4+W^4+B^4)+30(R^3W+RW^3+RB^3+BR^3+WC^3+CW^3)+60(R^2W^2+R^2C^2+W^2C^2)+24(RWCB)]$$. Since each term
corresponds to admissible color patterns, we equate coefficients to options provided in the problem statement. Upon
inspection, we see that option C matches perfectly: $$N=(R^4+W^4+B^4)+3(R^3W+RW^3+RB^3+BR^3+WC^3+CW^3)+6(R^2W^2+R^
The boundary value problem presented is a type of differential equation where the function value is specified at
the boundaries of the domain. In this case, we have the second order ordinary differential equation -y''(x) = 0,
where y(x) is the unknown function, defined on the open interval Omega := (0,π). Additionally, the boundary
conditions are given by y(0) = y(π) = 0. This particular problem falls under the category of Dirichlet boundary
problems. Solving such problems involves finding functions that satisfy both the differential equation and the
prescribed boundary values. For this specific problem, analytical solutions exist and can be readily found using
Write an educational piece suited for college students related to the following standard techniques. However, when faced with more complex equations or higher dimensions, numerical methods become
text snippet: "• Yesterday, 16:31 We consider the boundary problem \begin{align*}-& essential tools for obtaining approximate solutions. These methods often involve discretizing the underlying domain
y''(x)=0 \ \ \text{ for } \ x\in \Omega:=(0,\pi) \\ &y(0)=y(\pi)=0\end{align*} into smaller subintervals or elements, transforming the continuous problem into a discrete system of linear
Let's pick the grid... 16 replies | 396 view(s) • Yesterday, 14:11 Yes. And it also algebraic equations. In subsequent messages, there appears to be some discussion about applying iterative methods
assumes we take $\nu=2$ steps with the damped Jacobi method. (Nod) It's not quite like the damped Jacobi method to solve these systems of linear equations arising from discretization processes.
clear from the text to me. With a choice... 16 replies | 396 view(s) • When solving large sparse linear systems, iterative methods generally converge faster than direct methods.
Yesterday, 12:43 Ahh I thought that at the proposition 8.9.13 (pg. 365) they would Moreover, their memory requirements tend to scale better with increasing matrix size, making them particularly
mean that we can get a good approximation for the error, but now I read that suitable for high dimensional problems. The damping parameter ω plays a crucial role in ensuring convergence;
again... 16 replies | 396 view(s) • Yesterday, 12:19 We can freely choose to use $\ selecting appropriate values of ω may lead to accelerated convergence rates. As evidenced in the conversation, the
omega=2/3$ or a different $\omega$, or not use the damped Jacobi method at all. The user has the freedom to select any value between 0 and 2 for ω, including the commonly used choice of ω = 2/3.
chapter explains that it's giving... 16 replies | 396 view(s) • Yesterday, 11:47 Ah Alternatively, one could even opt out of utilizing the damped Jacobi method entirely. Proposition 8.9.13 mentioned
ok! Do we take $\omega=2/3$ only at the 2-grid or in general? (Wondering) Could you earlier seems to refer to estimates regarding the rate of convergence towards the true solution during the
give me a specific example where we can apply the method... 16 replies | 396 view iteration process. Specifically, propositions of this form provide bounds on either the absolute error ||e\_k|| = |
(s) • Yesterday, 06:19 Thanks steenis .." Do not just list concepts, but develop |x\_k - x*|| or relative errors r\_k = ||e\_k|| / ||x\_k|| after k iterations, enabling us to determine how rapidly
each one in detail before moving to the next, as we prioritize depth of our approximations improve. Here, e\_k represents the difference between the exact solution x* and the current
understanding and comprehensive exploration of the subject matter over breadth. estimate x\_k, while k denotes the number of completed iterations. Note that tighter bounds typically correspond to
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: sharper decreases in error per iteration, leading to quicker overall convergence. To illustrate the application of
Write with an academic, professional and engaging tone that captivates interest. - these ideas, let's examine a simple finite difference approach applied to the present boundary value problem. We
Application: Incorporate specific, practical examples, such as proofs in calculus begin by dividing the interval (0,π) into N equally sized segments, resulting in nodes x\_j = jΔx for j = 0,...,N,
or critical dates and figures in history. Do not include a title or an where Δx = π/(N+1) signifies the step length. Approximating the second derivative via central differences yields
introduction, simply write the content without headlines and introductory phrases. (-y\_{j+1} + 2y\_j - y\_{j-1})/(Δx)^2 ≈ -y''(x\_j), which holds provided y(x) possesses sufficient smoothness
Do not use images. properties near the node x\_j. Imposing the boundary condition results in a tridiagonal coefficient matrix A, whose
entries depend solely on N and Δx. Consequently, the task reduces to determining the vector b such that Ax = b
corresponds to the set of nodal values obtained through evaluating the known exact solution at each interior point
within the partitioned domain. Applying the damped Jacobi method then proceeds according to the recursion formula:
x\_k+1 = D^-1[(D-A)x\_k + b], where D denotes the diagonal part of A, representing the sum of off-diagonal terms
along each row. At every stage, the algorithm updates its current guess x\_k based on the previous approximation
until satisfactory accuracy levels are reached or predetermined stopping criteria are fulfilled.
The given text presents an algorithm for generating a random permutation of a set containing N elements. A
permutation is an arrangement of items in a particular order, so a permutation of size N consists of N distinct
integers arranged in some sequence. This concept has various applications across different fields like computer
science, mathematics, physics, and statistics. Here, you will learn about the importance of randomness,
permutations, and their generation using the provided algorithm. Furthermore, we will delve into its
implementation, analyze why it generates a uniform random permutation, and discuss potential extensions. ### Random
Numbers and Permutations Random numbers play a crucial role in several areas of research and development. By
creating sequences free from patterns or biases, they enable us to model complex systems more accurately, perform
better statistical analysis, design robust cryptographic protocols, simulate real-world phenomena, and much more.
Permutations lie at the heart of combination problems, which involve arranging objects within groups while
preserving uniqueness. These ideas find application in combinatorics, graph theory, optimization, machine learning,
and data processing tasks. When faced with choosing a single item out of multiple possibilities, studying
permutations can provide valuable insights regarding overall selection probabilities. ### Algorithm Description The
Write an educational piece suited for college students related to the following proposed algorithm revolves around two nested loops iterating through the range 0 to N-1. At first glance, the
text snippet: "시간 제한메모리 제한제출정답맞힌 사람정답 비율 60 초 (추가 시간 없 outer loop builds up the identity permutation – a simple array filled with sequential integers starting from zero.
음) 512 MB0000.000% ## 문제 A permutation of size N is a sequence of N numbers, However, things become interesting when examining the inner loop, responsible for shuffling these values randomly.
each between 0 and N-1, where each number appears exactly once. They may appear in For every iteration i of the outer loop, ranging from 0 to N-1, the algorithm selects another index j uniformly at
any order. There are many (N factorial, to be precise, but it doesn't matter in random from interval i to N-1. It then swaps the contents of both indices in the working array. As demonstrated by
this problem) permutations of size N. Sometimes we just want to pick one at random, the example below, this process eventually leads to a completely shuffled version of the initial identity
and of course we want to pick one at random uniformly: each permutation of size N permutation. Consider a toy example where N equals four: Initial state: [0, 1, 2, 3] Iteration 1: p = randint
should have the same probability of being chosen. Here's the pseudocode for one of (1..3), let's say p=2; swap(a[1], a[2]); -> [0, 2, 1, 3] Iteration 2: p = randint(2..3), suppose p=3; swap(a[2], a
the possible algorithms to achieve that goal (we'll call it the good algorithm [3]); -> [0, 2, 3, 1] Iteration 3: p = randint(3..3); since there's only one option left, no swap occurs; thus, the
below): for k in 0 .. N-1: a[k] = k for k in 0 .. N-1: p = randint(k .. N-1) swap(a final result remains unchanged after three iterations: [0, 2, 3, 1]. ### Uniformity Analysis To ensure our
[k], a[p]) In the above code, randint(a .. b) returns a uniform random integer algorithm produces truly random permutations, we need to prove that every valid reordering has equal chances of
between a and b, inclusive. Here's the same algorithm in words. We start with the appearing during execution. Consider analyzing the behavior based on the moment right before performing the last
identity permutation: all numbers from 0 to N-1 written in increasing order. Then, swap operation in the inner loop. Let K denote the current value of the outer loop variable (i.e., K ranges from 0
for each k between 0 and N-1, inclusive, we pick an indepen" Do not just list to N-1). Also, define P_K as the remaining segment length yet to be processed in the inner loop (initially P_N=N,
concepts, but develop each one in detail before moving to the next, as we then gradually decreasing down to P_1=1). Now, focus on counting the total number of unique ways to reach each
prioritize depth of understanding and comprehensive exploration of the subject intermediate step, including the residual subarray lengths still open for further shuffles. At stage K, observe
matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the concepts/ that the residual subarray contains K contiguous slots filled with values {K, ..., N-1}. Since the algorithm picks
sections. - Engagement: Write with an academic, professional and engaging tone that a random position inside this region, namely P_K+1, the number of available choices corresponds directly to P_K
captivates interest. - Application: Incorporate specific, practical examples, such itself! Moreover, notice how the choice made at iteration I impacts subsequent stages J > I because the former
as proofs in calculus or critical dates and figures in history. Do not include a determines the latter's domain limits. Consequently, each consecutive phase inherits the previous ones' degrees of
title or an introduction, simply write the content without headlines and freedom minus one, yielding respective counts F_J = P_J - (J-1). With this recursive relationship established,
introductory phrases. Do not use images. compute the product of all individual factors F_J contributing to the current configuration under inspection: Total
Count = F_N * F_(N-1) * ... * F_(K+1) = (P_N-(N-1)) * (P\_{N-1}-(N-2)) * ... * (P\_K-(K-1)) Recall that P_N=N, P\_J
=(N-J+1), implying: Total Count = (N-(N-1)) * ((N-(N-1))-((N-1)-1)) * ... * (((N-K)+1)-(K-1)) = 1 * 2 * ... * (N-K)
This expression turns out to be equivalent to the formula computing combinations of n objects taken r at a time,
denoted as C(n,r)=n!/(r!(n-r)!) – precisely what we aimed for demonstrating the desired uniform distribution
property! ### Extensions & Applications Besides the direct applicability mentioned earlier, similar techniques form
the backbone behind numerous advanced methods, e.g.: * Fisher–Yates Shuffle: Generalizes the described approach by
allowing arbitrary access patterns rather than strictly sequential scans. * Knuth Shuffle: An alternative name for
the Fisher–Yates Shuffle emphasizing Donald E. Knuth’s contribution towards popularization and formalization of the
methodology. * Sattolo’s Algorithm: Variant favoring cyclic shifts instead of generic element exchanges, leading to
simplified implementations especially appealing for circular lists or arrays representing rings. Overall, mastering
fundamental principles underlying seemingly straightforward procedures opens doors to appreciating intricate
machinery lurking beneath surface appearances, thereby fostering deeper comprehension levels capable of supporting
future endeavors requiring creative adaptations tailored specifically to novel contexts.
Write an educational piece suited for college students related to the following The concept of uniform continuity is often a source of confusion for many students studying analysis, and your
text snippet: "## confusion about the definition of Uniform Continutiy A function question highlights some common misconceptions regarding its definition and application. Let's address these issues
is uniformly continuous iff for every epsilon>0 there exists delta>0 such that for step by step to clarify the concept. First, let's revisit the definition of uniform continuity: A function `f`
all x in the domain of f and for all y in the domain of f, |x-y|<delta =>|f(x)-f(y) defined on a set `E` is said to be uniformly continuous if and only if for every positive number `ε`, there exists
|<epsilon. Here is what confuses me. How can there be a delta such that |x-y|<delta another positive number `δ` such that for all pairs of points `x` and `y` in `E`, whenever `|x - y| < δ`, it
for ALL x and y. Since epsilon depends on delta, we can pick epsilon such that follows that `|f(x) - f(y)| < ε`. This statement encompasses two main ideas: 1. For any given `ε > 0`, no matter
delta is small. Then we can surely pick x and y such x-y is bigger than delta. For how small, you should be able to find a corresponding `δ > 0`. 2. That particular `δ` must work for all pairs of
example, x^2 is uniformly continuous on [-5,5] because for epsilon>0, when delta= points `x` and `y` within the domain `E`. Now, let's examine the concern raised in your question: "How can there be
epsilon/10, |x^2-y^2|<|x+y||x-y|< or = 10|x-y|<10*delta=epsilon. Right here delta= a `δ` such that `|x - y| < δ` for *all* `x` and `y`?" Actually, this interpretation is incorrect. Instead, consider
epsilon/10. The definition states that for ANY x,y in domain f, |x-y|<delta. If we the correct phrasing: For a fixed `ε > 0`, we aim to find a suitable `δ > 0` so that `|x - y| < δ` implies `|f(x) -
pick x=5 and y=1 we have 4<epsilon/10 for any epsilon>0. But that is impossible f(y)| < ε` holds true *for any pair* of `x` and `y` from our chosen domain. We do not require `δ` to satisfy `|x -
since I can pick epsilon=1. 4<(1/10) is not true. PhysOrg.com science news on y| < δ` simultaneously for all possible choices of `x` and `y`. In light of this, let us analyze the provided
PhysOrg.com >> Leading 3-D printer firms to merge in $403M deal (Update)>> LA to example involving the function `f(x) = x^2` on the interval `[-5, 5]`. You correctly demonstrated finding a value
give every student an iPad;$30" Do not just list concepts, but develop each one in of `δ` based on a given `ε` using the inequality: ```scss |x^2 - y^2| = |(x + y)(x - y)| <= 10|x - y| ``` Given `ε
detail before moving to the next, as we prioritize depth of understanding and > 0`, setting `δ = min{1, ε / 10}` guarantees that if `|x - y| < δ`, then `|f(x) - f(y)| = |x^2 - y^2| < ε`.
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: However, this does not imply that `|x - y| < δ` for arbitrary `x` and `y` in the interval. It merely indicates that
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an when selecting any two numbers satisfying `|x - y| < δ`, their respective squares will differ by less than `ε`. To
academic, professional and engaging tone that captivates interest. - Application: summarize, uniform continuity requires finding a single `δ` that works universally across the entire domain for a
Incorporate specific, practical examples, such as proofs in calculus or critical given `ε`. Mistakenly believing that `δ` needs to restrict the distances between all pairs of `x` and `y`
dates and figures in history. Do not include a title or an introduction, simply concurrently leads to contradictions like those presented in the original question. By carefully examining the
write the content without headlines and introductory phrases. Do not use images. definitions and applying them methodically, clarity emerges, solidifying the foundation required to tackle more
complex problems in mathematical analysis.
Now that we have introduced the concepts of necessary and sufficient conditions, let us delve deeper into their
meaning and significance through some examples. We will also examine the concept of absolute continuity and its
relation to these conditions. To begin with, consider the logical statement "If I study hard, then I will pass the
exam." The "if" part of this statement, namely "I study hard," is known as the sufficient condition because it
guarantees the truth of the conclusion, i.e., passing the exam. However, it does not necessarily mean that passing
Write an educational piece suited for college students related to the following the exam is solely dependent on studying hard; there could be other factors at play as well. Thus, while satisfying
text snippet: "# Necessary vs. Sufficient? In sum, the sufficient condition (a.k.a. the sufficient condition ensures success, failing to satisfy it does not always imply failure. Conversely, consider
the "if" direction) allows you to get what you want. That is, if you assume the the statement "Only if I study hard will I pass the exam." Here, the word "only" implies that studying hard is a
sufficient condition, you'll obtain your desired conclusion. It's enough. It's necessary condition for passing the exam. If someone fails to meet this condition, they cannot possibly achieve the
sufficient. On the other hand, the necessary condition (a.k.a. the "only if" desired outcome of passing the exam. However, meeting the necessary condition alone does not guarantee success
direction) is the one you must assume in order to get what you want. In other since there may be additional requirements. For instance, even if one studies hard, they might still fail if they
words, if you don't have the necessary condition then you can't reach your desired do not manage their time effectively during the test. Now, let us turn our attention to absolute continuity, which
conclusion. It is necessary # Absolute Continuity (Part Two) There are two has applications in various fields including mathematics and physics. There exist two distinct yet interconnected
definitions of absolute continuity out there. One refers to an absolutely definitions of absolute continuity - one pertaining to functions and another concerning measures. An absolutely
continuous function and the other to an absolutely continuous measure. And although continuous function is characterized by its ability to accumulate area under its curve gradually without
the definitions appear unrelated, they are in fact very much related, linked experiencing any jumps or gaps. More formally, given ε > 0, no matter how many disjoint intervals {(ai, bi)}i=1^n
together by Lebesgue's Fundamental Theorem of Calculus. This is the second of a one chooses within the domain of the function, their total length must be greater than zero (i.e., ∑ni=1 (bi-ai)>0)
two-part series where we explore that relationship. Welcome to the final post in a to ensure that the corresponding variation of the function F, denoted by VF((a_i, b_i)), satisfies the inequality
six-part series where we prove that the fundamental grou" Do not just list VF((a_i, b_i)) < ε. On the other hand, an absolutely continuous measure is defined relative to a sigma-finite
concepts, but develop each one in detail before moving to the next, as we reference measure, typically the Lebesgue measure m. Specifically, a measure μ is said to be absolutely continuous
prioritize depth of understanding and comprehensive exploration of the subject with respect to m if for every measurable set E, the implication m(E)=0⇒μ(E)=0 holds true. A classic example of an
matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the concepts/ absolutely continuous measure is the Riemann integral of a nonnegative integrable function f, denoted by ∫f dm,
sections. - Engagement: Write with an academic, professional and engaging tone that which assigns a mass proportional to the height of the graph of f to each infinitesimal interval (x, x+dx). Despite
captivates interest. - Application: Incorporate specific, practical examples, such appearing unrelated initially, these two definitions share a profound connection facilitated by Lebesgue's
as proofs in calculus or critical dates and figures in history. Do not include a Fundamental Theorem of Calculus. According to this theorem, if F is an arbitrary absolutely continuous function,
title or an introduction, simply write the content without headlines and then it possesses an integrable derivative almost everywhere, implying that F'(x) = f(x) for almost all x belonging
introductory phrases. Do not use images. to the domain of F. Furthermore, F can be expressed as the antiderivative of its derivative plus a constant, i.e.,
F(x) = ∫f(t) dt + C, demonstrating the intimate link between absolute continuity of functions and measures.
Understanding the nuances of necessary versus sufficient conditions and their implications across diverse contexts
deepens our appreciation for mathematical structures and enables more robust problem-solving capabilities.
Similarly, grasping the intricacies of absolute continuity provides valuable insights into real analysis and
Write an educational piece suited for college students related to the following
text snippet: "# What is the energy dissipation through the wire? ## Homework
Statement A 3 m long copper wire that has a diameter of 6 mm is connected to a 9 V
battery. a) What is the current through the wire? b) If wire a was connected to a When calculating the cross-sectional area (A) of the copper wire, it seems that there was an error in converting
battery for 1 second, how much energy would the wire dissipate? ## Homework the diameter to radius. The given diameter of the wire is 6 mm, so the radius would be half of that, which is 3 mm
Equations A = $\pi$$r^{2}$ R = $\rho$$\frac{L}{A}$ i = $\frac{V}{R}$ $\rho$ = 1.72 or 3 x 10^-3 meters. Therefore, the calculation for the area should be: A = π(3 × 10^-3 m)^2 A = 2.83 x 10^-5 m^2
x $10^{-8}$ P = i * V W = P * t ## The Attempt at a Solution I got: R = 1.82mΩ i = Now let's calculate the resistance (R) using the formula R = ρ(L/A), where ρ is the resistivity of copper, L is the
4945 A p = 44505.45 I'm unsure of my answers. The values seem to be too high. Last length of the wire, and A is the cross-sectional area. Substituting the known values gives: R = (1.72 x 10^-8 Ω·m)
edited: Related Engineering and Comp Sci Homework Help News on Phys.org rude man (3 m / 2.83 x 10^-5 m^2) R = 0.182 Ω Next, let's find the current (i) flowing through the wire using Ohm's law, i =
Homework Helper Gold Member Your equations are all correct. Your R is too low by 2 V/R, where V is the voltage: i = 9 V / 0.182 Ω i = 49.45 A The calculated value for the current is still quite
orders of magnitude! Recompute A! Well, I tried to recalculate A and I got the same large, but it is now within a reasonable range compared to typical household circuits. Finally, let's compute the
answer. I even used an online calculator and I got the same answer A = $\pi$$r^{2}$ power (P) using the formula P = i*V: P = 49.45 A × 9 V P = 445 W And finally, to determine the amount of energy (W)
A = $\pi$ $(3 \times 10^{-3}) ^{2}$ A = $\pi$ (9 $\times$ $10^{-6}$) A = 2.83 $\ dissipated through the wire during one second, multiply the power by time (t): W = 445 W × 1 s W = 445 J Therefore,
times$ $10^{-5}$ rude " Do not just list concepts, but develop each one in detail if the 3 m long copper wire with a diameter of 6 mm is connected to a 9 V battery for 1 second, it will dissipate
before moving to the next, as we prioritize depth of understanding and approximately 445 joules of energy. This is equivalent to the kinetic energy of a baseball traveling at about 130
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: km/h (80 mph). It is important to note that this level of energy dissipation could lead to significant heating of
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an the wire and potential safety hazards, especially in confined spaces.
academic, professional and engaging tone that captivates interest. - Application:
Incorporate specific, practical examples, such as proofs in calculus or critical
dates and figures in history. Do not include a title or an introduction, simply
write the content without headlines and introductory phrases. Do not use images.
Now that you have grasped the concept of derivatives as they apply to products and quotients, it's time to expand
your knowledge to encompass additional types of functions. Specifically, we will delve into computing derivatives
of polynomials, rational functions, and powers and roots of rational functions. This newfound ability stems
primarily from mastering the derivative of x^n and employing both linearity and the product rule. Nevertheless,
another essential ingredient remains to be added to our mathematical repertoire - the Chain Rule. But before
tackling this final "rule", let us augment our collection of differentiable functions by introducing several other
types of well-behaved functions. These additions are crucial due to their seamless interaction with derivatives and
frequent appearances in real-world scenarios. Among these newly introduced functions is the exponential function.
Exploring the Exponential Function ---------------------------------- An exponential function is characterized by
its defining formula, f(x) = a^x, where a surpasses zero (a > 0). It may seem counterintuitive at first glance;
however, this seemingly simple equation encapsulates remarkable growth properties. Consider two distinct cases
Write an educational piece suited for college students related to the following based on whether 'a' exceeds or falls short of unity (i.e., when a > 1 or a < 1): ### Case I: a > 1 When the base a
text snippet: "## Section2.7Derivatives of Exponential Functions Now that we transcends one, the resulting graph ascends steeply, symbolizing rapid growth. As x increases, so does the value of
understand how derivatives interact with products and quotients, we are able to y, leading to a curve that perpetually heads towards positive infinity. Despite the ever-growing nature of this
compute derivatives of • polynomials, • rational functions, and • powers and roots function, noticeable undulations occur, creating inflection points that prove pivotal in determining directional
of rational functions. Notice that all of the above come from knowing changes along the curve. ### Case
1Differentiating powers and roots of functions is actually quite a bit easier II: 0 < a < 1 Conversely, when a resides between zero and one (0 < a < 1), the graph descends gradually,
once one knows the chain rule — which we will discuss soon. the derivative of $x^n$ representing diminishing growth. Although the pattern appears reversed compared to case I, some striking
and applying linearity of derivatives and the product rule. There is still one more similarities remain. Most notably, the presence of inflection points persists, signaling subtle shifts in curvature
“rule” that we need to complete our toolbox and that is the chain rule. However throughout the domain. In
before we get there, we will add a few functions to our list of things we can either scenario, the intrinsic behavior of the exponential function lends itself naturally to various applications
differentiate 2One reason we add these functions is that they interact very across disciplines like finance, physics, biology, chemistry, computer science, economics, statistics, probability
nicely with the derivative. Another reason is that they turn up in many “real theory, actuarial science, engineering, and mathematics education. Consequently, understanding the nuances of the
world” examples.. The first of these is the exponential function. Let $a \gt 0$ and exponential function becomes paramount for any aspiring scholar pursuing higher studies in these fields. Before
set $f(x) = a^x$ — this is what is known as an exponen" Do not just list concepts, diving deeper into the realm of exponential functions, recall the significance of the number e, approximately equal
but develop each one in detail before moving to the next, as we prioritize depth of to 2.71828. Commonly referred to as Euler's Number, e represents the unique base at which the slope of the tangent
understanding and comprehensive exploration of the subject matter over breadth. line equals the height of the curve. This intimate relationship implies that the rate of change coincides precisely
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: with the current value, yielding a fascinating self-similarity property unparalleled among other bases. Indeed,
Write with an academic, professional and engaging tone that captivates interest. - graphing the exponential function using e reveals an elegant symmetry centered around the origin: ![Graph of e^x]
Application: Incorporate specific, practical examples, such as proofs in calculus (https://www.desmos.com/calculator/bmdlwgkpdi?embed) As we progress further, bear in mind that subsequent
or critical dates and figures in history. Do not include a title or an discussions concerning the exponential function shall implicitly assume a base of e unless otherwise stated
introduction, simply write the content without headlines and introductory phrases. explicitly. With this foundation laid out, let us now formally derive the derivative of an exponential function
Do not use images. through first principles. Deriving the Derivative of an Exponential Function
--------------------------------------------------- To calculate the derivative of an exponential function, namely
f(x) = a^x, we must resort to the definition of a limit: f'(x) = lim(h->0) [(a^(x+h) - a^x)/h] Through algebraic
manipulation, isolate the term containing h within the numerator: f'(x) = lim(h->0) [a^x * (a^h - 1)/h] At this
juncture, recognize that the expression (a^h - 1)/h approaches the natural logarithmic constant, ln(a), as h tends
to zero. To see why, consider substituting h with increasingly smaller values: lim(h->0) [(a^h - 1)/h] ≈ (a^0.1 -
1)/0.1 ≈ (a^0.01 - 1)/0.01 ≈ ... ≈ ln(a) With this insight, rewrite the original limit as follows: f'(x) = a^x \*
lim(h->0) [(a^h - 1)/h] Since the remaining limit evaluates to ln(a), we arrive at the coveted result: f'(x) = a^x
\* ln(a) This fundamental relation elucidates the interplay between exponentiation and logarithms while providing a
succinct representation for the derivative of an arbitrary exponential function. Furthermore, setting a = e
simplifies the formula considerably owing to the special property mentioned earlier: If a = e, then f'(x) = e^x \*
ln(e) = e^x \* 1 = e^x Henceforth, whenever confronted with an exponential function having a base of e, remember
that its derivative reduces to the same form!
To accomplish your task of modeling a repeated measures growth curve using cumulative population totals from 2016
and 2017, you will need to follow several steps involving data preparation, selecting appropriate models, fitting
those models, and interpreting their results. This process involves various statistical techniques which require
careful consideration. Throughout this discussion, we will assume that you have basic knowledge of statistics and
some experience working with R software. Firstly, let us clarify what a *repeated measures* design entails. It
refers to situations where the same subjects (in this case, months) are measured under different conditions
(years), resulting in correlated measurements. Consequently, traditional methods assuming independence among
observations may lead to incorrect conclusions. Therefore, employing specialized approaches accommodating
Write an educational piece suited for college students related to the following dependencies within groups becomes essential. In terms of preparing your dataset, it seems adequate already since
text snippet: "# Modeling a repeated measures growth curve I have cumulative it includes necessary variables – year, month, time (month number), norm (proportion of year-end total), and
population totals data for the end of each month for two years (2016, 2017). I log-transformed total. However, combining two years into one dataset requires merging them intelligently so that
would like to combine these two years and treat each months cumulative total as a each observation corresponds appropriately across both periods. You can achieve this via either binding rows or
repeated measure (one for each year) and fit a non linear growth model to these joining columns depending on how they were initially stored; base R functions `rbind()` and `merge()`,
data. The goal is to determine whether our current 2018 cumulative monthly totals respectively, could be helpful here. Once data preprocessing concludes, choosing an appropriate growth model comes
are on track to meet our higher 2018 year-end population goal by increasing the next. Given that you aim to predict future trends based on past ones while incorporating random effects due to
model's asymptote to our 2018 year-end goal. I would ideally like to integrate a repeated measures, mixed-effects models appear suitable candidates. Specifically, nonlinear mixed-effects models
confidence interval into the model that reflects the variability between the two seem more befitting because they allow curvature specification directly in the structural form rather than relying
years at each month. My columns in my data.frame are as follows: - Year is year - solely on link functions used in generalized linear mixed models. One popular choice for such problems is the
Month is month - Time is the month's number (1-12) - Total is the month-end Richards growth model, characterized by its flexibility and capability to capture sigmoidal patterns common in
cumulative population total - Norm is the proportion of year-end total for that biological systems. Its mathematical representation takes the form: Total = Asym / (1 + exp(-k*(Time-t0))) Where
month - log is the Total log transformed Year Month Total Time Norm log 1 2016 'Asym', 'k', and 't0' denote parameters representing the upper limit (asymptote), rate of change, and inflection
January 3919 1 0.2601567 8.273592 2 2016 February 5887" Do not just list concepts, point, respectively. By allowing these parameters to vary randomly according to normal distributions centered
but develop each one in detail before moving to the next, as we prioritize depth of around overall estimates, we account for correlations arising from multiple measurements per unit. To implement
understanding and comprehensive exploration of the subject matter over breadth. this in R, you might consider packages like "nlme" or "mgcv". For instance, utilizing "nlme", you could define the
Focus on: - Rigor: Ensure in-depth coverage of the concepts/sections. - Engagement: nonlinear mixed-effects model as follows: ```R library(nlme) model <- nlme(Total ~ SSlogis(Time, Asym, t0, k),
Write with an academic, professional and engaging tone that captivates interest. - fixed = Asym + t0 + k ~ 1, random = Asym + t0 + k ~ 1 | Year, start = c(Asym=max(df$Total), t0=median(df$Time), k=
Application: Incorporate specific, practical examples, such as proofs in calculus log(4)), data = df) ``` Here, `SSlogis()` represents the logistic function embedded within the Richards model
or critical dates and figures in history. Do not include a title or an structure. Note that setting starting values close to expected final estimates helps ensure successful convergence
introduction, simply write the content without headlines and introductory phrases. during optimization. Lastly, integrating uncertainty quantification through confidence intervals forms part of
Do not use images. robust statistical practice. With mixed-effects models, obtaining standard errors for parameter estimators presents
challenges due to complex covariance structures. Nevertheless, packages such as "nlme" provide tools to calculate
approximate standard errors, enabling construction of corresponding confidence intervals. For example, after
fitting the above model, you can extract estimated parameters along with their associated standard errors using:
```R summary(model)$tParFixed summary(model)$tParRandom ``` From there, constructing confidence intervals relies on
multiplying these standard errors by multiples of the standard normal distribution (e.g., 1.96 for 95% CIs), adding
/subtracting these products to/from the estimated parameters, then reporting the resulting ranges. Remember,
though, that inferential procedures applied to nonlinear models should be approached cautiously due to potential
biases induced by departures from assumed functional forms. Thus, conducting sensitivity analyses exploring
alternative plausible shapes remains crucial when drawing definitive conclusions about underlying processes.
To begin, let's recall the definition of an equivalence relation. A relation ~ on a set X is said to be an
Write an educational piece suited for college students related to the following equivalence relation if it satisfies three properties: 1. Reflexivity: For every x in X, x ~ x. 2. Symmetry: If x ~
text snippet: "Find Equivalence Classes for the relation $S$ on the set $\mathbb{R} y in X, then y ~ x. 3. Transitivity: If x ~ y and y ~ z in X, then x ~ z. Once you find an equivalence relation on
^*$ is definied as $aSb \iff ab > 0$ I am trying to understand how to determine the a set, you can partition the set into distinct subsets called equivalence classes. Two elements belong to the same
equivalence classes. I have this question in the book The relation $$S$$ on the set equivalence class if they are equivalent according to the given relation. Specifically, for any element a in X, its
$$\mathbb{R}^*$$ is definied as $$aSb \iff ab > 0$$ The answer in the book is $$[1] equivalence class denoted by [a] contains all those elements in X which are equivalent to a. That is, [a] = {x in X
= \mathbb{R}^+$$ and $$[-1] = \mathbb{R}^-$$. I don't even know where to start with : x ~ a} Now, let us focus on your question involving the set of nonzero real numbers, denoted by ${\mathbb{R}}^*$,
this problem. I know that $$ab > 0$$ so then $$a \cdot b$$ always have to be and the relation S defined as follows: For a, b in ${\mathbb{R}}^*$, aSb if and only if ab > 0 To identify the
positive and I know that equivalence class means disjoint categories but why are $$ equivalence classes, we will construct two sets based on whether the number is positive or negative, since their
[1]$$ and $$[-1]$$ the equivalence classes? • "I know that ab>0 so then a⋅b always product should yield a positive result according to the definition of S. We proceed as follows: Let A be the subset
have to be positive". – user247327 Mar 29 at 12:59 First, for all $$a,b \in \mathbb containing all positive real numbers (excluding zero): A = ${\mathbb{R}}^{+}$. Notice that for any two elements a
{R}^-$$, $$ab > 0$$. So all the elements in $$\mathbb{R}^-$$ are in the same and b in A, both are greater than zero, meaning their product ab will also be greater than zero. This results in
equivalence class. Secondly, for all $$a,b \in \mathbb{R}^+$$, $$ab > 0$$. So all aSa for all a, b in A; hence, these elements form an equivalence class. Therefore, our first candidate for the
the elements in $$\mathbb{R}^+$$ are in the same equivalence class. Finally, if $$a equivalence class would be: [1] = A = ${\mathbb{R}}^{+}$ Next, consider B being the subset containing all negative
\in \mathbb{R}^-$$ an" Do not just list concepts, but develop each one in detail real numbers (excluding zero): B = ${\mathbb{R}}^{-}$. Likewise, when selecting arbitrary elements from B, say c
before moving to the next, as we prioritize depth of understanding and and d, both will be less than zero. Consequently, cd yields a positive value, resulting in cSc for all c, d in B.
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor: Thus, forming another equivalence class consisting entirely of negative real numbers: [-1] = B = ${\mathbb{R}}^{-}$
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an It remains to show that no other equivalence classes exist besides [1] and [-1]. Assume there exists some $c_0$
academic, professional and engaging tone that captivates interest. - Application: different from either -1 or 1. Clearly, neither ${{\mathbb{R}}^{+}}$ nor ${{\mathbb{R}}^{-}}$ could serve as the
Incorporate specific, practical examples, such as proofs in calculus or critical corresponding equivalence class because none of their members share the property of having the same sign as $c_0$.
dates and figures in history. Do not include a title or an introduction, simply Hence, no additional equivalence classes arise outside $[1]$ and $[-1]$. In conclusion, the nonzero reals ${\mathbb
write the content without headlines and introductory phrases. Do not use images. {R}}^*$ partitioned under the relation S consist precisely of the two equivalence classes mentioned above: [1] = $
{\mathbb{R}}^{+}$, and [-1] = ${\mathbb{R}}^{-}$.
Write an educational piece suited for college students related to the following
text snippet: "# Is AR(1) a Markov process? Is AR(1) process such as $y_t=\rho y_
{t-1}+\varepsilon_t$ a Markov process? If it is, then VAR(1) is the vector version
of Markov process? The following result holds: If $\epsilon_1, \epsilon_2, \ldots$
are independent taking values in $E$ and $f_1, f_2, \ldots$ are functions $f_n: F \
times E \to F$ then with $X_n$ defined recursively as the process $(X_n)_{n \geq 0}
$ in $F$ is a Markov process starting at $x_0$. The process is time-homogeneous if A fundamental concept in the study of stochastic processes is the Markov property, which states that the future
the $\epsilon$‘s are identically distributed and all the $f$-functions are state of a process depends only on its current state and not on its past states. More formally, a stochastic
identical. The AR(1) and VAR(1) are both processes given in this form with Thus process {XT} is said to be a Markov process if for every n and every set of states A, P[XT+1 ∈ A | XT = x, XT−1 =
they are homogeneous Markov processes if the $\epsilon$‘s are i.i.d. Technically, x′, ..., X1 = x(0)] = P[XT+1 ∈ A | XT = x] for all x, x′, ..., x(0), where P denotes probability. This means that
the spaces $E$ and $F$ need a measurable structure and the $f$-functions must be the conditional distribution of the future state of the process, given its entire history up to the present,
measurable. It is quite interesting that a converse result holds if the space $F$ depends only on the current state and not on the past states. One important class of Markov processes is the
is a Borel space. For any Markov process $(X_n)_{n \geq 0}$ on a Borel space $F$ autoregressive (AR) process of order 1, denoted by AR(1). An AR(1) process is defined as follows: yt=ρyt−1+εt,\
there are i.i.d. uniform random" Do not just list concepts, but develop each one in *-open*
detail before moving to the next, as we prioritize depth of understanding and
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor:
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an
academic, professional and engaging tone that captivates interest. - Application:
Incorporate specific, practical examples, such as proofs in calculus or critical
dates and figures in history. Do not include a title or an introduction, simply
write the content without headlines and introductory phrases. Do not use images.
When working with vector spaces, there are two approaches to presenting proofs and definitions: basis-dependent and
basis-free. A basis for a vector space is a linearly independent set of vectors that spans the entire space. While
every vector space has a basis, it may not always be unique or canonical, meaning there could be multiple bases,
and none might be considered preferred or standard. Understanding the difference between these two methods and
recognizing their respective merits is crucial for developing strong mathematical reasoning skills. This piece will
delve into the details of basis-dependent and basis-free approaches, highlighting the importance of being
basis-free and offering guidance on how to achieve this goal when writing proofs and definitions. ###
Basis-Dependent Proofs and Definitions A basis-dependent approach relies on choosing a particular basis for the
vector space under consideration and expressing objects and operations using components relative to that chosen
basis. For instance, consider the definition of a linear transformation T: V -> W, where V and W are
finite-dimensional vector spaces with bases {v\_i} and {w\_j}, respectively. In the basis-dependent approach, T
would be defined by specifying the images Ti(vi), written as column vectors, allowing us to represent T as a
matrix: [T](ij) = [T(v1), T(v2), ..., T(vd)] \* [[w1], [w2], ..., [wd]]^(-1) Here, [T](ij) represents the entries
of the matrix associated with T, while [T(vi)] denotes the coordinate vector of T(vi). The primary disadvantage of
this method lies in its dependence on the choice of bases; changing the bases used in either domain or codomain
Write an educational piece suited for college students related to the following results in a different matrix representation of T. Consequently, any conclusions drawn based on this representation
text snippet: "# What does it mean to work without a basis? When reading proofs or become tied to the selected bases, leading to potential issues if alternative choices were made instead. Proof
definitions on Wikipedia, I'm accustomed to seeing both a basis-dependent Example - Calculating Determinant of Linear Transformation Matrix: Let T: R³ -> R³ be a linear transformation given
discussion and basis-free discussion. Take, for example, this page on the tensor by the matrix: T = | 3 -4 5| | 6 -7 8| | 9 -10 11| Calculate det(T) using the Leibniz formula for determinants: det
product, which has a "free vector space" discussion alongside a matrix (T) = (-1)^(1+2)*3! * [T](1,1)\*(T)(2,2) - [T](1,2)*(T)(2,1) + ... = 3\*(3*8 - 4*6) - 4\*(6*8 - 7*6) + 5\*(6*7 -
representation discussion. While I understand not all vector spaces have a 8*4) = 0 This calculation demonstrates that although we arrived at the correct answer—namely, zero—the actual
canonical basis, it's not formally clear to me what it means for a proof or process depends heavily upon our initial selection of the standard basis. Thus, had another basis been chosen, the
definition to be basis-dependent vs. basis-free, and why it's better to be resulting calculations would differ accordingly. ### Advantages of Being Basis-Free The main advantage of adopting
basis-free in the first place. 1) If I'm writing a proof or defining an object, a basis-free perspective is avoiding unnecessary ties to specific bases, ensuring greater generality in arguments
what rules or criteria must I follow to be properly basis free? And, once I know and facilitating easier comparison across various contexts. By focusing on intrinsic properties inherent to vector
what a basis-dependent proof or definition looks like, what is the strategy from spaces themselves rather than those contingent upon arbitrary choices within them, mathematicians establish more
generalizing it to a basis-independent proof or definition? 2) If all vector spaces robust connections among diverse areas of study. Furthermore, basis-independence often simplifies notation,
can be represented by a (not necessarily canonical) basis, can't we always reducing visual clutter caused by explicit reference to components along particular axes. To illustrate this point,
represent operators and members of that that space with matrices and linear" Do not let's reconsider the earlier example of a linear transformation T: V -> W. Instead of representing T via matrices,
just list concepts, but develop each one in detail before moving to the next, as we a basis-free approach defines T directly through its action on vectors v∈V: T(αv₁ + βv₂) := αT(v₁) + βT(v₂) for
prioritize depth of understanding and comprehensive exploration of the subject scalars α, β and vectors v₁, v₂ ∈ V In doing so, we avoid committing ourselves to any fixed basis until absolutely
matter over breadth. Focus on: - Rigor: Ensure in-depth coverage of the concepts/ necessary, thereby preserving maximum flexibility throughout computations and enabling straightforward translation
sections. - Engagement: Write with an academic, professional and engaging tone that between distinct bases whenever required. Definition Example - Tensor Product Without Bases: Given two vector
captivates interest. - Application: Incorporate specific, practical examples, such spaces V and W over field F, define the tensor product V⊗W as follows: 1. As a vector space itself, span{v⊗w : v∈V,
as proofs in calculus or critical dates and figures in history. Do not include a w∈W} 2. With bilinearity satisfied: (a_1v\_1 + a_2v\_2) ⊗ w = a\_1(v\_1 ⊗ w) + a\_2(v\_2 ⊗ w) and w ⊗ (b\_1w\_1 + b
title or an introduction, simply write the content without headlines and \_2w\_2) = b\_1(w ⊗ w\_1) + b\_2(w ⊗ w\_2) for all a\_1, a\_2, b\_1, b\_2 ∈ F and v\_1, v\_2 ∈ V, w\_1, w\_2 ∈ W By
introductory phrases. Do not use images. employing abstract language devoid of references to specific bases, this definition emphasizes universal properties
shared by all tensor products regardless of underlying dimensionality or structure, making it applicable even
beyond traditional Euclidean settings. Transition Strategy: Generalizing From Basis-Dependent To Basis-Free
Representations Transforming a basis-dependent argument into a basis-free counterpart typically involves three
steps: 1. Identify key assumptions or constraints tied explicitly to chosen bases. 2. Replace these dependencies
with equivalent statements concerning invariant properties of relevant vector spaces and subspaces. 3. Revise
original proofs accordingly, appealing only to newly established foundational principles. For example, suppose you
wish to convert a basis-dependent proof regarding eigenvalues into a basis-free version. You would begin by noting
that eigenvalue equations rely crucially on scalar multiplication within vector spaces, thus shifting focus away
from individual components towards global algebraic structures. Subsequent manipulations then leverage fundamental
properties held in common by all eigenvectors sharing a common eigenvalue, culminating in a final result untethered
from any particular basis. Conclusion Familiarizing oneself with both basis-dependent and basis-free perspectives
enables deeper comprehension of advanced mathematics topics, promoting adaptability across myriad applications. By
mastering techniques for converting between these modes of thought, college students hone valuable problem-solving
abilities transferable across disciplines and careers alike. Ultimately, embracing rigorous yet accessible
explanatory frameworks fosters heightened engagement with complex material, empowering learners to tackle
increasingly sophisticated challenges confidently and enthusiastically.
Write an educational piece suited for college students related to the following
text snippet: "How to show this fraction is equal to 1/2? I have the fraction: $$\
frac{\left(2 \left(\frac {a}{\sqrt{2}}\right) + a \right) a} {2(1 + \sqrt{2})a^2}$$
Using Mathematica, I've found that this simplifies to $\frac{1}{2}$, but how did it
achieve the result? How can I simplify that fraction to $\frac12$? - Factor out an
$a$ on top, multiply top and bottom by $\sqrt 2$. It's really not complicated.
– Matt Samuel Jan 16 at 22:23 Just multiply everthing by $a$. Also $\frac 2{\sqrt
2} = \sqrt 2$ – Ant Jan 16 at 22:23 Despite it not being complicated I don't really
understand the downvotes. It doesn't necessarily mean the OP didn't try. – Matt
Samuel Jan 16 at 22:30 @MattSamuel Post that as an answer and I'll accept it.
– user305676 Jan 16 at 22:38 Of course there is the implicit assumption that $a\ne
0$ ... – Hagen von Eitzen Jan 17 at 12:05 Factor out an $a$ on top, multiply top
and bottom by $\sqrt 2$. You're left with $$\frac{(2+\sqrt2)a^2}{2(2+\sqrt2)a^2}$$
This is clearly equal to $\f" Do not just list concepts, but develop each one in
detail before moving to the next, as we prioritize depth of understanding and
comprehensive exploration of the subject matter over breadth. Focus on: - Rigor:
Ensure in-depth coverage of the concepts/sections. - Engagement: Write with an
academic, professional and engag | {"url":"https://huggingface.co/datasets/ajibawa-2023/Maths-College/viewer","timestamp":"2024-11-09T12:58:08Z","content_type":"text/html","content_length":"1049175","record_id":"<urn:uuid:544448b3-6b23-43df-86f9-5bc923288858>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00519.warc.gz"} |
The method to measure the plastic viscosity of cement paste
The method to measure the plastic viscosity of cement paste
In the previous post, I introduced the protocol to measure the yield stress of cement paste. I would like to share the method how to measure the plastic viscosity of cement paste.
The principle determine the plastic viscosity of cement paste is using the relationship between shear rate and shear stress. Applying a serials of shear rate on the cement paste would result in the
corresponding shear stress. After obtaining the set of shear rate and shear stress, the plastic viscosity of the paste is calculated by regression.
The measured shear rate and shear stress of a cement paste, Measuring device: HAAKE™ VT550 (ViscoTester VT550), CR (Controlled shear Rate): 0-100 1/s; t 120.00 s; T 20.00 °C.
The paste should be measured by increasing shear rate and then the decreasing rate. Typical shear rate applied on cement paste is linearly between 0 and 100 1/s. Since the rheological feature of most
of cement pastes is a kind of Bingham model, we can choose the liner part of the decreasing shear curve from the date set (shear rate vs. shear stress) for the regression. The slope from the
regression equation is the so-called plastic viscosity.
However, if the rheological cure of the paste behaves rather than Bingham model, selecting an appropriate model is necessary. | {"url":"https://www.cementscience.com/2015/03/the-method-to-measure-the-plastic-viscosity-of-cement-paste.html","timestamp":"2024-11-11T14:02:53Z","content_type":"text/html","content_length":"47719","record_id":"<urn:uuid:4134464e-3580-4989-8878-c9d1babdcde1>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00863.warc.gz"} |
607 research outputs found
The Kraichnan rapid advection model is recast as the stochastic dynamics of tracer trajectories. This framework replaces the random fields with a small set of stochastic ordinary differential
equations. Multiscaling of correlation functions arises naturally as a consequence of the geometry described by the evolution of N trajectories. Scaling exponents and scaling structures are
interpreted as excited states of the evolution operator. The trajectories become nearly deterministic in high dimensions allowing for perturbation theory in this limit. We calculate perturbatively
the anomalous exponent of the third and fourth order correlation functions. The fourth order result agrees with previous calculations.Comment: 14 pages, LaTe
We consider passive scalar convected by multi-scale random velocity field with short yet finite temporal correlations. Taking Kraichnan's limit of a white Gaussian velocity as a zero approximation we
develop perturbation theory with respect to a small correlation time and small non-Gaussianity of the velocity. We derive the renormalization (due to temporal correlations and non-Gaussianity) of the
operator of turbulent diffusion. That allows us to calculate the respective corrections to the anomalous scaling exponents of the scalar field and show that they continuously depend on velocity
correlation time and the degree of non-Gaussianity. The scalar exponents are thus non universal as was predicted by Shraiman and Siggia on a phenomenological ground (CRAS {\bf 321}, 279,
1995).Comment: 4 pages, RevTex 3.0, Submitted to Phys.Rev.Let
Kraichnan's model of passive scalar advection in which the driving velocity field has fast temporal decorrelation is studied as a case model for understanding the appearance of anomalous scaling in
turbulent systems. We demonstrate how the techniques of renormalized perturbation theory lead (after exact resummations) to equations for the statistical quantities that reveal also non perturbative
effects. It is shown that ultraviolet divergences in the diagrammatic expansion translate into anomalous scaling with the inner length acting as the renormalization scale. In this paper we compute
analytically the infinite set of anomalous exponents that stem from the ultraviolet divergences. Notwithstanding, non-perturbative effects furnish a possibility of anomalous scaling based on the
outer renormalization scale. The mechanism for this intricate behavior is examined and explained in detail. We show that in the language of L'vov, Procaccia and Fairhall [Phys. Rev. E {\bf 50}, 4684
(1994)] the problem is ``critical" i.e. the anomalous exponent of the scalar primary field $\Delta=\Delta_c$. This is precisely the condition that allows for anomalous scaling in the structure
functions as well, and we prove that this anomaly must be based on the outer renormalization scale. Finally, we derive the scaling laws that were proposed by Kraichnan for this problem, and show that
his scaling exponents are consistent with our theory.Comment: 43 pages, revtex
For a delta-correlated velocity field, simultaneous correlation functions of a passive scalar satisfy closed equations. We analyze the equation for the four-point function. To describe a solution
completely, one has to solve the matching problems at the scale of the source and at the diffusion scale. We solve both the matching problems and thus find the dependence of the four-point
correlation function on the diffusion and pumping scale for large space dimensionality $d$. It is shown that anomalous scaling appears in the first order of $1/d$ perturbation theory. Anomalous
dimensions are found analytically both for the scalar field and for it's derivatives, in particular, for the dissipation field.Comment: 19 pages, RevTex 3.0, Submitted to Phys.Rev. E, revised versio
Localized rain events have been found to follow power-law size and duration distributions over several decades, suggesting parallels between precipitation and seismic activity [O. Peters et al., PRL
88, 018701 (2002)]. Similar power laws are generated by treating rain as a passive tracer undergoing advection in a velocity field generated by a two-dimensional system of point vortices.Comment: 7
pages, 4 figure
Fusion rules in turbulence specify the analytic structure of many-point correlation functions of the turbulent field when a group of coordinates coalesce. We show that the existence of flux
equilibrium in fully developed turbulent systems combined with a direct cascade induces universal fusion rules. In certain examples these fusion rules suffice to compute the multiscaling exponents
exactly, and in other examples they give rise to an infinite number of scaling relations that constrain enormously the structure of the allowed theory.Comment: Submitted to PRL on July 95, 4 pages,
We consider turbulent advection of a scalar field T(\B.r), passive or active, and focus on the statistics of gradient fields conditioned on scalar differences $\Delta T(R)$ across a scale $R$. In
particular we focus on two conditional averages $\langleabla^2 T\big|\Delta T(R)\rangle$ and $\langle|abla T|^2\big|\Delta T(R) \rangle$. We find exact relations between these averages, and with the
help of the fusion rules we propose a general representation for these objects in terms of the probability density function $P(\Delta T,R)$ of $\Delta T(R)$. These results offer a new way to analyze
experimental data that is presented in this paper. The main question that we ask is whether the conditional average $\langleabla^2 T\big| \Delta T(R)\rangle$ is linear in $\Delta T$. We show that
there exists a dimensionless parameter which governs the deviation from linearity. The data analysis indicates that this parameter is very small for passive scalar advection, and is generally a
decreasing function of the Rayleigh number for the convection data.Comment: Phys. Rev. E, Submitted. REVTeX, 10 pages, 5 figs. (not included) PS Source of the paper with figure available at http://
The statistics of 2-dimensional turbulence exhibit a riddle: the scaling exponents in the regime of inverse energy cascade agree with the K41 theory of turbulence far from equilibrium, but the
probability distribution functions are close to Gaussian like in equilibrium. The skewness \C S \equiv S_3(R)/S^{3/2}_2(R) was measured as \C S_{\text{exp}}\approx 0.03. This contradiction is lifted
by understanding that 2-dimensional turbulence is not far from a situation with equi-partition of enstrophy, which exist as true thermodynamic equilibrium with K41 exponents in space dimension of $d=
4/3$. We evaluate theoretically the skewness \C S(d) in dimensions ${4/3}\le d\le 2$, show that \C S(d)=0 at $d=4/3$, and that it remains as small as \C S_{\text{exp}} in 2-dimensions.Comment: PRL,
submitted, REVTeX 4, 4 page
General turbulent mean statistics are shown to be characterized by a variational principle. The variational functionals, or ``effective actions'', have experimental consequences for turbulence
fluctuations and are subject to realizability conditions of positivity and convexity. An efficient Rayleigh-Ritz algorithm is available to calculate approximate effective actions within PDF closures.
Examples are given for Navier-Stokes and for a 3-mode system of Lorenz. The new realizability conditions succeed at detecting {\em a priori} the poor predictions of PDF closures even when the
classical 2nd-order moment realizability conditions are satisfied.Comment: 4 pages, LaTeX (Version 2.09), 3 figures, Postscript, Submitted to Phys. Rev. Let
We introduce a model for the turbulent energy cascade aimed at studying the effect of dynamical scaling on intermittency. In particular, we show that by slowing down the energy transfer mechanism for
fixed energy flux, intermittency decreases and eventually disappears. This result supports the conjecture that intermittency can be observed only if energy is flowing towards faster and faster scales
of motion.Comment: 4 pages, 3 figure | {"url":"https://core.ac.uk/search/?q=author%3A(R.%20H.%20Kraichnan)","timestamp":"2024-11-01T23:19:05Z","content_type":"text/html","content_length":"179745","record_id":"<urn:uuid:477e2a7d-78d9-4d4d-9749-d8b1d1eeaec6>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00396.warc.gz"} |
Quotient Rule Calculator: Simplify Derivative Quotients
Exploring the Efficiency of Quotient Rule Calculator in Mathematics
When it comes to simplifying complicated issues, knowing the Quotient Rule might feel like opening a secret door when you're just starting out in calculus. Nevertheless, do not be alarmed; we are
going to reveal the secrets of the Quotient Rule and its calculator.
What's the Quotient Rule?
Consider a fractional function in which the denominator and numerator are also functions.
This is where the Quotient Rule swoops in. The Quotient Rule is a mathematical technique used to differentiate such functions. In simple terms, it helps find the derivative of functions that are
Using the Quotient Rule Calculator
Now, let's talk convenience. Thanks to technology, we've got Quotient Rule calculators that make life easier for math enthusiasts. These calculators are designed to swiftly compute derivatives
following the Quotient Rule formula.
The Quotient Rule Formula
Mathematically, the Quotient Rule is articulated as:
\[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2} \]
This formula provides a structured approach to calculate the derivative of a quotient function. To apply this rule, one must follow a systematic procedure that involves identifying the numerator and
denominator functions, differentiating each, and applying the prescribed formula.
Step-by-Step Procedure
1. Identify \( f(x) \) and \( g(x) \)
Recognize the functions involved in the quotient.
2. Differentiate \( f(x) \) and \( g(x) \)
Compute the derivatives of the numerator and denominator individually.
3. Apply the Quotient Rule Formula
Utilize the formula to determine the derivative of the quotient.
Example 1: \( \frac{x^2 + 3x + 1}{x} \)
Let's delve into a practical example to illustrate the application of the Quotient Rule. Consider the function \( \frac{x^2 + 3x + 1}{x} \). To find its derivative, we apply the Quotient Rule:
Given: \( f(x) = x^2 + 3x + 1 \)
\( g(x) = x \)
1. Derivative of \( f(x) \) (\( f'(x) \)): \( f'(x) = 2x + 3 \)
2. Derivative of \( g(x) \) (\( g'(x) \)): \( g'(x) = 1 \)
Applying the Quotient Rule Formula:
\[ \frac{d}{dx} \left( \frac{x^2 + 3x + 1}{x} \right) = \frac{x \cdot (2x + 3) - (x^2 + 3x + 1) \cdot 1}{x^2} \]
Simplifying the expression yields the derivative of the given function.
Example 2: \( \frac{2x^3 - x}{x^2 + 1} \)
Given: \( f(x) = 2x^3 - x \) and \( g(x) = x^2 + 1 \)
1. Derivative of \( f(x) \) (\( f'(x) \)): \( f'(x) = 6x^2 - 1 \)
2. Derivative of \( g(x) \) (\( g'(x) \)): \( g'(x) = 2x \)
Applying the Quotient Rule Formula:
\[ \frac{d}{dx} \left( \frac{2x^3 - x}{x^2 + 1} \right) = \frac{(x^2 + 1) \cdot (6x^2 - 1) - (2x)(2x^3 - x)}{(x^2 + 1)^2} \]
Example 3: \( \frac{\sin(x)}{x^2} \)
Given: \( f(x) = \sin(x) \) and \( g(x) = x^2 \)
1. Derivative of \( f(x) \) (\( f'(x) \)): \( f'(x) = \cos(x) \)
2. Derivative of \( g(x) \) (\( g'(x) \)): \( g'(x) = 2x \)
Applying the Quotient Rule Formula:
\[ \frac{d}{dx} \left( \frac{\sin(x)}{x^2} \right) = \frac{(x^2)(\cos(x)) - (\sin(x))(2x)}{(x^2)^2} \]
Example 4: \( \frac{e^x - x^2}{\sqrt{x}} \)
Given: \( f(x) = e^x - x^2 \) and \( g(x) = \sqrt{x} \)
1. Derivative of \( f(x) \) (\( f'(x) \)): \( f'(x) = e^x - 2x \)
2. Derivative of \( g(x) \) (\( g'(x) \)): \( g'(x) = \frac{1}{2\sqrt{x}} \)
Applying the Quotient Rule Formula:
\[ \frac{d}{dx} \left( \frac{e^x - x^2}{\sqrt{x}} \right) = \frac{(\sqrt{x})(e^x - 2x) - (e^x - x^2)(\frac{1}{2\sqrt{x}})}{(\sqrt{x})^2} \]
How to Use Quotient Rule Calculator?
Input Functions: Enter your numerator and denominator functions.
Hit Calculate: Let the calculator do its magic.
Advantages of Utilizing a Quotient Rule Calculator
Time-Saving: No more tedious manual calculations. With a Quotient Rule calculator, you save time by swiftly obtaining derivatives, allowing you to focus on understanding concepts rather than number
Error Reduction: Humans make errors, it's a fact. But calculators operate with precision. Using a calculator reduces the chances of miscalculations or transcription mistakes in differentiating
Why Opt for Quotient Rule Calculators?
Imagine solving lengthy problems involving multiple functions manually.
Quotient Rule calculators streamline the process, ensuring accuracy and efficiency in derivative calculations.
The Quotient Rule calculator serves as a remarkable tool, simplifying the differentiation process for complex functions. It's a time-saver, an error reducer, and a valuable asset for anyone diving
into the world of calculus. Embrace this technological aid and enhance your mathematical journey!
Remember, mastering the Quotient Rule and its calculator not only simplifies problem-solving but also empowers your understanding of calculus. Happy calculating!
The Quotient Rule in calculus is a method used to find the derivative of a function that is expressed as the quotient of two other functions. It provides a systematic approach to differentiate
functions involving division.
Yes, the Quotient Rule Calculator can handle complex functions involving polynomials, trigonometric functions, exponential functions, radicals, and more. It simplifies the process of finding
derivatives of complex quotient functions.
The accuracy of results from applying the Quotient Rule depends on the correct application of the formula. Verifying solutions manually is recommended to ensure accuracy.
Online resources, educational websites, mathematical software, and calculators offer assistance in understanding and applying the Quotient Rule. These tools aid in practicing and solving problems
related to differentiation using this rule. | {"url":"https://www.calculatestudy.com/quotient-rule-calculator","timestamp":"2024-11-04T04:36:03Z","content_type":"text/html","content_length":"52986","record_id":"<urn:uuid:8f641d5a-dfb7-4b98-97f8-b5c2042b485f>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00411.warc.gz"} |
interpretability उदाहरण वाक्य
1. The ambiguity and interpretability of these abstract concepts may drive them to decouple from each other over time ."
2. It must be pointed out that interpretability of the Mamdani-type neuro-fuzzy systems can be lost.
3. They showed that the Bi-interpretability Conjecture is equivalent to there being no nontrivial automorphism of the Turing degrees.
4. Generally sparse multiple kernel learning is particularly useful when there are many kernels and model selection and interpretability are important.
5. Of course, the interpretability of the formula is lost, but this approach is very useful for repetitive numerical calculations.
6. At the same time, interest is already growing in the interpretability of the results of such consumption-based approaches.
7. Pragmatic factors such as appropriateness, comprehensibility and interpretability justified the use of English as an international and intra-national language.
8. Other research in provability logic has focused on first-order provability logic, interpretability logics intended to capture the interaction between provability and interpretability.
9. Other research in provability logic has focused on first-order provability logic, interpretability logics intended to capture the interaction between provability and interpretability.
10. A final reason that data can be transformed is to improve interpretability, even if no formal statistical analysis or visualization is to be performed. | {"url":"https://m.hindlish.in/m.aspx?q=interpretability&p=2&l=en","timestamp":"2024-11-10T02:41:42Z","content_type":"text/html","content_length":"20759","record_id":"<urn:uuid:8784a38c-c42d-4bc9-a46d-46aaba269474>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00554.warc.gz"} |
Which of the following gives the correct definition and SI unit for electrical power (P)?
• A
Electrical power is voltage (V) divided by current (I) (P = V/I). It is measured in watts (W).
• B
Electrical power is energy multiplied by time (t) (P = Energy x t). It is measured in watts (W).
• C
Electrical power is energy is divided by time (t) (P = Energy/t). It is measured in watts (W).
• D
Electrical power is current (I) divided by voltage (V) (P = I/V). It is measured in amperes (A).
Power is the rate of "doing work". It can also be described as the rate of energy transfer, which is the same thing.
Power = Energy Transferred / Time
Power is therefore a measure of how much energy is being transferred into another form per second. It is measured in Watts (W). It is sometimes useful to rearrange this equation to read:
Energy Transferred = Power x Time
In electrical terms, and this applies to any transfer of energy which goes through electrical energy (generators, motors, etc.), the power can also be calculated using the equation:
Power = Voltage x Current
Your Notes (not visible to others)
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Graphical convergence
Graphical convergence --- Introduction ---
A function may define a sequence (u[n]) with a starting point u[1] by the formula
u[n+1] = f(u[n] ) .
Graphical convergence is a graphical exercise on recursive sequences. It randomly draws the graph of a function and a starting value u[1], then asks you to determine the convergence of the sequence
according to the graph.
Other exercises on: sequences Convergence Limit
The most recent version
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Multiplication Fact Worksheets
Math, especially multiplication, creates the foundation of numerous academic disciplines and real-world applications. Yet, for several learners, grasping multiplication can posture a difficulty. To
resolve this obstacle, instructors and parents have actually embraced an effective tool: Multiplication Fact Worksheets.
Intro to Multiplication Fact Worksheets
Multiplication Fact Worksheets
Multiplication Fact Worksheets -
You may select between 12 and 30 multiplication problems to be displayed on the multiplication worksheets These multiplication worksheets are appropriate for Kindergarten 1st Grade 2nd Grade 3rd
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These multiplication facts worksheets provide various exercise to help students gain fluency in the multiplication facts up to 12 x 12 Jump to your topic Multiplication facts review times tables
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10 Best Images Of Multiplication Worksheets 1 12 Multiplication Worksheets 1 10 100 Division
10 Best Images Of Multiplication Worksheets 1 12 Multiplication Worksheets 1 10 100 Division
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40 Multiplication Worksheets These multiplication worksheets extend the Spaceship Math one minute timed tests with the x10 x11 and x12 facts Even if your school isn t practicing multiplication past
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Multiplication Facts Worksheets
Multiplication Facts Worksheets
Multiplication is the leading resource for helping kids learn the times tables and multiplication facts Play free multiplication games take auto scored quizzes drill flashcards and access worksheets
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How to Produce Engaging Multiplication Fact Worksheets
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Associating multiplication to day-to-day scenarios includes importance and usefulness to exercises.
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Practice Routine practice strengthens multiplication skills, advertising retention and fluency. Stabilizing Rep and Range A mix of recurring exercises and diverse problem layouts maintains interest
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Multiplication facts worksheets K5 Learning
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Regular method is vital. Normal sessions, preferably a few times a week, can yield considerable improvement.
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Polar moment of inertia of circle: Formula, Derivation, Examples
Hi friends, In the preceding article, we discussed the polar moment of inertia. In this post, we’ll go through the polar moment of inertia of a circle in detail.
In this article, we’re going to discuss:
1. Polar moment of inertia of a circle:
2. Polar moment of inertia circle formula:
3. Polar moment of inertia of circle derivation:
3.1. For solid circle:-
3.2. For hollow circle:-
4. Polar moment of inertia of circle examples:
Polar moment of inertia of a circle:
The polar moment of inertia of circle is used for the analysis of objects with circular profiles subjected to the torsional or twisting load.
It helps to find the shear stresses across the cross-section of the circular shaft, axles, couplings, etc.
Polar moment of inertia circle formula:
The polar moment of the solid circle is given by,
J = `\frac{\pi }{32} D^{4}`
J = `\frac{\pi R^{4}}{2}`
D = Diameter of circle
R = Radius of circle
The polar moment of inertia for hollow circular profile is given by,
J = `\frac{\pi }{32}[D_{o}^{4}-D_{i}^{4}]`
J = `\frac{\pi }{2}[R_{o}^{4}-R_{i}^{4}]`
D[o] = Outer diameter
D[i] = Inner diameter
R[o] = Outer radius
R[i] = Inner radius
Polar moment of inertia of circle derivation:
A] For solid circle:-
The above figure shows a circle of radius ‘R’. Consider a smaller portion of thickness ‘dr’ located at radius ‘r’ from the center.
The area of this smaller portion is given by,
dA = Perimeter x dr
dA = 2πr.dr
As per the definition of polar moment of inertia, the polar moment of inertia of a smaller portion is given by,
`J_{dA}` = `r^{2}.dA`
The total polar moment of inertia of the circle can be found by integrating the above equation.
J = `\int_{0}^{R}r^{2}.dA`
By putting the value of dA, the equation becomes,
J = `\int_{0}^{R}r^{2}.2\pi r.dr`
J = `2\pi\int_{0}^{R}r^{3}.dr`
J = 2π`[\frac{r^{4}}{4}]_{0}^{R}`
J = `2\pi[\frac{ R^{4}}{4}-0]`
J = `\frac{\pi R^{4}}{2}`
To find the polar moment of inertia in the form of diameter, put R = D/2 in the above equation.
J = `\frac{\pi (\frac{D}{2})^{4}}{2}`
J = `\frac{\pi D^{4}}{32}`
B] For hollow circle:-
The above figure shows the hollow circle with an inner radius of `R_{i}` and an outer radius of `R_{o}`.
Consider a smaller portion of thickness ‘dr’, located at a distance ‘r’ from the center of the hollow circle.
The area of this small portion is given by,
dA = Perimeter x Thickness
dA = (2πr).dr
As per the definition, the polar moment of inertia of a smaller portion is given by,
`J_{dA}` = `r^{2}.dA`
The total polar moment of inertia of the circle can be found by integrating the above equation.
J = `\int_{Ri}^{Ro}r^{2}.dA`
By putting the value of dA, the equation becomes,
J = `\int_{Ri}^{Ro}r^{2}(2\pi r).dr`
J = `2\pi \int_{Ri}^{Ro}r^{3}.dr`
J = `2\pi [\frac{r^{4}}{4}]_{Ri}^{Ro}`
J = `2\pi [\frac{R_{o}^{4}}{4}-\frac{R_{i}^{4}}{4}]`
J = `\frac{2\pi[R_{o}^{4}-R_{i}^{4}]}{4}`
J = `\frac{\pi }{2}[R_{o}^{4}-R_{i}^{4}]`
To find the polar moment of inertia in the form of diameter, put Ro = Do/2 and Ri = Di/2 in the above equation.
J = `\frac{\pi }{2}[(\frac{D_{o}}{2})^{4}-(\frac{D_{i}}{2})^{4}]`
J = `\frac{\pi }{32}[D_{o}^{4}-D_{i}^{4}]`
Polar moment of inertia of circle examples:
1] A machine has a hollow circular shaft of an outer diameter of 80 mm and an inner diameter of 65 mm. Find the polar moment of inertia about the axis of the shaft.
D[o] = 80 mm
D[i] = 65 mm
The polar moment of inertia for the hollow circular profile is given by,
J = `\frac{\pi }{32}[D_{o}^{4}-D_{i}^{4}]`
J = `\frac{\pi }{32}[80^{4}-65^{4}]`
This is the polar moment of inertia for the hollow circular shaft.
2] A solid shaft of length 1 m has a diameter of 40 mm and a hollow shaft of length 1.2 m has an outer diameter of 60 mm. If both the shafts are made by using the same amount of material then compare
the polar moment of inertia of both shafts.
D[solid] = 40 mm = 0.04 m
L[solid] = 1 m
D[Hollow(o)] = 60 mm = 0.06 m
L[Hollow]= 1.2 m
Step 1] Inner diameter of hollow shaft (`\mathbf{D_{\text{Hollow(i)}}}`):
As the both shafts are made up by using same amount of material.
∴ `V_{\text{solid}}` = `V_{\text{hollow}}`
∴ `A_{\text{solid}}` x `L_{S}` = `A_{\text{hollow}}` x `L_{H}`
`\frac{\pi }{4}D_{s}^{2}` x `L_{S}` = `\frac{\pi }{4}[D_{\text{H(o)}}^{2}-D_{\text{H(i)}}^{2}]` x `L_{H}`
`\frac{\pi }{4}\times 0.04^{2}` x 1 = `\frac{\pi }{4}[0.06^{2}-D_{\text{H(i)}}^{2}]` x 1.2
`\mathbf{D_{\text{H(i)}}}` = 0.047 m
Step 2] Polar moment of inertia of solid circle:
Polar moment of inertia of a solid circle is given by,
`J_{S}` = `\frac{\pi }{32}D_{S}^{4}`
`J_{S}` = `\frac{\pi }{32}0.04^{4}`
`\mathbf{J_{S}}` = 2.513 x 10⁻⁷ m⁴
Step 3] Polar moment of inertia of hollow circle:
Polar moment of inertia of a hollow circle is given by,
`J_{H}` = `\frac{\pi }{32}[D_{\text{H(o)}}^{4}-D_{\text{H(i)}}^{4}]`
`J_{H}` = `\frac{\pi }{32}[0.06^{4}-0.047^{4}]`
`\mathbf{J_{H}}` = 7.932 x 10⁻⁷ m⁴
Step 4] Compare polar moment of inertia of solid and hollow shaft:
`\frac{J_{H}}{J_{S}}` = `\frac{7.932\times 10^{-7}}{2.513\times 10^{-7}}`
`\frac{J_{H}}{J_{S}}` = 3.156
`\mathbf{J_{H} = 3.156 J_{S}}`
Thus the hollow shaft has a polar moment of inertia equal to 3.156 times of polar moment of inertia of the solid shaft.
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e46 330i manual conversion
Been looking at purchasing a 6-cylinder e46 recently, but most listings on TradeMe are automatics and the manuals are generally priced out of my budget.
How much would it cost to convert an automatic to a manual approximately? this seemed like a far better option to me as ill most likely be able to find an automatic that hasn't been thrashed too much
in its past and is in good shape.
Vass 829
Depends if you're doing the work yourself or taking it to a shop. If the latter, then I suspect the labour will be the killer part in it.
Parts wise, I've seen conversion kits sold locally for around $4.5k of late. That's with a ZF box. Personally, I got my kit shipped from the UK and came out to around $3.3k with the full driveshaft
and a 2.93 diff (not strictly necessary). Add to that something like $750 for the certification if you want to do it properly and get it fully compliant. Plus all sorts of tools, consumables like
oils and whatever bits you might want to replace preventatively (clutch, flywheel, slave cylinder etc.) whilst you're at it. Depends on how far you want to go.
At a minimum I'd say you'd want to budget for $4-5k.
• 1
• 1
Eagle 1661
+1. Id say 5-6k if you doing in properly with ZF box, new shifter parts, driveshaft, cylinders and clutch\flywheel bits etc. Good portion of the converted ones are done on the cheap with Getrag box
and old\sh*t parts used. Rarely see a properly converted ones for sale as id imagine people tend to keep them these days.
Msport factory manuals are around ~15k if you can find one with standard spec manual being more common and cheaper at around ~10k. If that out of budget then id probably look at something else
because buying these cars is the cheapest part (unless you buying one with all the work done and paying good money for it)
• 1
• 1
smiddy 76
I find it differcult to understand the percieved gain and over the top cost in a manual converstion of a 230 odd hp torque based motor.
Vass 829
12 minutes ago, smiddy said:
I find it differcult to understand the percieved gain and over the top cost in a manual converstion of a 230 odd hp torque based motor.
You're not alone.
Gain - for me, personally, grasping onto a bit of analog driving experience in an ever-increasingly automated world.
Over the top cost - basic supply and demand. For whatever reason, manuals are a rather rare breed in NZ, unfortunately, hence people being able to ask silly money for something that's fairly abundant
in the old world.
I wouldn't be able to justify it to myself if I merely saw it as a mode of transportation. It's more of a hobby project at this point. That's why they call us enthusiasts I guess.
• 3
Deffed 10
You can pick up so called conversion kit. Box,clutch fluid res,hose,2nd hand 5speed and clutch $1200,the hard part is finding the shifter. | {"url":"https://bimmersport.co.nz/topic/67002-e46-330i-manual-conversion/?tab=comments#comment-706873","timestamp":"2024-11-12T00:10:51Z","content_type":"text/html","content_length":"123602","record_id":"<urn:uuid:72070641-c9e3-4ee7-a1e2-a290f1d94ff0>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00006.warc.gz"} |
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Duality and hidden equilibrium in trans
SciPost Submission Page
Duality and hidden equilibrium in transport models
by Rouven Frassek, Cristian Giardina, Jorge Kurchan
This is not the latest submitted version.
This Submission thread is now published as
Submission summary
Authors (as registered SciPost users): Rouven Frassek
Submission information
Preprint Link: https://arxiv.org/abs/2004.12796v2 (pdf)
Date submitted: 2020-06-04 02:00
Submitted by: Frassek, Rouven
Submitted to: SciPost Physics
Ontological classification
Academic field: Physics
• Probability
• Condensed Matter Physics - Theory
Specialties: • Mathematical Physics
• Quantum Physics
• Statistical and Soft Matter Physics
Approach: Theoretical
A large family of diffusive models of transport that has been considered in the past years admits a transformation into the same model in contact with an equilibrium bath. This mapping holds at the
full dynamical level, and is independent of dimension or topology. It provides a good opportunity to discuss questions of time reversal in out of equilibrium contexts. In particular, thanks to the
mapping one may define the free-energy in the non-equilibrium states very naturally as the (usual) free energy of the mapped system.
Current status:
Has been resubmitted
Reports on this Submission
Report #2 by Anonymous (Referee 3) on 2020-8-8 (Invited Report)
• Cite as: Anonymous, Report on arXiv:2004.12796v2, delivered 2020-08-08, doi: 10.21468/SciPost.Report.1902
Establishes a novel mapping between transport and "equilbrium" models for a large familiy of stochastic processes.
The authors show that are fairly wide class of classical stochastic
processes that describe transport are related to equilibrium
models. The class considered contains much studied interacting
many-particle models like the symmetric exclusion process.
It is well known that stochastic processes on lattices can be
represented as imaginary time Schroedinger equations for certain
lattice spin "Hamiltonians". In this context the authors may want to
cite an old work by Alacraz et al Annals Phys. 230 (1994) 250-302.
The authors consider a number of such "Hamiltonians" studied in
transport settings transport and show that they are related by
similarity transformations to "Hamiltonians" that look like they
describe equilibrium situations. I think this work is very interesting
as it generalizes results in the literature to a very large class of
stochastic processes, and I therefore recommend publication. However,
I have a number of questions and comments the authors should consider.
(1) In case of the integrable symmetric exclusion process the similarity
transformation has been explicitly constructed by one of the authors
in a recent work. While this is mentioned above (4.23) it is left unclear
what parts of the following discussion is new, and what is quoted from
said earlier work. I think the authors should be more explicit about
this. As far as I can tell (4.31) is old, but (4.36) is new.
(2) The authors may want to add a reference to the work by Alcaraz et
al mentioned above to establishing the isospectrality of
"Hamiltonians" for stochastic processes differing by operators that
are upper triangular in the right basis, as is the case for (4.21) and
(3) My main question has to do with the authors' statement that their
class of transport models are "in fact hidden equilibrium models". My
understanding is that the authors have established that the evolution
operators of the processes they study can be mapped, by means of a
similarity transformation, to "Hamiltonians" that look like they
describe equilibrium situations. I am confused about the authors
statement quoted above because these "Hamiltonians" are generally not stochastic
and therefore cannot be thought of as generators of stochastic
processes. The steady state of the original stochastic process can of
course by construction be obtained from the "ground state" of the
transformed "Hamiltonian" (now viewed as a quantum spin chain at zero
temperature), but this looks like a purely technical observation to me
(in the sense that expectation values in the quantum model will not
generally be related to averages in the original stochastic
process). I think it would be very helpful if the authors expressed
more precisely in which sense their class of transport models are "in
fact hidden equilibrium models".
Report #1 by Anonymous (Referee 2) on 2020-7-28 (Invited Report)
• Cite as: Anonymous, Report on arXiv:2004.12796v2, delivered 2020-07-28, doi: 10.21468/SciPost.Report.1865
This is a very interesting paper, a long awaited (for me at least) follow up to work by some of the authors over a decade ago on mapping between driven stochastic systems and equilibrium ones. The
paper thus addresses a very important question in statistical mechanics, that of the fundamental distinction between non-equilibrium and equilibrium.
The setting is that of transport processes connected to leads, like the SEP and its generalisations. Bulk conservation allows to describe these models in hydrodynamic terms in the appropriate scaling
limit via MFT. This is the usual framework to study non-equilibrium transport and associated questions. In such settings what is being transported is conserved (let's say carries energy) so
individual transitions in the bulk are reversible (have equal probabilities per link that is connected) and is created and destroyed at the leads. When these creation/destruction rate pairs are the
same in all leads the system obeys detailed balance (aka is an equilibrium problem), when at least one of these rate pairs is different in one lead the system is driven (eg. inject at one end and
eject at the other).
The key result of the paper is that generically in this setting a driven system (leads at different temperatures so the rate pairs not all equal) can be mapped to an equilibrium system (leads all at
the same temperature) with the (i) same state space and (ii) same connectivity between states, including leads. This is a non-trivial result. For integrable systems (eg. SEP chain with boundary
injection/ejection) the mapping can be shown explicitly. The logic is very simple and easy to follow - below I give my perspective on it and ask a couple of questions. The paper, maybe with some of
my questions addressed, definitively should be published in SciPost.
The logic of how one arrives to the result is straightforward and that is one thing I like here. I explain if from the perspective I see it most clearly, see refs.[1-3]. What is exploited in Sec.4 is
what is sometimes called a "gauge transformation" in Markov chains (the most used one is the so-called Doob transform). This is borrowed from terminology in matrix product states (MPS) which
represent hidden Markov chains (and thus also explicit Markov chains).
The probability of a particular trajectory is a product of operators (e.g. exponential of the escape rate during a waiting time, times the probability for a particular jump, and so on). The
probability for all trajectories, i.e. the ensemble of trajectories, can thus be collected in an MPS in time. MPS have a gauge invariance, since one can insert identities between every pair of the
operators above. It is a gauge symmetry as the operators can be time dependent. The gauge transformation is a change of basis or alternatively a change of generators of the trajectory. Only certain
choices of gauge give stochastic generators.
Equation 4.11 is one such gauge transformation *in the long-time limit* where the transformation becomes time-independent. It differs crucially with the Doob transforms used e.g. in large deviation
theory in that it is not a point transformation (which trivially leaves the allowed transitions invariant while modifying their rates). Nevertheless, the authors find transformation operators which
while not diagonal (as in Doob) still maintain the transition network (thus mapping the problem into itself with different rates). The transformation maps one stochastic generator to a different
stochastic generator which an equivalent ensemble of trajectories.
So these are my questions/comments:
- The full equivalence of trajectories at all times (not just long times) requires the time-dependent form of the transformation. Could the authors please comment on this.
- Specifically, the initial probability in the original problem (generated by H) gets mapped to a different probability in the transformed problem (generated by H_eq) by an unpaired gauge operator P.
How do these probabilities relate to each other? Furthermore, for the overall probability over trajectories to be well behaved the usual condition is that the transformation reverts eventually to the
identity at the final time, see e.g. [3]. Does not the same need to occur here?
- If one is interested in steady state dynamics, the initial state becomes irrelevant. However, in light of the intriguing comment by the authors towards the end about the free-energy of H_eq problem
being the non-equilibrium free-energy of the driven problem there is a related consideration. To obtain the probability over states one needs to sandwich (using the MPS language) the train of
operators with a configuration vector. But again there is another unpaired transformation operator that maps one basis to the other. I think one then needs to explain more clearly how the steady
state probabilities in one problem map to the steady state probabilities in the other. I do not think it is just p(x) = p_eq(x), and thus it is not totally obvious to me what the statement about the
free-energy precisely means. Can the authors please comment?
I apologise if some or all of these are already answered in the MS, but in that case I would suggest making those statements more prominent and explicit.
[1] Haegeman J, Cirac J I, Osborne T J and Verstraete F 2013 Phys. Rev. B 88 085118
[2] Chetrite R and Gupta S 2011 J. Stat. Phys. 143 543
[3] Chetrite R and Touchette H 2015 Ann. Henri Poincaré 16 2005
Report #3 by Anonymous (Referee 1) on 2020-7-17 (Invited Report)
• Cite as: Anonymous, Report on arXiv:2004.12796v2, delivered 2020-07-17, doi: 10.21468/SciPost.Report.1913
Dear Editor, dear Authors
I read with interest the manuscript Duality and hidden equilibrium in transport models. The Authors discuss several examples of transport models that admit a similarity transformation of the
generator into an operator satisfying detailed balance, arguing that this transformation is possible whenever a general notion of “duality” is present - rather than the more restrictive integrability
usually considered in the specialized literature.
It is a nicely written remarkable paper. The ideas presented in the introductory parts of the paper are very broad and intriguing, and the deployment of technical ability impressive. However, the
development is challenging to the non-expert, and it is not clear to what degree the ideas proposed in the introduction are proven and motivated in full generality, given that the paper mostly
focuses on specific models.
More precisely:
A general notion of duality is mentioned but never introduced in a general setting, apart from stating that it is a well-known duality in probability theory. Also, it is not clear how the existence
of a symmetry group for the generator interplays with such duality. Is there a general way to introduce group-based diffusive models and duality independently of the specific cases analyzed?
The technical knowledge displayed of the rich mathematics of interacting particle models is impressive, but at times it looks like a piece of virtuosism (for example: what is the actual need for eq.
4.9? Also: in Sec: 4.3 it is mentioned that, quite intriguingly, integrability allows to solve for W whenever Q_0 is easier to diagonalize than H_0: but it appears that in the following treatment an
expression for W is found without diagonalizing H_0)
In the introduction it is mentioned that there exist a time-reversed mapping between \rho_2 and \rho_1 even when the system is driven out of equilibrium. But how is this special of equilibrium
systems? Given a MC propagator e^tH one can always define the inverse map e^-tH. Detailed balance has more to do with correlation functions than with probabilities.
It is mentioned in passing that the new detailed balanced dynamics has absorbing sites. But then, the stationary state of the new equilibrium dynamics will be different from the stationary state of
the original dynamics. So, while there is a mapping between generators, it appears that there is no mapping between the time-evolved probabilities, and that stationary states map to specific
transient states. Maybe the Authors should explain better what exactly the mapping consists of.
Best regards
The referee | {"url":"https://scipost.org/submissions/2004.12796v2/","timestamp":"2024-11-13T09:44:44Z","content_type":"text/html","content_length":"46973","record_id":"<urn:uuid:f7d85b68-7d67-4488-9d73-ee98dbdf2ca3>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00664.warc.gz"} |
Suppressing quantum errors by scaling a surface code logical qubit – Google AI Blog
Many years from today, scientists will be able to use fault-tolerant quantum computers for large-scale computations with applications across science and industry. These quantum computers will be much
bigger than today, consisting of millions of coherent quantum bits, or qubits. But there’s a catch — these basic building blocks must be good enough or the systems will be overrun with errors.
Currently, the error rates of the qubits on our 3rd generation Sycamore processor are typically between 1 in 10,000 to 1 in 100. Through our work and that of others, we understand that developing
large-scale quantum computers will require far lower error rates. We will need rates in the range of 1 in 10^9 to 1 in 10^6 to run quantum circuits that can solve industrially relevant problems.
So how do we get there, knowing that squeezing three to six orders of magnitude of better performance from our current physical qubits is unlikely? Our team has created a roadmap that has directed
our research for the last several years, improving the performance of our quantum computers in gradual steps toward a fault-tolerant quantum computer.
Roadmap for building a useful error-corrected quantum computer with key milestones. We are currently building one logical qubit that we will scale in the future.
Today, in “Suppressing Quantum Errors by Scaling a Surface Code Logical Qubit”, published in Nature, we are announcing that we have reached the second milestone on our roadmap. Our experimental
results demonstrate a prototype of the basic unit of an error-corrected quantum computer known as a logical qubit, with performance nearing the regime that enables scalable fault-tolerant quantum
From physical qubits to logical qubits
Quantum error correction (QEC) represents a significant shift from today’s quantum computing, where each physical qubit on the processor acts as a unit of computation. It provides the recipe to reach
low errors by trading many good qubits for an excellent one: information is encoded across several physical qubits to construct a single logical qubit that is more resilient and capable of running
large-scale quantum algorithms. Under the right conditions, the more physical qubits used to build a logical qubit, the better that logical qubit becomes.
However, this will not work if the added errors from each additional physical qubit outweigh the benefits of QEC. Until now, the high physical error rates have always won out.
To that end, we use a particular error-correcting code called a surface code and show for the first time that increasing the size of the code decreases the error rate of the logical qubit. A
first-ever for any quantum computing platform, this was achieved by painstakingly mitigating many error sources as we scaled from 17 to 49 physical qubits. This work is evidence that with enough
care, we can produce the logical qubits necessary for a large-scale error-corrected quantum computer.
Quantum error correction with surface codes
How does an error-correcting code protect information? Take a simple example from classical communication: Bob wants to send Alice a single bit that reads “1” across a noisy communication channel.
Recognizing that the message is lost if the bit flips to “0”, Bob instead sends three bits: “111”. If one erroneously flips, Alice could take a majority vote (a simple error-correcting code) of all
the received bits and still understand the intended message. Repeating the information more than three times — increasing the “size” of the code — would enable the code to tolerate more individual
Many physical qubits on a quantum processor acting as one logical qubit in an error-correcting code called a surface code.
A surface code takes this principle and imagines a practical quantum implementation. It has to satisfy two additional constraints. First, the surface code must be able to correct not just bit flips,
taking a qubit from |0⟩ to |1⟩, but also phase flips. This error is unique to quantum states and transforms a qubit in a superposition state, for example from “|0⟩ + |1⟩” to “|0⟩ – |1⟩”. Second,
checking the qubits’ states would destroy their superpositions, so one needs a way of detecting errors without measuring the states directly.
To address these constraints, we arrange two types of qubits on a checkerboard. “Data” qubits on the vertices make up the logical qubit, while “measure” qubits at the center of each square are used
for so-called “stabilizer measurements.” These measurements tell us whether the qubits are all the same, as desired, or different, signaling that an error occurred, without actually revealing the
value of the individual data qubits.
We tile two types of stabilizer measurements in a checkerboard pattern to protect the logical data from bit- and phase-flips. If some of the stabilizer measurements register an error, then
correlations in the stabilizer measurements are used to identify which error(s) occurred and where.
Surface-code QEC. Data qubits (yellow) are at the vertices of a checkerboard. Measure qubits at the center of each square are used for stabilizer measurements (blue squares). Dark blue squares check
for bit-flip errors, while light blue squares check for phase-flip errors. Left: A phase-flip error. The two nearest light blue stabilizer measurements register the error (light red). Right: A
bit-flip error. The two nearest dark blue stabilizer measurements register the error (dark red).
Just as Bob’s message to Alice in the example above became more robust against errors with increasing code size, a larger surface code better protects the logical information it contains. The surface
code can withstand a number of bit- and phase-flip errors each equal to less than half the distance, where the distance is the number of data qubits that span the surface code in either dimension.
But here’s the problem: every individual physical qubit is prone to errors, so the more qubits in a code, the more opportunity for errors. We want the higher protection offered by QEC to outweigh the
increased opportunities for errors as we increase the number of qubits. For this to happen, the physical qubits must have errors below the so-called “fault-tolerant threshold.” For the surface code,
this threshold is quite low. So low that it hasn’t been experimentally feasible until recently. We are now on the precipice of reaching this coveted regime.
Making and controlling high-quality physical qubits
Entering the regime where QEC improves with scale required improving every aspect of our quantum computers, from nanofabrication of the physical qubits to the optimized control of the full quantum
system. These experiments ran on a state-of-the-art 3rd generation Sycamore processor architecture optimized for QEC using the surface code with improvements across the board:
• Increased qubit relaxation and dephasing lifetimes through an improved fabrication process and environmental noise reduction near the quantum processor.
• Lowered cross-talk between all physical qubits during parallel operation by optimizing quantum processor circuit design and nanofabrication.
• Reduced drift and improved qubit control fidelity through upgraded custom electronics.
• Implemented faster and higher-fidelity readout and reset operations compared with previous generations of the Sycamore processor.
• Reduced calibration errors by extensively modeling the full quantum system and employing better system-optimization algorithms.
• Developed context-aware and fully parallel calibrations to minimize drift and optimize control parameters for QEC circuits.
• Enhanced dynamical decoupling protocols to protect physical qubits from noise and cross-talk during idling operations.
Running surface code circuits
With these upgrades in place, we ran experiments to compare the ratio (𝚲[3,5]) between the logical error rate of a distance-3 surface code (ε[3]) with 17 qubits to that of a distance-5 surface code
(ε[5]) with 49 qubits — 𝚲[3,5] = ε[3] / ε[5].
Comparison of logical fidelity (defined as 1-ε) between distance-3 (d=3) and distance-5 (d=5) surface codes. The distance-5 code contains four possible distance-3 arrangements, with one example shown
in the red outline (left). As improvements were made, the d=5 fidelity increased faster than that of the d=3, eventually overtaking the distance-3 code, as shown in the top-right data points (right),
whose average lies slightly to the left of the ε[3] = ε[5] line.
The results of these experiments are shown above on the right. Continued improvements over several months allowed us to reduce the logical errors of both grids, leading to the distance-5 grid (ε[5] =
2.914%) outperforming the distance-3 grids (ε[3] = 3.028%) by 4% (𝚲[3,5] = 1.04) with 5𝛔 confidence. While this might seem like a small improvement, it’s important to emphasize that the result
represents a first for the field since Peter Shor’s 1995 QEC proposal. A larger code outperforming a smaller one is a key signature of QEC, and all quantum computing architectures will need to pass
this hurdle to realize a path to the low errors that are necessary for quantum applications.
The path forward
These results indicate that we are entering a new era of practical QEC. The Google Quantum AI team has spent the last few years thinking about how we define success in this new era, and how we
measure progress along the way.
The ultimate goal is to demonstrate a pathway to achieving the low errors needed for using quantum computers in meaningful applications. To this end, our target remains achieving logical error rates
of 1 in 10^6 or lower per cycle of QEC. In the figure below on the left, we outline the path that we anticipate to reach this target. As we continue improving our physical qubits (and hence the
performance of our logical qubits), we expect to gradually increase 𝚲 from close to 1 in this work to larger numbers. The figure below shows that a value of 𝚲 = 4 and a code distance of 17 (577
physical qubits with good enough quality) will yield a logical error rate below our target of 1 in 10^6.
While this result is still a few years out, we have an experimental technique to probe error rates this low with today’s hardware, albeit in limited circumstances. While two-dimensional surface codes
allow us to correct both bit- and phase-flip errors, we can also construct one-dimensional repetition codes that are only able to solve one type of error with relaxed requirements. On the right
below, we show that a distance-25 repetition code can reach error rates per cycle close to 1 in 10^6. At such low errors, we see new kinds of error mechanisms that are not yet observable with our
surface codes. By controlling for these error mechanisms, we can improve repetition codes to error rates near 1 in 10^7.
Left: Expected progression as we improve performance (quantified by 𝚲) and scale (quantified by code distance) for surface codes. Right: Experimentally measured logical error rates per cycle versus
the distance of one-dimensional repetition codes and two-dimensional surface codes.
Reaching this milestone reflects three years of focused work by the entire Google Quantum AI team following our demonstration of a quantum computer outperforming a classical computer. In our march
toward building fault-tolerant quantum computers, we will continue to use the target error rates in the figure above to measure our progress. With further improvements toward our next milestone, we
anticipate entering the fault-tolerant regime, where we can exponentially suppress logical errors and unlock the first useful error-corrected quantum applications. In the meantime, we continue to
explore various ways of solving problems using quantum computers in topics ranging from condensed matter physics to chemistry, machine learning, and materials science.
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Data Mining (HSE)
Section outline
• On this page, we will publish short reminders about topics covered. Instead of endless lists of scientific papers, we will give you short and friendly articles. We won't shy away from directing
you to Wikipedia, too.
Every section below contains a few papers (or even Wikipedia pages) that easy to read, without much math. Some students asked more advanced questions (e.g. relation between Bayesian modeling
and logistic regression), so here is a list of some more advanced books.
• The purpose of visualization could be to:
□ provide nice illustrations for newspaper articles;
□ help us discover patterns in the data.
In data mining, we use them for the latter (except when trying to impress our bosses). We have not evolved as number crunching animals, so our brains are specialized for visual patterns. Given a
sequence of numbers, we are not able to find patterns, like trends and correlations. Seeing a picture, we are. So the main purpose of insight-giving visualizations are to present the data in such
a form that our brains can spot things that we would not be able to spot if the data was presented as just numbers (or whatever human-unfriendly form the data has).
Here is a brief summary of what we covered.
□ Box Plot: shows basic statistics, good for observing a single variable. We can split it into multiple box plots by a categorical variable. In this case, we can compute chi squares or ANOVAs
and order the variables by how well they variable splits the categories.
□ Distribution: shows distributions as histograms. We can again split them by a categorical variable. We emphasized the problem of choosing the number of bins, which can strongly affect the
visualization. Curves also have to be considered carefully since they are either result of some smoothing, which has some parameters that are hidden or difficult to set and interpret, or they
result of fitting of parameters to a shape that we have to decide -- and often have no good argument for a particular shape. Bins are essentially better because they do not hide anything --
unless the criteria for deciding the number of bins or boundaries are manipulated.
□ Scatter Plot: plots the relationship between two numeric variables. To show other quantities in the same plot, we can use colors, sizes, shapes or other fancy properties, such as angles. Out
of these three, colors are the only one that really works. It is easy to see that a certain region has a prevalence of certain colors. (Sometimes this is too easy to see. We are very able to
spot patterns where there are none. We'll see some examples in due time.)
A note about using colors for showing numeric quantities. It is a very good idea to use discrete scale with not too many colors. An average human can distinguish roughly five different
colors, except for women who can distinguish about a dozen thousands. If the scale of colors is continuous, it is impossible to compare a color in a legend with a color of the spot on the
graph. While continuous scales appear to be more exact than discrete scales with, say, eight different colors, the latter are probably better.
You have learned about other traps (such as using sizes) in the other course.
□ Mosaic Display: splits the data by 1-4 variables into ever smaller subgroups, such that the size of the area corresponds to the size of the group. The area can be split further to show the
target variables distribution inside the group. This visualization is useful for finding relations between categorical features in a similar fashion as the scatter plot does for numeric.
□ Sieve Diagram: essentially a visualization of chi-square, it plots actual frequencies of two variables against their expected frequencies. Unlike in the Mosaic display, the area is split
independently along each axis, so sizes correspond to the expected number of instances assuming that the variables are independent. Sieve then shows - using a grid and colors - the violations
of independence, that is, combinations of values that are more common or rare than expected.
(If the video doesn't play: it happens to me, too. There seem to be something wrong on YT's side. I hope it resolves, otherwise I'll reupload.)
A blog post about how statistical tests work and why we have to be very careful when using them in data mining.
A shorter and drier version of the same.
A famous, juicy paper with general arguments against null-hypothesis testing.
• We considered a simple machine learning algorithm: induction of classification trees. Classification trees are models that can predict the value of the target variable (outcome, class) for a
given data instance. Trees can be induced from data using a suitable algorithm. We used a top-down approach that divides the data into ever smaller subsets, at each step using the most
informative variable as a splitting criterion.
Models must generalize, that is, they must be able to make predictions for data instances such that were never seen before. In order to generalize, they must "smooth" over the data, ignoring
potential errors or specific cases. If we take trees as an example: they must be neither too large (overfitting to every specific data instance) nor too small (too general).
The amount of smoothing is regulated using certain parameters of the fitting (learning) algorithm. In case of trees, we set the minimal number of data instances in leaves, maximal tree depth, the
proportion of majority class at which we stop dividing etc.
Pioneers of AI liked classifications trees because they believed they mimic human reasoning. While they may be interesting for historic reasons, they are of little practical importance today.
They are useful, though, to illustrate some basic principles: they introduced us to some basic ideas that we will keep encountering as we proceed on, to more complex models.
The page contains the crux of the lecture. Its title and the fact that the first link on the page points to a completely unrelated kind of decision trees demonstrate why classification tree
is a better term than decision tree. [Mandatory reading, with grain of salt; you most certainly don't need to know that "It is also possible for a tree to be sampled using MCMC." :) ]
We have spent a lot of time explaining the concept of entropy. Wikipedia page is rather mathematical and dry, but may be a good antidote to less formal and less organized exposition at the
lecture. :)
Quinlan is the author of one of the first and most influential algorithm for induction of classification trees. The article is more of historical interest, but it shows the thinking of the
pioneers of AI. After some philosophy in the first two sections, it explains the reasoning behind the tree-induction algorithms.
• We saw a model that was able to predict randomly assigned data labels. Since this is clearly impossible, something must had been wrong in our procedure.
We discovered that the tree essentially memorized the data: it split the space of data instances (you can imaging this space as a scatter plot) so that each part (a leaf of the tree) contained
just a few data instances, all belonging to the same class. When making predictions, the tree just checked the region to which the data instance belongs and recalled the correct class.
To avoid such situations, we should never ever estimate the quality of the tree by measuring its performance of the data that was used for learning the tree (also called training data), but
always retain separate data for testing. We discussed different sampling techniques, the most known of which is cross validation.
Next we focused on measuring the quality of models. We can use them for (at least) two purposes to make predictions, or to get some insight into the data.
For the latter, a good model is a model that gives us some useful insight. We should treat models as hypothesis, as ideas, and test them against our background knowledge (does it make sense? does
the tree use attributes that are related to the problem? ...)
For the former, we have different performance scores. Classification accuracy is the simplest, but it ignores the types of mistakes; classifying a sick person as healthy is (usually, at least for
most observers :) worse than vice versa. Hence we have derived a number of scores that measure probabilities of various types of mistakes, such as the probability that a sick person will be
classified as sick (sensitivity or recall), and the probability that a person classified as sick is actually sick (precision). There are dozens of such combinations. It is less important to know
their names than to be aware that the score we are using should correspond to our goals.
Many of the above measures appear in pairs: by changing the decision threshold, we can improve one at a cost of another. To visualize this, we use curves, such as ROC curve.
Everybody should draw a ROC curve manually at least once in his life. We've done so in the lecture. This led us to think about other interesting properties of the curve and especially the
interpretation of the area under it. All that we've done and said can be found in the paper we cite below An introduction to ROC analysis. To test our understanding, we solved a problem faced by
Sara, a hamster veterinarian.
The most important message here was that when making predictions, we have to set a decision threshold and the threshold usually balances between two opposing quantities (sensitivity-specificity,
precision-recall), and can also take costs into consideration. In practice we'd often construct other types of curves to show the possible operating points.
Use this page as a list of different sampling techniques.
A very accessible paper about ROC curves.
• We discussed models that work by assigning a certain number of points to each possible feature/variable, for instance, a certain number of points for being male (or female), a certain number of
points for every year the person has and so forth. The total can be expressed as a sum $\beta^\intercal \mathbf{x}$. If the total exceeds 0, we predict the positive, otherwise the negative class.
For easier understanding, we can imagine that all variables are numeric. The points for which $\beta^\intercal x = 0$ lie on a (hyperplane), which is called a decision boundary.
The modelling problem can thus be reimagined as follows: we have a room of red (positive) and blue (negative) points, and the modelling task is to draw a plane that separates them (as well as
possible). The plane is defined by the vector $\beta$.
Models of this kind are linear models.
One of them is logistic regression, which defines the probability of the class $y$ given data $\textbf{x}$ as $p(y|x) = 1 / (1 + e^{-\beta^\intercal \textbf{x}})$. It uses the logistic function
to transform the distance from the plane into probabilities. Logistic regression tries to find such a plane that all points from one class are as far away from the boundary (in the correct
direction) as possible. Formally, it maximizes the product of probabilites that the model will assign to the correct class. Such product is called likelihood and the process of finding the
optimal decision boundary by optimizing the likelihood is called maximum likelihood estimation. You will surely encounter it in other classes, too. More in the paper below.
Another common linear model is (linear) support vector machine (SVM), which optimizes a slightly different criteria: it maximizes the distances between the plane and the closest points, with some
punishment for points lying on the wrong side. We have not spent much time on this since SVMs will become more interesting later.
Our final linear model was the Naive Bayesian classifier. It is derived differently than the other two. We want to predict the probability of class $c$ given some attributes $\mathbf{x} = (x_1,
x_2, \ldots, x_n)$, that is $P(c|\mathbf{x})$. By applying Bayesian rule (twice), we discover that $P(c|\mathbf{x}) \sim P(c) \prod_i P(c|x_i)$, if we naively (thence the name) assume that the
attributes are independent. With some clever manipulation (read about it in the paper about nomograms), we can see that this model can also be expressed with an equation of the same form as
linear regression. The only difference is (again) in how the hyperplane (expressed by $\beta$) is fit to the data.
Naive Bayesian classifier and logistic regression differ in important aspects you should remember. Difference stem from the fact that Naive Bayesian classifier is univariate (it considers a
single variable at a time, independent of others), while logistic regression is multivariate.
Naive Bayesian classifier does not take correlations into account, because it assumes the attributes are independent. Considering one variable at a time, $\mathbf{w}$ contains the importance of
each attribute separately. We will use it when we want to know the importance of each attributes.
Logistic regression observes all variables at once and takes the correlation into account. If some variables are correlated, their importance will be spread among them. With proper regularization
(we'll talk about this later), LR can be used to find a subset of non-redundant (non-correlated, "non-overlapping") variables sufficient for making predictions.
Probabilities returned by Naive Bayesian classifier are not well-calibrated, because the method is univariate and considers the same piece of evidence multiple times. Logistic regression is
usually well-calibrated (logistic function is actually used for calibrating other classifiers sometimes).
Being univariate and simpler Naive Bayesian classifier needs less data than logistic regression. It can also handling missing data: if a value is unknown, its contribution is zero. Logistic
regression cannot do this.
Finally, we observed naive Bayesian classifier and logistic regression in a nomogram, which shows regression coefficients assigned to individual values. The nomogram can be used for making
predictions or exploring the model. To make a prediction, we drag the points for each variable to its corresponding value and the axes at the bottom convert the sum into a prediction. In case of
Bayesian classifier, we can also leave a point at the center if the value is unknown. In terms of exploring the model, the lenghts of lines in the nomogram tell us how many points a data instance
can get or lose based on each variable. Bayesian classifier can also show us the impact of the value on the decision.
These as well as other differences (e.g. the nomogram for logistic regression does not tell us importance of individual features) come from the general differences between the two techniques.
A more mathematical (compared to our lecture), but still friendly explanation of logistic regression. Read the first 6 pages, that is, section 12.1 and the (complete) section 12.2. You don't
have to know the formulas, but you need to understand their meaning.
(This is Chapter 12 from Advanced Data Analysis from an Elementary Point of View. Download the draft from the author's site while it's free.)
A quick derivation of the Naive Bayesian classifier, and derivation and explanation of nomograms.
• In the lecture, we gretaly improved the algorithms we have seen so far.
□ Support vector machines (SVM) become non-linear. The problem with linear support vector machines is that they are linear, that is, the boundary they draw is always a hyperplane. We observed
that we can use a similar trick as in linear regression: add squares of variables and their products. The trick (which actually has a name: kernel trick) is how to expand the data into
higher-dimensional, perhaps even infinite-dimensional space, without explicitly transforming it. It turns out the SVM (and other linear methods) can be reformulated in such way that they use
the data instances (e.g. $x_i$, $x_j$) only to compute scalar products - in this context, they are usually written as $\left<x_i, x_j\right>$. If data is expanded using a certain function $\
phi(x)$ that, for instance, adds products, we need to compute $\left<\phi(x_i), \phi(x_j)\right>$ instead. It is often possible to compute this product without explicitly computing $\phi$,
but instead compute another function of $x_i$ and $x_j$, that is, we compute some (sufficiently simple) $K(x_i, x_j)$ whose value is equal to $\left<x_i, x_j\right>$. Such function $K$ is
called a kernel.
Kernel function can be interpreted as a function that computes the similarity between the two data instances. If $\phi(x)=x$, the kernel is just an ordinary scalar product that, if data rows
are normalized, computes whether two vector point into the same direction (1), in the opposite (-1) or anywhere in between.
One can thus go from the opposite end: define some similarity measure $K$ and use it as a kernel without explicitly define the mapping $\phi$. The magic of SVM (and other methods that can use
kernels, and are thus called kernel methods) is that they will implicitly "invent" a transformation into a (usually infinite-dimensional) space, in which the distances between objects are
such as prescibed by the kernel, and draw a hyperplane in this space.
Abstract talking aside, SVM with different kernels can split the data not by ordinary hyperplanes, but with more complex curves. The complexity of the curve is decided by the kernel type and
by the arguments given to the algorithm, like the degree and coefficients, and the penalty for misclassifications.
□ Classification trees become forests. The basic limitation of trees (in particular compared to linear models) is that they cannot sum up the evidence. When the decision requires taking into
account hundreds or even thousands of small factors, trees won't do. Consider even ten binary variables: if each decision needs to take into account all of them, then every path from the root
to a leave will have 10 steps, hence the tree will have 210=1024 leaves, which will require 1024 data instances just to have a single one in each leave (and we know that's far from enough).
With 20 binary variables, we have 220≈1million leaves and with 30 we have 230≈1billion. With multiple values instead of binary, or with numeric features ... you get the picture.
Instead of a single tree, we construct many of them by somehow randomizing their induction. We can use different sample of the data or of variables, or change the algorithm to pick one of the
best-rated attributes instead of the best one. We sometimes build entire trees, and sometimes severely limit them, like allowing just a few levels. Due to the algorithm's instability, we'll
have very different trees. We create thousands of them and call it a random forest. We classify by voting, and we can also make better probability estimates by counting the number of trees
voting in favour and against a particular decision.
□ K-nearest neighbours. With a bit of a stretch, we related this to the naive Bayesian classifier. It's problem is naivety: it computes $P(x_1, x_2, \ldots, x_n|c)$ as $\prod_i P(x_i|c)$
because the latter can be easily estimated from the data and the former can't since the data probably does not contain any (or, at least, not enough) instances with values exactly equal to $
(x_1, x_2, \ldots, x_n)$. The idea of k-nearest neighbours is to find the $k$ instances that are the most similar to $(x_1, x_2, \ldots, x_n)$. We make the prediction or estimate
probabilities based on the classes of these $k$ instances.
Unlike all algorithms we have seen so far, this one does not construct a model but just stores the data. This kind of learning is called lazy learning.
Which algorithm to choose?
The first question is whether you want to be able to explain the decision.
□ If your primary aim is to construct a model and observe it to find some patterns in your data, you can use linear models (with nomograms) or trees.
□ If you need the model to help you make decisions, you probably want to know what the model's predictions are based on - you want the model to explain the prediction. In this case, use any of
the above or maybe random forest (checking which trees voted for some class and which variables they used may provide some insight) or k-nearest neighbours (the explanation consists of the
algorithm showing you the most similar data instances).
□ If you aim at high accuracy, compute the accuracy of all models with some kind of cross validation (as in Test & Score widget), and pick the best model. Linear models (Bayes, Logistic
regression) should work best for some simple data, kNNs may sometimes surprise, SVMs are good if its parameters are fit properly, and random forests are the method to bet on when there are
many variables that give small pieces of information. An additional advantage of random forests is that they often don't need a lot of fitting - just construct a lot of (like one thousand)
The best-known book about kernel methods like SVM. Warning: lots of mathematics. Not a required reading for this class.
Contrary from SVM, random forests are so easy to explain and understand that they don't require additional literature. But if anybody is interested, here's the definitive paper about them.
• Linear regression can be made non-linear by adding columns with products of variables. This increases it capacity to fit the data, but at the same time its capacity to overfit. We have used a
widget Polynomial regression in Orange's add-on Educational, where then curve through the points we draw in Paint goes through all the data at a cost of wildly steering up and down; we also
observed the coefficients in a data table and saw that they get huge.
To counter this, we change the function that is optimized by the regression. Regression fits a model $\hat{y} = \sum_i\beta_i x_i$ by optimizing a loss function such as mean square error, $\sum_j
(y_i - \hat{y_i})^2$. To this, we add a punishment for large coefficients, so we optimize $\sum_j (y_i - \hat{y_i})^2 + \alpha\sum_i \beta_i^2$. The first term punishes bad fits, the second
punishes large coefficient. The factor $\alpha$ balances between the two parts: larger $\alpha$ forces the model to fit the data less, that is, to keep all coefficients small.
Alternatively, we can take absolute values of coefficients instead of squaring them, $\sum_j (y_i - \hat{y_i})^2 + \alpha\sum_i |\beta_i|$. This will encourage the model to set coefficients to
zero instead of keeping them small (see a youtube video with an explanation why this happens). This kind of regularization is thus useful for selecting subsets of features.
The first type of regularization is called ridge regression or L2, and the second is LASSO or L1.
Similar regularization can be used in classification, in particular in logistic regression.
We should take this as a concrete example of a general problem. In our first lecture, where we said it is impossible to decide what is the right number of bins in distribution, or the appropriate
amount of smoothing. In modelling, we balance between overfitting and overgeneralizing (or undergeneralizing and underfitting). If we have a rather smooth decision boundary and some data
instances are on the wrong side, do we curve the boundary (less generalizing, more fitting) or do we treat the data on the wrong side as pure noise in the data (and thus generalize more and fit
Practically every model has a parameter for setting the balance, like the above $\alpha$.
□ Classification trees: setting the requires number of data instances in leaves or sufficient proportion of majority class decides when to stop splitting the data and thus "smoothing" the
decision boundary. Smaller trees are more general and less (over?)fit.
□ Logistic regression: we have L1 and L2, just as described above.
□ SVM: there are a bunch of parameters that set the complexity of the boundary and the misclassification penalty.
□ k Nearest neighbours: the larger the number of neighbours, the more the model moves towards the mean/majority.
□ Naive Bayesian classifier is usually implemented without balancing, though one can essentially regularize probabilities $P(c|x_1)$. We are not going deeper here since you will have a whole
course dedicated to it.
□ Random forest: this method is strange because there is no clear consensus about how and when it can overfit. It is surely a bad idea to have a small forest with large trees, but with small
trees it seems that you can't overfit by increasing the number of trees.
In summary, with power comes responsibility. The more complex the model, the more it can overfit and the more you have to be careful avoid overfitting it.
We are just telling you about this book because we must do it at some point. It is too difficult for this course, but it provides an overview of machine learning methods from statistical
point of view. Chapters 4 and 5 should not be too difficult, and you can read them to better understand linear models and regularization.
You can download the book for free.
• We use predictive modelling (supervised learning) when our goal is to predict the value of dependent variable (also called target, outcome, class, label) based on independent variables (features,
attributes, variables, properties). In a different setup, we do not have any particular target but just some objects with known properties (essentially independent variables) and our goal is to
find some general patterns. Such methods belong to the field of unsupervised modelling.
One such task is finding separate groups in the data, if those exist. We call such groups clusters, and the procedure is called clustering.
We learned about two approaches. In hierarchical clustering, every data point represents a cluster and we iteratively merge clusters until arriving at a single cluster. The process results in a
hierarchy of clusters. The hierarchy is presented with a dendrogram, a tree-like visualization in which the lengths of branches correspond to distances between clusters. Dendrogram also offers an
insight into the number of clusters in the data: we usually set a cut-off at a point where the distances begin to grow more rapidly. One of the decisions we have to make is the definition of
distances between two clusters of points: we may consider the closest pair (single linkage), the farthest (complete linkage), the average distance (average linkage) or another measure based on
intra-cluster variance (Ward linkage). We will usually use the latter, unless there are some principal arguments for others.
The other approach we considered is k-means clustering. We first set the number of clusters ($k$). The method puts $k$ centroids at random position and assigns each data instance to the closest
centroid, thus forming some clusters. Next it moves the centroid to the center of this cluster. Then it reassigns the data instances according to the new positions of the centroids. Then it again
moves the centroids...
The two methods - hierarchical clustering and k-means clustering - differ in several imporant aspects.
Hierarchical clustering starts from a matrix of distances that do not necessarily correspond to any observable coordinates; they may represent, say, subjectively determined dissimilarities
between people. K-means works with coordinates and computes (Euclidean) distances between them.
Hierarchical clustering provides a nice visualization, which can be used to inspect the quality of clustering and determine the number of clusters. K-means does none of this.
Hierarchical clustering consumes a lot of memory and may be slow. K-means is often faster, although it may get stuck in suboptimal positions. In practice, we restart it multiple times and take
the best result.
Hierarchical clustering results in a hierarchy, and we decide the number of clusters after the procedure. In k-means we set it in advance. In practice, we tried with different numbers of clusters
and take the one with the highest silhouette score - another topic of the lecture.
The chapter below describes all of this in more detail.
Obligatory reading: sections 8.2 (you may skip 8.2.6), 8.3 (skip 8.3.3), The Silhouette Coefficient (pg. 541). Everything else is also quite easy to read, so we recommend it.
• Text mining is a field of machine learning with some specific for dealing with text. As texts are complex objects, we need to transform them into a numeric structure before the analysis.
Preprocessing. Preprocessing is the first and a very important step and we can never go without it. For example, say that we have a sentence:
"I like working with computers."
Now, we can transform it in the following way:
□ transform to lowercase: "i like working with computers."
□ split into analytical units (called tokens): "i", "like", "working", "with", "computers", "."
□ normalize words (transform them into base words): "i", "like", "work", "with", "computer", "."
□ filter stopwords, punctuation: "i", "like", "work", "computer"
□ filter by document frequency: remove words that appear in more than X and less than Y % of documents\
Some tokenization procedures can already discard punctuation. Otherwise we have to do it manually. Filtering by document frequency can be relative (word should appear in a certain percent of
documents) or absolute (word should appear in a certain number of documents).
Bag of Words. The second step is Bag of Words, which transforms text (and the prepared tokens) into document vectors. A simple way to do it is to count the words, but a more elegant approach is
term frequency - inverse document requency (TF-IDF), which ecreases the count of words which appear frequently across all documents and increases the count for those that are significant for a
small number of documents.
$$\mathrm{TF} = occurences\ of\ word\ in\ doc, \;\; \mathrm{IDF} = log\frac{\mathrm{number\ of\ docs}}{\mathrm{docs\ that\ contain\ word}}$$
TF-IDF measure is the product of the two, $$\mathrm{TFIDF} = \mathrm{TF}\times \mathrm{IDF}.$$
Afterward we can do clustering, classification or any other analysis we wish.
Word Enrichment. Word enrichment is a nice way to inspect a data subset. It computes those words that are significant for a subset compared to the entire corpus.
Sentiment Analysis. Sentiment analysis is a popular approach for analyzing user opinion, product reviews, and so on. The most simple methods are lexicon-based, which means there is a list in the
background that defines positive and negative words, then counts the occurrences of each and sums them. The sum is the final sentiment score.
Comprehensive overview of text mining techniques and algorithms. [obligatory]
Why regular expression can be very helpful. [optional read]
Why using TF-IDF is a good idea. [technical, interesting read]
• Hierarchical clustering shows us hierarchical grouping, yet it does not tell us that much about the structure of the data. Some points from one cluster may be close to points from another, yet
they fall into a different part of the hierarchy.
Multidimensional scaling (MDS) is technique that puts data instance onto a two dimensional plane (or, in principle, into a higher number of dimensions), so that the distances between points
correspond to the given distances between objects as closely as possible. The optimization works by starting from a random (or a not-so-random, but precomputed by another method) position, and
the iteratively move the points towards positions that more closely match the prescribed distances. It may get stuck in local optima, so we can restart the procedure. The resulting visualization
allows us to manually observe any structure in the data, like groups, and relations between points. MDS is a method of (multivariate) data projections, but it is non-parametric: the position of
the point does not depend on values of some variables, but on its relation with other points.
Principal component analysis is a method that finds a linear projection of data into a lower dimensional space. Just as three-dimensional data may actually lie on a plane (or within a relative
small distance above or below it), 100-dimensional data may actually lie on, say, 13-dimensional hyperplane (with small distances from it). You will learn more about PCA in another class, so we
presented it just intuitively and showed its application in Orange.
PCA is unsupervised: it does not take the target value (if there is any) into account. In the sense of the scatter plot: it is color blind. It may be the case that some 3d data lies more or less
on the plane, so PCA would select these two dimensions and discard the third as explaining too little variance. The problem may be that we actually have a classification problem and the class
depends exactly on whether the point is (slightly) below or above the plane. This may be solved by the partial least squares.
A funny little method for this is FreeViz, invented by yours truly. It performs a similar optimization as MDS, but the projection is linear, so a point can only move if the corresponding axes are
moved. See the paper or, better still, play with it in Orange. (The whole thing is rather naive, though, and partial least squares probably does the same job.)
Self-organizing map is another projection method. It's unsupervised, like PCA and MDS (and unlike Freeviz), it's based on coordinates (variables, space, vectors), like PCA and FreeViz (and unlike
MDS), and it's not linear like MDS (and unlike PCA and FreeViz)... As the name says: it's a map. It splits the region into rectangular or hexagonal cells and assigns a "profile" (a vector of the
same dimensionality as data instances) to each cell. Then every data instance is assigned to the cell whose profile matches it best. Next, profiles of cells are updated to match the instances
that contain; each data instance does not affect only its cell, but also its neighbouring cells, with some weight. Then it repeats both steps - (re)assigning instances and then recomputing
weights - until convergence. The effect is that data instances are spread across the map, usually forming clusters with similar properties.
The chapter is particularly interesting because of some nice examples at the end.
See the example in the introduction. You can also read the Methods section, if you're curious.
• We concluded with a brief introduction to deep neural networks as another form of projection, usually called embedding. This method is used for large homogenous data in which variables represent
the same measurement at different coordinates, times or similar. Typical examples are texts (variables are letters), sounds (variables are amplitudes) and, most notably, images (variables are
pixel values). A trained neural networks takes such data and computes a smaller number, for instance 2000 features that describe an object (e.g. an image) in a way that is not interpretable, but
describes the object well. For fun, we took photos from Moscow, computed an embedding, measured distances between images and put them into hierarchical clustering, which successfully
distinguished between different themes like parks, buildings (further split into churches and non-churches), statues...
For the second part of our final lecture, we showed a practical analysis of Human Development Index data, in which we used various techniques we learned in this course. | {"url":"https://ucilnica.fri.uni-lj.si/course/view.php?id=294","timestamp":"2024-11-15T03:29:12Z","content_type":"text/html","content_length":"689465","record_id":"<urn:uuid:59bd2f50-fa16-49ab-b32b-3114bca6a1ff>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00870.warc.gz"} |
When quoting this document, please refer to the following
DOI: 10.4230/LIPIcs.MFCS.2018.28
URN: urn:nbn:de:0030-drops-96107
URL: http://dagstuhl.sunsite.rwth-aachen.de/volltexte/2018/9610/
Clementi, Andrea ; Ghaffari, Mohsen ; GualĂ , Luciano ; Natale, Emanuele ; Pasquale, Francesco ; Scornavacca, Giacomo
A Tight Analysis of the Parallel Undecided-State Dynamics with Two Colors
The Undecided-State Dynamics is a well-known protocol for distributed consensus. We analyze it in the parallel PULL communication model on the complete graph with n nodes for the binary case (every
node can either support one of two possible colors, or be in the undecided state).
An interesting open question is whether this dynamics is an efficient Self-Stabilizing protocol, namely, starting from an arbitrary initial configuration, it reaches consensus quickly (i.e., within a
polylogarithmic number of rounds). Previous work in this setting only considers initial color configurations with no undecided nodes and a large bias (i.e., Theta(n)) towards the majority color.
In this paper we present an unconditional analysis of the Undecided-State Dynamics that answers to the above question in the affirmative. We prove that, starting from any initial configuration, the
process reaches a monochromatic configuration within O(log n) rounds, with high probability. This bound turns out to be tight. Our analysis also shows that, if the initial configuration has bias
Omega(sqrt(n log n)), then the dynamics converges toward the initial majority color, with high probability.
BibTeX - Entry
author = {Andrea Clementi and Mohsen Ghaffari and Luciano Gual{\`a} and Emanuele Natale and Francesco Pasquale and Giacomo Scornavacca},
title = {{A Tight Analysis of the Parallel Undecided-State Dynamics with Two Colors}},
booktitle = {43rd International Symposium on Mathematical Foundations of Computer Science (MFCS 2018)},
pages = {28:1--28:15},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-95977-086-6},
ISSN = {1868-8969},
year = {2018},
volume = {117},
editor = {Igor Potapov and Paul Spirakis and James Worrell},
publisher = {Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik},
address = {Dagstuhl, Germany},
URL = {http://drops.dagstuhl.de/opus/volltexte/2018/9610},
URN = {urn:nbn:de:0030-drops-96107},
doi = {10.4230/LIPIcs.MFCS.2018.28},
annote = {Keywords: Distributed Consensus, Self-Stabilization, PULL Model, Markov Chains}
Keywords: Distributed Consensus, Self-Stabilization, PULL Model, Markov Chains
Collection: 43rd International Symposium on Mathematical Foundations of Computer Science (MFCS 2018)
Issue Date: 2018
Date of publication: 27.08.2018
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Math Games
Scotland Numeracy And Mathematics
Multiplication is an essential math skill with an abundance of uses in everyday life. Using Math Games, children can make sure they get the hang of this skill and actually enjoy reviewing it. Our
free games, PDF worksheets, apps, and digital textbook help make that possible!
Using our suite of instructional tools, kids can practice:
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Teachers and parents can feel confident that students are learning the relevant material for their grade. They can also track pupils’ progress using reports sent to them by Math Games. Practicing
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C++ Program help
I have a task to write a program in C ++ with the following condition: To write a program for obtaining all natural numbers not exceeding a given n, which represent a strictly increasing sequence of
digits. I will be very grateful if you help me with the code.
Tell you what. You explain HOW you are going to do this and we'll help you with the code.
#include <iostream>
int main()
using namespace std;
int n;
cin >> n;
for (int i = 1; i <= n; i++)
return 0;
But I don't know how to write code about "which represent a strictly increasing sequence of digits"
There are various ways of doing the problem. For your code, however, the simplest would be to write a function
bool isStrictlyIncreasing( int x )
which returns true or false for a test number x.
Within your main() loop over i you could then just use
if ( isStrictlyIncreasing( i ) ) cout << i << '\n';
So, it's that function isStrictlyIncreasing( int x ) that you need to write.
Note that you can peel off digits from the end of a number as x % 10 (followed by x /= 10 to move to the left). Each time you get a final digit compare it with the previous one (and return false if
it fails). If you successfully reach x=0 without failing then you can return true.
Keep a variable stored as the previous digit to compare with (it can be initialised as something higher than a digit, e.g. 10).
Last edited on
This function if ( isStrictlyIncreasing( i ) ) cout << i << '\n'; should be inside the for loop right?
And bool isStrictlyIncreasing(int x) is like operator?
I have no idea how to declared and defined outside main ().
how big is N going to be? If this is a big N problem, you need to code the 'strictly increasing' efficiently. possibly a lookup table of the first 1000 integers (0-999) and then you can process upt
to groups of 3 at a time instantly, eg 12345679 would check 679, 345, 12, logical and the 3 results, done in a jiffy. the largest value is 123456789 so anything larger than that you can stop
iterating. 0 must be set to true; if single digits are not valid (unclear) then you can special case an actual 0 as input but it must be true in the table.
whole thing becomes
if( tbl[num%1000] && tbl[(num/1000)%1000] && tbl[(num/1000000)%1000])
its one you want
its not.
Last edited on
I would argue that I made it easier to code and more efficient at one stroke :P
Yes, but jonnin’s lookup table is a mature solution requiring more CS/algorithms knowledge than a first- or second-year student is expected to have.
At OP’s homework question level, the classwork is designed to get him thinking about how numbers and numeric operators work and apply simple flow control logic.
That is fair. This thing is going to take all day if you check every number one digit at a time, though. Ok, computers being what they are ... maybe just a few min, but still. And mine isnt the
ultimate fast answer; I believe you can generate the numbers flat out but *that* would take some time to code up for a beginner for sure. And, the real answer .. get the 511 values into a table and
everything else is false... :)
I didn't verify 511 ... did you count the little ones like 12 13 etc?
Last edited on
511 is correct, @jonnin.
Every one of the digits 1...9 is either in or out, which gives 2^9 possibilities, but obviously the single case with none in isn't much use, so you have
2^9-1 or 511 possibilities. The code lists them.
Checking every number doesn't take as long as you think, either - there is a greater than 50% chance of returning false when you've only peeled off two digits. It took about a second with -O3
Last edited on
Hmm. Ill just admit that I don't see the 50% thing and look at it a bit more (don't tell me, it will be fun). You eliminated zero, a for all the numbers loop would have to look at it. 10234 10235 ...
etc. A 'for the ones in the digit list' loop could avoid them. (see OP's for loop in his 2nd post).
Last edited on
Heh, it’s a fun homework prompt. I did my version using a recursive function to compute the next integer.
Topic archived. No new replies allowed. | {"url":"https://cplusplus.com/forum/general/280962/","timestamp":"2024-11-12T20:10:49Z","content_type":"text/html","content_length":"30878","record_id":"<urn:uuid:9119e95f-9f86-49be-a886-5552cf321aab>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00058.warc.gz"} |
Half quantized Hall effect | Coshare Science
1. E.H. Hall, On a new action of the magnet on electric currents, Am. J. Math. 2, 287 (1879).
2. A.W. Smith, The variation of the Hall effect in metals with change of temperature, Phys. Rev. (Series I) 30, 1 (1910).
3. K.v. Klitzing, G. Dorda, and M. Pepper, New method for high-accuracy determination of the fine-structure constant based on quantized Hall resistance, Phys. Rev. Lett. 45, 494 (1980).
4. D.C. Tsui, H.L. Stormer, and A.C. Gossard, Two-dimensional magnetotransport in the extreme quantum limit, Phys. Rev. Lett. 48, 1559 (1982).
5. F.D.M. Haldane, Model for a quantum Hall effect without Landau levels: condensed-matter realization of the "parity anomaly", Phys. Rev. Lett. 61, 2015 (1988).
6. C.Z. Chang, J.S. Zhang, X. Feng, J. Shen, Z.C. Zhang, M.H. Guo, K. Li, Y.B. Ou, P. Wei, L.L. Wang, Z.Q. Ji, Y. Feng, S.H. Ji, X. Chen, J.F. Jia, X. Dai, Z. Fang, S.C. Zhang, J. He, Y.Y. Wang, L.
Lu, X.C. Ma, and Q.K. Xue, Experimental observation of the quantum anomalous Hall effect in a magnetic topological insulator, Science 340, 167 (2013).
7. X.L. Qi and S.C. Zhang, Topological insulators and superconductors, Rev. Mod. Phys. 83, 1057 (2011).
8. A.N. Redlich, Parity violation and gauge noninvariance of the effective gauge field action in three dimensions, Phys. Rev. D 29, 2366 (1984).
9. G.W. Semenoff, Condensed-matter simulation of a three-dimensional anomaly, Phys. Rev. Lett. 53, 2449 (1984).
10. S.Q. Shen, Topological Insulators (Springer, 2012).
11. D.J. Thouless, M. Kohmoto, M.P. Nightingale, and M. den Nijs, Quantized Hall conductance in a two-dimensional periodic potential, Phys. Rev. Lett. 49, 405 (1982).
12. Q. Niu, D.J. Thouless, and Y.S. Wu, Quantized Hall conductance as a topological invariant, Phys. Rev. B 31, 3372 (1985).
13. H.Z. Lu, W.Y. Shan, W. Yao, Q. Niu, and S.Q. Shen, Massive Dirac fermions and spin physics in an ultrathin film of topological insulator, Phys. Rev. B 81, 115407 (2010).
14. K.G. Wilson, Quarks: From paradox to myth. In: A. Zichichi, (eds) New phenomena in subnuclear physics. The Subnuclear Series, vol 13 (Springer, Boston, MA, 1977).
15. B. Fu, J.Y. Zou, Z.A. Hu, H.W. Wang, and S.Q. Shen, Quantum anomalous semimetals, npj Quantum Mater. 7, 94 (2022).
16. E. Fradkin, Field theories of condensed matter physics (Cambridge University Press, 2013).
17. J.Y. Zou, R. Chen, B. Fu, H.W. Wang, Z.A. Hu, and S.Q. Shen, Half-quantized Hall effect at the parity-invariant Fermi surface, Phys. Rev. B 107, 125153 (2023).
18. Y.L. Chen, J.H. Chu, J.G. Analytis, Z.K. Liu, K. Igarashi, H.H. Kuo, X.L. Qi, S.K. Mo, R.G. Moore, D.H. Lu, M. Hashimoto, T. Sasagawa, S.C. Zhang, I.R. Fisher, Z. Hussain, and Z.X. Shen, Massive
Dirac fermion on the surface of a magnetically doped topological insulator, Science 329, 659 (2010).
19. M. Mogi, Y. Okamura, M. Kawamura, R. Yoshimi, K. Yasuda, A. Tsukazaki, K.S. Takahashi, T. Morimoto, N. Nagaosa, M. Kawasaki, Y. Takahashi, and Y. Tokura, Experimental signature of the parity
anomaly in a semi-magnetic topological insulator, Nat. Phys. 18, 390 (2022).
20. H. Wang, B. Fu, and S.Q. Shen, Recent progress of transport theory in Dirac quantum materials, Acta. Phys. Sin. 72(17), 177303 (2023).
21. R. Chen and S.Q. Shen, On the half-quantized Hall conductance of massive surface electrons in magnetic topological insulator films, Sci. China - Phys. Mech. Astron. (2024).
22. E. Prodan, Disordered topological insulators: a non-commutative geometry perspective, J. Phys. A Math. Theor. 44, 113001 (2011).
23. J. Li, R.L. Chu, J.K. Jain, and S.Q. Shen, Topological Anderson insulator, Phys. Rev. Lett. 102, 136806 (2009).
24. J.Y. Zou, B. Fu, H.W. Wang, Z.A. Hu, and S.Q. Shen, Half-quantized Hall effect and power law decay of edge-current distribution, Phys. Rev. B 105, L201106 (2022).
25. B. Fu, K.Z. Bai, and S.Q. Shen, Half quantum mirror Hall effect, arXiv: 2402.02654 (2024).
26. K.Z. Bai, B. Fu, Z.Y. Zhang, and S.Q Shen, Metallic quantized anomalous Hall effect without chiral edge states, Phys. Rev. B 108, L241407 (2023). | {"url":"https://www.cosharescience.com/articles/cs.202402.128","timestamp":"2024-11-05T13:25:51Z","content_type":"text/html","content_length":"497065","record_id":"<urn:uuid:a606f80e-c838-4896-9057-736cdabed9e2>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00215.warc.gz"} |
YY 0775-2010 PDF in English
YY 0775-2010 (YY0775-2010) PDF English
│Standard ID │ Contents │USD│ STEP2 │ [PDF] delivered in │ Name of Chinese Standard │Status│
│ │ [version] │ │ │ │ │ │
│YY 0775-2010│ English │150│Add to Cart│ 0-9 seconds. │Teleradiotherapy treatment planning system accuracy of dosimetric calculation and test methods for high energy X(γ) │ │
│ │ │ │ │ Auto-delivery. │ beam │ │
Standards related to (historical): YY 0775-2010
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YY 0775-2010: PDF in English
YY 0775-2010 YY PHARMACEUTICAL INDUSTRY STANDARD OF THE PEOPLE’S REPUBLIC OF CHINA ICS 11.040.60 C 43 Teleradiotherapy treatment planning system accuracy of dosimetric calculation and test methods
for high energy X (γ) beam 远距离放射治疗计划系统 ISSUED ON. DECEMBER 27, 2010 IMPLEMENTED ON. JUNE 1, 2012 Issued by. China Food and Drug Administration Table of Contents Foreword ... 3 1 Scope ...
4 2 Normative references ... 4 3 Terms and definitions ... 4 4 Requirements ... 5 4.1 Simple geometric conditions ... 5 4.2 Complex geometric conditions ... 5 4.3 Combination of complex
geometric conditions ... 5 4.4 The outer edge of radiation field ... 6 4.5 The outer edge of radiation field, complex geometric conditions and the central axis is blocked ... 6 5 Test methods ...
6 5.1 General test conditions ... 6 5.2 Test example... 9 5.3 Simple geometric conditions ... 9 5.4 Complex geometric conditions ... 9 5.5 Combination of complex geometric conditions ... 10 5.6
The outer edge of radiation field ... 10 5.7 The outer edge of radiation field, complex geometric conditions and the central axis is blocked ... 10 Appendix A ... 11 Appendix B ... 12 References
... 23 Foreword All technical content of this Standard is mandatory. This Standard was drafted in accordance with the rules given in GB/T 1.1-2009. The test data and examples in this Standard adopt
the test data (machine configuration data) and examples in IAEA No.1540 Technical Report. Please note that some of the content of this document may involve patents; the issuing agency of this
document does not undertake the responsibility for the identification of these patents. This Standard was proposed by and shall be under the jurisdiction of Sub- committee on Radiotherapy, Nuclear
medicine and Radiation Dosimetry Equipment of National Technical Committee on Medical Electrical Equipment of Standardization Administration of China (SAC/TC 10/SC 3). Drafting organizations of this
Standard. Beijing Institute for Medical Device Quality Supervision and Testing of State Food and Drug Administration. Main drafters of this Standard. Zhang Xin, Zhang Zhaoyuan, Yan Xu, Hu Jia and
Qing Hou. Teleradiotherapy treatment planning system accuracy of dosimetric calculation and test methods for high energy X (γ) beam 1 Scope This Standard specifies the accuracy of dosimetric
calculation and test methods of Radiotherapy Treatment Planning Systems (hereinafter referred to as RTPS). This Standard applies to RTPS with dosimetric calculation and display function of high
energy X rays (4 MV ~ 25 MV), 60Co γ rays remote radiotherapy treatment on photon radiation. This Standard does not apply to the radiation treatment planning systems that are used by stereotactic
radiotherapy, intensity modulated radiation therapy (IMRT) or other specialized radiotherapy techniques. However, the developers are encouraged to use this Standard in development, and use as
reference standard in verifying dosimetric calculation algorithm. Test packets in this Standard cannot be used for clinical treatment planning. 2 Normative references The following referenced
documents are indispensable for the application of this document. For dated references, only the edition cited applies. For undated references, the latest edition of the referenced document
(including any amendments) applies. GB/T 17857-1999 Medical radiology - Terminology (Equipment for radiotherapy, nuclear medicine and radiation dosimetry) GB/T 18987-2003 Radiotherapy equipment -
Coordinates, movements and scales 3 Terms and definitions The terms and definitions defined by GB/T 17857-1999 and the following ones apply to this document. The test requires to calculate point dose
of a series of positions in a cube water body model. Users are recommended to specify a cube water body model of which the side length is 40 cm for RTPS, before test. Most RTSP allows to determine
the dose measurement points by inputting coordinate values. However, if RTPS do not have the aforementioned functions, an easy way is to use beam’s central axis, off-axis distance (1cm, 2cm, 2.5cm,
3cm, 5cm, 6cm, 7cm, 9cm and 19cm) and calculation depth (dmax, 1cm, 3cm, 5cm, 10cm, 11cm, 15cm, 20cm, 25cm, 30cm and 35cm) to represent for each measurement point. Each test example has specified the
aforementioned off-axis distance and depth. For each test example, the operator shall place radiation beam at the position stipulated in the test example; and mark the calculation point of depth and
off- axis distance required by test. It shall use a great enough radiation beam weighting to ensure that calculation result has enough accuracy, i.e., not using too small weighting may be because of
that the rounding or data truncation causes the accuracy of calculation result less than 1%. 5.1.3 Beam dose setting In some RPTS, beam dose setting is corresponding to the dose when beam is
transferred to dmax. While in some other systems, beam dose setting is corresponding to the dose when beam is transferred to reference radiation field size (i.e., 10 cm × 10 cm). Some other systems
use different setting ways to set beam dose. Users shall be familiar with beam dose setting ways and ensure that calculation result is consistent to the conditions of obtaining corresponding test
data. 5.1.4 Coordinate system In order to more-clearly express the geometric relationship between beam data and test examples, the following coordinate system is stipulated related to water body
model. Water body model’s coordinate system is consistent to the fixed coordinate system stipulated in GB/T 18987-2003. Coordinate origin is at the isocenter of therapy equipment. Except isocenter
test example (test example 3); isocenter of other test examples all locates at the surface of water body model. Z-axis is perpendicular to the upper surface and points upward from the surface. Z-axis
coincides with the central axis of the radiation beam and points to radiation source. Oblique incidence test example is excluded. 5.5 Combination of complex geometric conditions Test shall conduct
test example 12; calculate error according to formula (1). Test results shall comply with requirements of 4.3. 5.6 The outer edge of radiation field Test on measurement point of the outer edge of
radiation field shall conduct test example 1, 2, 3, 6; calculate error according to formula (2). Where, δ2 - Normalized relative error of measurement point of the outer edge of radiation field, %; Dc
- Calculated values; Dr - Reference values; Da - Reference values of same depth on radiation beam. Test results shall comply with requirements of 4.4. 5.7 The outer edge of radiation field, complex
geometric conditions and the central axis is blocked Test on central beam measurement point of the outer edge of radiation field under complex geometric conditions and the central axis is blocked
shall conduct test example 5, 7; calculate error according to formula (3). Where, δ3 - Normalized relative error of the outer edge of radiation field, complex geometric conditions and the central
axis is blocked, %; Dc - Calculated values; Dr - Reference values; Do - Reference values of same depth in open field. Test results shall comply with requirements of 4.5. Appendix A (Normative) Test
data This Standard introduces data package in IAEA No.1540 Technical Report [1] as predefined test conditions and test data of this Standard. The data package is divided into four parts, respectively
including input data and test results of 6 MV, 10 MV, 18 MV and 60Co γ. When using data package, it shall select suitable data sets, input data by combining corresponding RTPS requirements, select
appropriate radiation beam model, build up test’s geometric conditions, calculate dose of measurement point and input calculated values into evaluation table for error calculation. Test data package
includes the following four parts. - IAEApack06MV NCS data package for 6 MV X; - IAEApack10MV NCS data package for 10 MV X; - IAEApack18MV NCS data package for ... ......
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Using math websites
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1. Introduction to NumPy
By Bernd Klein. Last modified: 01 Feb 2022.
NumPy is a module for Python. The name is an acronym for "Numeric Python" or "Numerical Python". It is pronounced /ˈnʌmpaɪ/ (NUM-py) or less often /ˈnʌmpi (NUM-pee)). It is an extension module for
Python, mostly written in C. This makes sure that the precompiled mathematical and numerical functions and functionalities of Numpy guarantee great execution speed.
Furthermore, NumPy enriches the programming language Python with powerful data structures, implementing multi-dimensional arrays and matrices. These data structures guarantee efficient calculations
with matrices and arrays. The implementation is even aiming at huge matrices and arrays, better know under the heading of "big data". Besides that the module supplies a large library of high-level
mathematical functions to operate on these matrices and arrays.
SciPy (Scientific Python) is often mentioned in the same breath with NumPy. SciPy needs Numpy, as it is based on the data structures of Numpy and furthermore its basic creation and manipulation
functions. It extends the capabilities of NumPy with further useful functions for minimization, regression, Fourier-transformation and many others.
Both NumPy and SciPy are not part of a basic Python installation. They have to be installed after the Python installation. NumPy has to be installed before installing SciPy.
The image above is the graphical visualisation of a matrix with 14 rows and 20 columns. It's a so-called Hinton diagram. The size of a square within this diagram corresponds to the size of the value
of the depicted matrix. The colour determines, if the value is positive or negative. In our example: the colour red denotes negative values and the colour green denotes positive values.
NumPy is based on two earlier Python modules dealing with arrays. One of these is Numeric. Numeric is like NumPy a Python module for high-performance, numeric computing, but it is obsolete nowadays.
Another predecessor of NumPy is Numarray, which is a complete rewrite of Numeric but is deprecated as well. NumPy is a merger of those two, i.e. it is build on the code of Numeric and the features of
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Comparison between Core Python and Numpy
When we say "Core Python", we mean Python without any special modules, i.e. especially without NumPy.
The advantages of Core Python:
• high-level number objects: integers, floating point
• containers: lists with cheap insertion and append methods, dictionaries with fast lookup
Advantages of using Numpy with Python:
• array oriented computing
• efficiently implemented multi-dimensional arrays
• designed for scientific computation
A Simple Numpy Example
Before we can use NumPy we will have to import it. It has to be imported like any other module:
But you will hardly ever see this. Numpy is usually renamed to np:
Our first simple Numpy example deals with temperatures. Given is a list with values, e.g. temperatures in Celsius:
cvalues = [20.1, 20.8, 21.9, 22.5, 22.7, 22.3, 21.8, 21.2, 20.9, 20.1]
We will turn our list "cvalues" into a one-dimensional numpy array:
C = np.array(cvalues)
[20.1 20.8 21.9 22.5 22.7 22.3 21.8 21.2 20.9 20.1]
Let's assume, we want to turn the values into degrees Fahrenheit. This is very easy to accomplish with a numpy array. The solution to our problem can be achieved by simple scalar multiplication:
[68.18 69.44 71.42 72.5 72.86 72.14 71.24 70.16 69.62 68.18]
The array C has not been changed by this expression:
[20.1 20.8 21.9 22.5 22.7 22.3 21.8 21.2 20.9 20.1]
Compared to this, the solution for our Python list looks awkward:
fvalues = [ x*9/5 + 32 for x in cvalues]
[68.18, 69.44, 71.42, 72.5, 72.86, 72.14, 71.24000000000001, 70.16, 69.62, 68.18]
So far, we referred to C as an array. The internal type is "ndarray" or to be even more precise "C is an instance of the class numpy.ndarray":
In the following, we will use the terms "array" and "ndarray" in most cases synonymously.
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Graphical Representation of the Values
Even though we want to cover the module matplotlib not until a later chapter, we want to demonstrate how we can use this module to depict our temperature values. To do this, we us the package pyplot
from matplotlib.
If you use the jupyter notebook, you might be well advised to include the following line of code to prevent an external window to pop up and to have your diagram included in the notebook:
The code to generate a plot for our values looks like this:
import matplotlib.pyplot as plt
The function plot uses the values of the array C for the values of the ordinate, i.e. the y-axis. The indices of the array C are taken as values for the abscissa, i.e. the x-axis.
Memory Consumption: ndarray and list
The main benefits of using numpy arrays should be smaller memory consumption and better runtime behaviour. We want to look at the memory usage of numpy arrays in this subchapter of our turorial and
compare it to the memory consumption of Python lists.
To calculate the memory consumption of the list from the above picture, we will use the function getsizeof from the module sys.
from sys import getsizeof as size
lst = [24, 12, 57]
size_of_list_object = size(lst) # only green box
size_of_elements = len(lst) * size(lst[0]) # 24, 12, 57
total_list_size = size_of_list_object + size_of_elements
print("Size without the size of the elements: ", size_of_list_object)
print("Size of all the elements: ", size_of_elements)
print("Total size of list, including elements: ", total_list_size)
Size without the size of the elements: 96
Size of all the elements: 84
Total size of list, including elements: 180
The size of a Python list consists of the general list information, the size needed for the references to the elements and the size of all the elements of the list. If we apply sys.getsizeof to a
list, we get only the size without the size of the elements. In the previous example, we made the assumption that all the integer elements of our list have the same size. Of course, this is not valid
in general, because memory consumption will be higher for larger integers.
We will check now, how the memory usage changes, if we add another integer element to the list. We also look at an empty list:
lst = [24, 12, 57, 42]
size_of_list_object = size(lst) # only green box
size_of_elements = len(lst) * size(lst[0]) # 24, 12, 57, 42
total_list_size = size_of_list_object + size_of_elements
print("Size without the size of the elements: ", size_of_list_object)
print("Size of all the elements: ", size_of_elements)
print("Total size of list, including elements: ", total_list_size)
lst = []
print("Emtpy list size: ", size(lst))
Size without the size of the elements: 104
Size of all the elements: 112
Total size of list, including elements: 216
Emtpy list size: 72
We can conclude from this that for every new element, we need another eight bytes for the reference to the new object. The new integer object itself consumes 28 bytes. The size of a list "lst"
without the size of the elements can be calculated with:
64 + 8 * len(lst)
To get the complete size of an arbitrary list of integers, we have to add the sum of all the sizes of the integers.
We will examine now the memory consumption of a numpy.array. To this purpose, we will have a look at the implementation in the following picture:
We will create the numpy array of the previous diagram and calculate the memory usage:
a = np.array([24, 12, 57])
We get the memory usage for the general array information by creating an empty array:
e = np.array([])
We can see that the difference between the empty array "e" and the array "a" with three integers consists in 24 Bytes. This means that an arbitrary integer array of length "n" in numpy needs
96 + n * 8 Bytes
whereas a list of integers needs, as we have seen before
64 + 8 len(lst) + len(lst) 28
This is a minimum estimation, as Python integers can use more than 28 bytes.
When we define a Numpy array, numpy automatically chooses a fixed integer size. In our example "int64". We can determine the size of the integers, when we define an array. Needless to say, this
changes the memory requirement:
a = np.array([24, 12, 57], np.int8)
print(size(a) - 96)
a = np.array([24, 12, 57], np.int16)
print(size(a) - 96)
a = np.array([24, 12, 57], np.int32)
print(size(a) - 96)
a = np.array([24, 12, 57], np.int64)
print(size(a) - 96)
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Time Comparison between Python Lists and Numpy Arrays
One of the main advantages of NumPy is its advantage in time compared to standard Python. Let's look at the following functions:
import time
size_of_vec = 1000
def pure_python_version():
t1 = time.time()
X = range(size_of_vec)
Y = range(size_of_vec)
Z = [X[i] + Y[i] for i in range(len(X)) ]
return time.time() - t1
def numpy_version():
t1 = time.time()
X = np.arange(size_of_vec)
Y = np.arange(size_of_vec)
Z = X + Y
return time.time() - t1
Let's call these functions and see the time consumption:
t1 = pure_python_version()
t2 = numpy_version()
print(t1, t2)
print("Numpy is in this example " + str(t1/t2) + " faster!")
0.0010614395141601562 5.2928924560546875e-05
Numpy is in this example 20.054054054054053 faster!
It's an easier and above all better way to measure the times by using the timeit module. We will use the Timer class in the following script.
The constructor of a Timer object takes a statement to be timed, an additional statement used for setup, and a timer function. Both statements default to 'pass'.
The statements may contain newlines, as long as they don't contain multi-line string literals.
A Timer object has a timeit method. timeit is called with a parameter number:
The main statement will be executed "number" times. This executes the setup statement once, and then returns the time it takes to execute the main statement a "number" of times. It returns the time
in seconds.
import numpy as np
from timeit import Timer
size_of_vec = 1000
X_list = range(size_of_vec)
Y_list = range(size_of_vec)
X = np.arange(size_of_vec)
Y = np.arange(size_of_vec)
def pure_python_version():
Z = [X_list[i] + Y_list[i] for i in range(len(X_list)) ]
def numpy_version():
Z = X + Y
#timer_obj = Timer("x = x + 1", "x = 0")
timer_obj1 = Timer("pure_python_version()",
"from __main__ import pure_python_version")
timer_obj2 = Timer("numpy_version()",
"from __main__ import numpy_version")
for i in range(3):
t1 = timer_obj1.timeit(10)
t2 = timer_obj2.timeit(10)
print("time for pure Python version: ", t1)
print("time for Numpy version: ", t2)
print(f"Numpy was {t1 / t2:7.2f} times faster!")
time for pure Python version: 0.0021230499987723306
time for Numpy version: 0.0004346180066931993
Numpy was 4.88 times faster!
time for pure Python version: 0.003020321993972175
time for Numpy version: 0.00014882600225973874
Numpy was 20.29 times faster!
time for pure Python version: 0.002028984992648475
time for Numpy version: 0.0002098319964716211
Numpy was 9.67 times faster!
The repeat() method is a convenience to call timeit() multiple times and return a list of results:
print(timer_obj1.repeat(repeat=3, number=10))
print(timer_obj2.repeat(repeat=3, number=10))
[0.0030275019962573424, 0.002999588003149256, 0.0022120869980426505]
[6.104000203777105e-05, 0.0001641790004214272, 1.904800592456013e-05]
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Teaching Issues and Experiments in Ecology - Volume 5, July 2007
- 1 -
Teaching Issues and Experiments in Ecology - Volume 5, July 2007
Assessing gains in undergraduate students’ abilities to analyze graphical data
Chris Picone
Department of Biology
FitchburgState College
Fitchburg, MA01420
Jennifer Rhode
Department of Biological and Environmental Science,
GeorgiaCollege & StateUniversity
Milledgeville, GA31061
Laura Hyatt
Department of Biology
Lawrenceville, NJ08648
Tim Parshall
Department of Biology
WestfieldState College
Westfield, MA01086
Analytical and graphing skills are critical to scientific understanding. While these skills are usually taught in ecology and environmental science courses, rarely are they assessed.How much do
students’ analytical skills improve over the course of a semester?What areas provide the most challenges for students?In this study we assessed analytical and graphing abilities in 240 students at
four colleges and universities. Over the course of a semester, we integrated graphing and data analysis throughout our lecture and lab courses, using active-learning exercises that we developed.We
assessed student skills before, during, and after the courses.In post-tests, most students (75-90 %) were adept at interpreting simple bar graphs and scatterplots, and their skills in making graphs
from raw data improved considerably. However, little improvement was found in their understanding of independent and dependent variables, and most students (> 50-75 %) had difficulty properly
summarizing trends from data with variation.Students also did not improve in their abilities to interpret complex bar graphs with interactions.These challenges indicate areas that may deserve
attention from those who teach analytical skills at the college level.We recommend strategies to teach these skills and strategies to assess whether our teaching is effective.
Graphing, analytical skills, assessment, TIEE, pre- and post-test
The content of courses and the methods by which students learn are crucial in teaching the life sciences (NRC 1999, NRC 2003).Skills in data analysis and graph interpretation are particularly
critical, not only in training future scientists (Mathewson 1999) but for all students.As members of the general public, all students must make informed decisions about scientific issues and
controversies (von Roten 2006).However, graph presentation and interpretation are difficult skills that require cognitive steps that may be new to college students (Preece and Janvier 1992; Bowen et
al 1999; Roth et al 1999: Bowen and Roth 2002; Roth 2004; Roth and McGinn 1997).Faculty teaching ecology and environmental courses should assess whether our courses are improving critical skills such
as graph interpretation and should evaluate the most effective practices (D’Avanzo 2000; 2003a; Handelsman et al. 2004).In this study, we assessed changes in graph interpretation skills demonstrated
by undergraduate students in our courses at four colleges.
Our study had two goals.The first was to use a variety of quantitative materials to train students to interpret ecological data.We developed analytical and graphing exercises to improve analytical
skills, and we integrated these exercises into lectures and labs.The exercises were adapted from the ESA’s electronic publication, Teaching Issues and Experiments in Ecology (TIEE).TIEE provides
teachers with case studies, graphs, data sets, and essays that encourage active learning and impart a greater understanding of the science behind ecology.We developed exercises that would engage and
challenge students with the material through student-active learning and other strategies demonstrated to be effective for teaching difficult content and scientific skills (Ebert-May and Brewer 1997;
McNeal and D'Avanzo 1997; D’Avanzo 2003a,b; Brewer 2004; Handelsman et al. 2004).Our exercises required students to interpret scatterplots, line graphs and bar graphs, and to produce their own graphs
from data.Several of these exercises are appended as tools for faculty to adopt in their own courses (see Resources).
Our second goal was to develop assessment tools to measure students’ abilities to create and interpret graphical information.At the beginning, during, and end of our courses we tested students’
analytical skills in order to assess the impacts of our teaching and to reveal which skills were most challenging to our students.Our study was not designed to assess the effectiveness of any
particular teaching method we used (lectures, labs, or analytical exercises), but rather the effectiveness of each course as a whole.As such, our study provides tools and recommendations for outcomes
assessment, which is increasingly required by state and regional accrediting agencies.Despite extensive experience doing research, most ecologists have little background in educational research and
assessment of their teaching (D’Avanzo 2003a,b).Such assessment, however, is an important first step to improve the quality of our teaching and to develop more scientific approaches to teaching
(D’Avanzo 2000; 2003a; Handelsman et al. 2004).An example assessment tool is appended (see Pre-Post Test in Resources).
Most previous work on graph interpretation has focused on middle and secondary students (reviewed in Phillips 1997).Our assessment research contributes to the field of pedagogical research by adding
to the few studies that have addressed analytical skills at the tertiary level of education (Bowen et al 1999; Bowen and Roth 2002).By assessing large populations of undergraduates from two different
student populations (science majors and non-majors) at four different institutions, we can draw general conclusions about analytical skills and methods of teaching these skills at this level.
We assessed skills and progress of 240 students at four institutions: Fitchburg State College (MA), Georgia College & State University (GA), Rider University (NJ) and Westfield State College
(MA).Most students tested (66%) were non-science majors in introductory Environmental Science or Life Science courses, and the remainder (33%) were science majors in introductory Ecology courses
(Table 1).
Each investigator used several strategies to teach analytical and graphing skills.First, we began with a single lecture or lab that provided background on interpreting and creating graphs.While we
each developed this background material independently, it was based on the “Step-One, Step-Two” strategy (TIEE 2005).In “step-one,” students describe how the graph is set up: the variables, axes,
legend, and patterns in the data.In “step-two,” students interpret the graph and the relationships among variables.An example handout from this presentation is appended (see How To Read A Graph in
Second, we created exercises in which students interpreted data and graphs as a means to learn course content.We included graphs and data sets available from the TIEE site, supplemented with graphs
from primary literature.Because our courses covered different content, we did not use identical exercises, although some exercises were shared among two or three investigators (Table 1).Example
exercises from four topics are appended (see Examples in Resources). Exercises were presented every few weeks when appropriate, given the schedule of lecture and lab topics.Most exercises only
occupied 20-30 minutes within a single lecture or lab, while a few required a 2-3 hour lab period, and a few were assigned as homework.Exercises were designed as small group, collaborative activities
in which students presented their work orally in class or as a written assignment.Students received oral and written feedback from class discussions and assignments. In addition to these exercises,
every week’s lectures included graphs to reinforce principles covered in both the background material and analytical exercises.
Five of the six courses in this study also included a lab (Table 1).In most labs, students created graphs from raw data, including data the students collected.Skills included generating scatterplots
and bar graphs of means with error bars, and most importantly, interpreting the trends to test hypotheses.To improve understanding, we required students to first plan their graphs by sketching them
out by hand before plotting the data with Microsoft Excel.
To assess whether our courses improved student’s skills, we compared responses to test questions before, during, and after each course.Three investigators emphasized pre- and post-tests (see Pre-Post
Test in Resources for an example).Two of these researchers used pre- and post-tests with identical questions, and one changed the questions in the post-test (Table 1).The fourth researcher monitored
skills throughout the course with a pre-course survey and analytical questions incorporated into course exams every few weeks.Because we used different assessment strategies and may have worked with
different types of students, we analyzed the results from each researcher separately.
Despite differences in testing design, we generally assessed similar skills in our students:
• Interpreting simple bar graphs and scatterplots
• Interpreting scatterplots with multiple independent and dependent variables
• Distinguishing independent and dependent variables
• Interpreting bar graphs with interactions
• Choosing the correct type of graph (bar vs. scatterplot) to summarize data
• Using a mean to summarize categorical data
• Designating and precisely labeling axes
We developed rubrics to determine whether answers in post-tests could be categorized as “Improved,” “No change, satisfactory,” “No change, unsatisfactory” or “Worsened” compared to the pre-test.The
rubric depended on the skill assessed and the test question.Specific rubrics are provided with their corresponding test questions in the Results.
Areas where students’ analytical skills improved
At all four institutions our courses and exercises improved students’ abilities to interpret graphs (Figure 1). Students were presented graphs and asked to explain the patterns among variables.Test
questions were either open-ended (short-answer) or multiple-choice (e.g., see Example #1 in Pre-Post Test in Resources).The percent of correct answers varied with the complexity of the graph and with
the school or instructor (Figure 1). Prior to our courses, only 25-60 percent of students could correctly describe the patterns among variables in a graph (Figure 1).For instance, students’
descriptions often omitted trends in a complex graph, or they used imprecise language to describe trends (e.g., “this graph describes effects of…”, “the variables are related” or “the variables are
linear”).Sometimes students confused cause and effect, or indicated poor understanding of the figure.After our courses, over 75-90 percent of students at each institution were proficient in
interpreting graphs (Figure 1).Students were more thorough in their descriptions, and they used more precise language e.g., “nitrogen and phosphorous are positively correlated.”Their descriptions
indicated they had increased their understanding of the ecology depicted in the graphs.
Our courses also improved students’ ability to create graphs, and therefore interpret data.In one example, students were presented with data that should be summarized as a scatterplot (Example #4 in
Pre-Post Test).By the end of each course, more than 75 percent of students could create a proper scatterplot, with the axes correctly placed and labeled, and with accurate descriptions of trends
(Figure 2).The number of proficient students increased 35-45 percent compared to the pre-test.To assess skills in making bar graphs, students at FitchburgState were also asked to plot categorical
data (Example #3 in Pre-Post Test).Almost 50 percent of students improved in this basic skill (Figure 3).
Areas where students’ analytical skills did not improve
Identifying independent and dependent variables. Our results also indicated several areas where most undergraduates continued to struggle despite our lectures, labs and exercises.First, we tested for
both superficial and deeper understandings of independent and dependent variables.This concept may be important for students to understand experimental design and to interpret data.Our students could
easily identify independent and dependent variables in simple graphs, but not in graphs with more than two variables.For example, when exam questions asked students to identify the independent/
dependent variables in simple graphs, 80-90 percent of students answered correctly at Rider University (Figure 4) and at Fitchburg State (N=43; data not presented because it was from a single test.)
However, when complex graphs included multiple independent or dependent variables, far fewer students were successful.For instance, Example #1 in the Pre-Post test presents a scatterplot with two
dependent variables (nitrogen and phosphorus concentrations) and one independent variable (biomes tested).When the post-test asked students to list all dependent and independent variables in this
figure, only 30-40 percent correctly listed and categorized all three variables.Earlier in the semester at FitchburgState, only a few more students (50-57 percent) had accomplished this task with
similarly complex graphs on exams, when the definitions of these variables had been recently learned and were easier to recall.Therefore, this concept seems to have been understood by only half the
students and retained by even fewer.
Likewise, half of the students struggled with the following multiple-choice question from the pre- and post-test (see Pre-Post Test in Resources):
In a graph, the dependent variable….
A.is plotted on the x axis
B.is measured in the same units as the independent variable
C.is hypothesized to respond to changes in the independent variable
D.describes the experimental treatments
(correct answer: C)
In the post-test, only 51 % answered correctly (Figure 5). This represents only a slight improvement from the 43 % who answered correctly in the pre-test.
Detecting trends in data. A second area in which undergraduates struggled was the ability to discern general trends amid statistical “noise” in data.Many students believed that any variation in the
data resulted from important factors worth emphasizing.In one example, students were presented the number of days of lake ice on Lake Mendota, WI over the last 150 years (see Climate Change in
Resources).An especially warm or cold year (outlier) often distracted them from seeing more important, long-term trends.Similarly, most students graphed every data point in a bar graph, rather than
summarize the trends with a mean value.In the post-test, students were given categorical data on the number of eggs laid by Daphnia fed from two sources, and they were asked to summarize the pattern
with an appropriate graph (Example # 3 in Pre-Post Test).The “replicate number” was listed in the first column of data as a distracter.Most students (57 %) plotted the replicate number as the
independent variable on the x-axis (Figure 6A), and most (67 %) did not use a mean to summarize the trends (Figure 6B).Similar results were obtained from questions incorporated into course exams
(data not presented).These data from bar graphs and scatterplots suggest that our students generally emphasized individual data points rather than overall trends.
Interpreting interactions among variables. Finally, students seemed to have difficulty interpreting interactions among variables.To test this skill, we presented a bar graph from an experiment with a
3x3 factorial design (Example #2 in Pre-Post Test).Frog survival was measured in relation to exposure to three predator treatments crossed with three pesticide treatments.Answers were only considered
correct (“Improved” or “Satisfactory”) if students recognized that – according to the graph – malathion increased frog survival in the presence of beetles, and therefore should not be banned to
protect frogs. This required students to recognize the significant interaction between pesticides and predators.Answers were unsatisfactory if they were unclear, confused, or incomplete, including
statements such as “pesticides decreased frog populations” or “there is little effect of pesticides,” or if students recognized that malathion “killed beetles” while also recommending that it should
be banned.In the post-test only 23 of 74 students recognized a likely benefit of malathion, and there was no net improvement in the post-test answers (Figure 7).
Teaching analytical skills
Our assessment tools revealed some analytical skills that can be taught to undergraduates with relative ease and other areas where students continued to struggle despite our efforts to include
extensive data analysis and interpretation in our courses. In post-tests, 75-90 % of students were capable of creating and interpreting simple bar graphs, scatterplots and line graphs (Figures
1-3).Success with simple graphs has also been found in studies of middle and secondary school students (e.g., Phillips 1997; Tairab & Khalaf Al-Naqbi 2004).
Our study was designed to determine whether our courses as a whole improved analytical skills, so we cannot compare the relative effectiveness of any particular strategy we used. However, at the end
of their courses, students at FitchburgState were asked to comment if there were any activities, exercises, labs or concepts that helped them with the post-test. All of the strategies we used were
praised in their responses. The most commonly cited strategy was the background introduction to graphing (e.g., “when to use a line graph vs. a bar graph, and which axes are which”). Some students
cited the graphs we discussed from group exercises and lectures. Others noted the benefits from plotting data from their labs as a way to better design and interpret graphs.Several recalled that
using Microsoft Excel helped them, “even though Excel is very frustrating.”A few students noted how “everything combined helped” or that “it takes repetition when it comes to understanding graphs.”
Although our courses improved some analytical skills, students continued to struggle in several specific areas.First, most students lacked a profound understanding of dependent and independent
variables: most could define these variables from simple graphs but not from complex graphs with more than two variables.
We thought that the ability to define and identify independent and dependent variables would be essential to understanding experimental design and the graphs.However, our results suggest that
misapplying these terms does not necessarily inhibit general analytical skills.While only 30-40 percent of students were able to identify these variables from a complex graph in the post-test, most
(75 %) could clearly describe the relationships among those same variables (Figure 1A).Because our goal was to help students improve broad analytical understanding, and to apply rather than memorize
definitions, perhaps their understanding of these variable types was sufficient. | {"url":"https://docest.com/doc/563323/teaching-issues-and-experiments-in-ecology-volume-5-july-2007","timestamp":"2024-11-06T17:50:42Z","content_type":"text/html","content_length":"41039","record_id":"<urn:uuid:82493d7f-4fc5-46dc-a9ba-d57a032dc008>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00802.warc.gz"} |
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Direct Quadrature for Delay Renewal Equation
I'm trying to solve a delay renewal equation with a quadratic nonlinearity using direct quadrature in MATLAB. Here's the code I'...
18 days ago | 0 answers | 0
How to Extract Delayed State Terms in a Model with Distributed Delay?
I'm working on a model with distributed delays, and I'm using the ddesd function to solve the delay differential equations. My s...
3 months ago | 2 answers | 0
how to modify code for distributed delay
I have a code, which gives a solution of a delay logistic equation with discrete delay. tau = 1; tspan = [0 20]; y0 = 0.5;...
3 months ago | 1 answer | 0
Stopping Criteria in bayesopt for Early Termination
I'm working with bayesopt for Bayesian optimization in MATLAB. I want to implement a stopping criterion that terminates the opti...
7 months ago | 1 answer | 0
how to construct a batch for multiple trajectories in neural network
If we have more than one trajectory and want to construct a batch for all these trajectories what will be the possible way to us...
9 months ago | 1 answer | 0
How to write diagonal matrix in deep learning array
I have this matrix D = zeros(M, M + 1); D(1:end-1, 1:end-2) = diag((1/(2*h)) * ones(M-1, 1)); D(1:end-1, 3:end) = diag((-1/...
10 months ago | 1 answer | 0
error in neural ODE simulation
I am discretizing DDE to ODE and constructing Neural ODE for this discretized ODE. But when I run this neural ODE using dlode45 ...
10 months ago | 1 answer | 0
Matlab code of Neural delay differential equation NDDE
I have written a code of NDDE, but it is not correct and I am not able to simulate this code, because I am getting errors from i...
10 months ago | 1 answer | 0
can we apply ode45 on .mat file
I have a data of 100x5 columns. can i apply ode45 solver on this data against my specfic initial conditions? or what is the bes...
11 months ago | 1 answer | 0
adding noise to the solution obtained from ode45/dde23
what is the best way to add noise to the solution obtained from the solver like ode45/dde23?
11 months ago | 2 answers | 0
how to calculate the drivative of discretized ODE
I am discretizing DDE to ODE using pseudospectral method. I want to compute derivative of its solution for training state and wa...
11 months ago | 1 answer | 0
Conversion to cell from double is not possible.
I am trying to solve delay logistic equation with multiple delay terms but for finding delayed states i am getting error and my ...
12 months ago | 1 answer | 0 | {"url":"https://uk.mathworks.com/matlabcentral/profile/authors/29694118","timestamp":"2024-11-10T06:35:10Z","content_type":"text/html","content_length":"81740","record_id":"<urn:uuid:8108ac9e-3677-497c-a462-15120a873452>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00584.warc.gz"} |
Sin Function
Sayiinii kofa deebisi.Kofni raadiyaaniin ibsama.Gatiin gidduu -1 fi 1 oola.
Using the angle Alpha, the Sin function returns the ratio of the length of the opposite side of an angle to the length of the hypotenuse in a right-angled triangle.
Sin(Alpha) = side opposite the angle/hypotenuse
Sin (Number As Double) As Double
Lakkoofsa: Himannoon lakkoofsa kofa raadiyaaniin kan mul'isu Saayinii shallaguu barbaaddef.
To convert degrees to radians, multiply degrees by Pi/180, and to convert radians to degrees, multiply radians by 180/Pi.
Pi is approximately 3.141593.
REM Fakkeenya kana keessaatti, aseentuun rog-sadee sirrii armaan gadiif ni danda'ama.
REM rogni faallaa kofaa fi kofni(diigriidhaan)dheerina hayipootinasii shallaguuf.
Sub ExampleSine
REM Pi = 3.1415926 is a predefined variable
Dim d1 As Double
Dim dAlpha As Double
d1 = InputBox("Enter the length of the opposite side: ","Opposite Side")
dAlpha = InputBox("Enter the angle Alpha (in degrees): ","Alpha")
Print "The length of the hypotenuse is"; (d1 / sin (dAlpha * Pi / 180))
End Sub | {"url":"https://help.libreoffice.org/latest/om/text/sbasic/shared/03080103.html","timestamp":"2024-11-15T04:28:03Z","content_type":"text/html","content_length":"11321","record_id":"<urn:uuid:738931dd-a572-4179-ace8-a71e8c283a07>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00073.warc.gz"} |
Class 10 Maths Chapter 7 Assertion and Reason Questions - Coordinate Geometry
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion (A): The ordinate of a point A on y-axis is 5 and B has coordinates (–3, 1). Then the length of AB is 5 units.
Reason (R): The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6, 5) and Q(0, –4).
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The area of the triangle with vertices (-5 , -1), (3, -5), (5, 2), is 32 square units.Reason : The point (x, y) divides the line segment joining the points (x[1], y[1]) and (x[2], y[2])
in the ratio k : 1 externally then
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The co-ordinates of the point which divides the join of A(-5, 11) and B(4,-7) in the ratio 7 : 2 is (2, -3) Reason : The coordinates of the point P(x, y) which divides the line segment
joining the points A(x[1], y[1]) and B(x[2], y[2]) in the ratio m[1] : m[2] is
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion (A): △ABC with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to △DEF with vertices D(–4, 0), E(4, 0) and F(0, 4).
Reason (R): A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle.
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The points (k, 2 -2k), (- k+ 1,2k) and (- 4 - k, 6 - 2k) are collinear if k = 1/2. Reason : Three points A,B and C are collinear in same straight line, if AB + BC = AC.
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : If the points A(4, 3) and B(x, 5) lies on a circle with the centre O(2,3) then the value of x is 2.
Reason : The mid-point of the line segment joining the points P(x[1], y[1]) and Q(x[2], y[2]) is
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion (A): If the distance between the point (4, p) and (1, 0) is 5, then the value of p is 4.
Reason (R): The point which divides the line segment joining the points (7, – 6) and (3, 4) in ratio 1 : 2 internally lies in the fourth quadrant.
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : Centroid of a triangle formed by the points (a, b) (b, c), and (c, a) is at origin, Then a + b + c = 0 .
Reason : Centroid of a △ABC with vertices A (x[1], y[1]), B(x[2], y[2]) and C (x[3], y[3]) is given by
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The possible values of x for which the distance between the points A(x, -1) and B(5, 3) is 5 units are 2 and 8.
Reason : Distance between two given points A(x[1], y[1]) and B(x[2], y[2]) is given by,
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : Mid-point of a line segment divides line in the ratio 1 : 1.
Reason : If area of triangle is zero that means points are collinear.
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in the ratio 2 : 7 internally.
Reason : Three points A,B and C are collinear if AB + BC = AC
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in the ratio 2 :7 internally.
Reason : Three points A,B and C are collinear if area of △ ABC = 0 .
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The value of y is 6, for which the distance between the points P(2,- 3) and Q(10, y) is 10.
Reason : Distance between two given points A(x[1], y[1]) and B(x[2], y[2]) is given by,
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : If A (2a, 4a) and B (2a, 6a) are two vertices of a equilateral triangle ABC then the vertex C is given by (2a + a√3, 5a).
Reason : In equilateral triangle all the coordinates of three vertices can be rational.
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The point (0, 4) lies on y -axis.
Reason : The x co-ordinate on the point on y -axis is zero.
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The value of y is 6, for which the distance between the points P(2, -3) and Q(10, y) is 10.
Reason : Distance between two given points A (x[1], y[1]) and B(x[2], y[2]) is given 6,
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : C is the mid-point of PQ, if P is (4, x), C is (y,- 1) and Q is (- 2, 4), then x and y respectively are -6 and 1.
Reason : The mid-point of the line segment joining the points P(x[1], y[1]) and Q(x[2], y[2]) is
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The point (0, 4) lies on y -axis.
Reason : The x co-ordinate on the point on y -axis is zero.
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself: Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : Ratio in which the line 3x + 4y = 7 divides the line segment joining the points (1, 2) and (- 2, 1) is 3 : 5Reason : The coordinates of the point P(x, y) which divides the line segment
joining the points A(x[1], y[1]) and B(x[2], y[2]) in the ratio m[1] : m2[ ]is
Question for Assertion & Reason Type Questions: Coordinate Geometry
Try yourself:Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion: The point which divides the line joining the points A(1, 2) and B(- 1, 1) internally in the ratio 1: 2 is
Reason : The coordinates of the point P(x, y) which divides the line segment joining the points A(x[1], y[1]) and B(x[2], y[2]) in the ratio m[1] : m[2 ]is | {"url":"https://edurev.in/studytube/Class-10-Maths-Chapter-7-Assertion-and-Reason-Questions-Coordinate-Geometry/9ca8c4f5-301e-4dbe-a114-fefb009b2127_t","timestamp":"2024-11-10T01:08:39Z","content_type":"text/html","content_length":"1049251","record_id":"<urn:uuid:bfaf1937-79c6-40e7-a09d-1e5db54ab346>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00507.warc.gz"} |
How do you solve:
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Mathematics 63 Online
OpenStudy (anonymous):
How do you solve: A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true. Sn: 12 + 42 + 72 + . . . + (3n - 2)2 = (n(6n^
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OpenStudy (anonymous):
replace \(n\) by 1 to get \[S_1\] which will be \[12=\frac{1(6\times 1^2-3\times 1-1)}{2}\]which does not seem to be true does it?
OpenStudy (anonymous):
no it does not it equals 0
OpenStudy (anonymous):
and sorry it was 1^2+4^2+7^2 then (3n-2)^2
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efractive index
Optical constants of CdTe (Cadmium telluride)
Marple 1964: n 0.86–2.5 µm
Wavelength: µm
Complex refractive index (n+ik)
n k LogX LogY eV
Derived optical constants
Dispersion formula
Room temperature
D. T. F. Marple. Refractive index of ZnSe, ZnTe, and CdTe, J. Appl. Phys. 35, 539-542 (1964)
Cadmium telluride, CdTe
Cadmium telluride (CdTe) is an inorganic compound that typically crystallizes in a cubic zincblende structure. It is a direct bandgap semiconductor with a bandgap energy of approximately 1.5 eV,
making it highly suitable for photovoltaic applications, especially thin-film solar cells. CdTe solar cells are commercially significant due to their lower manufacturing costs and competitive
efficiency levels compared to silicon-based cells. Besides photovoltaics, CdTe is also employed in infrared detectors, radiation detectors, and electro-optic modulators. However, the presence of
cadmium, a toxic heavy metal, calls for stringent handling and disposal protocols and has prompted research into recycling methods. Synthesis methods for CdTe include physical vapor deposition,
chemical vapor deposition, and close-spaced sublimation, among others, each offering unique advantages in terms of film quality and manufacturing scalability. Despite the environmental concerns
related to its toxicity, CdTe remains a highly relevant material in renewable energy technologies and continues to attract research efforts aimed at mitigating its environmental impact.
External links | {"url":"https://refractiveindex.info/?shelf=main&book=CdTe&page=Marple","timestamp":"2024-11-07T04:31:59Z","content_type":"text/html","content_length":"39758","record_id":"<urn:uuid:e5465d56-0790-4963-9a47-75e607759ecc>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00762.warc.gz"} |
general math – Walking Randomly
Archive for the ‘general math’ Category
October 4th, 2021
The MATLAB community site, MATLAB Central, is celebrating its 20th anniversary with a coding competition where the only aim is to make an interesting image with under 280 characters of code. The 280
character limit ensures that the resulting code is tweetable. There are a couple of things I really like about the format of this competition, other than the obvious fact that the only aim is to
make something pretty!
1. No MATLAB license is required! Sign up for a free MathWorks account and away you go. You can edit and run code in the browser.
2. [DEL:Stealing:DEL] reusing other people’s code is actively encouraged! The competition calls it remixing but GitHub users will recognize it as forking.
Here’s an example. I wrote a piece of code to render the Sierpiński triangle – Wikipedia using a simple random algorithm called the Chaos game.
This is about as simple as the Chaos game gets and there are many things that could be done to produce a different image. As you can see from the Remix tree on the right hand side, I’ve already done
one of them by changing a from 3 to 4 and adding some extra code to ensure that the same vertex doesn’t get chosen twice. This is result:
Someone else can now come along, hit the remix button and edit the code to produce something different again. Some things you might want to try for the chaos game include
• Try changing the value of a which is used in the variables t and c to produce the vertices of a polygon on an enclosing circle.
• Instead of just using the vertices of a polygon, try using the midpoints or another scheme for producing the attractor points completely.
• Try changing the scaling factor — currently p=(p+s)/2;
• Try putting limitations on which vertex is chosen. This remix ensures that the current vertex is different from the last one chosen.
• Each vertex is currently equally likely to be chosen using the code idx=randi(a); Think of ways to change the probabilities.
• Think of ways to colorize the plots.
Maybe the chaos game isn’t your thing. You are free to create your own design from scratch or perhaps you’d prefer to remix some of the other designs in the gallery.
The competition is just a few hours old and there are already some very nice ideas coming out.
June 13th, 2016
I was in a funk!
Not long after joining the University of Sheffield, I had helped convince a raft of lecturers to switch to using the Jupyter notebook for their lecturing. It was an easy piece of salesmanship and a
whole lot of fun to do. Lots of people were excited by the possibilities.
The problem was that the University managed desktop was incapable of supporting an instance of the notebook with all of the bells and whistles included. As a cohort, we needed support for Python 2
and 3 kernels as well as R and even Julia. The R install needed dozens of packages and support for bioconductor. We needed LateX support to allow export to pdf and so on. We also needed to keep up to
date because Jupyter development moves pretty fast! When all of this was fed into the managed desktop packaging machinery, it died. They could give us a limited, basic install but not one with
batteries included.
I wanted those batteries!
In the early days, I resorted to strange stuff to get through the classes but it wasn’t sustainable. I needed a miracle to help me deliver some of the promises I had made.
Miracle delivered – SageMathCloud
During the kick-off meeting of the OpenDreamKit project, someone introduced SageMathCloud to the group. This thing had everything I needed and then some! During that presentation, I could see that
SageMathCloud would solve all of our deployment woes as well as providing some very cool stuff that simply wasn’t available elsewhere. One killer-application, for example, was Google-docs-like
collaborative editing of Jupyter notebooks.
I fired off a couple of emails to the lecturers I was supporting (“Everything’s going to be fine! Trust me!”) and started to learn how to use the system to support a course. I fired off dozens of
emails to SageMathCloud’s excellent support team and started working with Dr Marta Milo on getting her Bioinformatics course material ready to go.
TL; DR: The course was a great success and a huge part of that success was the SageMathCloud platform
Giving back – A tutorial for lecturers on using SageMathCloud
I’m currently working on a tutorial for lecturers and teachers on how to use SageMathCloud to support a course. The material is licensed CC-BY and is available at https://github.com/mikecroucher/
If you find it useful, please let me know. Comments and Pull Requests are welcome.
December 23rd, 2015
Some numbers have something to say. Take the following, rather huge number, for example:
This number wants to tell you ‘Happy Holidays’, it just needs a little code to help it out. In Maple, this code is:
n := 18532529104068264480353131238404133659515101876112780772576330806424607039523076495646885634139967048751461005248758632306757568791464282975763655513845614593843019187655175699232981800640177552230121901623724542589154403221854439086181827152684585874764890938291566599716051702867105827305295569713835061785617174899049034655848488352249531058730460687733248824488684969031964141214711866905054239875930383262767247976845232997188307342087743859641917976242185446451606034726912968063437466250120212904972794971185874579656679344857677824:
modnew := proc (x, y) options operator, arrow; x-y*floor(x/y) end proc:
tupper := piecewise(1/2 < floor(modnew(floor((1/17)*y)*2^(-17*floor(x)-modnew(floor(y), 17)), 2)), 0, 1):
points := [seq([seq(tupper(x, y), y = n+16 .. n, -1)], x = 105 .. 0, -1)]:
plots:-listdensityplot(points, scaling = constrained, view = [0 .. 106, 0 .. 17], style = patchnogrid, size = [800, 800]);
The result is the following plot
Thanks to Samir for this one!
The mathematics is based on a generalisation of Tupper’s self-referential formula.
There’s more than one way to send a message with an equation, however. Here’s an image of one I discovered a few years ago — The equation that says Hi
December 13th, 2015
Way back in 2008, I wrote a few blog posts about using mathematical software to generate christmas cards:
I’ve started moving the code from these to a github repository. If you’ve never contributed to an open source project before and want some practice using git or github, feel free to write some code
for a christmas message along similar lines and submit a Pull Request.
November 17th, 2014
Given a symmetric matrix such as
What’s the nearest correlation matrix? A 2002 paper by Manchester University’s Nick Higham which answered this question has turned out to be rather popular! At the time of writing, Google tells me
that it’s been cited 394 times.
Last year, Nick wrote a blog post about the algorithm he used and included some MATLAB code. He also included links to applications of this algorithm and implementations of various NCM algorithms in
languages such as MATLAB, R and SAS as well as details of the superb commercial implementation by The Numerical algorithms group.
I noticed that there was no Python implementation of Nick’s code so I ported it myself.
Here’s an example IPython session using the module
In [1]: from nearest_correlation import nearcorr
In [2]: import numpy as np
In [3]: A = np.array([[2, -1, 0, 0],
...: [-1, 2, -1, 0],
...: [0, -1, 2, -1],
...: [0, 0, -1, 2]])
In [4]: X = nearcorr(A)
In [5]: X
array([[ 1. , -0.8084125 , 0.1915875 , 0.10677505],
[-0.8084125 , 1. , -0.65623269, 0.1915875 ],
[ 0.1915875 , -0.65623269, 1. , -0.8084125 ],
[ 0.10677505, 0.1915875 , -0.8084125 , 1. ]])
This module is in the early stages and there is a lot of work to be done. For example, I’d like to include a lot more examples in the test suite, add support for the commercial routines from NAG and
implement other algorithms such as the one by Qi and Sun among other things.
Hopefully, however, it is just good enough to be useful to someone. Help yourself and let me know if there are any problems. Thanks to Vedran Sego for many useful comments and suggestions.
• NAG’s commercial implementation – callable from C, Fortran, MATLAB, Python and more. A superb implementation that is significantly faster and more robust than this one!
March 12th, 2014
A recent Google+ post from Mathemania4u caught my attention on the train to work this morning. I just had to code up something that looked like this and so fired up Mathematica and hacked away. The
resulting notebook can be downloaded here. It’s not particularly well thought through so could almost certainly be improved on in many ways.
The end result was a Manipulate which you’ll be able to play with below, provided you have a compatible Operating System and Web browser. The code for the Manipulate is
circArray[circrad, theta, pointsize, extent, step, phase,
, {{showcirc, True, "Show Circles"}, {True, False}}
, {{theta, 0, "Dot Angle"}, 0, 2 Pi, Pi/10, Appearance -> "Labeled"}
, {{pointsize, 0.018, "Dot Size"}, 0, 1, Appearance -> "Labeled"}
, {{phase, 2, "Phase Diff"}, 0, 2 Pi, Appearance -> "Labeled"}
, {{step, 0.25, "Circle Separation"}, 0, 1, Appearance -> "Labeled"}
, {{extent, 2, "Plot Extent"}, 1, 5, Appearance -> "Labeled"}
, {{circrad, 0.15, "Circle Radius"}, 0.01, 1, Appearance -> "Labeled"}
, Initialization :>
dotCirc[{x_, y_, r_, theta_, pointsize_, showcirc_}] := If[showcirc,
{Circle[{x, y}, r], PointSize[pointsize],
Point[{x + r Cos[theta], y + r Sin[theta]}]}
Point[{x + r Cos[theta], y + r Sin[theta]}]}]
circArray[r_, theta_, pointsize_, extent_, step_, phase_,
showcirc_] := Module[{},
Flatten[Table[{x, y, r, theta + x*phase + y*phase, pointsize,
showcirc}, {x, -extent, extent, step}, {y, -extent, extent,
step}]], 6]
If you can use the Manipulate below, I suggest clicking on the + icon to the right of the ‘Dot Angle’ field to expose the player controls and then press the play button to kick off the animation.
I also produced a video – The code used to produce this is in the notebook.
February 28th, 2014
A lot of people don’t seem to know this….and they should. When working with floating point arithmetic, it is not necessarily true that a+(b+c) = (a+b)+c. Here is a demo using MATLAB
>> x=0.1+(0.2+0.3);
>> y=(0.1+0.2)+0.3;
>> % are they equal?
>> x==y
ans =
>> % lets look
>> sprintf('%.17f',x)
ans =
>> sprintf('%.17f',y)
ans =
These results have nothing to do with the fact that I am using MATLAB. Here’s the same thing in Python
>>> x=(0.1+0.2)+0.3
>>> y=0.1+(0.2+0.3)
>>> x==y
>>> print('%.17f' %x)
>>> print('%.17f' %y)
If this upsets you, or if you don’t understand why, I suggest you read the following
Does anyone else out there have suggestions for similar resources on this topic?
August 23rd, 2013
In a recent blog-post, John Cook, considered when series such as the following converged for a given complex number z
z[1] = sin(z)
z[2] = sin(sin(z))
z[3] = sin(sin(sin(z)))
John’s article discussed a theorem that answered the question for a few special cases and this got me thinking: What would the complete set of solutions look like? Since I was halfway through my
commute to work and had nothing better to do, I thought I’d find out.
The following Mathematica code considers points in the square portion of the complex plane where both real and imaginary parts range from -8 to 8. If the sequence converges for a particular point, I
colour it black.
LaunchKernels[4]; (*Set up for 4 core parallel compute*)
ParallelEvaluate[SetSystemOptions["CatchMachineUnderflow" -> False]];
convTest[z_, tol_, max_] := Module[{list},
list = Quiet[
NestWhileList[Sin[#] &, z, (Abs[#1 - #2] > tol &), 2, max]];
Length[list] < max && NumericQ[list[[-1]]]
, 1, 0]
step = 0.005;
extent = 8;
data = ParallelMap[convTest[#, 10*10^-4, 1000] &,
Table[x + I y, {y, -extent, extent, step}, {x, -extent, extent,
, {2}];]
I quickly emailed John to tell him of my discovery but on actually getting to work I discovered that the above fractal is actually very well known. There’s even a colour version on Wolfram’s
MathWorld site. Still, it was a fun discovery while it lasted
Other WalkingRandomly posts like this one:
February 6th, 2013
The University of Manchester has uploaded some videos of Cornelius Lanczos the (re-)discoverer of the fast Fourier transform and singular value decomposition among other things. Recorded back in
1972, these videos discuss his life and mathematics. Worth taking a look.
Update: Manchester’s Nick Higham has written a more detailed blog post about these videos – http://nickhigham.wordpress.com/2013/02/06/the-lanczos-tapes/
August 29th, 2012
While on the train to work I came across a very interesting blog entry. Full LaTeX support (on device compilation and .dvi viewer) is now available on iPad courtesy of TeX Writer By FastIntelligence
. Here is the blog post telling us the good news http://litchie.com/blog/?p=406
At the time of writing, the blog is down (Update: working again), possibly because of the click storm that my twitter announcement caused..especially after it was picked up by @TeXtip. So, here is
the iTunes link http://itunes.apple.com/us/app/tex-writer/id552717222?mt=8
I haven’t tried this yet but it looks VERY interesting. If you get a chance to try it out, feel free to let me know how you get on in the comments section.
Update 1: This version of TeX writer(1.1) cannot output to .pdf. Only .dvi output is supported at the moment. | {"url":"https://walkingrandomly.com/?cat=6","timestamp":"2024-11-02T06:09:58Z","content_type":"application/xhtml+xml","content_length":"90407","record_id":"<urn:uuid:c2d94a58-902f-4694-a7fb-803f7923b4ab>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00247.warc.gz"} |
What is a ray?
This article will discuss how rays are defined and how they're related to wavefronts. We will show how rays are used in OpticStudio and will discuss ray characteristics such as amplitude, phase, and
optical path, as well as their interactions with media.
Authored By Ken Moore
Rays are used to model the propagation of light through optical systems. For homogeneous, isotropic media, such as common optical glass, rays are straight lines. Rays are normal to the local
wavefront and point along the direction of energy flow. Rays have position, direction, amplitude, phase, and possibly polarization data associated with them.
At surfaces between different types of media, rays can refract, reflect, or diffract. The interface between different media will generally alter the direction and other properties of the ray.
This article will show how rays are defined and used in OpticStudio.
Rays in OpticStudio
OpticStudio uses ray tracing to model the propagation of light from the source point, through the optical system, and on to the final image surface. The resulting distribution of amplitude, phase,
and polarization of a collection of rays can be used to predict a wide range of optical phenomena.
The ray model is a very convenient, powerful, and accurate means for propagating light.
When tracing through inhomogeneous media, such as gradient index glass, rays will in general follow curved paths. Some materials, such as uniaxial crystals, are not isotropic, and within these types
of media rays are not normal to the wavefront. Both of these cases are handled by OpticStudio and are described in other articles. Polarization ray tracing is a complex subject and is also described
in a separate article.
Rays and wavefronts
Rays are closely related to wavefronts. A wavefront is an imaginary surface where the electric field of an optical beam has a uniform phase with respect to some reference point. For example, imagine
a coherent point source radiating into a sphere. At any distance away from the source, there is a spherical wavefront with a uniform phase.
It is numerically difficult to propagate wavefronts directly. Fortunately, direct propagation of the wavefront is often only required near the focus of a beam. Far from focus, it is much easier to
convert the wavefront to rays and then propagate the rays.
At any point on a wavefront, some small area may be selected. The small area can be thought of as a piece of a plane wave. The ray position and coordinates are defined by the position of the small
area and the orientation of the normal vector to this small plane wave, respectively. The integrated energy or flux of the beam over this small area may be assumed to be entirely represented by the
ray. For computations near focus, such as the Huygens Point Spread Function, the rays can be converted back into wavefronts.
Coordinates, cosines, and propagation
The two most fundamental properties of a ray are position and direction.
Position is defined by:
The coordinates are measured in units of length commonly called "lens units".
Direction is defined by:
The values are the direction cosines of the unit vector that points along the ray.
Both r and k may be measured in surface local coordinates, or in global coordinates relative to some reference frame.
If a ray is propagated a distance t, where t is a length in lens units, the new coordinates are given by:
For a discussion of how rays are traced to specific types of optical surfaces, see reference 1.
OpticStudio can trace a single ray and create a table of the ray coordinates and cosines, as well as other data, using the Analysis...Calculations...Ray Trace feature.
The coordinates of the ray are listed at each surface, starting from the object. The direction cosines listed for a surface are always after refraction, reflection, or diffraction into the next
Refraction, reflection, and diffraction
Rays are traced to the point of intersection with a surface that defines the boundary between different media. At this interface, rays will refract according to Snell's famous law, here written in
the vector form:
where N is the unit normal vector of the surface at the point of intersection, and k is the ray direction cosine vector. Primed quantities are after refraction, and unprimed quantities are prior to
For reflection, the index of refraction is unimportant, and Snell's law simplifies to:
This expression may be derived by setting n' = -n, and this convention is often used by ray tracing programs to eliminate the distinction between refraction and reflection.
Some optical surfaces, such as diffractive gratings, also bend rays. The general form of the diffraction expression, which includes refractive or reflective effects, is
where M is the diffraction order, l is the wavelength, p is the local grating period (length per 2p period), q is a unit vector tangent to the surface and parallel to the local grating lines. Note if
M is zero, or p is infinite, the general grating expression simplifies to Snell's law.
For a more comprehensive discussion of this topic, see reference 2.
Amplitude, phase, and optical path
A ray models propagation of an electric field, and therefore has both magnitude and phase. The ray amplitude is a complex number of the form
The intensity of a ray is given by the square of the ray amplitude, or A*A. The intensity of the ray can be measured in units of power, energy, power per unit time, or energy per unit time. What
units are appropriate to use for rays depends on the details of the specific computation being performed. Note the term intensity used here is not the same as is commonly used in radiometry, where
intensity is power per solid angle.
The ray amplitude is generally reduced whenever the ray crosses an index boundary, such as refracting from air into glass. The presence of thin films on the glass alters the amount of amplitude loss.
The amplitude of the ray is also reduced when traveling through glass due to bulk absorption. All these effects are modeled as part of the polarization ray trace feature in OpticStudio, which are
discussed in a separate article.
As a ray propagates, the phase of the ray changes. The phase change due to propagation of a distance t is given by
where n is the index of refraction in the media, λ[0] is the vacuum wavelength, and φ is measured in radians.
For homogeneous, isotropic media the optical path length (OPL) of a ray is the product of the distance traveled by the ray multiplied by the refractive index through which the ray travels:
OPL = nt
where n is the index of refraction and t is the distance traveled. The OPL may be computed for a complete optical system, from source point to image surface, by summing the OPL for each media between
each optical surface.
For best image quality, every ray should intercept the image at the same point, and with the same phase. For this reason, it is more convenient to keep track of the optical path length of a ray as
the difference between the ray and some reference ray, usually the chief ray. It is also convenient, for reasons described in reference 1, to measure the OPL not from the source to the image, but
from the source to the reference sphere. The reference sphere is a spherical surface centered on the point where the chief ray intersects the image surface. The radius of the reference sphere is
defined by the distance from the image surface to the paraxial exit pupil. This quantity is called optical path difference (OPD):
OPD = OPL[ray] - OPL[chief]
where both OPL values are measured from the source to the reference sphere. The OPD is what is displayed on the OPD fan:
The OPD can be used to reconstruct the wavefront from the ray data. OpticStudio uses OPD as the primary means of computing and displaying wavefront aberrations.
Ray sampling
A single ray is generally not adequate to model a wavefront. Instead, a collection of rays is used. The most common collection of rays is a rectangular grid.
Many OpticStudio features, such as PSF or MTF analysis, use grids of rays to sample the entire pupil. Usually, these grids are specified by N×N, where N is an integer power of 2. For example, the
grid sizes are typically 32 x 32, 64 x 64, 128 x 128, etc.
The higher the pupil sampling, the greater the accuracy of the computation and the longer the data will take to compute.
The amount of energy associated with each ray is proportional to the pupil area associated with each ray. If more rays are traced in the pupil, each ray will have less energy.
1. Shannon, R. R., The Art and Science of Optical Design, Cambridge University Press.
2. Bass, Michael, Handbook of Optics Volume I, McGraw Hill | {"url":"https://support.zemax.com/hc/en-us/articles/1500005488381-What-is-a-ray","timestamp":"2024-11-14T13:30:21Z","content_type":"text/html","content_length":"52721","record_id":"<urn:uuid:e54b7106-e9db-4bf6-a10f-4606e4df4c4d>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00665.warc.gz"} |
How do you write the equation in standard form for a circle with center (-3,7) for a circle and tangent to the x axis? | Socratic
How do you write the equation in standard form for a circle with center (-3,7) for a circle and tangent to the x axis?
1 Answer
${\left(x + 3\right)}^{2} + {\left(y - 7\right)}^{2} = 49$
The standard form is ${x}^{2} + {y}^{2} = {r}^{2}$. First, let's determine $r$. The center is $7$ above the x-axis and the circle is tangent to the x-axis, so the radius $r$ should be equal to $7$.
The center moved $3$ to the left, so substitute $x$ by $\left(x + 3\right)$.
The center moved $7$ up, so substitute $y$ by $\left(y - 7\right)$.
You can determine these numbers by filling in the center coördinates, the outcome must be zero. [$x + 3 = - 3 + 3 = 0$]
Maybe you've seen the standard equation ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ for a circle with center $\left(a , b\right)$.
So, the equation becomes ${\left(x + 3\right)}^{2} + {\left(y - 7\right)}^{2} = {7}^{2} = 49$
Impact of this question
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The First-Order Theory of Sets with Cardinality Constraints is Decidable
Data structures often use an integer variable to keep track of the number of elements they store. An invariant of such data structure is that the value of the integer variable is equal to the number
of elements stored in the data structure. Using a program analysis framework ports abstraction of data structures as sets, such constraints can be expressed using the language of sets with
cardinality constraints. The same language can be used to express preconditions that guarantee the correct use of the data structure interfaces, and to express invariants useful for the analysis of
the termination behavior of programs that manipulate objects stored in data structures. In this paper we show the decidability of valid formulas in one such language. Specifically, we examine the
first-order theory that combines 1) Boolean algebras of sets of uninterpreted elements and 2) Presburger arithmetic operations. Our language allows relating the cardinalities of sets to the values of
integer variables... | {"url":"https://www.sciweavers.org/publications/first-order-theory-sets-cardinality-constraints-decidable","timestamp":"2024-11-09T04:31:59Z","content_type":"application/xhtml+xml","content_length":"38900","record_id":"<urn:uuid:c4c005a1-e521-4631-9001-b369cfa9a200>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00005.warc.gz"} |
Wealth inequality denial
In the week a report established that 97.1% of scientists publishing on the subject have concluded that man-made climate change exists, it seems the right have re-opened an old front in their war on
reality. In Saturday’s Financial Times the normally sober paper appears to have fancied a day masquerading as Fox News as it piled into Thomas Piketty’s Capital in the 21^st Century, arguing that its
main thesis — broadly that wealth inequality is rising at Belle Époque rates — was undercut by flawed data.^1http://www.ft.com/intl/cms/s/2/e1f343ca-e281-11e3-89fd-00144feabdc0.html#axzz32oya0S2a
Picking apart the analysis is fair enough you might argue, and Piketty would agree; that’s why he put all his data on his website so it could be checked by third parties (compare the secrecy of
right-wing think tanks, especially about who funds them). The problem is that the FT’s criticisms are laughable. I mean some of them are seriously embarrassing — what on earth was their editor
thinking? The FT turned their splash into a video for good measure^2see link in above ref. and I have to wonder if I am the only one seeing unease on Economics Editor Chris Giles’ face as he presents
his ‘findings’?
For example, the FT leads its video assault^3About the 1 min. mark with a strange appearance of an inexplicable “2” in a formula that reads “2+J11*(FS12/FS11)” allegedly because Piketty did not think
the numbers high enough. The FT’s analysis corrects for these sorts of inclusions. I slightly choked down my tea^4Yorkshire if you’re asking. at this. Was Giles seriously suggesting that the addition
of a constant to a function was fatal to an analysis of its trend — its rate of change or slope? If you take a simple function like f(x)=x then it’s slope is f´(x)=1. If you add a constant c=2 then f
(x)=x+2 and the slope is f´(x)=1. Let’s set c=42 billion, then f(x)=x+42 billion and its slope is, yes you guessed it, f´(x)=1. The point is the slope does not change if you move the same function up
and down AND Piketty is interested in the slope — he is interested in how inequality is changing over time NOT where we are starting from.
Think of it another way: imagine Piketty is a geographer. The monetarists all claim that we live on a flat planet and Piketty is trying to show that there is at least one hill or valley on this dull
plain. His data shows that at some point on the plain there is a slope (a sure sign of a hill or a valley). The FT is arguing that because his slope is slightly higher than their slope Piketty’s
analysis is flawed and that consequently there are no hills or valleys (quod erat demonstrandum). Not only is the argument flawed, it actually insults not only the competence of the FT’s readership
but the Painean common sense of everyone who can see all the bloody hills and valleys out their window. As a commenter wittily noted on Paul Krugman’s New York Times blog, the argument runs that if
the statistical plot of the course of hurricane Katrina was off by 10 metres, there was no hurricane.
Applying this exciting new methodology of ontological critique I can now bring to you CLT’s very own splash which knocks the FT’s blustering into a hat. I can conclusively show that Newton’s
arguments are undercut by flawed analysis. It turns out that his old second law of motion fails to take into account a reciprocal factor in the order of the square root of [1 – velocity²/speed of
light²].^5Which in classical case is 1 minus a really really small number. Therefore — hold onto everything not nailed down — Newton was a fraud and gravity doesn’t exist.
The FT then goes on to criticise Piketty for using available data and for having holes where data is not available. You can see why they led with the additive constant. Piketty did his best, was open
about it and has put his sources online for all to see. As Piketty notes, subsequent analyses using new data confirm his thesis. Furthermore, he has tried to obtain data which is of its nature
difficult to obtain. Turning the FT’s argument on its head Piketty has responded (see endnote) that not only is he open about holes in the data, but that this is a key reason why he is arguing for a
wealth tax: because taxes mean record-taking and record-taking means data to assess wealth inequality. Given that James Henry has calculated that there is US$21 trillion of assets hidden offshore,
even a wealth tax of the French type would not provide sufficient data as to real inequality.^6http://www.theguardian.com/business/2012/jul/21/global-elite-tax-offshore-economy
Paul Krugman sums up the FT’s Chris Giles’ position as follows:
The point is that Giles is proving too much; if his attempted reworking of Piketty leads to the conclusion that nothing has happened to wealth inequality, what that really shows is that he’s
doing something wrong.^7http://krugman.blogs.nytimes.com/2014/05/24/is-piketty-all-wrong/?_php=true&_type=blogs&_r=0
Quite. And from the other side of the debate Scott Winship, an economist at the right-leaning Manhattan Institute for Policy Research, and notable critic of Piketty’s analysis, said the FT’s
allegations weren’t:
significant for the fundamental question of whether Piketty’s thesis is right or not … It’s hard to think Piketty did something unethical when he put it up there for people like me to delve into
his figures and find something that looks sketchy … Piketty has been as good or better than anyone at both making all his data available and documenting what he does generally.^8http://
People will provide more detailed comebacks on the economic arguments and the new data will be integrated. This is not what spurred me to blog. Nor any particular love for Piketty; not only are the
exponential growth demands of capital and its manner of reproduction self-evident to me at least, but Piketty’s solutions are unlikely to make the slightest dent. This is indeed why he is more
popular with the centre-left of the UK and East Coast US than in France.
No. What struck me was the manner in which the FT engaged in all the classical tactics of climate change denial but in turbocharged form. From behind its paywall it created a storm by suggesting
Piketty’s findings were “undercut by errors” knowing full well that on the internet this would be amplified by right-wing bloggers as “Piketty completely wrong” and “wealth inequality does not exist
(save as [insert bête noire here] propaganda)”. I remember The Spirit Level and to a lesser degree Le nouvel ésprit du capitalisme receiving assaults, some of which were heavy artillery attacks
funded by think-tank sponsors, and they still held out. Yet in those cases the usual suspects engaged in the campaign to discredit. Here the FT has risked what currency it had even with its own fans
on the right to bare its teeth and destroy Piketty’s reputation.
To have brought upon himself the very anger of the celestial realm; perhaps Piketty really has brought us a spark of fire.
Piketty’s preliminary response to the FT is here.
Ida Ince is an independent researcher in critical legal finance and has previously worked for many years in international finance.
• 1
• 2
see link in above ref.
• 3
About the 1 min. mark
• 4
Yorkshire if you’re asking.
• 5
Which in classical case is 1 minus a really really small number.
• 6
• 7
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Generalising the mean of a list
30 Jan 2022
The free semigroup on a set X, denoted X^+, is the set of nonempty ordered lists whose items are elements of X. For example, ℤ^+ is the set of nonempty lists of integers, such as [3, 1] and [5, -6,
12]. For all sets X, there is a “length” function Λ : X^+ → ℕ[>0] that calculates the length of a list. (Here, ℕ[>0] is the set of natural numbers excluding 0.)
If (X, +) is a semigroup, then there is also a “sum” function ∑ : X^+ → X that combines every item in a list using the semigroup operation. For example, ∑[x, y, z] = x + y + z. Because (X, +) can be
any semigroup, the + symbol does not necessarily denote addition. It could also be multiplication, function composition, maximum, minimum, greatest common divisor, etc. Any associative operation will
Every semigroup (X, +) admits a binary function p : ℕ[>0] × X → X, which I will denote by juxtaposition; nx, short for p(n, x), is the result of combining x with itself n times. For example, 3x = p
(3, x) = x + x + x.
In summary, for any semigroup (X, +), you can calculate the length of a list in X^+ with Λ and the “sum” of a list in X^+ with ∑. You can also “multiply” nonzero natural numbers with the elements of
X. Using these three operations, I will define a binary relation ⤚ between X^+ and X. If a is an element of X^+ and x is an element of X, then a ⤚ x if and only if ∑(a) = Λ(a)x. In this case, we say
that x is a mean of a.
That’s a very general definition (it works for any semigroup), so here are some concrete examples:
1. If the semigroup in question is (ℝ, +), then a ⤚ x means that x is the arithmetic mean of the list a.
2. If the semigroup is (ℝ[>0], ·), the set of positive real numbers under multiplication, then a ⤚ x means that x is the geometric mean of the list a.
3. If the semigroup is (ℤ, max), then a ⤚ x means that x is the maximum of a. This is because ∑(a) calculates the maximum of a, and nx = x for all n (i.e. the semigroup is idempotent). Similarly, in
(ℤ, min), a ⤚ x means that x is the minimum of a.
4. If the semigroup is itself a free semigroup, i.e. X^+ for some set X, then ∑ : X^++ → X^+ is a function that flattens a list of lists. Unlike the previous examples, ⤚ in this case is not a total
relation, or in other words, not every list of lists has a mean. Those that do have a mean are of a very special variety. For example, the mean of [[x, y], [z, x, y, z], [x, y, z]] is [x, y, z],
because flattening the former gives the same result as repeating the latter 3 times.
5. If the semigroup is (𝔹, ∧), the set of boolean values under logical conjunction, then a ⤚ 1 if every item in a is 1; otherwise, a ⤚ 0. Conversely, in (𝔹, ∨), a ⤚ 1 if any item in a is 1.
6. If the semigroup operation is modular addition, then means are not always unique. For example, in the cyclic group ℤ/3ℤ, the mean of [1, 2, 0] can be either 0, 1, or 2, because 1 + 2 + 0 = 3(0) =
3(1) = 3(2). This shows why ⤚ has to be defined as a relation and not a function. A list doesn’t always have exactly one mean; it could have 0, or 3, or even an infinite number.
7. If the semigroup is (ℝ → ℝ, ∘), then a ⤚ f means that performing every function in the list a is equivalent to repeating f, Λ(a) times. Finding a mean of a list of functions amounts to composing
every function in the list and then taking the functional root.
A semigroup (X, +) is meaningful if the ⤚ relation can be made into a function; for every a ∈ X^+, there is a unique μ(a) ∈ X such that a ⤚ μ(a). Instead of saying that μ(a) is a mean of a, we can
now confidently say that μ(a) is the mean of a. In examples 1, 2, 3, and 5 above, the semigroup in question is meaningful, but in examples 4, 6, and 7, it is not. A semigroup that is not meaningful
is called meaningless. In a meaningless semigroup, the ⤚ relation is either non-total or multi-valued, so it cannot be made into a function.
A mean-preserving homomorphism from (X, +) to (Y, ∗) is a function h : X → Y with the following properties:
1. h is a semigroup homomorphism: h(x + y) = h(x) ∗ h(y).
2. h respects the mean relation: a ⤚ x implies h(a) ⤚ h(x), where h(a) denotes the application of h to every item in a.
One can prove that all semigroup homomorphisms are mean-preserving. Assume h is a homomorphism from (X, +) to (Y, ∗), a ∈ X^+, and x ∈ X. Then a ⤚ x ⇔ ∑(a) = Λ(a)x ⇒ h(∑(a)) = h(Λ(a)x). Because h is
a homomorphism, h(nx) = nh(x), and h(∑(a)) = ∑(h(a)). The latter can be demonstrated by replacing the list a with its components: h(∑(a)) = h(a[0] + a[1] + … a[n]) = h(a[0]) ∗ h(a[1]) ∗ … h(a[n]) = ∑
(h(a)). Therefore, the equality h(∑(a)) = h(Λ(a)x) is equivalent to ∑(h(a)) = Λ(a)h(x). Applying h to a list does not change its length, so we get that ∑(h(a)) = Λ(h(a))h(x) ⇔ h(a) ⤚ h(x). In
summary, for any homomorphism h, a ⤚ x implies h(a) ⤚ h(x).
I will use the term divisible semigroup for a semigroup X satisfying the following property: for every x ∈ X and n ∈ ℕ[>0], there is a unique y ∈ X such that ny = x. The element y is called the nth
factor of x, and is written x / n. It isn’t difficult to see that all divisible semigroups are meaningful, with μ(a) = ∑(a) / Λ(a). But the reverse implication (that all meaningful semigroups are
divisible) is not true.
One example of a semigroup which is meaningful but not divisible is (ℝ[>2], ·). This is a valid semigroup, because the product of numbers greater than 2 will also be greater than 2. It is meaningful,
because the geometric mean of a list of numbers greater than 2 will also be greater than 2. But it is not divisible, because the nth root of a number greater than 2 is not necessarily greater than 2.
(nth roots are the equivalent of nth factors in multiplicative semigroups.) Therefore, the concept of “meaningfulness” can be thought of as a generalisation of divisibility.
Meaningfulness does imply divisibility if the semigroup has an identity element. To find the factor x / n, you can construct a list with length n and sum x, and then calculate the mean. In a monoid,
it is always possible to construct a list with these properties by appending the identity: [x], [x, e], [x, e, e], and so on. So for monoids, meaningfulness and divisibility are equivalent.
Given two meaningful semigroups (X, +) and (Y, ∗), their direct product is also meaningful. For any list of pairs c ∈ (X × Y)^+, we can split c into two lists a ∈ X^+ and b ∈ Y^+ by putting the first
element of each pair in a and the second element of each pair in b. Then the pair (μ(a), µ(b)) is a mean of c, because Λ(c)(μ(a), μ(b)) = (Λ(c)μ(a), Λ(c)μ(b)) = (Λ(a)μ(a), Λ(b)μ(b)) = (Σ(a), Σ(b)) =
Σ(c). It is also clearly the only mean of c, because μ(a) and μ(b) are the only means of a and b respectively. This means that the direct product of any two meaningful semigroups is meaningful.
Many other constructions also preserve meaningfulness.
• Given a meaningful semigroup (X, +), the opposite semigroup (X, (x, y) ↦ y + x) is also meaningful, with μ(a) in the latter being equivalent to μ(reverse(a)) in the former.
• The semigroup formed by augmenting a meaningful semigroup (X, +) with an absorbing element ∞ (such that ∞ + x = x + ∞ = ∞) is meaningful. The mean of any list not containing ∞ is already defined,
and the mean of a list containing ∞ is just ∞.
• If a meaningful semigroup (X, +) can be divided into two disjoint subsets Y and Z such that both are closed under the + operation and Y ∪ Z = X, then both (Y, +) and (Z, +) are meaningful. The
mean of a list in Y^+ must be in Y because Z is closed, and vice versa. For example, the meaningful semigroup (ℝ[>0], ·) can be divided into two subsemigroups (ℝ[(0, 1]], ·) and (ℝ[>1], ·). Both
are meaningful by the preceding theorem.
One construction that does not preserve meaningfulness is the addition of an identity element. For example, (ℝ[>1], +) is meaningful, but (ℝ[>1] ∪ {0}, +) is not, even though it is a valid semigroup.
The arithmetic mean of [0, 2] is not contained in ℝ[>1] ∪ {0}. | {"url":"https://owenbechtel.com/blog/generalising-mean-of-list/","timestamp":"2024-11-01T22:01:27Z","content_type":"text/html","content_length":"13291","record_id":"<urn:uuid:84f62d70-2b1a-4076-90db-488e6f2ad456>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00597.warc.gz"} |
Fill the grid in every level by dragging on the tiles from the side you are allowed in this new puzzle game
by Logical Cell.
48 comments:
Nice concept, but the resulting symbol of stage 13 reminds me of something...
Stuck on level 8 :/
me too stuck on 8...
Hints for stage 8: The trick is to push the left starting block into the corner without letting the path from the right block cross into the left side of the puzzle. Each block then fills in one
half of the puzzle, split down the middle. The right starting block doesn't move.
yup, I can see that that brings the solution...just haven't yet managed to move that left block in that way..
Spoiler: level 8
AHhhhhh, thanks for that. Okay, so you can build an entire framework, and then shove THE WHOLE THING to the side. That makes sense and I wasn't getting that.
General hint: if you can only push them, how can you reshape the (in particular level 8) shapes so that you can move the shapes where you need them...
Still don't get how to push the left block into the corner..
you have to first draw the left box then as you draw the right box you push the whole left box and everything you already drawn to the left.
I'm stuck in level 13 :(
Ahhhhh! Finally got it! Whew!
am i the only one stuck on 11?
Level 15?.. Stumped..
Spoiler level 15: bit difficult to explain. First expand by one square, then push the two squares beyond an "entry" so you can get in.
Any hints on level 14?
stuck on stage 24 at newgrounds.
SPOILER for 14:
Move UULLDDDRR, then move the 1 up, and finish.
17 looks like a lot of fun...
#15 push out one square from each block starting with the top and going to the right, and follow with the others in a clockwise pattern, one square. then starting with the top and going left push
the left block down, do the same with the others going in a counterclockwise pattern, then fill in the rest.
anyone can help me with stage 24?
here comes help for stage 13:
start with the one on the top left, draw it 1x down and 1x to the right, so you push two numbers in the 'outer circle'.
Then, continue with the one at the very right side, draw it 1x up and 1x left.
the rest is easy...
and some help for stage 15:
draw the one on the left 1x up and push it 1x down with the one on the top.
draw the one on the bottom 1x left and push it 1x right with the one on the left.
draw the one on the right 1x down and push it 1x up with the one on the bottom.
finsh wirh the one on the top, then the one on the right, etc. finishing should be easy... ;-)
Stage 24: There are several different solutions that I've seen (and I'm sure many more). I'll spell out the general principle for my solutions: imagine the blocks are numbered 1-9, left-right,
top-bottom, like a phone. Move 7 right, then 9 left, so that 8 is centered again. Move 5 right, then 8 up (this moves 3 up to the corner where it can spread out). Move 3 down and 6 left, then 2
left, 4 up, and 1 right to fill in the spaces.
17 looks less fun.
Kind of aggravating at this point.
Stage 17: Use up the long block before you touch anything else - the idea is to push the 2s toward the top of the screen. The left part of the large block should end in the lower left slot (after
traveling 8 spaces). That leaves 6 spaces, so the right part of the large block can end in the right slot. From there you should be able to fill in the gaps (it takes a 2 block nudging another 2
Argh, Stage 18 is KILLING me -- because there are about a million solutions that all leave me exactly one space long OR short. :P
I'll get it eventually, though . . .
I'm having the same issue with 18 - always one blank square!
I can confirm...
18 is tough.
Here's my thoughts on 18.
In order to reach the square all by itself, one of these dudes has to get where only the guy that starts top right of the four of them can get to. Those two squares are two up and two right from
where that square starts. The reason only it can reach them is because no one else can be pushed there.
Right? I'll take a hint in the form of "that block has to be pushed up/right twice" so I know. If I'm wrong on the above, tell me that, too.
I got 18. Here's a hint: The block in the corner needs to make an S-shape -- right, up, right.
Working on 19 now. :P
And finished 25... Not without hints on some levels though. And 25 is easy once you've cracked 24...
HELP on level 16!!
For lvl 16.
Move: #4 left over #1
Move #1 (far right)up.
Move#2(far left) over one-pushing #3 under now second #2.
Move top #2 up one.
Rest should be evident...
23 looks like it should be easier.
Now I can't get lvl 18!! That above "S" solution isn't working for me...
Still can't get 18 - that s solution? Just not working for me, so having a brain fart on this level, lol.
For anon @ 19:44 today -
For level 18:
Name the 3-cubes this way: A B
C D
C pushes D once to the right. A pushes B once to the right. Then right edge of C pushes
A up once. That same edge then pushes B over to the right. D then moves once to the right and then
pushes B up twice. A fills in the blocks above and below it. C fills in above and to the left of itself.
Hope that helps!
TO Anon at 19 May 2015 at 22:07
Thank you for replying finally got it! :)
These Logical Cell games are really addictive. This is the only one I have managed to finish.
help me level 20 please
For level 20: Go right 1 (which pushes the blank box over 1 space). Go up 1, right 1 and then down twice (pushes the blank box). Go left 1, down 1 and right twice (pushing the blank box all the
way over). Go down 1, left twice and then down. Go right 3 times (all the way over) and then up 3 times (pushing the blank box up, level with the empty space on the right). Go left 1 and up 1. Go
right 1 (pushing blank box into empty space) then go up and then left!!
Hope that helps!
Level 23 help?
LEVEL 23 is solved by halving the board:
Right side: Move 2-block to the left once and up once. Move 3-block to the left once (pushes former 2-block). Move 4-block over twice and down twice (pushes former 2-block down). From here,
finishing that side is obvious!
Left side: Move 2-block to the right once and up once. Move 1-block to the right (pushes former 2-block). Again, from here the finish is obvious!
Hope it helps!
LEVEL 24 is hard, but possible. I can give you some hints for now. The down right 1 should be extended up and it goes to the most right place on the board. You have several moves when you'll
encounter a situation for the middle one to push the central left 1 to the left, the 1 above it pushes it down and you extend the pushed 1 to the left. In this situation you should have the down
left 1 over the middle pocket on the board, after which it should be pushed down into the whole and then extended up.
that it for the hints, it is a little bit difficult to explain, but the trick is that the bottom left one goes into the middle whole on the board!
As I said LEVEL 24 is difficult but LEVEL 25 I did it in 3-th try.
here is an image of it http://imgur.com/q7qaHyx
For Level 11: You need to get one of the '2' blocks in the bottom left corner that forms an 'L' | {"url":"https://www.bontegames.com/2015/05/spreadset.html","timestamp":"2024-11-11T01:22:45Z","content_type":"text/html","content_length":"215748","record_id":"<urn:uuid:0ce7154d-6315-4636-aa4a-0ccc4b62fe01>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00657.warc.gz"} |
Topics: Fixed-Point Theorems
In General > s.a. posets.
* Simple example: If I hold a map of an area including the point where I stand, there must be a point on the map which is exactly above the corresponding point of the real world (even if I deform or
crumple the map).
* Motivation: If A is any differential operator, the existence of solutions of the equation A f = 0 is equivalent to the existence of fixed points for A + I; We are interested in equations like df =
0 for the study of critical points (> see morse theory, etc).
* Applications: The earliest one was to the existence of periodic orbits for 3 bodies.
@ General references: Granas & Dugundji 03; Farmakis & Moskowitz 13 [and applications].
@ Various types: Prykarpatsky a0902 [Leray-Schauder, Borsuk-Ulam type generalization].
> Online resources: see Wikipedia page.
Simple Case: Real Functions of One Variable
$ Def: Given any interval [a, b] ⊂ \(\mathbb R\), any f : [a, b] → [a, b] must have at least one fixed point (the graph must cross at least once the line x = y).
Brouwer Fixed-Point Theorem
$ Def: Any continuous f : D^n → D^n has (at least) one fixed point (D^n is the n-dimensional ball).
* Special cases: For n = 2, it can be proved using the fundamental group, through the fact that S^1 is not a retract of D^2 (in general, it is done using higher-order homotopy groups, or considering
the index of the vector field v(x):= x − f(x), at different points with respect to different loops).
Lefschetz Fixed-Point Theorem
* Idea: Let S be an n-dimensional manifold (or "almost"), and f a map f : S → S; Then f induces a map f[*] of the homology groups,
f[*]: H[i] → H[i], for i = 1, ..., n,
and from these we can define some numbers n[i], which are something like the trace of f[*], and
L = ∑[i] (−1)^i n[i] , the Lefschetz number ;
The theorem states that this number can also be obtained as a sum of contributions of all the fixed points of f.
* Relationships: It is like a finite (as opposed to infinitesimal) generalization of the concept of Euler characteristic.
* Example: Consider the inversion map on S^2, A: S^2 → S^2; We know that this does not have any fixed points; H[0] = \(\mathbb R\), H[1] = 0, H[2] = \(\mathbb R\), and
L = ∑[i] (−1)^i n[i] = 1 · 1 + (−1) · 0 + 1 · (−1) = 0 ;
If the map had been orientation-preserving, there would have been fixed points and L ≠ 0.
* Example: Every Lorentz transformation fixes at least one null direction.
@ References: van Lon MS-a1509 [quantum mechanical path integral methods, and other index theorems].
Borel Fixed-Point Theorem
$ Def: A connected solvable linear algebraic group over an algebraic closed field, when acting on a complete variety, has a fixed point.
@ References: Borel AM(56).
main page – abbreviations – journals – comments – other sites – acknowledgements
send feedback and suggestions to bombelli at olemiss.edu – modified 6 feb 2016 | {"url":"https://www.phy.olemiss.edu/~luca/Topics/f/fixed_point.html","timestamp":"2024-11-02T14:30:55Z","content_type":"text/html","content_length":"7582","record_id":"<urn:uuid:8279edd4-75fd-4b3e-8687-57df3c17affa>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00131.warc.gz"} |
Bug in Excel's (not VBA's) MOD function
The value 12,345,678,000 is too large to store as a long integer but well
within the range of integers that a double precision floating point real can
store. Excel's worksheet MOD function seems to be able to work with reals,
e.g., MOD(2.25,1/3) returns 0.25 and MOD(2.5,1/3) returns 0.166666666666667.
However, there seems to be a glitch in it when it comes to large integer
values divided by small integer values.
For instance, MOD(12345678000,64) returns #NUM! even though the equivalent
(per online help) expression 12345678000-64*INT(12345678000/64) returns 48
as expected. Clearly Excel's worksheet MOD function isn't just argument
checking wrapped around a call to standard C's fmod(3) call. [No, Microsoft
must have decided they needed to 'enhance' it.]
Maybe this is just a glitch on this particular machine. Does anyone else get
#NUM! from the preceding MOD call? I'm running Excel 97 SR-2 and 2000 SP-3
on this particular machine.
I get #NUM! in XL01 and XLv.X.
J.E. McGimpsey said:
I get #NUM! in XL01 and XLv.X.
So will Microsoft consider this a bug or a feature?
Harlan Grove said:
For instance, MOD(12345678000,64) returns #NUM! . . .
FWIW, the Works 2000 (ver 5) spreadsheet also returns an error, but
OpenOffice Calc 1.0 returns 48 as expected. While I haven't tested this
under other spreadsheets (yet), I think it's safe to say this, er,
functionality is unique to Microsoft.
Win98: Excel2000 - the same result. It looks like the error is returned
whenever quotient reaches the value 134217728
Arvi Laanemets
Excel 2000 (with latest patches) returns #NUM
Lotus 1-2-3 97 returns 48
Lotus 1-2-3 Release 5 returns 48
I agree with your analysis. 134217728 is 2^27, but I have no clue why
2^27 as the integer part of the division (regardless of divisor) would
be a problem.
Arvi said:
Win98: Excel2000 - the same result. It looks like the error is returned
whenever quotient reaches the value 134217728
Arvi Laanemets
The value 12,345,678,000 is too large to store as a long integer but well
within the range of integers that a double precision floating point real
store. Excel's worksheet MOD function seems to be able to work with reals,
e.g., MOD(2.25,1/3) returns 0.25 and MOD(2.5,1/3) returns
However, there seems to be a glitch in it when it comes to large integer
values divided by small integer values.
For instance, MOD(12345678000,64) returns #NUM! even though the equivalent
(per online help) expression 12345678000-64*INT(12345678000/64) returns 48
as expected. Clearly Excel's worksheet MOD function isn't just argument
checking wrapped around a call to standard C's fmod(3) call. [No,
must have decided they needed to 'enhance' it.]
Maybe this is just a glitch on this particular machine. Does anyone else
#NUM! from the preceding MOD call? I'm running Excel 97 SR-2 and 2000 SP-3
on this particular machine.
Jerry W. Lewis said:
I agree with your analysis. 134217728 is 2^27, but I have no clue why
2^27 as the integer part of the division (regardless of divisor) would
be a problem.
The only explanation I can think of - by calculation is the quotient as
intermediate value temporarily stored into some variable with upper limit as
2^27, and it's overflow causes error.
Arvi Laanemets
Arvi said:
Win98: Excel2000 - the same result. It looks like the error is returned
whenever quotient reaches the value 134217728
Arvi Laanemets
The value 12,345,678,000 is too large to store as a long integer but well
within the range of integers that a double precision floating point real
store. Excel's worksheet MOD function seems to be able to work with reals,
e.g., MOD(2.25,1/3) returns 0.25 and MOD(2.5,1/3) returns
However, there seems to be a glitch in it when it comes to large integer
values divided by small integer values.
For instance, MOD(12345678000,64) returns #NUM! even though the equivalent
(per online help) expression 12345678000-64*INT(12345678000/64) returns 48
as expected. Clearly Excel's worksheet MOD function isn't just argument
checking wrapped around a call to standard C's fmod(3) call. [No,
must have decided they needed to 'enhance' it.]
Maybe this is just a glitch on this particular machine. Does anyone else
#NUM! from the preceding MOD call? I'm running Excel 97 SR-2 and 2000 SP-3
on this particular machine.
Given their big "mea culpa" on statistical functions,
I'm hopeful.
OK, so how does one submit a bug report for Excel to Microsoft without having to
call their support number and pay for the priviledge of reporting it?
FWIW, the Works 2000 (ver 5) spreadsheet also returns an error, but
OpenOffice Calc 1.0 returns 48 as expected. While I haven't tested this
under other spreadsheets (yet), I think it's safe to say this, er,
functionality is unique to Microsoft.
And FWIW, Lotus 123 releases 5 and 9.7, Quattro Pro 10 and OpenOffice Calc 1.1
all give the correct/expected result of 48. I'm not going to bother testing
various windows shareware/freeware or Linux spreadsheets.
One way:
Email (e-mail address removed)
make sure Excel is in the subject line so that it can be routed to
the appropriate product managers.
Mac users have a Feedback item in Office v.X's Help menus that take
them to
Don't remember if there's a dedicated site for WinOffice.
Did you get 134217728 from the following Microsoft kb article?
XL: MOD() Function Returns #NUM! Error Value
Dana DeLouis
Using Windows XP & Office XP
= = = = = = = = = = = = = = = = =
Arvi Laanemets said:
Win98: Excel2000 - the same result. It looks like the error is returned
whenever quotient reaches the value 134217728
Arvi Laanemets
Harlan Grove said:
The value 12,345,678,000 is too large to store as a long integer but well
within the range of integers that a double precision floating point real can
store. Excel's worksheet MOD function seems to be able to work with reals,
e.g., MOD(2.25,1/3) returns 0.25 and MOD(2.5,1/3) returns 0.166666666666667.
However, there seems to be a glitch in it when it comes to large integer
values divided by small integer values.
For instance, MOD(12345678000,64) returns #NUM! even though the equivalent
(per online help) expression 12345678000-64*INT(12345678000/64) returns 48
as expected. Clearly Excel's worksheet MOD function isn't just argument
checking wrapped around a call to standard C's fmod(3) call. [No, Microsoft
must have decided they needed to 'enhance' it.]
Maybe this is just a glitch on this particular machine. Does anyone else get
#NUM! from the preceding MOD call? I'm running Excel 97 SR-2 and 2000 SP-3
on this particular machine.
Did you get 134217728 from the following Microsoft kb article?
XL: MOD() Function Returns #NUM! Error Value
I did try searching the KB, but I was lazy and just fed Excel and MOD to basic
search rather than using advanced search restricted to Excel-related articles.
Interesting they don't call this a bug. So I guess it's a feature. Also
interesting that Microsoft seems to believe they needed to improve upon either
the FPREM1 FPU (yes, I'm being very Wintel-centric) instruction's or standard C
fmod(3) call's results. Of course the code for MOD may never have been touched
since the original Mac version was released, and that would have preceded both
wide-spread hardware floating point support and C language standardization, and
it could possibly explain the odd 2^27 value. Was there something magic about
three nine-bit words or maybe 28-bit signed values on really old Macs?
It'd be interesting to find out whether Excel 5/95 and 2003 also work like this.
On Excel 2003 Beta, I got #NUM! with =MOD(12345678000,64)
On Excel 2003 Beta, I got #NUM! with =MOD(12345678000,64)
Seems pretty conclusive that Microsoft considers this a 'feature'.
Harlan Grove said:
Seems pretty conclusive that Microsoft considers this a 'feature'.
I tried it on 2003 commercial version and it returns the same..
I got it by experimentation, and then saw that Arvi had also determined
the same limit. I find it interesting that MS makes no attempt to
explain such an unusual limit.
An additional unusual limit that applies, is that MOD returns #NUM!
regardless of the quotient if the first argument exceeds
2.68873542664192E14 = 2^20+2^19+2^18+2^17+2^15+2^12+2^8+2^6
which is within the range of exact DP representation of whole numbers by
more than an order of magnitude. | {"url":"https://www.pcreview.co.uk/threads/bug-in-excels-not-vbas-mod-function.1017091/","timestamp":"2024-11-08T15:42:28Z","content_type":"text/html","content_length":"167720","record_id":"<urn:uuid:fa84d134-ffad-4ec4-9bff-f497a52ff867>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00753.warc.gz"} |
Out-of-Distribution Detection in Deep Neural Networks
Making AI systems Robust and Reliable
Deep neural networks are often trained with closed-world assumption i.e the test data distribution is assumed to be similar to the training data distribution. However, when employed in real-world
tasks, this assumption doesn’t hold true leading to a significant drop in their performance. Though this performance drop is acceptable for tolerant applications like product recommendations, it is
dangerous to employ such systems in intolerant domains like medicine and home robotics as they can cause serious accidents. An ideal AI system should generalize to Out-of-Distribution (OOD) examples
whenever possible and flag the ones that are beyond its capability to seek human intervention. In this article, we’ll dive deeper into the concept of OOD and explore various techniques for its
detection. This is a self-contained article yet expects familiarity with the basic Machine Learning concepts.
1. A bit on OOD
— Why OOD Detection is important?
— Why Models have OOD brittleness?
— Types of Generalizations
— Plausible Reasons for Higher Robustness of pre-trained models (like BERT) than Traditional Models
— Other Related Problems
2. Approaches to Detect OOD instances
— Maximum Softmax Probability
— Ensembling of Multiple Models
— Temperature Scaling
— Training a Binary Classification model as a Calibrator
— Monte-Carlo Dropout
Check out my publications on the topic of robustness and reliability of NLP systems:
A bit on OOD
The term “distribution” has slightly different meanings for Language and Vision tasks. Consider a dog breed image classification task, here the images of dogs would be in-distribution while images
like bike, ball, etc. would be out-of-distribution.
For Language Tasks, some associate “change in author, writing style, vocabulary, dataset, etc.” with distribution shift while some correlate it with reasoning skill. Example: For a Question-Answering
model trained on Maths questions, a question from History is OOD.
Why is OOD Detection important?
In real-world tasks, the data distribution usually drifts over time, and chasing an evolving data distribution is costly. Hence, OOD detection is important to prevent AI systems from making
prediction errors.
Why Models have OOD brittleness?
1. Neural Network models can rely heavily on spurious cues and annotation artifacts present in the training data while OOD examples are unlikely to contain the same spurious patterns as
in-distribution examples.
2. The training data can’t cover all the facets of a distribution hence limiting the model’s generalization ability.
Types of Generalizations:
• In-Distribution Generalization — Generalization to examples that are novel but drawn from the same distribution as the training set.
• Out-of-Distribution Generalization — Generalization to examples that are drawn from a different distribution than the training set.
Plausible Reasons for Higher Robustness of pre-trained models (like BERT) than Traditional Models:
• Pre-training with diverse data.
• Pre-training with self-supervised learning objectives.
Note: Interested readers can get a quick refresher on BERT here.
Other Related Problems:
1. Success and Error prediction — Predicting whether the model will solve an example correctly or not.
2. Selective Prediction — Making a prediction only for those examples where the model is sufficiently confident. This is especially useful in settings where errors are costly but abstention (not
making a prediction) is acceptable.
3. Domain Adaptation —Extrapolating from training data to test data from a different distribution. We try to generalize a model to new distributions but assume some knowledge about the test
distribution, such as unlabeled examples or a few labeled examples.
There are several other related tasks like outlier detection, anomaly detection, etc.
Approaches to Detect OOD instances:
One class of OOD detection techniques is based on thresholding over the prediction confidence i.e computing prediction confidence and labeling the example as OOD if the confidence is below a certain
threshold. Another class of techniques trains to optimize both the task (classification or regression) and OOD detection simultaneously.
1. Maximum Softmax Probability (MaxProb):
For classification problems, neural network model outputs a vector known as logits. The logits vector is passed through a softmax function to get class probabilities.
Softmax Function
The maximum softmax probability (MaxProb) (i.e maximum softmax across all classes) is used as the prediction confidence. This is one of the simplest yet strong OOD detection techniques. The intuition
behind this approach is:
Correctly classified examples tend to have greater MaxProb than erroneously classified and out-of-distribution examples. In other words, more confident predictions indeed tend to be more
2. Ensembling of Multiple Models:
In Ensemble Learning, multiple models (trained using the same learning algorithm) are used to make predictions for each data point and the decisions from these models are combined to improve the
overall performance. There are various ways of combing decisions:
• Max Voting — The prediction which we get from the majority of the models is used as the final prediction.
• Averaging — Averaging the predictions of all models is straightforward for regression tasks but for classification tasks, we can average the softmax probabilities.
• Weighted Averaging — In this technique, models are assigned different weights and a weighted average is taken to compute the final prediction.
Here, combining the decisions means to compute the prediction confidence from multiple models.
Let’s learn the concept of calibration before proceeding to the remaining approaches.
Calibration is the problem of predicting probability estimates representative of the ground truth correctness likelihood. For this, a model should output is prediction along with the confidence
measure. Let’s try to understand this mathematically.
Let ‘f’ be a neural network with f(x) = (y, p) where y is the prediction and p is the prediction confidence i.e probability of correctness.
We would like p to be calibrated i.e p represents a true probability.
Perfect Calibration:
For instance, if a perfectly calibrated predictor makes 100 predictions each with a confidence of 0.8 then 80 predictions should be correct.
Reliability Diagram:
A reliability diagram is used to represent model calibration. It plots expected sample accuracy as a function of confidence.
Sample Reliability Diagrams. Source: Guo, Chuan, et al.
Examples are grouped into various bins based on the prediction confidence value and the accuracy (i.e ratio of correctly answered examples) for each bin is calculated. Let there be a total of M bins,
Bₘ represents the mᵗʰ bin i.e it contains samples with confidence in the range: ((m-1)/M, m/M].
In case of perfect calibration, the diagram should plot the identity function. Any deviation from a perfect diagonal (up or down) represents miscalibration.
Note: Reliability diagrams do not display the proportion of samples in a given bin (as it just depicts the ratio of correctly answered samples) and thus can not be used to estimate how many samples
are calibrated.
Measuring Calibration Error:
• Expected Calibration Error
where n is the total number of samples.
ECE is a weighted average of the bins’ accuracy/confidence difference.
• Maximum Calibration Error
MCE is the maximum of the bins’ accuracy/confidence difference.
• Negative Log-Likelihood
This is same as the cross-entropy loss.
We discuss other relevant evaluation metrics in the supplementary section.
3. Temperature Scaling
For MaxProb, the prediction confidence is computed using the softmax function. Temperature Scaling is an extension of Platt scaling that uses a single scalar parameter T > 0. Here, the prediction
confidence is computed using a function ‘q’ shown below.
Temperature Scaling
It “softens” the softmax (i.e. raises the output entropy) with T > 1. This makes the network slightly less confident, which makes the confidence scores reflect true probabilities.
• As T → ∞, the probability approaches 1/J, which represents maximum uncertainty.
• With T = 1, the original softmax probability is recovered.
• Parameter T is learned with respect to Negative Log-Likelihood on the validation set.
• Since the parameter T does not change the maximum of the softmax function (i.e the class prediction as the probability of all classes is scaled), temperature scaling does not affect the model’s
4. Training a Binary Classification model as a Calibrator:
It requires evaluating the trained model on a held-out dataset and annotating correctly answered examples as positive and incorrectly answered examples as negative (Note that this step is independent
of the actual label of the examples). Then a binary classification model can be trained on this annotated dataset to predict new examples as belonging to positive or negative class. Though this
approach is more suitable for Success and Error Prediction task, it can be easily modified for OOD detection by incorporating OOD examples while training the calibrator.
5. Monte-Carlo Dropout or Test-Time Dropout
Dropout is a way of preventing overfitting. Usually, dropout is disabled at test time, but it is observed that dropout gives good confidence estimates on OOD data. In this approach, the input is
passed through the network with K different dropout masks. Two statistics are commonly used as confidence measures: Mean and Variance. This is similar to ensembling but with K different dropout
This approach has a few drawbacks:
• It requires access to internal model representations as different dropout masks are used.
• It requires K forward passes of the model, thus increasing runtime by K times.
Other methods like SelectiveNet, ODIN, etc. will soon be added to this article.
Evaluation Metrics
In this section, we’ll describe the popular metrics: AUROC, AUPR, FAR95, and the Risk-Coverage Curve.
AUROC (Area under Receiver Operating Characteristics):
• ROC curve is a graph showing the performance of a classification model at all classification thresholds.
• It’s a plot of False Positive Rate (x-axis) and True Positive Rate (y-axis).
• Measures how much model is capable of distinguishing between classes.
• Typically used for binary classification problems but can be extended for multi-class settings as well.
Demonstrating ROC Curve. Source: Glass Box.
This curve (especially its name) seems daunting but let’s dive deeper and try to understand it.
• A higher X-axis value indicates a higher number of False positives than True negatives.
• A higher Y-axis value indicates a higher number of True positives than False negatives.
• There is a point for every classification threshold on this curve. (assume that the model labels all the instances with a score above the threshold as positive (class 1) while the ones below as
negative (class 0)). Hence, lowering the classification threshold classifies more items as positive, thus increasing both False Positives and True Positives.
• AUC provides an aggregate measure of performance across all possible classification thresholds.
• AUC value ranges from 0 to 1. A model whose predictions are 100% wrong has an AUC of 0 while the one whose predictions are 100% correct has an AUC of 1.
• The curves of different models can be compared directly in general or for different thresholds. Higher the AUC, better the model is at predicting 0s as 0s and 1s as 1s or negatives as negative
and positives as positive.
Hence, Blue > Yellow > Green. , we can compare different points for the same model (i.e different thresholds) depending on how many false positives and false negatives we want to tolerate for our
• AUC close to 1 → the model has a good measure of separability.
• AUC of 0 → the model is reciprocating the result i.e predicting 0s as 1s and 1s as 0s.
• AUC of 0.5 → the model has no discrimination capacity to distinguish between positive class and negative class.
• All points above the diagonal line (Random Classifier — red line) correspond to the situation where the proportion of correctly classified points belonging to the Positive class is greater than
the proportion of incorrectly classified points belonging to the Negative class.
AUROC is not ideal when the positive class and negative class have greatly differing base rates, and the AUPR adjusts for these different positive and negative base rates.
AUPR (Area under Precision-Recall Curve):
• ROC curves are appropriate when the observations are balanced between each class, whereas precision-recall curves are appropriate for imbalanced datasets.
PR curve. Source: towardsdatascience.com
• A no-skill classifier is one that cannot discriminate between the classes and would predict a random class or a constant class in all cases. The no-skill line changes based on the distribution of
the positive to negative classes. It is a horizontal line with the value of the ratio of positive cases in the dataset. For a balanced dataset, this is 0.5.
FAR95 (False Alarm Rate):
• The FAR95 is the probability that an in-distribution example raises a false alarm, assuming that 95% of all out-of-distribution examples are detected.
• Hence a lower FAR95 is better.
Risk-Coverage Curve:
A popular metric in selective answering literature.
Sample Risk-Coverage Curve.
It’s a plot between Coverage (x-axis) and Risk (y-axis). Coverage is the fraction of examples the model makes a prediction on while Risk is the error on that fraction of examples. For any choice of
threshold, a model has an associated coverage and risk. Area under the curve (AUC) is used to compare the performance of different models.
Check out my related articles:
1. Hendrycks, Dan, and Kevin Gimpel. “A baseline for detecting misclassified and out-of-distribution examples in neural networks.”
2. McCoy, R. Thomas, Junghyun Min, and Tal Linzen. “Berts of a feather do not generalize together: Large variability in generalization across models with similar test set performance.”
3. Hendrycks, Dan, et al. “Pretrained transformers improve out-of-distribution robustness.”
4. Geifman, Yonatan, and Ran El-Yaniv. “Selectivenet: A deep neural network with an integrated reject option.”
5. Guo, Chuan, et al. “On calibration of modern neural networks.” | {"url":"https://medium.com/analytics-vidhya/out-of-distribution-detection-in-deep-neural-networks-450da9ed7044?sk=ee83987a0af68ad8202afe15676710fc","timestamp":"2024-11-15T05:05:41Z","content_type":"text/html","content_length":"225402","record_id":"<urn:uuid:ef35297f-8bc0-44ff-892f-539cbca9c0fb>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00662.warc.gz"} |
OS2080: Probability and Statistics I - NPS Online
Probability and Statistics I
Course #OS2080
Est.imated Completion Time: 3 months
Fundamentals of probability and statistics useful in military modeling. Topics include probability laws and calculation methods, conditional probability, Bayes' Theorem, discrete and continuous
random variables, the binomial, geometric, Poisson, exponential, and normal distributions, expectation, variance, and covariance, confidence intervals, hypothesis testing, and simple linear
regression. Emphasis is on understanding uncertainty and developing computational skills for military systems analysis.
Included in degrees & certificates
Learning Outcomes
· Learn basic probability concepts and counting techniques. Compare and contrast probability and statistics. Present counting techniques, permutations, combinations, Venn diagrams and
conditional probability, independence, disjoint events, law of total probability and Bayes’ theorem.
· Use concept of discrete and continuous random variables to discuss probability mass functions, probability density functions, cumulative distribution functions, expected value and variance. Learn
about joint and marginal distributions, condition distributions and expectation for jointly distributed random variables, covariance and correlation.
· Specific distributions discussed are Bernoulli, binomial, multinomial, geometric, Poisson discrete distributions. Continuous distributions discussed are uniform, exponential, gamma, normal,
chi-squared, F and t distributions.
· Discuss sampling distributions, central limit theorem, statistical terminology for location (mean, median, trimmed mean), for variability (variance, standard deviation, range) and
categorical data (mode, sample proportions, indicator variables).
· Use numerical, graphical and tabular summaries to describe data. Find point estimates using method of moments and maximum likelihood.
· Find standard error for sample data then construct confidence and prediction intervals for population parameters using sample data. Also conduct suitable one- or two-sided hypotheses tests and also
paired hypotheses tests and tests for proportions.
Offerings database access
Application Deadlines
No upcoming deadlines.
Academic Calendar
No upcoming events. | {"url":"https://online.nps.edu/-/os2080-probability-and-statistics-i","timestamp":"2024-11-12T08:47:41Z","content_type":"text/html","content_length":"253890","record_id":"<urn:uuid:8ce71c30-fbab-4652-aba0-936d90fb8fef>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00185.warc.gz"} |
The Quantum Mechanical Oscillator as a Possible Source of 1/f Fluctuations
We investigate consecutive absorption or emission of photons of the quantum mechanical harmonic oscillator as a possible source of 1/f fluctuations. Separating the absorption and emission process, we
show that consecutively absorbed or emitted photons give rise to an intermittent stochastic process; thereby fluctuating clusters of photons are intermitted by distinct breaks. Let the number of
photons in a cluster be m and the cluster size distribution be pm. We find that the intermittent process with a cluster size distribution pm proportional to m-2 generates a pure 1/f spectrum. We show
that 1/f fluctuations are present in thermal equilibrium but average out to zero. As an example we investigate phonons as a possible origin of 1/f fluctuations in an extrinsic semiconductor. Acoustic
phonons always produce a change in the volume; this affects the donor ionization energy modulating also the g-r process. We calculate the spectrum of such a modulated g-r process; thereby the
intermittent character of phonon activity is identified as the origin of 1/f fluctuations. The Hooge parameter is found to depend on the mean number of phonons comprised in a cluster, on the
modulation depth of the g-r process and on a factor which describes the temperature dependence of 1/f noise.
arXiv e-prints
Pub Date:
January 2012
□ Physics - Data Analysis;
□ Statistics and Probability
12 pages, 4 figures This paper has been withdrawn by the author due to an erroneous application of the equipartition theorem in the absorption and emission processes | {"url":"https://ui.adsabs.harvard.edu/abs/2012arXiv1201.4162G/abstract","timestamp":"2024-11-12T16:22:33Z","content_type":"text/html","content_length":"38711","record_id":"<urn:uuid:dd9d4b64-2a26-46ae-ba22-746075cd3ad6>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00378.warc.gz"} |
The mean of ungrouped data and the mean calculated when the same data is grouped are always the same. Do you agree with this statement? Give a reason for your answer
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
The mean of ungrouped data and the mean calculated when the same data is grouped are always the same. Do you agree with this statement? Give reason for your answer
When the mean of grouped data is calculated, it is assumed that the frequency of each class is centred at the mid point
So the two values of the mean, i.e. from grouped and ungrouped data, are the same.
Therefore, the statement is not true.
The mean of ungrouped data and the mean calculated when the same data is grouped are always the same. Do you agree with this statement? Give reason for your answer.
✦ Try This: Someone is asked to take a number from 1 to 30. The probability that it is a prime is
Probability can be defined as the ratio of the number of favorable outcomes to the total number of outcomes of an event.
We know that
Total number of outcomes = 30
Prime numbers between 1-30 are
2, 3, 5, 7, 11, 13, 17, 19, 23 and 29
So the probability that it is a prime = Favourable outcomes/ Total outcomes
Substituting the values
= 10/30
= 1/3
Therefore, the probability that it is a prime is 1/3.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 14
NCERT Exemplar Class 10 Maths Exercise 13.2 Sample Problem 1
The mean of ungrouped data and the mean calculated when the same data is grouped are always the same. Do you agree with this statement? Give a reason for your answer
The statement “The mean of ungrouped data and the mean calculated when the same data is grouped are always the same” is not true
☛ Related Questions:
Math worksheets and
visual curriculum | {"url":"https://www.cuemath.com/ncert-solutions/the-mean-of-ungrouped-data-and-the-mean-calculated-when-the-same-data-is-grouped-are-always-the-same-do-you-agree-with-this-statement-give-a-reason-for-your-answer/","timestamp":"2024-11-03T07:24:07Z","content_type":"text/html","content_length":"203256","record_id":"<urn:uuid:4bd0df3e-9cc4-49ea-aa88-de832a5510d8>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00392.warc.gz"} |
Introduction to Binary Tree
1. What is Binary Tree?
A binary tree is a tree-type non-linear data structure with a maximum of two children for each parent. Every node in a binary tree has a left and right reference along with the data element. The node
at the top of the hierarchy of a tree is called the root node. The nodes that hold other sub-nodes are the parent nodes.
A parent node has two child nodes: the left child and the right child. Hashing, routing data for network traffic, data compression, preparing binary heaps, and binary search trees are some of the
applications that use a binary tree.
2. Terminologies related to a Binary Tree
• Node: A node is a basic unit of a data structure, such as a linked list or tree data structure. Nodes contain data and also may link to other nodes.
• Root: A tree’s topmost node.
• Parent: Each node (apart from the root) in a tree that has at least one sub-node of its own is called a parent node.
• Child: A node that straightway came from a parent node when moving away from the root is the child node.
• Leaf Node: These are external nodes. They are the nodes that have no children.
• Internal Node: As the name suggests, these are inner nodes with at least one child.
• Depth of a Tree: The depth of a node in a binary tree is the total number of edges from the root node to the target node. Similarly, the depth of a binary tree is the total number of edges from
the root node to the most distant leaf node.
• Height of a Tree: It is the number of edges from the node to the deepest leaf. The tree height is also considered the root height.
3. Types of Binary Tree
• Full Binary Tree: A full Binary tree is a special type of binary tree in which every parent node/internal node has either two or no children.
• Complete Binary Tree: A complete binary tree is just like a full binary tree but with the following major differences-
1. Every level must be completely filled
2. All the leaf elements must lean towards the left.
3. The last leaf element might not have a right sibling i.e. a complete binary tree doesn't have to be a full binary tree.
• Perfect Binary Tree: A perfect binary tree is a type of binary tree in which every internal node has exactly two child nodes and all the leaf nodes are at the same level
• Balanced Binary Tree: It is a type of binary tree in which the difference between the height of the left and the right subtree for each node is either 0 or 1.
• Degenerate or Pathological Binary Tree: A degenerate or pathological tree is a tree having a single child either left or right. This means that the tree will behave like a linked list data
Interesting Fact: The height of a Degenerate Binary Tree is equal to the total number of nodes in that tree.
• Skewed Binary Tree: A skewed binary tree is a pathological/degenerate tree in which the tree is either dominated by the left nodes or the right nodes. Thus, there are two types of skewed binary
tree: left-skewed binary tree and right-skewed binary tree.
In the next blog, we'll discuss the basic operations of a binary tree. | {"url":"https://safiulkabir.com/introduction-to-binary-tree","timestamp":"2024-11-07T12:29:34Z","content_type":"text/html","content_length":"118090","record_id":"<urn:uuid:6aed0f50-a482-41ba-8125-167d3e0fef0e>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00416.warc.gz"} |
Journal of the European Optical Society-Rapid Publications
Issue J. Eur. Opt. Society-Rapid Publ.
Volume 19, Number 1, 2023
EOSAM 2022
Article Number 15
Number of page(s) 5
DOI https://doi.org/10.1051/jeos/2023011
Published online 07 April 2023
J. Eur. Opt. Society-Rapid Publ. 2023,
, 15
Short Communication
Scalable sub-cycle pulse generation by soliton self-compression in hollow capillary fibers with a decreasing pressure gradient
Grupo de Investigación en Aplicaciones del Láser y Fotónica, Departamento de Física Aplicada, Universidad de Salamanca, 37008, Salamanca, Spain
^* Corresponding author: marinafergal@usal.es
Received: 27 January 2023
Accepted: 13 March 2023
Advances in the generation of the shortest optical laser pulses down to the sub-cycle regime promise to break new ground in ultrafast science. In this work, we theoretically demonstrate the potential
scaling capabilities of soliton self-compression in hollow capillary fibers with a decreasing pressure gradient to generate near-infrared sub-cycle pulses in very different dispersion and
nonlinearity landscapes. Independently of input pulse, gas and fiber choices, we present a simple and general route to find the optimal self-compression parameters which result in high-quality
pulses. The use of a decreasing pressure gradient naturally favors the self-compression process, resulting in shorter and cleaner sub-cycle pulses, and an improvement in the robustness of the setup
when compared to the traditional constant pressure approach.
Key words: Ultrafast nonlinear optics / Hollow capillary fibers / Soliton-self compression / Sub-cycle pulses
© The Author(s), published by EDP Sciences, 2023
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
In a continuous effort to access the briefest and most fundamental phenomena in nature, intense ultrashort laser pulses have become indispensable tools for ultrafast science, breaking new ground in
time-resolved spectroscopy and strong-field physics [1, 2]. At present, few-cycle femtosecond pulses in the optical spectral region are routinely generated by nonlinear post-compression in gas-filled
hollow capillary fibers (HCFs) [3], which stand out among other compression schemes for their simplicity, high-damage threshold and the possibility of tuning their nonlinearity and dispersion by
modifying the filling gas or its pressure. However, although HCF post-compression experiments have been greatly optimized [4], they are currently reaching their limit in terms of the shortest
achievable pulse duration due to the complexity of dealing with uncompensated high-order dispersion in octave-spanning spectra. Overcoming this problem, parametric light-field synthesizers have
succeeded in generating the shortest optical laser pulses well down into the sub-cycle regime, offering new opportunities for advancing real-time observation and precision control of electron
dynamics at the atomic scale [5, 6]. As a promising alternative to these extremely complex systems, high-energy soliton dynamics in HCFs is attracting a great interest as a direct route to extreme
pulse self-compression down to the sub-cycle regime [7, 8]. As opposed to more conventional post-compression techniques, soliton self-compression relies on the simultaneous nonlinear spectral
broadening and phase compensation arising from the interplay between the negative group-velocity dispersion (GVD) of the waveguide and self-phase modulation (SPM). Still, for a practical
implementation, the complexity of this nonlinear interaction calls for theoretical investigations on scaling rules and design guidelines to identify the optimal experimental parameters which result
in high-quality self-compression [9]. So far, extreme soliton self-compression in HCFs in the near-infrared (NIR) has only been demonstrated for pre-compressed (~10 fs) pump pulses [7, 10], or
otherwise in configurations which ensure a strong anomalous response, i.e., working with longer wavelengths [8, 11], high-order modes [12], or small-core photonic crystal fibers [9, 13]. However, the
unexpected applicability of sub-cycle self-compression to standard experimental setups driven by NIR multi-cycle pulses propagating in the fundamental mode of large-core HCFs has been recently
demonstrated in negatively pumped fibers, i.e., HFCs filled with a decreasing pressure gradient [14]. Pressure gradients are routinely implemented by sealing the fiber into a gas cell at each end,
which can be independently evacuated and filled with gas, yielding a longitudinal pressure distribution with a square root-type profile when the gas flows from highest to lowest pressure [10].
In this communication we further investigate the scalability of soliton self-compression down to the sub-cycle regime in HCFs filled with a decreasing pressure gradient. Varying the filling gas and
choosing between atomic (Ne) and molecular (N[2]) species, we study the compression process in two completely different dispersion and nonlinearity landscapes, which are of critical importance in
soliton dynamics. Our results demonstrate that nearly identical self-compression performance can be achieved in very distinct HCF scenarios, and provide a surprisingly simple universal route to find
the optimal parameters for generating high-quality NIR sub-cycle pulses.
Our work is based on one-dimensional numerical simulations of nonlinear pulse propagation [15, 16], including the complete linear response of the gas-filled HCF [17], SPM, stimulated Raman scattering
modeled in a damped harmonic oscillator approximation [18], and self-steepening. This theorical model accurately describes ultrashort pulse propagation down to the single-cycle limit in a regime of
moderate intensities [19].
In order to identify a route towards high-quality self-compression, we have followed the procedure detailed in [14]. In brief, we have systematically simulated the propagation of a transform-limited
30 fs gaussian pump pulse at 800 nm in the fundamental mode of a 3 m long, 100 μm core radius HCF, filled with either Ne or N[2], and with both constant gas pressure and a longitudinal decreasing
pressure gradient ending in vacuum. The latter two situations are fairly compared by matching the integrated nonlinear phase shift acquired by the pulse peak during its propagation, which is often
referred to as B-integral. Neglecting the fiber losses, a system with a decreasing pressure gradient from p[0] to vacuum can then be compared to that with a constant pressure p[eq] simply if p[0] =
(3p[eq])/2 [10, 12]. The main difference lies in that non-uniform pressure allows for a dynamic tuning of the dispersion and nonlinearity experienced by the pulse during its self-compression, as the
propagation constant and the nonlinear parameter scale linearly with the gas density and, thus, with pressure.
For the parameters considered in our study, Figure 1 shows the GVD and instantaneous nonlinear coefficient (related only to instantaneous Kerr effect or SPM [15]) of a HCF filled with Ne or N[2] at
different pressures. As we can see, a Ne-filled HCF has a weaker instantaneous nonlinearity and displays anomalous dispersion (GVD < 0) over a larger pressure range than an identical fiber filled
with N[2]. An inner fiber radius of 100 μm was chosen because it offers a good balance between acceptable losses at 800 nm and a sufficiently strong anomalous response in N[2]. Furthermore, a pulse
propagating in N[2] might experience a delayed molecular contribution to the optical Kerr effect which vanishes in noble gases like Ne. Therefore, owing to their very distinct linear and nonlinear
nature, the optimal self-compression in either Ne or N[2] is expected to occur for different input pulse and fiber parameters.
Fig. 1
GVD (top) and instantaneous nonlinear coefficient (bottom) at 800 nm of the fundamental mode of a 100 μm core radius HCF filled with Ne or N[2] as a function of gas pressure. Labels indicate the
zero-dispersion pressure (p[ZD]) in each case.
Following [14], we have simulated the soliton self-compression of the aforementioned 30 fs pulse while varying its initial energy and the equivalent gas pressure in the HCF. For each energy-pressure
pair in the resulting bi-dimensional parameter space, we have plotted the intensity full width at half-maximum (FWHM) duration and the ratio of output to input peak power of the self-compressed
pulses, as shown in Figure 2. In these plots, the optimal region for high-quality self-compression can be readily identified as the intersection between the areas of shortest output pulse duration
and largest peak power enhancement. Surprisingly, the results for Ne and N[2] show an identical behavior, which also follows that previously reported for Ar [14], except for the fact that they are
displaced to different input energy and gas pressure ranges as mentioned earlier. In both cases, it is clear that, in the whole parameter ranges considered here, the self-compression process is
substantially enhanced when the fiber is negatively pumped rather than statically filled, resulting in the generation of self-compressed pulses with extremely short durations well down into the
sub-cycle regime (~1 fs) and high peak powers, which in turn implies a clean temporal profile. The most outstanding feature is that there is not just a single pair of input energy and gas pressure
values that allow for a high-quality compression, but there is a whole parameter region which yields similar results. This optimal region is found to always appear towards the same corner of the
contour line in the energy-pressure map where the fixed fiber length (L) matches an average compression length (L[av]), that we defined as:$L av = L sc + L fiss 2 ,$(1)where L[sc] and L[fiss]
represent, respectively, the characteristic self-compression and soliton fission lengths, which are given by [20]:$L fiss = L D N ; L sc = L fiss 2 .$(2)
Fig. 2
FWHM duration (top row) and ratio of output to input peak power (bottom row) of the self-compressed pulses as a function of the input energy and the equivalent constant pressure (see text) in both
a statically filled or a negatively pumped 3 m long, 100 μm core radius HCF filled with Ne (left) or N[2] (right). The solid black lines represent the contour lines where L = L[av], which run along
the optimal region for self-compression in a decreasing pressure gradient. Note the one order of magnitude change in the pressure range from Ne to N[2] owing to their different dispersion and
N = (L[D]/L[NL])^1/2 being the soliton order, and $L D = T p 2 / ( 4 ln 2 | β 2 | )$ the dispersion and $L NL = 1 ( γ i P 0 ) = π ( 4 ln 2 ) T p ( γ i E 0 )$ the nonlinear lengths, which describe the
characteristic length scales of GVD and SPM, respectively. Here T[p] represents the intensity FWHM duration of the gaussian pump pulse, P[0] is its input peak power and E[0] its initial energy, β[2]
is the GVD coefficient of the HCF, and γ[i] is the instantaneous nonlinear parameter as defined elsewhere [15]. The constraint L = L[av] ensures that L[sc] < L <L[fiss] and, therefore, guarantees
that the self-compressing pulse reaches its maximum compression without entering in the soliton fission regime. In addition, the soliton order should be kept N < 15 to achieve a high-quality
compression [13], inevitably setting an upper limit to the achievable pulse energy. Independently of input pulse, gas and fiber parameters, the condition L = L[av] always describes a contour line in
the energy-pressure plane which, when falling inside the space with N < 15, can be used to identify the optimal region for high-quality self-compression in a universal way. A detailed inspection of
the conditions L = L[av] and N < 15 suggests that upscaling our results towards millijoule pump pulses should become possible, even in practical short fibers (~1 m), by pushing the central wavelength
into the mid-infrared spectral region.
As an example of the high-quality sub-cycle waveforms that can be generated from the negatively pumped fiber, in Figure 3 we have plotted the self-compressed pulses obtained for two different pairs
of input pulse energy and equivalent gas pressure which lie towards the same area of the optimal self-compression regions in Figure 2, corresponding to Ne and N[2], respectively. When compared to the
output pulses in the equivalent constant pressure situations, it is clear that those generated with a decreasing gradient are much better, displaying shorter durations, higher peak powers, a cleaner
temporal profile with a higher contrast, and a broader spectrum spanning from the NIR to the mid-ultraviolet. The self-compressed pulses from the negatively pumped HCF reach sub-cycle FWHM durations
of 1.1 and 1.2 fs, and output peak powers of 8.8 and 10.7 GW in Ne and N[2], respectively. However, in the equivalent constant pressure situations, the output pulses where only 2.2 and 2.3 fs in
duration, and 4.6 and 5.4 GW in peak power. The improvement with the decreasing pressure gradient has been attributed to an effective suppression of higher-order dispersion and self-steepening in the
last stages of the pulse compression, together with a continuous blue-shift of the zero-dispersion frequency at the same time as the pulse spectrum broadens by SPM [12, 14]. When propagating in the
anomalous dispersion regime, it is straightforward to understand from the trends shown in Figure 1 that a decreasing pressure gradient is the most natural way to emphasize and favor the
characteristic dynamics of the self-compression process [11]. In short, at the fiber entrance the higher pressure enhances the accumulation of nonlinear phase shift and the spectral broadening of the
input pulse by SPM. In later stages, the larger spectral extent combined with an increase in the magnitude of the anomalous GVD and a reduction of third-order dispersion due to the drop in pressure,
assist the phase compensation for pulse self-compression and delay the fission process beyond the maximum compression point that can be reached with constant pressure. Altogether, the decreasing
pressure gradient enables unprecedent compression ratios (≿25) which had remained out of reach due to detrimental high-order effects [13]. Furthermore, the great similarities between the pulses in
Figure 3, generated with different gases, energies and pressures, demonstrates the promising scaling capabilities of HCF self-compression down to the sub-cycle regime in different configurations.
Another interesting point is that the optimal self-compressed pulse is accompanied by the onset of resonant dispersive wave (RDW) emission, when the strongly nonlinear self-compressing soliton
transfers its excess energy to a linear wave propagating in the normal dispersion regime [7, 8, 10, 16]. This is manifested by the isolated peak around 200 nm in the output spectra of the lower
panels in Figure 3. At this point, RDW emission has just started and the energy transfer to the ultraviolet is still low, resulting in conversion efficiencies quite below saturation. This fact could
be used to experimentally predict the best sub-cycle pulse parameters based on RDW spectral content at the fiber output.
Fig. 3
Temporal intensity profile (top row) and spectrum (bottom row) of the self-compressed sub-cycle pulses obtained after propagation through a HCF filled with Ne (left) or N[2] (right), at both
constant or decreasing pressure, for two different pairs of input pulse energy and equivalent gas pressure which lie towards the same area of the optimal self-compression regions in Figure 2.
In summary, we have demonstrated that broadly similar high-quality NIR sub-cycle pulses can be generated by extreme soliton self-compression in negatively pumped HCFs in different configurations.
Independently of input pulse, gas and fiber choices, the optimal self-compression parameters can always be found by matching the fiber length to and average compression length, providing a simple
design guideline for experiments. Furthermore, the decreasing pressure gradient can help to improve the robustness of HCF self-compression and the quality of the generated sub-cycle pulses when
compared to the equivalent constant pressure situations, also preventing the onset of undesirable high-order effects. We believe that these findings will pave the way towards a new generation of
ultrafast experiments which might benefit from the availably of tailored sub-cycle waveforms, especially those which are carried out in vacuum chambers, like the synthesis of high-frequency isolated
attosecond pulses through high-order harmonic generation.
Conflict of interest
The authors declare no conflict of interest.
Authors acknowledge financial support from Ministerio de Ciencia e Innovación under grant PID2019-106910GB-I00 funded by MCIN/AEI/ 10.13039/501100011033. M.F.G. acknowledges support from Ministerio
de Universidades under grant FPU21/02916.
All Figures
Fig. 1
GVD (top) and instantaneous nonlinear coefficient (bottom) at 800 nm of the fundamental mode of a 100 μm core radius HCF filled with Ne or N[2] as a function of gas pressure. Labels indicate the
zero-dispersion pressure (p[ZD]) in each case.
In the text
Fig. 2
FWHM duration (top row) and ratio of output to input peak power (bottom row) of the self-compressed pulses as a function of the input energy and the equivalent constant pressure (see text) in both
a statically filled or a negatively pumped 3 m long, 100 μm core radius HCF filled with Ne (left) or N[2] (right). The solid black lines represent the contour lines where L = L[av], which run along
the optimal region for self-compression in a decreasing pressure gradient. Note the one order of magnitude change in the pressure range from Ne to N[2] owing to their different dispersion and
In the text
Fig. 3
Temporal intensity profile (top row) and spectrum (bottom row) of the self-compressed sub-cycle pulses obtained after propagation through a HCF filled with Ne (left) or N[2] (right), at both
constant or decreasing pressure, for two different pairs of input pulse energy and equivalent gas pressure which lie towards the same area of the optimal self-compression regions in Figure 2.
In the text
Current usage metrics show cumulative count of Article Views (full-text article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on
Vision4Press platform.
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Proofs of Standard Integrals - Maths Academy
Today’s resource contains proofs of standard integrals. You can download a complete sheet of the standard rules if you are a site member.
In mathematics, an integral is a way to describe a continuous quantity that can be represented as the area under a curve. It is a good idea to become familiar with the basic integrals as these can be
directly quoted, especially when tackling harder questions.
Basic Integration Rules
The first rule that you will probably learn is:
\int x^n d x=\frac{x^{n+1}}{n+1}+c \quad \text { where } n \neq-1
In the case above, the value of n cannot be equal to -1. If n = -1 then we would be dividing by zero in the result.
We can extend the above result and consider the following:
& \int a x^n \space d x=\frac{a x^{n+1}}{n+1}+c \quad \text { where } n \neq-1 \
Another typical rule that is taught fairly early is:
& \int \frac{1}{x} \space d x=\ln\left|x\right|+c
You may see this written as:
& \int \frac{dx}{x}=\ln\left|x\right|+c
Trigonometric Integrals
Before learning integration, you should have been taught differentiation. The following rules should, therefore, be self-explanatory. Some of the first rules you will learn are:
& \int \sin x \space d x=-\cos x+c \
& \int \cos x \space d x=\sin x+c
When first learning integration, you may be shown that coefficients can be taken outside the integral as follows:
\int 3 \sin x \space d x & =3 \int \sin x \space d x \\\\
& =3(-\cos x)+c \\\\
& =-3 \cos x+c
\int -5 \sin x \space d x & =-5 \int \sin x \space d x \\\\
& =-5(-\cos x)+c \\\\
& =5 \cos x+c
From your knowledge of differentiation, you should also be aware of the following rules:
\int \sec^2 x \space d x &=\tan x+c\\\\\\
\int \csc^2 x \space d x &=-\cot x+c\\\\\\
\int \sec x\tan x \space d x &=\sec x + c\\\\\\
\int \csc x\cot x \space d x &=-\csc x + c\\\\\\
Integration of \(\tan x\)
The next integral is derived through substitution:
\int \tan x \space d x & =\int \frac{\sin x}{\cos x} \space d x \\\\
\text { Let } u & =\cos x \\\\
\frac{d u}{d x} & =-\sin x \\\\
d u & =-\sin x \space d x
Now substitute
\int \frac{\sin x}{\cos x} \space d x & =-\int \frac{-\sin x}{\cos x} \space d x \\\\
& =-\int \frac{d u}{u} \\\\
& =-\ln |u|+c \\\\
& =-\ln |\cos x|+c
Note that there is an alternative form for this result:
\int \tan x \space d x & =-\ln |\cos x|+c\\\\
&=\ln |(\cos x)^{-1}|+c \quad \quad \text {by applying law of logs}\\\\
&=\ln \left|\frac{1}{\cos x}\right|+c\\\\
&=\ln |\sec x|+c
Integration of \(\cot x\)
You can use a similar method to the one above:
\int \cot x \space d x & =\int \frac{\cos x}{\sin x} \space d x \\\\
\text { Let } u & =\sin x \\\\
\frac{d u}{d x} & =\cos x \\\\
d u & =\cos x \space d x
Now substitute
\int \frac{\cos x}{\sin x} \space d x & =\int \frac{du}{u} \space d x \\\\
& =\ln |u|+c \\\\
& =\ln |\sin x|+c
Integration of \(\sec x\)
The first step involves multiplying by \(\sec x + \tan x \). Since this is done in the numerator and denominator, the result is equivalent to multiplying by 1 i.e. we are not changing the integral’s
\int \sec x \space d x&=\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \space d x \\\\
&=\int \frac{\sec ^2 x+\sec x \tan x}{\sec x+\tan x} \space d x \\\\
\text { Let } u&=\sec x+\tan x \\\\
\frac{d u}{d x}&=\sec x \tan x+\sec ^2 x \\\\
d u&=\left(\sec x \tan x+\sec ^2 x\right) d x \\\\
\int \frac{\sec ^2 x+\sec x \tan x}{\sec x+\tan x} \space d x&=\int \frac{d u}{u} \\\\
&= \ln |u|+c \\\\
&=\ln |\sec x+\tan x|+C
Integration of \(\csc x\)
We can use a similar method as above. You really need to know your rules for differentiation in order to follow the steps.
\int \csc x \space d x&=\int \frac{\csc x(\csc x+\cot x)}{\csc x+\cot x} \space d x \\\\
& =\int \frac{\csc ^2 x+\csc x \cot x}{\csc x+\cot x} \space d x \\\\
\text { Let } u&=\csc x+\cot x \\\\
\frac{d u}{d x}&=-\csc x \cot x-\csc ^2 x \\\\
d u&=-\left(\csc x \cot x+\csc ^2 x\right) d x \\\\
\int \frac{\csc ^2 x+\csc x \cot x}{\csc x+\cot x} \space d x&=-\int \frac{d u}{u} \\\\
& =-\ln |u|+c \\\\
& =-\ln |\csc x+\cot x|+c
Integration Rules Questions
If you need to practise some basic integration questions then try the integration worksheets with 100 questions and answers. | {"url":"https://www.mathsacademy.com.au/proofs-of-standard-integrals/","timestamp":"2024-11-08T02:13:56Z","content_type":"text/html","content_length":"81681","record_id":"<urn:uuid:1a963e74-1f65-4f54-bf4e-e5eb9d1e85f3>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00288.warc.gz"} |
What is extended euclidean algorithm
extended euclidean algorithm
n. An extension to the Euclidean algorithm, which computes the coefficients of Bézout's identity in addition to the greatest common divisor of two integers.
Extended Euclidean algorithm
In arithmetic and computer programming, the extended Euclidean algorithm is an extension to the Euclidean algorithm, which computes, besides the greatest common divisor of integers a and b, the
coefficients of Bézout's identity, that is integers x and y such that
It allows one to compute also, with almost no extra cost, the quotients of a and b by their greatest common divisor.
Extended Euclidean algorithm also refers to a very similar algorithm for computing the polynomial greatest common divisor and the coefficients of Bézout's identity of two univariate polynomials.
The extended Euclidean algorithm is particularly useful when a and b are coprime, since x is the modular multiplicative inverse of a modulo b, and y is the modular multiplicative inverse of b modulo
a. Similarly, the polynomial extended Euclidean algorithm allows one to compute the multiplicative inverse in algebraic field extensions and, in particular in finite fields of non prime order. It
follows that both extended Euclidean algorithms are widely used in cryptography. In particular, the computation of the modular multiplicative inverse is an essential step in RSA public-key encryption | {"url":"https://findwords.info/term/extended%20euclidean%20algorithm","timestamp":"2024-11-10T15:33:51Z","content_type":"text/html","content_length":"12592","record_id":"<urn:uuid:f1cde931-15ac-4c03-abc9-2ac2a40666bf>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00008.warc.gz"} |
What Is The Cost Of Debt? - It Business mind
What Is The Cost Of Debt?
What Is The Cost Of Debt?
Cost Of Debt: A company’s cost of debt is the effective interest rate a company pays on its debt obligations, including bonds, mortgages, and any other forms of debt the company may have. Because
interest expense is deductible, it’s generally more useful to determine a company’s after-tax cost of debt. along with the cost of equity, which makes up a company’s cost of capital.
debt can be useful when assessing a company’s credit situation, and when combined with the size of the debt, it can be a good indicator of overall financial health. For instance, $1 billion in debt
at 3% interest is actually less costly than $500 million at 7%, so knowing both the size and cost of a company’s debt can give you a clearer picture of its financial situation.
What Is The Cost Of Debt?
The cost of debts is the return that a company provides to its debtholders and creditors. These capital providers need to be compensated for any risk exposure that comes with lending to a company.
Since observable interest rates play a big role in quantifying it is relatively more straightforward to calculate the cost of debt than the cost of equity. Not only does the cost debt, as a rate,
reflect the default risk of a company, it also reflects the level of interest rates in the market. In addition, it is an integral part of calculating a company’s Weighted Average Cost of Capital or
Cost Of Debt Formula
The cost of debts is the minimum rate of return that the debt holder will accept for the risk taken. The cost of debt is the effective interest rate that the company pays on its current liabilities
to the creditor and debt holders. Generally, it is referred to as the after-tax cost of debts. The difference between the before-tax cost of debt and the after-tax cost of debt is depended on the
fact that interest expenses are deductible. It is an integral part of WACC i.e. weight average cost of capital. The cost of capital of the company is the sum of the cost of debt plus the cost of
equity. And Cost of debt is 1 minus tax rate into interest expense.
How To Calculate Cost Of Debt
In order to calculate a company’s cost debts, you’ll need two pieces of information: the effective interest rate it pays on its debt and its marginal tax rate.
Many companies publish their average debt interest rate, but if not, it’s fairly easy to calculate using the company’s financial statements. On the income statement, you can find the total interest
the company paid (note: If you’re looking at a quarterly income statement, multiply this figure by four in order to annualize the data). Then, on the balance sheet, you can find the total amount of
debt the company is carrying. Divide the annual interest by total debt and then multiply the result by 100, and you’ll get the effective interest rate on the company’s debt obligations.
Keep in mind that this isn’t a perfect calculation, as the amount of debt a company carries can vary throughout the year. If you’d like a more reliable result, then you can use the average of the
company’s debt load from its four most recent quarterly balance sheets.
Read Also: gross profit formula
Next, determine the company’s marginal tax rate (federal and state combined). For most large corporations, the federal marginal tax rate is 35%, as this rate applies to all income over $18.33
million. State corporate income taxes range from 0% to 12% as of 2016.
Finally, to calculate the after-tax cost of, simply subtract the company’s marginal tax rate from one and then multiply the result by the effective tax rate you found earlier.
After-Tax Cost Of Debt
Cost is the total amount of interest that a company pays on loans, credit cards, bonds, and other forms of debt. Since companies can deduct the interest paid on business debt. The most common cost
formula is:
Cost of Debt = Interest Expense (1 – Tax Rate)
The health of your business finances depends in large part on the cost of capital that your business takes on. By calculating the cost, you can figure out not only the true cost of a specific loan
but also whether you can justify taking on that debt given your business’s goals.
Whether you want to launch a new product, open another storefront, or hire a new employee, debt has to help your business grow and increase your company’s profits. Otherwise, the debt simply isn’t
worth the cost. The understanding cost will help you know whether to pull the trigger.
Find out the formula for cost, see how to use the formula with a few examples, and learn how to lower your borrowing costs and become a smarter borrower.
How To Find Cost Of Debt
Wondering how much that loan costs you? You should understand the cost. The cost of debt is the total amount of interest that a company pays over the full term of a loan. The cost accounts for tax
deductions that the company can claim on interest expenses. Business owners can use cost to evaluate how a loan can increase profits.
Determining the cost is important when you’re shopping around for a business loan. You might have heard of the saying, “You can’t make an omelet without breaking some eggs.” This is often true for
small businesses. In order to move the needle, business owners often need to rely on business loans. But there comes a tipping point where there’s too much debt and not enough growth. The cost helps
you identify this tipping point.
When you calculate the cost, you can compare that to the income growth that will come from the capital. For instance, if you can use a $10,000 low-interest-rate loan to create a new product that’ll
generate three times as much revenue, then the loan is probably worth the cost. But if the income potential of the loan doesn’t surpass the cost, you’re better off getting another loan or adapting
your expectations.
Prospective lenders might also evaluate your cost by looking at projections in your business plan. If your cost is too high, the lender might not approve you for the loan, unless you can propose a
different use of funds with higher growth potential.
Pretax Cost Of Debt
The other approach is to look at the credit rating of the firm found from credit rating agencies like S&P, Moody’s, and Fitch. A yield spread over US treasuries can be determined based on that given
rating. That yield spread can then be added to the risk-free rate to find the cost of the company. This approach is particularly useful for private companies that don’t have a directly observable
cost in the market. Simply put, a company with no current market data will have to look at its current or implied credit rating and comparable debts to estimate its cost of debt. When comparing, the
capital structure of the company should be in line with its peers.
When neither the YTM nor the debt-rating approach works, the analyst can estimate a rating for the company. This happens in situations where the company doesn’t have a bond or credit rating, or where
it has multiple ratings. We would look at the leverage ratios of the company, in particular, its interest coverage ratio. A higher number for this ratio means a safer borrower. | {"url":"https://itbusinessmind.com/cost-of-debt/","timestamp":"2024-11-10T14:45:14Z","content_type":"text/html","content_length":"83126","record_id":"<urn:uuid:1294f441-51ee-40e0-8c64-ac1d8bfa8104>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00415.warc.gz"} |
Using structural equation modelling to determine measurement error in different administrative data sources
In recent years, Structural Equation Modelling (SEM) has been used as a tool to quantify measurement error of multiple data sources. The current application is a way to investigate if this is a
method that could be used at the Office for National Statistics (ONS) for administrative data.
In this case study of SEM, we investigated the measurement error of floor area measured by two different administrative data sources, Valuation Office Agency (VOA) and Energy Performance Certificate
(EPC) data. Since the national dataset was too large to run in R, we generated a covariance matrix to use as input for the SEM.
Our model showed that the national-level results for the VOA floor area tended to have less measurement error than the EPC floor area. This was the same at local authority (LA) level apart from eight
London LAs and four rural LAs.
Future research could investigate what drives these differences in results, but our initial investigation into the characteristics of those LAs showed that the London LAs had a high proportion of
small apartments, or a very small count of properties. This finding is in line with new research by the Social Statistics Admin First Housing team in the Developing admin-based floor area statistics
for England and Wales: 2021 article, where measurement error would affect the prediction of smaller properties. This method can also be used for more complex models with several latent constructs and
more co-variates if available.
Nôl i'r tabl cynnwys
As a result of increasing use of administrative data sources in official statistics, there is a need to test and develop new methods to determine the quality of these data sources. Structural
equation modelling (SEM) is a useful technique for estimating measurement error where there are multiple data sources, each including a variable which measures the same (or nearly identical) concept.
In the case of continuous data, SEM can be seen as a combination of linear regression analysis and factor analysis where latent variables are constructed, and the relationships between observed
variables are measured. SEMs are a broad and flexible class of models and include latent class models, latent growth models and confirmatory factor analysis. Through estimating a latent (endogenous)
variable, the measurement errors of the measured (exogenous) variables used as the different indicators can be quantified. This is done by comparing the indicators with the latent variable. The
assumption is that the latent variable symbolises the true error-free variable. The benefit of estimating such a presumably error-free variable for administrative data is that you do not necessarily
need a gold standard measurement in your model. However, it is important to be aware that your results will reflect the specified model and could still potentially include bias.
The Methodological Research Hub in the Methodology and Quality directorate at the Office for National Statistics (ONS) has been working on an ongoing project: Estimating and communicating measurement
error in administrative data sources. We have been applying novel methods to assess measurement error in administrative data.
Previously, Professor Paul Smith and Dr James Dawber from Southampton University were contracted to investigate the available methods to determine the quality of administrative data sources. One
method proposed was SEM, which has primarily been applied in social sciences such as sociology and psychology, to evaluate hard-to-measure concepts through several indicator variables. They applied
different types of SEMs to synthetic data, as well as comparing clothing prices with web scraped data versus barcode data. The Methodological Research Hub aims to test SEM on data used in the ONS to
understand its applicability and to develop greater knowledge of the measurement error of administrative data sources. This research is part of a series of projects on latent class modelling methods
in the hub.
This research will contribute towards the evidence for the National Statistician's Recommendation to Government in 2023: towards assessing quality of administrative data sources. SEM provides a
recent method that can quantify the measurement error and contribute to evaluating the quality of administrative and other types of data sources.
Comparing measures of floor area
Official statistics about floor area are produced though surveys such as the Department for Levelling Up, Housing and Communities' (DLUHC) English Housing Survey. However, because of limited sample
sizes, analysis of floor space for sub-regional geographies has been limited.
The ONS Social Statistics Admin First (SSAF) Housing team have recently explored the feasibility of using two administrative data sources to produce a harmonised floor area measure to provide
sub-national estimates for England and Wales. These sources are Valuation Office Agency (VOA) data and Energy Performance Certificate (EPC) data. The goal of this work is to provide an alternative
measure of overcrowding that focusses on the available living space per person. Current overcrowding measures instead consider the number of rooms per person. This new measure has the potential to
provide policymakers with more accurate information to meet housing needs.
More information on VOA floor area has previously been published in the Admin-based statistics for property floor space, feasibility research: England and Wales methodology. A breakdown of the EPC
data can be found in the DLUHC Energy Performance of Buildings Data guidance.
The SSAF Housing team concluded that the best performing multiple regression model does not produce address-level floor area estimates of high enough statistical quality for further analysis (such as
into overcrowding). They expected to see a stronger correlation between VOA and EPC floor area (as discussed in the Developing admin-based floor area statistics for England and Wales: 2021 article),
and needed more information related to the measurement error in the two different data sources to judge their comparability. With SEM, this can be further investigated.
The purpose of this project was to test if SEM could be a useful method to apply at the ONS to administrative data. We used the SEM method to evaluate the measurement errors of the two administrative
data sources as measures of floor area.
The purpose was also to provide estimates of measurement error for the VOA and EPC floor area to provide the ONS SSAF Housing team with further insight on the quality and comparability between the
data sources and the predictability of the EPC data on VOA. Estimates are provided for national data (England and Wales) as well as for each local authority (LA). For practical purposes this
publication will provide the combined national estimates of the measurement error.
Research questions
1. What is the comparability (specifically measurement error) of the floor area measured by two different administrative data sources, VOA and EPC?
2. How well does SEM work as a tool to estimate measurement error in multiple administrative data sources for use in official statistics?
Nôl i'r tabl cynnwys
3. Structural equation models
Structural equation models (SEM) can be complex. For our research, we have used a model with only two covariates to estimate the measurement error. Had the dataset included more suitable variables
that would have improved the model, these could have been included. The following equation illustrates the basic SEM. In this model, we want to estimate the measurement error for the observed
variable 𝑦 where variable 𝑦 is
• μ is the latent construct variable estimated based on the exogenous indicators
• 𝜆 is the factor loading that measures the validity of 𝑦
• 𝜀 is the measurement error
In this case, the measurement error 𝜀 could be because of either invalidity or unreliability, and so it represents both. If 𝑦 and 𝜆 are both standardised to have variance 1 and μ and 𝜀 are
uncorrelated, then var(𝜀) = 1 – 𝜆^2. So, if 𝜆 =1, then 𝜀 = 0, which would indicate that there is no measurement error in the variable of interest.
A useful aspect of this model is that it does not rely on a gold standard dataset, and so all sources are assumed erroneous in some way. While the method does not require a gold standard measurement
in the model, there is always a risk to be considered if the included measurements are biased in the same direction. ESSnet KOMUSO (2019) included audit survey data to act as a gold standard
comparison, but another reason for this was to include enough parameters for the SEM to be identified.
More complex models have been investigated for the purpose of estimating measurement error in survey and administrative data sources, see for example the work of Bakker (2012) and Scholtus and Bakker
(2013). Other examples of combining survey and administrative data sources through SEM can also be seen in the work of Pavlopoulos and Vermunt (2015). They investigated whether survey or
administrative data from the Dutch labour force survey best captured whether a person had a permanent or temporary job.
Nôl i'r tabl cynnwys
This research uses linked Valuation Office Agency (VOA) and Energy Performance Certificate (EPC) data. The VOA data were collected in April 2021 and the EPC data in March 2021. The dataset covers
14.5 million addresses in England and Wales, of which 57.5% of VOA addresses in England were linked to an EPC address, and 53.7% in Wales. This means that the results of this study are only valid for
the proportion of properties that could be linked. The datasets contain a floor area variable, as well as a number of rooms variable, which were both used for the modelling. More detail on the
cleaning and linkage procedure can be found in the Developing admin-based floor area statistics for England and Wales: 2021 article.
The structural equation model (SEM) was estimated using the R package lavaan (v0.6-7; Rosseel, 2012). The national-level SEM was computed on a covariance matrix that had been derived from the
original linked VOA and EPC dataset. SEM is generally calculated based on a covariance matrix so if the original micro data are not accessible or too large, but the covariance matrix is available, it
is possible to fit the SEM without using a large dataset. This was the case with our national level analysis.
In our example we fit a SEM with two exogenous indicators for one endogenous latent variable, and two exogenous covariates. The observed variables in the model are:
• VOA floor area
• EPC floor area
• VOA number of rooms
• EPC number of rooms
The first indicator of the latent variable was the VOA floor area, and the second indicator was the EPC floor area, measured in square metres. Even though we were not interested in any specific
covariates predicting floor area, we needed to include covariates to identify the model. As covariates, we used number of rooms originally from the VOA data and number of rooms originally from the
EPC data. Number of rooms as a covariate in the model works conceptually since we can assume that the floor area of a building tends to increase with an increased number of rooms. To be able to
estimate the factor slopes, one of the indicators of the latent variable was fixed to one.
Nôl i'r tabl cynnwys
The structural equation model (SEM) was fitted on the covariance matrix from the national Valuation Office Agency (VOA) and Energy Performance Certificate (EPC) linked dataset. lavaan, like other SEM
software, produces several statistics that give different types of indicators on how well the model is fitted against the data, referred to as fit statistics.
If the model shows a good fit to the data, we can trust its estimates to a higher degree than if it does not. These indicators can be dependent on sample size and number of indicators of the latent
variables. Shi and others (2022) tested which fit statistics worked best for small models with few degrees of freedom, which is the case with our model. They found that the comparative fit index
(CFI) and the standardised root mean square residual (SRMR) were the most robust for small models, so we have chosen to show these statistics. For the final model on floor area in this publication
based on national level data, the CFI was 0.986 and SRMR was 0.008. Common cut-off values for CFI are more than 0.95 and SRMR less than 0.08, which indicates that our model fits the data well.
To find out the measurement errors, we first looked at the standardised factor slopes for the floor area indicators which are 0.975 for the VOA floor area and 0.935 for the EPC floor area (Diagram
1). 0 would mean that there would have been no consistency between the two measures and 1 would have meant that they are the same. We expect a number close to 1 since the aim of both measures is to
measure floor area. What is not explained in the factor slope is the measurement error (1 - factor loading).
Based on our model, the EPC floor area had a presumed larger measurement error of 6.5% (0.065), compared with the VOA floor area measurement error of 2.5% (0.025), which was not a "perfect" measure
either. These measurement errors can be seen as non-negligible since in theory both measures represent the same thing, and we would want them to be the same.
The covariate regression slopes were not of particular interest for this application. The main reason for including them was to have enough parameters for the model to be identified, which means that
there needs to be enough "known" parameters in the model to estimate the "unknown". The estimates for our covariates show that the VOA floor area increased on average 14.0 metres squared and EPC 11.4
metres squared per room. Tests of only one covariate were performed, but the fit statistics showed that the inclusion of both yielded the best model fit. The inclusion of covariates from both the VOA
and EPC datasets reduced the strength of either factor slope compared with a model with only a VOA covariate or only an EPC covariate.
Diagram 1: Structural equation model of Valuation Office Agency (VOA) and Energy Performance Certificate (EPC) floor area at a national level (England and Wales)
Source: Valuation Office Agency and Energy Performance Certificate 2021
Download this image Diagram 1: Structural equation model of Valuation Office Agency (VOA) and Energy Performance Certificate (EPC) floor area at a national level (England and Wales)
.png (49.6 kB)
The model was also fitted separately at local authority (LA) level. Overall, based on the 355 separate LAs, the individual LAs had a similar pattern of measurement error as national level. There
was slightly less error in the VOA floor area measure, and more measurement error in the EPC floor area measure. The only deviation from this pattern were eight LAs within London and four other LAs,
that had more measurement error in the VOA measure, compared with the full data.
When investigating the characteristics of these LAs, we found that the London LAs with deviating results patterns were those with the smallest average floor areas. We also found this pattern in two
island LAs (Isles of Scilly and Isle of Anglesey) as well as more rural Gwynedd and South Hams.
Future research could investigate what drives these differences in results patterns, but our initial investigation into the characteristics of those LAs showed that the London LAs had a high
proportion of small apartments, or a very small count of properties. This finding is in line with new research in the Developing admin based floor area statistics for England and Wales: 2021 article
, which shows measurement error would affect the prediction of smaller properties.
Nôl i'r tabl cynnwys
The main conclusion from this research is that structural equation modelling (SEM) should be further explored and can be used for quantifying measurement error in multiple administrative data
sources. We intend to continue to assess administrative data quality by using latent class modelling methods.
We are also looking into applying the method to prices data and comparing scanner and barcode prices.
If you have any questions, comments or would like to collaborate with the Methodological Research Hub, please contact Methods.Research@ons.gov.uk.
Nôl i'r tabl cynnwys
We want to thank Professor Paul Smith for his input and support for this research. We are also grateful to the Social Statistics Admin First Housing team for support and topic knowledge.
Nôl i'r tabl cynnwys
Bakker B F M (2012), 'Estimating the validity of administrative variables', Statistica Neerlandica, Volume 66, pages 8 to 17
ESSnet KOMUSO (2019), 'Quality Measure and Calculation Method QMCM_A_18 (PDF, 813KB)'
Office for National Statistics (ONS), released 26 October 2022, ONS website, Developing admin based floor area statistics for England and Wales: 2021
Pavlopoulos D and Vermunt J K (2015), 'Measuring temporary employment. Do survey or register data tell the truth? (PDF, 267KB)', Survey Methodology, Volume 41, pages 197 to 214
Rosseel Y (2012), 'lavaan: An R Package for Structural Equation Modeling', Journal of Statistical Software, Volume 48, Issue 2, pages 1 to 36
Shi D, DiStefano C, Maydeu-Olivares A, and Lee T (2022), 'Evaluating SEM Model Fit with Small Degrees of Freedom', Multivariate Behavioral Research, Volume 57, Issues 2 to 3, pages 179 to 207
Nôl i'r tabl cynnwys | {"url":"https://cy.ons.gov.uk/methodology/methodologicalpublications/generalmethodology/onsworkingpaperseries/usingstructuralequationmodellingtodeterminemeasurementerrorindifferentadministrativedatasources","timestamp":"2024-11-10T03:12:55Z","content_type":"text/html","content_length":"60118","record_id":"<urn:uuid:216ce50b-c210-4e41-be02-c3a62a41dc91>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00483.warc.gz"} |
Polynomials in Algebra
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What is the degree of the polynomial x^3 - 2x^2 + x - 1?
Which of the following is a binomial?
x^2 - 4
x + 2
x^2 + 3x - 2
x^2 + 3x (correct)
What is the leading coefficient of the polynomial 2x^2 + 3x - 1?
What is the result of adding the polynomials x^2 + 2x and x^2 - 3x?
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What is the result of multiplying the polynomials x + 2 and x + 3?
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What is the factored form of the polynomial 2x^2 + 4x?
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What is the solution to the quadratic equation x^2 + 4x + 4 = 0?
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What is the type of polynomial with three terms?
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What is the constant term in the polynomial x^2 + 3x - 2?
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What is the result of subtracting the polynomial x^2 - 2x from the polynomial x^2 + 3x?
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Study Notes
• A polynomial is an expression consisting of variables (such as x or y) and coefficients (numbers) combined using only addition, subtraction, and multiplication.
• The variables are raised to non-negative integer powers.
Types of Polynomials
• Monomial: A polynomial with only one term (e.g., 3x^2 or 5y)
• Binomial: A polynomial with two terms (e.g., x^2 + 3x or 2y^2 - 4y)
• Trinomial: A polynomial with three terms (e.g., x^2 + 2x + 1 or y^3 - 2y^2 + y)
Properties of Polynomials
• Degree: The highest power of the variable(s) in a polynomial (e.g., x^2 + 3x has a degree of 2)
• Leading Coefficient: The coefficient of the term with the highest degree (e.g., in x^2 + 3x, the leading coefficient is 1)
• Constant Term: The term with no variables (e.g., in x^2 + 3x + 2, the constant term is 2)
Operations with Polynomials
• Addition and Subtraction: Combine like terms (e.g., (x^2 + 2x) + (x^2 - 3x) = 2x^2 - x)
• Multiplication: Distribute each term in one polynomial to each term in the other polynomial (e.g., (x + 2) × (x + 3) = x^2 + 5x + 6)
Factoring Polynomials
• Factoring out the Greatest Common Factor (GCF): Divide each term by the GCF (e.g., 2x^2 + 4x = 2x(x + 2))
• Factoring Quadratic Expressions: Use the formula x^2 + bx + c = (x + d)(x + e) to factor quadratic expressions
Solving Polynomial Equations
• Linear Equations: Solve for the variable by isolating it on one side of the equation (e.g., 2x + 3 = 5 → x = 1)
• Quadratic Equations: Use factoring or the quadratic formula (x = (-b ± √(b^2 - 4ac)) / 2a) to solve quadratic equations
• An expression consisting of variables and coefficients combined using addition, subtraction, and multiplication.
• Variables are raised to non-negative integer powers.
Types of Polynomials
• Monomial: One-term expression (e.g., 3x^2 or 5y).
• Binomial: Two-term expression (e.g., x^2 + 3x or 2y^2 - 4y).
• Trinomial: Three-term expression (e.g., x^2 + 2x + 1 or y^3 - 2y^2 + y).
Properties of Polynomials
• Degree: Highest power of variables in a polynomial (e.g., x^2 + 3x has a degree of 2).
• Leading Coefficient: Coefficient of the term with the highest degree (e.g., in x^2 + 3x, the leading coefficient is 1).
• Constant Term: Term with no variables (e.g., in x^2 + 3x + 2, the constant term is 2).
Operations with Polynomials
• Addition and Subtraction: Combine like terms (e.g., (x^2 + 2x) + (x^2 - 3x) = 2x^2 - x).
• Multiplication: Distribute each term to each term in the other polynomial (e.g., (x + 2) × (x + 3) = x^2 + 5x + 6).
Factoring Polynomials
• Factoring out the GCF: Divide each term by the Greatest Common Factor (e.g., 2x^2 + 4x = 2x(x + 2)).
• Factoring Quadratic Expressions: Use the formula x^2 + bx + c = (x + d)(x + e) to factor quadratic expressions.
Solving Polynomial Equations
• Linear Equations: Isolate the variable on one side of the equation (e.g., 2x + 3 = 5 → x = 1).
• Quadratic Equations: Use factoring or the quadratic formula (x = (-b ± √(b^2 - 4ac)) / 2a) to solve quadratic equations.
Studying That Suits You
Use AI to generate personalized quizzes and flashcards to suit your learning preferences.
Understand the definition and types of polynomials, including monomials, binomials, and trinomials, and their properties in algebra. | {"url":"https://quizgecko.com/learn/polynomials-in-algebra-wblbuy","timestamp":"2024-11-08T22:17:43Z","content_type":"text/html","content_length":"319513","record_id":"<urn:uuid:e41c12b1-70bb-440f-bd23-87fc983d76e5>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00293.warc.gz"} |
GameObject with the CurvySpline component attached
Control Point
A point controlling a spline's curvation (sometimes called “Node” by others). Control Points are regular GameObjects with the CurvySplineSegment component attached, always child objects of a spline
and automatically managed.
The Tangent vector is the (normalized) direction of a spline point
The Orientation- or Up-Vector defines the Upside of a spline point. Together with the Tangent it forms the rotation used by controllers for a given spline point
Total Fragment
Also known as TF. A relative position on the spline, ranging from 0 (start) to 1 (end)
The TF of a control point is shown in it's inspector.
This is the “time” parameter used in the splines' formulas. A point's TF is not proportional to its distance from the spline's start. Depending on the spline, a value of 0.5 does not always mean the
middle, distance wise, of the spline.
A relative position on a spline segment, ranging from 0 (start of segment) to 1 (end of segment)
The F of a control point is shown in it's inspector.
F is not proportional to its distance from the segment's start. Depending on the spline, a value of 0.5 does not always mean the middle, distance wise, of the segment.
A distance on the spline, ranging from 0 (start) to spline's length (end)
The Distance of a control point is shown in it's inspector.
Relative Distance
A normalized distance on the spline, ranging from 0 (start) to 1 (end)
It is defined as: Distance / Spline_length
The Relative Distance of a control point is shown in it's inspector.
Orientation Anchor
Used by dynamic Orientation mode, this Control Point option tells Curvy Splines to use the transform rotation of the Control Point as an orientation target. Think of “orientation” waypoints. | {"url":"https://curvyeditor.com/documentation/glossary","timestamp":"2024-11-13T08:09:49Z","content_type":"application/xhtml+xml","content_length":"16763","record_id":"<urn:uuid:414579de-21ca-4833-a327-e31bfb58c707>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00463.warc.gz"} |
Force Ball Mill Operation In Laboratory
The right laboratory ball mill for each grinding task
A laboratory ball mill is an indispensable tool when it comes to the rapid, reproducible pulverizing, milling or grinding of a large variety of materials. Depending on the Laboratory Ball Mill
YouTube,71· https://911metallurgist/equipment/laboratory/comminution/mills/laboratory Process Engineering with Planetary Ball Mills ResearchGate,201327· The amount of energy transferred from the
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Process Engineering with Planetary Ball Mills
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Back to Basics Hammer Milling and Jet Milling
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jars, 16320oz Milling and Mixing Materials Research Institute,The working principle of the vibratory grinding mill is based on intense motion of the grinding balls, with both inertia and centrifugal
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Planetary Ball Mill 911Metallurgist
A typical planetary ball mill consists of one turn disc (sometimes called turn table) and two or four bowls. The turn disc rotates in one direction while the bowls rotate in the opposite direction.
The centrifugal forces created by How to Measure the Impact Forces in Ball Mills,116· Final ball diameter was 140mm. The test ball must be substantially larger than standard 75mm balls in the Climax
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methodTop Planetary Ball Mill Manufacturers For Laboratory In India.,Labindia Analytical manufactures the top quality Planetary Ball Mill For Laboratory uses.Planetary ball mill BM 1100+ also meets
the requirements of colloid grinding Interval operation; Powerful & quick grinding to nano fineness up to less than 0.1µm Feed Size less than 10mm; Two different grinding Models (Dry/Wet). GForce :
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mill principle of operation planetary ball mills are smaller than mon ball mills and mainly used in laboratories for grinding sample material down to very small sizes a planetary ball mill consists
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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7
You will find Chapter 13 Surface Areas and Volumes Exercise 13.7 NCERT Solutions for Class 9 Maths here that will help you in knowing the basic concepts of the chapter.
Class 9 Maths NCERT Solutions
prepared by Studyrankers experts that are accurate and detailed so students can understand concepts behind the questions. Through the help of NCERT questions and answers, you can score better marks
in the examinations.
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CAMERA Mathematicians Build an Algorithm to ‘Do the Twist’
New Approach Extracts Rotational Diffusion from X-ray Photon Correlation Spectroscopy Experiments
August 18, 2021
Mathematicians at the Center for Advanced Mathematics for Energy Research Applications (CAMERA) at Lawrence Berkeley National Laboratory (Berkeley Lab) have developed a mathematical algorithm to
decipher the rotational dynamics of twisting particles in large complex systems from the X-ray scattering patterns observed in highly sophisticated X-ray photon correlation spectroscopy (XPCS)
These experiments — designed to study the properties of suspensions and solutions of colloids, macromolecules and polymers — have been established as key scientific drivers to many of the ongoing
coherent light source upgrades occurring within the U.S. Department of Energy (DOE). The new mathematical methods, developed by the CAMERA team of Zixi Hu, Jeffrey Donatelli, and James Sethian, have
the potential to reveal far more information about the function and properties of complex materials than was previously possible.
Particles in a suspension undergo Brownian motion, jiggling around as they move (translate) and spin (rotate). The sizes of these random fluctuations depend on the shape and structure of the
materials and contain information about dynamics, with applications across molecular biology, drug discovery, and materials science.
XPCS works by focusing a coherent beam of X-rays to capture light scattered off of particles in suspension. A detector picks up the resulting speckle patterns, which contain several tiny fluctuations
in the signal that encode detailed information about the dynamics of the observed system. To capitalize on this capability, the upcoming coherent light source upgrades at Berkeley Lab's Advanced
Light Source (ALS), Argonne's Advanced Photon Source (APS), and SLAC's Linac Coherent Light Source are all planning some of the world's most advanced XPCS experiments, taking advantage of the
unprecedented coherence and brightness.
But once you collect the data from all these images, how do you get any useful information out of them? A workhorse technique to extract dynamical information from XPCS is to compute what’s known as
the temporal autocorrelation, which measures how the pixels in the speckle patterns change after a certain passage of time. The autocorrelation function stitches the still images together, just as an
old-time movie comes to life as closely related postcard images fly by.
Current algorithms have mainly been limited to extracting translational motions; think of a Pogo stick jumping from spot to spot. However, no previous algorithms were capable of extracting
“rotational diffusion" information about how structures spin and rotate — information that is critical to understanding the function and dynamical properties of a physical system. Getting to this
hidden information is a major challenge.
Twisting the Light Away
A breakthrough came when experts came together for a CAMERA workshop on XPCS in February 2019 to discuss critical emerging needs in the field. Extracting rotational diffusion was a key goal, and Hu,
a UC Berkeley math graduate student; Donatelli, the CAMERA Lead for Mathematics; and Sethian, Professor of Mathematics at UC Berkeley and CAMERA Director, teamed up to tackle the problem head on.
The result of their work is a powerful new mathematical and algorithmic approach to extract rotational information, now working in 2D and easily scalable to 3D. With remarkably few images (less than
4,000), the method can easily predict simulated rotational diffusion coefficients to within a few percent. Details of the algorithm were published August 18 in the Proceedings of the National Academy
of Sciences.
The key idea is to go beyond the standard autocorrelation function, instead seeking the extra information about rotation contained in angular-temporal cross-correlation functions, which compare how
pixels change in both time and space. This is a major jump in mathematical complexity: simple data matrices turn into 4-way data tensors, and the theory relating the rotational information to these
tensors involves advanced harmonic analysis, linear algebra, and tensor analysis. To relate the desired rotational information to the data, Hu developed a highly sophisticated mathematical model that
describes how the angular-temporal correlations behave as a function of the rotational dynamics from this new complex set of equations.
“There were lots of layered mysteries to unravel in order to build a good mathematical and algorithmic framework to solve the problem,” said Hu. “There was information related to both static
structures and to dynamic properties, and these properties needed to be systematically exploited to build a consistent framework. Taken together, they present a wonderful opportunity to weave
together many mathematical ideas. Getting this approach to pick up useful information out of what seems at first glance to be awfully noisy was great fun.”
However, solving this set of equations to recover the rotational dynamics is challenging, as it consists of several layers of different types of mathematical problems that are difficult to solve all
at once. To tackle this challenge, the team built on Donatelli's earlier work on Multi-Tiered Iterative Projections (M-TIP), which is designed to solve complex inverse problems where the goal is to
find the input that produces an observed output. The idea of M-TIP is to break a complex problem into subparts, using the best inversion/pseudoinversion you can for each subpart, and iterate through
those subsolutions until they converge to a solution that solves all parts of the problem.
Hu and his colleagues took these ideas and built a sister method, “Multi-Tiered Estimation for Correlation Spectroscopy (M-TECS),” solving the complex layered set of equations through systematic
“The powerful thing about the M-TECS approach is that it exploits the fact that the problem can be separated into high-dimensional linear parts and low-dimensional nonlinear and nonconvex parts, each
of which have efficient solutions on their own, but they would turn into an exceedingly difficult optimization problem if they were instead to be solved for all at once,” said Donatelli.
“This is what enables M-TECS to efficiently determine rotational dynamics from such a complex system of equations, whereas standard optimization approaches would run into trouble both in terms of
convergence and computational cost.”
Opening the Door to New Experiments
"XPCS is a powerful technique that will feature prominently in the ALS upgrade. This work opens up a new dimension to XPCS, and will allow us to explore the dynamics of complex materials such as
rotating molecules inside water channels," said Alexander Hexemer, Program Lead for Computing at the ALS.
Hu, who won UC Berkeley's Bernard Friedman Prize for this work, has joined CAMERA — part of Berkeley Lab’s Computational Research Division — as its newest member. “This sort of mathematical and
algorithmic co-design is the hallmark of good applied mathematics, in which new mathematics plays a pivotal role in solving practical problems at the forefront of scientific inquiry," said Sethian.
The CAMERA team is currently working with beamline scientists at the ALS and APS to design new XPCS experiments that can fully leverage the team’s mathematical and algorithmic approach to study new
rotational dynamics properties from important materials. The team is also working on extending their mathematical and algorithmic framework work to recover more general types of dynamical properties
from XPCS, as well as apply these methods to other correlation imaging technologies.
This work is supported by CAMERA, which is jointly funded by the Office of Advanced Scientific Computing Research and the Office of Basic Energy Sciences, both within the U.S. Department of Energy’s
Office of Science.
About Berkeley Lab
Founded in 1931 on the belief that the biggest scientific challenges are best addressed by teams, Lawrence Berkeley National Laboratory and its scientists have been recognized with 16 Nobel Prizes.
Today, Berkeley Lab researchers develop sustainable energy and environmental solutions, create useful new materials, advance the frontiers of computing, and probe the mysteries of life, matter, and
the universe. Scientists from around the world rely on the Lab’s facilities for their own discovery science. Berkeley Lab is a multiprogram national laboratory, managed by the University of
California for the U.S. Department of Energy’s Office of Science.
DOE’s Office of Science is the single largest supporter of basic research in the physical sciences in the United States, and is working to address some of the most pressing challenges of our time.
For more information, please visit energy.gov/science. | {"url":"https://crd.lbl.gov/news-and-publications/news/2021/camera-mathematicians-build-an-algorithm-to-do-the-twist/","timestamp":"2024-11-07T23:42:21Z","content_type":"text/html","content_length":"32962","record_id":"<urn:uuid:8ba83f7c-70eb-47b5-99b0-9597e909ac95>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00196.warc.gz"} |
Metaphors and Mathematics 2 - International Maths Challenge
Metaphors and Mathematics 2
Robert Recorde, author the first English textbook on algebra (published in 1557), chose to give his book the metaphorical title The Whetstone of Witte to encourage people to take up the new and
difficult practice of algebra. The metaphor of a whetstone, or blade-sharpener, suggests that algebra is not only useful, but also good mental exercise. In the verse that he included on its title
page, he writes,
Its use is great, and more than one. Here if you lift your wits to wet, Much sharpness thereby shall you get. Dull wits hereby do greatly mend, Sharp wits are fined to their full end.
Mathematics, and algebra in particular, according to The Whetstone of Witte is like a knife-sharpener for the brain. Four hundred years later, in his book Mathematician’s Delight (1961), W.W. Sawyer
takes up a similar metaphor, suggesting that “Mathematics is like a chest of tools: before studying the tools in detail, a good workman should know the object of each, when it is used, how it is
used.” Whether they describe mathematics as a sharpener or other tool, these mechanical metaphors are commonly used to emphasize the practicality and versatility of mathematics, particularly when
employed in engineering or science, and suggest that it should be used thoughtfully, and with precision.
An often quoted mechanical metaphor that suggests a more frantic and less precise process of mathematical creation is often attributed to Paul Erdos: “a mathematician is a machine for turning coffee
into theorems.”
For more such insights, log into www.international-maths-challenge.com.
*Credit for article given to dan.mackinnon* | {"url":"https://international-maths-challenge.com/metaphors-and-mathematics-2/","timestamp":"2024-11-09T16:17:59Z","content_type":"text/html","content_length":"144903","record_id":"<urn:uuid:d26c196f-955c-4979-a299-35c8c60e3e34>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00238.warc.gz"} |
Angular Momentum - Physics | Socratic
Angular Momentum
Key Questions
• Explanation:
For linear motion we have the conservation of linear momentum(mass $\times$linear velocity): that is in any system, providing no external forces act, the total linear momentum is always constant.
For rotational bodies we have another conservation law. Providing NO external Torques act, the total angular momentum
(moment of inertia $\times$ angular velocity) is constant.
• Explanation:
Angular momentum is denoted by $\vec{L}$.
Definition :-
The instantaneous angular momentum $\vec{L}$ of the particle relative to the origin $O$ is defined as the cross product of the particle’s instantaneous position vector $\vec{r}$and its
instantaneous linear momentum $\vec{p}$
$\vec{L}$= $\vec{r}$$\times$$\vec{p}$
For a rigid body having fixed axis rotation , the angular momentum is given as $\vec{L} = I \vec{\omega}$ ; where $I$ is the Moment of Inertia of the body about the axis of rotation.
The net torque $\vec{\tau}$ acting on a body is give as the rate of change of Angular Momentum.
$\therefore \sum \vec{\tau} = \frac{\mathrm{dv} e c L}{\mathrm{dt}}$ | {"url":"https://socratic.org/physics/rotational-motion/angular-momentum","timestamp":"2024-11-05T07:06:06Z","content_type":"text/html","content_length":"396701","record_id":"<urn:uuid:c117b7c6-ea6c-4d0c-ab69-1bd1c08ae665>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00533.warc.gz"} |
groundSet -- (internal) ground set
Returns the internal representation of the ground set.
Important: read the following if you encounter warnings/errors when specifying subsets of a matroid (e.g. restriction/deletion/contraction, ranks of subset, etc.)
For a matroid M, there are 2 main differences between M.groundSet and the elements of M (given by M_*). First is data types: M.groundSet is a Set, and M_* is a List. Second, M.groundSet always
consists of integers from 0 to n-1, where n is the number of elements of M; on the other hand, the elements of M themselves can be arbitrary (e.g. symbols, matrices, edges in a graph, etc.).
Thus, one can think of M.groundSet as the set of indices of the elements in the list M_*: the first element of M has index 0, corresponding to the element 0 in M.groundSet; the second element of M
has index 1, etc.
The key point is that all sets associated to the structure of a matroid - bases, circuits, flats, etc. - are subsets of M.groundSet (not M_*). In particular, they are also of class Set (although a
collection of them is usually a List), and are also indexed from 0 to n-1. (An exception here is loops and coloops, which are given as a list of indices, rather than single-element sets).
A recommended way to circumvent this distinction between indices and elements is to use $\{0, ..., n-1\}$ as the actual elements of M, in which case an element is equal to its index in M.groundSet.
Most methods in this package will accept either a list of elements or a set of indices, and if the elements of M are $\{0, ..., n-1\}$, then functionally there will be no difference between inputting
lists or sets.
In summary: lists are used for elements in M, and given as sublists of M_*, while sets are used for indices, and given as subsets of M.groundSet.
i1 : M = matroid({a,b,c,d},{{a,b},{a,c}})
o1 = a "matroid" of rank 2 on 4 elements
o1 : Matroid
i2 : peek M
o2 = Matroid{bases => {set {0, 1}, set {0, 2}}}
cache => CacheTable{...1...}
groundSet => set {0, 1, 2, 3}
rank => 2
i3 : M.groundSet
o3 = set {0, 1, 2, 3}
o3 : Set
i4 : M_*
o4 = {a, b, c, d}
o4 : List
i5 : bases M
o5 = {set {0, 1}, set {0, 2}}
o5 : List
i6 : (bases M)#0
o6 = set {0, 1}
o6 : Set
i7 : circuits M
o7 = {set {1, 2}, set {3}}
o7 : List
i8 : flats M
o8 = {set {3}, set {1, 2, 3}, set {0, 3}, set {0, 1, 2, 3}}
o8 : List
i9 : loops M
o9 = {3}
o9 : List
i10 : coloops M
o10 = {0}
o10 : List
Note in particular the types of the various outputs above.
The following illustrates how to perform operations with a specified subset of M.groundSet. In the final example, a list of indices is given, which goes against the conventions above, but the
elements of the list are treated (correctly) as indices, and if debugLevel is greater than 0, then a warning is printed.
i11 : N1 = M | {a,c,d}
o11 = a "matroid" of rank 2 on 3 elements
o11 : Matroid
i12 : N2 = M | set{0,2,3}
o12 = a "matroid" of rank 2 on 3 elements
o12 : Matroid
i13 : N1 == N2
o13 = true
i14 : debugLevel = 1
o14 = 1
i15 : N3 = M | {0,2,3} -- gives a warning, but attempts to treat 0 as an index
-- indicesOf: 0 is not a member of {a, b, c, d}. Treating 0 as an index (cf. 'help groundSet')...
o15 = a "matroid" of rank 2 on 3 elements
o15 : Matroid
i16 : N3 == N2
o16 = true | {"url":"https://macaulay2.com/doc/Macaulay2/share/doc/Macaulay2/Matroids/html/_ground__Set.html","timestamp":"2024-11-04T05:59:02Z","content_type":"text/html","content_length":"11683","record_id":"<urn:uuid:51bf1ad0-cb5c-483e-b58f-adb37db8519c>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00676.warc.gz"} |
CGAL 5.5.2 - STL Extensions for CGAL: Member List
This is the complete list of members for CGAL::Project_opposite_prev< Node >, including all inherited members.
argument_type typedef CGAL::Project_opposite_prev< Node >
operator()(argument_type &n) const CGAL::Project_opposite_prev< Node >
operator()(const argument_type &n) const CGAL::Project_opposite_prev< Node >
Project_opposite_prev() CGAL::Project_opposite_prev< Node >
result_type typedef CGAL::Project_opposite_prev< Node > | {"url":"https://doc.cgal.org/5.5.2/STL_Extension/structCGAL_1_1Project__opposite__prev-members.html","timestamp":"2024-11-12T16:21:58Z","content_type":"application/xhtml+xml","content_length":"9587","record_id":"<urn:uuid:e2d4bbe4-7d8e-4c29-b965-3a49d2b408eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00713.warc.gz"} |
Computational Complexity
Every paper has a story but Sunny Daniel's
Arxiv paper
from yesterday deserves a blog post.
We begin in 1982 when Ravi Kannan proved that Σ
(the problems computable in NP with an NP oracle) cannot have n
size circuits. Kannan's result hold for n
-size circuits but for this story we'll keep it simple.
Kannan had an ingeniously simple proof. By diagonalization you can create a language L in Σ
[4 ]
that does not have n
-size circuits. Now there are two cases:
1. SAT doesn't have n^2-size circuits. Since SAT is in Σ[2] we are done.
2. SAT has n^2-size circuits. Then by Karp-Lipton Σ[4] = Σ[2] so L is in Σ[2] and we are done.
Kannan's proof is non-constructive and doesn't give an explicit Σ[2] language that we can show doesn't have n^2-size circuits. Either SAT or L but one can't be sure which one.
In 1998, Sunny Daniels, a PhD student at Rutgers, took Eric Allender's complexity course. Eric offered up a constructive example of Kannan as an open problem. Sunny came up with a solution. He wrote
up a draft in LaTeX but for personal reasons dropped out of academics and never published the paper.
In 2003, Jin-Yi Cai and Osamu Watanabe, not aware of Daniels' paper, came up with their own independent construction and presented
their paper
at the COCOON conference in Montana. Word got back to Sunny but he thought he had lost the LaTeX file and didn't want to retypeset the whole proof.
Sunny had
Iomega Zip Drive
cartridges from his Rutgers days. Recently he found someone who had a Zip Drive reader and managed to recover the files. In there he discovered the original LaTeX, cleaned the paper up, and sixteen
years after his proof put
the paper
on ArXiv. Even if you don't care about the math, read the introduction for the complete version of this story.
Kannan's proof actually shows Σ
does not have n
-size circuits and this was later
to S
. Whether we have any constructive language in Σ
or S
without n
-size circuits still remains open.
Better algorithms can lead to better medicine and save lives. Just today Tim Gowers discusses Emmanuel Candès' ICM Plenary Lecture, which among other things describes how Candès' work on compressed
sensing leads to shorter MRI scans for children, greatly reducing the risk of oxygen deprivation. Prove P = NP with a practical algorithm, and you'll conquer that worst of our diseases. Sounds great
until you realize what we can't do.
I was talking to a cancer researcher recently and he points out that many of their challenges are indeed algorithmic. But he also brings up the contrapositive. Since we don't have great algorithms
now, we don't know how to make sense of DNA sequences and in particular don't know how to map genetic markers to an appropriate cancer treatment. He works with cancer patients, knowing he can't give
them the best possible treatment, not because of lack of data, but due to lack of ways to analyze that data. People die because we don't have the ability to break through the complexity of these
algorithmic challenges.
My daughter had a summer project to read and summarize some popular science articles. Having heard me talk about Alan Turing more than a few times, she picked a cover story from a recent New
Scientist. The cover copy says "Turing's Oracle: Will the universe let us build the ultimate thinking machine" sounds like an AI story but in fact more of an attack on the Church-Turing thesis. The
story is behind a paywall but here is an excerpt:
He called it the "oracle". But in his PhD thesis of 1938, Alan Turing specified no further what shape it might take...Turing has shown with his universal machine that any regular computer would
have inescapable limitations. With the oracle, he showed how you might smash through them.
This is a fundamental misinterpretation of Turing's oracle model. Here is what Turing said in his paper Systems of Logic Based on Ordinals, Section 4.
Let us suppose we are supplied with some unspecified means of solving number-theoretic problems; a kind of oracle as it were. We shall not go any further into the nature of the oracle apart from
saying it cannot be a machine. (emphasis mine)
The rest of the section defines the oracle model and basically argues that for any oracle O, the halting problem relative to O is not computable relative to O. Turing is arguing here that there is no
single hardest problem, there is always something harder.
If you take O to be the usual halting problem then a Turing machine equipped with O can solve the halting problem, just by querying the oracle. But that doesn't mean that you have some machine that
solves the halting problem for, as Turing has so eloquently argued in Section 9 of his On Computable Numbers, no machine can compute such an O. Turing created the oracle model, not because he thought
it would lead to a process that would solve the halting problem, but because it allowed him to show there are problems even more difficult.
Turing's oracle model, like so much of his work, has played a major role in both computability and computational complexity theory. But one shouldn't twist this model to think the oracle could lead
to machines that solve non-computable problems and it is sacrilege to suggest that Turing himself would think that.
In recent years, I've heard complaints from my complexity colleagues that FOCS and STOC are mostly algorithms and from the algorithm buddies that STOC and FOCS are mostly complexity. What exactly
counts as a complexity or algorithms paper has become quite blurred in recent years. So let's try an experiment. Below is a poll I've created using titles from the upcoming FOCS conference. Which of
these papers do you consider complexity? Does complexity in the title make them a complexity paper?
If you are interested, you can find the manuscripts for most of these papers on the FOCS accepted papers list.
survey tools
Disclaimer: This is a completely non-scientific poll solely for the interest of the readers of this blog. The results will have no effect on future conference papers.
When can limited randomness act as well as true random bits?
Polylogarithmic independence fools AC^0 circuits by Mark Braverman (JACM 2010)
To explain this result consider choosing uniformly from among the following four strings:
If we look at any two of the bits, say the first and third, all four possibilities 00 10 11 01 occur. The sequence is thus 2-wise independent. We can get 2-wise independence using only two random
bits to choose one of the four strings. In general one can get k-wise independent in n-bit strings using O(k^2 log n) random bits.
Braverman shows that any polylogarithmic independent distribution fools polynomial-size constant-depth circuits (AC^0), or more precisely for a size s depth d circuit C, for k=(log (m/ε))^O(d^2), the
probability that C will output 1 with uniformly-random inputs will differ at most ε from the probability C outputs 1 from inputs chosen from a k-wise random distribution. Braverman's result followed
after Bazzi and Razborov proved a similar result for depth-2 circuits (CNF formulas).
Another nice result along these lines: Ilias Diakonikolas, Parikshit Gopalan, Ragesh Jaiswal, Rocco Servedio and Emanuele Viola show that Bounded Independence Fools Halfspaces. A half-space is just a
weighted threshold function (is the sum of w[i]x[i] at most some given θ). Diakonikolas et al. show that one can fool halfspaces with k=O(ε^-2log^2(1/ε)), in particular for constant ε, k is a
constant independent of the number of variables.
At the opening ceremonies of the International Congress of Mathematicians in 2014, Subhash Khot was awarded the Rolf Nevanlinna Prize, given every four years to an under-40 researcher for outstanding
contributions in Mathematical Aspects of Information Sciences. Subhash's citation reads
Subhash Khot is awarded the Nevanlinna Prize for his prescient definition of the “Unique Games” problem, and leading the effort to understand its complexity and its pivotal role in the study
of efficient approximation of optimization problems; his work has led to breakthroughs in algorithmic design and approximation hardness, and to new exciting interactions between computational
complexity, analysis and geometry.
Khot's work has indeed generated a large research agenda over the last decade. I highlighted his work in March's favorite theorems post.
In other big news, we have our first female Fields Medalist Maryam Mirzakhani for contributions to the dynamics and geometry of Riemann surfaces and their moduli spaces. Still no female winners among
the nine Nevanlinna winners. Artur Avila, Manjul Bhargava and Martin Hairer also received Fields medalists. Stanley Osher won the Gauss Prize, Phillip Griffiths the Chern Medal and Adrián Paenza the
Leelavati Prize.
Pictures and press releases and citations of all the prize winners.
I taught a 3-week, 3-hours-a-day course to High School student titled
Computer Science: A Hands Off Approach.
Given that time constraints and the fact that some already know (say) Java and some don't know any language, this seemed like a good choice.
I decided to teach mostly pre-RSA crypto with the following theme: Alice and Bob want to pass secret messages. How do they do it? I cover Shift, affine, general sub, Vigenere, Matrix, 1-time pad,
Diffie-Helman (a highpoint of the course since Alice and Bob don't have to meet in a dark alley). In also did secret sharing with polynomials, error correcting codes (elementary), Huffman codes, and
some applications of mod arithmetic.
While teaching this course some points of interest came up. I suspect most are know and I appreciate polite comments telling me so.
1. A student suggested this cipher: code a,b,c,...,z into a 100-letter alapahbet and map each letter to a set of symbols that is the size of the freq. For example, if e occurs 9% of the time then
map e to 9 letters. Then use those letters at random. This would seem to foil freq analysis? Does it? Has it been used? What are the downsides.
2. Many students suggested using Vigenere but instead of having every xth letter be done by a different shift, have it be affine or general. Of course this can be cracked the same way Vig is
cracked. But it does raise an interesting point: Which ciphers are used and not used can be the based on when things were discovered. Martians may very well have used some kind of Vig where every
xth letter is a different gen sub cipher.
3. Wikipedia and other sources say that the Vig cipher was unbroken for 300 years. A student pointed out that it might have been broken but the one who broke it didn't say. Jon Katz (crypto prof at
UMCP) can't believe it wasn't broken immediately, but of course hindsight is 20-20.
4. (I have commented on this before) A matrix cipher with a 10x10 matrix seems uncrackable using plaintext only. I have made this question rigorous here.
5. I made the comment that 1-time pads are not used much (is this even true? Might depend on the definition of ``must'') because getting a perfect source of randomness is hard. During WW II they
also would be hard to use because its hard to carry around a billion bits. But now that would not be a problem. Again--if history had been different we may use 1-time pads, or quasi-random ones
6. I told the students about arithmetic mod n. One of the students really really didn't like that (say) in mod 7 we have 1/3 = 5. He REALLY wants 1/3 to be between 0 and 1. I suspect he didn't care
much for discrete logs either. This was a GOOD student- so his objection was not that it was hard.
7. For some of the students their first exposure to matrices was matrix codes over mod 26. I hope they can recover from that.
8. Most of the students know what logs were, but weren't that comfortable with them. And here I go and teach them discrete logs! I hope they can recover from that.
9. I showed the students that there were quadratic equations over mod 12 with more than 2 roots and challenged them to see how many roots they could come for other mods. One of my students ran
massive computer tests and found stuff and in the end had a result that didn't need all of his computations: x^2 \equiv 0 mod n^2 has n roots. And I later had on a HW x^a \equiv 0 mod n^a. I am
sure none of this is new, but its new to him when he discovered it and of course new to the class when I taught it.
10. I taught the class the Baby-Step/Giant-Step Discrete Log algorithm which has sqrt(p) prepocessing an sqrt(p) running time. Its not used because it also takes sqrt(p) space; however, it was good
to show them that Discrete Log can be done in sqrt(p) time, much better than p time--- hence Alice and Bob need to pick their parameters larger than they may have thought when doing
Diffie-Helman. That night I easily worked out that it can be modified to do p^{2/3} preprocessing (and space) but just p^{1/3} time. HW was p^a prep, p^{1-a} time. One of the students inquired if
this algorithm has a name. I then looked over the web but couldn't find it anywhere so I told them to call it The Gasarch Variant of Baby-Step, Giant-Step. I also quickly told them to NOT be
impressed--- and this helped me make a point I had made often, that CS is a NEW field, so NEW that one can present new or newish results to HS students. I also made the point that I am sure this
variant is KNOWN to anyone who would care, but (1) they may not care since it takes even more space if x is larger than 0.5 and more time if x is less than 1/2 and (2) not everything is no the
web. That last point freaked them out more than the quadratic equation mod 12 that had more than two roots.
UPSHOT- teaching HS students CS can bring up some issues that you had not thought of. Or at least that I had not thought of.
Recall: PIR stands for Private Information Retrieval. Here is the model: A database is an n-bit string (my wife tells me this is not true). The user wants to find the ith bit without the database
knowing what bit the user wants. The user COULD just request ALL n bits. Can the user use less communication? See here for a website of many papers on PIR.
1. If there is just one copy of the DB and there are no comp. constraints on the computational power of the DB then ANY protocol requires n bits comm.
2. If there is one copy of the DB then, with some comp constraints, you can do better. I state one result: if quadratic residue is hard then there is a protocol using n^epsilon bits of comm.
(Kushilevitz and Ostrovsky). The rest of this post is about the info-theoretic case, so the DB has no comp. constraints.
3. If there are two copies of the DB then there is a protocol that uses n^{1/3} bits. (Chor, Kushilevitz, Goldreich, Sudan)
4. If there are three copies of the DB then there is a protocl that uses n^{1/32582658} bits. Really! (Yekhanin).
5. If a 2-server protocol is a bilinear-group protocol (which all prior constructions were) then it must take n^{1/3}.(Razborov and Yekhanin)
6. Most known constructions put into a geometric framework (Woodruff and Yekhanin).
Recently Dvir ahd Gopi posted a paper here which gives a 2-server PIR protocol that uses n^P bits where P= sqrt( (log log n)/log n)). That breaks the barrier proven by RY! How is that possible?
The barrier result was only for a certain type of PIR. It DID cover all the known PIR schemes at the time. But it does not cover this one.
This is how Barrier SHOULD work--- they should not discourage, but they should point to difficulties so that someone can overcome them.
Also note, the new result used the some of the Framework of Woodruff and Yekhanin. So good to unify and obtain new proofs of old results.
A few years ago I saw the movie Moneyball about how the Oakland A's used intelligent statistics to... win?
No, but to do better-than-expected. Even if I didn't know it was a true story I would have assumed it was because the following are very rare or odd in a fictional story:
1. At the end of the team doesn't win- it just does better than expected. In the typical sports movie the underdog pulls it all together and wins. In some the underdogs loses but they are now better
people or something. In an episode of Cheers where they were the underdog to Gary's Tavern in a bloody mary making contest the Cheers gang cheats and wins. But in Moneyball, and in NO other
sports (or contest) movie that I know of, does the underdog do better-than-expected in an undramatic matter. This is NOT a complaint- just note that its real life.
2. In Moneyball the General Manager wants to try out mathematical methods and the Manager resists. In most movies its the suits that are wrong and the people on the ground that are right. This is
even a theme of many articles about business that I read in magazines on airplanes. So this inversion is odd - but again, you can't argue that a true story is unrealistic or undramatic.
3. Bill Beane, the one who wants to use math techniques, thinks that what Baseball scounts look for is the wrong thing. In fact, they misjudged him when he was a young player. But in what direction?
--- they thought he was BETTER than he was. If this was a fictional story surely the scouts would think he was WORSE than he was.
I am glad that (as far as I know) neither the book nor the movie tried to make the story more satisfying or dramatic.
In academia we do clean things up for a better story line. If the true motivation for working on a problem doesn't really make sense when you see the final paper, we change our motivation. Our
original proof is intuitive but ugly, so we change it to be polished but less clear where it came from. Historians often simplfiy to make sense of things. I am NOT complaining- merely asking, do we
do it too much?
When I was in ninth grade and was told that you could solve a quadratic equation (I rederived the quadratic formula once a month to make sure I could), a cubic, a a quartic, but not quintic, I
immediately said "I want to goto College to learn why you can't solve a quintic" That sparked my interest in math.
Is the above story true? I am sure that in ninth grade I did learn that the quintic was unsolvable and that was of great interest to me, and I really did rederive the quadratic equation once a month.
And I was interested to learn that the quintic was not solvable. But I really doubt the story is as clean as presented above. Even so, the story is true in spirit. However, I would not want to push
the point.
How about you? Do you tell stories about yourself or about others that are just a little too polished? Not so much false, and not even to put yourself in a better light, but just a little to clean to
have really happened. | {"url":"https://blog.computationalcomplexity.org/2014/08/?m=0","timestamp":"2024-11-07T17:15:23Z","content_type":"application/xhtml+xml","content_length":"220096","record_id":"<urn:uuid:ed9c69ad-726f-422e-b17c-4a6091a495e6>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00775.warc.gz"} |
How to find gcd in java - Java Program to Find GCD of two Numbers - BTech Geeks
How to find gcd in java – Java Program to Find GCD of two Numbers
How to find gcd in java: Don’t miss the chance of Java programs examples with output pdf free download as it is very essential for all beginners to experienced programmers for cracking the
Program to Find GCD of two Numbers
GCD of two numbers in java: In this article, we will see multiple ways to find the GCD(Greatest Common Divisor) of two numbers.
In mathematics , the greatest common divisor of two or more integers, which are not all zero is the largest positive integer.
For example:
24 = 2*2*2*3
18 = 2*3*3
GCD = 2*3 = 6.
Method 1 : To find GCD of Two Numbers using a while loop with the if-else statement
GCD program in java: We can use while loop with if-else statement to find GCD of two number.
• First, assign the values for int n1 and n2 for which you want to find GCD.
• Then the smaller integer is subtracted from the larger integer, and the result is assigned to the variable holding the larger integer.
• This process is continued until n1and n2 are equal.
class Main
public static void main(String[] args) {
int n1 = 81, n2 = 153;
while(n1 != n2) {
if(n1 > n2) {
n1 -= n2;
else {
n2 -= n1;
System.out.println("GCD: " + n1);
GCD: 9
Method 2 : To find GCD of two numbers using for loop and if statement
Java gcd code: We can use for loop with if-statement to find GCD of two numbers.
• Two numbers whose GCD are to be found are stored in n1 and n2
• Then, a for loop is executed until i is less than both n1 and n2. This way, all numbers between 1 and the smallest of the two numbers are iterated to find the GCD.
• If both n1and n2 are divisible by i, gcd is set to the number. This goes on until it finds the largest number (GCD) which divides both n1 and n2 without remainder.
class Main
public static void main(String[] args)
int n1 = 81, n2 = 153;
int gcd = 1;
for (int i = 1; i <= n1 && i <= n2; ++i)
if (n1 % i == 0 && n2 % i == 0)
gcd = i;
System.out.println("GCD of " + n1 +" and " + n2 + " is " + gcd);
GCD of 81 and 153 is 9
Method 3: GCD for both positive and negative number
GCd function java: In this approach we will see GCD for both positive and negative number.
• first, assign the values for int n1 and n2 for which you want to find GCD.
• Then the smaller integer is subtracted from the larger integer, and the result is assigned to the variable holding larger integer.
• This process is continued until n1 and n2 are equal.
class Main
public static void main(String[] args)
int n1 = 81, n2 = -153;
n1 = ( n1 > 0) ? n1 : -n1;
n2 = ( n2 > 0) ? n2 : -n2;
while(n1 != n2) {
if(n1 > n2) {
n1 -= n2;
else {
n2 -= n1;
System.out.println("GCD: " + n1);
Method 4 : GCD of more than two (or array) numbers
Java gcd function: In this we will see how to get GCD of more than 2 numbers.
• A class named Demo contains the main function that takes in two values.
• If the first value is 0, the second value is returned as output. Otherwise, a recursive function is written that computes the greatest common divisor of the two elements.
• Next, another static function is defined that takes an array and another integer value as a parameter.
• The first element of the array is assigned to a variable named ‘result’ and a ‘for’ loop iterates over elements from 1 to the integer value that was passed as a parameter to the function.
• This output is assigned to the ‘result’ variable itself. If the value of ‘result’ is 1, then the output is 1, otherwise, the value of ‘result’ is returned.
public class Main
static int gcd_of_nums(int val_1, int val_2)
if (val_1 == 0)
return val_2;
return gcd_of_nums(val_2 % val_1, val_1);
static int find_gcd(int arr[], int no){
int result = arr[0];
for (int i = 1; i < no; i++){
result = gcd_of_nums(arr[i], result);
if(result == 1){
return 1;
return result;
public static void main(String[] args)
int my_arr[] = { 7, 49, 177, 105, 119, 42};
int no = my_arr.length;
System.out.println("The GCD of the elements in the array is ");
System.out.println(find_gcd(my_arr, no));
The GCD of the elements in the array is 1
Method 5: To find GCD using the modulo operator
gcd java: We can use for loop with if-statement to find GCD of two numbers.
• First, we have defined a recursive function named GCD().
• It parses two parameters a and b of type int.
• If the second number (b) is equal to 0, the method returns, and as GCD else returns a%b.
public class Main
public static void main(String[] args)
int a = 112, b = 543;
System.out.println("GCD of " + a +" and " + b + " is " + GCD(a, b));
static int GCD(int a, int b)
if (b == 0)
return a;
return GCD(b, a % b);
GCD of 112 and 543 is 1
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00230212: Intelligent Bio-computation (智能生物计算) (Fall 2024)
Course Information
When: Monday 19:20 pm – 20:55 pm.
Where: 清华大学法律图书馆B112.
Instructor: Jianzhu Ma (马剑竹) Email: majianzhu at tsinghua dot edu dot cn.
Office Hour: Tuesday. 19:30 pm – 20:30 pm.or by appointment (send e-mail) Where: TBD
Course description:
This course encompasses all the principal areas within bioinformatics and computational biology, making it an 'all-you-need’ program for those aspiring to craft algorithms for analyzing and
interpreting biological data. Students who complete this course will gain insights into a range of fundamental algorithmic strategies, drawn directly from seminal and highly referenced recent
literature. The course imparts foundational knowledge in statistics, mathematics, and machine learning, including deep learning aspects. Both lectures and student-led presentations are provided, as
detailed in the Syllabus. The curriculum advances beyond mere application of existing software; it endeavors to impart a thorough understanding of the core principles behind today's cutting-edge
techniques, equipping students with the skills to devise and pioneer the forthcoming wave of bioinformatics tools.
This course is intended to be a first course in Bioinformatics for both undergraduate and graduate students in Electronic Engineering, Computer Science, Mathematics, Statistics, and Life Sciences.
Students from non-biological disciplines may wonder about the level of prior biological knowledge required to pursue studies in bioinformatics. The answer is minimal; no advanced understanding beyond
what is typically taught in high school biology is necessary. But you will need a willingness to learn biological concepts needed to understand the computational problems we study. We will begin with
a quick review of some of the biological concepts needed for bioinformatics, and will learn more as needed.
Students are expected to have the following background:
• Basic programming skills to write a reasonably non-trivial computer program in Python.
• At least an undergraduate course in algorithms (a graduate course in algorithms is preferred).
• Undergraduate or graduate level machine learning courses.
• Students without this background should discuss their preparation with the instructor. The introduction session in the first week of the class will give an overview of the expected background.
Optional textbooks
“Biological sequence analysis” by Durbin, Eddy, Krogh, Mitchison
“An introduction to bioinformatics algorithms” by Jones, Pevzner
“Network Biology:” section in Topology of molecular interaction networks
“Deep Learning” by Ian Goodfellow, Yoshua Bengio, and Aaron Courville
Online book “Bioinformatics Algorithms” [link], Phillip Compeau and Pavel Pevzner
• Homework (50%)
• Final projects (50%)
There will be FOUR homework assignments including both theory and programming problems. Programming projects can be written in both R and Python. Deep learning final projects should be written in
either Tensorflow and PyTorch.
Late policy
Assignments are to be submitted by the due date listed. Assignments will NOT BE accepted if they are submitted late. Additional extensions will be granted only due to serious and documented medical
or family emergencies.
Final projects
Students are encouraged to work as teams of two or three. Each team can either choose the provided projects on the course website or a separate project related to Biology. If you choose to do your
own project, please contact me beforehand to discuss it. Here are the bioinformatics projects you might consider:
• Easy projects:
□ Using deep neural networks to predict protein secondary structures.
□ Using deep neural networks to predict protein contact/distance map.
□ Implement a RNA secondary structure prediction algorithm.
□ Implement a single cell analysis pipeline.
□ Predicting protein functions prediction using multiple protein interaction networks.
□ Implement an algorithm to correct the batch effect in single cell data.
□ Predicting enhancers using genomics features.
• Advanced projects:
□ Compression of protein multiple sequence alignment files.
□ Implement a (hierarchical) network community detection algorithms to recover Gene Ontology.
□ Protein function prediction from interaction network using graph.
Syllabus (tentative)
Sequence Analysis (3 classes)
Time Topic Contents
(1) Basic biology knowledge; (2) Syllabus and grading policy; (3) Final projects and presentations; (4) Dynamic programming; (5) Local alignment; (6) Global
09/09 Introduction & Sequence alignment
alignment Optional Reading:
What Is the Role for Algorithmics and Computational Biology in Responding to the COVID-19 Pandemic? [paper link]
09/23 Homology search (I) (1) Homology; (2) PAM; (3) BLOSUM; (4) Extreme Value Theory; (5) Blast; (6) PSIblast; (7) Multiple sequence alignment
09/30 Homology search (II) (1) Hidden Markov Models; (2) HMM-based sequence alignment; (3) HMMER; (4) HHpred/HHsearch; (5) HHblits
Introduction to Machine Learning (2 classes)
Time Topic Contents
(1) Introduction (2) AI & Neural Network history; (3) ML task, Overfitting, Underfitting; (4) Linear Regression; (5) MLE; (6) Neural Network; (7) Gradient Descent; (8)
Backpropagation algorithm Optional reading:
(1) Deep Learning in Bioinformatics [pdf]
10/ Introduction to Machine (2) Deep learning for computational biology [pdf]
14 Learning (3) Machine Learning in Genomic Medicine [pdf]
(4) Awesome DeepBio [github]
(5) PyTorch Performance Tuning Guide [youtube]
(6) The graph neural network model [pdf]
(7) Graph Representation Learning Book [website]
(1) Convolutional Neural Network; (2) Recurrent Neural Network; (3) Graph Neural Network; (4) Transformer
10/ Introduction to Deep Optional reading:
21 Learning (1) Attention Is All You Need [pdf]
(2) Transformer from scratch using pytorch [link]
Structure Biology (3 classes)
Time Topic Contents
(1) Template-based modelling; (2) Co-evolutation; (3) Template-free folding; (4) RFdiffusion & ProteinMPNN; (5) Protein structure alignment
Optional Reading:
(1) Fold proteins by playing games [link]
10/28 Protein structure prediction (2) Fold proteins by deep learning [link]
(3) TM-align: a protein structure alignment algorithm based on the TM-score [pdf]
(4) Protein structure alignment beyond spatial proximity [pdf]
(5) Matt: Local Flexibility Aids Protein Multiple Structure Alignment [pdf]
(1) RNA structures and functions; (2) Nussinov Algorithm; (3) Zucker algorithm; (4) Stochastic context free grammars (SCFG)
11/04 RNA structure prediction Optional Reading:
RNA Secondary Structure Prediction By Learning Unrolled Algorithms [link]
(1) MSA transformer; (2) Structural Module; (3) FAPE loss; (4) Triangle Multiplication
Optional Reading:
11/11 AlphaFold (1) Improved protein structure prediction using potentials from deep learning [pdf]
(2) Highly accurate protein structure prediction with AlphaFold [pdf]
(3) Accurate structure prediction of biomolecular interactions with AlphaFold 3 [pdf]
AI Drug Development (1 classes)
Time Topic Contents
11/18 CADD & AIDD (1) Small Molecule Drug; (2) Antibody; (3) Mathematical Representation; (3) Structure-based drug design; (4) ADMET prediction; (5) Retrosynthesis algorithm; (6) Drug Repositioning
Single Cell in Bioinformatics (2 classes)
Time Topic Contents
(1) Intro to single cell technology; (2) Differential expression analysis; (3) Dimension reduction: PCA, TSNE; (4) Missing data imputation: MAGIC, DCA algorithms; (5) Batch
effect correction: MNN algorithm
Optional reading:
11/ Single cell analysis (1) Single-cell RNA sequencing technologies and bioinformatics pipelines [pdf]
25 (I) (2) Comprehensive integration of single cell data [pdf]
(3) Single-cell RNA sequencing for the study of development, physiology and disease [pdf]
(4) A Systematic Evaluation of Single-cell RNA-sequencing Imputation Methods [pdf]
(5) A benchmark of batch-effect correction methods for single-cell RNA sequencing data [pdf]
(1) Trajectory embedding; (2) RNA velocity; (3) Bulk cell deconvolution; (4) Multi-omics integration; (5) Doublets and multiplet detection
12/ Single cell analysis Optional reading:
02 (II) (1) Lineage tracing meets single-cell omics: opportunities and challenges [pdf]
(2) Sparse Inverse Covariance Estimation with the Graphical Lasso [pdf]
(3) Foldit computer game [link]
Network in Bioinformatics (2 classes)
Time Topic Contents
12/09 Network motif detection (1) Introduction of Network Biology (2) Basic concepts about graphs (3) Random graphs (4) Network motifs in biological networks (25) G-tries algorithm
(1) Network visualization (2) PathBLAST (3) IsoRank (4) Representation-based network alignments
12/16 Network Alignment Optional Reading:
(1) REGAL: Representation Learning-based Graph Alignment [pdf]
(2) Deep Adversarial Network Alignment [pdf]
Project presentations (1 classes)
Time Topic Presentors
12/23 Project presentations TBD | {"url":"http://majianzhu.com/teaching_bioAI.html","timestamp":"2024-11-12T08:18:06Z","content_type":"application/xhtml+xml","content_length":"18760","record_id":"<urn:uuid:d6d511d7-d25c-46c7-9d04-9030c11a573e>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00197.warc.gz"} |
Is KB Bigger Than MB? Find Out Here!
• 6 min read
Is MB bigger than KB?
KB stands for Kilobyte, which is equal to 1024 Bytes or 1,000,000 B.
MB stands for Megabytes, which is equal to 1,000,000,000 B or 1,000,240,000 Bytes.
So, if you were wondering whether MB was bigger than KB, the answer is YES. One MB is larger than one KB.
But why does this matter? Well, if you’re trying to figure out how big a file size should be before uploading it to your website, you might find yourself asking, “What’s the difference between KB and
Well, here’s a quick explanation of the differences between KB and MB.
A kilobyte is equal to 1000 bytes, which is equal to 1000 KB.
A megabyte is equal to 1 million bytes, which is equal to 1,000,000 KB.
So if you’re working with files, you may notice that there’s a huge difference between KB and MB. But don’t worry, it doesn’t matter. You can still upload files that are smaller than 1 MB without any
However, if you’re dealing with large files, you will probably want to keep them under 1 MB. Otherwise, you won’t be able to view them properly on your computer screen.
Information Units In Computer Systems
MB stands for megabytes or 1 million bytes. KB stands for kilobyte or 1000 bytes. GB stands for gigabytes or 1000 megabytes. TB stands for terabytes or 1000 gigabytes. PB stands for petabyte or 1000
terabytes. And finally, YB stands for yottabyte or 10^24 bytes.
These units of information are commonly used in computing systems today. However, there are several others that are rarely used, including exabytes, zettabytes, and yottabytes.
1. Bit
A single bit is the smallest unit of information that can exist. You might think of a bit as being similar to a switch that can either be turned on or off. But there’s actually a lot more to it than
In computer science, a bit is represented by a 0 or 1. And a byte is made up of 8 bits. So if we were to count how many bits are in a single byte, we’d find out that there are eight.
So why does this matter? Well, it turns out that the number of bits needed to store any given piece of information depends on the size of the information itself.
For example, let’s say I wanted to write down my phone number. My phone number would be 7 digits long, so I’d need seven bits to store it. However, if I had a list of names and addresses, my address
book would be longer, so I’d need fewer bits to store it.
This is where the concept of multiple bits comes in. Because a byte is made up entirely of 8 bits, it can hold 2^8 256 different values.
That means that a byte can store anywhere between 0 and 255 different pieces of information. And that’s because bytes are made up of 8 bits, meaning that they can store 2^8 256 unique values.
2. Byte
A byte is the smallest unit used to store data. You may think of a byte as being similar to a kilobyte or megabyte, but it’s actually smaller. There are 8 bits in a byte, which makes it equal to 1/
256th of a kilobyte.
Bytes are commonly used to represent binary numbers. Binary numbers are made up of ones and zeros. Each number consists of two digits, called the base 2 logarithms of the number. For example, the
decimal value of 1001 is 10 in base 2. The binary equivalent of 1001 is 1101001.
In computing, a byte is usually referred to as a character. However, a byte is not necessarily a character; it can be any group of bits that represents a specific quantity of data. For example, a
byte can hold 256 values, whereas a character can only hold 128 values.
There are several types of bytes, including ASCII, EBCDIC, Unicode, UTF8, etc.
3. Kilobyte
A unit kilobyte (KB) is a multiple of the unit bit, which is a measurement for computer data storage and RAM. One KB contains 1024 bits and usually has a size of 1,024 bytes. However, it can be any
number of bytes.
Kilo means one thousand; byte means eight bits. So, 1Kb = 1000 bytes.
4. Megabyte
The size of megabyte is a measurement of data storage capacity. It’s equal to 1,000,000,000 bytes. One megabyte is equal to 1,048,576 bits. And if you’re wondering how big a megabyte really is,
here’s a handy conversion chart.
So, if you’re trying to figure out whether a file size is larger than a megabyte, you should know that it’s actually smaller than a gigabyte. But, if you’re comparing two file sizes, you should
compare them in terms of megabytes instead of gigabytes. Its unit representation is “MB”
5. Gigabyte
A gigabyte is a unit of measurement for computer storage capacity. It represents 1,073,741,824 bytes. One gigabyte equals 1,000 megabytes, which is equal to approximately 1,024 gigabytes.
In terms of computer memory, a gigabyte is equivalent to 1,024 megabytes. However, a gigabyte is usually measured in binary numbers, whereas a megabyte is measured in decimal numbers.
A gigabit is equal to 1,037,534,976 bytes. A megabit is equal to 1 million bits. A kilobit per second is equal to 1,000 bits per second. A kilobyte per second is equal to 1000 bits per second. A bit
per second is equal to 10 bits per second.
6. Terabyte
A terabyte is a lot of data. And if you think that there isn’t enough space on your computer hard drive to store all that data, you’d be wrong. You can fit almost 2 million songs onto a single
terabyte of storage.
So if you’re wondering how big a terabyte really is, here’s a fun fact for you: a single terabyte consists roughly of 1,024 gigabits or 1,024 billion bits. That’s a whole lot of data!
To give you an idea of how large a terabyte is, here’s a little fun fact for you: one terabyte equals 1,024 gigabytes. That’s a lot of data!
7. Petabyte
MB stands for megabit, which means 1 million bits. And a gigabit is 1 billion bits. But how does the number of bits relate to the size of memory? Well, if you were to have 10^15 bytes of memory, that
would mean 10^15 * 1024 10^21 bits. So if you had 1 petabyte of memory, you’d have 10^21 bits.
But what happens after the petabyte? How big is the next larger storage unit symbol? You guessed it, the exabyte. An exabyte is equal to 10^24 bits. So if you wanted to have 1 exabyte of memory, you
would have 10^24 * 1024 10^28 bits.
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Aryan Benedict
I'm Aryan, welcome to my profile and website TechLoved.com. My aim is to answer all your tech and gadget related questions in one, easy-to-navigate, website. I love technology and a lot of my
interest lies in gadgets of today. There are many common questions I am constantly asked about various products - hence the birth of Tech Loved. My awesome team and I will answer all the questions
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Dimensionality Reduction: Linear methods
Juan Manuel Ortiz de Zarate
In the realm of data analysis and machine learning, dealing with high-dimensional data can often be challenging. High-dimensional datasets can be difficult to visualize, process, and analyze
effectively. To address this issue, we use dimensionality reduction techniques. These techniques aim to project high-dimensional data onto a lower-dimensional space while retaining as much important
information as possible. This process simplifies the data, making it easier to work with and understand.
When performing dimensionality reduction, our objective is not only to reduce the number of dimensions but to do so while preserving some form of optimality. This optimality refers to maintaining
critical properties of the original dataset, such as variance, distances between data points, or class separability, depending on the specific method used.
There are primarily two main approaches to dimensionality reduction:
Feature Selection: This approach involves selecting a subset of the original features (dimensions) of the data based on certain criteria. The goal is to retain the most informative features while
discarding the less important ones. Feature selection methods include techniques such as filter methods, wrapper methods, and embedded methods. Each of these methods has its own way of evaluating the
importance of features and selecting the optimal subset.
Feature Extraction: Unlike feature selection, feature extraction creates new features by transforming the original data. These new features are combinations or projections of the original features
and are designed to capture the most significant variations in the data.
This second approach is what we will be diving into in this article.
Why would I want to reduce the dimensionality?
These techniques serve several important purposes in data analysis and machine learning. The main objectives of employing these techniques include:
Visualization: High-dimensional data is challenging to visualize directly. By reducing the number of dimensions, we can project the data into 2D or 3D spaces, making it easier to visualize and
understand the underlying patterns and relationships. Effective visualization helps in exploratory data analysis and gaining intuitive insights.
Noise Reduction: Huge datasets often contain noise, which can obscure the underlying structure of the data. Dimensionality reduction can help filter out this noise, focusing on the most significant
features and thus improving the quality of the data.
Data Regularization: Reducing the dimensionality can act as a form of regularization, preventing overfitting in machine learning models. By simplifying the data, we reduce the risk of models
capturing noise and irrelevant details, leading to better generalization of new, unseen data.
Information Compression: They allow for compressing the information content of a dataset. This is particularly useful for storage efficiency and speeding up data processing. Compressed data retains
essential information while occupying less space, making it easier to handle and transmit.
Computational Efficiency: Complex datasets data require significant computational resources for processing and analysis. By reducing the number of dimensions, we decrease the computational load,
making it more feasible to apply complex algorithms and models. This efficiency is crucial for handling large datasets and real-time applications.
Beyond these primary objectives, they can aid in other tasks such as feature engineering, anomaly detection, and improving the performance of clustering algorithms. Each application might emphasize
different aspects of dimensionality reduction based on the specific requirements and goals of the analysis.
World Maps
A familiar example of dimensionality reduction is the creation of world maps. The Earth is a three-dimensional sphere, and projecting this 3D surface onto a 2D plane inevitably involves some form of
dimensionality reduction. In this process, certain distortions or loss of information are unavoidable. These distortions manifest in various ways, such as the sizes and shapes of countries and
A world map is a dimensional reduction of Earth.
For instance, in many commonly used world maps, countries closer to the poles, such as Greenland and Russia, appear significantly larger than they are in reality [1]. This distortion occurs because
the projection methods used to flatten the globe into a 2D map cannot perfectly preserve all geographic properties. As world maps are created through dimensionality reduction, they strive to retain
important characteristics while reducing dimensions. So, projections aim to preserve essential geographical features but must balance compromises in area, shape, distance, and direction.
We will now delve into some of the most popular techniques used to achieve this. These methods each offer unique approaches to reducing dimensions, balancing the trade-offs between preserving
important data characteristics and simplifying the dataset. Let’s explore these techniques in detail, highlighting their principles, applications, and advantages.
Principal Component Analysis
Principal Component Analysis (PCA) is one of the most widely used techniques for dimensionality reduction. Given a variable X = (x1, x2,…,xp), the goal is to find a set of orthogonal linear
projections of X such that these projections are ordered in descending order according to their variance. This means that the first projection captures the maximum variance in the data, the second
projection captures the next highest variance, and so on.
• Variance as a Measure of Information: Here, variance is used to quantify the amount of information in the data. The higher the variance of a projection, the more information it retains.
Therefore, variance serves as a criterion for determining the optimality of the projections.
• Orthogonal Projections: The output involves a rotation or change of the coordinate system such that the projections (principal components) are uncorrelated. This orthogonality ensures that each
principal component captures a unique aspect of the data's variance.
The core idea behind this technique is to rotate the dataset in a way that maximizes the variance of the data in orthogonal projections. By doing so, PCA identifies the directions (principal
components) along which the data varies the most. These directions are then used to re-express the data in a lower-dimensional space while retaining as much information as possible.
Goals of PCA at inferring dimension to project on. Source [2]
PCA operates under the assumption that the data predominantly lies in a lower-dimensional linear subspace. This means that while the original data may exist in a high-dimensional space, the essential
structure and variability of the data can be captured in a space with fewer dimensions.
In the following sections, we will explore the mathematical foundations of PCA, the process of computing the principal components, and its practical applications.
The math behind this involves identifying the directions in which the variance of the data is maximized. These directions are found by analyzing the covariance matrix of the data. First, compute the
covariance matrix, Σ, of the dataset. It captures the pairwise relationships between the variables in the data. Then, to find the directions that maximize variance, solve the eigenvalue problem for
Σ. This involves finding the eigenvectors and corresponding eigenvalues of Σ.
The direction in which the variance is maximized is given by the eigenvector u of Σ with the largest associated eigenvalue λ. This is because the eigenvalue λ represents the amount of variance in the
data when projected onto the direction of its corresponding eigenvector u. Therefore, the largest eigenvalue indicates the direction where the data has the highest variance, making its eigenvector
the optimal direction for capturing the most significant variance in the data.
So, u represents the first principal component, and λ quantifies the amount of variance explained by this principal component.
Eigenvectors and eigenvalues represents the direction and magnitude of the principal components
Once the eigenvectors and eigenvalues are obtained, the eigenvectors (principal components) are ordered by their corresponding eigenvalues in descending order. The first principal component has the
highest eigenvalue, thus explaining the most variance, followed by the second principal component, and so on.
After obtaining all principal components, select the top N components that explain the most variance. The number N is chosen based on how much total variance needs to be retained. Typically, this
involves retaining enough components to capture a significant percentage of the total variance, such as 95% or 99%.
Finally, project the original data onto the selected N principal components. This projection transforms the data into a lower-dimensional space, preserving the maximum variance and thus retaining as
much of the original information as possible.
Projection and Reconstruction
After performing Principal Component Analysis (PCA) and identifying the principal components, the next steps involve projecting the original data onto these components and potentially reconstructing
the original data from the reduced dimensions.
Consider our data matrix X, which has dimensions n x p (where n is the number of samples and p is the number of features). Apply PCA to obtain the k principal components. Let V be the resulting
matrix of eigenvectors, with dimensions p x k.
To project the original data X onto the k-dimensional subspace spanned by the principal components, the projected data matrix Z is given by Z = XV. Here, Z has dimensions n x k, representing the data
in the reduced k-dimensional space.
To reconstruct the original data matrix X from the reduced data matrix Z, use the transpose of the eigenvector matrix V. The reconstructed data matrix X’ is given by
Here, transposed V has dimensions k x p, and X’ will have dimensions n x p, matching the original data matrix.
PCA projection and reconstruction
If the total variance captured by the selected k components is less than 100% of the original variance (which is usually the case), the reconstruction X’ will not be perfect. Some information is
inevitably lost during dimensionality reduction, leading to an approximation rather than an exact reconstruction. The quality of the reconstruction depends on how much of the total variance is
captured by the chosen principal components.
Singular Value Decomposition
Singular Value Decomposition (SVD) is another powerful technique for dimensionality reduction that is widely used in data analysis and machine learning. It decomposes a matrix into three other
matrices, capturing the essential patterns and structures within the original data. While PCA is a statistical technique focused on maximizing variance, SVD is a linear algebra technique that
directly decomposes the original data matrix. It does not require the calculation of the covariance matrix.
The goal of SVD is to maximize the "information" captured by the singular values, where the “information” is defined as the sum of the squared singular values. The singular values represent the
magnitude of the data along the corresponding singular vectors. Therefore, selecting the top k singular values and vectors ensures that the reconstruction retains the maximum possible energy of the
original matrix within the kk-dimensional subspace.
First, we perform matrix decomposition. Given the data matrix X, the goal is to find the matrices U,Σ, and transposed V. This can be done through a factorization process, the most famous algorithm to
do it is the Golub-Reinsch[3], which is a direct method that doesn’t do any iteration. Once the decomposition is complete, matrix Σ contains the singular values on its diagonal, sorted in descending
order and the matrices U and
contain the left and right singular vectors, respectively.
Next, we select the top k singular values from Σ and their corresponding singular vectors from U and V, similar to how we choose eigenvalues and vectors in PCA. The value of k is chosen based on
the desired level of dimensionality reduction and the amount of information to retain. This involves retaining the largest singular values, which capture the most significant patterns in the data.
So, to form the reduced representation of the data, we use the top k singular values and vectors. The reduced data matrix X_k can be constructed as:
where U_k contains the first k columns of U and Σ_k is a k x k diagonal matrix with the top k singular values.
Projection and Reconstruction
Once the data is reduced using SVD, it can be reconstructed by multiplying the calculated matrices. The reconstruction process is like projection but we also multiply it by the third (k-truncated)
As with PCA, if k is less than the original number of dimensions, the reconstruction will not be perfect but will retain the most significant information captured by the top k components. In the
following figure you can see an example of an image reduced and reconstructed with SVD where we only keep the first 50 singular components.
SVD reduction and reconstruction over an image
Both methods are linear dimensionality reduction, do they share many aspects. Indeed, PCA can be viewed as a special case of SVD. Specifically, PCA is typically performed on the covariance matrix of
the data, which is a symmetric matrix. When you center the data matrix X (subtract the mean of each feature) and then perform SVD on this centered data matrix, the left singular vectors U correspond
to the principal component directions, and the singular values in Σ are related to the eigenvalues of the covariance matrix.So, these methods have several strengths and weaknesses in common that make
them suitable for specific applications while limiting its utility in others.
• No Need to Set Hyperparameters: they do not require the user to set hyperparameters, which simplifies its application and reduces the risk of suboptimal parameter choices.
• No Iterations: Unlike many optimization algorithms, these techniques do not involve iterative procedures. This means they can be computationally efficient and straightforward to implement.
• No Local Optima: They guarantees a global optimum since it relies on solving an eigenvalue problem. This avoids the pitfalls of algorithms that may converge to local optima, ensuring consistent
and reliable results.
• Restriction to Linear Projections:
□ The primary limitation of PCA and SVD is their restriction to linear projections. They can only capture linear relationships in the data. If the underlying structure of the data is
non-linear, these methods might fail to effectively reduce the dimensionality while preserving important features.
□ For instance, if the data lies on a curved manifold in the high-dimensional space, linear methods like PCA and SVD will not be able to uncover this structure. They would only find the best
linear approximation, which might not be sufficient for capturing the essential characteristics of the data.
We have examined the two most widely used linear methods for dimensionality reduction: PCA and SVD. PCA is primarily used when the goal is to maximize the variance captured by the principal
components, making it ideal for data with linear relationships where a straightforward interpretation of the importance of each component is desired. It is commonly applied in fields such as
financial analysis [4] for risk management and portfolio optimization, as well as in data visualization to simplify the complexity of high-dimensional data.
On the other hand, SVD is a more general approach that applies directly to the data matrix and is particularly useful in natural language processing (NLP). For instance, SVD is used in topic analysis
and latent semantic analysis (LSA)[5] to uncover the underlying structure in text data by identifying patterns and relationships between terms and documents.
Although both methods are powerful, they have limitations when dealing with data that exhibit non-linear relationships. In our next article, we will explore non-linear dimensionality reduction
methods that allow us to overcome these limitations and capture more complex structures in the data. These methods will provide additional tools for effectively and robustly working with
high-dimensional data.
[2] Alwan, A., Cogranne, R., Beauseroy, P., Grall-Maës, E., Belloy, N., Debelle, L., ... & Almagro, S. (2021). A Fully Automatic and Efficient Methodology for Peptide Activity Identification Using
Their 3D Conformations. IEEE Access, 9, 92143-92156. | {"url":"https://www.transcendent-ai.com/post/dimensionality-reduction-linear-methods","timestamp":"2024-11-02T05:50:03Z","content_type":"text/html","content_length":"1050531","record_id":"<urn:uuid:cdf6fd7d-874a-4a33-b7df-c3dba42c25fc>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00302.warc.gz"} |
Multiplying Negative Numbers Worksheet
Multiplying Negative Numbers Worksheet function as foundational tools in the world of maths, offering an organized yet flexible platform for learners to explore and understand numerical ideas. These
worksheets supply an organized method to comprehending numbers, nurturing a solid structure upon which mathematical effectiveness flourishes. From the easiest checking workouts to the intricacies of
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A worksheet where you need to multiply pairs of negative numbers Number of problems 10 problems 20 problems 40 problems Answer sheet Include answer sheet Save worksheet Worksheet name SAVE Save
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At their core, Multiplying Negative Numbers Worksheet are cars for conceptual understanding. They envelop a myriad of mathematical concepts, guiding learners via the labyrinth of numbers with a
collection of interesting and purposeful workouts. These worksheets transcend the boundaries of conventional rote learning, encouraging active interaction and cultivating an instinctive understanding
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Multiplying Negative Numbers Worksheet
Multiplying Negative Numbers Worksheet
Multiplication subtraction negative numbers Textbook Exercise Previous Negatives Addition and Subtraction Textbook Exercise Next Factors Textbook Exercise The Corbettmaths Textbook Exercise on
Welcome to our Negative Numbers Worksheets hub page On this page you will find links to all of our worksheets and resources about negative numbers Need help practicing adding subtracting multiplying
or dividing negative numbers You ve come to
The heart of Multiplying Negative Numbers Worksheet hinges on cultivating number sense-- a deep comprehension of numbers' definitions and interconnections. They urge exploration, inviting learners to
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From Theory to Real-World Application
Negative Numbers Multiplication Worksheet
Negative Numbers Multiplication Worksheet
A negative number multiplied by a positive number results in a negative number and multiplying two negative numbers results in a positive number To help students practice multiplication with negative
numbers we have designed
Click here for Answers Negative Numbers Answers Corbettmaths Watch on adding subtracting arithmetic negative numbers addition subtraction add subtract multiplying multiplication dividing division
Practice Questions Previous Real life Negatives Practice Questions
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Versatility is inherent in Multiplying Negative Numbers Worksheet, employing an arsenal of instructional devices to deal with varied discovering designs. Aesthetic aids such as number lines,
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Adding And Subtracting Integers Worksheet 7th Grade With Ans
Multiply And Divide Integers Worksheet
Check more of Multiplying Negative Numbers Worksheet below
Negative Numbers
The Multiplying Integers Positive Times A Negative A Math Worksheet Negative Numbers
Adding And Subtracting Negative Numbers Worksheets Negative Numbers Worksheet AliviaWeeks14
20 Best Multiplying Negative Numbers Worksheet Ks3
Multiplying Negative Numbers Worksheet
Multiplying Negative Numbers Worksheet
Multiply Negative Numbers WorksheetMath
A worksheet where you need to multiply pairs of negative numbers Number of problems 10 problems 20 problems 40 problems Answer sheet Include answer sheet Save worksheet Worksheet name SAVE Save
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Negative Numbers DadsWorksheets
16 Negative Numbers Worksheets The worksheets in this section introduce negative numbers integers in multiplication and division math problems All problems deal with smaller integers that can be
solved without multi digit multiplication or long division Multiplication and Division Facts
A worksheet where you need to multiply pairs of negative numbers Number of problems 10 problems 20 problems 40 problems Answer sheet Include answer sheet Save worksheet Worksheet name SAVE Save
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16 Negative Numbers Worksheets The worksheets in this section introduce negative numbers integers in multiplication and division math problems All problems deal with smaller integers that can be
solved without multi digit multiplication or long division Multiplication and Division Facts
20 Best Multiplying Negative Numbers Worksheet Ks3
The Multiplying Integers Positive Times A Negative A Math Worksheet Negative Numbers
Multiplying Negative Numbers Worksheet
Multiplying Negative Numbers Worksheet
Mixed Problems Negative Numbers Worksheets Multiplication Worksheets Math Worksheets 5th
Multiplying Negative Numbers Worksheet
Multiplying Negative Numbers Worksheet
49 Multiplying Negative Numbers Worksheet | {"url":"https://szukarka.net/multiplying-negative-numbers-worksheet","timestamp":"2024-11-14T04:23:19Z","content_type":"text/html","content_length":"29421","record_id":"<urn:uuid:32c8976f-8851-4836-9556-f259515c2122>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00166.warc.gz"} |
Plot the given point (
Step 1
The rectangular coordinate system consists of two real number lines that intersect at a right angle.
The horizontal number line is called the x-axis, and the vertical number line is called the y-axis.
These two number lines define a flat surface called a plane, and each point on this plane is associated with an ordered pair of real numbers (x, y). The first number is called the x-coordinate, and
the second number is called the y-coordinate.
The intersection of the two axes is known as the origin, which corresponds to the point (0, 0)
Step 2
An ordered pair (x, y) represents the position of a point relative to the origin.
The x-coordinate represents a position to the right of the origin if it is positive and to the left of the origin if it is negative. The y-coordinate represents a position above the origin if it is
positive and below the origin if it is negative.
Using this system, every position (point) in the plane is uniquely identified. For example, the we have pair (-4, 0) denotes the position relative to the origin as shown:
Thus, point A is shown in the above graph. | {"url":"https://plainmath.org/college-statistics/2129-plot-the-given-point-4-0-in-a-rectangular-coordinate-system","timestamp":"2024-11-04T17:02:37Z","content_type":"text/html","content_length":"154800","record_id":"<urn:uuid:38ff16df-5e21-43ae-b78a-a660c1001bbd>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00568.warc.gz"} |
is the Wrong Debate
It’s time to wave the truce flag over one of personal finance’s greatest battles.
One of the most spirited debates in personal finance circles is the classic “earn more” vs. “save more” standoff.
The “earn more” crowd argues that raising your income is the best way to reach financial independence. There’s no upper limit to this potential, especially if you’re an entrepreneur.
The “save more” crowd argues that, left unchecked, income and spending naturally scale together. Focusing on frugality, therefore, is a better method for creating financial independence.
Here’s what neither side admits:
Both hold unstated assumptions. The “earn more” crowd believes that they’ll spend a fixed amount, regardless of their income. The “save more” crowd believes that they’ll spend a percentage of their
The debate boils down to this difference in worldview.
The classic argument “if you earn more, you’ll spend more” is based on behavior and mindset. It argues that people spend a proportion of their income — 90 percent, 80 percent, 50 percent, whatever —
regardless of how much they earn.
The counterargument “the more you earn, the easier it is to save” focuses on math and logic. It argues that you need a baseline foundation to cover your cost-of-living plus reasonable comforts. The
surplus goes directly to savings; therefore, higher surpluses equal more savings.
Both sides, though, focus on growing the gap between income and spending. And that’s where this conversation needs to turn.
The gap should be our one and only focus. Here’s how I define this gap:
You earn $X. You spend $Y. The gap is the difference, $X-$Y.
Your job is simple: make this gap as large as possible. You can push $X higher. You can drive $Y lower. You can attack both simultaneously. The size of the gap matters more than the method you
Once you’ve grown the gap, invest this into anything that creates cash flow. “Cash flow” are the critical words. Your home is not an investment. It’s cash-flow-negative. Rental properties, index
funds, and owning a business, however, enhance your account balances rather than deplete it. These are great uses for the gap.
Let’s look at the power of the gap.
You earn $100,000 annually after taxes. You spend $60,000 and create a gap of $40,000 per year, which you invest in index funds that grow at 7 percent compounding annually. After 19 years, you’ve
grown $1.5 million.
If you quit working and withdraw 4 percent of this, you’ll have $60,000 per year before taxes for the rest of your life. That’s financial independence in less than two decades (or faster, if you
fling raises into investments).
Can we pause for a moment? We just outlined $0 to retirement in 19 years. Nineteen years. And this happened while living on a comfortable $60,000 of annual spending, which isn’t a fringe extreme.
In fact, let’s take a step back. Let’s imagine a different situation. You want to retire to a balmy beach in Thailand on $40,000 per year. How much money will you need? Let’s figure this out in our
heads, within a few seconds, using easy back-of-the-napkin rules-of-thumb.
If you prefer index funds:
The 4 Percent Rule: You can sustainably live on 4 percent of your investments every year. In other words, $1 million in your portfolio means you can live on $40,000, adjusted for inflation, every
The 25x Rule: Calculate the amount you want to live on. Multiply by 25. This is the size your portfolio needs to be. Notice the relationship to the 4 percent rule — $40,000 x 25 equals $1 million.
If you prefer rental properties:
The 1 Percent Rule: Your rental properties should collect one percent of their value in monthly rent. In other words, $1 million worth of rentals should collect $10,000 per month in top-line revenue.
(“There’s nothing in my area!” — Go where the money is.)
The 50 Percent Rule: Half of the rent will get gobbled up by operating costs, such as repairs, maintenance, management, etc. The implication: if you hold free-and-clear properties that meet the 1
percent rule, you can expect to pocket 6 percent of the home’s value every year, after expenses.
How soon can you retire to a tropical Thai island, based on these general parameters? If you invest in index funds, you’ll need $1 million. If you invest in rental properties, you’ll need $660,000 in
properties free-and-clear.
Your target, therefore, is to build $660,000 – $1 million worth of investments. Reach this singular goal, and you’ll receive $40,000 in passive income for the rest of your life. If that
income feels too low, double everything. You’ll need $1.3 – $2 million to create $80,000 in passive income per year.
Some of you might be thinking: “Really? That’s it?,” while others might think these numbers sound daunting. To those in the latter group: everything worthwhile feels daunting at the start, and even
the most complex situations feel easy in hindsight.
You’ve mastered driving a car, memorizing all 50 states, filing taxes, and all types of other cognitively complex behaviors. You can manage this, too. You’ve already taken the first step. You’ve
built the map. You know the goal. Now the focus is growing the gap.
Financial independence boils down to one simple focus: the gap. Grow it. Invest it. Repeat. There’s not much more to say. We can draw diagrams and build models under a hailstorm of variables, juicing
ourselves with pure nerd-fun along the way. But this spreadsheet tinkering isn’t necessary.
The only road to financial independence comes from focusing on the gap. Do this one thing, and everything else falls into place. | {"url":"https://affordanything.com/financial-independence/","timestamp":"2024-11-11T15:12:39Z","content_type":"text/html","content_length":"77928","record_id":"<urn:uuid:d9b5cbd0-d25a-4eef-8512-08587a7dc27e>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00489.warc.gz"} |
ACO Seminar
The ACO Seminar (2012-2013)
Feb 28
, 3:30pm, Wean 8220
Patrick Bennett
, CMU
A greedy algorithm for finding a large 2-matching on a random cubic graph
A 2-matching of a graph G is a spanning subgraph with maximum degree two. The size of a 2-matching U is the number of edges in U, and this is at least n - k(U) where n is the number of vertices of G
and k denotes the number of components. The algorithm 2GREEDY is a greedy algorithm due to Frieze which finds a maximal 2-matching in a graph. This algorithm is similar to the well known Karp-Sipser
algorithm, which finds ordinary matchings in graphs. We analyze the performance of 2GREEDY on a random 3-regular graph. Our analysis yields (more-or-less) matching upper and lower bounds on the final
number of components. In particular, we prove that with high probability, the algorithm outputs a 2-matching U with k(U) = n^(1/5 + o(1)). | {"url":"https://aco.math.cmu.edu/abs-12-02/feb28.html","timestamp":"2024-11-05T07:23:55Z","content_type":"text/html","content_length":"2160","record_id":"<urn:uuid:c86fc061-b4f3-4e88-9fc5-98a68dccfaac>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00037.warc.gz"} |
TCS Divisibility Question Answers 2 » PREP INSTA
TCS Divisibility Question Answers 2
Sarita Wants to gift the stationary kit. Which includes a pencil, an eraser and a sharpener to some kids in her village. She went to a wholesaler for purchase these items. The seller sells pencils in
packing of 12. Erasers in a packing 18 and Sharpener in a packing of 10.She purchases the minimum number of packets of pencils,eraser and sharpener.so that affix number of kits can be prepared
without any item being unused or short. What is the number of packet of pencils purchased by her ?
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What is the greatest four digit number divisible by 5 , 10 , 15 and 20?
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What is the greatest possible positive integer n if $8^{n}$ divides $(44)^{44}$ without leaving a remainder?
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Find the number of divisors of 1728.?
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what is the remainder of $\frac{(16937^{30})}{31}$
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16937 & 31 are coprimes so we can use Euler\'s theorem to find 16937^30 mod 31. By Euler\'s totient function, Phi (31) = 30. For a prime p, phi (p) = p-1. By Euler\'s theorem, 16937^[phi (31)] mod 31
= 1. So, 16937^30 mod 31 = 1. Read more on Brainly.in - https://brainly.in/question/5047203#readmore
The numbers 272738 and 232342, when divided by n, a 2 digit number, leave a remainder of 13 and 17 respectively. Find the sum of the digits of n?
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What is the remainder of $32^{31^{301}}$ when it is divided by 9?
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Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?
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If a number is divided by 357 the remainder is 5, what will be the remainder if the number is divided by 17?
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26ab5 is a four-digit number divisible by 25. If the number formed from the two digits ab is a multiple of 13, then ab =?
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In base 2, 19 is 10011. In base 1/2, 19 is 1.1001
In base 10, 19 is 19. In base 1/10, 19 is 9.1
The trick is to write the number in reverse, and in the fractional version, put a decimal point after the ones column. This is because decimals have negative exponents, so the fraction gets flipped.
Fractional bases that aren't 1/something seem like a real mess to use.
Also, I wanted to count in balanced base 3. I will use -, 0, and + for my digits.
+ +- +0 ++ +-- +-0 +-+ +0- +00 +0+
You'll notice that you *can* count be incrementing the one's column repeatedly. You just have to remember to reset things to - not 0, after they overflow.
Will count from -1 to -5 now:
- -+ -0 -- -++
Notice it's the same as positive, with the -'s and +'s reversed, and 0's untouched. And you can count by decrementing the one's column, and when it overflows, decrement the next column and reset
things to a +. I guess I should point out that you can add as many leading zeroes as you want, which is how decrementing a column that doesn't exist works.
Want to discuss this? Join my forum.
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The Rule of 70: A Simple Tool for Calculating Economic Growth - Neat Dollar
Reading Time: 3 minutes
The Rule of 70 is a mathematical formula used to estimate the years it takes for a quantity to double. It is a formula that divides the number 70 by the before-given fixed annual growth rate.
This rule of 70 formula is commonly used in finance and economics to calculate the doubling time of investments, population growth, or inflation, among other things.
This guide will dive deeper into the Rule of 70 and explain it with examples. So with all that said, let’s begin.
Rule of 70 Examples
The easiest way to explain the Rule of 70 is through examples. Here are a few financial, population, and inflation examples that explain the formula behind the rule.
Rule of 70 for Investment Growth
As mentioned, The Rule of 70 is often used in finance to estimate the years it will take for an investment to double. For example, if an investment is expected to grow at 7% per year, the Rule of 70
tells us that it will double in approximately 70/7 = 10 years.
Rule of 70 for Population Growth
The Rule of 70 is also used to estimate the years it will take for a population to double, given a fixed annual growth rate. For example, if a population grows at a rate of 1% per year, the Rule of
70 tells us that it will double in approximately 70/1 = 70 years.
Rule of 70 for Inflation
We can also use the Rule of 70 formula to estimate the number of years it will take for the purchasing power of money to be halved, given a fixed annual inflation rate. For example, let’s say the
inflation rate is 3% per year. In that case, the Rule of 70 formula tells us that the purchasing power of money will be halved in approximately 70/3 = 23 years.
Why 70 in the Rule of 70?
The number 70 is used in the Rule of 70 because of mathematics. Since the formula determines the years it takes for an investment to double, it looks at the number “2” to calculate.
First, we need to calculate the natural logarithm of 2, which is 0.69. We can round up the number to 0.7.
Then, we need to convert 0.7 to percentages, which is 70% or 70. This is why mathematicians use the number 70. Although it’s commonly accepted in finance to use the number 72 as they believe it is a
“nicer” number to divide into.
How to Use It?
To use the Rule of 70, you simply need to remember the formula, which is 70 divided by the annual growth rate expressed as a percentage.
For example, let’s say the annual growth rate is 7%. In that case, the doubling time will be approximately 70/7 = 10 years.
The rule is based on the exponential growth formula, which states that the quantity at any time can be calculated as the initial quantity multiplied by the exponential function of the growth rate.
This rule of 70 formula shows that the growth of a quantity depends on both the size of the initial quantity and the annual growth rate. It’s also important to understand the exponential function,
which has the property of compounding, which means that the growth rate multiplies the quantity over time.
The Rule of 70 isn’t accurate for very high or low growth rates or for quantities that grow at a variable or non-constant rate and is used to express a rough estimate.
However, it is a useful tool for making quick, rough estimates and comparing different quantities’ growth.
The Rule of 70 is a simple and useful formula for estimating the doubling time of a quantity given a fixed annual growth rate. It is widely used in finance and economics and can provide a rough
estimate for various applications. However, it is important that the rule is an estimate and may not be accurate for all situations.
Alternatively, the Rule of 72 and the Rule of 69 can also be used for the same purpose.
How Does the Rule of 70 Work?
The rule of 70 is a mathematical formula calculating the years it takes for a quantity to double. It works by dividing 70 by the annual growth rate.
What is the Rule of 70 in Population?
The rule of 70 in population is a formula that helps governments and institutions calculate the years it takes for a population to double.
What is Rule of 70 In Inflation?
The rule of 70 in inflation helps calculate the years it takes for the purchasing power of money to be halved. | {"url":"https://neatdollar.com/rule-of-70/","timestamp":"2024-11-06T08:33:47Z","content_type":"text/html","content_length":"205023","record_id":"<urn:uuid:47a2005c-7e46-4f4a-997e-f332c3cd4fbe>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00195.warc.gz"} |
multiply mul
Multiplies properties of objects with a specified factor.
Multiplies properties of objects with a specified factor.
multiply PointLoad_1 4 Multiplies components of PointLoad_1 with a factor of 4.
Number Number Factor by which the values of the object properties should be multiplied
Multiplies properties of objects with a specified factor.
multiply LineLoad_1_1 "qx_start" Phase_3 2
<{1,...}: Text'>' PlxBasePhase' Number
{1,...} 1 ∞ Sequence of one or more objects
Text Property name Name of property that should be multiplied
PlxBasePhase Phase reference Name of the phase in which the object is to be multiplied
Number Number Factor by which the values of the object properties should be multiplied
Multiplies properties of objects with a specified factor.
multiply LineLoad_1 "qy_start" 2 Multiplies qy_start component of LineLoad_1 with a factor of 2.
multiply PointLoad_1 "Fy" "Fz" 10 Multiplies Fy and Dz components of PointLoad_1 with a factor of 10.
<{1,...}: Text'>' Number
{1,...} 1 ∞ Sequence of one or more objects
Text Property name Name of property that should be multiplied
Number Number Factor by which the values of the object properties should be multiplied
Multiplies properties of objects with a specified factor.
multiply (LineLoad_1 PointLoad_2.Fy) 2 Multiplies LineLoad_1 with a factor of 2 and the Fy component of PointLoad_2 with a factor of 2.
multiply (PointLoad_1 PointLoad_2) "Fy" 2 Multiplies Fy component of PointLoad_1 and PointLoad_2 with a factor of 2.
multiply (PointLoad_1_1 PointLoad_2_1) "Fy" InitialPhase 2 Multiplies Fy component of PointLoad_1 and PointLoad_2 in the Initial phase with a factor of 2.
<{1,...}: PlxObject'>' <Everything>'
{1,...} 1 ∞ Sequence of one or more objects
PlxObject PlxObject Plaxis object to be multiplied
Everything Parameter One or more parameters for the command that should be applied
Multiplies properties of objects with a specified factor.
multiply (LineLoad_1 PointLoad_2) 2 Multiplies components of LineLoad_1 and PointLoad_2 with a factor of 2.
Listable' Number'
Listable Listable reference Name of the listable to which the command should be applied
Number Number Factor by which the values of the object properties should be multiplied
Multiplies properties of objects with a specified factor.
multiply (PointLoad_1 PointLoad_2) "Fy" 2 Multiplies Fy component PointLoad_1 and PointLoad_2 with a factor of 2.
Listable' <{1,...}: Text'>' Number
Listable Listable reference Name of the listable to which the command should be applied
{1,...} 1 ∞ Sequence of one or more objects
Text Parameter Parameter required to be multiplied
Number Number Factor by which the values of the object properties should be multiplied | {"url":"https://www.eatrice.cn/Plaxis3D/input_commands/commands_multiply.html","timestamp":"2024-11-14T05:06:34Z","content_type":"text/html","content_length":"28560","record_id":"<urn:uuid:9eee9a78-e183-4faf-9af8-ad16c539a8dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00631.warc.gz"} |
Investigation of the dependence of the error in the approximate solution of the Laplace equation on the mean minimum sine of the angle of the cells of the computational grid
Investigation of the dependence of the error in the approximate solution of the Laplace equation on the mean minimum sine of the angle of the cells of the computational grid
Incoming article date: 17.12.2021
The paper studies the issue of the influence of the quality of the computational triangular grid on the accuracy of calculations in various computational problems. There is a well-known example
of Schwartz, which shows that the approximation of a smooth surface by a polyhedral surface can give very large errors for calculating the surface area. This is due to the quality of the
constructed triangulation of the surface. Therefore, it is natural to expect that there is some connection between a certain triangulation characteristic and the accuracy of solving some
computational problem. In the presented article, as such a characteristic, a value is chosen - the average value of the minimum sine of the angle of all triangles of the computational grid. In
the course of numerical experiments, the Dirichlet problem for the Laplace equation in a circular ring was solved, in which the error of the approximate solution was calculated (the gradient
descent method was used to find a solution to the corresponding variational problem.). For the ring, a series of triangulations was constructed with a uniform division along the angle and a
non-uniform division along the radius in polar coordinates. In this example, a linear dependence of the error on was shown. The article presents both the results of the calculation with different
values and the calculation of the correlation coefficient of the studied quantities.
Keywords: boundary value problem, Delaunay triangulation, calculation accuracy, Dirichlet problem, mathematical modeling, triangular mesh, minimum triangle angle, piecewise linear approximation,
variational method, Laplace equation | {"url":"http://ivdon.ru/en/magazine/archive/n1y2022/7407","timestamp":"2024-11-11T03:35:27Z","content_type":"text/html","content_length":"17768","record_id":"<urn:uuid:5cdb5540-ee7b-44dd-9506-c8f534176ffa>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00692.warc.gz"} |
Advanced Order Of Operations Worksheet With Answers | Order of Operation Worksheets
Advanced Order Of Operations Worksheet With Answers
Order Of Operations PEDMAS With Integers 3 Worksheet
Advanced Order Of Operations Worksheet With Answers
Advanced Order Of Operations Worksheet With Answers – You may have listened to of an Order Of Operations Worksheet, however what exactly is it? In addition, worksheets are an excellent means for
trainees to practice new skills and also evaluation old ones.
What is the Order Of Operations Worksheet?
An order of operations worksheet is a kind of mathematics worksheet that calls for pupils to carry out math operations. These worksheets are divided into 3 main sections: multiplication, addition,
and also subtraction. They likewise include the examination of exponents and parentheses. Pupils that are still discovering how to do these jobs will certainly discover this kind of worksheet
The primary function of an order of operations worksheet is to assist pupils learn the right means to address math formulas. They can examine it by referring to a description web page if a trainee
doesn’t yet recognize the idea of order of operations. On top of that, an order of operations worksheet can be split right into numerous groups, based upon its problem.
One more important function of an order of operations worksheet is to instruct students how to perform PEMDAS operations. These worksheets begin with simple issues related to the fundamental policies
as well as build up to more complex problems including all of the rules. These worksheets are an excellent way to present young learners to the exhilaration of addressing algebraic equations.
Why is Order of Operations Important?
One of one of the most vital points you can find out in math is the order of operations. The order of operations guarantees that the math problems you address are consistent. This is crucial for
examinations and real-life calculations. When addressing a math issue, the order ought to start with parentheses or backers, followed by subtraction, addition, and multiplication.
An order of operations worksheet is a terrific way to instruct trainees the appropriate means to fix mathematics equations. Before trainees begin utilizing this worksheet, they may require to
evaluate ideas associated with the order of operations. To do this, they should examine the principle page for order of operations. This concept page will certainly provide pupils a summary of the
An order of operations worksheet can assist pupils create their abilities furthermore and also subtraction. Educators can use Prodigy as an easy means to set apart practice as well as supply engaging
web content. Prodigy’s worksheets are an ideal means to help trainees find out about the order of operations. Educators can begin with the standard concepts of multiplication, division, and also
addition to aid trainees construct their understanding of parentheses.
Advanced Order Of Operations Worksheet With Answers
12 Sample Order Of Operations Worksheets Sample Templates
Cobb Adult Ed Math Interger And Order Of Operation Worksheet Solutions
Study Guides Extra Practice Mrs Wilson 6th Grade AC Math
Advanced Order Of Operations Worksheet With Answers
Advanced Order Of Operations Worksheet With Answers offer a great resource for young students. These worksheets can be easily tailored for details requirements. They can be discovered in three levels
of trouble. The initial degree is basic, requiring pupils to exercise making use of the DMAS technique on expressions consisting of four or even more integers or 3 operators. The second degree
requires students to use the PEMDAS method to streamline expressions utilizing outer and internal parentheses, brackets, and also curly dental braces.
The Advanced Order Of Operations Worksheet With Answers can be downloaded totally free and can be printed out. They can after that be assessed using addition, multiplication, division, and
subtraction. Pupils can likewise utilize these worksheets to assess order of operations as well as using backers.
Related For Advanced Order Of Operations Worksheet With Answers | {"url":"https://orderofoperationsworksheet.com/advanced-order-of-operations-worksheet-with-answers/","timestamp":"2024-11-11T13:59:53Z","content_type":"text/html","content_length":"43209","record_id":"<urn:uuid:b7902562-9f58-481c-ab31-9a6737f068a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00532.warc.gz"} |
Equations estimate relative permeability of mixed wettability shaly sands
New equations for estimating the relative permeability curves for both water-oil and gas-oil systems have been developed for shaly sands in mixed wettability formations of the Western Desert in
These correlations also may be used in other regions of the world.
The developed equations have good agreement with laboratory data.
Permeability correlations
Click here to enlarge image
Relative permeability is a direct measure of a porous medium's ability to conduct one fluid when one or more fluids are present. These flow properties are the composite effect of pore geometry,
wettability, fluid saturation, and saturation history.^1 2 The direct determination of relative permeability from special core analysis data and theoretical considerations is valuable information for
reservoir analysis.
Laboratory methods for measuring the relative permeability curves include steady state and unsteady-state methods.^3 4 These methods are expensive and time consuming. With these methods, the relative
permeability curves of representative and sufficient core samples from a reservoir are measured and then the curves are averaged to obtain an average relative-permeability curve for that reservoir.
In the case in which the number of the available core samples is insufficient, the Khairy technique can be used to generate two-phase relative permeability curves.^5 In the absence of core analysis
data, the relative permeability can be determined from empirical equations. In the case of three-phase flow, three-phase relative permeability models such as those of Naar,^6 Blunt,^7 and Moulu^8 can
describe the flow in the reservoir.
For two-phase flow, two-phase relative permeability correlations, such as those of Wyllie,^9 Naar,^10 Pirson,^11 Honarpour,^12 and Mohamed,^13 predict flow in a porous media.
Some of these correlations have disadvantages. For instance, the Naar correlation does not specify the rock type and the Honarpour correlation does not distinguish between intermediate and mixed
wettability. Some of the other correlations also do not specify wettability conditions.
Additionally from the Mohamed correlation, the development of the two-phase relative permeability curves from normalized relative permeability vs. normalized saturation is impossible without knowing
the end points of the relative permeability curves and the absolute reservoir permeability.
Petroleum engineers are liable to use these models indiscriminately for predicting reservoir performance. One study compared the relative permeabilities estimated by ten different two-phase relative
permeability models with 12 experimental data sets gathered from Berea and Brown sandstone.^14
This study also used a two-dimensional, two-phase numerical simulator to compare the performance of four of the experimental relative permeability data sets and those predicted by the models used to
predict Buckley-Levert displacement and waterflooding performance.
The study indicated a varying degree of performance for the correlations in reservoir engineering calculations. This may have been due to each correlation having its own applicable conditions, such
as lithology, wettability, rock properties, and fluid saturations ranges.
The study showed that the Wyllie, Naar, Pirson, and Honarpour correlations have a low accuracy when applied to shaly sand formations of mixed wettability.
Data description
The data used as a basis for this study to derive new correlations were derived from core samples in the Western Desert in Egypt.^5 15 16 The data sets are for consolidated shaly-sandstone of mixed
wettability. The wettability classification is based on Amott and Amott-Harvey wettability indices, oil-recovery efficiency of water displacement and characteristics of relative permeability, and
capillary pressure curves.^5 15 16
The study included 57 core samples from 13 different formations. All the laboratory tests were made at room temperature and atmospheric pressure.
Equations of the oil-water system refer to displacement of oil by water, and the gas-oil system equations refer to displacement of oil by gas. The base permeability of the measured relative
permeability values is the effective oil permeability at irreducible water saturation, S[wi]. The data were normalized by use of published methods.^5 17
A comparison of the analyzed curve shape with the usual shape of the relative permeability curves checked the validity of the relative permeability analysis. This led to the exclusion of three core
samples for the water-oil system and eight core samples the for gas-oil system.
The new equations for water-oil system were based on 54 core samples and 627 data points, while the equations for gas-oil system were based on 49 core samples and 641 data points.
Tables 1 lists the range of the data used to develop the equations.
Click here to enlarge image
Equations 1-13 (see equation box) were derived by using linear regression analysis techniques to develop equations that would approximate the normalized data, (K[ro]*)[o-w] and (K[rw]*)[o-w] vs. S[w]
*, and (K[ro]*)[o-g] and (K[rg]*)[o-g] vs. S[g]*.
During the preliminary evaluation stage, one can estimate the critical gas saturation with Equation 13, and the critical gas saturation can be refined by history-matching production data.
The R^2 statistical concept determined the goodness of fit of the new equations. As R^2 approaches 1, it indicates an adequate fit.
Estimation of curves
Estimating the relative permeability curves for water-oil and gas-oil systems using the new equations requires the following steps:
1. Determine the effective oil permeability (K[o])[o-w] at S[wi] from transient well test.
2. Use the value of (K[o])[o-w] at S[wi] in Equations 10-13, estimate values to (K[rw])[o-w] at S[or], (K[rg])[o-g] at (1-S[wi]-S[or]), (K[ro])[o-g] at S[gc], and S[gc].
3. Estimate values to S[wi], and S[or] from well log data.
4. Use the estimated values in Steps 2 and 3 in Equations 5-8 to estimate the relative permeability curves for oil-water and gas-oil systems.
Click here to enlarge image
The results from these equations were compared to the results from correlations of Wyllie, Naar, Pirson, Honarpour, and Mohamed, as well as experimental data (Fig. 1).
Click here to enlarge image
The end points of the experimentally determined relative-permeability curves and the absolute permeability of the core were used to calculate Mohamed's normalized relative permeability curves and
covert them to denormalized relative permeability curves.
Click here to enlarge image
For water-oil systems, Figs. 2 and 3 show estimated oil and water relative permeability vs. measured oil-water relative permeability, respectively. From these figures, one can see the that oil
relative permeability lies between that of Equation A9, which is for intermediate, mixed wetting conditions, and Equations A1 and A5, which are for strong water wet and water wet, respectively.
The figures also show that the new equations for relative permeability are the closest to experimental data. The correlations of Wyllie , Naar, Pirson, and Honarpour have a low accuracy probably
because of wetting and lithological conditions.
Figs. 1 and 3 show that the new equation for water relative permeability lies between Equation A2, which is for oil and intermediate wetting conditions, and Equations A2, A6, A1, and A10, which are
for strong water wet, water wet, water wet, and intermediate, mixed wettability, respectively.
Click here to enlarge image
The new equation for water relative permeability is the closest to the experimental data except at a high water saturation. Fig. 4 shows oil and gas relative permeability curves generated by the new
equations, published correlations, and the experimental data from one core sample.
Figs. 5 and 6 illustrate the estimated gas and oil relative permeability vs. measured gas and oil relative permeability, respectively.
Click here to enlarge image
Figs. 4 and 5 indicate that the new equation for relative permeability gives higher values than Equation A18, which is for unknown wetting conditions. Also, the equation, except at high gas
saturation, has the same order of magnitude as Equation A5, which is for any wetting condition.
Click here to enlarge image
The new equation for gas relative permeability is the closest to the experimental data along the entire gas saturation range, and the new equation for oil relative permeability is almost equal to the
experimental data along the entire gas saturation range (Figs. 4 and 6).
The figures illustrate also that the new correlation for oil relative permeability lies between Equation A17, which is for unknown wetting conditions, and Equation A4, which is for any wetting
1. Craig, F.F., The Reservoir Engineering Aspects of Waterflooding, SPE Monograph Series, 1973.
2. Unalmiser, S., and Funk, J.J., "Engineering Core Analysis," JPT, April 1998.
3. Amyx, J.W., Bass, D.M., and Whiting, R.L., Petroleum Reservoir Engineering, McGraw-Hill, New York, 1960.
4. Honarpour, M., Koederitz, L.F., and Herbert, A.H., Relative Permeability of Petroleum Reservoirs, CRC Press Inc. Boca Raton, Fla., 1986.
5. Khairy, M., "Estimating Two-Phase Relative Permeability Using Well Test and Log Data," Journal of Engineering and Applied Science, Vol. 47, No. 4, August 2000, pp. 739-53.
6. Naar, J., and Wygal, R.J., "Three-Phase Imbibition Relative Permeability," SPE Journal, December 1961, pp. 254-58.
7. Blunt, J.M., "An Empirical Model for Three-Phase Relative Permeability," Paper No. SPE 56474, SPE Annual Technical Conference and Exhibition, Houston, Oct. 3-6, 1999.
8. Moulu, J.C., et al., "A New Three-Phase Relative Permeability Model for Various Wettability Conditions," Paper No. SPE 56477, SPE Annual Technical Conference and Exhibition, Houston, Oct. 3-6,
9. Wyllie, M.R.J., "Interrelationship Between Wetting and Nonwetting Phase Relative Permeability," Transactions, AIME, Vol. 192, 1950, pp. 381-82.
10. Naar, J. and Henderson, J.H., "An Imbibition Model-Its Application to Flow Behavior and the Prediction of Oil Recovery," SPE Journal, June 1961, pp. 61-70.
11. Pirson, S.J., Oil Reservoir Engineering," McGraw-Hill, New York, 1958.
12. Honarpour, M., Koederitz, L.F., and Herbert, A.H., "Empirical Equations for Estimating Two-Phase Relative Permeability in Consolidated Rock," JPT, December 1982, pp. 2905-08.
13. Mohamed, I.M.N., and Koederitz, L.F., "Two-Phase Relative Permeability Prediction Using a Linear Regression Model," Paper No. SPE 65631, SPE Eastern Regional Meeting, Morgantown, W.Va, Oct.
17-19, 2000.
14. Siddiqui, M., Ertekin, T., and Hicks, P.J. Jr., "A Comparative Analysis of the Performance of Two-Phase Relative Permeability Models in Reservoir Engineering Calculations," Paper No. SPE 26911,
Eastern Regional Conference and Exhibition, Pittsburgh, Pa., Nov. 2-4, 1993.
15. Khairy, M., "An Integrated Approach for Deciding Wettability of Shaly-Sand Reservoirs," Al-Azhar Engineering Sixth International Conference, Cairo, Sept. 1-4, 2000.
16. Khairy, M., "Technique determines average capillary pressure," OGJ, Jan. 15, 2001, pp. 37-43.
17. Chierici, G.L., "On the Normalization of Relative Permeability Curves," SPE Paper No.5962, 1976.
The author
Mohamed Khairy is an associate professor at King Saud University, Riyadh. He previously was an associate professor at Cairo University and also worked for Agiba Petroleum Co. Khairy has a BS and MS
in petroleum engineering from Cairo University and a PhD through a program between Cairo University and Haute Alsace University in France. | {"url":"https://www.ogj.com/home/article/17220570/equations-estimate-relative-permeability-of-mixed-wettability-shaly-sands","timestamp":"2024-11-04T12:23:17Z","content_type":"text/html","content_length":"414082","record_id":"<urn:uuid:dfff198d-f9ef-4b17-9550-96aa35b1e6a8>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00881.warc.gz"} |
Decimals Worksheets - Definition, Advantages, Practice
Decimal Worksheets
Decimal worksheets are educational resources designed to help students in lower grades practice and understand decimal numbers. These worksheets typically include a set of problems or exercises
related to decimal arithmetic, such as addition, subtraction, multiplication, and division.
By working through these exercises, students can improve their skills in using decimals in real-life situations and develop a deeper understanding of the concepts related to decimal numbers. Decimal
worksheets can be a valuable tool for teachers to assess their students' progress and identify areas where additional instruction may be needed. They can also be used by students for self-study or
Decimal numbers are an important part of mathematics and a key concept for students to understand in grades 4 to 8. Decimal numbers represent values greater than one in a compact and efficient way,
allowing for the expression of precise quantities. Understanding decimals is crucial for students as they prepare for higher mathematics and real-world applications. In this article, we will discuss
what decimal numbers are, the advantages of decimal worksheets, and provide decimal worksheets for practice.
What are decimal numbers?
Decimal numbers are a way to represent fractional values in a compact and efficient manner. They are written using a dot (.) to separate the whole number part from the fractional part. For example,
the decimal number 1.2 represents one whole number and two-tenths (0.2). Decimal numbers can be used to express a wide range of values, from small (e.g., 0.00001) to large (e.g., 1,000,000).
Advantages of Decimal Worksheets
Decimal worksheets offer several advantages for students:
1. Skill Development: By practicing decimal operations, students can improve their skills in using decimals in real-world situations, such as measuring, calculating, and problem-solving.
2. Conceptual Understanding: Completing decimal worksheets helps students develop a deeper understanding of the concepts related to decimal numbers, such as place value, rounding, and the
relationship between decimals and fractions.
3. Assessment and Feedback: Decimal worksheets allow teachers to assess their students' progress and identify areas where additional instruction may be needed. This feedback can help guide the
teaching and learning process.
4. Self-Study and Homework: Decimal worksheets can be used by students for independent practice or as part of their homework assignments, reinforcing the concepts learned in the classroom.
Decimal worksheets for practice
1. Rounding decimals: This worksheet provides practice in rounding decimal numbers to the nearest whole number, tenth, and hundredth.
2. Adding and subtracting decimals: This worksheet provides practice in adding and subtracting decimal numbers, including decimals with different numbers of decimal places.
3. Multiplying and dividing decimals: This worksheet provides practice in multiplying and dividing decimal numbers, including decimals with different numbers of decimal places.
4. Decimal word problems: This worksheet provides practice in solving real-world problems that involve decimal numbers, such as finding the cost of an item with tax included.
Decimal numbers are an important concept for students to understand as they prepare for higher mathematics and real-world applications. Decimal worksheets provide students with the opportunity to
practice working with decimal numbers and improve their understanding of this crucial concept.
Also Read -
Frequently Asked Questions on Decimal Worksheets
A decimal number is a fraction represented in base 10, while a fraction is a ratio of two numbers, such as 1/2 or 3/4.
Decimal numbers are used in a variety of real-world applications, including measuring length, weight, and time, calculating money and taxes, and more.
Decimal worksheets provide students with the opportunity to practice working with decimal numbers and reinforce their understanding of this important concept.
Decimal worksheets are usually designed for students in grades 4-6, but can be adapted for younger or older students depending on their level of understanding.
Decimal worksheets can be in various formats, including paper worksheets, digital worksheets, or interactive worksheets.
Yes, there are many free decimal worksheets available online. Some websites offer printable worksheets, while others offer digital worksheets that can be completed on a computer or tablet. | {"url":"https://www.home-tution.com/maths-topics/decimal-worksheets","timestamp":"2024-11-08T05:23:49Z","content_type":"text/html","content_length":"117686","record_id":"<urn:uuid:8070a761-6016-498f-91f8-124e51a2332f>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00526.warc.gz"} |
Reply To: AQWA Eigenvalue vs Free Decay Test Resulting in Different Heave Natural Period? | Ansys Learning Forum
Mike Pettit
Ansys Employee
Hi Danu,
No problem, I'm happy to help.
Just to clarify, Aqwa has two types of time domain solver:
• when the 'Analysis Type' is 'Irregular Wave Response with Slow Drift' or 'Slow Drift Only', the calculation includes drift forces, but only uses linear hydrostatic/incident/diffracted/radiated
wave pressures;
• when the 'Analysis Type' is 'Irregular Wave Response' or 'Regular Wave Response', the calculation does not include drift forces, but the hydrostatic and incident wave pressures are calculated
under the instantaneous water surface (diffracted/radiated pressures are still linear).
To answer your questions:
1. Ideally you should model at least some of the structure above the waterline, unless you are only going to run a Hydrodynamic Diffraction analysis. In the diffraction calculation the solver will
ignore any non-diffracting panels, so they do not make a difference to the calculation time. Otherwise, in your Hydrodynamic Response analysis the solver will assume that the cut waterplane area
is constant above the cone as it becomes fully submerged.
2. This is what the convolution method does for us - rather than using frequency-dependent added mass/radiation damping terms in the equation of motion, instead it calculates the acceleration
impulse function matrix for the structure, then we have:
Where m is structural mass; A_inf is added mass at infinite frequency (estimated by Aqwa); X is position, X˙ is velocity, X¨ is acceleration; c is damping excluding radiation damping; K is
stiffness; F is the total force; h is the acceleration impulse function, which is calculated by Aqwa from the frequency-dependent added mass/radiation damping.
If you turn off convolution, the Aqwa solver will use an RAO-based method to calculate radiation forces, then you will find that there is no damping (because you have defined no wave).
I hope this answers your questions.
Cheers, Mike | {"url":"https://innovationspace.ansys.com/forum/forums/reply/291823/","timestamp":"2024-11-04T20:32:47Z","content_type":"text/html","content_length":"196199","record_id":"<urn:uuid:ca479df6-ebcf-4c1d-83b1-a7c6da6f6e47>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00269.warc.gz"} |
New Attribution Functions for PortfolioAnalytics | R-bloggersNew Attribution Functions for PortfolioAnalytics
New Attribution Functions for PortfolioAnalytics
[This article was first published on
tradeblotter » R
, and kindly contributed to
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if you have a blog, or
if you don't.
Another Google Summer of Code (GSoC) project this summer focused on creating functions for doing returns-based performance attribution. I’ve always been a little puzzled about why this functionality
wasn’t covered already, but I think that most analysts do this kind of work in Excel. That, of course, has its own perils. But beyond the workflow issues, there have been a number of methodological
refinements through time that I’m guessing most analysts don’t take advantage of. Furthermore, FactSet, Morningstar, and other vendors provide attribution functionality embedded within their
reporting functions.
I’m of the opinion that R is a natural place for this kind of functionality to reside, and portfolio attribution has long been on the list of additions to the PortfolioAnalytics package. The
calculations are relatively straight forward, but the mechanics of specifying and managing a portfolio hierarchy can be inconvenient. They require an attention to calculation order and other details,
as well.
The mentors for this project included David Cariño, who is a Research Fellow at Russell Investments. He is also a co-author of the book that was used as a reference[1] for the project, and he teaches
a course on this topic at UW. Doug Martin, who is a Professor of Statistics, Adjunct Professor of Finance and Director of Computational Finance at the University of Washington, and former Chairman of
the Department of Statistics, was also a mentor.
Again, student interest in the project was strong. The mentors for GSoC decided that Andrii Babii, a student at Toulouse School of Economics, provided the strongest proposal and example code.
Working from Christopherson, Carino and Ferson (2009), Bacon (2008), and several other sources, Andrii proceeded to construct functions for calculating contribution and linking returns through time
both arithmetically and geometrically. He supports a variety of methods for multi-period linking including Carino, Menchero, Davies-Laker, Frongello, and GRAP.
Using his functions, arithmetic effects can be displayed either as suggested in Brinson, Hood and Beebower (1986), or as in Brinson and Fachler (1985). Priority can be switched around from top-down
to bottom-up, as well. Currency attribution is included with arithmetic effects handled according to Ankrim and Hensel (1992).
Andrii established a data format for specifying a portfolio hierarchy, which allows him to provide multi-level attribution. This is usually discussed within the Brinson model as
“asset-country-sector,” although the function is written generally. It then returns the total multi-period effects and attribution effects at each level. He provides functions for weight aggregation
to accomplish this, which I think will also be generally useful in PortfolioAnalytics.
Finally, Andrii also constructed functions for Fixed Income attribution and delta adjusting option returns for attribution.
All of these attribution-related functions will be moved into the PortfolioAnalytics package proper, shortly. I think you will find that all of these functions are well documented and come with good
But, wait, there’s more! Andrii also knocked out a function for estimating market timing attribution according to either the Treynor-Mazuy or Merton-Henriksson models, called MarketTiming. That
function is likely to end up in PerformanceAnalytics, since it is estimated using multiple regression and doesn’t require portfolio weights.
He also provided an AcctReturns function that extends the blotter package. This function takes the account data calculated from transactions and prices, along with external cash flows (such as
contributions or withdrawals) to calculate time-weighted returns or linked modifed Dietz returns. This will be a welcome addition to blotter.
Congratulations to Andrii on a very successful GSoC 2012! This is a substantial contribution to PortfolioAnalytics, and is one that I think will see a great deal of use through time. I’m also looking
forward to seeing where Andrii’s interests take him, and I hope they result in more contributions. Thanks also go to the mentors, David and Doug, and also to Google to making the summer possible.
[1] Jon A. Christopherson, David R. Carino, Wayne E. Ferson “Portfolio Performance Measurement and Benchmarking”, 2009. 466p. – (Wiley finance series) | {"url":"https://www.r-bloggers.com/2012/09/new-attribution-functions-for-portfolioanalytics/","timestamp":"2024-11-10T14:36:05Z","content_type":"text/html","content_length":"93778","record_id":"<urn:uuid:88041b35-fe69-4234-9eb1-4e2ec22b9e3b>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00549.warc.gz"} |
3. A powerful method in control flow is recursion. - Asksia.ai
3. A powerful method in control flow is recursion. A function acts recursively if it calls itself while being
executed. This is often useful when a large computation can be reduced to the same calculation on smaller
Here's an example:
def count_down(n)
if n > 0:
elif n ==0:
This function counts down from a given integer, and then prints Boom when it gets to zero.
Use recursion to write a function that returns n!, where
n!=n.(n-1).(n- . - 2) .
Answer from Sia
Posted 4 months ago
The student is asking for a recursive function to calculate the factorial of a number $n$.
Step-by-step Instruction
The function should call itself with a decremented value of $n$ until it reaches 1
Define concrete use cases and testing cases for the function: When the function is called with a positive integer, it should return the factorial of that integer
Here is the code to achieve this:
Time Complexity
The time complexity of the recursive factorial function is $O(n)$ because it makes $n$ recursive calls.
Space Complexity
The space complexity is also $O(n)$ due to the $n$ recursive calls that add to the call stack. | {"url":"https://www.asksia.ai/question-and-answer/3-A-powerful-method-in-control-flow-is-recursion-A-function-acts-recursively-if-it-calls-itself-while-beingexecuted-This-is-often-useful-when-a-large--DU3g4","timestamp":"2024-11-14T07:36:39Z","content_type":"text/html","content_length":"93776","record_id":"<urn:uuid:22ba7562-47fa-400a-b6db-6b07dc27194f>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00357.warc.gz"} |
Compare Fewer Same (solutions, examples, homework, worksheets, lesson plans)
Eureka Math/EngageNY Kindergarten, module 3, lesson 19 Common Core Worksheets
New York State Common Core Math Kindergarten, module 3, lesson 19
Worksheets for Kindergarten
Worksheets, solutions, and videos to help Kindergarten students learn how to Compare using fewer than and the same as.
Topic E: Is There Enough?
Lesson 19 Concept Development and Problem Set
Use your clay to make six little pretend pancakes. How many people could you serve with your pancakes if you were going to have a tiny pancake party? What if another person joined them? Put your clay
back together into a ball and make new tiny pancakes so there would be just enough. Talk about your “cooking” with your friend.
Count the objects. Circle the set that has fewer.
Draw more ladybugs so there is the same number of ladybugs as leaves.
Count the objects. Circle the set that has fewer.
Draw more watermelon slices so there is the same number of watermelon slices as peaches.
1. How did you figure out how many “pancakes” to make when you had an extra guest?
2. What strategies did you use to compare the cubes and the pennies on your desk?
3. How did you know when there were the same number of cubes as pennies?
4. In the Problem Set, how did you know which set had fewer than the other? How did you color to make the same number of ladybugs as leaves?
5. How many suns and stars did you draw on the back of your Problem Set? Were there fewer suns or stars?
6. Talk to your neighbor about how your drawings were different. Did your partner have more suns or stars? Did you have more suns or stars? Count all of your suns and stars. How many did you have?
Check with your partner. Who had fewer than their partner? Did anyone have the same number as their partner?
7. What important math vocabulary did we use to communicate precisely?
Lesson 19 Homework
Draw another bird so there is the same number of birds as birdcages.
Learning Goal
I can compare using fewer than and the same as.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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Prime and composite numbers Prime factors
The definition of a Prime number
Definition: a Natural number simpleif it has only two positive divisors — the number itself
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, ... — the Prime numbers.
Prime numbers are infinitely many.
The definition of composite numbers
Definition: a Natural number is called compositeif it has more than two positive divisors.
6, 15, 130, ... are composite numbers.
1 is neither Prime number nor composite.
Properties of Prime factors of natural numbers
1. Any natural number (greater than unity) or is divided by this Prime number coprime with it.
2. If the product of several factors is divisible by a Prime number
3. The smallest Prime divisor of a composite number
The main theorem of the theory of divisibility
Theorem: Any natural number greater than unity, can be decomposed into the product of primes, and this decomposition is unique up the order of the multiplicands.
The theorem on division with Stacey
Theorem: For any pair of numbers
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Lecture 28: Quicksort and Mergesort
Lecture 28: Quicksort and Mergesort
Divide-and-conquer algorithms and their big-O behaviors
In the last lecture we considered two simple sorting algorithms: the natural insertion-sort algorithm for our standard ILists, and the natural selection-sort for ArrayLists. Both algorithms were
straightforward applications of the design recipe to their respective data. But unfortunately, the performance of those algorithms was not very good: in the worst case, they took quadratic amounts of
time proportionate to the size of their inputs.
Can we do better? The problem with both algorithms is that they’re too “forgetful”: they will repeatedly compare the same pairs of numbers many times, and they don’t take advantage of a key property
about comparisons. If we know that \(a < b\) and \(b < c\), then we do not need to compare \(a\) and \(c\), because by transitivity we already know the answer. Let’s see if we can use this fact to
our advantage.
28.1 Quicksort: a usually fast way to sort
Suppose we could guess the median value (the “middle-most” value, such that half the values are less than it, and half are greater) from a list of values. We could use that fact to divide our data
into a list of “small values” and a list of “big values” Once we’ve done that, we never need to compare any “small” value to any “big” value ever again, thereby cutting out a potentially huge number
of wasteful comparisons!
28.1.1 Quicksort using swapspace
As our running example, let’s use the following list of strings, and say we want to sort them in increasing lexicographic order:
Index: 0 1 2 3 4 5 6 7 8
Data: [grape, cherry, apple, kiwi, watermelon, banana, honeydew, date, fig]
We need to guess a median value, which we’ll call the pivot value. We might as well use the very first value we find (in index 0), because it’s pretty likely for that value to be a “middle” value,
and not a maximum or minimum.
Once we’ve identified a pivot value, we need to find all the values less than it and move them to the front of the list, find all the values greater than it and move them to the back, and then place
the pivot between them. To do this, we’ll use a temporary array in which to rearrange the values:
Index: 0 1 2 3 4 5 6 7 8
Data: [grape, cherry, apple, kiwi, watermelon, banana, honeydew, date, fig]
== move all smaller values left, and all bigger values right ==>
Index: 0 1 2 3 4 5 6 7 8
Data: [cherry, apple, banana, fig, date, grape, kiwi, watermelon, honeydew]
Now indices 0 through 4 of the temporary array contain values less than the pivot, indices 6 through 8 contain values greater than the pivot, and the pivot is in its final, sorted position. Even
better, we know that since indices 0 through 4 contain values less than the pivot, we do not need to compare them to anything in indices 6 through 8. Essentially, we now have two smaller sub-lists
which need sorting. If only we had a technique to sort them...Of course: we’re writing one now!
Let’s try to implement this algorithm. We want to implement a method in our ArrayUtils class with signature
<T> void quicksortCopying(ArrayList<T> arr, IComparator<T> comp);
but this signature will not be general enough for our recursive calls.
Why not? What helper method will be needed?
We need to supply a temporary array to hold the temporary values, and we’ll need to supply the indices between which we are sorting:
// In ArrayUtils // EFFECT: Sorts the given ArrayList according to the given comparator <T> void quicksortCopying(ArrayList<T> arr, IComparator<T> comp) {
// Create a temporary array ArrayList<T> temp = new ArrayList<T>();
// Make sure the temporary array is exactly as big as the given array for (int i = 0; i < arr.size(); i = i + 1) {
quicksortCopyingHelp(arr, temp, comp, 0, arr.size());
<T> void quicksortCopyingHelp(ArrayList<T> source, ArrayList<T> temp, IComparator<T> comp,
int loIdx, int hiIdx) {
The helper method must:
1. Select the pivot element,
2. Partition: Copy all elements less than the pivot into the lower half of the temp list, copy all elements greater than the pivot into the upper half of the temp list, and place the pivot between
3. Copy the entire list back from temp to source, and
4. Sort the upper and lower halves of the list
Step 3 is crucial, to ensure that the source list ends up sorted, and not just the temp list!
// EFFECT: sorts the source array according to comp, in the range of indices [loIdx, hiIdx) <T> void quicksortCopyingHelp(ArrayList<T> source, ArrayList<T> temp, IComparator<T> comp,
int loIdx, int hiIdx) {
// Step 0: check for completion if (loIdx >= hiIdx) {
return; // There are no items to sort }
// Step 1: select pivot T pivot = source.get(loIdx);
// Step 2: partition items to lower or upper portions of the temp list int pivotIdx = partitionCopying(source, temp, comp, loIdx, hiIdx, pivot);
// Step 4: sort both halves of the list quicksortCopyingHelp(source, temp, comp, loIdx, pivotIdx);
quicksortCopyingHelp(source, temp, comp, pivotIdx + 1, hiIdx);
// Returns the index where the pivot element ultimately ends up in the sorted source // EFFECT: Modifies the source and comp lists in the range [loIdx, hiIdx) such that // all values to the left
of the pivot are less than (or equal to) the pivot // and all values to the right of the pivot are greater than it <T> int partitionCopying(ArrayList<T> source, ArrayList<T> temp, IComparator<T>
int loIdx, int hiIdx, T pivot) {
int curLo = loIdx;
int curHi = hiIdx - 1;
// Notice we skip the loIdx index, because that's where the pivot was for (int i = loIdx + 1; i < hiIdx; i = i + 1) {
if (comp.compare(source.get(i), pivot) <= 0) { // lower temp.set(curLo, source.get(i));
curLo = curLo + 1; // advance the current lower index }
else { // upper temp.set(curHi, source.get(i));
curHi = curHi - 1; // advance the current upper index }
temp.set(curLo, pivot); // place the pivot in the remaining spot // Step 3: copy all items back into the source for (int i = loIdx; i < hiIdx; i = i + 1) {
source.set(i, temp.get(i));
return curLo;
The interesting work happens in partitionCopying. We maintain two indices, curLo and curHi, that indicate where the next small value or next large value should be placed. When we’re done, curLo
equals curHi, and they both equal the index where the pivot should be placed. Finally, partitionCopying also copies the items back from temp into source into their appropriate locations.
28.1.2 Quicksort in place
All this copying back and forth seems wasteful. Can we eliminate the extra ArrayList? The key idea is to use the curLo and curHi indices more cleverly. Let’s look again at our example:
Index: 0 1 2 3 4 5 6 7 8
Data: [grape, cherry, apple, kiwi, watermelon, banana, honeydew, date, fig]
Once we select "grape" as our pivot, we can pretend that "grape" has been removed from the list and we have a “hole” where it used to be:
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, kiwi, watermelon, banana, honeydew, date, fig]
Remember that we need to move all the “large” values to the right, and all the “small” values to the left, but we don’t yet know where to place the pivot. So let’s use curLo and curHi to keep track
of indices of “values we know to be lower than the pivot” and “values we know to be higher than the pivot”, and if we ever find a pair of values out of place, we swap them. Concretely, we start curLo
at 1, and curHi at 8.
curLo curHi
v v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, kiwi, watermelon, banana, honeydew, date, fig]
Since "cherry" is less than "grape", we increase curLo:
curLo curHi
v v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, kiwi, watermelon, banana, honeydew, date, fig]
Since "apple" is less than "grape", we increase curLo again:
curLo curHi
v v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, kiwi, watermelon, banana, honeydew, date, fig]
Since "kiwi" is not less than "grape", we stop, and turn our attenion to curHi. Since "fig" is not greater than "grape", we stop. We now know that "kiwi" belongs to the right of the pivot, wherever
it winds up, and "fig" belongs to the left, so we swap them:
curLo curHi
v v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, fig, watermelon, banana, honeydew, date, kiwi]
And now we resume. Since "fig" is less than "grape", we advance "curLo".
curLo curHi
v v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, fig, watermelon, banana, honeydew, date, kiwi]
Since "watermelon" is not less than "grape", we stop, and turn our attention to curHi. Since "kiwi" is greater than "grape", we advance curHi:
curLo curHi
v v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, fig, watermelon, banana, honeydew, date, kiwi]
Since "date" is not greater than "grape", we stop, and swap the two values:
curLo curHi
v v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, fig, date, banana, honeydew, watermelon, kiwi]
Since "date" is less than "grape", we advance curLo. Since "banana" is less than "grape", we advance curLo again:
curLo curHi
v v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, fig, date, banana, honeydew, watermelon, kiwi]
Since "honeydew" is not less than "grape", we stop, and turn our attention to curHi. Since "watermelon" is greater than "grape", we advance curHi. Since "honeydew" is greater than "grape", we advance
curHi again.
curHi curLo
v v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, cherry, apple, fig, date, banana, honeydew, watermelon, kiwi]
Since curHi is now less than curLo, we must have found the crossover point. We need to place "grape" at index curHi, which means we need to move "banana" somewhere else. Conveniently, there’s room
for it at index 0, where "grape" used to be!
Let’s see if we can implement this algorithm in Java.
What kind of loop should be used?
Since we do not know in advance how many times to run the loop, but we do know to stop the loop once curLo is greater than curHi, we should use a while loop:
// Returns the index where the pivot element ultimately ends up in the sorted source // EFFECT: Modifies the source list in the range [loIdx, hiIdx) such that // all values to the left of the
pivot are less than (or equal to) the pivot // and all values to the right of the pivot are greater than it <T> int partition(ArrayList<T> source, IComparator<T> comp,
int loIdx, int hiIdx, T pivot) {
int curLo = loIdx;
int curHi = hiIdx - 1;
while (curLo < curHi) {
// Advance curLo until we find a too-big value (or overshoot the end of the list) while (curLo < hiIdx && comp.compare(source.get(curLo), pivot) <= 0) {
curLo = curLo + 1;
// Advance curHi until we find a too-small value (or undershoot the start of the list) while (curHi >= loIdx && comp.compare(source.get(curHi), pivot) > 0) {
curHi = curHi - 1;
if (curLo < curHi) {
swap(source, curLo, curHi);
swap(source, loIdx, curHi); // place the pivot in the remaining spot return curHi;
Now we can tweak our original quicksort method to use this, and avoid the temp list altogether:
// In ArrayUtils // EFFECT: Sorts the given ArrayList according to the given comparator <T> void quicksort(ArrayList<T> arr, IComparator<T> comp) {
quicksortHelp(arr, comp, 0, arr.size());
// EFFECT: sorts the source array according to comp, in the range of indices [loIdx, hiIdx) <T> void quicksortHelp(ArrayList<T> source, IComparator<T> comp,
int loIdx, int hiIdx) {
// Step 0: check for completion if (loIdx >= hiIdx) {
return; // There are no items to sort }
// Step 1: select pivot T pivot = source.get(loIdx);
// Step 2: partition items to lower or upper portions of the temp list int pivotIdx = partition(source, comp, loIdx, hiIdx, pivot);
// Step 3: sort both halves of the list quicksortHelp(source, comp, loIdx, pivotIdx);
quicksortHelp(source, comp, pivotIdx + 1, hiIdx);
28.1.3 Runtime analysis of quicksort
We’ve successfully avoided allocating any memory at all for quicksort, so we know for certain that \(M_{quicksort}(n) = 0\). But what about the runtime?
• The runtime for swap is simple: it performs two reads and two writes, so \(T_{swap}(n) = 4\), regardless of the size of the input.
• The runtime for partition depends on the size of the interval \(n = hiIdx - loIdx\). It checks every item in that range against the pivot, and potentially swaps all of them. So
\begin{equation*}\begin{aligned} T_{partition}^{best}(n) &= n \\ T_{partition}^{worst}(n) &= 3n \end{aligned}\end{equation*}
In other words, the performance is always linear in the size of the interval.
• The runtime for qucksortHelp depends on the size of the interval \(n = hiIdx - loIdx\) as well. But it has two recursive calls, whose runtimes depend on how good of a pivot we chose. In the best
case, the pivot is very close to the median value, and divides the list exactly in half:
\begin{equation*}T_{quicksortHelp}^{best}(n) = T_{partition}^{best}(n) + 2T_{quicksortHelp}^{best}(n/2)\end{equation*}
In the worst case, the pivot is either the minimum or maximum, and divides the list into an empty piece and a nearly-complete piece:
\begin{equation*}T_{quicksortHelp}^{worst}(n) = T_{partition}^{worst}(n) + T_{quicksortHelp}^{worst}(0) + T_{quicksortHelp}^{worst}(n-1)\end{equation*}
We’ve seen this latter recurrence before, when we determined the runtime of selection-sort: it’s \(O(n^2)\). The former recurrence is a trickier one. We can expand the recurrence a few times:
\begin{equation*}\begin{aligned} T_{quicksortHelp}^{best}(n) &= T_{partition}^{best}(n) + 2T_{quicksortHelp}^{best}(n/2) \\ &= n + 2(T_{partition}^{best}(n/2) + 2T_{quicksortHelp}^{best}(n/4)) \\
&= n + 2(n/2 + 2(T_{partition}^{best}(n/4) + 2T_{quicksortHelp}^{best}(n/8))) \\ &= n + 2(n/2) + 4(n/4) + 8(n/8) + \cdots + n(1) \end{aligned}\end{equation*}
How many terms are there in the final expanded line? It’s however many times we can divide \(n\) in half before reaching one. This is familiar: it’s \(\log_2 n\). Accordingly, our runtime is \(T_
{quicksortHelp}^{best}(n) = n\log_2 n\).
From the best and worst case results we can conclude that
Runtime for quicksort Best-case Worst-case
\(T_{quicksort}\) \(\Omega(n \log_2 n)\) \(O(n^2)\)
\(M_{quicksort}\) \(\Omega(1)\) \(O(1)\)
What inputs produce the best and worst case behaviors of quicksort?
Almost all inputs produce nearly-best-case behavior for quicksort. The edge case happens when the pivot that’s picked is repeatedly either the maximum or the minimum — in such cases, the “partition”
doesn’t split the list evenly at all. This occurs when the input list is already sorted (or reverse-sorted).
Can we do better? Can we create a sorting algorithm that never has any bad cases?
28.2 Mergesort: sorting in a guaranteed better worst-case time
If we allow ourselves the luxury of a temporary ArrayList again, we can do even better. Recall the merge method we implemented in Assignment 6: it allows us to build up a larger sorted list from two
smaller sorted lists. We can implement a similar method for ArrayLists. But how to use it? Let’s conceptually divide our ArrayList exactly at the middle index, into a low-index half and a high-index
half. If only we could sort those two halves, then we could use merge to combine them into a sorted ArrayList containing the entire contents of the original ArrayList, and we’d be done! How can we
sort those two halves? Well, we can divide them in half, at their middle indices, and magically sort their respective halves, and then merge their halves. We can clearly repeat this process, but when
does it stop? Conveniently, a list containing at most one item is always sorted — so we have a base case for our recursion. This approach defines the mergesort algorithm.
// In ArrayUtils // EFFECT: Sorts the provided list according to the given comparator <T> void mergesort(ArrayList<T> arr, IComparator<T> comp) {
// Create a temporary array ArrayList<T> temp = new ArrayList<T>();
// Make sure the temporary array is exactly as big as the given array for (int i = 0; i < arr.size(); i = i + 1) {
mergesortHelp(arr, temp, comp, 0, arr.size());
// EFFECT: Sorts the provided list in the region [loIdx, hiIdx) according to the given comparator.
// Modifies both lists in the range [loIdx, hiIdx) <T> void mergesortHelp(ArrayList<T> source, ArrayList<T> temp, IComparator<T> comp,
int loIdx, int hiIdx) {
Our helper routine must:
1. Find the middle index of the current range. (If the range contains at most one item, stop.)
2. Recursively sort the lower range, and the higher range.
3. Merge the two ranges.
Let’s elaborate this sketch into code:
// EFFECT: Sorts the provided list in the region [loIdx, hiIdx) according to the given comparator.
// Modifies both lists in the range [loIdx, hiIdx) <T> void mergesortHelp(ArrayList<T> source, ArrayList<T> temp, IComparator<T> comp,
int loIdx, int hiIdx) {
// Step 0: stop when finished if (hiIdx - loIdx <= 1) {
return; // nothing to sort }
// Step 1: find the middle index int midIdx = (loIdx + hiIdx) / 2;
// Step 2: recursively sort both halves mergesortHelp(source, temp, comp, loIdx, midIdx);
mergesortHelp(source, temp, comp, midIdx, hiIdx);
// Step 3: merge the two sorted halves merge(source, temp, comp, loIdx, midIdx, hiIdx);
Design the merge helper method to complete the algorithm above.
Now all we need to do is define merge. When we defined merge over IList, we used dynamic dispatch to give us the first items of each of the two lists, as needed. Here, we do not have Cons objects, so
we must maintain two indices, and advance them as needed. Recall our running example:
Index: 0 1 2 3 4 5 6 7 8
Data: [grape, cherry, apple, kiwi, watermelon, banana, honeydew, date, fig]
And suppose that at the next-to-last stage, we had sorted indices 0 through 4, and indices 5 through 8:
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, cherry, grape, kiwi, watermelon, banana, date, fig, honeydew]
We now need to merge those two sub-lists. We maintain three indices: curLo keeps track of the next item in the lower half-list, curHi keeps track of the next item in the upper half-list, and curCopy
keeps track of where to insert into the temp storage. We loop, until curLo runs off the end of the lower half-list and curHi runs off the end of the upper half-list, and copy the smaller of the two
current values into the temp storage, and advance the relevant index. Concretely, curLo starts off at loIdx, curHi starts off at midIdx, and curCopy starts off at loIdx:
curLo curHi
SOURCE v v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, cherry, grape, kiwi, watermelon, banana, date, fig, honeydew]
TEMP v
Index: 0 1 2 3 4 5 6 7 8
Data: [-----, ------, -----, -----, -----, ----, ----, ------, -----------]
Since "apple" is less than "banana", we copy "apple", and advance curLo and curCopy:
curLo curHi
SOURCE v v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, cherry, grape, kiwi, watermelon, banana, date, fig, honeydew]
TEMP v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, ------, -----, -----, -----, ----, ----, ------, -----------]
Since "cherry" is greater than "banana", we copy "banana", and advance curHi and curCopy:
curLo curHi
SOURCE v v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, cherry, grape, kiwi, watermelon, banana, date, fig, honeydew]
TEMP v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, banana, ------, -----, -----, ----, ----, ------, -----------]
Since "cherry" is less than "date", we copy "cherry", and advance curLo and curCopy:
curLo curHi
SOURCE v v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, cherry, grape, kiwi, watermelon, banana, date, fig, honeydew]
TEMP v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, banana, cherry, -----, -----, ----, ----, ------, -----------]
Continuing in this manner, we eventually get to the following state:
curLo curHi
SOURCE v v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, cherry, grape, kiwi, watermelon, banana, date, fig, honeydew]
TEMP v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, banana, cherry, date, fig, grape, honeydew, ----, ----------]
At this point, we have nothing left in the upper half-list, so we copy the remaining items from the lower half-list, because they are already sorted, and evidently are greater than everything in the
upper half-list:
curLo curHi
SOURCE v v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, cherry, grape, kiwi, watermelon, banana, date, fig, honeydew]
TEMP v
Index: 0 1 2 3 4 5 6 7 8
Data: [apple, banana, cherry, date, fig, grape, honeydew, kiwi, watermelon]
Since curLo now equals midIdx we’ve finished the lower half-list, and so we’re done. All that remains is to copy everything back from temp into source.
// Merges the two sorted regions [loIdx, midIdx) and [midIdx, hiIdx) from source // into a single sorted region according to the given comparator // EFFECT: modifies the region [loIdx, hiIdx) in
both source and temp <T> void merge(ArrayList<T> source, ArrayList<T> temp, IComparator<T> comp,
int loIdx, int midIdx, int hiIdx) {
int curLo = loIdx; // where to start looking in the lower half-list int curHi = midIdx; // where to start looking in the upper half-list int curCopy = loIdx; // where to start copying into the
temp storage while (curLo < midIdx && curHi < hiIdx) {
if (comp.compare(source.get(curLo), source.get(curHi) <= 0) {
// the value at curLo is smaller, so it comes first temp.set(curCopy, source.get(curLo));
curLo = curLo + 1; // advance the lower index }
else {
// the value at curHi is smaller, so it comes first temp.set(curCopy, source.get(curHi));
curHi = curHi + 1; // advance the upper index }
curCopy = curCopy + 1; // advance the copying index }
// copy everything that's left -- at most one of the two half-lists still has items in it while (curLo < midIdx) {
temp.set(curCopy, source.get(curLo));
curLo = curLo + 1;
curCopy = curCopy + 1;
while (curHi < hiIdx) {
temp.set(curCopy, source.get(curHi));
curHi = curHi + 1;
curCopy = curCopy + 1;
// copy everything back from temp into source for (int i = loIdx; i < hiIdx; i = i + 1) {
source.set(i, temp.get(i));
28.2.1 Runtime analysis of mergesort
What is the runtime of mergesort, in the best and worst cases?
First let’s consider the runtime of merge, as a function of the difference \(n = hiIdx - loIdx\). The three while loops together examine every item in the given range of indices and copy them exactly
once into the temp list. This takes \(O(n)\) time to complete. The final counted-for loop again iterates over the given range of indices, and copies every item back into source; this takes another \
(O(n)\) time to complete. Therefore, \(T_{merge} \in O(n)\) in all cases — there is no difference between best or worst-case behavior.
Next, let’s look at mergesortHelp, again as a function of the difference \(n = hiIdx - loIdx\). We can see from the code that it makes two recursive calls, followed by one call to merge. Crucially,
the computation of midIdx guarantees that both recursive calls are to subproblems that are at most half as big as the current one. This yields a recurrence
\begin{equation*}T_{mergesortHelp}(n) = 2T_{mergesortHelp}(n/2) + T_{merge}(n)\end{equation*}
This is the same recurrence as the best-case scenario for quicksort above, and by similar reasoning, we conclude that we can subdivide the list in half at most \(\log_2 n\) times before reaching a
base case, so our total running time is
\begin{equation*}T_{mergesortHelp}(n) = n\log_2 n\end{equation*}
There is no best or worst case to consider: all cases have the same performance.
Finally, we look at mergesort itself. It requires a temporary ArrayList of the same size as the input, and then makes one call to mergesortHelp. So we obtain
Runtime for mergesort Best-case Worst-case
\(T_{mergesort}\) \(\Omega(n \log_2 n)\) \(O(n\log_2 n)\)
\(M_{mergesort}\) \(\Omega(n)\) \(O(n)\)
28.3 Divide-and-conquer algorithms
The quicksort algorithm above is an example of what’s known as a divide-and-conquer algorithm. By splitting the input into nearly equally-sized chunks and processing them independently, we can
compute answers in much better performance than if we processed the inputs strictly via structural recursion. Divide-and-conquer algorithms are an example of generative recursion: they require a leap
of insight to figure out the best way to decompose the problem; once that insight is gained, the rest of the solution follows from the same design strategies we’ve been following all semester.
How much better, really, is \(O(n \log_2 n)\) than \(O(n^2)\)? For small values, there isn’t much difference. But when \(n = 256\), \(\log_2 n = 8\), which means \(O(n \log_2 n)\) is already over 60
times better than \(O(n^2)\). For \(n = 2^{10} = 1024\), that factor improves to over 100; for \(n = 2^{20} = 1048576\), that factor improves to over 50000, and it keeps getting better. Conversely,
for \(n = 2^{20}\), \(O(n \log_2 n)\) is only twenty times worse than \(O(n)\):
28.3.1 Time/space tradeoffs
Mergesort is another divide-and-conquer algorithm, and it gets a better worst-case behavior bound than quicksort. But to accomplish this, it requires the use of \(O(n)\) additional memory. This is a
classic example of what’s known as a time/space tradeoff: frequently we can speed up algorithms by allowing them addtional space in which to store temporary results. (Conversely, if we are
constrained for space, we can sometimes improve our memory usage by recomputing results as needed instead of storing them.) Many, many algorithms and their variations result from exploring this
tradeoff in various ways; for more details, there are entire courses devoted to studying algorithms!
In the next lecture, we will see one more sorting algorithm, that again achieves \(O(n\log_2 n)\) worst-case time performance, but uses no additional storage. To accomplish this, we’ll make use of a
novel data structure, which is of widespread utility (besides merely sorting efficiently!), and whose construction combines several of the data structures we’ve seen so far.
28.4 Discussion
At this point, after going through four sorting algorithms, several helper functions and even more recurrence relations, we might well ask, “Wait! Do we really need to keep coming up with more
algorithms? Isn’t one of these good enough?” In addition to using sorting algorithms to illustrate how to analyze algorithms, this sequence of four algorithms also illustrates how to come up with
improved algorithms in the first place.
• We arrived at insertion-sort pretty much by following the design recipe for the IList data type.
• We arrived at selection-sort by asking “could we improve things if only we allowed mutation?” The central insight for selection-sort was that we could build an in-place sorting algorithm that
didn’t require reallocating so many intermediate results.
• We arrived at quicksort by looking at selection-sort’s worst case behavior and asking, “couldn’t we avoid doing all those comparisons?” The central insight here is that because comparisons are
transitive, we could pick a “middle-most value” and let it partition the data for us, and then we’d never have to compare any of the smaller numbers to any of the larger numbers.
• We arrived at mergesort by looking at quicksort’s worst case behavior and asking, “could we ensure we never make a bad choice, and always split the list cleanly in half?” The central insight of
mergesort was to do away with the pivot altogether, because the key to quicksort’s good behavior was the divide-and-conquer approach.
In general, algorithm designers often look at the solutions we currently know to problems, and look at their worst cases and ask, “why is this the worst case? What’s triggering it?” The leap of
intuition here is precisely the same as the leap needed to come up with a generatively recursive algorithm, and can be just as elusive at first. By studying many related algorithms, and identifying
their central insights, we actually can get some guidance as to where to keep looking for improvements. | {"url":"https://course.khoury.northeastern.edu/cs2510h/lecture28.html","timestamp":"2024-11-04T08:22:10Z","content_type":"text/html","content_length":"149574","record_id":"<urn:uuid:301052eb-82ee-4505-a993-a770bbde6b80>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00337.warc.gz"} |
Computing the Polar Decomposition and the Matrix Sign Decomposition in Matrix Groups
Higham, Nicholas J. and Mackey, D. Steven and Mackey, Niloufer and Tisseur, Françoise (2004) Computing the Polar Decomposition and the Matrix Sign Decomposition in Matrix Groups. SIAM Journal On
Matrix Analysis and Applications, 25 (4). pp. 1178-1192. ISSN 1095-7162
Download (223kB)
For any matrix automorphism group $\G$ associated with a bilinear or sesquilinear form, Mackey, Mackey, and Tisseur have recently shown that the matrix sign decomposition factors of $A\in\G$ also lie
in $\G$; moreover, the polar factors of $A$ lie in $\G$ if the matrix of the underlying form is unitary. Groups satisfying the latter condition include the complex orthogonal, real and complex
symplectic, and pseudo-orthogonal groups. This work is concerned with exploiting the structure of $\G$ when computing the polar and matrix sign decompositions of matrices in $\G$. We give sufficient
conditions for a matrix iteration to preserve the group structure and show that a family of globally convergent rational Pad\'e-based iterations of Kenney and Laub satisfy these conditions. The
well-known scaled Newton iteration for computing the unitary polar factor does not preserve group structure, but we show that the approach of the iterates to the group is precisely tethered to the
approach to unitarity, and that this forces a different and exploitable structure in the iterates. A similar relation holds for the Newton iteration for the matrix sign function. We also prove that
the number of iterations needed for convergence of the structure-preserving methods can be precisely predicted by running an associated scalar iteration. Numerical experiments are given to compare
the cubically and quintically converging iterations with Newton's method and to test stopping criteria. The overall conclusion is that the structure-preserving iterations and the scaled Newton
iteration are all of practical interest, and which iteration is to be preferred is problem-dependent.
Item Type: Article
Uncontrolled automorphism group, bilinear form, sesquilinear form, adjoint, complex orthogonal matrix, symplectic matrix, perplectic matrix, pseudo-orthogonal matrix, polar decomposition, matrix
Keywords: sign decomposition, structure preservation, matrix iteration, Newton iteration, convergence tests
Subjects: MSC 2010, the AMS's Mathematics Subject Classification > 15 Linear and multilinear algebra; matrix theory
MSC 2010, the AMS's Mathematics Subject Classification > 65 Numerical analysis
Depositing Nick Higham
Date Deposited: 24 Nov 2005
Last Modified: 20 Oct 2017 14:12
URI: https://eprints.maths.manchester.ac.uk/id/eprint/49
Available Versions of this Item
• Computing the Polar Decomposition and the Matrix Sign Decomposition in Matrix Groups. (deposited 24 Nov 2005) [Currently Displayed]
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Harris-Benedict Calculator (Basal Metabolic Rate)
Last updated:
Harris-Benedict Calculator (Basal Metabolic Rate)
If you care about your health and weight, this Harris-Benedict calculator (BMR calculator), which runs on the Harris-Benedict equation, is the perfect tool for you! In the following article, we will
explain how this simple tool works and how you can use it to your advantage.
What is a BMR calculator?
Your basal metabolic rate (BMR) is equivalent to the amount of energy (in the form of calories) that your body needs to function if it were to rest for 24 hours.
How can you calculate your BMR?
Calculating your basal metabolic rate is easy; all you need to measure is your weight and height, although your age and sex also come into play. The calculators for both sexes are as follows:
• For men: BMR = 66.5 + (13.75 × weight in kg) + (5.003 × height in cm) - (6.75 × age)
• For women: BMR = 655.1 + (9.563 × weight in kg) + (1.850 × height in cm) - (4.676 × age)
What is the Harris-Benedict equation?
The Harris-Benedict Equation is a formula to calculate the BMR. Then the BMR can be multiplied by an activity factor to determine your total daily energy expenditure (given in calories).
Want to start at the very beginning? Check the basic calorie calculator!
How to find your total daily calorie needs using Harris-Benedict equation?
To determine your TDEE (Total Daily Energy Expenditure), multiply your BMR by the appropriate activity factor, as follows:
• Sedentary (little or no exercise): calories = BMR × 1.2;
• Lightly active (light exercise/sports 1-3 days/week): calories = BMR × 1.375;
• Moderately active (moderate exercise/sports 3-5 days/week): calories = BMR × 1.55;
• Very active (hard exercise/sports 6-7 days a week): calories = BMR × 1.725
• Extra active (very hard exercise/sports & a physical job): calories = BMR × 1.9; and
• If you train as a professional athlete: calories = BMR × 2.3
For more on BMR, check out the Mifflin St Jeor equation calculator and the Katch-McArdle formula calculations.
Keeping a healthy diet and using the Harris-Benedict calculator to keep an eye on your BMR will make it easier for you to stay in good health. Thanks to the Harris-Benedict calculator and
Harris-Benedict equation, you can know the perfect amount of calories (or maintenance calories) to eat if you want to keep your body weight as it is. If you are looking to gain or lose weight, you
can also use this number as a point to eat more or less then, respectively. Please remember to consult a medical expert if you want to gain or lose a lot of weight.
How do I calculate my BMR?
To calculate your BMR, you would usually use the Harris-Benedict equation:
1. For women, it's: BMR = 655.1 + (9.563 × weight in kg) + (1.850 × height in cm) - (4.676 × age).
2. For men, the formula is: BMR = 66.5 + (13.75 × weight in kg) + (5.003 × height in cm) - (6.75 × age).
What is BMR?
Your basal metabolic rate (BMR) is equivalent to the amount of energy, in calories, that your body needs to function if it were to rest for 24 hours.
What does BMR mean?
BMR stands for basal metabolic rate. It is the minimum number of calories that your body needs for basal functions like breathing, digesting, and keeping the body temperature steady over a day.
What is a good BMR?
The average BMR is about 1409 kcal (5900 kJ) for a woman and about 1696 kcal (7100 kJ) for a man.
How do I increase my BMR?
To increase your BMR (basal metabolic rate), you can try:
1. Eating more protein in every meal.
2. Doing more HIIT (high-intensity interval training) workouts.
3. Increasing your daily activity – choose stairs over the elevator, walk more, switch to a standing desk.
4. Eating more spicy foods.
5. Doing more resistance workouts – muscles burn more calories than fat tissue.
How do I calculate BMR in pounds?
To calculate your basal metabolic rate in imperial units, use the equations:
1. For men: 66.47 + (6.24 × weight in pounds) + (12.7 × height in inches) − (6.76 × age in years).
2. For women: BMR = 65.51 + (4.34 × weight in pounds) + (4.7 × height in inches) - (4.7 × age in years)
How do I use BMR to lose weight?
Counting your BMR can be a good starting point if you want to lose weight. First, it tells you how many calories your body needs to function properly, so you never want to eat less than that. Then,
if you multiply it by your physical activity level (PAL), you know how many calories you actually burn each day. Subtracting a reasonable number of calories (e.g., 500 kcal/day) will let you lose
weight gradually and steadily.
What is the Harris-Benedict equation?
The Harris-Benedict equation is a mathematical formula that allows you to count your basal metabolic rate – BMR. It needs your age, weight, and height. There are two separate formulas for men and
1. For women – BMR = 655.1 + (9.563 × weight in kg) + (1.850 × height in cm) - (4.676 × age).
2. For men – BMR = 66.5 + (13.75 × weight in kg) + (5.003 × height in cm) - (6.75 × age).
Is the Harris-Benedict equation accurate?
The Harris-Benedict equation is thought to be the most accurate of all the BMR equations. However, you need to remember that calculating your BMR with a calculator is only a rough estimation and may
differ from your actual bodily needs. You need to put that into practice; still, calculating your BMR can be an excellent place to start.
How do I calculate my BMR as a female?
To calculate your female BMR, you need to:
1. Know your exact weight and height. Don't cheat - it will only make the results inaccurate.
2. Use the Harris-Benedict equation for women: BMR = 655.1 + (9.563 × weight in kg) + (1.850 × height in cm) - (4.676 × age).
3. The result is your basal metabolic rate – BMR. You can then multiply it with your physical activity level – PAL – to find your total daily energy expenditure – TDEE. | {"url":"https://www.omnicalculator.com/health/bmr-harris-benedict-equation","timestamp":"2024-11-03T22:39:12Z","content_type":"text/html","content_length":"601996","record_id":"<urn:uuid:c319b427-1636-4a10-85be-50632e7ad8ed>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00310.warc.gz"} |
How does soh cah toa apply to a right triangle? | Socratic
How does soh cah toa apply to a right triangle?
1 Answer
S in = O pposite over H ypotenuse
C os = A adjacent over H ypotenuse
T an = O pposite over A djacent
These are a set of mnemonics to help remember how to derive the basic trig functions for a (non-right) angle of a right triangle. (They only apply to right triangles).
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Energy formulas | energyfaculty.com
Basic concepts and units related to work and energy (SI units)
Force – The unit for force is newton (N). 1 newton (N) = 1 kg*1m/s^2, which can be translated to the force necessary to accelelrate an object with the mass 1 kg to a velocity of 1 m/second during a
period of 1 second.
Work – The unit of work is joule (J). 1 joule = 1 Nm = 1 kg * 1m^2/s^2, which can be translated to the amount of work performed by a force of 1 newton (N) over a distance of 1 meter in the same
direction as the force (or a component of the force).
Energy – The unit for energy is joule (J). 1 joule = 1Nm = 1 kg * 1m^2/s^2. This is the amount of energy necessary to apply a force of 1 newton over a distance of 1 meter.
Power – The unit for power is watt (W). 1 watt (W) = 1J/s = 1Nm/s = 1 kg * 1m^2/s^3 or the power produced by a current of one ampere acting across a potential difference of one volt. This can be
translated to the amount of work performed per second or the rate of using energy.
Formula for relativistic kinetic energy
The formulas for kinetic energy are based on Einstein’s relationship for energy:
Total energy:
Rest energy (energy at rest):
m = m_0\gamma
m = relativistic mass
m[0] = rest mass
c = speed of light in vacuum
\gamma =\dfrac{1}{\sqrt{1-(v/c)^2}}
Total energy may also be expressed as:
E=E_0 + E_k
Relativistic kinetic energy may therefore be calculated from:
E_k=E - E_0
E_k=mc^2 - m_0c^2
E_k=m_0 \gamma c^2 - m_0c^2
E_k=\dfrac{m_0c^2}{\sqrt{1-(v/c)^2}} - m_0c^2
At very low speeds, compared to the speed of light, the classical formula for kinetic energy (below) aligns well with the formula for relativistic kinetic energy, something we can see by applying the
binomial theorem to the formula for relativistic kinetic energy (above).
E_k=\dfrac{m_0c^2}{\sqrt{1-(v/c)^2}} - m_0c^2
E_k \approx \dfrac{1}{2}m_0c^2 + \dfrac{3}{8 } \dfrac{m_0v^4}{c^2 } + \dfrac{5}{16 } \dfrac{m_0v^6}{c^4 } + \cdots - m_0c^2 = \dfrac{1}{2}m_0v^2
Binomial Theorem
(a+b)^n=a ^n+na^{n-1}b +\dfrac{n(n-1)}{2}a^{n-2} b^2+\cdots+b^n
For any value of n, the value of the n^th power of a binomial is given by the above equation
Classical formula for kinetic energy
Ek = E[t]+ E[r]
E[t] = Translational kinetic energy
E[r] = Rotational kinetic energy
Formula for translational kinetic energy
m = mass (Kg)
v = velocity (m/s metres per second)
E[k] = Resulting energy is measured in joules
Formula for rotational kinetic energy
I = Moment of inertia (around the axis of rotation)
ω = Angular velocity = 2πf
f = Revolutions/sec
Formula for gravitational potential energy.
E_{pg}= mgh
m = Mass
g = Gravitational acceleration (9,8 m/s2 close to earth)
h = Height above the reference point
Formula for elastic potential energy for a linear spring.
k = Spring constant
x = Amount of stretch or compression
Formula for thermal energy
Q = \int_{t_1}^{t_2}mc\Delta t = m\int_{t_1}^{t_2}c\Delta t
Q = Thermal energy of a substance or a system
m = The mass of the substance or system
c= The specific heat capacity of the substance or system.
T = the absolute temperature of the substance or system
Δt = Temperature difference
For practical purposes the average specific heat capacity (cm) may be used, the formula then is:
Q = mcm (t2 – t1)
Formula electrical energy
Energy (Joule) = Power (Watt) x Time (Second)
Power (Watt) = Energy(Joule) / Time(Second)
1 Watt = 1 Joule / Second.
Electrical energy may be defined by the work (W) carried out or needed to move electrically charged particles.
W = UIt (Joule)
U = Differential potential (Volt)
I = current (Ampere) (Columb per second)
R = Resistanse (Ohms Ω)
t = time (second):
P (Power) = W/t = UIt/t = Ui (volts x ampere) (Watt)
P = R x I2
P = U2/R
I = P/U (ampere)
I = U/R
I = (P/R)1/2
Electrical potential
U = RI
U = P/I
U = (PR)1/
R = U/I (ohm`s law)
R= P/I
R = P/I2
Energy of an electric field
The work done in establishing the electric field, and hence the amount of energy stored, is:
W= \dfrac{1}{2}CV^2 + \dfrac{1}{2}VQ
Q = Charge stored
V = Voltage across the capacitor
C = Capacitance
Formula for electromagnetic energy
The energy for one individual photon is:
E = hv = \dfrac{hc}{\lambda} (Joule)
or if angular frequency is used:
E = \hbar w
ω = 2πv
ν = Frequency (cycles/second)
λ = Wavelength (metres)
c = Speed of light (metres/second)
1 Hz = 1 hertz; cycle per second (frequency)
1 nm = 10-9 m, nanometre (for wavelength of IR, visible, UV and X-rays).
1 pm = 10-12 m, picometer (for X-rays and gamma rays).
To calculate the energy giving the result in everyday quantities we need to calculate the combined energy for larger number of particles.
Formula for sound energy
The total sound energy will equal the maximum kinetic energy:
E= \dfrac{1}{2}mv^2 = \dfrac{1}{2}m(A\omega)^2
m = density of the medium the sound waves travel through
Aω = the maximum transverse speed of particles
A = \dfrac{\nu}{2\pi}
A = amplitude
Formula for nuclear energy
Mass defect and nuclear binding energy
E= mc^2 or
m= \dfrac{c^2}{E}
We first need to calculate the mass defect to be able to calcutlate the potential for releasing energy when fission takes place.
Mass defect
Mass c (combined mass) = MP + MN (Mass Neutron)
MP =Mass Proton = nP*amuP
MN =Mass Neutron = nN*amuN
Dm = Mass c – MassBM
Mass defect into kg
Dm(amu) * 1.6606 x 10-27 kg/nucleus
1amu = 1.6606 x 10-27 kg
Mass defect into energy
c = 2.9979 x 108 m/s
E = mc2 = (Dm(amu) *1.6606* 10-27 kg/nucleus) * (2.9979 x 108 m/s)2
E = DM*1,4924483 *10-10 J/nucleus
E= DM*1,4924483 *10-10 J/nucleus * 6.022 x 1023 nuclei/mol* (1 kJ/1000 J) * = DM*8,9875 1010 kJ/mol of nuclei.
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Traceability of pH to the Mole
Centro de Química Estrutural, Department of Chemistry and Biochemistry, Faculty of Sciences of the University of Lisbon; C8- Campo Grande, Lisboa 1749-016, Portugal
Author to whom correspondence should be addressed.
Submission received: 1 November 2014 / Revised: 24 July 2015 / Accepted: 27 July 2015 / Published: 3 August 2015
Free acidity of aqueous solutions was initially defined in 1909 by Søren Peter Lauritz Sørensen as pH = −lgc[H+] (c/mol·dm^−3 or m/mol·kg^−1 of the free hydrogen ions in solution, H^+) soon (1910)
was changed to pH = pa[H+] = −lga[H+], integrating the new concepts of activity, a[i] and activity coefficient γ[i], for the ionic species i under concern, H^+ in this case; it is a[i] = −lg(m[i]γ[i]
). Since individual ions do not exist alone in solution, primary pH values cannot be assigned solely by experimental measurements, requiring extra thermodynamic model assumptions for the activity
coefficient, γ[H+], which has put pH in a unique situation of not being fully traceable to the International System of Units (SI). Also the concept of activity is often not felt to be as perceptible
as that of concentration which may present difficulties, namely with the interpretation of data. pH measurements on unknown samples rely on calibration of the measuring setup with adequate reference
pH buffers. In this work, the assignment of pH values to buffers closely matching the samples, e.g., seawater, is revisited. An approach is presented to assess the quantity pm[H+] = −lgm[H+]
profiting from the fact that, contrary to single ion activity coefficients, mean activity coefficients, $γ ± = ( γ H + γ C l − )$ can be assessed based on experimentally assessed quantities alone, $γ
± E x p$, thus ensuring traceability to the mole, the SI base unit for amount of substance. Compatibility between $γ ± E x p$ and mean activity coefficient calculated by means of Pitzer model
equations, $γ ± P t z$, validates the model for its intended use.
1. Introduction
Water, H
O, undergoes self-ionization giving origin to the hydronium ion, H
(or H
for simplicity of writing) and hydroxide ion, OH
, in equal concentrations of the order of 10
, depending on temperature,
, pressure,
, and ionic strength,
; H
confers free acidity to the system, commonly expressed in terms of pH = −lg
, where
stands for activity of the indicated chemical species, H
and γ
being its molality and activity coefficient respectively. For simplicity of writing, when calculating logarithms,
, and
, are used instead of the mathematically correct forms
$a i 0$
$m i 0$
, where the quantities with superscript “
” represent the standard state for the chemical species
under concern and equal the value 1 mol·kg
. Since, in general terms, water is likely to come in contact with almost every other substance and materials, chemical reactions tend to take place in aqueous solutions, therefore affecting and
being affected by H
concentration, e.g.,
, hence its activity,
. Activity,
, and concentration,
, of chemical species are equivalent quantities,
, the activity coefficient, γ
= 1 in the limiting condition of ideal dilute solutions, as it is the case of pure water [
]. Aqueous systems range from simple dilute solutions, e.g., distilled water, to complex mixtures of high ionic strength, e.g., seawater [
]. In the very dilute solutions, the concentration of chemical species, molecules or ions, is very small and the approximation γ
= 1 may be acceptable for some intended uses. Solutions of high complexity, both in terms of species concentration and diversity, introduce interactions that reflect upon deviations from ideality,
with activity coefficients becoming significantly different from 1.
Mean values for activity coefficients,
$γ ± = ( γ H + γ C l − )$
can be measured for solutions where the concentrations of the respective ions,
, are known. Nevertheless owing to the inevitable presence of both anions and cations in solution, experimental assessment of the activity coefficients of individual ions, γ
or γ
cannot be done. It can only be achieved through model approaches [
]. Experimentally assessed mean values,
$γ ± E x p$
, or modeled mean, γ
, or single values γ
or γ
introduce uncertainty to pH values assigned to the reference buffer solutions subject to measurement. Particularly complex mixtures of high ionic strength, as is the case of seawater,
≈ 0.7 mol·kg
, present major deviations from ideality as well as problems of chemical interferences, e.g., due to the presence of sulphate.
Under these various circumstances, covering a wide range of concentrations and matrices, calibration of the measuring setup requires calibration standards of compositions similar to those of the
sample solutions. Nevertheless, the common situation is that commercial pH meters come with two or three pH reference buffer solutions (pH (S) = 4,7 and 10) from the list of NIST (ex-NBS) buffers,
≤ 0.1 mol·kg
], which are used by the great majority of practitioners who are not fully aware of this potential source of uncertainty. This may eventually bring the result to an unacceptably low level of quality,
not fit for its intended purpose. The use of other more appropriate pH buffer solutions, when available, is advisable and is becoming more common practice among experts.
Although the conceptual definition of any chemical quantity is error free, its value obtained from a measurement, or calculated from a model equation, is affected by an uncertainty which is the
result of added contributions from the various steps, propagating through the traceability chain [
Figure 1
schematically represents the unbroken chain that links the various levels of metrological quality for pH, below the Primary Level, with the corresponding uncertainty target values [
] for aqueous solutions up to an ionic strength of 0.1 mol·kg
. Above this metrological level an additional source of uncertainty, 0.01, is introduced by the purely electrostatic Debye-Hückel model with the Bates-Guggenheim convention for the distance of
closest approach of chloride ions, adopted for the assessment of the individual activity coefficient, γ
Figure 1.
Metrological traceability scheme for pH.
denotes the expanded measurement uncertainty (coverage factor
= 2; the value of the measurand lies with a probability of approximately 95% within the interval of values) assigned to the pH values obtained for aqueous solutions up to an ionic strength of 0.1
For higher ionic strength aqueous solutions, e.g., seawater,
≈ 0.7 mol·kg
, a more elaborate model, as it is the Pitzer model, accounting for electrostatic and specific interactions [
], is bound to match reality more closely.
In this work, aiming at addressing seawater, a strategy is presented for the adoption of experimentally assessed mean activity coefficients,
$γ ± E x p$
. Primary potentiometric measurements were performed in a series of acidic aqueous solutions of known concentrat
on, consistently approaching a seawater matrix, ASW, for the assessment of experimental values of the acidity function, p(
) and mean activity coefficients,
$γ ± = ( γ H + γ C l − )$
. This allows evaluation of p
, with the corresponding uncertainty, without extra-thermodynamic assumptions, thus establishing full traceability to the mole. Mean activity coefficients can also be calculated,
$γ ± C a l c$
, upon taking into account a value of
derived from the literature value for the acidity constant of the hydrogen sulphate ion, Ka (HSO
) [
]. Compatibility with mean activity coefficients calculated from the semi empirical Pitzer model equations
$γ ± P t z$
, validates the use of the model [
], also available for single ion activity coefficients.
2. Method
Primary pH = p
values, are conventionally assigned [
] to primary standard pH buffer solutions (S) of ionic strength,
≤ 0.1 mol·kg
, through a primary method, pH (S). This includes measurement of Harned cell (H+ sensitive hydrogen gas electrode and Cl
sensitive silver, silver chloride electrode, without transference) potentials,
, calculations with the Nernst equation (Equation (1)) and adoption of extra-thermodynamic assumptions concerning models of electrolyte solutions, for the assignment of mean or single activity
$Pt |H 2 | H + buffer S, Cl − ( m c l − ) |AgCl| Ag E = E 0 A g , A g c l − ( R T F ln 10 ) l g ( a H + a C l − )$
is the standard cell potential, assessed from measurement in 0.01 mol·kg
is the gas constant,
is the absolute temperature and
is the Faraday constant.
This equation can be rearranged to give the acidity function, p(
), (Equation (2)):
−lg(𝑎[H+]𝛾[Cl−]) = p(𝑎[H+]𝛾[Cl−])
solely expressed in terms of experimentally available quantities:
$p ( a H + γ C l − ) = E – E 0 [ ( R T F ) ln 10 ] + lg ( m C l − )$
pH (S) values can be calculated from the acidity function by adoption of a conventional value to the activity coefficient of the chloride ion, γ[Cl−].
H (S) = p𝑎[H+] = − lg(𝑎[H+]𝛾[Cl−]) +lg𝛾[Cl−]
By adopting the purely electrostatic Debye-Hückel model with the Bates–Guggenheim convention for the chloride ion, γ
, valid to ionic strengths,
, below 0.1 mol·kg
, and assuming the value 0.01 in pH for the respective uncertainty, pH becomes traceable to the internationally accepted SI [
]. Above this value, the model is no longer valid and the use of the Pitzer model [
] has been debated.
The experimental procedure is, nevertheless, the basis for the assessment of the mean activity coefficient,
$γ ± = ( γ H + γ C l − )$
, for solutions of known
, independent of model assumptions:
$E = E 0 Ag , AgCl − ( R T F ln 10 ) lg ( m H + m Cl − γ H + γ Cl − )$
$E = E 0 Ag , AgCl − ( R T F ln 10 ) lg ( m H + m Cl − γ ± 2 )$
$γ ± = 10 − ( E − E 0 Ag , AgCl + ( R T F ln 10 ) lg ( m H + m Cl − ) ( 2 R T F ln 10 ) ) = 10 − ( E − E 0 Ag , AgCl ( 2 R T F ln 10 ) + lg ( m H + m Cl − ) )$
Since it is:
$p ( a H + γ Cl − ) = − lg ( m H + γ H + γ Cl − ) = − lg ( m H + γ ± 2 )$
p(𝑚[H+]) = − lg𝑚[H+] = p(𝑎[H+]𝛾[Cl−]) + 2lg𝛾[±] = p𝑚H (S)
the standard uncertainty of the acidity function,
)], can be calculated [
] by Equation (10).
$u [ p ( a H + γ C l − ) ] = a E 2 u 2 ( E ) + a E 0 2 u 2 ( E 0 ) + a T 2 u 2 ( T ) + a p H 2 2 u 2 ( p H 2 ) + a m C l − 2 u 2 ( m C l − ) + a ∆ E 2 u 2 ( ∆ E )$
is the partial pressure of hydrogen in the cell and ∆
is the bias potential of the Ag/AgCl electrodes.
The uncertainty of p
(S) is assessed through Equations (11)–(13).
$u [ p ( m H + ) ] = u p m H ( S ) = u [ p ( a H + γ Cl − ) + 2 lg γ ± ]$
$u [ p m H ( S ) ] = u 2 [ p ( a H + γ Cl − ) + u 2 [ 2 lg γ ± ] ]$
$u [ 2 lg γ ± ] = 2 u ( γ ± ) γ ± ln 10$
Through the application of this methodology to proper selected standard pH buffer solutions (S), for which chloride concentration, m[Cl−], and valid γ[±] values have been assessed, it is possible to
obtain pm[H](S). Making the reference buffers further available for the calibration of measuring devices, allows evaluation of pm[H+] values for unknown aqueous samples, based on experimental values,
hence fully traceable to the mole. This is of particular relevance for speciation studies, in environmental sciences and toxicological studies.
The two quantities, pa[H+] and pm[H+], are related through the activity coefficient of the single hydrogen ion, γ[H+].
Activity coefficients, γ
, γ
and γ
can also be calculated by means of Pitzer electrolyte solutions model equations [
], thus enabling assessment of the consistency between experimental and model approaches.
Assessed by either potentiometry with the glass combination pH electrode, or by other recommended instrumental method for evaluation of hydrogen ion signal, measured pH or pm[H+] values will always
be affected by an uncertainty equal or bigger than the one respectively associated to the reference pH (S), or pm[H](S) values assigned to the calibrating solutions, S.
In the particular case of seawater, the methodology can be applied to, e.g., a Tris-TrisHCl reference buffer prepared in a matrix of artificial seawater, ASW [
] which is becoming widely adopted by the oceanographic community. UV-Vis spectrophotometry, with m-cresol purple indicator, has gained support as an alternative to potentiometry with pH electrodes [
Traceability of results to common references is a key issue for ensuring their comparability and enabling interpretation in terms of the physical-chemical meaning of the quantities, pH or p
, hence on the properties of the aquatic systems under study. This is illustrated by a double traceability chain, represented schematically by
Figure 2
, which enables assessment of compatibility of measured results [
Figure 2. Schematic representation of a double traceability chain, showing traceability of two different types of analytical signals pertaining to hydrogen ion, H^+, to common references of pH or pm
[H+], and higher up to the conceptually defined pH value.
3. Results and Discussion
Harned cell potentials were measured in acidic saline solutions approaching seawater composition, 0.01 mol·kg
HCl + (NaCl + KCl + Na
+ CaCl
+ MgCl
), at
= 0.67 mol·kg
]. All measurements were performed at 25 °C [
Solid reagents were of the highest purity grade (≥99.99%). Hydrochloric acid stock solutions were prepared from Merck Suprapur
30% HCl solution and were reassessed by titration against tris(hydroxymethyl) methylamine (Sigma-Aldrich ≥99.8% purity, supplied by Química, S.L., Sintra, Portugal) according to reference procedures
]. All solutions were prepared using high purity Millipore
Milli-Q Advantage water, with resistivity higher than 18.2 MΩ·cm.
Recommended procedures have been followed [
] ensuring experimentally assessed values of acidity function, p(
), mean activity coefficients,
$γ ± E x p$
, hence also p(
). Mean activity coefficients calculated from experimental data,
$γ ± E x p$
, for sulphate-free solutions were co-opted for solutions of the same ionic strength where sodium sulphate substituted an equivalent amount of sodium chloride. Mean activity coefficients can also be
$γ ± C a l c$
, upon taking into account a value of
based on the literature value for the acidity constant of the hydrogen sulphate ion, Ka (HSO
Mean activity coefficients have also been calculated by means of Pitzer model equations, $γ ± P t z$ at 25 °C.
Assessment of compatibility between
$γ ± E x p$
$γ ± C a l c$
$γ ± P t z$
, for approximately 95% confidence level, supports consistency of the Pitzer model for the different involved interactions [
The standard uncertainties for
$γ ± E x p$
were obtained for three different background solutions of different complexity approaching seawater composition,
Table 1
, respectively:
• Solution A—0.01 mol·kg^−1 HCl + NaCl
• Solution B—0.01 mol·kg^−1 HCl + (NaCl + KCl + CaCl[2] + MgCl[2])
• Solution C—0.01 mol·kg^−1 HCl + (NaCl + KCl + Na[2]SO[4] + CaCl[2] + MgCl[2])
The value of
$γ ± E x p$
for solution B has been imported into solution C, which allowed calculation of the corresponding
. In parallel, as an alternative procedure for this later, Ka = 1.2 × 10
], was used to calculate
, which on its turn led to
$γ ± C a l c$
Values obtained for the uncertainty components of p
(S), (Equation (12)), for HCl in saline background, are also presented in
Table 1
. These values lead to estimation of the combined uncertainty
Table 1. Acidity function p(a[H+]γ[Cl−]) and pm[H+] values with their respective uncertainty, u, budgets assessed in different background solutions: A, B and C.
Table 1. Acidity function p(a[H+]γ[Cl−]) and pm[H+] values with their respective
uncertainty, u, budgets assessed in different background solutions: A, B and C.
Solution γ[±] u(γ[±]) u(2lgγ[±]) p(a[H+]γ[Cl−]) u(pa[H+]γ[Cl−]) pm[H+] u(pm[H+])
A * 0.7440 0.0085 0.010 2.2939 0.0038 2.037 0.011
B * 0.7284 0.020 0.024 2.2919 0.0038 2.017 0.024
C * 0.7284 0.020 0.024 2.2628 0.0038 1.988 0.024
C ** 0.7249 0.046 0.056 2.2628 0.0038 1.983 0.056
Notes: * γ[±] = $γ ± E x p$; ** γ[±] = $γ ± C a l c$.
The considerably higher uncertainty value u(2lgγ[±]) for the sulphate containing solution (C**) is due to the high impact of the acidity constant, Ka(HSO[4]^−), on the calculation of the uncertainty
of $γ ± C a l c$.
The contribution of the uncertainty introduced by γ
to p
has been calculated for the two values of mean activity coefficient,
$γ ± E x p$
$γ ± C a l c$
, and for the possible scenario of an order of magnitude lower uncertainty level, C′,
Table 2
Table 2. Uncertainty introduced by γ[±] to the calculation of pm[H+] for acidic solutions in seawater background, function of its uncertainty level.
Table 2. Uncertainty introduced by γ[±] to the calculation of pm[H+] for
acidic solutions in seawater background, function of its uncertainty
Solution γ[±] uγ[±] u(2lgγ[±]) u(pm[H+]) Upm[H+] (Expanded, k = 2)
C^Exp 0.7284 0.020 0.024 0.024 0.048
C′^Exp 0.7284 0.0020 0.0024 0.0024 0.0048
C^Calc 0.7249 0.046 0.056 0.056 0.12
C′^Calc 0.7249 0.0046 0.0056 0.0056 0.012
C^Ptz 0.76998 0.00518 0.0058 0.0058 0.012
The major influence on the estimation of u(pm[H+]) comes from the uncertainty of the mean activity coefficient, uγ[±]. As shown, its effect, u(2lgγ[±]), can be lowered upon improvement of the quality
of the raw data .
4. Conclusions
A series of acidic aqueous solutions in single and complex chloride media of known concentration, consistently approaching a seawater matrix, ASW, were prepared and subject to primary potentiometric
measurements for the assessment of experimental values of the acidity function, p(aH+γCl−) and mean activity coefficients, $γ ± = ( γ H + γ C l − )$.
Assessment of the mean activity coefficient, γ[±], for reference pH buffer standards, S, allows further calculation of their respective concentrations, m[H+], hence −lgm[H+] = pm[H+] = pm[H](S).
Compatibility of experimental values, $γ ± E x p$, $γ ± C a l c$ and model assessed, $γ ± P t z$, enables adoption of either approach with equivalent quality.
Adding to the uncertainty of the acidity function, (pa[H+]γ[Cl−]) the major uncertainty component largely contributing to the combined uncertainty of pm[H+], u(pm[H+]), comes from (γ[±]).
Application of the developed methodology to adequately chosen reference buffer solutions, S, leads to assignment of pm[H](S) values with their respective uncertainties. This enables calibration of
measuring devices in terms of pm[H+], thus ensuring fully traceable sample pm[H+]values.
This work was financially supported by the EMRP- ENV 05 OCEAN and the associated REG2. The EMRP is jointly funded by the EMRP participating countries within EURAMET and the European Union. The
authors wish to acknowledge the Portuguese funding institution Fundação para a Ciência e a Tecnologia-FCT, for supporting their research, namely through project PEst-OE/QUI/UI0612/2013.
Author Contribution
Maria Filomena Camões conceived the research program and supervised the work of Bárbara Anes; both authors had active roles in the calculations, critical analysis and elaboration of the conclusions.
Conflicts of Interest
The authors declare no conflict of interest.
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Camões, M.F.; Anes, B. Traceability of pH to the Mole. Water 2015, 7, 4247-4255. https://doi.org/10.3390/w7084247
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Camões MF, Anes B. Traceability of pH to the Mole. Water. 2015; 7(8):4247-4255. https://doi.org/10.3390/w7084247
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Camões, Maria Filomena, and Bárbara Anes. 2015. "Traceability of pH to the Mole" Water 7, no. 8: 4247-4255. https://doi.org/10.3390/w7084247
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