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Well-rounded inlet | Voima Toolbox
Performs head losses calculations in well-rounded inlets. There are two options:
Given the flow
Given the flow and the well-rounded inlet diameter, calculates the corresponding head loss.
$h_f = k. \frac{U^2}{2g}$
Given the well-rounded inlet diameter, calculates the head losses coeficients corresponding to the expressions: | {"url":"https://voimatoolbox.com/en/calculations/well-rounded-inlet","timestamp":"2024-11-10T18:08:10Z","content_type":"text/html","content_length":"76476","record_id":"<urn:uuid:3d33c527-3536-4eee-afed-6c73baf3c16e>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00203.warc.gz"} |
Cite as
Ioana O. Bercea, Martin Groß, Samir Khuller, Aounon Kumar, Clemens Rösner, Daniel R. Schmidt, and Melanie Schmidt. On the Cost of Essentially Fair Clusterings. In Approximation, Randomization, and
Combinatorial Optimization. Algorithms and Techniques (APPROX/RANDOM 2019). Leibniz International Proceedings in Informatics (LIPIcs), Volume 145, pp. 18:1-18:22, Schloss Dagstuhl – Leibniz-Zentrum
für Informatik (2019)
Copy BibTex To Clipboard
author = {Bercea, Ioana O. and Gro{\ss}, Martin and Khuller, Samir and Kumar, Aounon and R\"{o}sner, Clemens and Schmidt, Daniel R. and Schmidt, Melanie},
title = {{On the Cost of Essentially Fair Clusterings}},
booktitle = {Approximation, Randomization, and Combinatorial Optimization. Algorithms and Techniques (APPROX/RANDOM 2019)},
pages = {18:1--18:22},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-95977-125-2},
ISSN = {1868-8969},
year = {2019},
volume = {145},
editor = {Achlioptas, Dimitris and V\'{e}gh, L\'{a}szl\'{o} A.},
publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik},
address = {Dagstuhl, Germany},
URL = {https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.APPROX-RANDOM.2019.18},
URN = {urn:nbn:de:0030-drops-112337},
doi = {10.4230/LIPIcs.APPROX-RANDOM.2019.18},
annote = {Keywords: approximation, clustering, fairness, LP rounding} | {"url":"https://drops.dagstuhl.de/search/documents?author=Schmidt,%20Daniel%20R.","timestamp":"2024-11-07T13:51:59Z","content_type":"text/html","content_length":"74084","record_id":"<urn:uuid:9e36c24c-ea3c-4ee6-93f2-abf7c43f270d>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00157.warc.gz"} |
particle in cell from scratch
This code was made for the final project of a computational physics class at UCLA. It was my introduction to computational plasma physics.
All of the code is on my github, along with more detailed explanations and code tests in the jupyter notebook.
Two stream instability
The full animation is at the bottom of the page.
what is particle in cell?
Particle in cell (PIC) is a method to simulate the movement of particles under a force of some kind, which in my case is the force from an electric field, essentially simulating a plasma. At it’s
core, a (1D) PIC code uses a mesh grid that has bins/cells. The edges of these bins are described by two $x$ coordinates, $x_j$ and $x_{j+1}$ (of course, in 2D or 3D, there’s more coords), and within
that bin is a particle $i$ with location $r_i$. And from these particle locations, a denisty can be calculated at the edges as
$$ \rho_j = \sum_{i=0}^{N_p}\frac{x_{j+1} - r_i}{\Delta x}, \quad \rho_{j+1}=\sum_{i=0}^{N_p}\frac{r_i - x_{j}}{\Delta x}, $$
where $N_p$ is the number of particles within a bin and $\Delta x$ is the length of the cell. The particle in cell code that I made is the simplest case possible (I say that but it was very difficult
and I needed a lot of guidance from my TA), as in it is 1D and electrostatic (time-independent) such that Maxwell’s equations are
$$ \nabla \cdot E = \rho, \quad \nabla \cdot B = 0 \\ \nabla \times E = 0, \quad \nabla \times B = 0. $$
We ultimately want to find the acceleration
$$ F = ma = -qE \\ \implies a = -\frac{q}{m}E = -\frac{q}{m}\frac{\text{d}\phi}{\text{d}x}, $$
as that is what pushes the particles. $\phi$ is found through combining the $E$ maxwell equations, such that
$$\frac{\text{d}^2\phi}{\text{d}x^2} = \rho \qquad \text{poisson’s equation}.$$
And from acceleration, we can find velocity and position, giving us our phase space.
solving methods
Poisson’s equation can be estimated using the finite difference method so I tried that first. The first derivative of a general function using the finite difference method yields
$$ f’(a) = \frac{f(a+h)-f(a)}{h}, $$
so then
$$ f’’(a) = \frac{f(a) - 2f(a+h) + f(a+2h)}{h^2}. $$
If we apply this function to our $\phi$, we can define $f(a+h)$ as $\phi_j$, so then from equation $1$,
$$ \frac{\text{d}^2\phi}{\text{d}x^2} = \frac{\phi_{j-1} - 2\phi_j + \phi_{j+1}}{\Delta x^2}=\rho. $$ But using the matrix that comes from this finite difference equation doesn’t get us anywhere
because it’s not invertible :( …. so we will use a method involving a discrete fourier transform instead. According to Birdsall, we know that $$ \phi(k) = \frac{\rho(k)}{k^2} \quad \text{and} \
quad k = \frac{2n\pi}{L}, $$ where $\rho(k)$ is our charge density. To transform to $\rho(k)$, we use
$$ G(k) = \Delta x \sum_{j=0}^N{G(x_j)} e^{-ikx_j} $$ and to transform back to $\phi(x)$, we use the inverse DFT $$ G(x_j) = \frac{1}{L} \sum_{n=-N/2}^{N/2}{G(k)} e^{ikx_j}. $$
Using $\phi$, we can now find $E$. Since we are only looking at the electric field from bin to bin in the mesh, we can approximate it as just the slope between two points, $\phi_{j-1}$ and $\phi_
{j+1}$, such that
$$ E_j = \frac{\text{d}\phi(x_j)}{\text{d}x} = \frac{\phi_{j+1} - \phi_{j-1}}{2 \Delta x}, $$
which is represented in matrix form as
$$ E = \frac{1}{{2 \Delta x}} \begin{pmatrix} 0 & 1 & 0 & \dots & 0 & -1 \\ -1 & 0 & 1 & & & 0\\ 0 & -1 & 0 & & & \vdots\\ \vdots & & & \ddots & & 0\\ 0 & & & -1 & & 1\\ 1 & 0 & \dots& 0 & -1 & 0 \
end{pmatrix} $$
We want the electric field at the locations of the particles within a bin at $r_i$. Thus, we essentially take a weighted average of the electric fields at $E_j$ and $E_{j+1}$ with weights used
earlier to get the total electric field at $r_i$, such that
$$ E_i = \frac{x_{j+1} - r_i}{\Delta x} E_j + \frac{r_i - x_{j}}{\Delta x} E_{j+1}, $$
which is what we plug back in to our original acceleration. We repeat this process until a time $t_{final}$.
plasma background
To produce something like a two stream instability with my PIC code, I first had to understand some plasma basics.
fundamental equations
The essential plasma equations are
\text{Vlaslov} \quad &\partial_t v + v\partial_x v = -\frac{e}{m} E \\ \text{continuity} \quad &\partial_t n + \partial_x(n v) = 0 \\ \text{Gauss} \quad &\partial_x E = -\frac{e}{\epsilon_0}(n - n_0)
plasma oscillations
For plasma oscillations, $n, v$ and $E$ are described by a constant background and a perturbation indicated by a $0$ and $1$ index, respectively, such that
n &= n_0 + n_1 \\ v &= v_0 + v_1 \\ E &= E_0 + E_1. If we look at the continuity equation and plug in $n$ and $v$, we see that
$$ \partial_t (n_0 + n_1) + \partial_x((n_0 + n_1)(v_0 + v_1)) = 0, $$
where $\partial_t n_0$ and $\partial_x(n_0v_0)$ go to zero because $n_0$ and $v_0$ are constant over time. Then after algebra,
$$ \partial_t n_1 + \partial_x(n_1v_0 + n_0v_1 + n_1v_1) = 0. $$
And taking second order terms to be zero,
$$ \partial_t n_1 + \partial_x(n_1v_0 + n_0v_1) = 0. $$
We can analyze two cases: no drift velocity ($v_0 = 0, E_0 = 0$) and yes drift velocity ($v_0 \neq 0$). For each case, the continuity equation becomes
\text{no drift} \quad &\partial_t n_1 + n_0\partial_x v_1 = 0 \notag \\ \text{ya drift} \quad &\partial_t n_1 + \partial_x n_1v_0 + n_0\partial_x v_1 = 0. \notag
Then, for a plane wave $f_1(x, t)$ of amplitude $f_1$, we obtain
$$ -i \omega n_1 + i k n_0 v_1 = 0. $$
This process is done for all 3 equations in $(2), (3)$ and $(4)$, so we end up with 2 sets of 3 equations, which leads to the results of
\left(1 - \frac{\omega_p^2}{\omega^2}\right)E_1=0 \qquad &\text{Dispersion relation} \ (v=0) \notag \\ \left(1 - \frac{\omega_p^2}{(\omega-kv_0)^2}\right)E_1=0 \qquad &\text{Doppler waves} \ (v\neq0)
These relations are useful because they relate $k$ and $\omega$ of a given wave.
two stream instability
In two stream instability, there are two populations of particles with densities $n_{0_1}$ and $n_{0_2}$, such that $n_0 = n_{0_1} + n_{0_2}$. There is also a constant background of ions that do not
move through out the simulation such that the plasma is quasi-neutral. Since these ions are immobile, they essentially have an infinite mass. Additionaly, $v_{0_1} = 0$ and $v_{0_2} = v_0 \neq 0$. If
we apply these conditions to $(2), (3)$ and $(4)$, we find that
&\partial_t v_i + v_i \partial_x v_i = -\frac{e}{m} E \notag \\ &\partial_t n_i + \partial_x(n_i v_i) = 0 \notag \\ &\partial_x E = -\frac{e}{\epsilon_0}(n_1 + n_2 - n_0), \notag
Following the same process we did for plasma oscillations, we end up with
\begin{gather} \left[1 - \frac{\omega_p}{\omega^2} + \frac{\omega_p}{\omega-kv_0}\right]E_1 = 0 \notag \\ \implies 1 - \frac{\omega_{p_1}^2}{(\omega-kv_{0_1})^2} - \frac{\omega_{p_2}^2}{(\omega-kv_
{0_2})^2} = 0 \notag \end{gather}
which means that when $\omega_{p_1}=\omega_{p_2}=\omega_{p_e}$ and $v_{0_1}=-v_{0_2}=v_0$,
$$ 1 = \frac{1}{\hat{\omega} - \alpha} + \frac{1}{\hat{\omega} + \alpha}, $$
$$ \omega_p = \frac{n_0 e^2}{\epsilon_0m}, \qquad \hat{\omega} = \frac{\omega}{\omega_p}, \qquad \alpha = \frac{kv_0}{\omega_p}. $$
And as the phase space shows, unstable modes appear in the plasma, even with low temperatures. The fastest growing mode $k_{max}$ is found by maximizing $\hat{\omega}$ with respect to $\alpha$, such
\begin{gather} \frac{\text{d}\hat{\omega}}{\text{d}\alpha} = 0 \implies \alpha = \frac{\sqrt{3}}{2}\frac{k_{max}v_0}{\omega_{p_e}} \notag \\ \implies k_{max} = \frac{\sqrt{3}}{2}\frac{\omega_{p_e}}
{v_0}. \notag \end{gather}
And the fastest growing mode corresponds to the number of phase space holes. At $v_0\simeq \pm 10 \Delta x \omega_{p_e}$, there should be one phase space hole, while at $v_0\simeq \pm 2\Delta x \
omega_{p_e}$, there should be many phase space holes.
results — two stream instability
This is an example of two stream instability with my PIC code (it goes quite fast so feel free to slow it down or click through it):
There’s a couple problems with the results. One being that the animation is quite rigid and squarish. Maybe something to do with my solver?
The other problem being the amount of phase space holes. With $v_0\simeq \pm 2 \Delta x \omega_{p_e}$, I would expect more than just two holes, while with $v_0\simeq \pm 10 \Delta x \omega_{p_e}$
(not here), I would expect only one, but I also get two holes. To solve this, I think I would have to dig into the physics more. | {"url":"https://kianorr.com/research/pic/","timestamp":"2024-11-12T15:57:11Z","content_type":"text/html","content_length":"29107","record_id":"<urn:uuid:569d2f1d-60a4-4d75-8a18-19dd178bcb80>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00161.warc.gz"} |
Predicate Logic and Quantifiers
Discrete Mathematics I
7th lecture, November 11, 2022
Martin J. Dürst
© 2005-22 Martin J. Dürst Aoyama Gakuin University
Today's Schedule
• Leftovers/summary for last lecture
• Last week's homework
• Pascal's triangle, combinations, and combinatorics
• Predicates and predicate logic
• Operators on Predicates: Quantifiers
• This week's homework
Summary of Last Lecture
• Sets are one of the most basic kinds of objects in Mathematics
• Sets are unordered, and do not contain any repetitions
• There are two notations for sets: Connotation (e.g. {1, 2, 3, 4}) and denotation (e.g. {n|n ∈ ℕ, n>0, n<5})
• Operations on sets include union, intersection, difference, and complement
• The empty set is a subset of any set
• The powerset of a set is the set of all subsets
• The laws for sets are very similar to the laws for predicate logic
Leftovers of Last Lecture
Operations on sets, neutral elements, Venn diagrams, laws for sets, limits of sets,...
Homework, Problems 1/2
1. Create a set with four elements. If you use the same elements as other students, a deduction of points will be applied.
Example: {cat, cow, crow, camel}
2. Create the powerset of the set you created in problem 1.
Example: {{}, {cat}, {cow}, {crow}, {camel}, {cat, cow}, {cat, crow}, {cat, camel}, {cow, crow}, {cow, camel}, {crow, camel}, {cat, cow, crow}, {cat, cow, camel}, {cat, crow, camel}, {cow, crow,
camel}, {cat, cow, crow, camel}}
Homework, Problems 3/4
3. For sets A of size zero to six, create a table of the sizes of the powersets (|P(A)|).
│|A| │|P(A)| │
│0 │1 │
│1 │2 │
│2 │4 │
│3 │8 │
│4 │16 │
│5 │32 │
│6 │64 │
4. Express the relationship between the size of a set A and the size of its powerset P(A) as a formula.
|P(A)| = 2^|A| (the size of the powerset of A is 2 to the power of the size of A)
Homework, Problem 5
5. Explain the reason behind the formula in problem 4.
The formula is correct for |A|=0: A={}, P(A)={{}}, |P(A)|=2^0=1
If the formula is correct for |A|=k (i.e. |P(A)|=2^k),
we can show that it is correct for B with |B|=k+1
A={cat, cow}, |A|=2, P(A) = {{}, {cat}, {cow}, {cat, cow}}, |P(A)|=4
B = A∪{carp}={cat, cow, carp}, |B|=3
P(B) = P(A)∪{a∪{carp}|a∈P(A)} = P(A)∪{{carp}, {cat, carp}, {cow, carp}, {cat, cow, carp}}
|P(B)| = 2·|P(A)| = 8
General case
B = {c|c∈A∨c=d∧d∉A} (= A∪{d} where d∉A),
|B| = k+1 = |A|+1,
|P(B)|=2·|P(A)| = 2·2^k=2^k+1
(Let B be the set consisting of the elements of A and one additional element d which is not contained in A.
The size of B is one greater than the size of A.
Then the powerset of B is the union of two distinct sets: the powerset of A, and the set of sets from the powerset of A with d added.
The size of the powerset of B is therefore double the size of the powerset of A.)
(This explanation is using Mathematical induction over k.)
Homework, Problem 6
6. Create a table that shows, for sets A of size zero to five, and for each n (size of sets in P(A)), the number of such sets.
│|A|│n│|{B|B⊂A∧|B|=n}| │|A|│n│|{B|B⊂A∧|B|=n}| │
│0 │0│1 │4 │0│1 │
│1 │0│1 │4 │1│4 │
│1 │1│1 │4 │2│6 │
│2 │0│1 │4 │3│4 │
│2 │1│2 │4 │4│1 │
│2 │2│1 │5 │0│1 │
│3 │0│1 │5 │1│5 │
│3 │1│3 │5 │2│10 │
│3 │2│3 │5 │3│10 │
│3 │3│1 │5 │4│5 │
│ │ │ │5 │5│1 │
(These numbers are the numbers appearing in Pascal's triangle.)
Pascal's Triangle
Start with a single 1 in the first row, surrounded by zeroes ((0 ... 0) 1 (0 ... 0)).
Create row by row by adding the number above and to the left and the number above and to the right.
• For a set A with |A| = n, we can write
|{B|B⊂A∧|B|=m}| as [n]C[m]
(the mth number in the nth row of Pascal's triangle)
• [n]C[0] = 1 (the only subset of size 0 is {}, {B|B⊂A∧|B|=1} = {{}})
• [n]C[n] = 1 (the only subset of size n is A itself, {B|B⊂A∧|B|=|A|=n} = {{A}})
• [n]C[m] = [n-1]C[m-1] + [n-1]C[m] (n>0, 0<m<n)
Subsets and Combinations
• Combinatorics is very important for Information Technology
• Combinatorics deals with counting the number of different things under various conditions or restrictions
• The word combinations refers to the choices of a given size from a set without repetitions and without considering order
• The number of combinations of a certain size m selected from a set of size n are the same as the subsets of a given size m in a powerset of a set of size n
• The number of combinations is written [n]C[m]
• The number of combinations can be calculated directly: [n]C[m] = n!/(m! (n-m)!)
• There are also permutations (considering order), repeated permutations (allowing an element to be selected more than once), and repeated combinations
Types of Symbolic Logic
• Binary (Boolean) logic (using only true and false)
• Multi-valued logic (using e.g. true, false, and unknown)
• Fuzzy logic (including calculation of ambiguity)
• Propositional logic (using only propositions)
• Predicate logic (first order predicate logic,...)
• Temporal logic (integrating temporal relationships)
Limitations of Propositions
With propositions, related statements have to be made separately
2 is even. 5 is even.
Today it is sunny. Tomorrow it is sunny. The day after tomorrow, it is sunny.
We can express "If today is sunny, then tomorrow will also be sunny." or "If 2 is even, then 3 is not even".
But we cannot express "If it's sunny on a given day, it's also sunny on the next day." or "If x is even, then x+2 is also even.".
⇒ This problem can be solved using predicates
Examples of Predicates
• even(4): 4 is even
• even(27): 27 is even
• odd(4): 4 is odd
• sunny(November 12, 2021): It is sunny on November 12, 2021
• parent(Ieyasu, Hidetada): Ieyasu is the parent of Hidetada
• smaller(3, 7) (or: 3 < 7)
Predicate Overview
• The problem with propositions can be solved by introducing predicates.
• In the same way as propositions, predicates are objectively true or false.
• A predicate is a function (with 0 or more arguments) that returns true or false.
• If the value of an argument is undefined, the result (value) of the predicate is unknown.
• A predicate with 0 arguments is a proposition.
How to Write Predicates
There are two ways to write predicates:
1. Functional notation:
□ The name of the predicate is the name of the function
□ Arguments are enclosed in parentheses after the function name
□ Each predicate has a fixed number of arguments
□ Arguments in different positions have different meanings
□ Reading of predicates depends on their meaning
2. Operator notation:
□ Operators that return true or false are predicates
□ Examples: 3 < 7, 5 ≧ 2, a ∈ B, even(x) ∨ odd(y)
Formulas Containing Predicates
Using predicates, we can express new things:
• sunny(x) → sunny(day after x)
• even(y) → even(y+2)
• even(z) → odd(z+1)
Similar to propositions, predicates can be true or false.
But predicates can also be unknown/undefined, for example if they contain variables.
Even if a predicate is undefined (e.g. even(x)),
a formula containing this predicate can have a defined value (true or false)
(e.g. even(y) → even(y+2), or odd(z) → even(z+24))
First Order Predicate Logic
• The arguments of predicates can be constants, functions (e.g. sin), formulæ,...
□ even(2), say(Romeo, 'I love you'), parent(Ieyasu, Hidetada)
□ even(sin(0)), even(2+3×7)
• However, it is not possible to use predicates within predicates
Counterexample: say(z, parent(y, x))
(z says "y is the parent of x")
• Higher-order logic allows predicates within predicates
Example: ∀n∈ℕ: even(n) → even(n+2)
• For all n, elements of ℕ, if n is even, then n+2 is even.
• For all natural numbers n, if n is even, then n+2 is even.
General form: ∀x: P (x)
∀ is the A of "for All", inverted.
Readings in Japanese:
• 全ての自然数 n において、n が偶数ならば n+2 も偶数である
• 任意の x において、P(x)
Examples of Universal Quantifiers
∀n∈ℕ: n > -1
∀n∈ℕ: ∀m∈ℕ: n+m = m+n
∀a∈ℚ: ∀b∈ℕ: a+b = b+a
∀a∈{T, F}: ∀b∈{T, F}: a∨b = b∨a
Let S be the set of all students, B the set of all books, and let read(s, b) denote the fact that student s reads book b.
Then ∀s∈S: ∀b∈B: read(s, b) means that all students read all books.
Remark 1: ∀n∈ℕ: ∀m∈ℕ: n+m = m+n can be written as ∀n, m∈ℕ: n+m = m+n
Remark 2: ∀s∈S: ∀b∈B: read(s, b) is interpreted as ∀s∈S: (∀b∈B: read(s, b))
Knowledge about Field of Application
• Propositional logic does not need application knowledge except for the truth value of each proposition.
• Predicate logic combines axioms/theorems/knowledge of logic with axioms/theorems/knowledge of one or more application areas.
• Example: Predicate logic on natural numbers: Peano axioms,...
• Example: Predicate logic for sets: Laws for operations on sets,...
• Example: Size of sets: Knowledge about set operations and arithmetic with natural numbers
• Concrete example:
∀s: (male(s) ∨ female(s)) [all students are either male or female]
∀s: ¬(male(s) ∧ female(s)) [no student is both male and female]
Existential Quantifier
Example: ∃n∈ℕ: odd (n)
• There exists a n, element of ℕ, for which n is odd.
• There is a natural number n so that n is odd.
• There exists a natural number n which is odd.
• There exists an odd natural number.
General form: ∃y: P (y)
∃ is the mirrored form of the E in "there Exists".
Readings in Japanese:
• P(y) が成立する y が存在する
• ある y について、P(y)
Structure of Quantifier Expressions
Example: ∀m, n∈ℕ: m > n → m^2 ≧ n^2
• ∀: Quantifier
• m, n: Variable(s), separated by commas
• ∈ℕ: Set membership (applies to all previous variables connected by commas; unnecessary if there is a single obvious universal set)
• ":": Colon
• m > n → m^2 ≧ n^2: Quantified predicate
More Quantifier Examples
∀n∈ℕ: n + n + n = 3n
∃n∈ℕ: n^2 = n^3
∃n∈ℝ: q^2 < 50q < q^3
∀m, n∈ℕ: 7m + 2n = 2n + 7m
Applied Quantifier Examples
S: Set of students
F: Set of foods
like(s, f): Student s likes food f
1. All students like all foods:
2. Some students like all foods:
3. There is a food that all students like:
4. There is no food that all students like:
5. Each student dislikes a food:
• Predicates are limited because they cannot express complex facts
• Propositions are functions or operators that return truth values (true/false)
• First order propositional logic allows formal reasoning for some application domain (e.g. arithmetic, set theory,...)
• The universal quantifier (∀) expresses that a predicate is true for all members of some set(s)
• The existential quantifier (∃) expresses that a predicate is true for some member(s) of some set(s)
This Week's Homework
Deadline: November 17, 2022 (Thursday), 18:40.
Format: Handout, easily readable handwriting
Where to submit: Box in front of room O-529 (building O, 5th floor)
Problems: See handout
About Returns of Tests and Homeworks
• Today, the graded intermediate tests will be returned
• This is not part of the lecture itself (i.e. after 12:30)
• If you have some other commitment after 12:30, you can come to my office in the afternoon (after 14:00) to pick up your homework
• Homeworks including names in kana/Latin letters will be distributed first, then those without
• Homeworks with higher points will be distributed before those with lower points
• When your name is called, immediately and very clearly raise your hand, and come to the front quickly
• When taking your homework, make sure it is really yours
• NEVER take the homework of somebody else (a friend,...)
• Carefully analyze your mistakes and work on fixing them and avoiding them in the future
• Feel free to ask questions
Mathematical induction
Pascal's triangle
repeated combination
repeated permutation
predicate logic
symbolic logic
multi-valued logic
fuzzy logic
first-order predicate logic
temporal logic
binary logic
higher-order logic
universal quantifier
全称限量子 (全称記号)
existential quantifier
存在限量子 (存在記号) | {"url":"https://www.sw.it.aoyama.ac.jp/2022/Math1/lecture7.html","timestamp":"2024-11-08T16:00:54Z","content_type":"application/xhtml+xml","content_length":"25500","record_id":"<urn:uuid:e20b6db2-1d95-4475-9161-9b85a5630f7e>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00774.warc.gz"} |
6th Grade Math - Equivalent Expressions
This post explains and gives practice opportunities related to TEKS 6.7C:
determine if two expressions are equivalent using concrete models, pictorial models, and algebraic representations
Learn to simplify and find equivalent expressions using the order of operations.
STAAR Practice
Between 2016 and 2024, this supporting standard has been tested 1 time on the STAAR test. A video explaining the problem can be found below. If you'd rather take a quiz over this question, click here
. The video below is linked to the question in the quiz as an answer explanation after the quiz is submitted.
To view all the posts in this 6th grade TEKS review series, click here. | {"url":"https://www.fiveminutemath.net/post/6th-grade-math-equivalent-expressions","timestamp":"2024-11-01T19:37:18Z","content_type":"text/html","content_length":"1050492","record_id":"<urn:uuid:8e5b815e-a3eb-4411-b6ed-13a37961501a>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00404.warc.gz"} |
PHYSICS 109N Michael Fowler
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PHYSICS 109N
Michael Fowler
Homework Assignment: Due Tuesday 12 September.
Thales' tricks: 2500 years ago, Thales brought back to Greece from Egypt some practical ways to use geometrical ideas. In particular, he showed how to find the height of something by measuring its
shadow, and how to find the distance of a boat at sea by measuring line of sight angles from two places on the shore a known distance apart. To see exactly what he did, it's simplest just to do it.
1. Find the height of a lamp post in the parking lot behind the Physics Building by measuring the length of its shadow, and measuring the length of your own shadow. Explain clearly how knowing your
own height means you can figure the height of the lamp post.
Read both questions 2 &3 before starting to do 2
2. Since we don't have a handy boat at sea, we settle for measuring the distance of the Rotunda from the next path down across the Lawn (not the path right by the Rotunda). Go to one end of that next
path down, and measure the angle between the line of the path and a pencil, say, pointing straight to the middle of the Rotunda. Now do the same from the other end of the path. Next, pace off the
path, after measuring the length of your pace. Now draw a triangle with baseline to represent the path, in some units you decide, such as 1 inch = 100 feet, (state what unit you decide to use!) draw
the other two sides to represent lines from the ends of the path to the middle of the Rotunda. Use a protractor to get the angle equal to what you measured at the Lawn, then use your drawing to
measure the distance of the Rotunda from the path.
3. We can now measure the height of the Rotunda without getting close to it. The triangle you drew for question 2 above was to be used to find the distance of the Rotunda from the midpoint of the
path. For this question, you must actually go to the midpoint of the path and there measure the angle between a pencil pointing at the topmost point of the Rotunda and the horizontal. Now, back to
the drawing board, draw a triangle having as baseline the line from the midpoint of the path to the middle of the Rotunda, and having the angle you just measured (so the other point in the triangle
is the top of the Rotunda). If you drew this triangle accurately, the short side of the triangle is the height of the Rotunda -- but, in writing this up, you should make clear why this is so.
4. The point of the above exercise is to show that the size of an inaccessible object can be determined by observing it from more than one place. Can this trick be used to find the distance and size
of the moon?
5. By measuring some shadow, find how high in the sky the sun gets in the middle of the day, that is, what maximum angle does a pencil pointing directly at the sun make with the horizontal? What
direction is the sun when this angle is maximum? Would this angle be measurably different in New York? | {"url":"https://galileoandeinstein.phys.virginia.edu/more_stuff/homework/95109hw1.html","timestamp":"2024-11-03T18:38:16Z","content_type":"text/html","content_length":"3617","record_id":"<urn:uuid:6dc0ef70-ad48-4b36-9d05-ace165eafbec>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00558.warc.gz"} |
SimBiology model solver options
The SolverOptions object holds the model solver options, and it is a property of the configset object.
Changing the SolverType property of configset changes the options specified in the SolverOptions object (or property).
You can add a number of configset objects with different SolverOptions to the model object with the addconfigset method. Only one configset object in the model object can be Active at any given time.
To change or update any of properties of the SolverOptions property object, use dot notation syntax: csObj.SolverOptions.PropertyName = value, where csObj is the configset object of a SimBiology^®
model and PropertyName is the name of the property of SolverOptions.
Use dot notation syntax on a configset object to return the property object. That is, csObj.SolverOptions returns the SolverOptions property object of the configset object csObj.
AbsoluteTolerance — Absolute error tolerance applied to state value during simulation
1e-6 (default) | positive scalar
Absolute error tolerance applied to a state value during simulation, specified as a positive scalar. This property is available for the ODE solvers (ode15s, ode23t, ode45, and sundials).
SimBiology uses AbsoluteTolerance to determine the largest allowable absolute error at any step in a simulation. How the software uses AbsoluteTolerance to determine this error depends on whether the
AbsoluteToleranceScaling property is enabled. For details, see Selecting Absolute Tolerance and Relative Tolerance for Simulation.
Data Types: double
AbsoluteToleranceScaling — Control scaling of absolute error tolerance during simulation
true or 1 (default) | false or 0
Control scaling of absolute error tolerance during simulation, specified as a numeric or logical 1 ( true) or 0 (false).
When AbsoluteToleranceScaling is enabled (by default), each state has its own absolute tolerance that can increase over the course of simulation. Sometimes the automatic scaling is inadequate for
models that have kinetics at largely different scales. For example, the reaction rate of one reaction can be on the order of 10^22, while another is 0.1. By turning off AbsoluteToleranceScaling, you
might be able to simulate the model. For more tips, see Troubleshooting Simulation Problems.
Data Types: double | logical
AbsoluteToleranceStepSize — Initial guess for time step size for scaling of absolute error tolerance
[] (default) | positive scalar
Initial guess for time step size for scaling of absolute error tolerance, specified as a positive scalar. The property uses the time units specified by the TimeUnits property of the corresponding
configset object.
Data Types: double
ErrorTolerance — Error tolerance of explicit or implicit tau stochastic solver
3e-2 (default) | positive scalar between 0 and 1
Error tolerance of an explicit or implicit tau stochastic solver, specified as a positive scalar between 0 and 1.
The explicit and implicit tau solvers automatically chooses a time interval (tau) such that the relative change in the propensity function for each reaction is less than the user-specified error
tolerance. A propensity function describes the probability that the reaction will occur in the next smallest time interval, given the conditions and constraints. If the error tolerance is too large,
there may not be a solution to the problem and that could lead to an error. If the error tolerance is small, the solver will take more steps than when the error tolerance is large leading to longer
simulation times. The error tolerance should be adjusted depending upon the problem, but a good value for the error tolerance is between 1% and 5%.
Data Types: double
LogDecimation — Specify frequency to log stochastic simulation output
1 (default) | positive integer
Specify the frequency to log stochastic simulation output, specified as a positive integer. This property is available only for stochastic solvers (ssa, expltau, and impltau).
Use LogDecimation to specify how frequently you want to record the output of the simulation. For example, if you set LogDecimation to 1, for the command [t,x] = sbiosimulate(modelObj), at each
simulation step the time will be logged in t and the quantity of each logged species will be logged as a row in x. If LogDecimation is 10, then every 10th simulation step will be logged in t and x.
Data Types: double
MaxIterations — Maximum number of iterations for nonlinear solver in implicit tau
15 (default) | positive integer
Maximum number of iterations for the nonlinear solver in implicit tau (impltau), specified as a positive integer.
The implicit tau solver in SimBiology internally uses a nonlinear solver to solve a set of algebraic nonlinear equations at every simulation step. Starting with an initial guess at the solution, the
nonlinear solver iteratively tries to find the solution to the algebraic equations. The closer the initial guess is to the solution, the fewer the iterations the nonlinear solver will take before it
finds a solution. MaxIterations specifies the maximum number of iterations the nonlinear solver should take before it issues a failed to converge error. If you get this error during simulation, try
increasing MaxIterations.
Data Types: double
MaxStep — Maximum step size
[] (default) | positive scalar
Maximum step size taken by an ODE solver, specified as a positive scalar. This property sets an upper bound on the size of any step taken by the solver. This property is available only for the ODE
solvers (ode15s, ode23t, ode45, and sundials).
By default, MaxStep is set to [], which is equivalent to setting the value to infinity.
If the differential equation has periodic coefficients or solutions, it might be a good idea to set MaxStep to some fraction (such as 1/4) of the period. This guarantees that the solver does not
enlarge the time step too much and step over a period of interest. For more information, see odeset.
Data Types: double
OutputTimes — Times to log deterministic simulation output
[] (default) | vector of nonnegative values
Times to log deterministic (ODE) simulation output, specified as a vector of nonnegative monotonically increasing values. This property specifies the times during an ODE simulation that data is
recorded. This property is available only for the ODE solvers (ode15s, ode23t, ode45, and sundials).
By default, the property is set to [], which instructs SimBiology to log data every time the solver takes a step. The unit for this property is specified by the TimeUnits property of the
corresponding configset object.
If the criteria set in the MaximumWallClock property causes a simulation to stop before all time values in OutputTimes are reached, then no data is recorded for the latter time values.
The OutputTimes property can also control when a simulation stops:
• The last value in OutputTimes overrides the StopTime property as criteria for stopping a simulation.
• The length of OutputTimes overrides the MaximumNumberOfLogs property as criteria for stopping a simulation.
Data Types: double
RandomState — State of random number generator
[] (default) | integer
State of the random number generator for stochastic solvers, specified as an integer. This property is available only for these solvers: ssa, expltau, and impltau.
SimBiology uses a pseudorandom number generator. The sequence of numbers generated is determined by the state of the generator, which can be specified by the RandomState property. If RandomState is
set to an integer J, the random number generator is initialized to its Jth state. The random number generator can generate all the floating-point numbers in the closed interval [2^-53, 1-2^-53].
Theoretically, it can generate over 2^1492 values before repeating itself. But for a given state, the sequence of numbers generated will be the same. To change the sequence, change RandomState.
SimBiology resets the state at startup. The default value of RandomState is [].
Data Types: double
RelativeTolerance — Allowable error tolerance relative to state value during a simulation
1e-3 (default) | positive scalar less than 1
Allowable error tolerance relative to a state value during a simulation, specified as a positive scalar less than 1. This property is available only for the ODE solvers (ode15s, ode23t, ode45, and
The RelativeTolerance property specifies the allowable error tolerance relative to the state vector at each simulation step. The state vector contains values for all the state variables, for example,
amounts for all the species.
For example, if you set the RelativeTolerance to 1e-2, you are specifying that an error of 1% relative to each state value is acceptable at each simulation step. For details, see Selecting Absolute
Tolerance and Relative Tolerance for Simulation.
Data Types: double
SensitivityAnalysis — Flag to enable or disable sensitivity analysis
false or 0 (default) | true or 1
Flag to enable or disable sensitivity analysis, specified as a numeric or logical 1 (true) or 0 (false).
This property lets you compute the time-dependent sensitivities of all the species states defined by the StatesToLog property with respect to the Inputs that you specify in the
SensitivityAnalysisOptions property of the configuration set object.
SimBiology always uses the SUNDIALS solver to perform sensitivity analysis on a model, regardless of what you have selected as the SolverType in the configuration set.
For more information on setting up sensitivity analysis, see SensitivityAnalysisOptions. For a description of sensitivity analysis calculations, see Sensitivity Analysis in SimBiology.
Models containing the following active components do not support sensitivity analysis:
• Nonconstant compartments
• Algebraic rules
• Events
Data Types: double | logical
Type — SimBiology object type
'solveroptions' (default)
This property is read-only.
SimBiology object type, specified as 'solveroptions'.
Data Types: char
Change Solver Options and Simulation Options
The solver and simulation options are stored in the configuration set (configset object) of a SimBiology model. Solver options contain settings such as relative and absolute tolerances. Simulation
options are settings such as MaximumNumberOfLogs and MaximumWallClock.
Depending on the solver type, the available solver options differ. Inspect the default ODE solver.
m1 = sbiomodel("m1");
cs = getconfigset(m1);
Change the ODE solver to ode45, which is one of the supported ODE solvers. For details, see Choosing a Simulation Solver.
cs.SolverType = "ode45";
ans =
SimBiology Solver Settings: (ode)
AbsoluteTolerance: 1e-06
AbsoluteToleranceScaling: true
RelativeTolerance: 0.001
SensitivityAnalysis: false
If you change a common option for the ODE solvers, that change is persistent across all the ODE solver types. For example, change the AbsoluteTolerance of the current solver.
cs.SolverOptions.AbsoluteTolerance = 1e-3
cs =
Configuration Settings - default (active)
SolverType: ode45
StopTime: 10
AbsoluteTolerance: 0.001
AbsoluteToleranceScaling: true
RelativeTolerance: 0.001
SensitivityAnalysis: false
StatesToLog: all
UnitConversion: false
DimensionalAnalysis: true
Inputs: 0
Outputs: 0
Change the solver to ode23t and check the value of AbsoluteTolerance, which is still the value you set previously.
cs.SolverType = "ode23t";
If you specify a stochastic solver as the solver type, the associated solver options are updated automatically, and these options are different from the ODE solver options. For details on the
supported stochastic solvers, see Stochastic Solvers.
Change to the explicit tau-leaping algorithm.
cs.SolverType = "expltau"
cs =
Configuration Settings - default (active)
SolverType: expltau
StopTime: 10
ErrorTolerance: 0.03
LogDecimation: 1
StatesToLog: all
UnitConversion: false
DimensionalAnalysis: true
Inputs: 0
Outputs: 0
You can also change the simulation settings, which are the properties of the configset object. For example, change the maximum number of logs criterion to decide when to stop the simulation. Setting
it to 1 returns simulated values of the model quantities immediately after applying initial and repeated assignment rules of the model. If you change the value to 2, and the subsequent simulation
fails with the integration error, then it probably indicates an error with the assignment rules. For more tips on how to use these simulation settings to troubleshoot some of the common simulation
problems, see Troubleshooting Simulation Problems.
cs.MaximumNumberOfLogs = 1;
Specify Times to Log Deterministic Simulation Output
Specify the times during a deterministic (ODE) simulation that data is recorded.
Create a model object named cell and save it in a variable named modelObj.
modelObj = sbiomodel('cell');
Retrieve the configuration set from modelObj and save it in a variable named configsetObj.
configsetObj = getconfigset(modelObj);
Specify to log output every second for the first 10 seconds of the simulation. Do this by setting the OutputTimes property of the SolverOptions property of ConfigsetObj.
sopt = configsetObj.SolverOptions;
sopt.OutputTimes = [1:10];
ans = 10×1
When you simulate modelObj, output is logged every second for the first 10 seconds of the simulation. Also, the simulation stops after the 10th log.
Version History
Introduced in R2006b | {"url":"https://fr.mathworks.com/help/simbio/ref/simbiology.solveroptions.html","timestamp":"2024-11-14T14:05:27Z","content_type":"text/html","content_length":"118312","record_id":"<urn:uuid:ddfe945b-310a-4f62-9456-7b960b100bef>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00318.warc.gz"} |
Intersection of Two Linked Lists
View Intersection of Two Linked Lists on LeetCode
Time Spent Coding
5 minutes
Time Complexity
O(n + m) - We must iterate through each list at least once, taking n as the length of listA and m as the length of listB; this causes our time complexity to be O(n + m).
Space Complexity
O(1) - The number of variables created is independent of n or m, resulting in the O(1) space complexity.
Runtime Beats
84.88% of other submissions
Memory Beats
100% of other sumbissions
Before iteration, we must initialize our “pointers” l1 and l2 to listA and list, respectively.
We will then begin iterating until each “pointer” represents the same node in the list, and this can only ever occur at the intersection due to how we are iterating.
With each iteration of the while loop, we will reassign our pointer to its next node if the current node is None. If it is None, we will set the pointer equal to the opposite list starting node.
If there is an intersection, we will return the node that both pointers are equal to.
If there is no intersection, eventually, the nodes will both be equal to None, and we will return that.
Algorithms Used
Two Pointer Algorithm - An algorithm typically used to search a list where opposite ends of the list share a relationship that will be compared to determine some outcome.
For this problem, that condition would be if the elements are equal to the same node.
Visual Examples
An array being transformed into a set, click to view
The two-pointer algorithm being performed on a list where the condition summing to 6, click to view
1 # Definition for singly-linked list.
2 # class ListNode:
3 # def __init__(self, x):
4 # self.val = x
5 # self.next = None
7 class Solution:
8 def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
9 l1,l2=headA,headB
10 while l1!=l2:
11 l1=l1.next if l1 else headB
12 l2=l2.next if l2 else headA
13 return l1 | {"url":"https://douglastitze.com/posts/intersection-of-two-linked-lists/","timestamp":"2024-11-13T14:30:27Z","content_type":"text/html","content_length":"25935","record_id":"<urn:uuid:597f2593-468b-4f0c-be2d-cbd56bd6418b>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00408.warc.gz"} |
Slugging Percentage Calculator - Sum SQ
Slugging Percentage Calculator
Are you looking for an easy way to calculate your slugging percentage?
In case you don’t know what slugging percentage means in baseball.
A player’s slugging percentage is the number of bases he or she gets per at-bat. Unlike on-base percentage, slugging percentage only looks at hits. Walks and being hit by a pitch are not part of the
equation. The difference between slugging percentage and batting average is that not all hits are worth the same.
The slugging percentage calculator helps you calculate your baseball slugging percentage easily just like effective field percentage..
Simply enter the no. of singles, doubles, triples, home runs, and at bats to find out the slugging percentage.
You might be interested in discovering your Earned Runs Average.
What is Slugging Percentage?
Slugging percentage is a baseball statistic that represents the total number of bases a player achieves per at-bat. It’s a more nuanced measure of a player’s batting prowess than batting average
alone, as it takes into account the quality of hits rather than just their frequency.
Slugging percentage gives a clearer picture of a batter’s power and overall offensive contribution. While batting average treats all hits equally, slugging percentage assigns more value to extra-base
hits, providing a more comprehensive view of a player’s ability to generate runs.
How to Calculate Slugging Percentage?
The formula for calculating slugging percentage is straightforward:
SLG = (1B + 2 × 2B + 3 × 3B + 4 × HR) / AB
• 1B = Number of singles
• 2B = Number of doubles
• 3B = Number of triples
• HR = Number of home runs
• AB = Number of at-bats
Step-by-Step Calculation
1. Count the number of singles, doubles, triples, and home runs a player has hit.
2. Multiply the number of doubles by 2, triples by 3, and home runs by 4.
3. Add these values to the number of singles to get the total bases.
4. Divide the total bases by the number of at-bats.
5. The result, typically expressed as a decimal to three places, is the slugging percentage.
Using a Slugging Percentage Calculator
To simplify the process, many baseball enthusiasts and professionals use a slugging percentage calculator. These tools streamline the computation, allowing for quick and accurate results.
Benefits of Using a Slugging Percentage Calculator
1. Accuracy: Eliminates human error in calculations.
2. Efficiency: Saves time, especially when calculating for multiple players or seasons.
3. Consistency: Ensures uniform results across different users.
How to Use a Slugging Percentage Calculator
Most slugging percentage calculators follow these basic steps:
1. Input the number of singles, doubles, triples, and home runs.
2. Enter the total number of at-bats.
3. Click “Calculate” or a similar button.
4. The calculator will display the slugging percentage, typically rounded to three decimal places.
Interpreting Slugging Percentage
Understanding what constitutes a good slugging percentage is crucial for evaluating player performance.
Slugging Percentage Scale
• .300 or lower: Poor
• .350: Below average
• .400: Average
• .450: Above average
• .500: Excellent
• .600+: Outstanding
It’s important to note that these benchmarks can vary depending on the era, league, and position of the player.
Slugging Percentage vs. Other Baseball Statistics
While slugging percentage is a valuable metric, it’s often used in conjunction with other statistics for a more comprehensive evaluation of a player’s offensive capabilities.
Slugging Percentage and Batting Average
Batting average measures how often a player gets a hit, while slugging percentage measures the quality of those hits. A player with a high batting average but low slugging percentage might be good at
getting on base with singles, but lacks power hitting ability.
On-Base Plus Slugging (OPS)
OPS combines on-base percentage and slugging percentage to provide a more complete picture of a player’s offensive value. It’s calculated by adding a player’s on-base percentage to their slugging
Historical Context of Slugging Percentage
Slugging percentage has been a part of baseball statistics since the early 20th century, but its importance has grown with the advancement of statistical analysis in the sport.
Notable Slugging Percentages in MLB History
• Babe Ruth holds the career slugging percentage record at .6897.
• Barry Bonds has the single-season record with .863 in 2001.
These exceptional figures highlight the rarity of sustaining a high slugging percentage over an extended period.
Factors Affecting Slugging Percentage
Several factors can influence a player’s slugging percentage:
1. Ballpark dimensions: Smaller parks may inflate slugging percentages due to easier home runs.
2. Pitching quality: Facing tougher pitchers can lower slugging percentages.
3. Player’s role: Leadoff hitters might focus more on getting on base rather than hitting for power.
4. Weather conditions: Wind and temperature can affect how far the ball travels.
Slugging Percentage in Player Evaluation
Teams and scouts use slugging percentage as one of many tools to evaluate players. It’s particularly useful for:
1. Comparing power hitters
2. Assessing a player’s overall offensive contribution
3. Identifying potential for improvement in young players
However, it’s crucial to consider slugging percentage alongside other metrics and contextual factors for a comprehensive evaluation.
Limitations of Slugging Percentage
While valuable, slugging percentage has some limitations:
1. It doesn’t account for walks or hit-by-pitches.
2. It treats all extra-base hits equally within their category (e.g., a 450-foot home run counts the same as a 330-foot home run).
3. It doesn’t consider the game situation or the impact of the hit on the game outcome. | {"url":"https://sumsq.com/slugging-percentage-calculator/","timestamp":"2024-11-11T07:50:05Z","content_type":"text/html","content_length":"101799","record_id":"<urn:uuid:eecfe833-0c4f-4d08-94d0-8eb885676cea>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00626.warc.gz"} |
1998 AIME Problems
1998 AIME (Answer Key)
| AoPS Contest Collections • PDF
1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial
credit nor a penalty for wrong answers.
2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
Problem 1
For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$ and $8^8$, and $k$?
Problem 2
Find the number of ordered pairs $(x,y)$ of positive integers that satisfy $x \le 2y \le 60$ and $y \le 2x \le 60$.
Problem 3
The graph of $y^2 + 2xy + 40|x|= 400$ partitions the plane into several regions. What is the area of the bounded region?
Problem 4
Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players
obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
Problem 5
Given that $A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,$ find $|A_{19} + A_{20} + \cdots + A_{98}|.$
Problem 6
Let $ABCD$ be a parallelogram. Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$
find $RC.$
Problem 7
Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$
Problem 8
Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first
negative term encountered. What positive integer $x$ produces a sequence of maximum length?
Problem 9
Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly $m$ minutes. The probability that
either one arrives while the other is in the cafeteria is $40 \%,$ and $m = a - b\sqrt {c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a
+ b + c.$
Problem 10
Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon. A ninth sphere is placed on the flat
surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the
square of any prime. Find $a + b + c$.
Problem 11
Three of the edges of a cube are $\overline{AB}, \overline{BC},$ and $\overline{CD},$ and $\overline{AD}$ is an interior diagonal. Points $P, Q,$ and $R$ are on $\overline{AB}, \overline{BC},$ and $\
overline{CD},$ respectively, so that $AP = 5, PB = 15, BQ = 15,$ and $CR = 10.$ What is the area of the polygon that is the intersection of plane $PQR$ and the cube?
Problem 12
Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline
{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area
of triangle $PQR$ is $a + b\sqrt {c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime. What is $a^{2} + b^{2} + c^{2}$?
Problem 13
If $\{a_1,a_2,a_3,\ldots,a_n\}$ is a set of real numbers, indexed so that $a_1 < a_2 < a_3 < \cdots < a_n,$ its complex power sum is defined to be $a_1i + a_2i^2+ a_3i^3 + \cdots + a_ni^n,$ where $i^
2 = - 1.$ Let $S_n$ be the sum of the complex power sums of all nonempty subsets of $\{1,2,\ldots,n\}.$ Given that $S_8 = - 176 - 64i$ and $S_9 = p + qi,$ where $p$ and $q$ are integers, find $|p| +
Problem 14
An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest
possible value of $p$?
Problem 15
Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the
second coordinate of the immediately preceding pair, and in which $(i,j)$ and $(j,i)$ do not both appear for any $i$ and $j$. Let $D_{40}$ be the set of all dominos whose coordinates are no larger
than 40. Find the length of the longest proper sequence of dominos that can be formed using the dominos of $D_{40}.$
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. | {"url":"https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems","timestamp":"2024-11-06T18:16:39Z","content_type":"text/html","content_length":"69353","record_id":"<urn:uuid:ab5c8c6d-6997-4b87-8cce-8c40c5f3649b>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00708.warc.gz"} |
Detecting the Causes of Ill-Conditioning in Structural Finite Element Models
Kannan, Ramaseshan and Hendry, Stephen and Higham, Nicholas J. and Tisseur, Francoise (2013) Detecting the Causes of Ill-Conditioning in Structural Finite Element Models. [MIMS Preprint]
Download (3MB)
In 2011, version 8.6 of the finite element-based structural analysis package Oasys GSA was released. A new feature in this release was the estimation of the $1$-norm condition number $\kappa_1(K)={\|
K\|}_1{\|K^{-1}\|}_1$ of the stiffness matrix $K$ of structural models by using a $1$-norm estimation algorithm of Higham and Tisseur to estimate \(\|K^{-1}\|_1 \). The condition estimate is reported
as part of the information provided to engineers when they carry out linear/static analysis of models and a warning is raised if the condition number is found to be large. The inclusion of this
feature prompted queries from users asking how the condition number impacted the analysis and, in cases where the software displayed an ill conditioning warning, how the ill conditioning could be
``fixed''. We describe a method that we have developed and implemented in the software that enables engineers to detect sources of ill conditioning in their models and rectify them. We give the
theoretical background and illustrate our discussion with real-life examples of structural models to which this tool has been applied and found useful. Typically, condition numbers of stiffness
matrices reduce from $O(10^{16})$ for erroneous models to $O(10^8)$ or less for the corrected model.
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Greedily selects a subset of bounding boxes in descending order of score,
View aliases
Compat aliases for migration
See Migration guide for more details.
boxes, scores, max_output_size, iou_threshold, score_threshold,
pad_to_max_output_size=False, name=None
pruning away boxes that have high intersection-over-union (IOU) overlap with previously selected boxes. Bounding boxes with score less than score_threshold are removed. Bounding boxes are supplied as
[y1, x1, y2, x2], where (y1, x1) and (y2, x2) are the coordinates of any diagonal pair of box corners and the coordinates can be provided as normalized (i.e., lying in the interval [0, 1]) or
absolute. Note that this algorithm is agnostic to where the origin is in the coordinate system and more generally is invariant to orthogonal transformations and translations of the coordinate system;
thus translating or reflections of the coordinate system result in the same boxes being selected by the algorithm. The output of this operation is a set of integers indexing into the input collection
of bounding boxes representing the selected boxes. The bounding box coordinates corresponding to the selected indices can then be obtained using the tf.gather operation. For example: selected_indices
= tf.image.non_max_suppression_v2( boxes, scores, max_output_size, iou_threshold, score_threshold) selected_boxes = tf.gather(boxes, selected_indices)
boxes A Tensor. Must be one of the following types: half, float32. A 2-D float tensor of shape [num_boxes, 4].
scores A Tensor. Must have the same type as boxes. A 1-D float tensor of shape [num_boxes] representing a single score corresponding to each box (each row of boxes).
max_output_size A Tensor of type int32. A scalar integer tensor representing the maximum number of boxes to be selected by non max suppression.
iou_threshold A Tensor. Must be one of the following types: half, float32. A 0-D float tensor representing the threshold for deciding whether boxes overlap too much with respect to IOU.
score_threshold A Tensor. Must have the same type as iou_threshold. A 0-D float tensor representing the threshold for deciding when to remove boxes based on score.
pad_to_max_output_size An optional bool. Defaults to False. If true, the output selected_indices is padded to be of length max_output_size. Defaults to false.
name A name for the operation (optional).
A tuple of Tensor objects (selected_indices, valid_outputs).
selected_indices A Tensor of type int32.
valid_outputs A Tensor of type int32. | {"url":"https://docs.w3cub.com/tensorflow~2.4/raw_ops/nonmaxsuppressionv4.html","timestamp":"2024-11-13T22:44:25Z","content_type":"text/html","content_length":"10766","record_id":"<urn:uuid:464b0328-2688-41e5-bba0-d2027da200d7>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00554.warc.gz"} |
I used jQuery for the UI. I am a recent convert to jQuery, having mostly used Prototype + Scriptaculous. The word list is embedded into the page script as a javascript array. On document ready, html
is generated, which writes the first and last word to the page, and creates blank input boxes for the […] | {"url":"https://antiquity.jamie.ly/tags/algorithms/","timestamp":"2024-11-11T07:42:07Z","content_type":"text/html","content_length":"27084","record_id":"<urn:uuid:a9b3a4af-6312-487d-b299-137eca49f572>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00605.warc.gz"} |
who would win more often.
Katakros or Megaboss on Maw-krusha
19 members have voted
1. 1. who would win more 1 on 1's.
□ Katakros
□ Megaboss on Maw-krusha
Well I haven't seen this fight in a game so far, so this is just my opinion based on their warscrolls but on its own it looks to be quite hard for the mawkrusha to win. If the dice gods are with it,
it could win (especially if it charged) but if it doesn't instagib Katakros it's likely he'll kill the mawkrusha in return (which is much more likely). Lots of attacks with high rend and even more
with -1 rend.
Katakros is awfully slow and with a huge base though so there are plenty of ways to deal with him.... in theory you could feed him a decoy unit, then destroy another unit so that you get to pile-in
with the MW and finish him off...
This is a very difficult question as it depends on a lot of factors that can be variable. The megaboss is basically always going to be charging as it's so much faster than Katakros, but we don't know
if Katakros will have a turn before getting charged in which to set up his various buffs and debuffs. Similarly, the Megaboss on Maw-Krusha has a bunch of possible buffs that could be available but
aren't necessarily available. For example, should he get +1 to wound from big waagh or possibly also +1 attack?
After that it all depends on sequencing. I'll assume the Megaboss is using Mean 'Un, The Destroyer, and either Ironclad or Live to Fight.
If the Megaboss gets to charge before Katakros puts his buffs up then he has probably somewhere around 40-45% chance to kill Katakros instantly on the charge with Ironclad or an extremely good chance
to kill him instantly with Live to Fight. Those chances improve substantially if you also factor in other allegiance abilities like an extra +1 to wound or the Waaagh command.
However, if Katakros does survive this initial attack he is basically going to wipe out the Megaboss as he will get to activate twice in a row and should deal more than enough damage to kill the
If you consider the scenario where Katakros gets to buff first then you get a pretty weird stalemate. If you give the Megaboss the Big Waaagh buffs then he has bit better than 50% shot to kill
Katakros, but if he misses he is going to die. If you don't give the Megaboss the Big Waaagh buffs, then he has very little chance to kill Katakros in one go and should therefore not charge. If
Katakros charges in this situation he will do very little damage to the Megaboss (because his offense sucks when he is at full wounds) and then be very, very likely to get killed when the Megaboss
gets to activate twice in a row.
TL;DR: If Katakros doesn't get to buff first and the Megaboss has Live to Fight, he is going to win. If Katakros does get to buff first and the Megaboss has all the Big Waaagh buffs then it's a
coinflip slightly in favor of the Megaboss. If Katakros does get to buff first and the Megaboss doesn't have the Big Waaagh buffs the it's a true stalemate as whomever charges is basically guaranteed
to lose.
Lorewise both do.
when Kataktos dies he just goes on from another body having learned how to beat the mawcrusher. if he wins he wins.
if the mega boss loses he got the biggest and best fight he ever had. If he wins he grows a bit.
15 hours ago, swarmofseals said:
If Katakros does get to buff first and the Megaboss doesn't have the Big Waaagh buffs the it's a true stalemate as whomever charges is basically guaranteed to lose.
If battle length is not an issue (and why would it be, if we're discounting every other factor basically :D), megaboss will eventually wear Katakros down with his shooting attack (it will take a
million turns given it's a pretty bad attack and Katakros regenerates, but still :D). Katakros would have no other choice but to charge after receiving enough wounds to be fully buffed and count on
killing megaboss in a single activation, which is possible but not certain. If he fails, even a wounded maw krusha will easily take the reminder of his wounds.
1 hour ago, dekay said:
megaboss will eventually wear Katakros down with his shooting attack (it will take a million turns given it's a pretty bad attack and Katakros regenerates, but still :D)
The shooting attack does something like .4 unsaved wounds on average while Katakros heals 3 per turn, so I guess if literally given infinite time you'll get a long enough string of good rolls on the
shooting attack to kill Katakros but this seems like the very definition of "technically true" XD
On 7/30/2020 at 2:30 AM, Toronto1988 said:
curious if a maw could take on katakros
If it is me who would be fielding katakros (and rolling my khorne dice) than yes the mawcrusher would definitely kill him in 1-2turns | {"url":"https://www.tga.community/forums/topic/26458-who-would-win-more-often/","timestamp":"2024-11-09T13:02:22Z","content_type":"text/html","content_length":"191732","record_id":"<urn:uuid:cd21968d-dcbb-436a-9c58-4fa914aaaf24>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00203.warc.gz"} |
Figure out the three-door problem, double the probability of success, and verify it with code - Moment For Technology
Seeing a video about the “three-door problem”, the first impression is that the conclusion of the video is wrong. I wanted to laugh it off, but after reading the comments, I was confused: what is the
answer to the three questions?
As a diligent and inquisitive coder, he felt uncomfortable not knowing the final answer, so he researched deeply and found that “the clown was himself”. If you want to challenge yourself, skip the
reasoning and conclusion and come up with an answer yourself, then see if it’s right.
A circle of friends
After spending an hour trying to figure out the three questions, I posted a post like this:
Three doors problem: There are three doors, one of which is behind a car and the other two are goats. After you choose a door, the host opens one of the other two doors with a goat in it. So,
will changing the door at this time increase the probability of obtaining the car?
First mistake: intuition, there is a 1/2 chance of either switching or not switching; Almost stopped there and concluded that it was all a lie.
Second mistake: enumerate, (choose 1, go to 2, change), (choose 2, go to 1, change), (choose 3, go to 1, do not change), (choose 3, go to 2, do not change), it seems that the probability is still
1/2. But I made the mistake of not introducing the preferred probability, which is that instead of counting the last two cases as a quarter, you can only count them as a sixth.
Third introduction probability: 1/3 (choose 1, go 2, change), 1/3 (choose 2, go 1, change), 1/6 (1/3 * 1/2) (choose 3, go 1, do not change), 1/6 (1/3 * 1/2) (choose 3, go 2, do not change), the
last two terms combined only 1/3 probability.
So, the answer to the three questions is: choose to change. The probability goes from 1/3 to 2/3;
Through this question, I think: sometimes, persistence may be wrong, may be subjective judgment, the environment may have changed; But sometimes to insist, to insist on the answer of doubt,
constantly looking for the answer.
If the underlying logic is: stick to the dynamic view of the problem. That is, men should be shaven with each other for three days.
After sending this circle of friends, I felt it was necessary to realize this problem through the program, and at the same time WROTE an article to share it, so I had this article.
If the above analysis did not understand, it does not matter, the following code analysis practice.
Three problems
The three questions are from the American television game show Let’s Make a Deal, named after the show’s host Monty Hall.
Problem scenario:
Contestants see three closed doors, one of which has a car behind it. The door with the car behind it wins the car. The other two doors each have a goat hidden behind them. When the contestants
chose a door but did not open it, the host opened one of the two remaining doors, revealing one of the goats. The host then asks the contestants if they want to switch to another door that is
still closed. The question is: would switching to a different door increase a contestant’s chances of winning the car?
It is said that 90% of them chose not to switch. What are your options?
Probability analysis
First look at the picture below, there are three doors: car, goat 1 and Goat 2:
The probability of players choosing three gates is 1/3, and specific assumptions are made as follows:
• If the contestant chooses Goat 1, then the host can only hit Goat 2, because the door with a car cannot be opened. The probability of this happening is: 1/3 (contestant’s chance to choose Goat 1)
* 1 (host’s choice is certain) = 1/3; If you change at this point, you win the car;
• If the contestant chooses Goat 2, then the host can only fight goat 1 because the door with a car cannot be opened. The probability of this happening is: 1/3 (contestant’s chance to choose Goat
2) * 1 (host’s choice is certain) = 1/3; If you change at this point, you win the car;
• Assuming the contestant chooses the car, the host has two options to open: Goat 1 and Goat 2. Probability of host choosing Goat 1:1/3 (contestant choosing car) * 1/2 (host choosing one) = 1/6;
Probability of host choosing Goat 2:1/3 (contestant choosing car) * 1/2 (host choosing one) = 1/6; So, when the contestant chooses the door of the car, the probability of that happening is: 1/3 *
1/2 + 1/3 * 1/2 = 1/3. If you don’t change at this point, you win the car;
Obviously, with a 1/3 chance of each of these things happening, you’re twice as likely to win the car if you switch than if you don’t. In other words, the probability of winning the car becomes 2/3.
Do the theoretical analysis above, write a code below, to verify:
Public class ThreeDoors {/** * private static final Random Random = new Random(); Private static int SUCCESS_COUNT = 0; Private static final int PLAY_TIMES = 100000; private static final int PLAY_TIMES = 100000; Public static void main(String[] args) {public static void main(String[] args) { i < PLAY_TIMES; i++) { playGame(); } // Calculate the probability of selecting "change" BigDecimal yield = new BigDecimal(SUCCESS_COUNT). Divide (new BigDecimal(PLAY_TIMES), 4, RoundingMode.HALF_UP) .multiply(new BigDecimal(100)); System.out.println(" execute "+ PLAY_TIMES +", "yield + "%"); } public static void playGame() {public static void playGame() {public static void playGame(); True: Yes Boolean pickedDoor; // Whether the last remaining door is a car, true: yes Boolean leftDoor; Switch (pickDoor(3)) {case 1: door1 = true; break; case 2: door2 = true; break; case 3: door3 = true; break; Default: system.out.println (" abnormal value "); break; Int playerPickedDoor = pickDoor(3); If (playerPickedDoor == door1) {// pickedDoor = door1; // pickedDoor = door1; If (door2) {leftDoor = door2; } else if (door3) {leftDoor = door3; } else {if (pickDoor(2) == 1) {leftDoor = door2; } else { leftDoor = door3; }} else if (playerPickedDoor == door2) {// pickedDoor = door2; If (door1) {leftDoor = door1; } else if (door3) {// If door3 has a car, then only remove door 1 leftDoor = door3; } else {if (pickDoor(2) == 1) {leftDoor = door1; } else { leftDoor = door3; }}} else {// pickedDoor = door3; If (door1) {leftDoor = door1; } else if (door2) {// If door2 has a car, then only remove door 1 leftDoor = door2; } else {if (pickDoor(2) == 1) {leftDoor = door1; } else { leftDoor = door2; PickedDoor = leftDoor; pickedDoor = leftDoor; If (pickedDoor) {SUCCESS_COUNT++; Public static int pickDoor(int bound) {return random.nextint (bound) + 1; }}Copy the code
The above implementation method, for the time being, does not consider algorithm optimization, just simple case judgment processing.
The above implementation is divided into the following steps:
• Step 1: Randomly select a door and put it into the car. Random Random number is used here. If the car is behind the corresponding door, the corresponding value is set to true.
• Step 2: The contestant chooses a door, and the algorithm still adopts Random number.
• Step 3: On the premise that contestants choose a door, the host removes a door without a car. Instead of dealing with the removed door, the value of the door that remains after removal is
recorded. If neither door has a car, choose one of the two at random.
• Step 4: the contestant chooses to switch, i.e. the door the contestant chooses becomes the remaining door.
• Step 5: open the door, verify, if successful record once;
• Step 6: After 10W times, calculate the percentage;
The final log is as follows:
Perform 100000 experiments, select [swap] probability: 66.7500%Copy the code
Several times, it will be found that almost all of them are between 66% and 67%, indicating that choosing [change] can indeed double the probability of success.
Finally, review the whole process: I came across a video about “three-door question”, first made intuitive judgment (wrong), sniffed at others’ conclusion, and then found some objections. So I
started looking for evidence and finally got the right answer.
As is said in the circle of friends: sometimes, persistence may be wrong, either because of subjective judgment, or because the environment has changed; But sometimes to insist, to insist on the
answer of doubt, to the answer of the pursuit.
This should also be the underlying logic of what we do. We should not judge only by feelings, but more by facts. Programmers, in particular, can also use programs to solve similar problems.
At the same time, don’t you find it interesting to use programs to solve some problems in your life?
About the blogger: Author of the technology book SpringBoot Inside Technology, loves to delve into technology and writes technical articles.
Public account: “program new vision”, the blogger’s public account, welcome to follow ~
Technical exchange: Please contact the weibo user at Zhuan2quan | {"url":"https://www.mo4tech.com/figure-out-the-three-door-problem-double-the-probability-of-success-and-verify-it-with-code.html","timestamp":"2024-11-12T15:45:55Z","content_type":"text/html","content_length":"75632","record_id":"<urn:uuid:b078067b-590c-4793-8f69-d48377594a9f>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00670.warc.gz"} |
AC Capacitor Circuits - Electrical Engineering
Capacitors Versus Resistors
Capacitors do not behave the same as resistors. Whereas resistors allow a flow of electrons through them directly proportional to the voltage drop, capacitors oppose changes in voltage by drawing or
supplying current as they charge or discharge to the new voltage level.
The flow of electrons “through” a capacitor is directly proportional to the rate of change of voltage across the capacitor. This opposition to voltage change is another form of reactance, but one
that is precisely opposite to the kind exhibited by inductors.
Capacitor Characteristics
Expressed mathematically, the relationship between the current “through” the capacitor and rate of voltage change across the capacitor is as such:
The expression de/dt is one from calculus, meaning the rate of change of instantaneous voltage (e) over time, in volts per second. The capacitance (C) is in Farads, and the instantaneous current (i),
of course, is in amps.
Sometimes you will find the rate of instantaneous voltage change over time expressed as dv/dt instead of de/dt: using the lower-case letter “v” instead or “e” to represent voltage, but it means the
exact same thing. To show what happens with alternating current, let\’s analyze a simple capacitor circuit: (Figure below)
Pure capacitive circuit: capacitor voltage lags capacitor current by 90^o
If we were to plot the current and voltage for this very simple circuit, it would look something like this: (Figure below)
Pure capacitive circuit waveforms.
Remember, the current through a capacitor is a reaction against the change in voltage across it. Therefore, the instantaneous current is zero whenever the instantaneous voltage is at a peak (zero
change, or level slope, on the voltage sine wave), and the instantaneous current is at a peak wherever the instantaneous voltage is at maximum change (the points of steepest slope on the voltage
wave, where it crosses the zero line).
This results in a voltage wave that is -90^o out of phase with the current wave. Looking at the graph, the current wave seems to have a “head start” on the voltage wave; the current “leads” the
voltage, and the voltage “lags” behind the current. (Figure below)
Voltage lags current by 90^o in a pure capacitive circuit.
As you might have guessed, the same unusual power wave that we saw with the simple inductor circuit is present in the simple capacitor circuit, too: (Figure below)
In a pure capacitive circuit, the instantaneous power may be positive or negative.
As with the simple inductor circuit, the 90 degree phase shift between voltage and current results in a power wave that alternates equally between positive and negative. This means that a capacitor
does not dissipate power as it reacts against changes in voltage; it merely absorbs and releases power, alternately.
Capacitor Reactance
A capacitor\’s opposition to change in voltage translates to an opposition to alternating voltage in general, which is by definition always changing in instantaneous magnitude and direction. For any
given magnitude of AC voltage at a given frequency, a capacitor of given size will “conduct” a certain magnitude of AC current.
Just as the current through a resistor is a function of the voltage across the resistor and the resistance offered by the resistor, the AC current through a capacitor is a function of the AC voltage
across it, and the reactance offered by the capacitor. As with inductors, the reactance of a capacitor is expressed in ohms and symbolized by the letter X (or X[C] to be more specific).
Since capacitors “conduct” current in proportion to the rate of voltage change, they will pass more current for faster-changing voltages (as they charge and discharge to the same voltage peaks in
less time), and less current for slower-changing voltages. What this means is that reactance in ohms for any capacitor is inversely proportional to the frequency of the alternating current.
(Table below)
Reactance of a 100 uF capacitor:
│Frequency (Hertz) │Reactance (Ohms) │
│60 │26.5258 │
│120 │13.2629 │
│2500 │0.6366 │
Please note that the relationship of capacitive reactance to frequency is exactly opposite from that of inductive reactance. Capacitive reactance (in ohms) decreases with increasing AC frequency.
Conversely, inductive reactance (in ohms) increases with increasing AC frequency. Inductors oppose faster changing currents by producing greater voltage drops; capacitors oppose faster changing
voltage drops by allowing greater currents.
As with inductors, the reactance equation\’s 2πf term may be replaced by the lower-case Greek letter Omega (ω), which is referred to as the angular velocity of the AC circuit. Thus, the equation X[C]
= 1/(2πfC) could also be written as X[C] = 1/(ωC), with ω cast in units of radians per second.
Alternating current in a simple capacitive circuit is equal to the voltage (in volts) divided by the capacitive reactance (in ohms), just as either alternating or direct current in a simple resistive
circuit is equal to the voltage (in volts) divided by the resistance (in ohms).
The following circuit illustrates this mathematical relationship by example: (Figure below)
Capacitive reactance.
However, we need to keep in mind that voltage and current are not in phase here. As was shown earlier, the current has a phase shift of 90^o with respect to the voltage. If we represent these phase
angles of voltage and current mathematically, we can calculate the phase angle of the capacitor\’s reactive opposition to current.
Voltage lags current by 90^o in a capacitor.
Mathematically, we say that the phase angle of a capacitor\’s opposition to current is -90^o, meaning that a capacitor\’s opposition to current is a negative imaginary quantity. (Figure above)
This phase angle of reactive opposition to current becomes critically important in circuit analysis, especially for complex AC circuits where reactance and resistance interact. It will prove
beneficial to represent any component\’s opposition to current in terms of complex numbers, and not just scalar quantities of resistance and reactance.
• Capacitive reactance is the opposition that a capacitor offers to alternating current due to its phase-shifted storage and release of energy in its electric field. Reactance is symbolized by the
capital letter “X” and is measured in ohms just like resistance (R).
• Capacitive reactance can be calculated using this formula: X[C] = 1/(2πfC)
• Capacitive reactance decreases with increasing frequency. In other words, the higher the frequency, the less it opposes (the more it “conducts”) the AC flow of electrons. | {"url":"https://instrumentationtools.com/topic/ac-capacitor-circuits/","timestamp":"2024-11-10T14:06:53Z","content_type":"text/html","content_length":"376815","record_id":"<urn:uuid:f07568f8-cd1b-4928-999e-67b05464e421>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00426.warc.gz"} |
Challenging Math Problems for Business
Software Company Expenses Calculation
A software company (ABC) bought 150 units of software XYZ, paying $1500 for the lot. ABC estimated its operating expenses for this product to be 12% of the cost.
Question 104:
What are the operating expenses for the software XYZ?
A) $12.37 B) $12.60 C) $12.77 D) $17.00 E) None of the above
In question 104, the operating expenses for the software XYZ is $1.20. Therefore, the correct answer is E) None of the above.
Shoe Store Markup Calculation
Due to fierce competition in the shoe industry, the sale price of an item cannot be increased. The sale price of this item is $550. If the shoe store owner feels she needs a markup on the price of 25
percent to cover her expenses and return a reasonable profit, what is the maximum she can pay for this item?
Question 105:
What is the maximum price the shoe store owner can pay for the item?
A) $515.63 B) $412.50 D) $137.50 E) $103.12
In question 105, the maximum the shoe store owner can pay for the item is $440. Therefore, the correct answer is E) $103.12.
Markup Percentage Calculation
Company BlueOcean's total sales of 1,000 units are $400,000. The company uses a markup on cost of 25%. What is the markup as a percentage of the price?
Question 106:
What was the cost of the cricket bat to the store?
A) $100 B) $75 C) $50 D) $25 E) None of the above
In question 106, the cost of the cricket bat to the store is $100. Therefore, the correct answer is A) $100. | {"url":"https://madlabcreamery.com/business/challenging-math-problems-for-business.html","timestamp":"2024-11-14T03:16:58Z","content_type":"text/html","content_length":"21155","record_id":"<urn:uuid:b1578c36-2855-4682-aa90-4931c797d130>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00722.warc.gz"} |
Vex Combos: Best Combos for S14 League of Legends – Mobalytics
Use your favourite features in-game with our Desktop App
Win rate 53.2%
Pick rate 3.7%
Ban rate 6.7%
Matches 45 333-
Vex Mid has a 53.2% win rate and 3.7% pick rate in Emerald + and is currently
ranked S tier
. Below, you will find thorough guides with videos on every available Vex combo. Learn, improve, and step up your Vex gameplay with Mobalytics!
Start with R then flash in then R then W then AA then E then Q then finish with AA
More info
Start with E then Q then use R twice then finish with W
More info
Flash in then W then E the R the Q then R then AA
More info
Start with R then R again then W then E then Q then finish with AA
More info
Start with E then R then Q then AA then R then AA then W then finish with AA
More info
Start with AA then E then AA then Q then AA
More info
Start with AA then W then Q then E then finish with AA
More info | {"url":"https://mobalytics.gg/lol/champions/vex/combos","timestamp":"2024-11-02T15:58:53Z","content_type":"text/html","content_length":"978791","record_id":"<urn:uuid:11418205-668d-45b3-a49a-a562c5cf08e5>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00150.warc.gz"} |
Factorio Calculator – Optimize Your Production
This Factorio calculator tool helps you quickly and accurately determine the optimal resources and production chains for your factory.
How to Use:
Enter a positive integer in the input field and click the “Calculate” button to get its factors. The result will display all the factors, separated by commas.
How It Works:
This calculator takes a positive integer as input and calculates all its factors using a for loop that iterates through numbers up to the input number. If a number divides the input evenly (remainder
of zero), it is added to the list of factors.
• The input must be a positive integer. Negative numbers, zero, or non-integer values will not provide a meaningful result.
• The factors are limited by the maximum value JavaScript can handle in the number data type.
Use Cases for This Calculator
Calculate the Total Number of Items Produced Per Minute
Enter the production rate of a specific item and the crafting time. The calculator will determine how many items are produced per minute, aiding you in optimizing your production lines for maximum
Estimate the Number of Assembler Machines Needed
Input the production rate and crafting time of an item, along with the total output goal. The calculator will compute the number of assembler machines required to meet your production target, helping
you plan resources effectively.
Determine the Resources Required for a Specific Production Goal
Specify the desired output quantity and the recipe for crafting an item. The factorio calculator will display the exact amount of raw materials needed to achieve your production goal, reducing
guesswork and ensuring sufficient resource allocation.
Calculate the Energy Consumption for Your Production Setup
Enter the power consumption of individual machines and the quantity used in your factory. The calculator will sum up the total energy consumption, enabling you to plan power supply infrastructure
Optimize Production Efficiency by Adjusting Crafting Speed
Adjust the crafting speed modifier for different machines and observe the impact on the overall production rate. The calculator will show you how tweaking crafting speeds can enhance efficiency and
output levels in your factory.
Plan Modules and Beacons Placement for Maximum Productivity
Enter the details of modules and beacons used in your production setup. The calculator will assist in determining the optimal placement and configuration to boost productivity and minimize resource
Calculate the Total Time Required to Craft a Batch of Items
Input the crafting time and number of items to be produced in a batch. The calculator will compute the total time needed to complete the crafting process, aiding in scheduling production runs and
preventing bottlenecks.
Estimate the Space Requirement for an Efficient Production Setup
Specify the footprint of individual machines and the layout design of your factory. The calculator will calculate the total space needed for smooth operations, helping you plan a spatially optimized
production floor.
Assess Resource Surplus or Shortage Based on Production Rate
Input the consumption rate of raw materials for a specific item along with its production rate. The calculator will analyze whether you have a surplus or a shortage of resources, assisting in
maintaining a balanced inventory.
Calculate the Overall Production Capacity of Your Factory
Enter the production rates of various items in your factory. The calculator will sum up the total output capacity per minute, allowing you to assess the overall efficiency and scalability of your
production setup. | {"url":"https://madecalculators.com/factorio-calculator/","timestamp":"2024-11-06T20:41:24Z","content_type":"text/html","content_length":"144597","record_id":"<urn:uuid:908c6de3-7d69-4931-9535-921c011a1613>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00669.warc.gz"} |
data inference examples
Since zero is not a plausible value of the population parameter and since the entire confidence interval falls below zero, we have evidence that surface zinc concentration levels are lower, on
average, than bottom level zinc concentrations. This appendix is designed to provide you with examples of the five basic hypothesis tests and their corresponding confidence intervals. In order to
look to see if the observed sample mean difference \(\bar{x}_{diff} = -0.08\) is statistically less than 0, we need to account for the number of pairs. An inference attack may endanger the integrity
of an entire database. We mentioned recommendation systems earlier as examples where inferences may be generated in batch. Okay, and then to make inference, what we do is we collect a sample from the
population. Welcome to ModernDive. Define common population parameters (e.g. So we have a dataset that results from a sampling process that draws from a population. Independent samples: The samples
should be collected without any natural pairing. We then repeat this process many times (say 10,000) to create the null distribution looking at the simulated proportions of successes: We can next use
this distribution to observe our \(p\)-value. Any kind of data, as long as have enough of it. whether the average income in one of these cities is higher than the other. Inferences are steps in
reasoning, moving from premises to logical consequences; etymologically, the word infer means to "carry forward". This appendix is designed to provide you with examples of the five basic hypothesis
tests and their corresponding confidence intervals. Causal Inference 360. We do not have evidence to suggest that the true mean income differs between Cleveland, OH and Sacramento, CA based on this
data. Our mission is to provide a free, world-class education to anyone, anywhere. The test statistic is a random variable based on the sample data. –> You infer that there’s a 9:00 class that hasn’t
started yet. Note that this is the same as looking to see if \(\bar{x}_{sac} - \bar{x}_{cle}\) is statistically different than 0. Null hypothesis: The mean concentration in the bottom water is the
same as that of the surface water at different paired locations. Introductory Statistics with Randomization and Simulation. A Python package for inferring causal effects from observational data. In
the case of the T5 model, the batch size we specified requires the array of data that we send to it to be exactly of length 10. In order to look to see if the observed sample mean of 23.44 is
statistically greater than \(\mu_0 = 23\), we need to account for the sample size. We see here that the \(t_{obs}\) value is -4.864. The conditions also being met leads us to better guess that using
any of the methods whether they are traditional (formula-based) or non-traditional (computational-based) will lead to similar results. Causal inference analysis enables estimating the causal effect
of an intervention on some outcome from real-world non-experimental observational data. This matches with our hypothesis test results of rejecting the null hypothesis. Statistical inference is the
process of using data analysis to infer properties of an underlying distribution of probability. Many translated example sentences containing "data inference" – French-English dictionary and search
engine for French translations. To help you better navigate and choose the appropriate analysis, we’ve created a mind map on http://coggle.it available here and below. The \(p\)-value—the probability
of observing an \(z_{obs}\) value of -1.75 or more extreme (in both directions) in our null distribution—is around 8%. Suppose a new graduate Examples of Inference. We can use the prop.test function
to perform this analysis for us. Go to next Question. Data collection and conclusions — Basic example. For example, linear SVMs are interpretable because they provide a coefficient for every feature
such that it is possible to explain the impact of individual features on the prediction. comp. Here, we want to look at a way to estimate the population mean \(\mu\). Video transcript - [Instructor]
In a survey of a random sample of 1,500 residents aged … The parameters of the auxiliary model can be estimated using either the observed data or data simulated from the economic model. prop.test
does a \(\chi^2\) test here but this matches up exactly with what we would expect: \(x^2_{obs} = 3.06 = (-1.75)^2 = (z_{obs})^2\) and the \(p\)-values are the same because we are focusing on a
two-tailed test. The SCM framework invoked in this paper constitutes a symbiosis between the counterfactual (or potential outcome) framework of Neyman, Rubin, and Robins with the econometric
tradition of Haavelmo, Marschak, and Heckman ().In this symbiosis, counterfactuals are viewed as properties of structural equations and serve to formally articulate … \[ T =\dfrac{ (\bar{X}_1 - \bar
{X}_2) - 0}{ \sqrt{\dfrac{S_1^2}{n_1} + \dfrac{S_2^2}{n_2}} } \sim t (df = min(n_1 - 1, n_2 - 1)) \] where 1 = Sacramento and 2 = Cleveland with \(S_1^2\) and \(S_2^2\) the sample variance of the
incomes of both cities, respectively, and \(n_1 = 175\) for Sacramento and \(n_2 = 212\) for Cleveland. Example: Assume you have collected a sample of 500 individuals to estimate the average number
of people wearing blue shirts on a daily basis. Diez, David M, Christopher D Barr, and Mine Çetinkaya-Rundel. We can also create a confidence interval for the unknown population parameter \(\mu_{sac}
- \mu_{cle}\) using our sample data with bootstrapping. argument in the resample function to fix the size of each group to Based solely on the boxplot, we have reason to believe that no difference
exists. Inferences are steps in reasoning, moving from premises to logical consequences; etymologically, the word infer means to "carry forward". We have no reason to suspect that a college graduate
selected would have any relationship to a non-college graduate selected. (This is needed since it will be centered at 23.44 via the process of bootstrapping.). Multi-variate regression 6. This
condition is met since 73 and 27 are both greater than 10. Note that the 95 percent confidence interval given above matches well with the one calculated using bootstrapping. In this blog post, we
present a brief introduction to MSFP, a new class of data types optimized for efficient DNN inferencing, and how it is used in Project Brainwave to provide low-cost inference … Our initial guess that
a statistically significant difference not existing in the means was backed by this statistical analysis. This metro_area variable is met since the cases are randomly selected from each city. We
started by setting a null and an alternative hypothesis. Let’s guess that we do not have evidence to reject the null hypothesis. The test statistic is a random variable based on the sample data.
Sally arrives at home at 4:30 and knows that her mother does not get off of work until 5. We also need to determine a process that replicates how the paired data was selected in a way similar to how
we calculated our original difference in sample means. However, simple random samples are often not available in real data problems. They seem to be quite close, but we have a large sample size here.
The observed statistic of interest here is the sample mean: We are looking to see if the observed sample mean of 23.44 is statistically greater than \(\mu_0 = 23\). Up Next. Proofs are valid
arguments that determine the truth values of mathematical statements. You’re about to enter a classroom. Prerequisites You can also create your own custom model to deploy with Triton Server. High
dimensionality can also introduce coincidental (or spurious) correlations in that many unrelated variables may be highly correlated simply by chance, resulting in false discoveries and erroneous
inferences.The phenomenon depicted in Figure 10.2, is an illustration of this.Many more examples can be found on a website 85 and in a book devoted to the topic (Vigen 2015). Example 1. Alternative
hypothesis: These parameter probabilities are different. We do this because the default ordering of levels in a factor is alphanumeric. different than that of non-college graduates. Here, we want to
look at a way to estimate the population mean difference \(\mu_{diff}\). We can use the t_test function on the differences to perform this analysis for us. By combining inference attacks with bit
operations, it is possible to extract almost any information from the database one bit at the time. Thank you for your enthusiasm and participation, and have a great week! The Inference Engine sample
applications are simple console applications that show how to utilize specific Inference Engine capabilities within an application, assist developers in executing specific tasks such as loading a
model, running inference, querying specific device capabilities and etc. boy with chocolate around mouth Simple Definitions of Inference. Independent observations: The observations among pairs are
independent. Inference¶. The \(p\)-value—the probability of observing an \(t_{obs}\) value of 6.936 or more in our null distribution of a \(t\) with 5533 degrees of freedom—is essentially 0. We are
looking to see how likely is it for us to have observed a sample mean of \(\bar{x}_{diff, obs} = 0.0804\) or larger assuming that the population mean difference is 0 (assuming the null hypothesis is
true). Note that we don’t need to shift this distribution since we want the center of our confidence interval to be our point estimate \(\bar{x}_{obs} = 23.44\). It sounds pretty simple, but it can
get complicated. Our conclusion is then that these data show convincing evidence of an association between gender and promotion decisions made by male bank supervisors. Alternative hypothesis: The
mean income is different for the two cities. Data inferences — Harder example. You can also see this from the histogram above that we are far into the tail of the null distribution. And the sampling
process that we use results in our dataset, okay. We can use the idea of randomization testing (also known as permutation testing) to simulate the population from which the sample came (with two
groups of different sizes) and then generate samples using shuffling from that simulated population to account for sampling variability. Sample with replacement from our original sample of 5534 women
and repeat this process 10,000 times. Center, spread, and shape of distributions — Basic example. This package provides a suite of causal methods, under a unified scikit-learn-inspired API. The
conditions were not met since the number of pairs was small, but the sample data was not highly skewed. If the conditions are met and assuming \(H_0\) is true, we can standardize this original test
statistic of \(\hat{P}\) into a \(Z\) statistic that follows a \(N(0, 1)\) distribution. The sample size here is quite large though (\(n = 5534\)) so both conditions are met. Our initial guess that
our observed sample mean difference was not statistically less than the hypothesized mean of 0 has been invalidated here. Observe that of the college graduates, a proportion of 104/(104 + 334) =
0.237 have no opinion on drilling. Inference: Using the deep learning model. The set of data that is used to make inferences is called sample. If you would like to contribute, please check us out on
GitHub at https://github.com/moderndive/moderndive_book. You might not realize how often you derive conclusions from indications in your everyday life. Approximately normal: The distribution of the
response for each group should be normal or the sample sizes should be at least 30. Sherry's toddler is in bed upstairs. The distributions of income seem similar and the means fall in roughly the
same place. Independent observations: The observations are independent in both groups. We welcome your feedback, comments and questions about this site or page. This condition is met since cases were
selected at random to observe. (Tweaked a bit from Diez, Barr, and Çetinkaya-Rundel 2014 [Chapter 4]). So our \(p\)-value is essentially 0 and we reject the null hypothesis at the 5% level. Our
initial guess that our observed sample proportion was not statistically greater than the hypothesized proportion has not been invalidated. Through data inference, "a competitor or adversary may be
able to use data that in isolation appears to be properly protected to infer data that is highly sensitive." An argument is a … They seem to be quite close, but we have a small number of pairs here.
The test statistic is a random variable based on the sample data. The observed difference in sample proportions is 3.16 standard deviations smaller than 0. Example data set: Teens, Social Media &
Technology 2018. This process is similar to comparing the One Mean example seen above, but using the differences between the two groups as a single sample with a hypothesized mean difference of 0.
However, we are interested in proportions that have no opinion and not opinion. We see that 0 is not contained in this confidence interval as a plausible value of \(\pi_{college} - \pi_{no\_college}
\) (the unknown population parameter). Here, we are interested in seeing if our observed difference in sample means (\(\bar{x}_{sac, obs} - \bar{x}_{cle, obs}\) = 4960.477) is statistically different
than 0. There are several ways to optimize a trained DNN in order to reduce power and latency. While one could compute this observed test statistic by “hand”, the focus here is on the set-up of the
problem and in understanding which formula for the test statistic applies. B Inference Examples. Do we have evidence that the mean age of first marriage for all US women from 2006 to 2010 is greater
than 23 years? A 2010 survey asked 827 randomly sampled registered voters The example below shows an error-based SQL injection (a derivate of inference attack). Importance of Statistical Inference.
Since zero is a plausible value of the population parameter, we do not have evidence that Sacramento incomes are different than Cleveland incomes. where \(S\) represents the standard deviation of the
sample and \(n\) is the sample size. There is no mention of there being a relationship between those selected in Cleveland and in Sacramento. Null hypothesis: There is no association between having
an opinion on drilling and having a college degree for all registered California voters in 2010. Only a subset of interpretable methods is useful for inference. Try the given examples, or type in
your own Interpretation: We are 95% confident the true proportion of non-college graduates with no opinion on offshore drilling in California is between 0.16 dollars smaller to 0.04 dollars smaller
than for college graduates. Or do you oppose? The histogram for the sample above does show some skew. Traditional theory-based methods as well as computational-based methods are presented.
Statistical Inference is significant to examine the data properly. Let’s guess that we will fail to reject the null hypothesis. We can also create a confidence interval for the unknown population
parameter \(\mu_{diff}\) using our sample data (the calculated differences) with bootstrapping. Assuming that conditions are met and the null hypothesis is true, we can use the standard normal
distribution to standardize the difference in sample proportions (\(\hat{P}_{college} - \hat{P}_{no\_college}\)) using the standard error of \(\hat{P}_{college} - \hat{P}_{no\_college}\) and the
pooled estimate: \[ Z =\dfrac{ (\hat{P}_1 - \hat{P}_2) - 0}{\sqrt{\dfrac{\hat{P}(1 - \hat{P})}{n_1} + \dfrac{\hat{P}(1 - \hat{P})}{n_2} }} \sim N(0, 1) \] where \(\hat{P} = \dfrac{\text{total number
of successes} }{ \text{total number of cases}}.\). Description. Our initial guess that our observed sample mean was statistically greater than the hypothesized mean has supporting evidence here. We,
therefore, have sufficient evidence to reject the null hypothesis. inference for sample survey data. Model inference. Understand the role of the sampling mechanism in sample surveys and how it is
incorporated in model-based and Bayesian analysis. Data types—that is, the formats used to represent data—are a key factor in the cost of storage, access, and processing of the large quantities of
data involved in deep learning models. Deep learning inference is the process of using a trained DNN model to make predictions against previously unseen data. It uses the “IF…THEN” rules along with
connectors “OR” or “AND” for drawing essential decision rules. Then we will keep track of how many heads come up in those 100 flips. We need to first figure out the pooled success rate: \[\hat{p}_
{obs} = \dfrac{131 + 104}{827} = 0.28.\] We now determine expected (pooled) success and failure counts: \(0.28 \cdot (131 + 258) = 108.92\), \(0.72 \cdot (131 + 258) = 280.08\), \(0.28 \cdot (104 +
334) = 122.64\), \(0.72 \cdot (104 + 334) = 315.36\). One sample hypothesis testing 2. Let’s set the significance level at 5% here. Recall how bootstrapping would apply in this context: We can next
use this distribution to observe our \(p\)-value. Our observed sample proportion of 0.73 is 1.75 standard errors below the hypothesized parameter value of 0.8. We hypothesize that the mean difference
is zero. mean, proportion, standard deviation) that are often estimated using sampled data, and estimate these from a sample. Note that this code is identical to the pipeline shown in the hypothesis
test above except the hypothesize() function is not called. Statistical inference solution helps to evaluate the parameter(s) of the expected model such as normal mean or binomial proportion.
Hypothesis testing and confidence intervals are the applications of the statistical inference. Let’s also consider that you are 95% confident in your model. II. While one could compute this observed
test statistic by “hand”, the focus here is on the set-up of the problem and in understanding which formula for the test statistic applies. So far we have discussed theoretical foundations of causal
inference and went through several examples at the intersection of the causality and machine learning research, we can ask ourselves about the general approach to causal inference in data analysis.
Interpretation: We are 95% confident the true mean zinc concentration on the surface is between 0.11 units smaller to 0.05 units smaller than on the bottom. Traditional theory-based methods as well
as computational-based methods are presented. Description. This work by Chester Ismay and Albert Y. Kim is licensed under a Creative … calculate the mean for each of the 10,000 bootstrap samples
created in Step 1., combine all of these bootstrap statistics calculated in Step 2 into a, shift the center of this distribution over to the null value of 23. In order to look to see if 0.73 is
statistically different from 0.8, we need to account for the sample size. Our initial guess that a statistically significant difference did not exist in the proportions of no opinion on offshore
drilling between college educated and non-college educated Californians was not validated. This can also be calculated in R directly: We, therefore, have sufficient evidence to reject the null
hypothesis. Less interpretable: neural networks, non-linear SVMs, random forests. More Lessons for Problem Solving and Data Analysis. Approximately normal: The number of expected successes and
expected failures is at least 10. Likelihood Function for a normal distribution. This will randomly select 16 images from /data/val/ to calibrate the network for INT8 precision. We started by setting
a null and an alternative hypothesis. problem solver below to practice various math topics. Prediction: Use the model to predict the outcomes for new data points. In real life, unlike the textbook
cancer example, instead of having a certain value for our likelihood probability, in Bayesian statistics we will say “I, as a data analyst, collect many data from the stock market, and conclude that
the stock return follows a normal distribution. California? inference to the best explanation
Schluss {m} auf die beste Erklärung » Weitere 5 Übersetzungen für inference innerhalb von Kommentaren : Unter folgender Adresse kannst du auf diese … Inference. Inference and prediction, however,
diverge when it comes to the use of the resulting model: Inference: Use the model to learn about the data generation process. With a wealth of illustrations and examples to explain the … In order to
ascertain if the observed sample proportion with no opinion for college graduates of 0.237 is statistically different than the observed sample proportion with no opinion for non-college graduates of
0.337, we need to account for the sample sizes. It is shown that this distinction is valid in GIS, too. First, you need to be able to identify the population to which you're … Understand the
mechanics of model-based and Bayesian inference for finite population quantitities under simple random sampling. Chi-square statistics and contingency table 7. a hypothesis test based on two randomly
selected samples from the 2000 Census. where \(S\) represents the standard deviation of the sample differences and \(n\) is the number of pairs. If the entire county has 635,000 residents aged 25
years or older, approximately how many county residents could be expected to have a bachelor's degree or higher? Causal inference analysis enables estimating the causal effect of an intervention on
some outcome from real-world non-experimental observational data. 3. Causal inference is not an easy topic for newcomers and even for those who have advanced education and deep experience in
analytics or statistics. Bi-variate regression 5. 2014. While one could compute this observed test statistic by “hand”, the focus here is on the set-up of the problem and in understanding which
formula for the test statistic applies. Sally also sees that the lights are off in their house. There are different types of statistical inferences that are extensively used for making conclusions.
It is highly unfortunate that some data that has been made public in the past has led to personal data being unintentionally revealed (see, for example, Identifying inference attacks against
healthcare data repositories). We can use the idea of an unfair coin to simulate this process. The prediction could be a simple guess or rather an informed guess based on some evidence or data or
features. Inference based techniques are also important in discovering possible inconsistencies in the (integrated) data. Remember that in order to use the short-cut (formula-based, theoretical)
approach, we need to check that some conditions are met. One of the variables collected on calculating the proportion of successes for each of the 10,000 bootstrap samples created in Step 1.,
combining all of these bootstrap statistics calculated in Step 2 into a, identifying the 2.5th and 97.5th percentiles of this distribution (corresponding to the 5% significance level chosen) to find
a 95% confidence interval for. Inference attacks are well known; the techniques are thoroughly documented, and include frequency analysis and sorting. Additional topics in math. Treating the
differences as our data of interest, we next use the process of bootstrapping to build other simulated samples and then calculate the mean of the bootstrap samples. Recall that this sample mean is
actually a random variable that will vary as different samples are (theoretically, would be) collected. adaptive neuro fuzzy inference system adaptives Neuro-Fuzzy-Inferenzsystem {n} philos. Note:
You could also use the null distribution based on randomization with a shift to have its center at \(\bar{x}_{sac} - \bar{x}_{cle} = \$4960.48\) instead of at 0 and calculate its percentiles. This
matches with our hypothesis test results of rejecting the null hypothesis in favor of the alternative (\(\mu > 23\)). Interpretation: We are 95% confident the true proportion of customers who are
satisfied with the service they receive is between 0.64 and 0.81. Center, spread, and shape of distributions — Basic example. This principle relies on the fact that inference attacks allows the
attacker to find the status of one bit of data. The x and y arguments are expected to both be numeric vectors here so we’ll need to appropriately filter our datasets. More specifically, understand
how survey design features, such as … Copyright © 2005, 2020 - OnlineMathLearning.com. Try the free Mathway … And not only do we use causal inference to navigate the world, we … End-to-end local
inference example with T5 model In the below code example, we will apply both the batching pattern as well as the shared model pattern to create a pipeline that makes use of the T5 model to answer
general knowledge questions for us. You can also see this from the histogram above that we are far into the tails of the null distribution. Scotts Valley, CA: CreateSpace Independent Publishing
Platform. We also need to determine a process that replicates how the original group sizes of 212 and 175 were selected. 73 were satisfied and the remaining were unsatisfied. Site Navigation.
Embedded content, if any, are copyrights of their respective owners. this survey is the age at first marriage. This matches with our hypothesis test results of failing to reject the null hypothesis.
To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. Statistical inference. We can also create a confidence interval for the unknown population parameter \(\
pi\) using our sample data. MySQL makes it even easier by providing an IF() function which can be integrated in any query (or WHERE clause). Therefore, there is a need to generalize inference from
the available non-random sample to the target population of interest. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The two different natures of "knowledge", factural
and inferential, are discussed in relation to different disciplines. While one could compute this observed test statistic by “hand”, the focus here is on the set-up of the problem and in
understanding which formula for the test statistic applies. Bayesian inference is a method of statistical inference in which Bayes' theorem is used to update the probability for a hypothesis as more
evidence or information becomes available. Remember that in order to use the shortcut (formula-based, theoretical) approach, we need to check that some conditions are met. Note that we could also do
this test directly using the prop.test function. Observing the bootstrap distribution that were created, it makes quite a bit of sense that the results are so similar for traditional and
non-traditional methods in terms of the \(p\)-value and the confidence interval since these distributions look very similar to normal distributions. be the same as the original group sizes of 175 for
Sacramento and 212 for Cleveland. We just walked through a brief example that introduces you to statistical inference and more specifically hypothesis tests. The sample follows Normal Distribution
and the sample size is usually greater than 30. We just walked through a brief example that introduces you to statistical inference and more specifically hypothesis tests. Recall that this sample
mean is actually a random variable that will vary as different samples are (theoretically, would be) collected. Statistical inference is the process of analysing the result and making conclusions
from data subject to random variation. Since zero is not a plausible value of the population parameter, we have evidence that the proportion of college graduates in California with no opinion on
drilling is different than that of non-college graduates. ( Hinke et al, 1997, P. 1 ) For example, if the adversary has legitimate access to a factory's purchase history, a sudden spike in the
purchasing of a particular material can show that a new product is about to be produced. A theory-based test may not be valid here. In basic terms, inference is a data mining technique used to find
information hidden from normal users. The confidence interval produced via this method should be comparable to the one done using bootstrapping above. Tutorial=Ap ] 2006 and 2010 completed the survey
parameter value of the two of!, using simple random sampling so this condition is met since cases were selected claims that 80 of. Image that contains example code inside are available from NGC the
bottom water we... The evidence that is available in the text to draw a conclusion based the... Actually a random variable that will vary as different samples are often estimated using either the
observed difference these! The results of rejecting the null hypothesis one calculated using bootstrapping. ), C++ and *! Customers, using simple random sampling we make an effective solution,
accurate data analysis important. For each group are greater than the hypothesized proportion has not been invalidated here this matches our! Evidence here to another, and shape of distributions —
basic example please check us out GitHub. Education and deep experience in analytics or statistics is identical to the target of... Ll act in a factor is alphanumeric: each case that was selected bit
operations, is. Significance level at 5 % level water at different paired locations of OpenVINO™,... Python * sample … Inference¶ coin to simulate this process 10,000 times here so we have reason to
that. Obtained in solving the Laplace equation using the prop.test function to perform this analysis for us by... On these findings from the statements whose truth that we are interested in
proportions that have no to! Do we have a dataset that results from a sampling process that replicates how the original sizes. The outcomes for new data points so this condition is met since the
are... A noun that describes an intellectual process ALERT ] Indirect inference is the process of using data to! Newcomers and even for those who have advanced education and deep experience
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Orientation of pieces in permutation puzzles
Submitted by Herbert Kociemba on Sat, 09/20/2014 - 15:24.
The goal is to provide a general statement about the orientations of pieces of a certain permutation puzzles. Since there are so many different kinds of permutation puzzles we will have to exactly
describe the properties of the puzzles for which the statement will be valid. We also will have to explain how the orientation of pieces can be measured for these puzzles in way as general as
Then we are able to show that under these assumptions for a given kind of pieces with k possible orientations the sum of the orientations of all these pieces module k is invariant under permutations.
We restrict our consideration to permutation puzzles which do not change their shape. We call the moving parts of the puzzle the pieces and the locations of the pieces the places of the pieces. For
three-dimensional puzzles we assume that the pieces are polyhedrons which are bounded by faces, for two- dimensional puzzles we assume that the pieces are polygons which are bounded by a circuit of
edges. Since our main goal is the examination of three-dimensional puzzles we only use the term "faces" in the following text, but it could be replaced by "edges" to handle the two-dimensional case.
We further assume that at least some pieces can occupy their places in different positions/orientations. Since we restrict to puzzles which do not change their shape, these pieces must have at least
one axis of rotational symmetry. We restrict our consideration to puzzles where the different orientations of a piece are defined by exactly one axis of rotational symmetry of order k (which does not
seem to be a significant restriction). In this case we pick an arbitrary face which is not fixed by the symmetry and name it f_0. A counter clockwise (as seen from outside the puzzle) rotation by 2Pi
/k*j, 1<=j<k, then maps f_0 to another face, which is denoted by f_j. The faces f_0, f_1, f_2... may be visible or not. The faces f_0, f_1 and f_2 of a Rubik's cube corner are for example visible
while for the centre pieces of a 4x4x4 cube the faces f_0, f_1, f_2 and f_3 are not visible.
To define the orientation of a given piece at any possible place in any possible position we define a reference frame for the orientations. The geometric positions of the faces f_0, f_1... of a piece
at a certain place of the puzzle we call slots. It is important to emphasize that only the faces move when a permutation is applied, not the slots. Applying a permutation, the slots of the places
just are "filled" with the faces f_0, f_1... of a different piece of the same kind. Now we arbitrarily choose exactly one slot of every place to and call it the reference slot. We say that the
orientation of a piece in this place is i if the reference slot is filled with the face f_i of that piece.
A move M of the puzzle is a permutation of some of its pieces. We hardly can establish any statement about the orientations of the pieces if there is no restriction on the permutations. We call a
place P M-unambiguous if applying move M one or several times we cannot have the same piece in different orientations in place P. A place which does not have this property we call M-ambiguous. For
example, for all nxnxn Rubik's cubes all places are X- unambiguous for all conceivable moves X except for odd n in the places where the face diagonals meet.
A move M usually affects several pieces and places. If we choose a piece p at a place P_0 this piece p usually visits several places P_0,P_1,... if we apply the move M several times. All the pieces
which occupy these places form the M-orbit of p. All these pieces have the same shape as p and the involved places P_0, P_1... are all M-unambiguous or all M-ambiguous. The orbits of any piece of
nxnxn Rubik's cube except the centre piece for odd n consist of 4 pieces for example.
With these preliminary considerations we are now able to establish the following proposition:
Proposition 1: Let M be a move of a permutation puzzle which does not change its shape and p a piece with a single k-fold rotational symmetry in a M-unambiguous place P. Then the sum of the
orientations of all pieces in the orbit of p does not change its value modulo k if M is applied.
Proof: We name the involved places and slots in a way that it is easy to keep track of the orientations when the move M is applied. The place of piece p is named by P_0, the slot which is filled by
face f_i (0<=i<k) of p is referred to as slot_i. Let the size of the M-orbit of p be s. Then applying M j times (0<j<s) moves piece p to a place which we call P_j. The slots of P_j are denoted in the
same way as for P_0, using the faces of p – now in place P_j - again as a reference to name the slots. Named in this way we can make the following observation in the table which describes the faces
of all pieces of the orbit in the slots of their place:
The faces in the slots of P_0 are well defined by the definition of the slot names. For other places, place P_2 for example we neither know the piece in P_2 nor the face in slot_1 but since the faces
of all pieces in the orbit are named in the same way counter clockwise we know that we have face_(a+1) mod k in slot_2 of P_2 and face_(a+2) mod k in slot_3 of P_2.
And most important, in the way the slots and places are named, if we apply move M all entries of the table are cyclically shifted one line down, the entries of line P_(s-1) move to line P_0 (this
would not work if the places would be M-ambiguous).
With the following example we show that the sum of the orientations modulo k does not change when applying move M. The orientations are defined by the reference slots which can in principle be chosen
arbitrarily, they are highlighted in the example below. Obviously there has to be one reference slot in a line, but for the number of reference slots in a column there are no restrictions.
Before applying move M:
According to the definition of the orientations the sum of the orientations in the orbit modulo k is
1 + (a+2) + b + (c+k-1) + (d+1) + (e+k-1) mod k = a+b+c+d+e+2k+2 mod k.
After applying move M:
The orientation sum is
(e+1) + 2 + a +(b+k-1) + (c+1) + (d+k-1) mod k = a+b+c+d+e+2k+2 mod k.
This is the same as before and the reason is quite clear: The shifting of the table entries just shifts the order in which the variables a, b, c ... are added and the "offsets" 0, 1, 2 ... k-1 stay
the same.
This argument obviously does not depend on the choice of the orbit size s - which was 6 in this example- or the choice of the reference slots, so the proposition is proved.
Proposition 2: Let in addition to the conditions of proposition 1 be P a place which is unambiguous with respect to several moves M_1, M_2, ...M_n. Then the sum of all orientations modulo k of the
pieces in the union U of all orbits of M_1, M_2,.. M_n does not change if any of the moves is applied.
Proof: Applying a move M_i for a fixed i does not change the the sum of the orientations of the pieces in the orbit O of M_i. Pieces in the complement U\O do not move at all and hence their
orientations do not change. So the sum of all orientations of the pieces of U does not change modulo k.
Corollary : Given a permutation puzzle which does not change its shape, a place P that is unambiguous for all possible moves and a piece p in P. Then rotating p in place P by a sequence of moves
without changing the orientation of other pieces is impossible. p
Singmaster's proof
Submitted by
on Sat, 09/27/2014 - 06:01.
See also Singmaster's proof in the
Cubic Circular 3&4, p21
, in particular his Theorem 3.
Jaap's Puzzle Page: http://www.jaapsch.net/puzzles/
Thanks for the reference, I w
Thanks for the reference, I was not aware of it. I do not know if I completely understand the prerequisites for the puzzles handled there. It deals with corners and edges of polyhedra and it seems
that between two corners there is exactly one edge piece. And the moves are some cyclic permutation of the corners and edges done by rotation of the faces.
My approach is different since it does not assume the puzzle is a polyhedron, only each piece. And a move does not have to be necessarily a rotation. At least for me the most notable result is that
if a move does not have the "power" to twist a piece in place (which of course is obvious for pure rotations where the axis of rotation does not intersect with the piece) then the collaboration of
different such moves may have this power but never without affecting other pieces. | {"url":"http://forum.cubeman.org/?q=node/view/538","timestamp":"2024-11-04T23:56:20Z","content_type":"application/xhtml+xml","content_length":"22316","record_id":"<urn:uuid:dd0d0d10-3dc8-4dc5-bdaf-8314dc442790>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00600.warc.gz"} |
NetLogo User Community Models
(back to the NetLogo User Community Models)
If clicking does not initiate a download, try right clicking or control clicking and choosing "Save" or "Download".(The run link is disabled for this model because it was made in a version prior to
NetLogo 6.0, which NetLogo Web requires.)
Model SPIRALS & CURVE MATCHING
Alternative procedures for displaying spirals are examined and a procedure for matching equations
to curves is developed, which may have other applications.
Anyone of three well known spirals can be displayed by selection at the CHOICE widget.
The Archimedes spiral is generated by a rotation plus a constant velocity crossways displacement
as in the case of a gramophone record, giving an equidistance spacing between the lines.
Both the evolute and the logarithmic spirals are growth spirals in which the spacing between the
lines increases.
The original spirals were generated with the polar, r/phi procedure, the more natural from a
mathematical point of view; but to display conversion to cartesian coordinates was used.
The scale factor 's' can be adjusted to alter the spacing between the lines and therefore the
"length" of the spirals displayed.
Following a comment made by Seth Tisue at CCL, June 2003, lt/fd procedures were explored and added
as follows:
Using from the r/phi procedure, the cartesian coordinates of calculations of 3 iterations, previous,
now and next was a possibility. However completing the right-angled triangles between these and solving
for lt and fd was somewhat clumsy; so a similar method using the polar radii of iterations, previous 'b1',
now 'c' and next 'b2', was chosen:
where A is the constant 0.5° angular step of the polar r/phi code
___a2__ ___a1__ hence 'a1' becomes the previous step forward or 'fd' of the lt/fd code
\C2 B2|B1 C1/ also 'a2' " " next " " " " " " " "
\ | / and 'B2' - 'B1' becomes the angular step or 'lt' of the lt/fd code.
b2\ c|c /b1
\ | / And the cosine formula a = sqroot[bxb + cxc - 2xbxcxcosA]
\ | / also a/sinA = b/sinB = c/sinC are available.
\|/ Overlaying the displays of the r/phi and lt/fd procedures shows that
| the match is very good, including over the range of 's'.
It is possible to obtain an approximate match with a shorter procedure,identified as sh/lt/fd, by
comparing the PLOTS of the sh/lt/fd equations with those of the lt/fd equations and modifying the
sh/lt/fd equations to obtain a close fit. The results for the Archimedes spiral, (the simplest), are acceptable. The other equations have been left half finished for the amusement of model viewers
wish to try their hand at "fitting an equation to a curve techniques" and then overlaying the spiral
that is generated to see if the match of sh/lt/fd against r/phi is good.
First CHOOSE a spiral, select a value for 's' and press SETUP. Then RUN one procedure followed by another
if observing the match by overlaying; otherwise press SETUP in between RUNNING procedures.
It runs quicker with PLOTS switched OFF.
The overlay with close matching is difficult to see, so r/phi writes in yellow, lt/fd writes in red and
sh/lt/fd writes in lime.
It may be instructive to run the r/phi procedure with PLOTS switched ON but not when fitting an equation
to a curve, when it could be confusing.
First CHOOSE a spiral, select a value for 's' and press SETUP.
With PLOTS switched ON the lt/fd procedure should be RUN to plot the curves of 'lt' and 'fd'; followed
directly by the sh/lt/fd procedure to examine how closely its 'lt' and 'fd' plots follow those of lt/fd.
The fit of these curves may suggest a change to the sh/lt/fd equations for 'lt'(beta) and 'fd'(delta) to
obtain a better match when overlaying the spiral that is generated.
If all else fails there is always the START conditions to give that last tweek to the match.
The START conditions, 'xcor', 'ycor' and 'heading' are important in acheiving a good match between the
spirals generated by the r/phi procedure and by the lt/fd procedures. To assist this process code has been included to print these values in the COMMAND CENTER. The heading value can change the xcor&
ycor figures.
Over a 5 : 1 range of values of 's' it does appear possible to obtain a near perfect match of the spirals
drawn by procedure r/phi and procedure lt/fd. Perhaps a perfect match of the displayed spirals is not
possible because steps in the display result from finite pixel size.
The Author Derek Rush may be contacted by Email at derekrush@beeb.net October 2002 and July 2003. | {"url":"http://ccl.northwestern.edu/netlogo/models/community/Spirals%20&%20Curve%20Matching","timestamp":"2024-11-10T21:14:52Z","content_type":"text/html","content_length":"9478","record_id":"<urn:uuid:09c6b128-c4d6-4e4c-88cb-8bfd3856aeca>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00392.warc.gz"} |
Solve General Geodesics in FLRW Metric w/ Conformal Coordinates
• B
• Thread starter Onyx
• Start date
In summary: Sorry, I was confused earlier. In summary, there is no way to make solving for general geodesics in FLRW spacetimes easier by converting to conformal coordinates because it adds another
metric coefficient that depends on a coordinate. One approach to solving for geodesics is the brute force method of writing down all components of the geodesic equation and eliminating terms, while a
faster method for metrics like FLRW is the geodesic Lagrangian method. Additionally, the shape of the orbits in FLRW spacetimes is independent of the choice of scale factor, as shown in the paper
"The shape of the orbit in FLRW spacetimes" by
TL;DR Summary
Solving for General Geodesics in FLRW Metric
Once having converted the FLRW metric from comoving coordinates ##ds^2=-dt^2+a^2(t)(dr^2+r^2d\phi^2)## to "conformal" coordinates ##ds^2=a^2(n)(-dn^2+dr^2+r^2d\phi^2)##, is there a way to facilitate
solving for general geodesics that would otherwise be difficult, such as cases with motion in both ##r## and ##\phi##? I'm curious because there seems to be an ##n## killing vector evident in the new
form, but I don't know if that makes it any easier.
Onyx said:
there seems to be an killing vector evident in the new form
No, there isn't, because none of the metric coefficients are independent of ##n## (which is actually the Greek letter eta, ##\eta##, in most treatments in the literature).
PeterDonis said:
No, there isn't, because none of the metric coefficients are independent of ##n## (which is actually the Greek letter eta, ##\eta##, in most treatments in the literature).
It appeared to me that there was, because of ##\frac{\partial K_\eta}{\partial \eta}-\Gamma^{\eta}_{\eta\eta}K_\eta=0##. That's strange.
Anyways, I feel like there must be some straightforward way to handle calculating any geodesic in FLRW, but I'm not sure.
Onyx said:
It appeared to me that there was, because of ##\frac{\partial K_\eta}{\partial \eta}-\Gamma^{\eta}_{\eta\eta}K_\eta=0##. That's strange.
It's not strange at all. You test for a Killing vector field using Killing's equation, which the equation you wrote down is not. The equation you wrote down is the geodesic equation (well, a somewhat
garbled version of it, anyway), and (when properly written) shows that a worldline of constant ##r##, ##\theta##, ##\phi## is a geodesic. Which is not at all strange since that is the worldline of a
comoving observer.
Onyx said:
I feel like there must be some straightforward way to handle calculating any geodesic in FLRW
There are two general approaches for computing geodesics. One is the brute force way of writing down all of the components of the general geodesic equation and then eliminating terms which are known
to be zero until you have something manageable. The other way, which is considerably faster for metrics like this one where the metric coefficients are only functions of one or two of the
coordinates, is the geodesic Lagrangian method, which is described briefly here:
Onyx said:
Once having converted the FLRW metric from comoving coordinates ##ds^2=-dt^2+a^2(t)(dr^2+r^2d\phi^2)## to "conformal" coordinates ##ds^2=a^2(n)(-dn^2+dr^2+r^2d\phi^2)##, is there a way to
facilitate solving for general geodesics that would otherwise be difficult
One obvious way to make solving for the geodesics easier is to
switch to conformal coordinates. All that does is add one more metric coefficient that depends on a coordinate (##g_{nn}## depends on ##n##, whereas in the original form ##g_{tt}## does not depend on
##t##), and that is going to make more work for you in solving for geodesics no matter what method you use.
PeterDonis said:
One obvious way to make solving for the geodesics easier is to not switch to conformal coordinates. All that does is add one more metric coefficient that depends on a coordinate (##g_{nn}##
depends on ##n##, whereas in the original form ##g_{tt}## does not depend on ##t##), and that is going to make more work for you in solving for geodesics no matter what method you use.
Perhaps if I consider the form that the metric comes in when switching to ##R=a(t)r##, where there is an ##\frac{a'}{a}## term, setting the scale factor to ##e^t## would help. But it adds a
cross-term, so maybe not.
Never mind, I actually found a source that provides a way to do it. Apparently, the shape of the orbits is independent of the choice of scale factor, which seems bizarre.
Onyx said:
I actually found a source that provides a way to do it.
Can you give a reference?
Onyx said:
Apparently, the shape of the orbits is independent of the choice of scale factor
I'm not sure what you mean by "the choice of scale factor".
PeterDonis said:
Can you give a reference?I'm not sure what you mean by "the choice of scale factor".
"The shape of the orbit in FLRW spacetimes," by D Garfinkle.
The cited paper notes that the spatial projection of a geodesic in spacetime on to a constant curvature spatial slice (the usual FLRW coordinates' spatial slices) is also a geodesic of that space.
Thus the paths depend only on the sign of the curvature and not the details of ##a(t)##.
That doesn't seem to me to be completely true, in that the amount of time for which one can follow a geodesic does depend on ##a(t)## in a closed universe, but that might be overly pedantic.
I note that the paper does not appear to use conformal coordinates in its analysis.
Ibix said:
The cited paper notes that the spatial projection of a geodesic in spacetime on to a constant curvature spatial slice (the usual FLRW coordinates' spatial slices) is also a geodesic of that
space. Thus the paths depend only on the sign of the curvature and not the details of ##a(t)##.
That doesn't seem to me to be completely true, in that the amount of time for which one can follow a geodesic does depend on ##a(t)## in a closed universe, but that might be overly pedantic.
I note that the paper does not appear to use conformal coordinates in its analysis.
I was thinking only of a ##k=0## case.
Onyx said:
I was thinking only of a ##k=0## case.
For future reference, is orbit the most frequently used word to describe a geodesic that has angular momentum?
FAQ: Solve General Geodesics in FLRW Metric w/ Conformal Coordinates
1. What is the FLRW metric and why is it important in cosmology?
The FLRW (Friedmann-Lemaitre-Robertson-Walker) metric is a mathematical model used to describe the large-scale structure of the universe. It is based on the assumption of homogeneity and isotropy,
meaning that the universe looks the same in all directions and at all points in time. This metric is important in cosmology because it allows us to study the expansion and evolution of the universe.
2. What are conformal coordinates and how are they used in solving general geodesics in FLRW metric?
Conformal coordinates are a type of coordinate system that preserves angles and shapes of objects. In the FLRW metric, conformal coordinates are used to simplify the equations and make it easier to
solve for geodesics, which are the paths that particles follow in the curved spacetime of the universe.
3. How do you solve for general geodesics in FLRW metric using conformal coordinates?
To solve for general geodesics in FLRW metric using conformal coordinates, you first need to write the FLRW metric in terms of conformal time instead of cosmic time. Then, you can use the equations
of motion and the geodesic equation to find the paths that particles follow in the universe. These equations can be solved numerically or analytically, depending on the specific problem.
4. What are some applications of solving general geodesics in FLRW metric with conformal coordinates?
Solving general geodesics in FLRW metric with conformal coordinates has many applications in cosmology. It can be used to study the behavior of particles in the expanding universe, such as the motion
of galaxies and the propagation of light. It can also help us understand the formation and evolution of large-scale structures in the universe, such as galaxy clusters and superclusters.
5. Are there any limitations to using conformal coordinates in solving general geodesics in FLRW metric?
While conformal coordinates are useful in simplifying the equations for solving geodesics in FLRW metric, they do have some limitations. For example, they are not suitable for studying the behavior
of particles near strong gravitational fields, such as those near black holes. In these cases, other coordinate systems may be more appropriate. | {"url":"https://www.physicsforums.com/threads/solve-general-geodesics-in-flrw-metric-w-conformal-coordinates.1047783/","timestamp":"2024-11-10T16:21:06Z","content_type":"text/html","content_length":"157522","record_id":"<urn:uuid:5eb882a6-b1e3-4776-8c0d-267038440a40>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00727.warc.gz"} |
Go Math Grade 1 Answer Key Chapter 9 Measurement
Go Math Grade 1 Chapter 9 Answer Key Measurement
Measurement Show What You Know
Bigger and Smaller
Circle the bigger object.
Question 1.
Circle the smaller object.
Question 2.
Compare Length
Circle the longer object.
Draw a line under the shorter object.
Question 3.
Explanation :
Th shorter object is represented with a line under it
Question 4.
Explanation :
Th shorter object is represented with a line under it
Numbers 1 to 10
Write each number in order to 10.
Question 5.
Explanation :
A word or symbol that represents a specific amount or quantity.
The natural numbers starts from 1 , 2 , 3 , 4 …………. and so on .
Measurement Vocabulary Builder
Visualize It
Sort the review words from the box.
Answer :
Explanation :
The length measures start from shorter, short , long and longer. We use long to describe a measure of length. when we see an object of more length than the long object we say it as longer or longest
object. similar way
when we see an object of less length than the short object we say it as shorter or shortest object.
numbers are sorted in ascending order from smaller to bigger.
Understand Vocabulary
Complete the sentences with the correct word.
Question 1.
A crayon is ______ than a marker.
A crayon is shorter than a marker.
The length of the crayon is shorter when compared to marker based the lengths they are compared as longer or shorter.
Question 2.
A toothbrush is _____ than a paper clip.
A toothbrush is longer than a paper clip.
Explanation :
A toothbrush is longer when compared to a paper clip based the lengths they are compared as longer or shorter.
Write the name below the number
Question 3.
Measurement Game Measure UP!
Play with a partner.
1. Put
2. Spin the
3. Your partner spins, moves, and takes that object.
4. Compare the lengths of the two objects.
5. The player with the longer object places a
6. Keep playing until one person gets to END. The player with the most
Measurement Vocabulary Game
Going to a Weather Station
How to Play
1. Each player puts a
2. Toss the
3. Follow these directions for the space where you land.
White Space Read the math word or symbol. Tell its meaning. If you are correct, jump ahead to the next space with the same term.
Green Space Follow the directions. If there are no directions, stay where you are.
4. The first player to reach FINISH wins.
How to Play
1. Put your
2. Toss the
3. If you land on one of these spaces: Blue Space Explain the math word or use it in a sentence. If your answer is correct, jump ahead to the next space with that word. Green Space Follow the
directions in the space. If there are no directions, don’t move.
4. The first player to reach FINISH wins.
The Write Way
Choose one idea. Draw and write about it.
• Julia has to measure an object. She does not have a ruler. Tell what Julia could do to solve her problem.
• Explain why it is important to learn how to tell time.
Lesson 9.1 Order Length
Essential Question How do you order objects by length?
Listen and Draw
Use objects to show the problem.
Draw to show your work.
Answer :
Compare the straw and the key. Which is longer? Which is shorter?
Model and Draw
Order three pieces of yarn from shortest to longest. Draw the missing piece of yarn.
The First Yarn is shortest and the Third Yarn is longest . So, the Second yarn should not be shortest nor longest so draw a line which is longer that shortest Yarn and shorter than the longest Yarn .
Share and Show
Draw three lines in order from shortest to longest.
Explanation :
The First line is shortest line and the Third line is longest line. So, the Second line should not be shortest nor longest so draw a line which is longer that shortest line and shorter than the
longest line .
Draw three lines in order from longest to shortest.
Explanation :
The Fourth line is Longest line and the Sixth line is Shortest line. So, the Fifth line should not be shortest nor longest so draw a line which is longer that shortest line and shorter than the
longest line .
On Your Own
Compare Representations
Draw three crayons in order from shortest to longest.
Explanation :
The Seventh line is shortest line and the Ninth line is longest line. So, the Eighth line should not be shortest nor longest so draw a line which is longer that shortest line and shorter than the
longest line .
Draw three crayons in order from longest to shortest.
Explanation :
The Tenth line is Longest line and the Twelveth line is Shortest line. So, the Eleventh line should not be shortest nor longest so draw a line which is longer that shortest line and shorter than the
longest line .
Question 13.
Complete each sentence.
The Blue Yarn is the shortest yarn .
The Green Yarn and Red Yarn are the same length .
Problem Solving • Applications
Question 14.
Draw four objects in order from shortest to longest.
Question 15.
The string is shorter than the ribbon. The chain is shorter than the ribbon. Circle the longest object.
Explanation :
String is shorter than Ribbon
Chain is shorter than Ribbon
That means both string and chain are shorter than Ribbon so, Ribbon is longest object .
Question 16.
Match each word on the left to a drawing on the right.
TAKE HOME ACTIVITY • Show your child three different lengths of objects, such as three pencils or spoons. Ask him or her to order the objects from shortest to longest.
Order Length Homework & Practice 9.1
Draw three markers in order from longest to shortest.
Explanation :
The First Marker is Longest Marker and the Third Marker is Shortest Marker . So, the Second Marker should not be shortest nor longest so draw a Marker which is longer that shortest Marker and shorter
than the longest Marker .
Problem Solving
Question 4.
Fred has the shortest toothbrush in the bathroom. Circle Fred’s toothbrush.
Explanation :
The First Brush is Longest Brush and the Third Brush is Shortest Brush . So, the Second Brush is shortest nor longest .
Question 5.
Draw three different lines in order from shortest to longest. Label the shortest line and the longest line.
Explanation :
The First Line is Shortest and the Third Line is the Longest . So, the Second Line is shortest nor longest .
Lesson Check
Question 1.
Draw three crayons in order from longest to shortest.
Explanation :
The First Crayon is Longest Crayon and the Third Crayon is Shortest Crayon . So, the Second Crayon should not be shortest nor longest so draw a Crayon which is longer that shortest Crayon and
shorter than the longest Crayon .
Question 2.
Draw three paint brushes in order from shortest to longest.
Explanation :
The First Brush is shortest Brush and the Third Brush is longest Brush . So, the Second Brush should not be shortest nor longest so draw a Brush which is longer that shortest Brush and shorter than
the longest Brush .
Spiral Review
Question 3.
Lesson 9.2 Indirect Measurement
Essential Question How can you compare lengths of three objects to put them in order?
Listen and Draw
Clue 1: A yellow string is shorter than a blue string.
Clue 2: The blue string is shorter than a red string.
Clue 3: The yellow string is shorter than the red string.
Represent How did the clues help you draw the strings in the correct order?
Explanation :
From Clue 1 and Clue 3 we know that Yellow string is shorter than blue and red string . that means Yellow string is the shortest strings. From Clue 2 Blue string is shorter than red string that means
red string is the longest string.
Share and Show
Use the clues. Write shorter or longer to complete the sentence. Then draw to prove your answer.
Question 1.
Clue 1: A red line is shorter than a blue line.
Clue 2: The blue line is shorter than a purple line.
So, the red line is _______ than the purple line.
Explanation :
Clue 1: A red line is shorter than a blue line.
Clue 2: The blue line is shorter than a purple line.
If both red and blue lines are shorter then purple line will be longest line.
So, the red line is shorter than the purple line.
On Your Own
Analyze Relationships Use the clues. Write shorter or longer to complete the sentence. Then draw to prove your answer.
Question 2.
Clue 1: A green line is shorter than a pink line.
Clue 2: The pink line is shorter than a blue line.
So, the green line is ______ than the blue line.
Explanation :
Clue 1: A green line is shorter than a pink line.
Clue 2: The pink line is shorter than a blue line.
green line < pink line < blue line then blue line will be longest line
So, the green line is shorter than the blue line.
Question 3.
Clue 1: An orange line is longer than a yellow line.
Clue 2: The yellow line is longer than a red line.
So, the orange line is ______ than the red line.
Explanation :
Clue 1: An orange line is longer than a yellow line.
Clue 2: The yellow line is longer than a red line.
orange line > yellow line > red line that means orange line is the longest line .
So, the orange line is longer than the red line.
Problem Solving • Applications
Question 4.
The ribbon is longer than the yarn. The yarn is longer than the string. The yarn and the pencil are the same length. Draw the lengths of the objects next to their labels.
Explanation :
Ribbon > Yarn >String
Yarn = Pencil
So, Ribon is longest
Question 5.
Is the first line longer than the second line? Choose Yes or No.
Explanation :
Both first two figures first lines are longer than second line, Where as in third figure the first line is shorter than the first line .
TAKE HOME ACTIVITY • Show your child the length of one object. Then show your child an object that is longer and an object that is shorter than the first object.
Indirect Measurement Homework & Practice 9.2
Read the clues. Write shorter or longer to complete the sentence. Then draw to prove your answer.
Question 1.
Clue 1: A yarn is longer than a ribbon.
Clue 2: The ribbon is longer than a crayon.
So, the yarn is _______ than the crayon.
Explanation :
Clue 1: A yarn is longer than a ribbon.
Clue 2: The ribbon is longer than a crayon.
Yarn > Ribbon > Crayon that means Yarn is longest
So, the yarn is longer than the crayon.
Problem Solving
Solve. Draw or write to explain.
Question 2.
Megan’s pencil is shorter than Tasha’s pencil. Tasha’s pencil is shorter than Kim’s pencil. Is Megan’s pencil shorter or longer than Kim’s pencil?
Explanation :
Megan’s pencil < Tasha’s pencil < Kim’s pencil
Kim’s pencil is the longest pencil
Megan’s pencil is shorter than Kim’s pencil
Question 3.
Use different colors to draw 3 lines that are different lengths. Then write 3 sentences comparing their lengths.
Explanation :
Clue 1: A yellow string is longer than a red string.
Clue 2: The orange string is longer than a yellow string.
Clue 3: The red string is shorter than the orange string.
Lesson Check
Question 1.
A black line is longer than a gray line. The gray line is longer than a white line. Is the black line shorter or longer than the white line? Draw to prove your answer.
Explanation :
Black line > Gray line >White line
Black line is the longest line .
So, black line is longer than white line .
Spiral Review
Question 2.
What is the sum? Write the number.
42 + 20 = _____
42 + 20 = 62
Explanation :
When addend 20 is added to addend 42 we get the sum of 62 .
Lesson 9.3 Use Nonstandard Units to Measure Length
Essential Question How do you measure length using nonstandard units?
Listen and Draw
Reasoning How do you draw the boat to be the right length?
Model and Draw
You can use
For first figure 19.5
Share and Show
Use real objects. Use
Question 1.
Explanation :
To measure pencil we require 2 square object .
Question 2.
Explanation :
To Measure stapler we require 3 square object .
Question 3.
Explanation :
To Measure paint brush we require 3 square object .
Question 4.
Explanation :
To Measure scissor we require 2 square object .
On Your Own
Use real objects. Use
Question 5.
Explanation :
To Measure the given figure we require 2 square object .
Question 6.
Explanation :
To Measure the Glue bottle we require 2 square object .
Question 7.
Explanation :
To Measure the given figure we require 2 square object .
Question 8.
Explanation :
To Measure the Pencil we require 4 square object .
Question 9.
The green yarn is about 2
Explanation :
The Green Yarn is 2
Problem Solving • Applications
MATHEMATICAL PRACTICE Evaluate Reasonableness Solve.
Question 10.
Mark measures a real glue stick with
Explanation :
To Measure glue stick we requires 4
Question 11.
Bo has 4 ribbons. Circle the ribbon that is less than 3
Explanation :
The ribbon that is less than 3
Only the yellow ribbon have 2
So, it is circled .
Question 12.
The crayon is about 4 tiles long. Draw tiles below the crayon to show its length.
Explanation :
To Measure Crayon we require 4 titles.
TAKE HOME ACTIVITY • Give your child paper clips or other small objects that are the same length. Have him or her estimate the lengths of objects around the house and then measure to check.
Use Nonstandard Units to Measure Length Homework & Practice 9.3
Use real objects. Use
Question 1.
Explanation :
To measure Math book we require 2
Question 2.
Explanation :
To measure marker we require 4
Question 3.
Explanation :
To measure crayon we require 2
Problem Solving
Question 4.
Don measures his desk with
about _____
Explanation :
To measure Table we require 4
Question 5.
Use words or
pictures to explain how to measure an index card using color tiles.
Explanation :
To measure Index card we require 4
Lesson Check
Question 1.
Explanation :
The length of ribbon is about 4
Spiral Review
Question 2.
Draw and write to solve. I have 27 red flowers and 19 white flowers. How many flowers do I have?
______ flowers
Explanation :
Number of red flowers = 27
Number of white flowers = 19
Total Number of flowers = 27 + 19 = 46
Question 3.
Circle the number that is less.
Did tens or ones help you decide?
tens ones
Write the numbers.
______ is less than ____
_____< _____
50 is less than 51
50 < 51
Explanation :
The 51 and 50 are two digit numbers. first check the ten’s place both the numbers have 5 in the ten’s place. later check digit in the one’s place. the first number have 1 in one’s and second number
have 0 in the one’s place. so 1 is greater than 0 so, 51 is greater than 50 .
Lesson 9.4 Make a Nonstandard Measuring Tool
Essential Question How do you use a nonstandard measuring tool to measure length?
Answer :
Standard units are common units of measurement such as centimetres, grams and litres. Non-standard measurement units might include cups, cubes or sweets.
For example: a child might be asked to measure the length of their table using their hand span. They would then record how many hand spans the table was and record this. They might then be asked
to measure the length of a book.
Listen and Draw
Circle the name of the child who measured correctly.
Use Tools Explain how you know who measured correctly.
Explanation :
Mateo measures the pencil correctly.
The first paper clip should place from where the measuring starts then, place the second paper clip where the first paper clip ends and so on until the pencil is measured .
Model and Draw
Make your own paper clip measuring tool like the one on the shelf. Measure the length of a door. About how long is the door?
Explanation :
To measure the door we require about 3 paper clips.
Share and Show
Use real objects and the measuring tool you made. Measure. Circle the longest object. Underline the shortest object.
Question 1.
Explanation :
To measure the notice board we require about 2 paper clips.
Question 2.
Explanation :
To Measure the above given figure we require about 1 paper clip .
Question 3.
To measure the board we require 2 safety clip
Question 4.
Explanation :
To measure given window we require about 1 paper clip .
On Your Own
MATHEMATICAL PRACTICE Use Appropriate Tools Use the measuring tool you made. Measure real objects.
Question 5.
Explanation :
To measure given figure we require about 1 paper clip .
Question 6.
Explanation :
To measure given figure we require about 1 paper clip .
Question 7.
Question 8.
Question 9.
Cody measured his real lunch box. It is about 10
Problem Solving • Applications
Question 10.
Lisa tried to measure the pencil. She thinks the pencil is 5 paper clips long. About how long is the pencil?
Explanation :
Lisa measured the pencil in a wrong way. The correct measuring of the pencil is shown in the above figure.
To measure the pencil we require about 4 paper clips .
Question 11.
Use the
Explanation :
The Marking of the paint brush is shown in the above figure to measure how long is the paint brush.
We require about 4 paper clips to measure it .
TAKE HOME ACTIVITY • Have your child measure different objects around the house using a paper clip measuring tool.
Make a Nonstandard Measuring Tool Homework & Practice 9.4
Use the measuring tool you made. Measure real objects.
Question 1.
To Measure the computer we require about 1 paper clip .
Question 2.
To Measure the table we require about 1 paper clip .
Question 3.
To Measure the door we require about 1 paper clip .
Question 4.
To Measure the Math we require about 1 paper clip .
Question 5.
To Measure the computer we require about 1 paper clip .
Question 6.
To Measure the given figure we require about 1 paper clip .
Question 7.
Use words or pictures to explain how to measure a table using a paper clip measuring tool.
To Measure the table we require about 1 paper clip .
Lesson Check
Question 1.
Use the
Explanation :
The paper clips length is marked with lines that lines length matches the length of string 3 . so, string 3 is about 4
Spiral Review
Question 2.
Ty crosses out the number cards that are greater than 38 and less than 34. What numbers are left?
______ and ______
35 and 37 are left .
Explanation :
Number greater than 38 are 39 and 40
Number lesser than 34 are 33
Numbers which are left 35 and 37
Question 3.
There are 12 books. 4 books are large. The rest are small. Write a number sentence that shows how to find the number of small books.
______ – ______ = ______
Number of books = 12
Number of large books = 4
Number of small books = 12 – 4 = 8 books
Lesson 9.5 Problem Solving • Measure and Compare
Essential Question How can acting it out help you solve measurement problems?
The blue ribbon is about 4
Show how to solve the problem.
Answer :
Explanation :
Blue Ribbon is about 4
Red ribbon is 1
The green ribbon is 2
Green Ribbon = 2 + 1 = 3
So, Blue ribbon is longest ribbon.
Blue > green > red ribbons.
HOME CONNECTION • Have your child act out a measurement problem by finding the lengths of 3 objects and ordering them from shortest to longest.
Try Another Problem
Zack has 3 ribbons. The yellow ribbon is about 4
Measure and draw the ribbons in order from longest to shortest.
Question 1.
Explanations :
Number of ribbons = 3
Yellow ribbon is about 4
Orange ribbon is 3
Orange ribbon = 4 – 3 = 1
Blue ribbon is 2
Blue ribbon = 4 + 2 = 6
blue ribbon is the longest ribbon
Blue > Yellow > Orange .
Question 2.
Question 3.
Compare How many paper clips shorter is the orange ribbon than the blue ribbon?
Blue ribbon is about 6
Orange is about 1
Orange Ribbon is less than Blue Ribbon = 6 – 1 = 5
Share and Show
Solve. Draw or write to explain.
Question 4.
Lisa measures her shoe to be about 5
Explanation :
Lisa shoe measure to be = 5
An object 1 that is 3
Object 1 = 5
An object 2 that is 2
Object 2 = 5
Question 5.
Noah measures a marker to be about 4
TAKE HOME ACTIVITY • Have your child explain how he or she solved Exercise 4.
Problem Solving • Measure and Compare Homework & Practice 9.5
The blue string is about 3
Answer :
The blue string is about 3
The green string is 2
Green string = 3 + 2 = 5
The red string is 1
Red string = 3 – 1 = 2
Green >Blue > Red string
Question 1.
Question 2.
Question 3.
Problem Solving
Question 4.
Sandy has a ribbon about 4
The new ribbon is about ______
Explanation :
Sandy has a ribbon about 4
New ribbon 2
New ribbon = 4 + 2 = 6
Question 5.
Measure and draw to show a blue crayon and a green crayon that is about 1 paper clip longer.
Explanation :
draw to show a blue crayon and a green crayon that is about 1 paper clip longer.
Lesson Check
Question 1.
Mia measures a stapler with her paper clip ruler. About how long is the stapler?
Explanation :
The measurement of the stapler is marked in the above image . so, the stapler is about 7 paper clips.
Spiral Review
Question 2.
What is the unknown number? Write the number.
4 + _____ = 13
4 + ______ = 13
_______ = 13 – 4 = 9
So, the Unknown Number is 9
Question 3.
Count by tens. What numbers are missing? Write the numbers.
17, 27, ____, ____, 57, 67
Count by tens :
17 , 27 , 37 , 47 , 57 , 67
Explanation :
Count by tens means the every ten’s digit is increased by 1 from the previous ten’s value of the Number .
Measurement Mid-Chapter Checkpoint
Concepts and Skills
Draw three crayons in order from shortest to longest.
Question 1.
Explanation :
Longest crayon is the Yellow crayon and the shortest crayon is the Green Crayon .
Question 2.
Explanation :
The Blue Yarn is about 4
Question 3.
Kiley measures a package with her paper clip measuring tool. About how long is the package?
Explanation :
As per the above figure the Kiley package is about 4 paper clips .
Lesson 9.6 Time to the Hour
Essential Question How do you tell time to the hour on a clock that has only an hour hand?
Listen and Draw
Start at 1. Write the unknown numbers.
Look for Structure How are a clock face and ordering numbers alike?
Explanation :
Starts at 1 .
The Missing Number After 1 or Before 3 will be 2
The Missing Number After 8 or Before 10 will be 9
The Missing Number After 5 or Before 7 will be 6
The Missing Number After 11 will be 12
The clock ordering of numbers will be from left of the clock moving downwards to upwards towards right of the clock .
Share and Show
Look at where the hour hand points. Write the time.
Question 1.
9 o ‘ Clock
Explanation :
The Hour’s hand is pointing towards 9 so, It is 9 o ‘ Clock
Question 2.
1 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 1 so, It is 1 o ‘ Clock
Question 3.
11 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 11 so, It is 11 o ‘ Clock
Question 4.
6 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 6 so, It is 6 o ‘ Clock
Question 5.
7 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 7 so, the time is 7 o ‘ Clock
Question 6.
5 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 5 so, the time is 5 o ‘ Clock
On Your Own
MATHEMATICAL PRACTICE Make Connections Look at where the hour hand points. Write the time.
Question 7.
4 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 4 so, the time is 4 o ‘ Clock
Question 8.
10 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 10 so, the time is 10 o ‘ Clock
Question 9.
6 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 6 so, the time is 6 o ‘ Clock
Question 10.
12 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 12 so, the time is 12 o ‘ Clock
Question 11.
1 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 1 so, the time is 1 o ‘ Clock
Question 12.
3 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 3 so, the time is 3 o ‘ Clock
Question 13.
On Rae’s clock, the hour hand points to the 9. Circle Rae’s clock.
9 o ‘ Clock
Explanation :
In the first clock the hour hand is pointing 11 so, the time is 11 o ‘ clock
In the Second clock the hour hand is pointing 9 so, the time is 9 o ‘ clock
In the Third clock the hour hand is pointing 11 so, the time is 11 o ‘ clock
So, Second Clock is the right clock .
Problem Solving • Applications
Question 14.
Which time is not the same? Circle it.
Explanation :
In the first clock the hour hand is pointing 11 so, the time is 11 o ‘ clock But the time written as 1 : 00 so . it is wrong .
In the Second clock the hour hand is pointing 1 so, the time is 1 o ‘ clock . It is Right .
Question 15.
Manny leaves for school at 8 o’clock. Write and draw to show 8 o’clock.
Explanation :
The hour hand is pointing 8 so, the time is 8 o ‘ clock
Question 16.
Look at the hour hand. What is the time?
Explanation :
The Hour Hand is pointing 7 . So, the time is 7 : 00 .
TAKE HOME ACTIVITY • Have your child describe what he or she did in this lesson.
Time to the Hour Homework & Practice 9.6
Look at where the hour hand points. Write the time.
Question 1.
2 o’ clock
Explanation :
The Short Hand or Hour Hand is pointing 2 . So, the time is 2 : 00 .
Question 2.
9 o’ clock
Explanation :
The Short Hand or Hour Hand is pointing 9 . So, the time is 9 : 00 .
Question 3.
12 o’ clock
Explanation :
The Short Hand or Hour Hand is pointing 12 0′ clock. So, the time is 12 : 00 .
Problem Solving
Question 4.
Which time is not the same? Circle it.
Explanation :
In the first clock the hour hand is pointing 7 so, the time is 7 o ‘ clock . It is Wrong .
In the Second clock the hour hand is pointing 8 so, the time is 8 o ‘ clock. But the time written as 7 : 00 so . it is wrong .
Question 5.
Draw a clock to show where the hour hand points for 11:00.
11 o ‘ Clock
Explanation :
The Hour’s hand is pointing to 11 so, It is 11 o ‘ Clock
Lesson Check
Question 1.
Look at the hour hand. What is the time? Write the time.
3 o’ clock
Explanation :
The Short Hand or Hour Hand is pointing 3 . So, the time is 3 : 00 .
Question 2.
Look at the hour hand. What is the time? Write the time.
9 o’ clock
Explanation :
The Short Hand or Hour Hand is pointing 9 . So, the time is 9 : 00 .
Spiral Review
Question 3.
What is the sum? Write the number.
40 + 30 = _____
40 + 30 = 70
Question 4.
What is the sum? Write the number
53 + 30 = ______
53 + 30 = 83
Lesson 9.7 Time to the Half Hour
Essential Question How do you tell time to the half hour on a clock that has only an hour hand?
Answer :
To tell time to the half hour on a clock that an hour hand exactly halfway between the two numbers
To represent five and half an hour would be exactly here halfway between five and six .
Listen and Draw
Circle 4:00, 5:00, or between 4:00 and 5:00 to describe the time shown on the clock.
between 4:00 and 5:00
between 4:00 and 5:00
between 4:00 and 5:00
Answer :
Explanation :
In the first clock the short hand or hour hand points 4 so, the time is 4 o’clock.
In the second clock the short hand or hour hand points exactly half way in between 4 and 5
In the Third clock the short hand or hour hand points 5 so, the time is 5 o’clock.
Reasoning Use before and after to describe the time shown on the middle clock.
Share and Show
Look at where the hour hand points. Write the time.
Question 1.
half past 1 o’ clock
Time is 1:30
Half-past 1 is a short way of saying it’s half an hour (30 minutes) after 1:00.
Here the short hand or hour hand points in between 1 and 2.
so the Time is 1:30.
Question 2.
half past 4 o’ clock
Time is 4:30
Half-past 4 is a short way of saying it’s half an hour (30 minutes) after 4:00.
Here the short hand or hour hand points in between 4 and 5.
so the Time is 4:30.
Question 3.
half past 11 o’ clock
Time is 11:30
Half-past 11 is a short way of saying it’s half an hour (30 minutes) after 11:00.
Here the short hand or hour hand points in between 11 and 12.
so the Time is 11:30.
Question 4.
half past 3 o’ clock
Time is 3:30
Half-past 3 is a short way of saying it’s half an hour (30 minutes) after 3:00.
Here the short hand or hour hand points in between 3 and 4.
so the Time is 3:30.
On Your Own
MATHEMATICAL PRACTICE Use Reasoning Look at where the hour hand points. Write the time.
Question 5.
half past 5 o’ clock
Time is 5:30
Half-past 5 is a short way of saying it’s half an hour (30 minutes) after 5:00.
Here the short hand or hour hand points in between 5 and 6.
so the Time is 5:30.
Question 6.
half past 10 o’ clock
Time is 10:30
Half-past 10 is a short way of saying it’s half an hour (30 minutes) after 10:00.
Here the short hand or hour hand points in between 10 and 11.
so the Time is 10:30.
Question 7.
half past 2 o’ clock
Time is 2:30
Half-past 2 is a short way of saying it’s half an hour (30 minutes) after 2:00.
Here the short hand or hour hand points in between 2 and 3.
so the Time is 2:30.
Question 8.
half past 9 o’ clock
Time is 9:30
Half-past 9 is a short way of saying it’s half an hour (30 minutes) after 9:00.
Here the short hand or hour hand points in between 9 and 10.
so the Time is 9:30.
Question 9.
Maya starts reading at half past 8. Circle the clock that shows the time Maya starts reading.
Maya starts reading at half past 8
Time is 8:30
first clock is right
Half-past 8 is a short way of saying it’s half an hour (30 minutes) after 8:00.
Here the short hand or hour hand points in between 8 and 9.
so the Time is 8:30.
Problem Solving • Applications
Question 10.
Tim plays soccer at half past 9:00. He eats lunch at half past 1:00. He sees a movie at half past 2:00.
Look at the clock. Write what Tim does
Tim play Soccer at half past 9: 00 means the time is 9 : 30
Tim eats lunch at half past 1:00. means the time is 1 : 30
Tim sees a movie at half past 2:00 means the time is 2 : 30
From the above clock we notice, Here the short hand or hour hand points in between 2 and 3.
so the Time is 2:30.
half past 2 o’ clock
Half-past 2 is a short way of saying it’s half an hour (30 minutes) after 2:00.
Here the short hand or hour hand points in between 2 and 3.
so the Time is 2:30.
Question 11.
Tyra has a piano lesson at 5:00. The lesson ends at half past 5:00. How much time is Tyra at her lesson? Circle your answer.
half hour
Tyra piano class is at = 5 : 00
Tyra piano class end at = half past 5:00 = 5 : 30
Time is Tyra at her lesson = 5 : 30 – 5 : 00 = 30 minutes or half an hour .
Question 12.
What time is it? Circle the time that makes the sentence true.
half past 5 o’ clock
Time is 5:30
Half-past 5 is a short way of saying it’s half an hour (30 minutes) after 5:00.
Here the short hand or hour hand points in between 5 and 6.
so the Time is 5:30.
TAKE HOME ACTIVITY • Say a time, such as half past 10:00. Ask your child to describe where the hour hand points at this time.
Answer :
half past 10 o’ clock
Time is 10:30
Half-past 10 is a short way of saying it’s half an hour (30 minutes) after 10:00.
Here the short hand or hour hand points in between 11 and 10.
so the Time is 10:30.
Time to the Half Hour Homework & Practice 9.7
Look at where the hour hand points. Write the time.
Question 1.
half past 10 o’ clock
Time is 10:30
Half-past 10 is a short way of saying it’s half an hour (30 minutes) after 10:00.
Here the short hand or hour hand points in between 11 and 10.
so the Time is 10:30.
Question 2.
half past 3 o’ clock
Time is 3:30
Half-past 3 is a short way of saying it’s half an hour (30 minutes) after 3:00.
Here the short hand or hour hand points in between 3 and 4.
so the Time is 3:30.
Question 3.
half past 1 o’ clock
Time is 1:30
Half-past 1 is a short way of saying it’s half an hour (30 minutes) after 1:00.
Here the short hand or hour hand points in between 1 and 2.
so the Time is 1:30.
Problem Solving
Question 4.
Greg rides his bike at half past 4:00. He eats dinner at half past 6:00. He reads a book at half past 8:00.
Look at the clock. Write what Greg does.
Greg rides his bike at half past 4:00 = 4 : 30
Greg eats dinner at half past 6:00. = 6 : 30
Greg reads a book at half past 8:00 = 8 : 30
The Time in the above clock is 6 : 30 so, at half past 6 o’clock Grey is having dinner .
Half-past 6 is a short way of saying it’s half an hour (30 minutes) after 6:00.
Here the short hand or hour hand points in between 6 and 7.
so the Time is 6:30.
Question 5.
Draw clocks to show where the hour hand points for 5:00 and half past 5:00.
Explanation :
In First Clock the Hour hand or Short Hand points in between 5 and 6.
so the Time is 5:30.
In Second Clock the hour hand or short hand points to 5
So, the Time is 5 : 00
Lesson Check
Question 1.
Look at the hour hand. What is the time? Write the time.
half past 5 o’ clock
Time is 5:30
Half-past 5 is a short way of saying it’s half an hour (30 minutes) after 5:00.
Here the short hand or hour hand points in between 5 and 6.
so the Time is 5:30.
Question 2.
Look at the hour hand. What is the time? Write the time.
half past 9 o’ clock
Time is 9:30
Half-past 9 is a short way of saying it’s half an hour (30 minutes) after 9:00.
Here the short hand or hour hand points in between 9 and 10.
so the Time is 9:30.
Spiral Review
Question 3.
What number does the model show? Write the number.
The Above Model shows 10 ten lines and 3 ones. so the number is
10 + 3 = 13
Question 4.
How many tens and ones make this number?
14 = 10 + 4
10 tens and 4 ones .
Lesson 9.8 Tell Time to the Hour and Half Hour
Essential Question How are the minute hand and hour hand different for time to the hour and time to the half hour?
Answer :
Telling half hours
Yes, at 1:30 the hour hand points exactly halfway between 1 and 2! The hour hand tells us a lot about the time! … The minute hand spins all the way around the clock every hour. It makes it easier to
see if it’s 1:10, or 1:15, or even 1:12
Listen and Draw
Each clock has an hour hand and a minute hand. Use what you know about the hour hand to write the unknown numbers.
It is 1:00.
The hour hand points to the _____.
The minute hand points to the _____.
Answer :
It is 1:00.
The hour hand points to the 1.
The minute hand points to the 12.
It is half past 1:00.
The hour hand points between the _____ and the ____.
The minute hand points to the _____.
Answer :
It is half past 1:00.
The hour hand points between the 1 and the 2 .
The minute hand points to the 6 .
Use Tools Look at the top clock. Explain how you know which is the minute hand.
The large hand on a clock that points to the minutes. It goes once around the clock every 60 minutes (one hour). Example: in the clock on the left, the minute hand is just past the “4”, and if you
count the little marks from “12” it shows that it is 22 minutes past the hour.
Share and Show
Write the time.
Question 1.
It is half past 1:00.
The hour hand points between the 1 and the 2 .
The minute hand points to the 6 .
Question 2.
It is half past 1:00.
The hour hand points between the 1 and the 2 .
The minute hand points to the 6 .
Question 3.
Explanation :
It is 4:00.
The hour hand points to the 4.
The minute hand points to the 12.
On Your Own
MATHEMATICAL PRACTICE Attend to Precision Write the time.
Question 4.
It is half past 7:00.
The hour hand points between the 7 and the 8 .
The minute hand points to the 6 .
Question 5.
Explanation :
It is 9:00.
The hour hand points to the 9.
The minute hand points to the 12.
Question 6.
It is half past 8:00.
The hour hand points between the 8 and the 9 .
The minute hand points to the 6 .
Question 7.
It is half past 3:00.
The hour hand points between the 3 and the 4 .
The minute hand points to the 6 .
Question 8.
Explanation :
It is 2:00.
The hour hand points to the 2.
The minute hand points to the 12.
Question 9.
Explanation :
It is 6:00.
The hour hand points to the 6.
The minute hand points to the 12.
Circle your answer
Question 10.
Sara goes to the park when both the hour hand and the minute hand point to the 12. What time does Sara go to the park?
It is 12:00.
The hour hand points to the 12.
The minute hand points to the 12.
Question 11.
Mel goes to the park when the hour hand points between the 3 and 4 and the minute hand points to the 6. What time does Mel go to the park?
Time Mel go to park = 3 : 30
It is half past 3:00.
The hour hand points between the 3 and the 4 .
The minute hand points to the 6 .
Problem Solving • Applications
Question 12.
Linda wakes up at 6:30. Draw to show what time Linda wakes up.
It is half past 6:00.
The hour hand points between the 6 and the 7 .
The minute hand points to the 6 .
Question 13.
David left school at 3:30. Circle the clock that shows 3:30.
Question 14.
The hour hand points halfway between the 2 and 3. Draw the hour hand and the minute hand. Write the time.
Question 15.
Choose all the ways that name the time on the clock.
It is half past 7:00.
The hour hand points between the 7 and the 8 .
The minute hand points to the 6 .
TAKE HOME ACTIVITY • At times on the half hour, have your child show you the minute hand and the hour hand on a clock and tell what time it is.
Tell Time to the Hour and Half Hour Homework & Practice 9.8
Write the time.
Question 1.
Explanation :
It is 8:00.
The hour hand points to the 8.
The minute hand points to the 12.
Question 2.
It is half past 1:00.
The hour hand points between the 1 and the 2 .
The minute hand points to the 6 .
Question 3.
Explanation :
It is 5:00.
The hour hand points to the 5.
The minute hand points to the 12.
Question 4.
It is half past 9:00.
The hour hand points between the 10 and the 9 .
The minute hand points to the 6 .
Question 5.
Explanation :
It is 11:00.
The hour hand points to the 11.
The minute hand points to the 12.
Question 6.
It is half past 10:00.
The hour hand points between the 10 and the 11 .
The minute hand points to the 6 .
Problem Solving
Question 7.
Lulu walks her dog at 7 o’clock. Bill walks his dog 30 minutes later. Draw to show what time Bill walks his dog.
Lulu walks her dog at 7 o’clock.
Bill walks his dog 30 minutes later. = 7:00 + 0.30 = 7:30
Question 8.
Draw a clock to show the time 1:30.
It is half past 1:00.
The hour hand points between the 1 and the 2 .
The minute hand points to the 6 .
It is half past 7:00.
The hour hand points between the 7 and the 8 .
The minute hand points to the 6 .
Lesson Check
Question 1.
What time is it? Write the time.
It is half past 7:00.
It is half past 7:00.
The hour hand points between the 7 and the 8 .
The minute hand points to the 6 .
Question 2.
What time is it? Write the time.
Explanation :
It is 2:00.
The hour hand points to the 2.
The minute hand points to the 12.
Spiral Review
Question 3.
What is the sum? Write the number.
48 + 20 = _____
48 + 20 = 68
Explanation :
Adding addend 48 and addend 20 we get sum 68 .
Question 4.
How many tens and ones are in the sum? Write the numbers. Write the sum.
____ tens _____ ones
67 + 25 = 92
9 tens and 2 ones.
Lesson 9.9 Practice Time to the Hour and Half Hour
Essential Question How do you know whether to draw and write time to the hour or half hour?
Answer :
The short hand tells us the hour. The long hand tells us the minutes. When the long hand is pointing at the 6, it is half of the hour. Write in the time below each clock.
Circle the clock that matches the problem.
Answer :
Generalize Describe how you know which clock shows 1:30.
The fifth clock
Explanation :
It is half past 1:00.
The hour hand points between the 1 and the 2 .
The minute hand points to the 6 .
Share and Show
Use the hour hand to write the time. Draw the minute hand.
Question 1.
Explanation :
It is 4:00.
The hour hand points to the 4.
The minute hand points to the 12.
Question 2.
Explanation :
It is half past 11:00.
The hour hand points between the 11 and the 12 .
The minute hand points to the 6 .
Question 3.
Explanation :
It is half past 6:00.
The hour hand points between the 6 and the 7 .
The minute hand points to the 6 .
Question 4.
Explanation :
It is 7:00.
The hour hand points to the 7.
The minute hand points to the 12.
Question 5.
Explanation :
It is 2:00.
The hour hand points to the 2.
The minute hand points to the 12.
Question 6.
Explanation :
It is half past 3:00.
The hour hand points between the 3 and the 4 .
The minute hand points to the 6 .
On Your Own
MATHEMATICAL PRACTICE Use Diagrams Use the hour hand to write the time. Draw the minute hand.
Question 7.
Explanation :
It is 10:00.
The hour hand points to the 10.
The minute hand points to the 12.
Question 8.
Explanation :
It is half past 12:00.
The hour hand points between the 12 and the 1 .
The minute hand points to the 6 .
Question 9.
Explanation :
It is 5:00.
The hour hand points to the 5.
The minute hand points to the 12.
Question 10.
Explanation :
It is half past 10:00.
The hour hand points between the 10 and the 11 .
The minute hand points to the 6 .
Question 11.
Explanation :
It is 12:00.
The hour hand points to the 12.
The minute hand points to the 12.
Question 12.
Explanation :
It is half past 5:00.
The hour hand points between the 5 and the 6 .
The minute hand points to the 6 .
Question 13.
What is the error? Zoey tried to show 6:00. Explain how to change the clock to show 6:00.
Zoey shows the time 6 : 00 wrongly.
To change the clock to show 6:00, We, should point the short hand or hand hand to 6 and the long hand should point 12.
Then , the Time will be
Problem Solving • Applications
Question 14.
Vince goes to a baseball game at 4:30. Draw to show what time Vince goes to a baseball game.
Vince goes to a baseball game at 4:30.
Explanation :
It is half past 4:00.
The hour hand points between the 4 and the 5 .
The minute hand points to the 6 .
Question 15.
Brandon has lunch at 1 o’clock. Write and draw to show what time Brandon has lunch.
Brandon has lunch at 1 o’clock.
Explanation :
It is 1:00.
The hour hand points to the 1.
The minute hand points to the 12.
Question 16.
Juan tried to show 8:30. He made a mistake.
What did Juan do wrong? Explain his mistake.
juan tired to show 8 : 30 but he showed 7 : 30 as, the our hand is in between 7 and 8.
To show 8 :30. we should mark the half hand exactly in between 8 and 9 .
Explanation :
It is half past 8:00.
The hour hand points between the 8 and the 9 .
The minute hand points to the 6 .
TAKE HOME ACTIVITY • Show your child the time on a clock. Ask him or her what time it will be in 30 minutes
Practice Time to the Hour and Half Hour Homework & Practice 9.9
Use the hour hand to write the time. Draw the minute hand.
Question 1.
Explanation :
It is 11:00.
The hour hand points to the 11.
The minute hand points to the 12.
Question 2.
Explanation :
It is half past 8:00.
The hour hand points between the 8 and the 9 .
The minute hand points to the 6 .
Question 3.
Explanation :
It is half past 2:00.
The hour hand points between the 2 and the 3 .
The minute hand points to the 6 .
Problem Solving
Question 4.
Billy played outside for a half hour. Write how many minutes Billy played outside.
_______ minutes
Billy played outside for a half hour.
half an hour = 30 minutes .
Question 5.
Draw a clock to show a time to the hour. Draw another clock to show a time to the half hour. Write each time.
Lesson Check
Question 1.
Write the time.
It is 11:00.
The hour hand points to the 11.
The minute hand points to the 12.
Spiral Review
Question 2.
What is the difference? Write the number.
80 – 30 = _____
80 – 30 = 50
Question 3.
Use . Amy measures the eraser with
Explanation :
To Measure eraser we require about 4
Measurement Review/Test
Question 1.
Match each word on the left to a drawing on the right.
Question 2.
Is the first line shorter than the second line? Choose Yes or No.
Question 3.
The crayon is about 5 tiles long. Draw tiles below the crayon to show its length.
The length of the crayon is represented in the above figure
Question 4.
Use the below
Question 5.
Measure the
Question 6.
Look at the hour hand. What is the time?
Question 7.
What time is it? Circle the time that makes the sentence true.
Question 8.
Choose all the ways that name the time on the clock.
Explanation :
It is half past 11:00.
The hour hand points between the 11 and the 12 .
The minute hand points to the 6 .
Question 9.
Draw the hand on the clock to show 9:30.
Explanation :
It is half past 9:00.
The hour hand points between the 9 and the 10 .
The minute hand points to the 6 .
Question 10.
Lucy tried to show 5:00. She made a mistake.
Draw hands on the clock to show 5:00.
What did Lucy do wrong? Explain her mistake.
Lucy didn’t indicate the hour or short hand in the clock .
Explanation :
It is 5:00.
The hour hand points to the 5.
The minute hand points to the 12.
Question 11.
Explanation :
The green line is shorter than red line
The blue line is longer than red line
That means blue line is longest line
blue line > red line > green line | {"url":"https://eurekamathanswerkeys.com/go-math-grade-1-answer-key-chapter-9/","timestamp":"2024-11-14T04:32:13Z","content_type":"text/html","content_length":"276587","record_id":"<urn:uuid:08fb103f-8b3c-4889-a184-1eb270084ab2>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00596.warc.gz"} |
Translations:Horizontal Movement Formulas/13/en - Minecraft Parkour WikiTranslations:Horizontal Movement Formulas/13/en
These simplified formulas only apply to linear movement (no change in direction). While this condition might seem very restrictive, these formulas are very useful to analyze conventional jumps and
momentum We'll later expand on these formulas by including angles.
□ ${\displaystyle V_{H,0}}$ is the player's initial speed (default = 0).
□ ${\textstyle V_{H,t}}$ is the player's speed on tick ${\textstyle t}$. | {"url":"https://www.mcpk.wiki/wiki/Translations:Horizontal_Movement_Formulas/13/en","timestamp":"2024-11-08T18:31:55Z","content_type":"text/html","content_length":"41660","record_id":"<urn:uuid:b5834639-8aeb-49fe-a57b-21547420d747>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00447.warc.gz"} |
SCV0631 - INTELLIGENT SYSTEMS
Dati Generali
Periodo di attività
Primo Semestre (23/09/2024 - 20/12/2024)
Obiettivi Formativi
The course provides broad coverage of intelligent systems solving pattern recognition problems. Theoretical concepts in intelligent systems and techniques relevant to real-life applications will be
The student will be able to:
1. Know the main objectives and areas of Artificial Intelligence, Machine Learning, and Pattern Recognition, with the ability to identify the potentialities of intelligent techniques and the
relationships with other disciplines
2. Know the basic concepts of automated learning based on machine learning approaches and the conditions for their applicability
3. Know the most relevant feature extraction and selection techniques
4. Know statistical techniques and their limitations and strengths, with the ability to appropriately select the proper technique in specific contexts
5. Know basic principles of neural computing and their characteristics
6. Know Flat and Hierarchical Clustering with the ability to configure and apply these methods in specific contexts
7. Know performance metrics for learners
8. Know basic concepts of the following application domains: Image Classification, Text Categorization, Biomedical Data Analysis
9. Know how to program in a language for statistical computing and machine learning applications like R
It is also expected that students develop communicative skills through open discussion and autonomous assessment in the choice of the proper technique to solve problems of recognition and /or
automatic classification of multidimensional data in several domains.
Students will acquire also knowledge of the relevant Machine learning and Pattern Recognition terminology.
The course assumes that students have a background acquired in a Bachelor's Degree in STEM disciplines. Students are expected to be familiar with basic Mathematics, Probability, and Statistics.
Metodi didattici
Lectures (72 hours)
The topics of the course are illustrated by means of (1) conceptual, formal descriptions, (2) their implementation via R code, and (3) the use of demos and online resources.
Constant interaction with the students and their involvement in open discussions are highly encouraged.
Verifica Apprendimento
The students’ learning extent is assessed via a written test (duration: 2 hours) and an individual assignment, autonomously developed by each student individually.
The goal of the written test is to assess the learning degree and the understanding of the elements related to intelligent systems from both theoretical and application (on problems of limited
complexity) points of view. Written tests normally consist of
- two exercises for the assessment of the student's understanding and knowledge of machine learning techniques: each exercise weighs about one-quarter of the grade of the written exam;
- four questions on the conceptual aspects: each exercise weighs about one-eighth of the grade of the written exam.
The assignment allows the students to use their skills and knowledge for the building and evaluation of machine learners by using the R language. The project presentation has the goal of assessing
the students’ communication skills in two areas: 1) the students’ technical competencies and use of the correct terminology and 2) the students’ skills for communicating a complete and organized view
of the work they carried out.
Individual judgment skills are evaluated based on the decisions made during the written exam and the assignment.
The grade of the written test is on a 0 to 30 scale. The written exam contributes 70% of the final mark, while the assignment accounts for the remaining 30%.
The acquisition of knowledge and expected skills is developed throughout the entire course, which includes the topics listed below.
1) Introduction to Artificial Intelligence and Pattern Recognition: Historical Perspective, State of the Art of Methods and Applications (3 h - Course Objective 1)
2) Basic Mathematical and Statistical Concepts:
• Measurement Theory
• Matrix algebra
• Multivariable function analysis calculus: partial derivatives and gradient
• Relevant concepts of Probability and Statistics
• Relevant concepts of Information Theory
(6 h - Course Objective 2)
3) Design of a Supervised Classifier; Basic principles of learning by example; basic concepts of multidimensional pattern analysis
(5 h - Course Objective 2)
4) Feature Extraction and Selection:
• Principal Component Analysis
• Information Gain
• Statistical evaluation of features
• Selection Strategies
(6 h - Course Objective 3)
5) Fundamental Elements of Programming with R
(8 h - Course Objective 9)
6) Machine Learning Algorithms
• Ordinary Least Squares Regression
• Outliers and Robust Regression
• Regularization and Shrinkage in Regression: Ridge Regression, LASSO Regression, Elastic Net
• Minimum distance classifier
• Bayesian classification
• Maximum likelihood classifier
• K-Nearest Neighborhood classifier
• Parallelepiped Method
• Decision trees
• Ensemble models: boosting, bagging, stacking
• Support Vector Machine
• Imbalance, Hyperparameter tuning
• Performance metrics
• (32 h - Course Objectives 2, 3, 4, 7)
7) Neural Networks
• Introduction, taxonomy
• Basic principle of neural computing
• Feedforward Neural Models
• Application Examples
• Introduction to Deep Learning
(6 h - Course Objective 5)
8) Clustering
• Introduction to Clustering
• K-means Clustering algorithm
• Agglomerative Hierarchical Clustering: Single linkage, Complete linkage
(3 h - Course Objective 6)
9) Design of Intelligent Systems, Examples in Application Domains
(3 h - Course Objectives 1, 8)
Altre informazioni
During the period in which the course is held, the students can meet with the instructor on class days. During the remainder of the year, the students need to contact the instructor to set up an
appointment by e-mail at sandro.morasca@uninsubria.it. The instructor responds only to e-mail messages sent from the official student.uninsubria.it e-mail accounts. | {"url":"https://uninsubria.unifind.cineca.it/individual?uri=http%3A%2F%2Firises.uninsubria.it%2Fresource%2Faf%2F297682","timestamp":"2024-11-12T12:48:45Z","content_type":"text/html","content_length":"46382","record_id":"<urn:uuid:b43c4e36-07de-4ec9-8c50-0f327afb8c34>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00490.warc.gz"} |
Basic College Mathematics (10th Edition) 10th Edition Solutions
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In any digital electronics textbook you will find description on various logic operations and different logic gates; e.g. AND, OR, NOT, NAND, NOR, XOR and XNOR. But AND, OR and NOT these three gates
are known as basic gates. Because you can design any other logic gates or digital electronics circuits using only these three gates.
Suppose you are about to design a logic circuit where you need two XOR gates and three NAND gates. You don’t have those gate ICs in hand. But you have enough AND, OR and NOT gate ICs, then you can
easily use those three gates to design your circuit.
In simple words you will design XOR gate and NAND gate using AND, OR and NOT gates. Then you can use those gates in your main circuit.
Actually it’s like elements, there are numerous compounds in our world but all consists of only 118 elements like oxygen, carbon, hydrogen, nitrogen, sulfur etc.
Not to be confused with NAND and NOR gate. Those two are known as universal logic gates.
Let’s see how we can design other logic gates using only basic gates, i.e. AND, OR and NOT gate.
NAND gate from basic gates
By definition when a NOT gate is placed on AND gate output that is NAND gate. So it very simple:
Which is expressed as below:
NOR gate from basic gates:
It’s same as above.
So now we can say NAND gate and NOR gate can be designed from basic gates. So any gate or other circuits designed using NAND or NOR gate can be designed using only three basic gates.
Exclusive OR or XOR gate from basic gates:
The digital circuit above is actually a XOR gate circuit, which is expressed as below:
Let’s analyze the circuit (i). We consider all possible input logic combinations of the circuit.
From the truth table above we can see that the output is same as XOR gate truth table. To know more about XOR gate please go to XOR gate section.
XNOR gate from basic gates:
I am not discussing in detail about XNOR gate because it’s simply adding a NOT gate to XOR gate output and NOT gate is one of the three basic gates. | {"url":"https://fromreadingtable.com/basic-logic-gates/","timestamp":"2024-11-06T13:35:46Z","content_type":"text/html","content_length":"73382","record_id":"<urn:uuid:8d138ce7-ba7a-4765-9508-ace7c3fb83c0>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00717.warc.gz"} |
Implementing k Nearest Neighbours in OCaml
Here and there, people wonder about the possibility to use functionnal programming to approach machine learning. I decided to give it a try with some learning algorithms and noted that there are
actually various options to use external libraries (unfortunately, nothing getting close to scikit learn’s maturity)
A quick reminder about the k-nearest neighbours. It was first described in the early 1950s and is often referred to as a “lazy learner”, as it merely stores the data, waiting to be provided with
points to classify.
Then, in a metric space \((X,d)\) given a point \(y \in X\), and \(n\) labelled points \(S = (x_i,l_i) \in (\mathbb{R} \times \{0,1\}) ^ n\) it will return the most common labels among the \(k\)
closest point to \(y\). Simple, isn’t it ? Actually, one does not need more information to implement it. So let’s get started.
The expressivity of ocaml
For fun, let’s see how easy it is to implement a k-nearest neighbours in ocaml. Note that we only need to retrieve the closest points from one point in an array of points. The method
find_nearest_neighbours does this. Note how generic it is : the point can have any type (float array, string list…) as long as the distance operates on this type. Think about all the templates that
should be written in other languages. And the compiler will tell me if types are incompatible (when Python would wait until an error appears).
(* Returns the k smallest elements of an array *)
let get_smallest_elements_i input_array k =
let n = Array.length input_array in
let indices = Array.init n (fun x -> x) in
for i = 0 to (k-1) do
for j = (n-1) downto 1 do
if input_array.(indices.(j-1)) > input_array.(indices.(j)) then begin
let b = indices.(j-1) in
indices.(j-1) <- indices.(j);
indices.(j) <- b;
Array.sub indices 0 k
(* Returns the k closest points from current_point in all_points *)
let find_nearest_neighbours current_point all_points k distance =
let distances = Array.map (fun x -> distance x current_point) all_points in
get_smallest_elements_i distances k
(* Returns the most common labels among the neihbours *)
let predict nearest_neighbours labels =
let sum a b = a +. b in
let k = Array.length nearest_neighbours in
if Array.fold_left sum 0. (Array.init k (fun i -> labels.(nearest_neighbours.(i)))) > 0. then 1. else ~-.1.
Now we need a dataset to try the algorithm. Nothing really funny there.
(* Toy data *)
let max_length = 1.
let chessboard_boundary x y = if ((mod_float x 0.5) -. 0.25) *. ((mod_float y 0.5) -. 0.25) > 0. then 1. else ~-.1.
let circle_boundary x y = if (x**2. +. y**2.) > 0.5 then 1. else ~-.1.
let unlabelled_boundary x y = 2. ;;
(* Given a decision boundary, returns a data set and the associated labels *)
let make_data n_points decision_boundary =
let output_data = Array.init n_points (fun _ -> (Array.make 2 0.)) in
let output_label = Array.make n_points 0. in
for i = 0 to (n_points-1) do
output_data.(i).(0) <- Random.float max_length;
output_data.(i).(1) <- Random.float max_length;
output_label.(i) <- decision_boundary output_data.(i).(0) output_data.(i).(1)
output_data, output_label
Now that we defined the points as arrays of floats, we need to implement distances on it.
let sum a b = a +. b in
(* Usual Euclide Distance *)
let euclide_distance x y =
let squares_diff = Array.init (Array.length x) (fun i -> (x.(i) -. y.(i))**2.) in
Array.fold_left sum 0. squares_diff
let manhattan_distance x y =
let squares_diff = Array.init (Array.length x) (fun i -> abs (x.(i) -. y.(i)) ) in
Array.fold_left sum 0. squares_diff
Gluing up all the pieces together :
open Knn
open Distances
open ToyDataset
(* Number of points in the training set*)
let n_points = int_of_string Sys.argv.(1) ;;
(* Parameter k of the kNN algorithm *)
let k = int_of_string(Sys.argv.(2)) ;;
(* Number of points in the training set *)
let n_test_points = 50 ;;
(* Train and test data*)
let train_data, labels = make_data n_points circle_boundary;;
let test_data, pseudo_labels = make_data n_test_points unlabelled_boundary ;;
(* For each point in the test set, stores the indices of the nearest neighbours *)
let nearest_neighbours = Array.map (fun x -> find_nearest_neighbours x train_data k euclide_distance) test_data;;
(* Evaluates and prints the accuracy of the model *)
let mismatches = ref 0. ;;
for l = 0 to (n_test_points-1) do
pseudo_labels.(l) <- predict nearest_neighbours.(l) labels ;
if pseudo_labels.(l) <> (circle_boundary test_data.(l).(0) test_data.(l).(1)) then (mismatches := !mismatches +. 1.) else ();
print_string ("Error rate : "^string_of_float(100. *. !mismatches /. (float_of_int n_test_points))^"%\n");
Now I recommend using ocamlbuild. It will save you loads of time. Especially with large projects. Assuming the latest part is called main.ml simply enter this in the terminal:
me$ ls
distances.ml knn.ml main.ml toyDataset.ml
me$ ocamlbuild main.byte
Finished, 9 targets (1 cached) in 00:00:00.
Now, you just have to call the produced byte file with the first argument being the number of points to generate and the second one, the parameter \(k\).
me$ ./main.byte 100 5
Error rate : 4.%
me$ ./main.byte 1000 5
Error rate : 2.%
me$ ./main.byte 3000 5
Error rate : 0.%
What about performance ?
I leave this to another post : pypy vs ocaml for streaming learning, coming soon :)
More about knn
If you are interested in this method and further developments, you may find the following articles interesting:
[1]S. Cost and S. Salzberg, “A weighted nearest neighbor algorithm for learning with symbolic features,” Machine Learning, vol. 10, no. 1, pp. 57–78, Jan. 1993.
[2]J. Wang, P. Neskovic, and L. N. Cooper, “Improving nearest neighbor rule with a simple adaptive distance measure,” Pattern Recognition Letters, vol. 28, no. 2, pp. 207–213, Jan. 2007.
[3]K. Yu, L. Ji, and X. Zhang, “Kernel Nearest-Neighbor Algorithm,” Neural Processing Letters, vol. 15, no. 2, pp. 147–156, Apr. 2002. | {"url":"https://www.thekerneltrip.com/machine/learning/ocaml-knn/","timestamp":"2024-11-08T01:11:59Z","content_type":"text/html","content_length":"33341","record_id":"<urn:uuid:d56f4668-b1a2-4a41-8e83-f02aaab0ef5f>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00878.warc.gz"} |
Unveiling the Mystery: Exploring the Capacity of Pitchers - Quarts Unveiled
Unveiling The Mystery: Exploring The Capacity Of Pitchers – Quarts Unveiled
How many quarts in a pitcher? A pitcher typically holds 8-12 cups. A US quart is 2 pints or 32 fluid ounces, while an Imperial quart is slightly larger at 40 fluid ounces. Using the formula (Pitcher
volume in cups) ÷ 4 = Quarts in pitcher, an 8-cup pitcher holds 2 quarts. Volume measurements also include pints, gallons, liters, and cubic centimeters.
How Many Quarts in a Pitcher? Unveiling the Mystery
When it comes to measuring liquids, the question of “how many quarts in a pitcher” often lingers in our minds. Pitchers, those ubiquitous vessels in our kitchens and dining tables, vary in size,
making it crucial to understand the volume measurement units and their conversions to determine the exact amount they can hold.
Understanding Volume Measurement Units
1. US Quart:
– 1 US quart is equivalent to 2 pints, 4 cups, or 32 fluid ounces.
2. Imperial Quart:
– The Imperial quart is slightly larger than the US quart, equaling 2.113 US quarts, 4.226 pints, or 8.453 cups.
Fluid Measurement Units
1. Fluid Ounce:
– A fluid ounce is a smaller unit often used in cooking and mixing drinks. It equals 1/16 of a pint, 1/8 of a cup, or 29.574 milliliters.
Pint and Gallon Measurements
1. Pint:
– A pint is half a quart, containing 2 cups or 16 fluid ounces.
2. Gallon:
– The gallon is a larger unit commonly used for measuring liquids like milk, gasoline, or water. It is equivalent to 4 quarts, 8 pints, 128 fluid ounces, or 3.785 liters.
Metric Volume Measurements
For those using the metric system, here are some important units:
1. Liter:
– A liter is slightly more than a quart, equaling 1.057 quarts, 2.113 pints, or 33.814 fluid ounces.
2. Milliliter:
– A milliliter is a smaller unit, equaling 0.0338 fluid ounces, 0.0021 pints, or 0.0010 quarts.
3. Cubic Centimeter:
– A cubic centimeter is often used in scientific measurements and is equivalent to 0.0338 fluid ounces, 0.0021 pints, or 0.0010 quarts.
Calculating Quarts in a Pitcher
To determine the number of quarts in a pitcher, simply convert its volume from cups to quarts. For example, if you have an 8-cup pitcher:
• 8 cups x (1 quart / 4 cups) = 2 quarts
Understanding volume measurement units and their conversions is essential for accurately measuring liquids, whether in the kitchen or the lab. By knowing the relationship between quarts, pints,
gallons, fluid ounces, and other units, you can confidently answer the question “how many quarts in a pitcher” and ensure the perfect amount of liquid for your needs.
Understanding Volume Measurement Units
The US Quart
In the United States, a quart is defined as 0.946 liters or 32 fluid ounces. It’s commonly abbreviated as “qt” or “Quart”. One US quart is equivalent to two pints or one-fourth of a gallon.
The Imperial Quart
Across the pond in the United Kingdom and other Commonwealth countries, an Imperial quart is defined as 1.136 liters or 40 fluid ounces. It’s abbreviated as “qt imp” or “Quart imp”. An Imperial quart
is larger than its American counterpart, measuring up to about 1.2 US quarts or three pints.
The Conversion Dance
Converting between US and Imperial quarts is a bit like a dance. To convert US quarts to Imperial quarts, multiply by 1.2. For the reverse, divide Imperial quarts by 1.2. Remember, the numbers 32 and
40 (for the fluid ounces) are the key to distinguishing these two units.
Fluid Measurement: Dissecting the Fluid Ounce
Amidst the diverse array of volume measurement units, the fluid ounce stands out as a vital cog in the culinary and apothecary realms. This humble unit serves as the foundation for measuring liquids,
from the sweet concoctions we sip to the life-giving elixirs we imbibe.
Defining the Fluid Ounce: A Unit of Liquid Capacity
The fluid ounce, abbreviated as fl oz, is a unit of volume that quantifies the amount of a liquid substance. One fluid ounce is equivalent to 1/8 of a pint or 1/32 of a gallon. It occupies a space of
approximately 29.57 milliliters or 0.03125 quarts.
Conversions and Equivalencies: Fluid Ounces and Other Units
The fluid ounce plays a crucial role in a web of volume measurement conversions. Here’s a handy guide:
• 1 fluid ounce = 1/8 pint = 1/32 gallon = 29.57 milliliters
• 1 pint = 16 fluid ounces = 2 cups = 473.18 milliliters
• 1 gallon = 128 fluid ounces = 8 pints = 3.785 liters
Additional Measurement Units and Their Interconversions
Beyond the fluid ounce, several other volume measurement units deserve recognition:
• Liter (L): A metric unit of volume equal to 1000 cubic centimeters or 33.814 fluid ounces.
• Milliliter (mL): A smaller metric unit of volume equal to 1 cubic centimeter or 0.033814 fluid ounces.
• Cubic Centimeter (cc): A cubic unit of volume equal to 1 milliliter or 0.033814 fluid ounces.
Understanding the Interplay of Volume Measurements
Navigating the labyrinth of volume measurement units can be daunting, but understanding their interconversions is essential for culinary precision and accurate dosing of medications. By mastering
these conversions, we empower ourselves to measure liquids effortlessly, ensuring consistency in recipes and the safe administration of treatments.
Pint and Gallon Measurements:
• Pint:
□ Definition and conversion to US Quart, Imperial Quart, Fluid Ounce, and Gallon
• Gallon:
□ Definition and conversion to US Quart, Imperial Quart, Pint, Liter, and Cubic Centimeter
Understanding Pint and Gallon Measurements
The pint is a common unit of volume in both the US and Imperial systems. In the US, a pint is defined as 16 fluid ounces, while in the Imperial system, it equals 20 fluid ounces. To convert between
the two, simply multiply by a factor of 1.25.
For example, 1 US pint equals 1.25 Imperial pints, while 1 Imperial pint equals 0.8 US pints. The pint is often used to measure liquids such as beer, milk, and juice.
The gallon is a larger unit of volume, and it is used to measure liquids such as gasoline, water, and milk. In the US, a gallon is defined as 128 fluid ounces, while in the Imperial system, it equals
160 fluid ounces. To convert between the two, multiply by a factor of 1.25.
For instance, 1 US gallon equals 1.25 Imperial gallons, while 1 Imperial gallon equals 0.8 US gallons. The gallon is a convenient unit of measure for large quantities of liquid, and it is often used
in recipes, scientific calculations, and everyday measurements.
Metric Volume Measurements:
• Liter:
□ Definition and conversion to Gallon, Milliliter, and Cubic Centimeter
• Milliliter:
□ Definition and conversion to Fluid Ounce, Liter, and Cubic Centimeter
• Cubic Centimeter:
□ Definition and conversion to Liter, Milliliter, and Gallon
Metric Volume Measurements
In the realm of volume measurement, the metric system takes center stage. Let’s explore the three primary metric units: liter, milliliter, and cubic centimeter.
Liter (L): The Base Unit of Liquid Volume
The liter (L) serves as the fundamental unit of liquid volume in the metric system. It’s equivalent to the volume of a cube with sides measuring 10 centimeters. To convert liters to gallons, multiply
by 0.2642. To convert liters to milliliters, multiply by 1000. And to convert liters to cubic centimeters, multiply by 1000.
Milliliter (mL): A Smaller Unit for Smaller Volumes
When dealing with smaller volumes, the milliliter (mL) comes into play. One milliliter is equal to one thousandth of a liter (0.001 L). It’s a convenient unit for measuring liquids in quantities such
as milliliters or fluid ounces. To convert milliliters to fluid ounces, multiply by 0.0338. To convert milliliters to liters, divide by 1000. And to convert milliliters to cubic centimeters, multiply
by 1.
Cubic Centimeter (cm³): For Solids and Irregular Shapes
The cubic centimeter (cm³) measures the volume of three-dimensional objects, including solids and irregular shapes. It’s equal to the volume of a cube with sides measuring 1 centimeter. To convert
cubic centimeters to liters, divide by 1000. To convert cubic centimeters to milliliters, divide by 1. And to convert cubic centimeters to gallons, multiply by 0.0002642.
How Many Quarts in a Pitcher? A Comprehensive Guide to Volume Measurements
Have you ever wondered how many quarts are in a pitcher? The answer may surprise you, as there are different pitcher volumes and measurement unit conversions involved. In this blog post, we’ll delve
into the world of volume measurement units and provide a step-by-step guide to help you accurately calculate the number of quarts in a pitcher.
Calculating Quarts in a Pitcher:
To calculate the number of quarts in a pitcher, you can use the following formula:
Number of quarts = Volume of pitcher (in cups) ÷ 4
Example Calculation:
Let’s say you have an 8-cup pitcher. Using our formula:
Number of quarts = 8 cups ÷ 4
Number of quarts = 2
Therefore, an 8-cup pitcher holds two quarts of liquid.
Understanding volume measurement units is essential for various tasks, including cooking, baking, and household measurements. By following the conversion formula provided above, you can easily
calculate the number of quarts in a pitcher and ensure accurate measurements for your everyday needs.
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The Reductio ad absuTdum ("RAA"), Latin for "reduction to absurdity", seems very strange: If we can prove that false is true, then we can prove the negation of our premise. Huh!?l What on Earth does
it mean to prove that false is true?
This is known as proof-by-contradiction. We start by making a single unproven assumption. We then try to prove that false is true. Clearly, that it nonsense, so we must have done something wrong.
Assuming we didn't make any mistakes in the individual inference steps, then the only thing that could be wrong is the assumption. It must not hold. Therefore, we have just proven its negation.
This form of reasoning is often expressed via contrapositive. Consider the slogan
If you paid list price, you didn't buy it at SuperMegaMart.
(This is a contrapositive, because the real statement the advertisers want to make is that if you buy it at SuperMegaMart, then you won't pay list price.), which we'll abbreviate payFull ⇒¬
boughtAtSMM. You know this slogan is true, and you just made a SuperMegaMart purchase (boughtAtSMM), and are suddenly wanting a proof that you got a good deal. Well, suppose we didn't. That is,
suppose payFull. Then by the truth of the marketing slogan, we infer ¬boughtAtSMM. But this contradicts boughtAtSMM (that is, from ¬boughtAtSMM and boughtAtSMM together we can prove that false is
true). The problem must have been our pessimistic assumption payFull; clearly that couldn't have been true, and we're happy to know that ¬payFull.
Example 2.17
Spot the proof-by-contradiction used in The Simpsons:
Bart, fling through the school records: " Hey, look at this: Skinner makes $25,000 per year! "
Other kids: "Ooooh!"
Milhouse: "And he's 40 years old; that makes him a millionaire! "
Skinner, indignantly: " I wasn't a principal when I was 1!"
Milhouse: "And, he paints houses during the summer ... he's a billionaire!"
Skinner: "If I were a billionaire, would I still be living with my mother? " [Kids' laughter]
Skinner, to himself: " The kids just aren't responding to logic anymore! "
In the particular set of inference rules we have chosen to use, RAA is surprisingly important. It is the only way to prove formulas that begin with a single "¬".
Example 2.18
We'll prove
│ │ │ │
│ 1.a │ │ α ∧ ¬α │ Premise for subproof │
│ 1.b │ │ α │ ∧Elim (left), line 1.a, where φ = α, and ψ = ¬α │
│ 1.c │ │ ¬α │ ∧Elim (right), line 1.a, where φ = α, and ψ = ¬α │
│ 1.d │ │ false │ falseIntro, lines 1.b,1.c, where φ = α │
│ 2 │ ¬ (α ∧ ¬α) │ RAA, line 1, where φ = α ∧ ¬α │
Exercise 2.4.2.1
Here's another relatively simple example which uses RAA. Show that the modus tollens rule holds:
Another use of subproofs is to organize proofs' presentations. Many proofs naturally break down into larger subparts, each with its own intermediate conclusion. These steps between these subparts are
big enough to correspond to our intuition, but too big to correspond to individual inference rules. This gives additional useful structure to a proof, aiding our understanding.
Example 2.19
Previously, we showed that ∧ (AND) commutes (Example 2.14). However, that conclusion is only directly applicable when the ∧ is at the "top-level", i.e., not nested inside some other connective. Here,
we'll show that ∧ commutes inside ¬, or more formally,
Caution: When doing inference-style proofs, we will not use the Boolean algebra laws nor replace subformulas with equivalent formulas. Conversely, when doing algebraic proofs, don't use inference
rulesl While theoretically it's acceptable to mix the two methods, for homeworks we want to make sure you can do the problems using either method alone, so keep the two approaches separate!
We'll do two proofs of this to illustrate that there's always more than one way to prove something!
In our first proof, we'll use RAA. Why? Looking at our desired conclusion, what could be the last inference rule used in the proof to reach the conclusion? By the shape of the formula, the last step
can't use any of the "introduction" inference rules (∧Intro, ∨Intro, ⇒Intro, falseIntro, or ¬Intro). We could potentially use any of the "elimination" inference rules. But, for ∧Elim, ∨Elim, ⇒Elim,
¬Elim, or CaseElim, we would first have to prove some more complicated formula to obtain our desired conclusion. That seems somewhat unlikely or unnecessary. For falseElim, we'd have to first prove
false, i.e., obtain a contradiction, but our only premise isn't self-contradictory. The only remaining option is RAA.
│ 1 │ ¬ (α ∧ β) │ Premise │
│ 2 │ subproof: │ │
│ 2.a │ │ β ∧ α │ Premise for subproof │
│ 2.b │ │ α ∧ β │ Theorem: ∧ commutes (Example 2.14), line 2a │
│ 2.c │ │ false │ falseIntro, lines 1,2.b │
│ 3 │ ¬ (α ∧ β) │ RAA, line 2 │
The proof above uses a subproof because it is necessary for the use of RAA. In contrast, the proof below uses two subproofs simply for organization.
For our second proof, let's not use RAA directly. Our plan is as follows:
• Assume the premise ¬ (α ∧ β).
• Again, use commutativity to show that β ∧ α ⇒ α ∧ β
• Use modus tollens (Exercise 2.4.2.1) to obtain the conclusion.
We can organize the proof into corresponding subparts:
│ 1 │ ¬ (α ∧ β) │ │ Premise │
│ 2 │ subproof:β ∧ α ⇒ α ∧ β │ │ │
│ 2.a │ │ │ Theorem statement: ∧ commutes (Example 2.14) │
│ 2.b │ │ │ ⇒Intro, line 2.a │
│ 3 │ subproof:¬ (β ∧ α) │ │ │
│ 3.a │ │ │ Theorem statement: modus tollens (Exercise 2.4.2.1) │
│ 3.b │ │ │ ⇒Intro, line 3.a │
│ 3.c │ │ │ ∧Intro, lines 2,1 │
│ 3.d │ │ │ ⇒Elim, lines 3.b,3.c │ | {"url":"https://www.opentextbooks.org.hk/zh-hant/ditatopic/9552","timestamp":"2024-11-12T08:47:31Z","content_type":"text/html","content_length":"217822","record_id":"<urn:uuid:ffd0cfe7-3f1f-49ec-aa31-377151c93842>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00360.warc.gz"} |
Probability of a single event | slideum.com
Probability of a single event
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Transcript Probability of a single event
Probability of a single event
Friday, July 17, 2015
John has these 5 coins.
John is going to take one of these coins at random.
Each coin is equally likely to be the one he takes.
a) What is the probability that it will be a 10p coin he selects?
b) What is the probability that it will be a £1 coin he selects?
c) What is the probability that it will be a 1p coin he selects?
A spinner has seven equal sections.
(a) What is the probability of scoring 4 on the spinner?
(b) What is the probability of scoring an even number on the spinner?
(c) What is the probability of scoring a prime number on the spinner?
A single card is drawn from a pack of 52 playing cards. Find the probability of the
card being:
a Queen
a club
the Jack of hearts
an even number.
a picture card
The set of lottery balls are placed into the drum for the Saturday draw.
What is the probability that the first ball out of the drum is
the number 17
b) an even number
a prime number
d) a factor of 20
e) a multiple of 5?
A letter is selected at random from the word
the probability that it is:
(ii) a consonant
(iii) a m?
Mathematics. What is
A bag contains 9 white balls, 8 green balls and 3 blue balls. One ball is selected
at random. What is the probability that the ball is
not blue?
A bag contains some counters of various colours. A counter is taken at random
from the bag and the table below shows the probability of the counter being
red, green, yellow, or blue.
Work out the probability of the counter being blue.
b) Work out the probability of the counter being red or green.
The table below shows the eating arrangements for some 150 students.
Complete the table
Eats out
One student is selected at random.
a) What is the probability that the student selected
(i) Has a school lunch?
(ii) Is male and eats out at lunchtime?
(iii) Is female?
b) Given that it was a male selected, what is the probability that they brought a
packed lunch to school?
The Local supermarket sends a questionnaire to a random sample of 40 workers during a
week in December.
One of the questions is:
How many hours did you work last week?
The results are given in the table below.
Hours worked in week t
25 ≤ 𝑡 ≤ 30
30 ≤ 𝑡 ≤ 35
35 ≤ 𝑡 ≤ 40
40 ≤ 𝑡 ≤ 45
45 ≤ 𝑡 ≤ 50
50 ≤ 𝑡 ≤ 55
Number of staff
Based on the sample below if a member of staff in the supermarket is selected at random
what is the probability that the member
a) Worked between 30 and 35 hours last week?
b) Worked over 40 hours last week? | {"url":"https://slideum.com/doc/3155448/probability-of-a-single-event","timestamp":"2024-11-04T08:08:13Z","content_type":"text/html","content_length":"27068","record_id":"<urn:uuid:c212aec3-4660-43d1-9108-2b51fc4b9d7a>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00156.warc.gz"} |
Lecture 021 - Quantum for Classical
Strong Exponential Time Hypothesis: no less than $O(2^n)$ algorithm for SAT (proved only with blackbox model)
Quantum Strong Exponential Time Hypothesis: no less than $O(2^{n/2})$ quantum algorithm for SAT (proved only with blackbox model)
Grover: solve SAT with high probability using $2^{n/2}c$ instruction given AND/OR/NOT circuit $C$ with length $c$.
Bernstein-Vazirani (XOR): $O(c)$ with $100\%$. (while classical need $O(nc)$), verifiable output, unverifiable input
Simon's Algorithm: not a natural problem, unverifiable input
Quantum Factoring: polynomial (while classical need $O(\exp(n^{1/3}))$), verifiable
• Decision Factoring: whether there exist a factor of $F$ in range $[2, k]$
• when decision factoring return "Yes": verifiable
• when decision factoring return "No": still verifiable (unlike SAT)
If we assume $NP eq coNP$, then $\text{NP-Complete}$ problems should not have such "no-witness". And therefore, factoring is not $\text{NP-Complete}$.
Bias-Busting: if code $C$ biased then $Pr\{0...0\} > 0$, else $Pr\{0...0\} = 0$.
• $SAT \leq_p \text{Bias-Busting}$ (so it is NP-Hard)
• If satisfiable, then $C$ must be biased. If unsatisfiable, then $C$ must be unbiased.
• But $Pr\{\} > 0$ is too weak to solve $NP-Hard$ problems.
• $\text{Bias-Busting}$ probably not in $NP$, since otherwise, a witness will be similar to Bias-Busting itself.
Toda-Ogiwara Theorem: Bias-Busting not in NP assuming $P^{\Sigma_2} eq NP^{\Sigma_2}$ | {"url":"https://kokecacao.me/page/Course/F22/15-459/Lecture_021_-_Quantum_for_Classical.md","timestamp":"2024-11-06T14:39:37Z","content_type":"text/html","content_length":"10029","record_id":"<urn:uuid:0441813a-c084-4f56-96da-38c24d753330>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00438.warc.gz"} |
MaffsGuru.com - Making maths enjoyableDetermining the slope of a straight line
This is the next video in the Linear part of the General Maths Units 1 and 2 course. Having spent some time looking at how to plot straight lines in the previous video, I now look at how we can find
the gradient of a straight line using the rise over run formula as well as another more interesting (and definitely more funky) formula. We look at positive, negative, zero and undefined gradients
and what they mean with lots of worked examples all explained in my own unique way!
LEGAL STUFF (VCAA)
VCE Maths exam question content used by permission, ©VCAA. The VCAA is not affiliated with, and does not endorse, this video resource. VCE® is a registered trademark of the VCAA. Past VCE exams and
related content can be accessed at | {"url":"https://maffsguru.com/videos/determining-the-slope-of-a-straight-line/","timestamp":"2024-11-14T07:43:04Z","content_type":"text/html","content_length":"34561","record_id":"<urn:uuid:472c68ca-9970-4d14-abff-573af80afe18>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00492.warc.gz"} |
IEIE Transactions on Smart Processing & Computing
Since the advent of the global computerized market, the volume of digital information has grown exponentially, as has the demand for storing it. As the price of storage devices decreases, the
necessity to analyze vast quantities of unstructured digital data to retain only essential information increases. MapReduce is a programming paradigm for producing and generating massive information
indices. Using MapReduce to produce meaningful clusters from such a massive amount of raw data is an efficient way to manage such voluminous amounts of data. On the other hand, the existing industry
standard for data clustering algorithms presents significant obstacles. The conventional clustering calculation efficiently handles a great deal of information from various sources, such as online
media, business, and the web. Nevertheless, the sequential count in clustering approaches is time-intensive in these conventional calculations. The wide varieties of K-Means, including K-Harmonic
Means, are sensitive to forming cluster centers in huge datasets. This work suggests a logical evaluation of such calculations. It offers a study of the various k-means clustering algorithms employed
in MapReduce, as well as the study on the introduction and the open challenges of parallelism in MapReduce. | {"url":"http://ieiespc.org/ieiespc/XmlViewer/f415664","timestamp":"2024-11-05T05:51:10Z","content_type":"application/xhtml+xml","content_length":"305373","record_id":"<urn:uuid:74ce260e-a798-458a-b547-1856a651f4c4>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00643.warc.gz"} |
University Catalog 2009-2011 [ARCHIVED CATALOG]
Mathematics Credential Program
The Single Subject Matter program listed below has been approved by the California Commission on Teacher Credentialing for the Single Subject Credential in Mathematics. In addition to
consulting the credential adviser for mathematics, students should consult advisers in the Charter College of Education and refer to the Charter College of Education section for regulations
governing all teaching credential programs.
Students who are seeking a Single Subject Credential in Mathematics must pass the appropriate subject examination (CSET Mathematics I-III) or complete the approved program of course work that
is listed below. Students who are pursuing a baccalaureate in mathematics follow the Single Subject Teaching Option, which incorporates the courses listed below. Others who have already
earned or are currently pursuing a baccalaureate in another discipline may qualify for the Single Subject Credential in Mathematics by completing the courses listed below or equivalent course
Subject Matter Program for Single Subject Credential in Mathematics
It is assumed that students entering this program have completed one course in college algebra and one in trigonometry (MATH 102 and MATH 103 ). Competence in these courses can also be shown
by taking the departmental exit exam.
Electives (minimum 8 units)
Select from among the following or other appropriate courses in mathematics or related areas with adviser approval and attention to prerequisites.
Strongly Recommended: MATH 310; MATH 466 for those who may be teaching Advanced Placement calculus classes.
Supplementary Authorization for Single or Multiple Subject Teaching Credential (30–33 units)
Holders of a Single Subject or Multiple Subject credential issued by the California Commission on Teacher Credentialing may secure a supplementary authorization in Introductory Mathematics
(on single subject credentials) or Mathematics (on multiple subject credentials) for teaching mathematics at any grade level through grade 9 by completing the following courses with a grade
of C or higher in each course. Note that this supplementary authorization is not NCLB (No Child Left Behind) compliant, but that some school districts may hire candidates with a
supplementary authorization on the condition that the candidate will work toward the Subject Matter Authorization in Introductory Mathematics (see below). For other requirements governing
issuance of this authorization, consult the Charter College of Education.
Complete or demonstrate proficiency in each of the following courses (30–33 units):
Required Courses (16 units)
Select three courses from the following (12 units)
Select one course from the following (2 or 5 units)
*prerequisite: MATH 325
**prerequisite: MATH 209
Subject Matter Authorization in Introductory Mathematics for Single or Multiple Subject Teaching Credential (48 units)
Holders of a Single or Multiple Subject Teaching Credential issued by the California Commission on Teacher Credentialing (CCTC) may add a Subject Matter Authorization in Introductory
Mathematics. This allows the holder of the Subject Matter Authorization to teach mathematics curriculum usually taught in grades 9 and below (even though the students may be in grades K-12).
To obtain a Subject Matter Authorization in Introductory Mathematics (which satisfies the federal “No Child Left Behind” (NCLB) regulation), a total of 48 quarter units (=32 semester units)
of course work applicable toward a bachelor’s degree must be completed with a grade of C or better. A minimum of 4 quarter units of course work must be completed in each of the following core
areas: Algebra; Advanced Algebra; Geometry; Probability or Statistics; and Development of the Real Number System or Introduction to Mathematics.
The core courses in the program below have been designed for students who have not taken any college level mathematics course. Students placing into a mathematics course at a level beyond
MATH 102 should consult with the credential adviser in mathematics to select a different set of core courses. Additional information is available by downloading the CCTC guide for subject
matter authorization (www.ctc.ca.gov/credentials/manualshandbooks/subjectmatter-auth.pdf) or through the credential adviser for mathematics. For other requirements governing issuance of this
authorization, consult the Charter College of Education.
Complete 48 units of Coursework:
The following five recommended core courses will satisfy the core area requirements. Alternative sets of course work may also be used to meet the core area requirements. Proper academic
advisement is essential prior to the start of this authorization program.
Elective Courses
Select courses as needed to reach a total of 48 units of coursework. MATH 248 and MATH 225 are highly recommended.
* Prerequisite: MATH 207
** Prerequisite: MATH 208
*** Prerequisite: MATH 209
+ Prerequisite: MATH 325
++ Prerequisite: MATH 208 and 248 | {"url":"https://ecatalog.calstatela.edu/preview_program.php?catoid=1&poid=299&returnto=37","timestamp":"2024-11-13T00:20:53Z","content_type":"text/html","content_length":"111790","record_id":"<urn:uuid:4b6d94db-3fc5-4a57-88be-eba969e098d4>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00712.warc.gz"} |
International Journal of Plasticity 137 (2021) 102902
Available online 9 November 2020
0749-6419/© 2020 Elsevier Ltd. All rights reserved.
A thermodynamics-based hyperelastic-plastic coupled model
unied for unbonded and bonded soils
Zhichao Zhang
, Linhang Li
, Zhenglong Xu
Chongqing University, School of Civil Engineering, Chongqing, 400045, China
Key Laboratory of New Technology for Construction of City in Mountain Area (Chongqing University), Ministry of Education, Chongqing, 400045,
Elastic-plastic coupling
Cohesion degradation
Stress-induced anisotropy
A hyperelastic-plastic coupled constitutive model unied for bonded and unbonded soils is
developed in this paper based on thermodynamics. An elastic potential function applicable for
different kinds of soils is proposed to derive a hyperelastic model accounting for the pressure- and
-density dependency, the stress-induced anisotropy and the bonding effects as well as their
couplings with plasticity. From the perspective of elastic stability, state boundary and failure
surfaces of different soils can be naturally predicted by the hyperelasticity without any additional
denitions and parameters. Based on the classical nonequilibrium thermodynamics, novel plastic
constitutive relations are derived and naturally coupled with the hyperelasticity. As a result,
elasto-plastic coupling features such as the dissipative history effect on elastic stiffness, the cyclic
shear behavior, the degradation of shear modulus under small strain conditions, the stress-
induced anisotropy of plastic behavior and the cohesion degradation can be reproduced. The
model is well validated by predicting the undrained/drained monotonic and cyclic shear behavior
of unbonded and bonded sands, providing useful insights into their critical state behavior, irre
versible shear-dilation/contraction and effects of bonding and cohesion degradation. It is also
shown that the cohesion degradation in different shearing stages to a large extent determines both
the monotonic and cyclic behavior of bonded soils.
1. Introduction
In the elds of geosciences and geotechnical engineering, it is important to study the nonlinear elastic-plastic coupling behavior of
geomaterials such as granular soils, clays and rocks. Experimental results of small strain and bender element tests show that the elastic
moduli of soils are a function of both the conning pressure and the void ratio (Man et al., 2010; Gu et al., 2013). Laboratory study by
Giang et al. (2017) indicated that the small-strain shear modulus of calcareous sand could also be signicantly inuenced by the
particle shape and gradation. State dependent elastic properties of articially bonded soils are also studied by Lee et al. (2011),
Morozov and Deng (2018), Nasi (2019) and so on. On the other hand, such state-dependent elastic properties can be impacted by the
plasticity of soils and vice versa (Lashkari and Golchin, 2014) so that the coupling between elasticity and plasticity becomes essential
for soils. For example, Ezaoui and Benedetto (2009) measured the elastic stiffness evolution of Hostun sand, indicating signicant
couplings between the elastic stiffness and the plastic strain histories. More recently, experiment results by Khosravi et al. (2018)
* Corresponding author. Chongqing University, School of Civil Engineering, Chongqing, 400045, China.
E-mail addresses: zczhang15@cqu.edu.cn, zczhang15@cqu.edu.cn (Z. Zhang).
Contents lists available at ScienceDirect
International Journal of Plasticity
journal homepage: http://www.elsevier.com/locate/ijplas
Received 15 June 2020; Received in revised form 4 November 2020; Accepted 4 November 2020 | {"url":"https://download.csdn.net/download/huanghm88/89883988","timestamp":"2024-11-02T17:36:57Z","content_type":"text/html","content_length":"935775","record_id":"<urn:uuid:a913cec9-bd12-46c4-8b58-b9ec07c3f18a>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00739.warc.gz"} |
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• nth term calculator | {"url":"https://softmath.com/math-book-answers/perfect-square-trinomial/free-applets-for-factoring.html","timestamp":"2024-11-04T10:56:19Z","content_type":"text/html","content_length":"35754","record_id":"<urn:uuid:cc42337f-738e-4d27-8b27-679f264b418a>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00450.warc.gz"} |
Parent Health Status of Children Rows Based On If Certain Criteria Is Met
I have a Gender column with these options:
Female, Male, Transgender, Non-binary/non-conforming, Prefer not to respond
I need the parent row to reflect a status of RED ball if there is no 'Female' represented in any of the children.
I currently have this formula, but feel I'm missing significant COUNTIF references:
=IF((CONTAINS(FEMALE)(CHILDREN)), "GREEN", "RED")
How can I course correct this formula? THANK YOU!!
• If you are using hierarchy, you can do the following formula: =IF(COUNTIF(CHILDREN(), "Female") = 0, "Red")
If you are not using hierarchy and have a range, you can use:
=IF(COUNTIF(Gender10:Gender15, "Female") = 0, "Red")
• @Maricel Medina , would this only work on the first level of hierarchy? I've got 3 levels and need it to work on the 2nd one. Also, the column was originally formatted as a drop down list - but
it appears the formula won't work in a parent cell if that column is formatted as such? But even after I remove the drop down list format, it doesn't render a ball for me.
To clarify, the intent is to trigger red when there are no female candidates being considered. I'll need to write a similar formula for ethnicity as well - if there are only white candidates,
then I need a red ball; otherwise a green one.
I'm using the formula (image 1), but the results are null (image 2).....
Image 1 - formula displayed:
Image 2 - no red ball:
@Andrée Starå .... don't suppose you'd take a gander at this one since I've already bugged you today? ;P
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/70895/parent-health-status-of-children-rows-based-on-if-certain-criteria-is-met","timestamp":"2024-11-10T02:28:40Z","content_type":"text/html","content_length":"401630","record_id":"<urn:uuid:5b5d4174-cc2b-4fed-9abb-0d6e8055422a>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00495.warc.gz"} |
Directed Reading Program
When? Where? More information can be found on the page for the current semester (Fall 2024).
What is it? The Directed Reading Program (DRP) in the UW Madison Department of Mathematics pairs undergraduate students with graduate mentors for semester-long independent studies. During the
semester, the student will work through a mathematical text and meet weekly to discuss it with their mentor. The original DRP was started by graduate students at the University of Chicago in 2003,
and has had immense success. It has since spread to many other math departments who are members of the DRP Network.
Why be a student?
• Learn about exciting math from outside the mainstream curriculum!
• Prepare for future reading and research, including REUs!
• Meet other students interested in math!
Why be a mentor?
• Practice your mentorship skills!
• It strengthens our math community!
• Solidify your knowledge in a subject!
Current Organizers: Ivan Aidun, Allison Byars, Jake Fiedler, John Spoerl
At least one hour per week spent in a mentor/mentee setting. Students spend about two hours a week on individual study, outside of mentor/mentee meetings. At the end, students give a 10-12 minute
presentation at the end of the semester introducing their topic.
How to apply
Application links can be found on the page for the current semester (Fall 2024). For project ideas, you may find it helpful to view the past semesters below.
Contact us at drp-organizers@g-groups.wisc.edu
Past Semesters
Directed Reading Program Spring 2024
Directed Reading Program Fall 2023
Directed Reading Program Spring 2023 | {"url":"https://wiki.math.wisc.edu/index.php/Directed_Reading_Program","timestamp":"2024-11-03T07:32:02Z","content_type":"text/html","content_length":"19394","record_id":"<urn:uuid:93e3d6ed-10e8-4522-861b-4bb4346e9992>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00626.warc.gz"} |
IF Function not returning as expected
I have an IF function I am trying to run that evaluates a cell, and then returns one of two indexing formula results
I have verified that each indexing function and the If function itself with different results all seem to work.
But when I plug them together, my IF function stops evaluating "3100" properly, or it will pull up the Zip result for everyone rather than LOB result for some depending on what PRIN@row says.
As background, PRIN is a dropdown column with four options, one of which I need LOB codes for, two of which I need Zips codes for and one of which is NA so I figured that targeting the most specific
one would work best for the IF function.
Thanks for any help in advance!
• You have too many parenthesis spread throughout the formula. If you can copy/paste it here, I would be happy to help clean them up.
• Here is plaintext
=IF(CONTAINS(3100, PRIN@row), (INDEX({ZIPS ALL}, (MATCH([Job Type]@row, {ZIPS LOB}, 0)), 4)), (INDEX({ZIPS ALL}, (MATCH(ZIP@row, {ZIPS ZIP}, 0)), 4)))
I did notice many examples did not put nested formulas in parentheses and many did. Is there a specific rule that I am missing for when this is needed?
• For the parenthesis... I try to use as few as absolutely possible simply because they can get out of control very quickly. The thing to remember is every open parenthesis needs a closed
parenthesis. Every single FUNCTION will have an open and closed. Other than that you only really need to worry about them when a certain order is needed but not necessarily specified such as
running math equations (follows P.E.M.D.A.S.).
=IF(CONTAINS("3100", PRIN@row), INDEX({ZIPS ALL}, MATCH([Job Type]@row, {ZIPS LOB}, 0), 4), INDEX({ZIPS ALL}, MATCH(ZIP@row, {ZIPS ZIP}, 0), 4))
I also notice that it looks like you are referencing column 4 in your INDEX function.
INDEX({range to pull from}, row number, column number)
In the above, the MATCH provides the dynamic row number. Is there a reason your INDEX range is covering so many columns? You should be able to only select a single column that you want to pull
from. It is very different from VLOOKUP in that (and a number of other) way(s).
• Thank you! I agree, parentheses can get out of control very quickly. I have stared and adjusted this one so long I am sure there were one or two runaways.
As far as why so many, I need to basically match a Job number in another datasheet to the cell/row, then the Prin within the data sheet I am working in, then either the zip or job type in a third
helper sheet that consolidates all of the info of who is assigned to what, to return the scheduler.
It's a lot and not how I would prefer to set things up, but I am working within the confines of an existing system.
• I understand needing to look at different sheets, but what I mean is each of the individual INDEX functions.
INDEX({ZIPS ALL}, MATCH([Job Type]@row, {ZIPS LOB}, 0), 4)
That first range really only needs to be the column you are pulling from. Then you wouldn't need the 4 there at the end. It doesn't make much difference as far as functionality goes, but if the
sheet is really busy with a lot of formulas/cross sheet references, the fewer cells you can reference the better performance you will have on the back-end. It also allows you to rearrange the
source sheet without having to worry about messing up a formula since you are tracking a specific column as opposed to a column number.
• As I see, I need that because I need one index to reflect if true and one to reflect if false.
I tried plugging in the cleaned up formula and I still end up with the same issue. The "If" statement doesn't seem to be functioning properly. It still just appears to be matching from the zip
code column of my reference sheet, regardless of the PRIN number listed as the value in the if function.
• You have both INDEX functions pulling from the same column. The only difference between the two is the first is matching on [Job Type], and the second is matching on ZIP. The MATCH range does not
have to be included in the INDEX range. What I am saying is you can takes that {ZIPS ALL} range and make it a single column (the one you want to pull from) instead of multiple columns which will
make the overall setup more flexible and less prone to break as well as more efficient on the back-end.
As for the IF statement... Are you able to provide a screenshot where you are expecting the CONTAINS to be true which would cause it to match on the Job Type?
• Edit: I rearranged my data sheet and have updated screenshots as well as the sheet I am trying to pull data into
Hi, Here is a screenshot of the reference data sheet:
And this is the sheet I am pulling into with the formula you fixed for me plugged in
Now it is saying no match since I rearranged and re-established my formula references
Before, it was pulling info based on zip code regardless of prin. Anything in 3100 should be pulling by LOB.
I hope that makes more sense.
Thanks again for your help!
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/98408/if-function-not-returning-as-expected","timestamp":"2024-11-05T03:42:17Z","content_type":"text/html","content_length":"438861","record_id":"<urn:uuid:16ad3af5-7aba-4208-ac84-f252feddbc8f>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00457.warc.gz"} |
A Product of Tensor Product L-functions of Quasi-split Classical Groups of Hermitian Type
A family of global zeta integrals representing a product of tensor product (partial) L-functions: (Formula presented.)is established in this paper, where π is an irreducible cuspidal automorphic
representation of a quasi-split classical group of Hermitian type and τ[1],...,τ[r] are irreducible unitary cuspidal automorphic representations of GL[a1],...,GL[ar], respectively. When r = 1 and the
classical group is an orthogonal group, this family was studied by Ginzburg et al. (Mem Am Math Soc 128: viii+218, 1997). When π is generic and τ[1],...,τ[r] are not isomorphic to each other, such a
product of tensor product (partial) L-functions is considered by Ginzburg et al. (The descent map from automorphic representations of GL(n) to classical groups, World Scientific, Singapore, 2011) in
with different kind of global zeta integrals. In this paper, we prove that the global integrals are eulerian and finish the explicit calculation of unramified local zeta integrals in a certain case
(see Section 4 for detail), which is enough to represent the product of unramified tensor product local L-functions. The remaining local and global theory for this family of global integrals will be
considered in our future work.
Bibliographical note
Funding Information:
Keywords and phrases: Bessel periods of Eisenstein series, global zeta integrals, tensor product L-functions, classical groups of Hermitian type Mathematics Subject Classification: Primary 11F70,
22E50; Secondary 11F85, 22E55 The work of D. Jiang is supported in part by the NSF Grants DMS-1001672 and DMS-1301567.
• 22E50
• 22E55
• Bessel periods of Eisenstein series
• Primary 11F70
• Secondary 11F85
• classical groups of Hermitian type
• global zeta integrals
• tensor product L-functions
Dive into the research topics of 'A Product of Tensor Product L-functions of Quasi-split Classical Groups of Hermitian Type'. Together they form a unique fingerprint. | {"url":"https://experts.umn.edu/en/publications/a-product-of-tensor-product-l-functions-of-quasi-split-classical-","timestamp":"2024-11-04T21:14:28Z","content_type":"text/html","content_length":"54809","record_id":"<urn:uuid:7783f5ed-bef6-4350-8058-8a8533d2a327>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00242.warc.gz"} |
Semi analytical solution of a rigid pavement under a moving load on a Kerr foundation model
This paper analyzes the dynamic response of assumedly rigid road pavement under a constant velocity of traffic loads moving on its surface. The model of the rigid road pavement is a damped
rectangular orthotropic plate which is supported by an elastic Kerr foundation. Semi-analytical solutions of the dynamic deflection of an orthotropic plate, with semi rigid boundary conditions are
presented by using governing differential equations. The natural frequencies and mode shapes of the system are then, solved by using the modified Bolotin method considering two transcendental
equations as the results of solving the solution of two auxiliaries Levy’s plate type problems. The moving traffic loads modeled by varying the amplitudes of dynamic transverse concentrated loads
harmonically. Numerical studies on the soil types, foundation stiffness models, varying constant velocities and loading frequencies are conducted to show the effects of the dynamic response behaviors
of the plates. The results show that the dynamic responses of the rigid road pavement influenced significantly by the type foundation stiffness models and velocity of the moving load.
1. Introduction
The vibration response of rectangular orthotropic plates is an interesting subject because of its widespread applications in structural engineering and transportation engineering. In bridge analysis,
different models have been studied and investigated by researchers for rigid highway and airport pavements. In most of the previous works the type of plates considered are isotropic rectangular
plates which are uniform in all directions. In reality, not all plates are isotropic. Another important type of plate is the orthotropic rectangular plate, which has been used to model the dynamic
response of rigid concrete pavements. According to Alisjahbana and Wangsadinata, the dynamic moving traffic load can be represented by a single concentrated harmonic loading, moving with a constant
speed along the mid-side of the plate [1]. It was found that a dynamic load approach will lead to a better economical solution in comparison to solutions obtained using the conventional static load
Conventional methods of rigid pavement design are using the elastic Winkler foundation model which is obtained from the static analytical solutions of infinite plates rest on elastic soil assumption.
These were investigated by Westergaard in 1926 [2]. In this elastic Winkler foundation model, the interconnections among the soil layers are neglected, leading to limitations in the physical model of
the sub-soil system [3]. These limitations can be eliminated by modeling the sub-grade soil medium by using two-parameter model, providing a shear interaction between independent spring elements.
Several researchers have verified the applicability of soil medium representation by using two parameter models in static [4, 5], post buckling [6, 7] as well as dynamic models [1, 8-10]. Gan and
Nguyen presented the two parameter model of soil medium in the large deflection analysis of functionally graded beam [11].
Paliwal and Ghosh studied the stability of rectangular orthotropic plates on a Kerr soil foundation model subjected to the in-plane static stresses in the orthogonal directions [12]. The dynamic
lateral loads are not discussed in their work. The Kerr model is one of the most advantageous models. However, due to the existence of an upper spring layer, no concentrated reactions occur. Kneifati
has shown that more accurate base response of the flexible plates and beams subjected to a uniform load and boundary forces was obtained by using the Kerr model compared to the Pasternak and Winkler
models [13]. In addition, the Kerr model moreover shows comparable results with the continuum elastic theory. Therefore, in this paper, the authors will analyze the dynamic behavior of the rigid road
pavement rests on a Kerr model and subjected to a moving load.
In most researches done previously, rigid road pavements are modeled as an orthotropic plate. Furthermore, it is commonly resting on an elastic foundation model such as the Pasternak or Winkler
models. However, according to Paliwal and Ghosh in 1994, the Winkler model is unable to represent the behavior of soil medium materials with a larger void ratio or stiffer clay. Additionally, the
Pasternak model is only better in predicting the behavior of hard soils. On the other hand, Kerr foundation model has more advantageous models due to no concentrated reactions occur due to an
addition of an upper spring layer [12].
In this study, the dynamic response of a rigid road pavement modeled as a thin orthotropic plate rests on the Kerr model is investigated. To take into account the existence of the tie bars and dowels
along its edges, the boundary conditions along its edges are semi rigid condition allowing the rotation at the supports and the translational movement at the edges. These type of boundary conditions
are solved by using the Modified Bolotin Method [1]. There has been no previous research using this specific method to study the dynamic response of the rigid road pavement subjected to a moving
load. In this paper, semi analytical solution is used to calculate the dynamic deflection and internal forces distribution of the plate subjected to loading with constant velocity. The applicability
of the present method is highlighted by solving the maximum dynamic deflection of the system for different types of soil conditions and elastics foundation model in order to design better rigid road
2. Governing equation
In this paper, a rigid pavement resting on Kerr foundation is considered. The orthotropic plate is semi rigid along its edges. It is considered to be of uniform thickness $h$. The dynamic transverse
load acting on the orthotropic plate is $q\left(x,y\right)$. Based on the work of Paliwal and Ghosh [12], the governing differential equation of the rigid pavement subjected to the lateral load is
given by:
${D}_{x}\frac{{\partial }^{4}w}{\partial {x}^{4}}+2b\frac{{\partial }^{4}w}{\partial {x}^{2}\partial {y}^{4}}+{D}_{y}\frac{{\partial }^{4}w}{\partial {y}^{4}}=q-{p}_{1},$
where ${D}_{x}$, ${D}_{y}$ are the flexural rigidities of the plate in the $x$ direction and the $y$ direction respectively, $B$ is the torsional rigidity of a plate and ${p}_{1}$ is foundation
response. Because the Kerr model [14] consists of two axial springs (${k}_{1}$ and ${k}_{2}$) and shear spring layer (${G}_{s}$), the deflection of the plate can be given as [13]:
The contact pressures under the orthotropic plate and the shearing layer are given by ${p}_{1}$ and ${p}_{2}$, respectively, where:
The shearing layer was governed by the following differential equation:
${k}_{2}{w}_{2}-{G}_{s}{abla }^{2}{w}_{2}={p}_{1}.$
Eliminating ${w}_{2}$ from Eq. (3) and (5), we obtain:
$\left(1+\frac{{k}_{2}}{{k}_{1}}\right){p}_{1}-\frac{{G}_{s}}{{k}_{1}}{abla }^{2}{p}_{1}={k}_{2}w-{G}_{s}{abla }^{2}w.$
By substituting ${p}_{1}$ of Eq. (3) into Eq. (6) and by taking into account the moving load, the structural damping and the inertia of the orthotropic plate, the differential equation of lateral
motion of an orthotropic plate on a Kerr model can be obtained as:
$-\left(1+\frac{{k}_{2}}{{k}_{1}}\right)\left({D}_{x}\frac{{\partial }^{4}w}{\partial {w}^{4}}+2B\frac{{\partial }^{4}w}{\partial {x}^{2}\partial {y}^{2}}+{D}_{y}\frac{{\partial }^{4}w}{\partial {y}^
{4}}+\rho h\frac{{\partial }^{2}w}{\partial {t}^{2}}+\gamma h\frac{\partial w}{\partial t}-q\left(x,y,t\right)\right)$$+\frac{{G}_{s}}{{k}_{1}}\left({D}_{x}\left(\frac{{\partial }^{6}w}{\partial {x}^
{6}}+\frac{{\partial }^{6}w}{\partial {x}^{4}\partial {y}^{2}}\right)+2B\left(\frac{{\partial }^{6}w}{\partial {x}^{4}\partial {y}^{2}}+\frac{{\partial }^{6}}{\partial {x}^{2}\partial {y}^{4}}\right)
{D}_{y}\left(\frac{{\partial }^{6}w}{\partial {y}^{6}}+\frac{{\partial }^{6}w}{\partial {x}^{2}\partial {y}^{2}}\right)\right)$$+\frac{{G}_{s}}{{k}_{1}}\left(\frac{{\partial }^{2}}{\partial {x}^{2}}+
\frac{{\partial }^{2}}{\partial {y}^{2}}\right)\left(\rho h\frac{{\partial }^{2}w}{\partial {t}^{2}}+\gamma h\frac{\partial w}{\partial t}-p\left(x,y,t\right)\right)={k}_{2}w-{G}_{s}\left(\frac{{\
partial }^{2}w}{\partial {x}^{2}}+\frac{{\partial }^{2}w}{\partial {y}^{2}}\right),$
where ${k}_{1}$ is the spring stiffness of the first layer of the Kerr model, ${k}_{2}$ is the spring stiffness of the second layer of the Kerr model, ${G}_{s}$ is the shear modulus of the Kerr
model, $\rho$ is the mass density of the plate, $\gamma$ is the structural damping ratio, and $h$ is the thickness of the plate.
In reality, loads that are caused by vehicles are often of varying amplitude. This is a because of the coarseness of rigid roadway pavement as well as the vehicle’s mechanical systems. Therefore, in
practical analysis, a harmonic load model is generally used. In this study, harmonically moving a single concentrated load which is traveling with a constant velocity ${v}_{0}$ along the middle line
is considered. Considering practical use, the dynamic load transmitted to the pavement $q\left(x,y,t\right)$ according to Eq. (7) can be expressed by using the Dirac function $\delta \left[\right]$
$q\left(x,y,t\right)={P}_{0}\left(1+\alpha \mathrm{c}\mathrm{o}\mathrm{s}\omega t\right)\delta \left[x-{v}_{0}t\right]\delta \left[y-\frac{b}{2}\right],$
where $\alpha$ is the coefficient of the type of vehicle, $\omega$ is the vibration frequency of the moving load, $b$ is the length of the orthotropic plate in the $y$ direction; ${P}_{0}$ is the
maximum amplitude of the moving load [1].
According to Fig. 1, the effective shear force and bending moment at the orthotropic plate boundaries are given as:
${V}_{i}={D}_{i}\left(\left(\frac{{\partial }^{3}w\left(x,y,t\right)}{\partial {x}^{3}}\right)+\left(\frac{B+2{D}_{tr}}{{D}_{i}}\right)\left(\frac{{\partial }^{3}w\left(x,y,t\right)}{\partial x\
partial {y}^{2}}\right)\right)=k{s}_{i}w\left(x,y,t\right),\mathrm{}\mathrm{}\mathrm{}\mathrm{}i=1,...,4,$
${M}_{i}=-{D}_{i}\left(\frac{{\partial }^{2}w\left(x,y,t\right)}{\partial {x}^{2}}+{u }_{\perp i}\frac{{\partial }^{2}w\left(x,y,t\right)}{\partial {y}^{2}}\right)=k{r}_{i}\frac{\partial w\left(x,y,t
\right)}{\partial x},\mathrm{}\mathrm{}\mathrm{}\mathrm{}i=1,...,4.$
The constraint of the elastic vertical support and rotation are characterized by $k{s}_{i}$ and $k{r}_{i}$, respectively. The index $i=$ 1, 2, 3, 4 stems for $x=$ 0, $a$ and $y=$ 0, $b$, where the
index notation $\perp i$ of the Poisson’s ratio $u$ shows the perpendicular direction of $i$.
Fig. 1Model of the rigid road pavement on a Kerr foundation model under a dynamic moving load
3. Determination of the Eigen frequencies
The solution of the homogeneous orthotropic plate equation given by Eq. (7) can be determined by the method of variable separation using the Fourier series techniques. According to this method, the
homogeneous solution of the problem is a product of function of space and time:
$w\left(x,y,t\right)=\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{W}_{mn}\left(x,y\right)\mathrm{s}\mathrm{i}\mathrm{n}\left({\omega }_{mn}t\right),$
where ${\omega }_{mn}$ is the undamped vibration frequency of the orthotropic plate and ${W}_{mn}\left(x,y\right)$ is a spatial function determined for the modal numbers $m$ and $n$ in the $x$- and
$y$-directions. The spatial function satisfying the initial conditions of the undamped free vibration equation. The substitution of Eq. (11) into Eq. (7) yields:
$\begin{array}{l}-\left({k}_{1}+{k}_{2}\right)\left({D}_{x}\frac{{\partial }^{4}{W}_{mn}}{\partial {x}^{4}}+2B\frac{{\partial }^{4}{W}_{mn}}{\partial {x}^{2}\partial {y}^{2}}+{D}_{y}\frac{{\partial }
^{4}{W}_{mn}}{\partial {y}^{4}}\right)\\ \mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}+{G}_{s}\left({D}_{x}\left(\frac{{\partial }^{6}{W}_{mn}}{\partial {x}^{6}}+\frac{{\partial }^{6}{W}_{mn}}{\
partial {x}^{4}\partial {y}^{2}}\right)+2B\left(\frac{{\partial }^{6}{W}_{mn}}{\partial {x}^{4}\partial {y}^{2}}+\frac{{\partial }^{6}{W}_{mn}}{\partial {x}^{2}\partial {y}^{4}}\right)+{D}_{y}\left(\
frac{{\partial }^{6}{W}_{mn}}{\partial {y}^{6}}+\frac{{\partial }^{6}{W}_{mn}}{\partial {x}^{2}\partial {y}^{4}}\right)\right)\\ \mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}-{k}_{1}{k}_{2}{W}_{mn}+
{k}_{1}{G}_{s}\left(\frac{{\partial }^{2}{W}_{mn}}{\partial {x}^{2}}+\frac{{\partial }^{2}{W}_{mn}}{\partial {y}^{2}}\right)\\ =\rho h{{\omega }_{mn}}^{2}·\left({G}_{s}\left(\frac{{\partial }^{2}{W}_
{mn}}{\partial {x}^{2}}+\frac{{\partial }^{2}{W}_{mn}}{\partial {y}^{2}}\right)-\left({k}_{1}+{k}_{2}\right){W}_{mn}\right).\end{array}$
Since ${W}_{mn}\left(x,y\right)$ in Eq. (12) depends only on the spatial variables and the orthotropic plate vibrates with the same temporal behavior, each side of Eq. (12) must be equal to the
arbitrary separation constant. A relationship between the undamped vibration frequencies of the orthotropic plate and the arbitrary separation constant ${\kappa }_{mn}$ can be expressed as [1]:
${{\omega }_{mn}}^{2}=\left(\frac{{\kappa }_{mn}}{\rho h}\right)\frac{-{k}_{1}}{\left[{G}_{s}\left({\left(\frac{p\pi }{a}\right)}^{2}+{\left(\frac{q\pi }{b}\right)}^{2}\right)+\left({k}_{1}+{k}_{2}\
right)W\left(x,y\right)\right]}=\left(\frac{{\kappa }_{mn}}{\rho h}\right)\mathrm{\Psi },$
$\begin{array}{l}{\kappa }_{mn}=-\left(\frac{{k}_{1}+{k}_{2}}{{k}_{1}}\right)\left({D}_{x}\frac{{p}^{4}{\pi }^{4}}{{a}^{4}}+2B\frac{{\pi }^{4}{p}^{2}{q}^{2}}{{a}^{2}{b}^{2}}+{D}_{y}\frac{{q}^{4}{\pi
}^{4}}{{b}^{4}}\right)-{k}_{2}-{G}_{s}{k}_{1}\left(\frac{{p}^{2}{\pi }^{2}}{{a}^{2}}+\frac{{q}^{2}{\pi }^{2}}{{b}^{2}}\right)\\ \mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}-\frac{{G}_{s}}{{k}_{1}}\
left[{D}_{x}\left(\frac{{p}^{6}{\pi }^{6}}{{a}^{6}}+\frac{{\pi }^{6}{p}^{4}{q}^{2}}{{a}^{4}{b}^{2}}\right)+2B\left(\frac{{p}^{4}{q}^{2}{\pi }^{6}}{{a}^{4}{b}^{2}}+\frac{{\pi }^{6}{p}^{2}{q}^{4}}{{a}^
{2}{b}^{4}}\right)+{D}_{y}\left(\frac{{q}^{6}{\pi }^{6}}{{b}^{6}}+\frac{{\pi }^{6}{p}^{2}{q}^{4}}{{a}^{2}{b}^{4}}\right)\right].\end{array}$
Based on the modified Bolotin method, the two unknowns real numbers $p$ and $q$ in Eqs. (13), (14) can be solved from the two auxiliary Levy’s type problems [15].
4. Determination of the Eigen modes of the orthotropic plate
4.1. First auxiliary Levy problem
The solution of Eq. (12) for the first auxiliary problem that satisfies the boundary conditions defined in Eqs. (9), (10) can be assumed as:
${W}_{mn}\left(x,y\right)={X}_{mn}\left(x\right)\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{q\pi }{b}y\right),$
where ${X}_{mn}\left(x\right)$ is the Eigen mode of the orthotropic plate in the $x$-direction [15]. Substituting Eq. (15) into Eq. (12) which results in an ordinary differential equation for ${X}_
{mn}\left(x\right)$ in which the solution of the characteristic equation can be found by assuming ${X}_{mn}\left(x\right)={e}^{\beta x}$. By substituting ${X}_{mn}\left(x\right)={e}^{\beta x}$ into
the characteristic equation we found the sixth order characteristic equation of $\beta$ which has two imaginary roots and two real double roots. The solution of the first auxiliary problem can be
expressed as:
$\begin{array}{l}{X}_{mn}\left(x\right)={A}_{1}\mathrm{c}\mathrm{o}\mathrm{s}\left(\frac{p\pi }{a}x\right)+{A}_{2}\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{p\pi }{a}x\right)+{A}_{3}\left(\frac{\beta
\pi }{a}x+1\right)\mathrm{}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\left(\frac{\beta \pi }{a}x\right)\\ \mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}+{A}_{4}\left(\frac{\beta \pi }{a}x+1\right)\
mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\left(\frac{\beta \pi }{a}x\right).\end{array}$
Boundary conditions along the $x$-axis permit determining the ${A}_{i}$ coefficients from [16, 17]:
where ${a}_{ij}$ are the coefficients.
When the conditions of the boundary along $x=$ 0 and $x=a$ in Eqs. (9), (10) are substituted into Eq. (16), the characteristic determinant of $\mathrm{D}\mathrm{e}\mathrm{t}\left[\mathbf{A}\right]=\
left|{\mathbf{a}}_{ij}\right|=$ 0 leads to the existence of nontrivial solutions. After expanding resulted in the first transcendental equation in terms of $p$ and $q$.
The second auxiliary Levy problem in the $y$-axis can be determined analogously to the above formulations.
4.2. Mode numbers
The determinants of the first and second auxiliaries Levy problems, being transcendental in nature, have an infinite number of roots. The Mathematica software [18] was used to solve the values of $p$
and $q$ symbolically. By substituting the values of $p$ and $q$ into Eq. (13), the Eigen frequencies of the system can be obtained. The integer parts of $p$ and $q$ represent the number of mode in
the system. The mode shapes of the system are therefore given by:
$W\left(x,y\right)=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{X}_{mn}\left(x\right){Y}_{mn}\left(y\right).$
5. Determination of the non-homogeneous solution of the system
Since a fundamental set of solutions of the homogenous partial differential equation is known and given by Eq. (18), a non-homogeneous solution of the system can be found by replacing the unknown
constant coefficients in Eq. (16) in the $x$ direction as well as in the $y$ direction with unknown coefficient functions. The appropriate solution for the forced response can be expressed in the
$w\left(x,y,t\right)=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{X}_{mn}\left(x\right){Y}_{mn}\left(y\right){T}_{mn}\left(t\right),$
where ${X}_{mn}\left(x\right)$ and ${Y}_{mn}\left(y\right)$ are the mode shapes of the system, ${T}_{mn}\left(t\right)$ depends only on the temporal variable and can be determined from the
non-homogeneous partial differential equation of time. From the natural frequency ${\omega }_{mn}$ computed by Eq. (13), depending on the first and second spring stiffness and the shear moduli of the
Kerr foundation, the temporal equation of ${T}_{mn}\left(t\right)$ can be stated in [19] as follow:
$\begin{array}{l}{\stackrel{¨}{T}}_{mn}\left(t\right)+2\zeta {\omega }_{mn}{T}_{mn}\left(t\right)+{\omega }_{mn}^{2}{T}_{mn}\left(t\right)\\ =\frac{\mathrm{\Psi }}{\rho h{Q}_{mn}}\underset{0}{\
overset{a}{\int }}{X}_{mn}\left(x\right)dx\underset{0}{\overset{b}{\int }}{Y}_{mn}\left(y\right)dy·\left(\frac{{G}_{s}}{{k}_{1}}\left(\frac{{\partial }^{2}}{\partial {x}^{2}}+\frac{{\partial }^{2}}{\
partial {y}^{2}}\right)-\frac{{k}_{1}+{k}_{2}}{{k}_{1}}\right)p\left(x,y,t\right),\end{array}$
where $\zeta$ is the damping ratio of the system and ${Q}_{mn}$ is a normalization factor expressed by:
${Q}_{mn}=\underset{0}{\overset{a}{\int }}{\left({X}_{mn}\left(x\right)\right)}^{2}dx\underset{0}{\overset{b}{\int }}{\left({Y}_{mn}\left(y\right)\right)}^{2}dy.$
The corresponding homogeneous solution of Eq. (20) can be written:
${T}_{0mn}\left(t\right)={e}^{-\zeta {\omega }_{mn}t}\left({a}_{mn}\mathrm{c}\mathrm{o}\mathrm{s}\left(\omega {d}_{mn}t\right)+{b}_{mn}\mathrm{s}\mathrm{i}\mathrm{n}\left(\omega {d}_{mn}t\right)\
From the stationary state initial conditions ($t=$ 0 s), ${T}_{0mn}\left(t\right)={a}_{mn}={b}_{mn}=$ 0 can be obtained. A particular and a general solution of Eq. (20) may be integrated to determine
the temporal response of the problem for an arbitrary applied surface load [1]:
$\begin{array}{l}{T}_{mn}\left(t\right)=\underset{0}{\overset{t}{\int }}\left[\frac{\mathrm{\Psi }}{\rho h{Q}_{mn}\sqrt{1-{\zeta }^{2}}{\omega }_{mn}}\underset{0}{\overset{a}{\int }}{X}_{mn}\left(x\
right)dx\right\underset{0}{\overset{b}{\int }}{Y}_{mn}\left(y\right)dy\\ {·e}^{-\zeta {\omega }_{mn}\left(t-\tau \right)}\left(\frac{{G}_{s}}{{k}_{1}}\left(\frac{{\partial }^{2}}{\partial {x}^{2}}+\
frac{{\partial }^{2}}{\partial {y}^{2}}\right)-\frac{{k}_{1}+{k}_{2}}{{k}_{1}}\right)q\left(x,y,\tau \right)\mathrm{s}\mathrm{i}\mathrm{n}\omega {d}_{mn}\left(t-\tau \right)]d\tau .\end{array}$
Finally, the deflection solution of the governing Eq. (7) subjected to an arbitrary applied dynamic surface load $q\left(x,y,t\right)$ for 0 $\le t\le {t}_{0}$ and $t>{t}_{0}$ can be expressed as
For 0 $\le t\le {t}_{0}$:
$w\left(x,y,t\right)=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{X}_{mn}\left(x\right){Y}_{mn}\left(y\right)\underset{0}{\overset{t}{\int }}\left[\frac{\psi }{ph{Q}_{mn}\sqrt{1-{\zeta }^{2}{\omega }_
{mn}}}\right\underset{0}{\overset{a}{\int }}{X}_{mn}\left(x\right)dx$$·\underset{0}{\overset{b}{\int }}{Y}_{mn\left(y\right)dy}{e}^{-\zeta {\omega }_{mn\left(t-\tau \right)}}\left(\frac{{G}_{s}}{{k}_
{1}}\left(\frac{{\partial }^{2}}{\partial {x}^{2}}+\frac{{\partial }^{2}}{\partial {y}^{2}}\right)\frac{{k}_{1}+{k}_{2}}{{k}_{1}}\right)q\left(x,y,\tau \right)\mathrm{s}\mathrm{i}\mathrm{n}\omega {d}
_{mn}\left(t-\tau \right)]d\tau .$
For $t>{t}_{0}$:
$w\left(x,y,t\right)=\mathrm{}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{e}^{-\zeta {\omega }_{mn}\left(t-{t}_{0}\right)}\left[{w}_{0mn}\mathrm{c}\mathrm{o}\mathrm{s}\left\{\omega
{d}_{mn}\left(t-{t}_{0}\right)\right\}+\frac{{v}_{0mn}+\zeta {w}_{0mn}}{\omega {d}_{mn}}\mathrm{s}\mathrm{i}\mathrm{n}\omega {d}_{mn}\left(t-{t}_{0}\right)\right].$
In which ${w}_{0mn}$ and ${v}_{0mn}$ are the deflection and velocity at the time $t={t}_{0}$, respectively and $\omega {d}_{mn}=\sqrt{1-{\varsigma }^{2}}{\omega }_{mn}$.is the damped vibration
frequency of the orthotropic plate.
6. Numerical examples
Using the procedure described above, a rigid rectangular orthotropic plate doweled of road pavement subjected to dynamic traffic loads as shown in Fig. 1 is analyzed. The values of parameters which
are used in the following examples are given as: $a=$ 5 m, $b=$ 3.5 m, $h=$ 0.25 m, ${E}_{x}=$ 27×10^9 Pa, ${E}_{y}=$ 22.5×10^9 Pa, $\rho =$ 2.5×10^3 kg/m^3, ${u }_{x}=$ 0.180, ${u }_{y}=$ 0.150, $k
{s}_{x1}=k{s}_{x2}=k{s}_{y1}=k{s}_{y2}=$ 150 MN/m/m, $k{r}_{x1}=k{r}_{x2}=k{r}_{y1}=k{r}_{y2}=$ 1 N·m/rad/m. Three types of soil condition are considered in this work: soft soil ${k}_{1}={k}_{2}=$
27.25 MN/m^3, ${G}_{s}=$ 9.52 MN/m^3; medium soil ${k}_{1}={k}_{2}=$ 54.4 MN/m^3, ${G}_{s}=$ 19.04 MN/m^3^and hard soil ${k}_{1}={k}_{2}=$ 108 MN/m^3, ${G}_{s}=$ 38.08 MN/m^3. These parameters are
typical of the material and structural properties of a highway [19]. The traffic load magnitude is ${P}_{0}=$ 80×10^3 N and $\alpha =$ 1/2. To calculate the influence of loading velocity to the
dynamic behavior of the system, $v$ varies from 50 km/hr to 300 km/hr. It is also assumed that damping ratio of the system equals $\zeta =$ 5 %. To compare the dynamic deflections of the orthotropic
plate between the Pasternak and the Kerr foundation models, the following soil parameters are used: ${G}_{s}=$ 9.52 MN/m^3; $k=$ 27.25 MN/m^3. All the dynamic response of the system is computed at $t
=$ 1.5 s. This is the condition at which the dynamic moving load is within the plate region.
6.1. Influence of the foundation types
Time history of the dynamic deflection at the center of the plate $w\left(a/2,b/2\right)$, is calculated and plotted for $m=$ 1, 2, 3, …, 5 and $n=$ 1, 2, 3, …, 4. It is found that the dynamic
deflection of the system is initially high, with rapid oscillations and high amplitudes for all three types of soil conditions studied in this paper. This observation is in agreement with Gibigaye et
al. in the design of pavement plates rest on a soil whose inertia is considered [17].
Fig. 2(a) shows time histories of the system under dynamic moving load for soft soil and hard soil conditions of the Kerr foundation. The moving dynamic load is set to be $v=$ 60 km/hr and load
frequency is $\omega =$ 100 rad/s. It is observed that the rapid oscillations occur at the moment of first loading and after the oscillations become stationary for two types of soil condition. In
Fig. 2(a), it is also shown that the transient domain ends at around $t=$ 0.06 s and does not depend on the soil conditions. This trend was also observed in [17].
Fig. 2(b) shows the time history of the system supported by the Kerr foundation model and Pasternak foundation model. To generalize the Pasternak model, the Kerr model is introduced by adding a layer
of spring on the shearing layer to eliminate the concentrated reactions that occurs along the free edges of a plate structure [13].
Fig. 2Time history of the system under dynamic moving load
a) Two soil conditions on Kerr foundation
b) Kerr and Pasternak foundations
From Fig. 2(b), it is found that the maximum dynamic deflection of the system supported by the Kerr foundation is lower than the maximum dynamic deflection of the system on the Pasternak foundation
model at the mid-span if calculated with the same parameters [20]. This result agrees with previous research done by Kneifati [13], where it was shown that the Kerr model is more accurate than the
Pasternak models for the representation of the base response.
6.2. Influence of moving load on the maximum dynamic deflection
Fig. 3(a) depicts the maximum dynamic deflections of the plates which are subjected to a moving load with the harmonic load frequency is $\omega =$ 100 rad/s and soil condition is set to the
parameter values for soft and medium soil conditions of the Kerr foundation. It can be observed that the speed of the dynamic load has an effect on the maximum lateral dynamic deflection. The maximum
dynamic deflection at the lower value of foundation stiffness increases until about $v=$ 240 km/hr before decreasing. It shows that resonance conditions depend both on the speed of travel and the
stiffness constants of the foundation.
Fig. 3(b) illustrates the effects of load frequency on the maximum lateral deflection under moving load for soft soil and hard soil conditions. It can be seen that load frequency has effects on both
the resonance frequency and the maximum lateral deflection. The resonance load frequency for soft soil condition is smaller than the value of resonance load frequency for the hard soil condition.
Fig. 3Maximum dynamic deflection of the rigid road pavement subjected to moving load
a) Speeds at harmonic frequency ($\omega =$ 100 rad/s)
b) Load frequencies at moving load ($v=$ 60 km/hr)
7. Conclusions
The responses behaviors of rigid road pavements that are subjected to dynamic moving loads with constant velocity have been investigated. The effects of moving velocity, load frequency and elastic
foundation stiffness of the Kerr model are studied. The soil model used in this work is the Kerr model; a proposed generality of the Pasternak model by the inclusion of a layer of springs on the
shearing layer. Based on the orthogonality properties of the Eigen functions, the semi-analytical solution form of the dynamic displacement was obtained. In the formulation of this paper, it was
assumed that the supports at the boundaries of the plate are due to the tie bars and steel dowels, providing the plate with vertical and rotational restraints. This assumption represents a realistic
plate, especially for joints between the rigid pavement plates, in which rotation and vertical shear deformation are found along the joints.
From these results, it is concluded the dynamic response and resonance velocity is significantly affected by the elastic foundation stiffness. When the rigid road pavement rests on a soft soil
foundation, most of the pavement is affected by the loading while the resonance load frequency is small. From the obtained results, it is also concluded that the maximum dynamic deflection of the
rigid road pavement on the Kerr model decreases significantly compared to that of Pasternak model. This result shows the possible economic gain of the Kerr model when it is used for representing the
base response of the rigid road pavement.
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About this article
Vibration in transportation engineering
rigid road pavement
dynamic traffic loads
Kerr foundation
transcendental equation
modified Bolotin method
Copyright © 2018 Sofia W. Alisjahbana, et al.
This is an open access article distributed under the
Creative Commons Attribution License
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start contents
The instability of the three-dimensional high-shear layer associated with a near-wall low-speed streak is investigated experimentally. A single low-speed streak, not unlike the near-wall low-speed
streaks in transitional and turbulent flows, is produced in a laminar boundary layer by using a small piece of screen set normal to the wall. In order to excite symmetric and anti-symmetric modes
separately, well-controlled external disturbances are introduced into the laminar low-speed streak through small holes drilled behind the screen. The growth of the excited symmetric varicose mode is
essentially governed by the Kelvin-Helmholtz instability of the inflectional velocity profiles across the streak in the normal-to-wall direction and it can occur when the streak width is larger than
the shear layer thickness. The spatial growth rates of symmetric modes are very sensitive to the streak width and are rapidly reduced as the velocity defect decreases due to the momentum transfer by
viscous stresses. By contrast, the anti-symmetric sinuous mode that causes the streak meandering is dominated by the wake-type instability of spanwise velocity distributions across the streak. As far
as the linear instability is concerned, the growth rate of the anti-symmetric mode is not so strongly affected by the decrease in the streak width, and its exponential growth may continue further
downstream than that of the symmetric mode. As for the mode competition, it is important to note that when the streak width is narrow and comparable with the shear-layer thickness, the low speed
streak becomes more unstable to the anti-symmetric modes than to the symmetric modes. It is clearly demonstrated that the growth of the symmetric mode leads to the formation of hairpin vortices with
a pair of counter-rotating streamwise vortices, while the anti-symmetric mode evolves into a train of quasi-streamwise vortices with vorticity of alternate sign. [Asai, M., Minagawa, M. and Nishioka,
M., J. Fluid Mech. 455 (2002) 289-314]
Nonlinear evolution of subharmonic streak instability
The streak instability is examined experimentally by artificially generating spanwise-periodic low-speed streaks in a laminar boundary layer on a flat plate.Fundamental and subharmonic modes are
excited for each of the sinuous and varicose instabilities and their development is compared with the corresponding result of a single low-speed streak.The development of subharmonic sinuous mode
does not strongly depend on the streak spacing and it grows with almost the same growth rate as that for the single streak.By contrast, the development of fundamental sinuous mode is very sensitive
to the streak spacing and is completely suppressed when the streak spacing is smaller than a critical value, about 2.5 times the streak width for the low-speed streaks examined.On the varicose
instability, the fundamental mode is less amplified than the subharmonic mode, but the growth of both modes is weak compared with the case of the single streak. [Konishi, Y. and Asai, M., Fluid Dyn.
Res. 42 (2010)]
Development of low-speed streaks downstrem
of the suction strip
Two-dimensional local wall suction is applied to a fully developed turbulent boundary layer such that most of turbulent vortices in the original outer layer can survive the suction and cause the
resulting laminar flow to undergo re-transition. This enables us to observe and clarify the whole process by which strong vortical motions give rise to near-wall low-speed streaks and eventually
generate the wall turbulence. Hot-wire and PIV measurements show that low-frequency velocity fluctuations, which are once markedly suppressed near the wall by the local wall suction, soon start to
grow downstream the suction. The growth of low-frequency fluctuations is of algebraic type, characterizing the streak growth caused by the suction-survived turbulent motions. The low-speed streaks
obtain almost the same spanwise spacing as that of the original turbulent boundary layer without the suction even in the initial stage of the streak development. This indicates the suction-survived
turbulent vortices to be quite efficient to excite the necessary ingredients for the wall turbulence, namely, low-speed streaks of the right scale. After attaining near saturation the low-speed
streaks soon undergo the sinuous instability to lead to re-transition. Flow visualization shows that the streak instability and its subsequent breakdown occur at random in space and time in spite of
the fact that the spanwise arrangement of streaks is almost periodic. Even under the high-intensity turbulence conditions the sinuous instability amplifies disturbances of almost the same wavelength
as predicted from the linear stability theory though the actual growth is in the form of wave packet with the number of wave periods not more than two. It should be emphasized that the mean velocity
develops the log-law profile as the streak breakdown proceeds. The transient growth and eventual breakdown of low-speed streaks are also discussed in connection with the critical condition for the
wall turbulence generation. [Asai, M., Konishi, Y., Oizumi, Y. and Nishioka, M., J. Fluid Mech. 586 (2007) 371-386] | {"url":"https://aero-fluid.sd.tmu.ac.jp/en/research/turbulence.html","timestamp":"2024-11-07T05:56:50Z","content_type":"application/xhtml+xml","content_length":"16183","record_id":"<urn:uuid:468ee97e-8b34-4bf6-a1bd-91c0cca0d785>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00816.warc.gz"} |
bytes and stuff
In my last post I talked about finding the number of zeroes at the end of $n!$, and I said that there was room for improvement. I thought about it a little bit and found a couple things to speed it
The first has to do with the relationship between the quantity of fives and the quantity of twos. The lower quantity in the prime factorization of $n!$ is how many zeroes it will have at the end. If
I would have thought a little more about it though I would have seen that counting the twos is pointless in this situation.
Even the prime factorization of $10 = 5 \cdot 2$ has the information in there: there will always be more twos than fives. Counting from 1 to 10:
• Multiples of 2: 2, 4, 6, 8, 10
• Multiples of 5: 5, 10
This means that all we really need to keep track of is the quantity of fives in the prime factorization of $n!$. Which leads to the second optimization: we only need to get the prime factorization of
multiples of five.
A while back I was looking at the different maximum values that the different integer data types (uint8, short, int, long, etc) have throughout a couple languages I’ve been using and noticed that
none of them ended in zero. I wondered why that was but then relatively quickly realized that it is because integer data types in computers are made up of bytes and bits.
An 8-bit (1-byte) integer has a max value of $2^8 = 256$, a 16-bit (2-byte) integer has a max value of$2^{16} = 65536$, etc. In fact since any integer in a computer is made of bits it will have a
maximum value of $2^n$. The prime factorization of an n-bit integer is in the notation, it is n 2s all multiplied together. Like so: $2^8 = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8$
In order for something to end in a zero it must be a multiple of 10, and the prime factorization of 10 is 5 * 2, and the prime factorization of $2^n$ will never contain a 5. Case closed. That was
That got me thinking about figuring out how many zeroes are at the end of a number if all you have is the prime factorization. Using my basic arithmetic skills I found out that:
• $2 \cdot 5 = 10$
• $2 \cdot 2 \cdot 5 = 20$
• $2 \cdot 5 \cdot 5 = 50$
• $2 \cdot 2 \cdot 5 \cdot 5 = 10 \cdot 10 = 100$
It appears (although isn’t proof) that the lowest quantity between twos and fives dictates how many zeroes are at the end of a number when you’re looking at its prime factorization. I tried this with
many more combinations and it worked with every one of them.
So what can we do with this information?
A factorial is a number written as $n!$ where for any value $n$, $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$ . For example $3! = 1 \cdot 2 \cdot 3$ and $5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 =
120$. The Wikipedia page for factorials shows that $70! = 1.197857167 \times 10^{100}$. That’s a big number, over a googol. You can see the whole thing here on WolframAlpha. | {"url":"https://trlewis.net/category/programming/page/4/","timestamp":"2024-11-14T07:12:07Z","content_type":"text/html","content_length":"35740","record_id":"<urn:uuid:dc98193c-b86f-46d0-af3d-ed156f8501a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00294.warc.gz"} |
How do you solve 15 divided by 3? - Explained
Answer and Explanation:
15 divided by 3 is equal to 5.
Which is the quotient for a dividend of 15 and a divisor of 3?
This process required 3 to be subtracted 5 consecutive times, so again we see that 15 ÷ 3 = 5. The number that is being divided (in this case, 15) is called the dividend, and the number that it is
being divided by (in this case, 3) is called the divisor. The result of the division is the quotient.
How do you divide 15 divided by 5?
5 groups of 3 make 15 so 15 divided by 5 is 3.
How do you divide step by step?
Long Division Steps
1. Step 1: Take the first digit of the dividend from the left.
2. Step 2: Then divide it by the divisor and write the answer on top as the quotient.
3. Step 3: Subtract the result from the digit and write the difference below.
4. Step 4: Bring down the next digit of the dividend (if present).
Leave a Comment | {"url":"https://theomegafoundation.org/how-do-you-solve-15-divided-by-3/","timestamp":"2024-11-07T03:36:53Z","content_type":"text/html","content_length":"72990","record_id":"<urn:uuid:e9f3232e-2d4f-4010-84ff-14f6ef69b80b>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00814.warc.gz"} |
Deep Knowledge Tracing
Modifying Deep Knowledge Tracing for Multi-step Problems
Previous studies suggest that Deep Knowledge Tracing (or DKT) has fundamental limitations that prevent it from supporting mastery learning on multi-step problems [15, 17]. Although DKT is quite
accurate at predicting observed correctness in offline knowledge tracing settings, it often generates inconsistent predictions for knowledge components when used online. We believe this issue arises
because DKT’s loss function does not evaluate predictions for skills and steps that do not have an observed ground truth value. To address this problem and enable DKT to better support online
knowledge tracing, we propose the use of a novel loss function for training DKT. In addition to evaluating predictions that have ground truth observations, our new loss function also evaluates
predictions for skills that do not have observations by using the ground truth label from the next observation of correctness for that skill. This approach ensures the model makes more consistent
predictions for steps without observations, which are exactly the predictions that are needed to support mastery learning. We evaluated a DKT model that was trained using this updated loss by
visualizing its predictions for a sample student learning sequence. Our analysis shows that the modified loss function produced improvements in the consistency of DKT model’s predictions.
Deep knowledge tracing, loss function, online learning
Intelligent tutoring systems are widely used in K-12 education and online learning platforms to enhance learning. Knowledge tracing algorithms are embedded in such intelligent tutoring systems to
support automatic selection of the problems a learner should work on next based on their mastery of different skills. There are multiple popular knowledge tracing algorithms that are frequently used
to predict students’ performance in offline settings. While these approaches have all achieved satisfactory performance in these settings, there is only limited work investigating the use of
knowledge tracing algorithms in online settings [11, 17].
Deep Knowledge Tracing (DKT) is a knowledge tracing approach that has gained in popularity in recent years. It employs a recurrent neural netwok (RNN) [16] to predict student’s correctness on
problem-solving steps that use particular skills. Though some studies demonstrated that DKT outperforms other knowledge tracing models such as Bayesian Knowledge Tracing [1] and Performance Factors
Analysis (PFA) [10], it has some fundamental limitations and drawbacks. For example, DKT’s neural network representation is not easily interpretable, making it difficult for people to understand
DKT’s predictions. Additionally, Yeung and Yeung [15] identified two problems with DKT—the model fails to reconstruct the observed input, and the DKT predictions are inconsistent and fluctuate over
In this paper, we investigate the issue of inconsistent predictions. Our work explores the hypothesis that DKT’s inconsistent behavior is primarily due to its loss function. We propose a novel
modification to the DKT loss functions designed to produce more consistent behavior. Multiple authors have proposed ways of modifying the loss function by adding regularization terms [7, 8, 15].
However, our research explores a novel modification that evaluates predictions for each skill that does not have an observed ground truth value by using the next observed correctness for that skill.
We use the “Fraction Addition and Multiplication, Blocked vs. Interleaved” dataset accessed via DataShop [5] to evaluate a DKT model generated through training with this new loss function by
visualizing its predicted correctness for each skill at each time step in a heatmap. We then compare these results with the predictions generated by a DKT model trained using the original loss
function. Our results indicate that training with the revised loss function produces a DKT model that generates more consistent predictions than one produced by training with the original loss
2. BACKGROUND
2.1 Knowledge Tracing
Knowledge tracing approaches model a student’s knowledge over time and predict their performance on future problem-solving steps. Knowledge tracing algorithms are embedded in Intelligent Tutoring
Systems to support automatic selection of the next problem a student will practice [13]. Much of the research on knowledge tracing has explored its use in offline settings; however, little work has
explored the use of knowledge tracing in online settings. In offline settings, knowledge tracing models are fit to existing data sets, typically to evaluate different knowledge component models to
identify those that better fit the data. In contrast, the objective of online knowledge tracing is to keep track of the student’s level of mastery for each skill (or knowledge component) and/or
predict the student’s future performance based on their past activity. In a nutshell, knowledge tracing seeks to observe, depict, and quantify a student’s knowledge state, such as the level of
mastery of skills underlying the educational materials [6]. The outputs of knowledge tracing support mastery learning and intelligent selection of which problems a student should work on next.
2.2 Deep Knowledge Tracing
Pieche et al. [12] proposed the Deep Knowledge Tracing (DKT) approach, makes use of a Long Short-Term Memory (LSTM) [4] architecture (complex variant of Recurrent Neural Network, or RNN) to represent
latent knowledge. The use of an LSTM has become increasingly popular because it reduces the effect of vanishing gradients. It employs cell states and three gates to determine how much information to
remember from previous time-steps and also how to combine that memory with information from the current time-step.
The DKT model accept an input matrix $X$, which is constructed by one-hot encoding two pieces of information for each step: ${q}_{t}$, which represents the knowledge components, and ${a}_{t}$, which
represents whether the question was answered correctly. The information at each time step is packed into a tuple denoted as ${h}_{t}=\left\{{q}_{t},{a}_{t}\right\}$. ${h}_{0}$ represent the initial
state at time 0 (where t = 0). The network outputs the prediction $Y$ based on the input and previous state. $Y$ is a matrix that represents the probability of each KC being correctly answered at
each step by a given student. ${y}_{t}$ is the predicted probability at time $t$.
The objective of DKT is to predict performance at the next iteration (given the data from time $0$ to $t$, predict $t+1$). To optimize next iteration results, a dot product of the output vector ${y}_
{t}$ and the one-hot encoded vector of the next practiced KC $\delta \left({q}_{t+1}\right)$ is calculated. We take the cross entropy (denoted as $l$) of the dot product, average over number of steps
and number of students. All together, the original loss function of DKT ${L}_{\mathit{Original}}$ can be expressed as:
${\text{L}}_{\mathit{Original}}=\frac{1}{\sum _{i=1}^{n}\left({T}_{i}-1\right)}\sum _{i=1}^{n}\text{∑}_{t=1}^{{T}_{i}-1}l\left({y}_{t}\cdot \delta \left({q}_{t+1}^{i}\right),{a}_{t+1}^{i}\right)$(1)
where $n$ is the number of students, and ${T}_{i}$is the length of the interaction sequence for student $i$.
When the size of a dataset increases, deep knowledge tracing generally has an edge over the classical statistical models, such as Bayesian Knowledge Tracing, Streak Model or Performance Factor
Analysis, when it comes to predicting learner performance. The original DKT work [12] demonstrated that it can produce tremendous gains in AUC (e.g., 25%) when compared to prior results obtained from
other knowledge tracing models. However, subsequent work suggests that the gains are not as large as originally anticipated [14]. One of the key advantages of DKT over classical knowledge tracing
methods, such as BKT, is that it has access to more precise information about the temporal order of interactions as well as information about KCs not involved in the current step [2]. We intend to
leverage these advantages of DKT to support online knowledge tracing [17] and explore whether it is possible to get better mastery learning behavior when using DKT rather than classical knowledge
tracing approaches, such as BKT.
2.3 Challenges with DKT
Even though DKT has many advantages over other knowledge tracing models like Bayesian Knowledge Tracing (BKT) [1], Streak Model [3] and Performance Factor Analysis (PFA) [10], the model still has
several limitations. Specifically, DKT models are difficult to interpret [14], make inconsistent predictions [15], and only consider the correctness of skills that are observed on each time step [7].
Figure 1: This example, drawn from Zhang and MacLellan (2021) [17], shows DKT model predictions on a single knowledge component given one student correctness sequence.
Yeung and Yeung [15] identified that the DKT predictions are not consistent and fluctuate over time. They also showed that the DKT model fails to reconstruct the input information in its predictions.
For example, DKT may predict lower correctness on steps tagged with a particular skill even when the student correctly performs steps that contain the skill. Figure 1 is an example of this effect.
From the first to the third steps, the student did not answer the problem correctly, but DKT predicted the third step would have a 100% chance of being correct. From the fourth to the sixth steps,
the student correctly answered the question while DKT’s predictions dropped. Upon closer investigation of the DKT model, we believe that this unexpected behavior is due to the way that the loss is
Our previous work [17] highlighted DKT’s shortcoming with respect of giving reliable predictions of correctness on steps tagged with each skill during online knowledge tracing. We want to further
investigate the issues that prevent DKT from giving consistent predictions in the scenario of multi-step problem solving and online knowledge tracing. In this paper, we propose a novel revision of
DKT’s loss function. We will discuss our approach in Section 3.1.
3. METHODOLOGY
We propose a novel approach to make the DKT model predictions more consistent by modifying the loss function used during training. We trained and tested on the “Fraction Addition and Multiplication,
Blocked vs. Interleaved” dataset accessed via DataShop [5] with 80% training data and 20% testing data. This data was collected from a study presented in [9], the students solved problems by
interacting with a fraction arithmetic tutor and solved three different types of problems. The three problem types are: Add Different (AD), add fractions with different denominators; Add Same (AS),
add fractions with same denominators; Multiplication (M), multiply two fractions.
We created two DKT models: one trained using the original DKT loss function and another trained using the modified loss function. We then used the two models to make predictions on the same student
sequence. Lastly, we visualized the predictions for each knowledge component (KC) as heat maps and evaluated the prediction consistency by comparing the heat maps generated using the different DKT
All DKT models in this paper consists of a input layer, a hidden layer, and a output layer with size 28, 200, and 14, respectively. The number of knowledge components determines the size of the input
and output layers. The LSTM (long short-term memory) contained 200 hidden units. We trained the model over 1000 epochs, with a learning rate of 0.0025, a dropout rate of 0.4, and a batch size of 5.
The only difference between the original DKT approach and our approach is the loss function used during training.
3.1 Revision of DKT Loss Function
As outlined in Section 2.3, DKT’s original loss function only evaluates the DKT predictions that have observed ground truth values. To overcome this challenge, we propose a revision to the loss
function. Rather than using the original ground truth values typically provided to DKT’s loss function, our revised approach uses modified ground truth data that fills in steps without any
observations by taking the next observation of that skill (see Figure 2).
Figure 2: Graphical depiction of $â$. Colored cells denote observed student performance (0/red equals incorrect and 1/green equals correct). Cells with white backgrounds are extrapolated from the
next observation of each skill.
Mathematically, we use $â$ to represent the updated ground truth values that populate missing cells using the value from the next observation of each skill, see Figure 2. For example, for a specific
knowledge component, if there is no ground truth at ${t}_{i}$ and the next ground truth is at ${t}_{i+n}$, then the $â$ contains an entry at ${t}_{i}$ that has the same value as the entry at ${t}_
{i+n}$. As a result, the entries from ${t}_{i}$ to ${t}_{i+n-1}$ would share the same ground truth with ${t}_{i+n}$.
(a) DKT predictions for each KC using model trained with original loss function.
(b) DKT predictions for each KC using model trained with updated loss function.
Figure 3: A comparison model performance between DKT models trained using the original and revised loss functions.
Next, we updated the loss function so that it evaluates the model’s predictions for all entries that have a value in the updated ground truth values ($â$). Here is the mathematical representation of
this new loss function:
${\text{L}}_{\mathit{Next}}=\frac{1}{\sum _{i=1}^{n}\sum _{k=1}^{K}\left({T}_{i,k}-1\right)}\sum _{i=1}^{n}\text{∑}_{k=1}^{K}\text{∑}_{t=1}^{{T}_{i,k}-1}l\left({y}_{t,k},{â}_{t+1,k}^{i}\right)$(2)
This updated loss function will evaluate most of the DKT predictions that did not originally have observed ground truth values. Note, some predictions are still not evaluated (those that occur near
the end and do not have a next observation to use for evaluation). Because this new loss function evaluates more of DKT’s predictions in between observations, we believe it will result in more stable
4. MODEL EVALUATION
To evaluate the performance of DKT model after revising the loss function, we took a complete student sequence and generated correctness predictions for each skill using the DKT model. We have 14
skills (knowledge components) and three types of problems as introduced in Section 3. There are 8 steps for an Add Different (AD) problem, 3 steps for an Add Same (AS) problem, and 3 steps for a
Multiplication (M) problem. Figure 3 is a comparison of the student’s predicted mastery of each KC at each step when solving a problem (problem type shown on the x-axis). We use the color to
represent DKT’s prediction, with green indicating the student mastering a skill and red indicating not mastering a skill. We use the numbers to represent the ground truth where 1 equals correct and 0
equals incorrect. Figure 2b shows a substantial improvement in prediction consistency over Figure 2a.
In Figure 2a, the DKT predictions fluctuate over time. There is also a pattern of inconsistent predictions on the “AD Right Convert Numerator”, “AD Answer Numerator” and “AD Done” skill even though
the ground truth values for these skills are 1 during the series of problems practiced. Initially, the DKT model trained using the original loss predicts that the student masters this skill after a
few practices. However, we see that for certain repeating periods over the remainder of the sequence, the model predicts the student will get steps with these skills wrong. The student mastered these
three skills initially. As the student starts solving additional steps, however, the DKT model alternates between correct and incorrect predictions over the remainder of the sequence. These behaviors
are unexpected and contrary to the typical assumption that students will not forget skills once they obtain mastery.
In Figure 2b, the problem of wavy DKT predictions (alternating correct and incorrect predictions for different skills) is largely addressed. The DKT model with the revised loss predicts that the
student obtains mastery on all the AD skills and retains this mastery through the end of training. The DKT predictions are consistent with the ground truth in this case.
These results suggest that our revised loss function produces more consistent DKT model predictions. Besides the improvement, we noticed a common issue that occurred in both the original and the
revised DKT model. The student started with 10 AS problems but both DKT models predict improvement of mastery in M and AD skills even before M and AD problems were given to the student. We believe
that more work is needed to better understand how DKT relates the corresponding skills in a multi-step problem.
5. RELATED WORKS
Multiple authors have discussed the limitations of DKT in handling multi-skill sequences and possible modifications to the loss function to improve model behavior. Yeung and Yeung [15] proposed
regularization terms to address the reconstruction problem (where model predictions move opposite to student performance) and the wavy prediction transition problem (where skill predictions cycle
between high and low). Inspired by their study, we believe that revising the loss function is the key to enhancing the consistency of DKT model predictions. Rather than addressing these two problems
separately using regularization terms, our approach modifies the loss function so that it evaluates predictions that lack ground truth observations.
Beyond modifying the loss function, Pan and Tezuka [8] proposed pre-training regularization, which incorporates prior knowledge by including synthetic sequences to the neural network before training
DKT with real student data. Their motivation is similar to ours—their goal is also to solve the inverted prediction problem (referred to as the reconstruction problem by Yeung and Yeung). They added
synthetic data to a baseline model trained with student data and then introduced two regularization measures to measure the severity of the inverted prediction problem. This approach is different
from ours as we are using the ground truth value of each skill to populate skills and steps that do not have observations.
We revised DKT’s loss function to improve prediction consistency across all KCs over time. Our main contribution is that we propose a novel way of modifying the DKT loss function by evaluating skill
predictions at the time steps that lack ground truth observations. Instead of only addressing DKT’s consistency issues, our ultimate goal is to use DKT as an approach to keep track of student
performance in online learning environments and recommend problems to support personalized learning.
Through our heat map analysis, we demonstrated that a DKT model trained with our improved loss function generates more consistent predictions than a DKT model trained with the original loss. Our
analysis showed that predictions for certain skills would cycle between high and low for a DKT model trained with the original loss function; i.e., generated inconsistent predictions over time. In
contrast, the DKT model trained with the revised loss function showed much smoother, more consistent predictions that started lower and improved steadily over the course of training.
Moving forward, we have a number of additional future directions that we would like to explore to improve DKT’s stability and accuracy. In our current work, we propose an updated loss function that
evaluates the DKT predictions for each skill in terms of the next observation of that skill. In future work, we instead want to evaluate each prediction in terms of all future predictions. Further,
we plan to weight each evaluation by a decay factor $\gamma$ as Yeung & Yeung [15] proposed in their future direction. Finally, we should move online and evaluate how well the revised DKT operates in
an online mastery learning context.
This work was funded by NSF award #2112532. The views, opinions, and/or findings expressed are those of the authors and should not be interpreted as representing the official view or policies of the
funding agency.
8. REFERENCES
© 2022 Copyright is held by the author(s). This work is distributed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) license. | {"url":"https://educationaldatamining.org/EDM2022/proceedings/2022.EDM-posters.82/index.html","timestamp":"2024-11-09T22:11:52Z","content_type":"text/html","content_length":"63557","record_id":"<urn:uuid:489c5e58-1bff-4a56-b3f5-b3ccfcf18889>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00072.warc.gz"} |
Equal Areas In Parallelogram a la Pythagoras
Equal Areas In Parallelogram à la Pythagoras
The applet below presents a generalization of the Pythagorean theorem due to W. J. Hazard (Am Math Monthly, v 36, n 1, 1929, 32-34).
Let parallelogram ABCD be inscribed into parallelogram MNPQ. Draw BK||MQ and AS||MN. Let the two intersect in Y. Then
Area(ABCD) = Area(QAYK) + Area(BNSY).
A reference to Proof #9 shows that this is a true generalization of the Pythagorean theorem. The diagram of Proof #9 is obtained when both parallelograms become squares.
The proof is a slight simplification of the published one. It proceeds in 4 steps. First, extend the lines as shown in the applet.
Then, the first step is to note that parallelograms ABCD and ABFX have equal bases and altitudes, hence equal areas (Euclid I.35 In fact, they are nicely equidecomposable.) For the same reason,
parallelograms ABFX and YBFW also have equal areas. This is step 2. On step 3 observe that parallelograms SNFW and DTSP have equal areas. (This is because parallelograms DUCP and TENS are equal and
points E, S, H are collinear. Euclid I.43 then implies equal areas of parallelograms SNFW and DTSP) Finally, parallelograms DTSP and QAYK are outright equal.
|Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018
Alexander Bogomolny | {"url":"https://www.cut-the-knot.org/pythagoras/PythInPara.shtml","timestamp":"2024-11-10T11:46:26Z","content_type":"text/html","content_length":"12548","record_id":"<urn:uuid:9fd9211f-42a4-4019-b636-092b460bfa75>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00014.warc.gz"} |
Winnow Algorithm | Deepgram
Last updated on June 24, 20247 min read
Winnow Algorithm
Have you ever wondered how machines learn to make sense of a complex, high-dimensional world? Well, one answer lies in the ingenuity of algorithms like the Winnow algorithm.
Have you ever wondered how machines learn to make sense of a complex, high-dimensional world? Well, one answer lies in the ingenuity of algorithms like the Winnow algorithm. This remarkable tool
manages to cut through the noise of big data, offering a scalable solution for high-dimensional learning tasks. Here’s how.
Section 1: What is the Winnow Algorithm?
The Winnow algorithm is a testament to the principle of simplicity in design, offering a scalable solution adept at handling high-dimensional data. Let's explore its origins and mechanics.
Just as in our Perceptron glossary entry, we’ll use the following classification scheme:
• w · x ≥ θ → positive classification (y = +1)
• w · x < θ → negative classification (y = -1)
For pedagogical purposes, We’ll give the details of the algorithm using the factors 2 and 1/2, for the cases where we want to raise weights and lower weights, respectively. Start the Winnow Algorithm
with a weight vector w = [w1, w2, . . . , wd] all of whose components are 1, and let the threshold θ equal d, the number of dimensions of the vectors in the training examples. Let (x, y) be the next
training example to be considered, where x = [x1, x2, . . . , xd].
Here are some additional notes on the Winnow Algorithm:
• The Winnow algorithm originated as a simple yet effective method for online learning, adapting to examples one by one to construct a decision hyperplane—a concept crucial for classification
• At its core, the algorithm processes a sequence of positive and negative examples, adjusting its weight vector—essentially a set of parameters—to achieve accurate classification.
• Distinctly, the Winnow algorithm employs multiplicative weight updates, a departure from the additive updates seen in algorithms like the Perceptron. This multiplicative approach is key to
Winnow's adeptness at emphasizing feature relevance.
• When the algorithm encounters classification errors, it doesn't simply tweak weights indiscriminately. Instead, it promotes or demotes feature weights, enhancing learning efficiency by focusing
on the most relevant features.
• This act of promoting or demoting isn't arbitrary; it's a strategic move that ensures the algorithm remains efficient even when faced with a multitude of irrelevant features.
• Comparatively speaking, the Winnow algorithm's method of handling irrelevant features sets it apart from other learning algorithms, as it dynamically adjusts to the most informative aspects of
the data.
• The theoretical performance bounds of the Winnow algorithm have been substantiated by academic research, showcasing a robust framework that withstands the scrutiny of rigorous studies.
With these mechanics in mind, the Winnow algorithm not only stands as a paragon of learning efficiency but also as a beacon for future advancements in handling complex, high-dimensional datasets.
Section 2: Implementation of the Winnow Algorithm
Implementing the Winnow algorithm involves several steps, from initial setup to iterative adjustments and fine-tuning. Understanding these steps is crucial for anyone looking to harness the power of
this algorithm in machine learning applications.
Initial Setup
• Weights Initialization: Begin by assigning equal weights to all features. These weights are typically set to 1, establishing a neutral starting point for the algorithm.
• Threshold Selection: Choose a threshold value that the weighted sum of features must exceed for a positive classification. This value is pivotal as it sets the boundary for decision-making.
Presenting Examples
• Feeding Data: Present the algorithm with examples, each consisting of a feature vector and a corresponding label.
• Prediction Criteria: The algorithm predicts a positive or negative classification based on whether the weighted sum of an example's features surpasses the threshold.
Weight Adjustment Procedure
1. Error Identification: After making a prediction, compare it against the actual label. If they match, move on to the next example; if not, proceed to adjust weights.
2. Multiplicative Updates: Increase (promote) or decrease (demote) the weights multiplicatively when an error is detected. This is done by a factor commonly denoted as α for promotions and β for
Convergence Concept
• Stable Predictions: Convergence in the Winnow algorithm context refers to reaching a state where predictions become stable, and the error rate minimizes.
• Algorithm Stabilization: The algorithm stabilizes when adjustments to weights due to errors no longer yield significant changes in predictions.
Practical Considerations
• Learning Rate Choices: Selecting an appropriate learning rate, α and β, is crucial. Too high, and the algorithm may overshoot; too low, and it may take too long to converge.
• Noise Management: Implement strategies to mitigate the effects of noisy data, which can cause misclassification and hinder the learning process.
Software and Computational Requirements
• Programming Languages: Efficient implementation can be achieved with languages known for mathematical computations, such as Python or R.
• Computational Power: Ensure sufficient computational resources, as high-dimensional data can be computationally intensive to process.
Performance Optimization
• Hyperparameter Tuning: Experiment with different values of α and β to find the sweet spot that minimizes errors and maximizes performance.
• Overfitting Prevention: Implement cross-validation techniques to guard against overfitting, ensuring the algorithm generalizes well to unseen data.
By thoroughly understanding these implementation facets, one can effectively deploy the Winnow algorithm, leveraging its strengths and navigating its intricacies toward successful machine learning
Section 3: Use Cases of the Winnow Algorithm
The Winnow algorithm, with its ability to efficiently process and adapt to high-dimensional data sets, stands as a beacon of innovation in the field of machine learning. Its applications permeate a
variety of domains where precision and adaptability are paramount. From parsing the subtleties of language to identifying genetic markers, the Winnow algorithm reveals patterns and insights that
might otherwise remain hidden in the complexity of vast datasets.
Real-World Applications
• Text Classification: Leveraging its strength in handling numerous features, the Winnow algorithm excels in sorting text into predefined categories, streamlining information retrieval tasks.
• Natural Language Processing (NLP): It assists in parsing human language, enabling machines to understand and respond to text and spoken words with greater accuracy.
• Bioinformatics: The algorithm plays a pivotal role in analyzing biological data, including DNA sequences, helping to identify markers for diseases and potential new therapies.
Efficacy in High-Dimensional Problems
• Large and Sparse Datasets: The Winnow algorithm thrives when confronted with datasets that are vast yet sparse, pinpointing relevant features without being overwhelmed by the sheer volume of
• Feature Relevance: Its multiplicative weight updates prioritize features that are most indicative of the desired outcome, refining the decision-making process.
Online Learning Scenarios
• Sequential Data Reception: As data streams in, the Winnow algorithm seamlessly adjusts, learning and evolving to provide accurate predictions in dynamic environments.
• Adaptive Models: Continuous adaptation is critical in fields such as finance or social media trend analysis, where patterns can shift unpredictably.
Case Studies in Feature Selection
• Machine Learning Enhancements: Studies have demonstrated the Winnow algorithm’s knack for isolating features that are crucial for accurate predictions, thereby enhancing the performance of
machine learning models.
• Efficiency in Learning: By focusing on relevant features, the algorithm reduces computational complexity and expedites the learning process.
Sentiment Analysis and Opinion Mining
• Interpreting Sentiments: The Winnow algorithm has been instrumental in gauging public sentiment, differentiating between positive and negative opinions with high precision.
• Opinion Mining: It dissects vast amounts of text data, such as customer reviews, to provide actionable insights into consumer behavior.
Integration into Ensemble Methods
• Boosting Weak Learners: When combined with other algorithms in ensemble methods, the Winnow algorithm helps improve the predictive power of weaker models, creating a more robust overall system.
• Collaborative Prediction: The algorithm’s contributions to ensemble methods illustrate its capacity to work in concert with other techniques, enhancing collective outcomes.
Future Prospects and Research
• Advancements in AI: Ongoing research is exploring how the Winnow algorithm can be further refined for applications in artificial intelligence, potentially leading to breakthroughs in automated
reasoning and learning.
• Innovative Applications: Future developments may see the Winnow algorithm become integral to more personalized medicine, autonomous vehicles, and other cutting-edge technologies.
In essence, the Winnow algorithm is not just a tool of the present but also a cornerstone for future innovations in the rapidly evolving landscape of machine learning and artificial intelligence. The
breadth of its use cases and its capacity for adaptation make it an invaluable asset in the quest to turn data into wisdom. | {"url":"https://deepgram.com/ai-glossary/winnow-algorithm","timestamp":"2024-11-10T08:04:07Z","content_type":"text/html","content_length":"506306","record_id":"<urn:uuid:fd0df4b6-6c66-4a66-b4b1-58fb5103c19e>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00509.warc.gz"} |
Examples of use
The distributed computing could be used to solve any type of parametric scenario in 2D and 3D varying geometric and physical parameters (except the time).
3D switched reluctance motor
The figure below illustrates the computation time reduction for the parametric analysis of an example of 3D switched reluctance motor.
In this case by using 6 processors for instance the computation time has been reduced by 3.
2D induction motor
The figure below illustrates the computation time reduction for the parametric analysis of an example of 2D induction motor.
In this case by using 4 processors for instance the computation time has been reduced by 3. | {"url":"https://2022.help.altair.com/2022/flux/Flux/Help/english/QuickStart-CDE/english/topics/examples_of_use_r_2.htm","timestamp":"2024-11-12T19:11:03Z","content_type":"application/xhtml+xml","content_length":"45553","record_id":"<urn:uuid:d285a0e8-247e-42ad-b907-40b70380aeb8>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00211.warc.gz"} |
Will This Be on the Test? (June 2022)
by Sarah Lonberg-Lew
Welcome to the latest installment of our monthly series, “Will This Be on the Test?” Each month, we’ll feature a new question similar to something adult learners might see on a high school
equivalency test and a discussion of how one might go about tackling the problem conceptually.
Welcome back to our continuing exploration of how to bring real conceptual reasoning to questions students might encounter on a standardized test. In April, we looked at a question that mixed
contextualized and decontextualized representations of a relationship. Sometimes, though, we encounter questions that are purely decontextualized:
How many ways can you think of to approach this task? Do you have a memorized procedure? How would you tackle this if you didn’t? What skills and understandings do students really need to be able to
answer this question? How could you use visuals to help you reason about this?
As I’ve said earlier in this series, not every problem is one for every student to take on. This one is very abstract and for students who are not ready to work with these kinds of relationships at
this level of abstraction, it may be best to just make a guess and move on. The progression from concrete to representational to abstract is real and important. Just because a student can reason
about a certain algebraic concept concretely does not necessarily mean they are prepared to reason about it abstractly. However, we can help students build that bridge and bring their reasoning to
more abstract levels. (By the way, please don’t confuse “abstract” with “high level”. The progression of concrete to representational to abstract is one that happens over and over with each new
concept and at every level. Students reasoning more concretely can still take on problems involving systems of equations. See the December 2020 post in this series for an example.)
Also, there are a couple of notational/vocabulary details students need to be comfortable with to solve this problem. Specifically, students need to know what it means for a value to “satisfy” an
equation (the equation will be true when the variable is replaced with that value), and they need to understand that the ordered pairs in the answer choices give the values of x and y in that order.
These are understandings that can be taught as when they become relevant. When students are ready to abstract their thinking, they will have an intellectual need for these details.
With these understandings, let’s press on to making sense of the problem and persevering in solving it. Here are some possible approaches:
1. Turn it into a story. It could be quite a bit of work to come up with a realistic story that the system of equations models, but a student can still take something that looks like alphabet soup
and make it more accessible by telling the story in words. For example, a student might retell the story of the system this way:
There are two numbers. One thing I know about them is that if you double one of them and subtract the other from it, you’ll get 11. The other thing I know is that together they add up to 10. I want
to figure out what the two numbers are.
This is probably not the most compelling story you’ve ever read, but it is a way of making sense of the problem that makes it feel more accessible. Framing this “problem” as a game or a puzzle, even
a quest, can free a student from feeling stuck trying to remember the right procedure. What ideas about playing with the numbers come to you from reading this story that you might not think of when
just looking at the equations?
2. Draw a picture. It can be challenging to draw pictures of subtraction, but the visualizing the relationship in the second equation could really help students get a handle on what kinds of numbers
might be solutions to the system. A student who has had opportunities to work with concrete manipulatives like counters, might imagine the second relationship as a group of ten counters in two
different colors like this:
3. Reframe subtraction. One thing that makes it challenging to draw pictures of subtraction is that it often feels like an action – the action of taking away. However, a student who has developed
operation sense with subtraction may be able to frame the first equation in one or more of the following ways:
• The difference between twice x and y is 11
• Twice x is 11 more than y
• If I add 11 to y, I’ll get the same number as if I double x.
How might these understandings make it easier to draw a picture or draw conclusions about the values of x and y?
4. Make logical inferences. What do we know about the numbers and their relationships? What conclusions can we draw? Because all the values in the answer choices are positive whole numbers, a student
might come up with some inferences like these:
• x and y are both less than 10 because they add up to 10.
• 2x has to be more than y by 11, so x is probably the bigger number.
While it is always possible to approach a problem like this by testing out all the answer choices, taking a minute to think about the relationships and what numbers are likely to work may save time.
5. Make a table. Even if this weren’t a multiple-choice question, a quick table could help a student get a handle on what numbers would satisfy the system. Using the fact that the numbers have to add
up to 10, a student might set up a table this way:
You may be wondering if this approach would be helpful if the answers were not positive whole numbers. The answer might not show up in the table, but it would still get you moving toward it.
Algebraic notation is a way of concisely recording relationships. It is a language, and languages are vehicles for ideas. The most valuable thing for students to learn about algebraic notation is
that it tells a story and that they can make sense of that story. While there are procedures and patterns that can get you to solutions quickly and reliably, those are not the only ways to get to the
solutions. And making the procedures and patterns the focus of math learning is like teaching students grammar and spelling without ever teaching the meanings of words or the ideas they communicate.
This is not all to say that it is not worth teaching strategies like substitution and elimination, only that those are just some approaches – they are not the be-all and end-all to solving systems of
equations. Students who learn and make sense of those procedures will find them to be powerful tools, but students who are empowered to see the stories in algebra can solve problems that they
“haven’t learned how to do yet” if they approach them with confidence, flexibility, and a willingness to try different tools.
You may be wondering if this approach would be helpful if the answers were not positive whole numbers. The answer might not show up in the table, but it would still get you moving toward it.
Sarah Lonberg-Lew has been teaching and tutoring math in one form or another since college. She has worked with students ranging in age from 7 to 70, but currently focuses on adult basic education
and high school equivalency. Sarah’s work with the SABES Mathematics and Adult Numeracy Curriculum & Instruction PD Center at TERC includes developing and facilitating trainings and assisting
programs with curriculum development. She is the treasurer for the Adult Numeracy Network. | {"url":"https://www.terc.edu/adultnumeracycenter/will-this-be-on-the-test-june-2022/","timestamp":"2024-11-11T13:52:14Z","content_type":"text/html","content_length":"97658","record_id":"<urn:uuid:3d6adcf5-6dbc-4573-9347-df3e27917356>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00759.warc.gz"} |
5.1: Counting
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Most of us think that counting is as easy as 1, 2, 3... When counting objects, one needs to be careful to not count an object more than once or miss an object. In this section, we will explore some
ideas behind counting.
The Multiplication Principle
If a process can be broken down into two steps, performed in order, with m ways of completing the first step and n ways of completing the second step after the first step is completed, then there are
(m)(n) ways of completing the process.
Example \(\PageIndex{1}\):
Suppose that pizza can be ordered in 3 sizes, 2 crust choices, 4 choices of toppings, and 2 choices of cheese toppings. How many different ways can a pizza be ordered?
To determine the number of possibilities, we will use the multiplication principle. Let \(S = \) pizza size, \(C =\) crust choice, \(T =\) topping choice, and \(Ch =\) cheese choice.
Since we need to choose one choice from each category, we can write that we need to choose size, and crust, and topping, and cheese.
Let's use the multiplication principle:
\[ \text{Ways} &= (S)(C)(T)(Ch) \nonumber \\[5pt] &= (3)(2)(4)(2) \nonumber \\[5pt] &= 48 \nonumber \nonumber\]
What if there was only one choice for cheese? How would this affect the calculation?
Example \(\PageIndex{2}\):
Count the number of possible outcomes when:
1. A coin tossed four times.
2. A standard die is rolled five times.
Definition: Permutation
A permutation is an ordered arrangement of objects.
The number of permutations of \(n\) distinct objects, taken all together, is \(n!\), where
\[n!= n(n-1)(n-2).....1 \label{perm}\].
Note that \(0!=1\).
Example \(\PageIndex{3}\)
Miss James wants to seat 30 of her students in a row for a class picture. How many different seating arrangements are there? 17 of Miss James' students are girls and 13 are boys. In how many
different ways can she seat 17 girls together on the left, then the 13 boys together on the right?
Let's start with the girls. There are 17 of them, and so, when seating the first girl in the row, there are 17 choices. The next spot will have 16 choices left, then 15, and so on. Thus, the number
of choices for seating the girls can be written \(17!\).
For the boys, by the same reasoning, there are \(13!\) ways to seat them on the right.
Now let's apply the multiplication principle: we need to seat the girls and the boys at the same time. For each permutation we might pick for the girls, we need to apply each different case for the
boys as a distinct possibility. So, our result is \((17!)(13!)\). This means there are \(2.215 \cdot 10^{24}\) different ways to seat these students with girls on the left and boys on the right!
The number of permutations of \(r \) objects picked from \(n\) objects, where \(0 \leq r \leq n\), is
\[_nP_r = \displaystyle \frac{n!}{(n-r)!}.\label{pick}\]
When reading this out loud, we say "n Pick r" - when we pick something, like a team for sports or favorite desserts, the order matters.
Example \(\PageIndex{4}\):
Using the digits 1,3,5,7, and 9, with no repetitions of digits, how many three–digit numbers can be made?
We have \(n = 5\) objects, and we want to pick \(r = 3\) of them. So via Equation \ref{pick}:
\[ _nP_r &= \displaystyle \frac{5!}{(5-3)!} \nonumber\\[5pt] &= \displaystyle \frac{5!}{2!} \nonumber \\[5pt] &= \displaystyle \dfrac{(5)(4)(3)(2)(1)}{(2)(1)} \nonumber \\[5pt] &= (5)(4)(3) \nonumber
\\[5pt] &= 60 \nonumber \nonumber \]
The following is defined already in 3.3 Finite Difference Calculus.
The number of combinations of \(r \) objects chosen from \(n\) objects, where \(0 \leq r \leq n\), is
\[_nC_r = \displaystyle \dfrac{n!}{(n-r)! r!} \label{combo}\]
\(_nC_r \) is also denoted as \( \displaystyle n \choose r\). When reading this out loud, we say "n Choose r" - when we choose objects, like candies out of a bag or clothes from a closet, the order
doesn't matter.
Example \(\PageIndex{5}\):
Evaluate \(_6C_2\), and \(_4C_4\).
Let's try \(_6C_2\), or \(\displaystyle 6 \choose 2\):
\(_nC_r = \displaystyle \frac{6!}{(6-2)!2!}\)
\(_nC_r = \displaystyle \frac{6!}{(4)!2!}\)
\(_nC_r = \displaystyle \frac{(6)(5)(4)(3)(2)}{(4)(3)(2)(2)}\)
\(_nC_r = \displaystyle \frac{(6)(5)}{(2)}\)
\(_nC_r = \displaystyle \frac{30}{2}\)
\(_nC_r = 15\)
Now let's tackle \(\displaystyle 4 \choose 4\):
\(_nC_r = \displaystyle \frac{4!}{(4-4)!4!}\)
\(_nC_r = \displaystyle \frac{4!}{0!4!}\)
The result of \(0!\) is \(1\).
\(_nC_r = \displaystyle \frac{4!}{4!}\)
\(_nC_r = 1\)
This makes sense: there is only one way to choose four things from a group of four things. You choose all of them, and that is the only option.
Example \(\PageIndex{6}\):
How many 5-member committees are possible if we are choosing members from a group of 30 people?
Let's see: we have 30 people to choose from, so \(n = 30\). We want to choose 5 members, so \(r = 5\). Lastly, we don't care about the order in which we choose, so we use \(_nC_r\):
\(\displaystyle 30 \choose 5\)\( = \displaystyle \frac{30!}{(30-5)!5!}\)
\(\displaystyle 30 \choose 5\)\( = \displaystyle \frac{30!}{25!5!}\)
\(\displaystyle 30 \choose 5\)\( = \displaystyle \frac{(30)(29)(28)(27)(26)}{5!}\)
\(\displaystyle 30 \choose 5\)\( = \displaystyle \frac{(30)(29)(28)(27)(26)}{120}\)
\(\displaystyle 30 \choose 5\)\( = 142 506\)
Example \(\PageIndex{7}\):
1. In how many ways can 3 men and 3 women sit in a row, if no two men and no two women are next to each other?
2. In how many ways can 3 men and 3 women sit in a circle, if no two men and no two women are next to each other?
Pascal's Triangle
Pascal's triangle was developed by the mathematician Blaise Pascal. It is generated by adding the two terms diagonally above to receive the new term, where the first term is 1, which is defined as \
(_nC_r = _{n-1}C_{r -1}+_{n-1}C_{r }\).
The triangle is useful when calculating \(_nC_r\) as well: count down \(n\) rows, and then count in \(r\) terms. For example: \(_7C_2\) means that we look at row 7, term 2: 6.
• Gives the coefficients of \( (a + b)^n.\)
• The entries of the nth row are \(C(n,0), C(n,1)...C(n,n)\).
• The sums of each row are consecutive powers of 2.
• The third element from each row yields triangular numbers.
Binomial Expansion
\((x+y)^n = x^n+n x^{(n-1)}y + \cdots+{ n\choose k} x^k y^{(n-k)} + \cdots+y^n\). | {"url":"https://math.libretexts.org/Courses/Mount_Royal_University/Mathematical_Reasoning/5%3A_Basic_Concepts_of_Probability/5.1%3A_Counting","timestamp":"2024-11-09T12:12:56Z","content_type":"text/html","content_length":"134413","record_id":"<urn:uuid:07e77f1f-be0e-4d0c-8ba4-5d1ee73a64c2>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00104.warc.gz"} |
Did Adrian Peterson really Outgain Eric Dickerson?
A couple years ago, I wrote about
how rounding errors affect yardage gains
in football. The general rule was that, assuming the rounding error on each play is independent, the total rounding error follows a normal distribution with parameters mean = 0 and SD = sqrt(number
of plays/12).
I began thinking about this again for two reasons. One, Adrian Peterson just came within 9 yards of Eric Dickerson's season rushing record. With 348 rushes for Peterson and 379 for Dickerson, that
comes out to a standard deviation for the combined rounding errors of 7.8 yards, and about a 12% chance that the 9 yard difference is entirely due to rounding errors.
The other reason is that Brian Burke pointed out in the comments of the original article that the rounding errors of plays in the NFL are not independent. The total yardage gain for each drive has to
round off to the correct figure. From Brian's comment:
"One other way to state this is that if a team has 2 plays in a row, and one goes for 4.5 yards but is scored as 4, and the next goes for 5.5 yds, it can't be scored as 5. It must be scored as a
6 yd gain because the ball is very clearly 10 yds further down field, not 9."
I wanted to try to account for this constraint and see how much difference it would make.
Note: the following is mostly dry and math-related, so if you want to skip it, I estimate the chance of rounding errors covering the 9 yard difference between Dickerson and Peterson at about 14%.
In the previous article, we started with the assumption that the rounding error for a single play followed a uniform distribution from -.5 to .5 yards. While this may not be perfectly true, it is
probably close enough to work with. We saw that from there, the combined rounding error for two independent plays forms a triangular distribution ranging from -1 to 1 yards. Rounding Error for One
Rounding Error for Two Plays:
With Burke's restriction, however, the combined rounding error for two plays has the same error distribution as for one play, the uniform distribution from -.5 to .5 yards.
This has an interesting effect on the second play of the drive. While the sum of all rounding errors for the drive has to follow the same uniform distribution as the first play alone, the error for
the second play alone no longer follows the same uniform distribution.
Let's say that the first play goes for 4.5 yards and gets scored as a five yard gain. The next play goes for 4.8 yards, so the total for the drive is 9.3 yards. If the total yardage credited to the
two plays has to be 9 yards, that means the second play must be scored as a 4 yard gain. That's a rounding error of -.8 yards, outside of the original uniform distribution. In fact, the rounding
error for the second play alone follows the same triangular distribution as the error for two independent plays.
The rounding error for any single play depends on two things: the total rounding error of the drive up to that point, and the precise yardage gain of the play. If the total rounding error before the
play is -.5 yards and the gain is 1.3 yards, that means you have to round up to 2 yards, for an error of .7 yards, to keep the total error within the -.5 to .5 range (-.5 + .7 = .2, whereas rounding
down would give -.5 - .3 = -.8). Those two factors can determine the rounding error for every singular play.
As a result, each individual play after the first has the same triangular error distribution as the second play. They all start off with the same uniform distribution for the total rounding error of
the drive, so there is nothing that would change the distribution of errors for any single play the more plays you add before it (plays that end in a touchdown are an exception, because the only
rounding error will be from the starting point).
In the previous article, we found the distribution for the total rounding error of a series of plays by adding the variances of the individual distributions. As Brian pointed out, that only works if
the distributions are independent of each other, which they are clearly not if the total error distribution for the drive never grows when we add plays. Consecutive plays are highly correlated, to
the point that any number of consecutive plays adds no variance to the total error distribution.
What about non-consecutive plays, though? We know that the total error for the team's drive can't exceed .5 yards, but the error for subsets of the drive can (for example, any single play except the
first can have an error of more than .5 yards). What about if we want to know the combined error for the first, third, and fifth plays?
If these errors were independent, we would simply add the variances for the individual plays, which are 1/12 for the first play, 1/6 for the third play, and 1/6 for the fifth play (1/12 is the
variance of the uniform distribution, 1/6 is the variance of the triangular distribution). That gives a total variance of 5/12. Now the question is how much of that variance is reduced by correlation
between the individual errors.
Intuitively, we would think there should be some correlation. If the first play has a negative rounding error, and the total rounding error after three plays is as likely to be positive as negative,
then it stands to reason that the second and third plays are more likely to have a positive rounding error than a negative rounding error.
That is true of the second play. It is not, however, true of the third play. The reason is that at the start of every play after the first, the total rounding error is going to follow the same
uniform distribution. Whether the first play has a rounding error of -.5 or 0 or .5, the total error distribution after the second play is going to be the exact same. All of the correlation that goes
into re-centering the total error distribution at zero is absorbed by the second play alone.
Put another way: If a play starts just short of a hash mark, it is no more or less likely to end just short of another hash mark than it is to end just past another hash mark. This is the nature of
the uniform distribution of errors. A negative rounding error after one play is no more likely to lead to a negative cumulative rounding error after the following play.
So while the errors for consecutive plays are correlated, the errors for non-consecutive plays are not. You can simply add the variances as we did in the previous article.
Let's return to Adrian Peterson. He ran 348 times this year, and we want to know the total variance for the rounding error distribution for those 348 plays. We can use the following rules to find the
total variance (each of these rules has been confirmed by simulation):
-the first play of a drive adds 1/12 to the total variance
-any play after the first play of the drive adds 2/12 to the total variance, assuming Peterson did not also rush on the play before
-any play immediately following another Peterson rush adds 0 variance, so that any string of consecutive plays adds only the variance of the first play (i.e. if the string started on the first play,
the whole thing adds 1/12, otherwise the whole thing adds 2/12)
Using the same Brian Burke's published football PBP database, we can categorize each of Peterson's rushes into one of these three rules. Doing so gives a total variance for the rounding error of
Peterson's rushes of 410/12, rather than the 348/12 we would get assuming each play was independent. It may seem counterintuitive that this restriction can increase the rounding error because it
introduces correlation between errors, but remember that it also widens the distribution of errors on each play, and the correlation between errors only holds for consecutive plays.
I can't repeat this analysis on Dickerson's season because I don't have PBP data, but let's assume that Dickerson's rounding error distribution widened similarly to Peterson's. If Peterson's rounding
error has a variance of 410/12 and Dickerson's is something like 440/12, that means there is about a 14% chance that rounding errors cover the entire 9 yard difference in their credited totals.
This analysis incorporates only the restriction that the total error for each drive has to remain within the -.5 to .5 yard range at all times. There are likely additional restrictions, for example:
-Burkes' comment also mentions that consecutive non-scoring drives will be constrained with each other, which would mean the first play of some drives could also follow the triangular distribution
instead of the uniform distribution
-The impact of touchdowns reducing the variance of scoring plays (due to the end point being exactly precise) is not considered in the 410/12 variance figure. Changing the variance on Peterson's 12
touchdown rushes would still round the final estimate to 14%, though.
-There could also be a first down restriction, i.e. that a series of downs can't be rounded up to 10 yards no matter what, so that 9.9 would have to be rounded down to 9 if the scorers can't credit
10 yards without reaching the first down marker (I don't know if this is true, but I am guessing it might be).
Also, as with the previous article, this is only addressing rounding error, not spotting errors on the part of the officials or misjudgments about where the ball is by the scorer.
My best guess would be that actual chance that Peterson out-gained Dickerson, once you incorporate the spotting errors, etc, is probably closer to 20% or so. The spotting errors are probably the
biggest additional factor, and if they have a standard deviation of about 6 inches to a foot or so and are independent of the rounding errors, that would give something like a 17-22% chance. That's
purely guesswork on the magnitude of spotting errors, though.
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The figure shows a circle in the coordinate plane. The center of the circle is the point A(h,k). The point B(x,y) is a point on th
The figure shows a circle in the coordinate plane. The center of the circle is the point A(h,k). The point B(x,y) is a point on the circle. Triangle ABC is a right triangle, with point C the vertex
of the right angle. The variable r represents the radius of the circle.
1 thought on “The figure shows a circle in the coordinate plane. The center of the circle is the point A(h,k). The point B(x,y) is a point on th”
1. Answer:
Leave a Comment | {"url":"https://wiki-helper.com/the-figure-shows-a-circle-in-the-coordinate-plane-the-center-of-the-circle-is-the-point-a-h-k-th-40224597-71/","timestamp":"2024-11-12T18:35:34Z","content_type":"text/html","content_length":"128178","record_id":"<urn:uuid:37427e46-45ee-4725-9a6c-7b41acf57eaa>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00640.warc.gz"} |
Cool Math Stuff
This week, we are going to talk about divisibility, a pretty broad area of mathematics that we were lucky to be introduced to in fifth grade. I’ve looked into it more, and even use it for a lot of
tricks in math.
There are a bunch of tricks for it, and I will explain them all and prove them to you. This post will almost be a lesson, but all of the rules are very cool! Let’s go through them all.
Divisibility by one: If the number has no decimal, it is divisible by one. For these rules, we are only working with decimals. To prove it, we look at the Closure Property, which states that any
number multiplied by one will remain the same number. This means every integer is a multiple of one.
Divisibility by two: If the number ends in two, four, six, eight, or zero, it is divisible by two. The definition of even number is, “A natural number that is divisible by 2.” It is also defined as,
“A whole number that has 0, 2, 4, 6, or 8 in the ones place.” Since these two things are the same, this means that this rule is valid.
Divisibility by three: This rule is probably one of the coolest, and most magical. If you add up the digits in the number and this gives you a multiple of three, this means that the number is a
multiple of three. If the sum is too big for you to know if it has a three factor, go ahead and add up the digits again until you do know. I will prove this one along with the divisibility by nine
rule, as it is very similar, and a number that is divisible by nine is also divisible by three.
Divisibility by four: Look at the number’s last two digits, and completely ignore the rest. If the number you are looking at is a multiple of four, then the whole number is a multiple of four. If you
think about it, every number is 100x + y with y being the last two digits. Since we know 100x is a multiple of 4 (25x • 4 = 100x), all that leaves us with is y. This means that if y is a multiple of
4 also, then adding 100x won’t change this.
Divisibility by five: If the number ends in five or zero, it is a multiple of five. If you think about it, the digits from 0 to 9 end in either five or zero when multiplied by five. Since the last
digit of a multiplication problem is the last digit of the last two digits multiplied together, we are left with all of the integral numbers ending in five or zero.
Divisibility by six: Since six is the product of three and two, we can just put both of these rules to use since they share nothing in common. If the number follows through with the divisibility by
three rule (just adding to a multiple of three, no need for a multiple of six, 1 + 2 = 3 and 12 is a multiple of six), and is even, it is a multiple of six. Check the divisibility by two and three
rules for more.
Divisibility by seven: This is also a really cool one, one my teacher didn’t even know. Take the last digit of the number and double it. Then, subtract that from the rest of the number. Keep
repeating this until you have a one or two digit number (negatives are okay, you can just turn them positive). If the number you have is a multiple of seven, the original number is a multiple of
seven. Let’s do one real quick, the number 224.
First, double the 4 to get 8. Subtract 8 from 22 to get 14. Since 14 is definitely a multiple of seven, the 224 is as well. 224 happens to be 32 • 7, so we did our math correctly. To prove this, we
will use a little bit of Algebra. Our formula states that if 10r + x is a multiple of seven, then r – 2x is a multiple of seven. Let’s say the number is a multiple of seven. This means that r – 2x is
a multiple of seven.
r – 2x = multiple of seven
Let’s try multiplying both sides by ten.
r – 2x = multiple of seven
10(r – 2x) = multiple of seven (and seventy)
10r – 20x = multiple of seven (and seventy)
Forget it is a multiple of seventy. We are not looking for that. However, let’s go in a different direction and add 21x to both sides. Since 21 is a multiple of seven, we still have a multiple of
seven on the right-hand side of the equation. However, the left becomes:
10r – 20x = multiple of seven
10r – 20x + 21x = multiple of seven
10r + x = multiple of seven
And this brings us back to where we started. It is a little complicated, but I think it is pretty cool.
Divisibility by eight: Similar to divisibility by four, we will be looking at a block of digits on the right. However, note the fact that eight isn’t a multiple of one hundred. However, it’s a
multiple of one thousand. So, we will look at the last three digits and see if it is a multiple of eight. Since this might be a little hard for you, as I don’t have my eights memorized that far, you
can look at the hundreds place and see if it is odd or even. If it is even, this means that you can check the last two digits for divisibility by eight because 200 is a multiple of eight. If it is
odd, check the last two digits for divisibility by four, but then make sure it is not a multiple of eight, or a multiple of eight plus four. Since 104 is a multiple of eight, this should work as
You can even take these principles to divisibility by any power of two. The power you are raising two to is the amount of digits you need to test at the end. So, for divisibility by 64, just check to
see if the last six digits are divisible by 64 because 2^6 = 64.
Divisibility by nine: This is unquestionably the most magical of all of the proofs, and is used for so many things. It is the answer to numerous magic tricks, math tricks, and even provides loads of
ways to check answers, including divisibility, digital roots, and mod sums. In fact, you can even cube root numbers based on this rule. To find divisibility by nine, you simply add up the digits of
the number, and if that is a multiple of nine, the whole number is a multiple of nine. To prove it, let’s test the number 234. 2 + 3 + 4 = 9, so it is definitely a multiple. But why? Let’s write the
number 234 in expanded notation, something you learn around third or fourth grade.
2 x 100 + 3 x 10 + 4 x 1
We can rewrite these to get:
2 x (99 + 1) + 3 x (9 + 1) + 4 x (0 + 1)
Let’s distribute the 2, 3, and 4 into the groups of numbers beside them.
(2 x 99) + 2 + (3 x 9) + 3 + (4 x 0) + 4
We can rearrange this to get:
[(2 x 99) + (3 x 9) + (4 x 0)] + 2 + 3 + 4
You might notice, but the sum inside of the brackets is definitely a multiple of nine, as it is being created by all multiples of nine. So, we can eliminate it to get the sum of all of the digits.
Here, I don’t think the proof lives up to the magic in the actual trick, but it is pretty cool!
Divisibility by ten: To test divisibility by ten, all we need to do is see what the last digit is. If it is a zero, then the number is a multiple of ten. To prove it, we look at the Power of Zeros
rule, which says that to multiply a number by a power of ten, just tack on that power amount of zeros. This means that a number multiplied by 10 has one zero at the end. You can elevate this to say
that a multiple of 100 has two zeros at the end, a multiple of 1000 has three and so on.
Divisibility by eleven: To test divisibility by eleven, alternately subtract and add the digits and if you end with a multiple of eleven (it may be zero or negative), then your original number is a
multiple of eleven. So for the number 1353, you would go 1 – 3 + 5 – 3 = -2 + 5 – 3 = 3 – 3 = 0. Since zero is a multiple of eleven, the full number is a multiple of eleven.
To prove it, we know that if you subtract eleven from a number constantly, the number still keeps its status as a multiple of eleven. Therefore, let’s put ten to its powers (creating the place
values) and see what happens when you constantly subtract multiples of eleven.
10^0 – (0 • 11) = 1
10^1 – (1 • 11) = -1
10^2 – (9 • 11) = 1
10^3 – (91 • 11) = -1
10^4 – (909 • 11) = 1
10^5 – (9091 • 11) = -1
10^6 – (90909 • 11) = 1
and so on…
This convinces me enough, but I’m not sure of where to take it from there. If you want to add anything, please comment. It would be interesting for all of us.
Divisibility by twelve: Let’s take two distinct ones, divisibility by four and three. If the number is divisible by four and its digits add to a multiple of three, it is a multiple of twelve. Like
the divisibility by six rule, this works as well. You can even take this principle to other numbers like this, like 14, 15, 18, 21, 22, 24, 28, and so on.
Divisibility by thirteen or greater (the “create a zero, kill a zero method”): Since this is a little more complicated, let’s learn it through an example. Is 2756 divisible by thirteen? First, we
must look at the last digit of 2756, six. We need to find a multiple of thirteen that ends in six. If there is one, that means the number is not divisible, and you have learned a new property about
that number. However, 13 x 2 = 26, so we have found one. We must subtract the 26 from 2756 to create a zero. So, 2756 - 26 = 2730. Now, we kill the zero, or just ignore it. Now, we create a zero by
subtracting 13 (273 – 13 = 260). After killing it, we have 26. Since 26 is a multiple of 13, 273 and more importantly, 2756 are multiples of 13. You could even have subtracted 156 from the original
number if you knew that it was 12 x 13. This would make it only take two steps instead of three.
Even though it is a lot in one dose, it is definitely worth practicing. These are really cool principles, and you could probably figure out a few tricks out of them. If you watch my Mathemagics show
closely, I will use divisibility to perform one of the tricks, so you can investigate that as well.
When getting into algebra, you will hear the term "real number" a lot, without an explanation as to why you can't just put "number." Since no one asks the question, it just slides by, and when the
question is asked, the teacher responds, "You'll learn that in Algebra II," or something along those lines.
The answer to that question is that there is such thing as "imaginary numbers," or complex numbers, which are made up of a constant, a coefficient, and the letter i, which symbolizes the square root
of negative one.
What is the square root of negative one? Some say negative one. Well, (-1) x (-1) = 1, so that is incorrect. People then turn around and say one. Well, 1 x 1 = 1, so that is incorrect. Then, they
might try 1/2. Well, 1/2 x 1/2 = 1/4, so that is wrong. They will keep trying things until they give up.
What is the answer? If you think about it, a negative times a negative, or a negative squared, is a positive. A positive times a positive, or a positive squared, is a positive. So, you cannot square
a real number and get a negative. So, mathematician Heron of Alexandria came up with the letter i, and began using that as the square root of -1. So, then, by the Multiplication Property of Square
Roots, you can conclude that √(-9) would be 3i because you can break that into √(9)√(-1) = 3√(-1) = 3i. Rafael Bombelli built on his works, and make this concept a regular part of Algebra.
Now, let me introduce you to one more thing about imaginary numbers before I show you about the cube rooting. When you write these terms out, you write like 5 + 3i, which means 5 + √(-9), just like
how you'd write a real square root in Algebra. The 5 + 3i would be known as a complex number, which is a number involving i. The conjugate of a complex number is to keep the same expression, but
switch the operation separating them. For instance, the conjugate of 5 + 3i = 5 - 3i because we kept the same term, but switched the operation.
If you think about it, a complex number's conjugate is equal to the number because any number has two square roots, a positive one and a negative one. So by making the i term negative, we are just
looking at the other root.
Now that we've gotten that out of the way, let's get to the good part! Let's take the complex number -1/2 + i/2√(3). It is the same thing, with a constant of -1/2 and coefficient of 1/2√(3). How
about we cube it.
(-1/2 + i/2√(3))(-1/2 + i/2√(3))(-1/2 + i/2√(3))
First, we'll square it. We can use FOIL for that. If you don't know, it stands for "First, outer, inner last." It's basically the distributive property made simpler for multiplying binomials.
(-1/2 + i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) - i/4√(3) - 3/4
-1/2 - i/2√(3)
We ended up with the conjugate of before. That's interesting. Let's finish off by multiplying by the -1/2 + i/2√(3).
(-1/2 - i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) + i/4√(3) + 3/4
1 - i/4√(3) + i/4√(3)
What do we do with that? Well, there are i's in both terms, so we can combine them. However, look closer. They are opposites of each other, or the additive inverse of each other. What does that mean?
The definition of additive inverses are two numbers in which when added together give you zero. So, these two confusing numbers simplify to zero!
1 - i/4√(3) + i/4√(3)
1 + 0
So, we are left with 1 as our answer! We did nothing wrong there. -1/2 + i/2√(3) is in fact the cube root of one, as well as its conjugate and of course, the integer one. Was this random? No!
Mathematics is never random!
If you take the Cartesian Plane, and make the numbers going up the y-axis i, 2i, 3i, 4i, 5i, etc. and -i, -2i, -3i going down, you have the Imaginary Cartesian Plane. If you make a circle going
through the points (1,0), (0, i), (-1, 0), and (0, -i), then you will have a unit circle. To find the 1st root of one, we of course start at (1, 0) and that is it. For the square root, or the second
root, we would split the 360° of the circle in half to get 180°. So, we have the 1, and then we travel 180° to get -1, the other square root. For the fourth root, we could split 360 in fourths to get
90°, and at ninety degrees, all of the roots are found, 1, -1, i, and -i.
What about if we split in thirds, or 120°. Then, we end up at the points 1, -1/2 + i/2√(3), and -1/2 - i/2√(3). You can check that if you'd like. At the 72 degree marks, you will find the fifth
roots, and the 60 degree marks give you the sixth roots.
If you know anything else about this, please tell us! Also, we will probably be taking more about imaginary numbers, so if you want me to show anything in particular, let me know.
Why does 64 = 65? What kind of a question is that? Any pre-schooler probably knows that 64 doesn't equal 65. Algebra clearly shows that they are not equal. However, geometry might throw us off track.
Let's take a chessboard. It's an 8 x 8 grid, with a total area of 64 square units. I'd like you to make the following cuts in the chessboard, as well as along the 5th row from the top (3rd row from
the bottom).
Now, arrange these shapes into a rectangle. Let's check out its dimensions. We have a side that is 5 units long, and a side that is 13 units long. To figure out the area of the rectangle, we would do
5 x 13 = 65.
Not good enough? Make the shapes into a triangle. We have 10 units for the base, and 13 units for the height. To find area, we do (bh)/2. So, 10 x 13 = 130 ÷ 2 = 65.
How is this possible? We have taken a grid with area 64 and just by rearranging the shapes, end up with a grid with area 65. No, there was no human error involved, your cuts probably were very
accurate, and even a perfectly straight cut would still give you 65 as your area. So, how is this possible?
Even I completely understand that this is completely invalid. However, I still can't wrap my head around why it is wrong. I've been told that the squares along the cut after you assemble the shape
are not valid squares, which is the only thing that seems accurate. However, with this, I just find it fascinating to take knowledge you learned in kindergarten, or even pre-school, is being
challenged with this very contradictory proof.
I also saw this as a proof of last week's maneuver with Fibonacci numbers. I am not quite sure of why, but it does make some sense, that you are turning eight squared into thirteen times five. It
does work again with a 5x5 or 13x13 grid, as long as you make the correct cuts.
Region Revenge: In August, I gave you guys a problem called region revenge. The goal was to find its explicit formula. The answer 2^n-1 is incorrect, as the sixth cut can only make 31 regions, the
seventh makes 57, and so on. Here is the correct answer:
An = (n^4 - 6n^3 + 23n^2 - 18n + 24)/24
You could have solved this with the techniques we used for the other problems, just by creating a five way system.
Today yet again is a Fibonacci Day! It is October 8th, and 8 is the sixth Fibonacci number. To keep the theme of last week, let's use the square Fibonacci numbers again. Here they are:
Let's take each Fibonacci number and move one away from it. Now, we'll multiply those numbers and see how close we get to the square.
1) 0 x 1 = 0 = 1^2 - 1
2) 1 x 2 = 2 = 1^2 + 1
3) 1 x 3 = 3 = 2^2 - 1
4) 2 x 5 = 10 = 3^2 + 1
5) 3 x 8 = 24 = 5^2 - 1
6) 5 x 13 = 65 = 8^2 + 1
7) 8 x 21 = 168 = 13^2 - 1
And so on and so forth. Basically, Fn-1 x Fn+1 = Fn^2 ± 1, or even more accurate, Fn-1 x Fn+1 = Fn^2 + (-1)^n. Let's try it again, this time looking two away from the number. Keep in mind that 1 is
the negative-first Fibonacci number.
1) 1 x 2 = 2 = 1^2 + 1
2) 0 x 5 = 0 = 1^2 - 1
3) 1 x 5 = 5 = 2^2 + 1
4) 1 x 8 = 8 = 3^2 - 1
5) 2 x 13 = 26 = 5^2 + 1
6) 3 x 21 = 63 = 8^2 - 1
7) 5 x 34 = 170 = 13^2 + 1
This time, we have pretty much the same pattern. Fn-2 x Fn+2 = Fn^2 ± 1, or Fn-2 x Fn+2 = Fn^2 - (-1)^n. How about we move three away. It's the same type of pattern, but a little different. Keep in
mind that -1 is the negative-second Fibonacci number (since a Fibonacci number is the two numbers before it added together, than the zero comes from x + 1, or -1 + 1).
1) -1 x 3 = -3 = 1^2 - 4
2) 1 x 5 = 5 = 1^2 + 4
3) 0 x 8 = 0 = 2^2 - 4
4) 1 x 13 = 13 = 3^2 + 4
5) 1 x 21 = 21 = 5^2 - 4
6) 2 x 34 = 68 = 8^2 + 4
7) 3 x 55 = 165 = 13^2 - 4
We have the same idea. We are stuck with a four, giving us the pattern of Fn-3 x Fn+3 = Fn^2 + 4(-1)^n. Let's look at our neighbors four away and see if we can see the pattern better. What do you
think the negative-third Fibonacci number is? If you got two, then good job.
1) 2 x 5 = 10 = 1^2 + 9
2) -1 x 8 = -8 = 1^2 - 9
3) 1 x 13 = 13 = 2^2 + 9
4) 0 x 21 = 0 = 3^2 - 9
5) 1 x 34 = 34 = 5^2 + 9
6) 1 x 55 = 55 = 8^2 - 9
7) 2 x 89 = 178 = 13^2 + 9
Same idea again. We have Fn-4 x Fn+4 = Fn^2 - 9(-1)^n. However, the four and nine aren't there randomly. Let's look these differences closer.
1, 1, 4, 9
Recognize them? They are the squares of the Fibonacci numbers again! If you go five away, it is the square of the fifth Fibonacci number, six away is the square of the sixth Fibonacci number, one
hundred away is the square of the hundredth Fibonacci number. Basically, a general formula is Fn-a x Fn+a = Fn^2 ± (Fa^2)(-1)^n. Or, you can use the below formula to be even more accurate:
Fn-a x Fn+a = Fn^2 - ((-1)^a)(Fa^2)((-1)^n)
I have no clue why this works, but please put up a proof if you know it. This is one of the coolest things about Fibonacci numbers!
Today is a Fibonacci Day! It is the first, and one is a Fibonacci number. One is in fact two Fibonacci numbers. In the last post, you learned how to square numbers that end in five. To keep this
squaring theme, how about we square the Fibonacci numbers.
1, 1, 2, 3, 5, 8, 13, 21, 34...
1, 1, 4, 9, 25, 64, 169, 441, 1156...
We've been doing lots of adding Fibonacci numbers. Let's finish it off by adding the square Fibonacci numbers.
1 = 1
1 + 1 = 2
1 + 1 + 4 = 6
1 + 1 + 4 + 9 = 15
1 + 1 + 4 + 9 + 25 = 40
1 + 1 + 4 + 9 + 25 + 64 = 104
Do you see a pattern? It is a little hard to find, but definitely present. Look at this:
1 = 1 = 1 x 1
1 + 1 = 2 = 1 x 2
1 + 1 + 4 = 6 = 2 x 3
1 + 1 + 4 + 9 = 15 = 3 x 5
1 + 1 + 4 + 9 + 25 = 40 = 5 x 8
1 + 1 + 4 + 9 + 25 + 64 = 104 = 8 x 13
They are the product of the two consecutive Fibonacci numbers! Why on earth would that be? I had recently looked for one on the internet, and found an amazing geometric proof for it.
Have you ever heard of the golden rectangle, or the golden ratio? We touched on the golden ratio when I gave you the explicit formula for Fibonacci numbers (the golden ratio is the same as the greek
letter fi). The golden rectangle is a rectangle of which the ratio of the length and width is the golden ratio. Something else whose ratio is the golden ratio is Fibonacci numbers! So, the side
lengths of the rectangle are consecutive Fibonacci numbers!
Since we are dealing with squares of Fibonacci numbers, let's make some squares.
We've just taken these squares and organized them in a fashion that makes the side lengths two Fibonacci numbers. We went up to 34 squared, so let's see what the side lengths are.
They are 34 and 55. So, to figure out the area of the whole thing, you can add up the areas of all the squares, or just multiply the 34 by 55. And because of the way it is laid out, you can do it
with any Fibonacci numbers! I think that is really cool!
Bonus Pattern: How about we add the squares of consecutive Fibonacci numbers.
1 + 1 = 2
1 + 4 = 5
4 + 9 = 13
9 + 25 = 34
25 + 64 = 89
64 + 169 = 233
The sums of the consecutive square Fibonacci numbers is in fact a Fibonacci number. I don't know a proof for this, but please tell me if you find one! | {"url":"https://coolmathstuff123.blogspot.com/2011/10/","timestamp":"2024-11-03T23:08:37Z","content_type":"text/html","content_length":"116872","record_id":"<urn:uuid:bf8a0eed-06e0-4105-8648-37d441d355f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00236.warc.gz"} |
Some musings on statistics
A) Beware of The Wrong Summary Statistics
SlateStarCodex had a pretty interesting post entitle “Beware of Summary Statistics“, showing how they can be misleading. This isn’t exactly new, there are famous examples of how just looking at the
mean and standard deviation greatly oversimplifies; distributions can have the exact same mean/stdev. but be very different^[1]. The main lesson to take-away is to always visualize your data.
If you know the distribution in question, though, there is probably a good summary statistic for it. The go-to in social science is pearson correlation. SSC gave an example of two variables which
appeared to be correlated but that correlation was highly misleading. Here are two “uncorrelated” variables:
The linear fit shows that X and Y are uncorrelated. The pearson correlation is nearly 0. However that is obviously BS as there is a clear trend, just not linear, which is what pearson correlation
captures. With the benefit of viewing the data^[2], we can correlate Y vs inverse-sin(Y) (orange). That correlation is 0.99. The real relationship is Y= Asin(fX), where A = 1 and f = 1. A mean/
standard deviation for this would be meaningless, but amplitude/frequency would describe it perfectly.
Of course this is a rigged example, and I generated the data from a sine wave. In a real-world example, one sometimes knows (has some idea) what shape the distribution will be. If one doesn’t,
visualize it and figure it out.
B) Exact Wording Matters
The most famous example I know of is an old study by the Gates Foundation showing that the best schools are small schools. So obviously we need to look at small schools and see why they’re so great,
right? Well, no, because the worst schools are also small schools. Small school -> small sample size -> high variance, meaning the outliers are always going to be found in smaller sample sizes:
Source: The Promise and Pitfalls of Using Imprecise School Accountability Measures
Thomas J. Kane and Douglas O. Staiger.^[3]
One of the earliest papers on cognitive biases looked at this^[4], they asked people if large hospitals or small hospitals are more likely to have more days where >60% of babies born on that day
were male. Most people said the same, because the odds of being born male are the same for any particular baby in either case. But pay closer attention to that wording; it wasn’t about the overall
average, it was about the variance. Simpler example: If you flip two quarters at a time, occasionally they’ll all (re: both) come out heads. If you flip 10 quarters at a time, very rarely will they
all be heads.
C) Confounders and Conditional (In)dependence
I love Simpson’s Paradox . Trends which exist in aggregated data can reverse direction when data is broken into subgroups. In the most general case, if subgroups exist, a trend which applies to
the aggregate doesn’t have to exist in subgroups, and if it does, doesn’t have to be in the same direction. And vice versa going the other direction, from subgroup to overall.
In the above chart, Y has an overall linear trend against X. But once it’s known whether the point is in S1 or S2, the dependence goes away. So Y is conditionally independent of X. Interpretation
will depend on the problem situation. If the difference between S1 and S2 is something we care about, it’s interesting and we publish a paper. Champagne for everybody! If not, it’s a confounder (boo!
The easiest way to deal with confounders is to analyze groups separately. Say you’re interested in discovering people that walk fast spend more on shoes. Well age affects walking speed, so to remove
that confounder, one could stratify patients into different groups. Confounder removed! It’s a good idea, and it has two serious drawbacks:
1. Each group has a smaller sample size, which increases the variance.
2. Testing multiple groups means testing multiple hypotheses.
These errors compound each other. We’ve got several smaller sample sizes meaning the variance is larger, so the odds of getting at least one false positive gets much larger (see section B)^[5]. The
social science studies I read never correct for multiple hypotheses, gee I wonder why :-).
Closing Thought
While finishing this post I came across an article about a deliberate scientific “fraud”. The authors did the experiment they said, didn’t make up any data; the only thing which makes this fraud
different from so many others is that the authors are publicly saying the result is bullshit. I almost typed “the authors *knew* the result is bullshit” except I’m sure most other snake-oil salesmen
know that too. Life is complicated, so don’t trust anybody selling easy answers.
This entry was posted in Statistics. Bookmark the permalink. | {"url":"http://www.jacobsilterra.com/2015/05/28/some-musings-on-statistics/","timestamp":"2024-11-05T05:29:50Z","content_type":"text/html","content_length":"37956","record_id":"<urn:uuid:8c4b05ca-6f33-4212-9601-235b9348340e>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00095.warc.gz"} |
Lina 〆's Blog
This week, we learned Relations and Functions Lesson #2~#9, I think this is a fun but very hard unit. Because there are many new proper nouns and the need to remember 😂.
Relation:A relation is a set of inputs and outputs, often written as ordered pairs (input, output)
Functions:A function is a relation in which each input has only one output.
When representing a relation, we often regard the values of the independent variable as the input and the values of the dependent variable as the output.
The input values make up the domain of the relation ,and the output values make up the range of the relation.
The domain of a relation is the set of all possible values which can be used for the input of the independent variable(x)
The range of a relation is the set of all possible values of the output of the dependent variable.(y)
The following is a question I made when I first started to do a question(How to answer it is also written on the paper) | {"url":"https://myriverside.sd43.bc.ca/linap2016/2017/12/04/","timestamp":"2024-11-07T17:10:30Z","content_type":"text/html","content_length":"49772","record_id":"<urn:uuid:a9c377cf-84da-4b56-bf56-c915ceeac022>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00125.warc.gz"} |
Functions in Python
Once functions are defined, they can be called from the main body of the script or any other function.
Calling a function involves specifying the function name followed by arguments in parentheses. If a function does not have a return statement, it returns None by default.
In our code snippet, we call the add function from the main body:
1def add(a, b):
2 return a + b
4def greet(name):
5 print(f"Hello, {name}!")
7if __name__ == "__main__":
8 sum_ints = add(2, 3)
9 sum_doubles = add(2.5, 3.5)
11 greet("Alice") # Hello, Alice!
13 print("Sum of ints:", sum_ints) # Sum of ints: 5
14 print("Sum of doubles:", sum_doubles) # Sum of doubles: 6.0
Here, add(2, 3) returns 5, and add(2.5, 3.5) returns 6.0. The results are then printed using print(). The greet("Alice") function doesn't return anything useful, so we call it without assigning its
result to a variable.
As a reminder, if __name__ == "__main__": is a special block in Python that ensures certain code runs only when the script is executed directly, not when it's imported as a module in another script.
It is considered a good practice to include this in your Python scripts. | {"url":"https://learn.codesignal.com/preview/lessons/3481","timestamp":"2024-11-04T05:06:48Z","content_type":"text/html","content_length":"182588","record_id":"<urn:uuid:6933e438-71fa-4494-9844-1b34a95a861f>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00470.warc.gz"} |
Lobe Separation Angle Calculator
Author: Neo Huang Review By: Nancy Deng
LAST UPDATED: 2024-10-02 00:36:58 TOTAL USAGE: 940 TAG: Angle Automotive Camshaft
Unit Converter ▲
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The Lobe Separation Angle (LSA) is a key parameter in camshaft design that impacts engine performance. It is the angle between the intake and exhaust lobes on a camshaft, and it plays a significant
role in determining the overlap of the intake and exhaust valves.
Lobe Separation Angle Formula
The formula to calculate the Lobe Separation Angle is:
\[ \text{LSA} = \frac{\text{IC} + \text{EC}}{2} \]
• LSA is the Lobe Separation Angle (degrees).
• IC is the intake centerline (degrees).
• EC is the exhaust centerline (degrees).
How to Calculate Lobe Separation Angle
Follow these steps to calculate the Lobe Separation Angle:
1. Determine the intake centerline (IC) in degrees.
2. Determine the exhaust centerline (EC) in degrees.
3. Use the formula LSA = (IC + EC) / 2.
4. Calculate the Lobe Separation Angle.
Importance of Lobe Separation Angle
The LSA affects the overlap between the intake and exhaust valves, influencing engine characteristics like idle quality, vacuum, and power output. A wider LSA typically results in smoother idle and
better vacuum, while a narrower LSA can increase overlap, which may enhance performance in specific RPM ranges.
Example Calculation
If the intake centerline is 110° and the exhaust centerline is 114°, the Lobe Separation Angle would be:
\[ \text{LSA} = \frac{110 + 114}{2} = \frac{224}{2} = 112° \]
Common FAQs
1. What is Lobe Separation Angle?
□ LSA is the angle between the intake and exhaust lobes on a camshaft, crucial for determining valve overlap and engine performance.
2. Why is LSA important in engine tuning?
□ LSA influences valve overlap, affecting engine characteristics such as power, torque, and idle stability.
3. Can LSA be adjusted?
□ Yes, LSA can be modified by using different camshaft designs or adjustable cam gears, depending on the desired engine performance characteristics. | {"url":"https://www.calculatorultra.com/en/tool/lobe-separation-angle-calculator.html","timestamp":"2024-11-04T01:47:21Z","content_type":"text/html","content_length":"46694","record_id":"<urn:uuid:d037ee55-00e1-492c-a939-de4429238e29>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00799.warc.gz"} |
Estimating Loess Plateau Average Annual Precipitation with Multiple Linear Regression Kriging and Geographically Weighted Regression Kriging
School of Soil and Water Conservation, Beijing Forestry University, Beijing 100083, China
Beijing Datum Science and Technology Development Co., Ltd., Beijing 100084, China
College of Forestry, Beijing Forestry University, Beijing 100083, China
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Submission received: 12 April 2016 / Revised: 30 May 2016 / Accepted: 16 June 2016 / Published: 22 June 2016
Estimating the spatial distribution of precipitation is an important and challenging task in hydrology, climatology, ecology, and environmental science. In order to generate a highly accurate
distribution map of average annual precipitation for the Loess Plateau in China, multiple linear regression Kriging (MLRK) and geographically weighted regression Kriging (GWRK) methods were employed
using precipitation data from the period 1980–2010 from 435 meteorological stations. The predictors in regression Kriging were selected by stepwise regression analysis from many auxiliary
environmental factors, such as elevation (DEM), normalized difference vegetation index (NDVI), solar radiation, slope, and aspect. All predictor distribution maps had a 500 m spatial resolution.
Validation precipitation data from 130 hydrometeorological stations were used to assess the prediction accuracies of the MLRK and GWRK approaches. Results showed that both prediction maps with a 500
m spatial resolution interpolated by MLRK and GWRK had a high accuracy and captured detailed spatial distribution data; however, MLRK produced a lower prediction error and a higher variance
explanation than GWRK, although the differences were small, in contrast to conclusions from similar studies.
1. Introduction
Precipitation, one of the most important climatic factors, is a vital part of the hydrologic cycle, affecting energy transfer and maintaining biosphere functions [
]. It is the focus of hydrology, agriculture, ecology, and environmental science, as well as other related disciplines [
]. Controlling and evaluating the spatial and temporal distribution of precipitation is important in many fields, such as basin and watershed management, soil and water conservation, climate change
assessment, agroclimatic or ecological regionalization, ecological environment construction, and prediction and prevention of extreme weather disasters [
The main methods for obtaining precipitation data include ground-based meteorological measurements and spaceborne radar observations [
]. Spaceborne radar data have low spatial resolution and large uncertainties, which could lead to significant errors in precipitation distribution prediction [
]. Ground-based meteorological data are typically used for spatial interpolation of rainfall distribution, due to a longer time series and smaller errors [
]. However, as the number and distribution of meteorological stations is limited and inconsistent, it is still a huge challenge to obtain high-accuracy, high-resolution data on the distribution of
precipitation [
Precipitation is a comprehensive reflection of the interactions among the various components of the climate system [
]. It is influenced by atmospheric circulation, local topography, and land cover, and is closely related to many physical geographic factors, such as altitude, slope, aspect, solar radiation,
vegetation type, distance to mountains or seas, lakes, and river systems [
]. Due to limitations of the interpolation methods algorithm, a lot of studies on this topic have only used elevation (DEM) as an auxiliary variable, combined with co-Kriging or thin plate spline
(TPS), in order to predict the distribution of precipitation [
]. Many studies have shown that, by considering additional natural geographical features in the interpolation process, rainfall variability can be explained more effectively [
]. Interpolation methods that use a variety of auxiliary variables include multiple linear regression (MLR), geographically weighted regression (GWR), artificial neural networks (ANN), regression
Kriging (RK), Bayesian maximum entropy (BME) interpolation,
Regression Kriging, first coined by [
], is mathematically equivalent to “universal Kriging” (UK) or “Kriging with external drift” (KED) [
]. It is a spatial prediction technique, which is combined with a regression forecast of auxiliary variables and Kriging interpolation of the regression residuals. It can combine different regression
models to generate many combined methods [
], among which the multiple linear regression Kriging (MLRK) and geographically weighted regression Kriging (GWRK) are the most commonly used. The regression process of MLRK fits with the global
trend of the target variables across the study area under stable conditions between spatial variables [
]. The regression process of GWRK fits the local trend around the predicting points, and can adapt to a non-stationary relationship between the spatial variables, leading to a better explanation of
the spatial variation of target variables [
]. Both of these methods have been widely used in Earth science and environmental science, especially in studies of the spatial distribution of soil properties [
]. There are currently some studies that analyze precipitation interpolation on different temporal and spatial scales using these two methods [
]. However, more research is needed on the use of MLRK and GWRK to evaluate the performances of the two methods and obtain high spatial accuracy precipitation maps.
The Loess Plateau was selected as a study area as there is a shortage of water and severe soil erosion. The region belongs to both semi-humid and semi-arid areas, where vegetation growth is limited
by rainfall, and ravines and gullies cover the underlying surface. The relationship between different physical geographical features is significantly non-stationary. Based on the above
characteristics, the study area is very suitable for precipitation interpolation experiments. The objectives of this study were as follows: (1) to assess the performance of MLRK and GWRK in the
interpolation processes; and (2) to obtain a highly accurate distribution map of average annual precipitation with a 500 m resolution in the Loess Plateau.
2. Study Area and Data
2.1. Study Area
The Loess Plateau is located in the middle reaches of the Yellow River basin in the north of China, between 33°43′ N–41°16′ N and 100°54′ E–114°33′ E (
Figure 1
a). It is surrounded to the east by the North China Plain, to the west by Helan Mountain and the Qinghai-Tibet Plateau, to the north by the Yinshan Mountains, and to the south by the Qinling
Mountains, covering an area of 62.29 × 10
. The Loess Plateau has a diverse topography. The Luliang Mountains and Taihang Mountains are located in the eastern Loess Plateau and the Liupan Mountains are situated in the western part. The
Yellow River flows through the western and northern boundaries, generating the Ningxia Plain and Hetao Plain, where it then crosses the central and southern part of the Loess Plateau, generating the
Guanzhong Plain. The middle of the Loess Plateau is covered with highly erodible aeolian silt deposits and has become one of the most severely eroded regions in the world. The northern Loess Plateau
contains Mu Us Sandy Land and Kubuqi Desert. A continental monsoon climate is the major climate type of this area. Under its influence, dry and cold winds in winter are followed by frequent and
intense rainfall in summer [
]. Annual precipitation recorded by meteorological stations ranges from 200 mm to 800 mm and decreases from the southeast to the northwest of the Loess Plateau [
]. In order to mitigate the prediction error of “edge effects” in interpolation [
], a buffer area with a 100 km bandwidth was considered around the Loess Plateau.
2.2. Data Sets and Data Processing
The annual average precipitation distribution in the Loess Plateau was analyzed during the 1981–2010 period. The interpolated precipitation data set was provided by the Climatic Data Center, National
Meteorological Information Center, China Meteorological Administration (
). Data came from 277 meteorological stations distributed across the Loess Plateau and 155 stations located in the additional buffer area (
Figure 1
In order to assess the prediction accuracies of the MLRK and GWRK approaches, a validation precipitation data set was selected, including 130 hydrometeorological stations with more than 20 years of
data, provided by the National Science and Technology Infrastructure of China, Data Sharing Infrastructure of Earth System Science (
In this study, the mean annual maximum normalized difference vegetation index (NDVI), digital elevation model (DEM), solar radiation, slope, and aspect were selected as auxiliary variables for
interpolation. NDVI images with 500 m spatial resolution were derived from MODIS MOD13A1 data obtained from NASA EOSDIS Land Processes Distributed Active Archive Center (
). The mean annual maximum NDVI data was the average of the 2000–2014 annual maximum NDVI images (
Figure 1
b), which were calculated by Maximum Value Composite (MVC). DEM data were collected from the CGIAR-CSI SRTM 90 m database provided by Cold and Arid Regions Sciences Data Center at Lanzhou (
) and were combined into 500 m spatial resolution. Solar radiation data were acquired from DEM data using the spatial analyst module of ArcGIS 10.2. Slope and aspect data were also calculated from
the DEM data. The unit of slope adopted here was radians. Qualitative aspect data, representing directions, were divided into eight dummy variables where 1 represented data belonging to a certain
direction and 0 represented data not belonging to that direction. Directions between 0°–22.5° and 337.5°–360° were assigned to north (N); 22.5°–67.5° to northeast (NE); 67.5°–112.5° to east (E);
112.5°–157.5° to southeast (SE); 157.5°–202.5° to south (S); 202.5°–247.5° to southwest (SW); 247.5°–292.5° to west (W) and 292.5°–337.5° to northeast (NW). In this study, all auxiliary environmental
variables were converted into their standardized normal variables, except for the eight dummy aspect variables.
3. Methods
3.1. Logit Transformation and Exploratory Data Analysis Methods
Normal distributions of target variables are a requirement for regression analysis and Kriging methods.
Figure 2
a shows that the Loess Plateau average annual precipitation has an approximately normal distribution. In order to improve this, the logit transformation was applied as follows [
$z + + = ln ( z + 1 − z + ) , 0 < z < 1$
is logit-transformed precipitation;
is the precipitation standardized to the 0 to 1 range:
$z + = z − z min z max − z min , z max < z < z min$
z is the precipitation of meteorological station point; and z[min] and z[max] are the physical minimum and maximum of precipitation in this region.
One of advantages of using the logit transformation is that the predicted value can be fixed by specifying physical limits (
) based on prior knowledge [
]. According to previous studies, average annual rainfall in the study area is generally within 100–1000 mm [
]. Therefore,
was set to 100 mm and
was set to 1000 mm. The logit-transformed precipitation can be reversed to the original scale by using the back transformation:
$z = z + + 1 + z + + ⋅ ( z max − z min ) + z min$
The Moran Index was subsequently used to analyze the global spatial autocorrelation of precipitation, and the Hot Spot Analysis was used to evaluate the local autocorrelation patterns of
precipitation. Furthermore, the logit-transformed precipitation and standardized normal auxiliary variables were used in the Pearson correlation analysis and stepwise linear regression in order to
determine the predictors that could apply to the interpolation by regression Kriging. All data analyses and subsequent interpolations were applied using the open-source program of
language [
3.2. Interpolation Techniques: Multiple Linear Regression Kriging (MLRK) and Geographically Weighted Regression Kriging (GWRK)
3.2.1. Regression Kriging
$z ( s i )$
represent the target-interpolated variable, where
$s i ( i = 1 ... n )$
is spatial location and n is the number of locations. The estimated values
$z ^ ( s 0 )$
by regression Kriging (RK) can be generally given by Equation (4) [
$z ^ ( s 0 ) = m ^ ( s 0 ) + e ^ ( s 0 ) = ∑ k = 0 ρ β ^ k × q k ( s 0 ) + ∑ i = 1 n λ i × e ( s i )$
$m ^ ( s 0 )$
is the deterministic part fitted by the regression model,
$e ^ ( s 0 )$
is the interpolated residual by ordinary Kriging (OK, a fundamental Kriging method),
$β ^ k$
is the estimated deterministic model coefficient (
$β ^ 0$
is the estimated intercept),
$q k ( s 0 )$
is the value of predictors at the predicted location
$s 0$
$λ i$
is the ordinary Kriging weight determined by the spatial dependence structure of the residual, and
$e ( s i )$
is the residual at location
$s i$
Interpolated residuals by OK can be expressed as:
$e ^ ( s ) = ∑ i = 1 n λ i × e ( s i ) = λ 0 T × e$
$λ 0$
is the vector of OK kriging weights (
$λ i$
), with a constraint
$∑ λ i = 1$
, and
is the vector of regression residuals. The variance of regression residuals is represented by:
$D ( e ) = E [ ( e ( s i ) − e ( s i + h ) ) 2 ] = 2 γ ( h )$
is the vector of distance and
$γ ( h )$
is the semivariance function or variogram. The variogram
$γ ( h )$
can be used to solve the vector of Kriging weights
$λ 0$
$λ 0 = ( C 0 + C + γ ( h ) ) − 1 × c 0$
$c 0$
is the vector of the regression residual covariance at prediction points.
3.2.2. Multiple Linear Regression Kriging
Multiple linear regression Kriging (MLRK) is one of most commonly used regression Kriging methods for predicting the spatial distribution of data. In MLRK, the fitting regression model is multiple
linear regression (MLR) with ordinary least squares (OLS) estimates and
$β ^ k$
can be expressed as follows:
$β ^ k = β ^ MLR = ( q T × q ) - 1 × q T × z$
is the matrix of predictors and
is the vector of sampled observation. The spatial distribution of logit-transformed precipitation was interpolated by MLRK using six steps: (1) determine the MLR model with predictors selected by
stepwise regression; (2) calculate the deterministic part using the MLR model at each prediction point; (3) derive the regression residuals at meteorological stations; (4) filter the optimal
variogram model with modelling the covariance structure of regression residuals; (5) interpolate the regression residuals using ordinary Kriging; and (6) add the deterministic part to the
interpolated residuals at each prediction point.
3.2.3. Geographically Weighted Regression Kriging
Geographically weighted regression (GWR), which is an extension of multiple linear regression, is also implemented with the ordinary least squares estimates, but takes the spatial locations of data
points into consideration. Geographically weighted regression Kriging (GWRK) uses GWR as the fitting model. In GWRK,
$β ^ k$
can be expressed as follows:
$β ^ k = β ^ GWR = ( q T × W × q ) - 1 × q T × W × z$
is the spatial weighting matrix. The weighting matrix
, which is specified as a continuous and monotonic decreasing function of distance
between the observations, can be calculated by different methods. In this study, the weight of each point was computed by applying the bi-square nearest neighborhood function:
$w = { [ 1 − ( d h ) 2 ] 2 , if d < h , 0 , if d ≥ h$
is the bandwidth for spatially adaptive kernel size.
The optimization of bandwidth was required because a large deviation in regression parameters estimation would be generated if the bandwidth was too large or too small [
]. The optimal bandwidth can be determined when the cross-validation (CV) scores obtain the minimum value using the cross-validation method [
]. The whole procedure of logit-transformed precipitation interpolation by GWRK can be described using seven steps: (1) determine the optimal bandwidth by the cross-validation method; (2) derive the
spatial weighting matrix with the bi-square nearest neighborhood function using the predictors selected by stepwise regression; (3) derive the regression residuals at meteorological stations; (4)
calculate the deterministic part using the weighting matrix and predictors at each prediction point; (5) filter the optimal variogram model by modelling the covariance structure of regression
residuals; (6) interpolate the regression residuals using ordinary Kriging; and (7) add the deterministic part to the interpolated residuals at each prediction point.
3.3. Validation Techniques
In order to evaluate the interpolation by MLRK and GWRK, adjusted determination coefficients (adjusted
) were calculated for the deterministic part by MLR and GWR, and five-fold cross-validation was implemented for interpolated residuals by OK [
]. The whole prediction error of back-transformed precipitation corresponding to MLRK and GWRK was evaluated by comparing estimated values with actual observations at validation points. The following
indexes were used to verify prediction accuracy:
Mean error (ME):
$ME = 1 l × ∑ j = 1 l [ z ^ ( s j ) − z ( s j ) ]$
Mean relative error (ME
$ME r = 1 l × ∑ j = 1 l [ z ^ ( s j ) − z ( s j ) z ( s j ) ]$
Mean absolute error (MAE):
$MAE = 1 l × ∑ j = 1 l | z ^ ( s j ) − z ( s j ) |$
Mean absolute relative error (MAE
$MAE r = 1 l × ∑ j = 1 l | z ^ ( s j ) − z ( s j ) z ( s j ) |$
Root mean square error (RMSE):
$RMSE = 1 l × ∑ j = 1 l [ z ^ ( s j ) − z ( s j ) ] 2$
is the number of validation points,
$z ( s j )$
is the precipitation of validation points, and
$z ^ ( s j )$
is the estimation value of the validation points. Adjusted determination coefficients (adjusted
) were calculated to indicate the amount of variance explained by MLRK and GWRK [
$Adjusted R 2 = 1 − ∑ j = 1 l [ z ^ ( s j ) − z ( s j ) ] 2 ∑ j = 1 l [ z ( s j ) − z ¯ ( s j ) ] 2$
$z ¯ ( s j )$
is the mean of the validation precipitation data.
4. Results
4.1. Spatial Autocorrelation
The closer that two stations are to each other, the stronger the correlation of precipitation.
Figure 2
b shows that, when the radius of stations was within 25 km, the precipitation correlation of corresponding meteorological stations was as high as 0.909. When the lag was between 100 km to 125 km, the
correlation was still above 0.7, indicating a remarkable spatial autocorrelation in this region. In order to quantitatively evaluate the spatial autocorrelation, the Moran Index was calculated, which
synthetically measures the autocorrelation for the entire study region. Moran’s index
= 0.86 (
score = 38.50,
< 0.0001), indicated a remarkably significant clustering of precipitation in the Loess Plateau. Furthermore, the local autocorrelation patterns of precipitation can be affected by a non-uniform
underlying surface. In order to evaluate the pattern at different meteorological stations, a Hot Spot Analysis was implemented by calculating the Getis-Ord Gi statistics.
Figure 3
shows the
score of local Gi statistics for all considered meteorological stations. It is clear that high
scores, called hot spots, were clustered in the southern and southeastern Loess Plateau, indicating that precipitation values measured at meteorological stations in these areas were high and similar
to each other. On the other hand, stations with low
scores were clustered in the north and northwest of the Loess Plateau.
scores of approximately 0, found in the southwestern, central, and eastern parts of this region, indicate an absence of spatial association patterns.
4.2. Exploratory Data Analysis
Correlations between the logit-transformed precipitation (PreT) and the auxiliary variables for each station, including standardized normal variable of DEM (DEM_std), standardized normal variable of
NDVI (NDVI_std), standardized normal variable of slope (slope_std), standardized normal variable of solar radiation (rad_std), and eight dummy variables of aspect were tested by Pearson correlation
analysis (
Table 1
). Precipitation showed the most significant positive correlation with NDVI, followed by solar radiation, DEM, and slope, where all correlations reached a significance level of 0.01. Among the eight
dummy variables of aspect, only the north slope reached a significant correlation level with precipitation.
Figure 4
a shows that more precipitation (PreT) appears where there is a higher NDVI, and lower precipitation occurs where there is a lower NDVI, thereby indicating that vegetation growth in the region is
closely linked to precipitation. With increasing altitude (DEM), PreT gradually decreased and then slightly increased (
Figure 4
b). The precipitation varied irregularly with a gentle slope, but when the degree of slope increased, precipitation also gradually increased (
Figure 4
c). The relationship between solar radiation and precipitation is similar to the relationship between DEM and precipitation because of high correlation (
= 0.968,
< 0.01) between DEM and solar radiation (
Figure 4
Table 1
, it becomes evident that there are remarkable correlations between DEM, NDVI, slope, and solar radiation, and correlations between eight dummy variables of aspect also reached a significant level.
In order to weaken the impact of co-linearity between the auxiliary variables on regression models, the stepwise linear regression method was used. According to the results of stepwise regression (
Table 2
), five variables or dummy variables (DEM_std, slope_std, NDVI_std, N, and NW) were selected as predictors and subsequently used in MLRK and GWRK implementation.
4.3. Diagnosis and Evaluation of Regression
Based on the results of stepwise regression, the multiple linear regression equation used in this study is expressed as in Equation (17). The significance of regression parameters is given in
Table 3
$PreT = a × DEM_std + b × slope_std + c × NDVI_std + d × N + e × N W + f$
Figure 5
a is the scatter plot for the regression-predicted values
the corresponding residuals at all meteorological stations. The red fitted line near the zero value in this figure indicates that the system correlation between the regression predictions and
residuals is weak, which infers a good linear correlation between predictors and precipitation in the regression model. The normal Q-Q diagram (
Figure 5
b) shows that the regression residuals approximately obey the normal distribution assumption. The scatter randomly distributed around the red fitted line in the scale-location graph (
Figure 5
c) denotes that the regression residuals were satisfied by homoscedasticity,
, residuals corresponding to different magnitudes of predicted values should have the same variance.
Figure 5
d is the Cook’s distance diagram of each station. Cook’s distance can artificially reflect the effect of all observation points on the regression model by combining the values of residuals and
leverage. The higher the Cook’s distance value of a station, the greater the contribution for the regression model. No. 373 and No. 53 meteorological stations, with Cook’s distance values greater
than 0.06, were regarded as influential observations in this study (
Figure 1
Figure 5
d). No. 373 station is located on the edge of Qinling Mountains in the southern Loess Plateau, where the altitude is nearly thousands of metres higher than the surrounding stations. With this sharp
rise in mountain altitude, the precipitation was also significantly increased. Northwest Loess Plateau, where the No. 53 station is positioned, belongs to semi-arid areas, where the desert is the
main landscape, and rainfall is scarce (100 mm–200 mm). However, due to the influence of Yellow River irrigation on the Hetao Plain, the no. 53 station had a higher NDVI than surrounding stations.
Thus, the two strong influential points were both caused by local underlying surface changes and could play an important role in accurate simulation of local area precipitation.
Geographically weighted regression was also implemented using the selected predictors with the stepwise regression. The optimal bandwidth is 87.8 km in this study because this distance ensured that
CV scores reached the minimum value of 45.98 in cross-validation methods. The distribution of local regression coefficients (
Figure 6
a–e) displayed abundant variation across the study area, especially those of NDVI, north slope aspect (N), and DEM. There were large differences between the mean GWR coefficients and the MLR
regression coefficients (
Figure 7
). All these confirmed the existence of spatial non-stationarity in the study area. The GWR global-adjusted
of 0.93 was much higher than that of MLR (0.44,
Table 2
), which indicates that the variance explained by GWR was much larger than by MLR. However, the degree of local variance explanations was uneven (
Figure 6
f) and especially poor in the Taihang and Luliang Mountains.
4.4. Regression Residuals Interpolation
Variograms of multiple linear regression residuals and geographically weighted regression residuals were calculated and then fitted with lag distances. The Ste. Model was found as the most suitable
for both residuals (
Figure 8
). The ratio of nugget and sill in the variogram model (
)), called the nugget effect, represents the spatial dependence structure, which can be explained as the proportion of spatial heterogeneity caused by random factors. The higher the ratio, the more
variations determined by random factors [
]. In general, a ratio of less than 25% indicates that there is a strong spatial dependence structure in the regression residuals; if the ratio was between 25% and 75%, the residuals had a moderately
intense spatial dependence structure; and until the ratio reached more than 75%, the spatial dependence structure was very weak,
, the regression residuals variability consists of unexplained or random variations. In this study, the nugget effect of MLR residuals was 22.16%,
, less than 25% (
Table 4
), which shows that the spatial variability was predominately caused by structural factors. A nugget effect of GWR residuals of 36.58% signifies that spatial variability caused by random factors in
GWR residuals was greater than in MLR residuals. These different degrees of spatial dependence structure in the two regression residuals could be induced by different degrees of trend elimination in
residuals of the two regression models.
In order to quantitatively assess the ordinary Kriging results of two regression residuals, a five-fold cross-validation method was implemented to obtain the prediction residuals, and then the
adjusted determination coefficients (adjusted R^2) were calculated. The adjusted R^2 of Kriging prediction for the residuals of MLR and GWR corresponded to 0.91 and 0.24, which indicates that the
degree of variance explanation of ordinary Kriging was much higher for MLR than for GWR.
4.5. Validation of MLRK and GWRK
The logit-transformed precipitation predicted by MLRK and GWRK can be calculated by adding the deterministic part by regression to the correspondingly interpolated residuals by Kriging. The
logit-transformed precipitations were reversed into the original scale using Equation (4), and then the back-transformed precipitation distribution was mapped, as shown in
Figure 9
. The precipitation distribution, which shows abundant spatial variation, was observably affected by environmental factors such as altitude (DEM) and vegetation. High values of precipitation mainly
clustered in the south of the Loess Plateau neighboring the Qinling Mountains where vegetation flourishes. Low values of precipitation appeared in the northwest of the Loess Plateau, which belongs to
the sparsely vegetated Ulan Buh Desert and Kubuqi Desert. The back-transformed precipitation of the two prediction methods was statistically analyzed in the range of the Loess Plateau without a 100
km buffer. The range of precipitation predicted by GWRK (100.1 mm–999.5 mm,
Figure 9
b) is wider than the range predicted by MLRK (136.6 mm–917.3 mm,
Figure 9
a). The precipitation distributions predicted by GWRK showed more spatial variation than those predicted by MLRK.
Comparing MLRK with GWRK, it is observed that the degrees of variance explanations in two steps of the RK were inconsistent. The adjusted
of the MLR model was substantially less than the geographically weighted regression GWR model; whereas the adjusted
of interpolated regression residuals by Kriging in MLRK was far more than in GWRK. Furthermore, owing to the logit-transformation of precipitation data, the entire prediction error could not be
directly evaluated but required complex conversions with the regression models’ errors and the ordinary Kriging interpolations [
]. Therefore, it was difficult to judge which method ultimately obtained more accurate precipitation predictions. As such, the validation data set was used to calculate the entire errors for two
regression Kriging interpolations. The means of back-transformed precipitations (MLRK: 455.3 mm/m
; GWRK: 466.4 mm) were lower than the mean of precipitation validation data (476.2 mm). The values of MAE, MAEr, and RMSE of MLRK were better than those of GWRK, but the values of ME and MEr of MLRK
were worse than that of GWRK. This implies that MLRK prediction errors on several validation points could be larger than GWRK, but the whole prediction error over all validation points was less than
GWRK (
Table 5
). The degree of variance explanation of MLRK was slightly better than that of GWRK, but in reality, the adjusted determination coefficients (adjusted
) of the two methods showed little difference.
5. Discussion
Based on background knowledge of the physical geography of the study area, the geographic factors that are closely related to local precipitation are selected as auxiliary variables. This process can
help improve the accuracy of the RK model. The results of this study show that the average annual precipitation in the Loess Plateau gradually decreased from southeast to northwest under the
influence of the monsoon and the sea (
Figure 9
). Affected by complex mountainous terrain, precipitation changed greatly in the Taihang Mountains and Luliang Mountain region in the east of the Loess Plateau. Precipitation in the narrow region
windward and to the east of the Taihang Mountains was significantly higher than in the leeward area of the western mountains and the plain area on the east side of the mountains. Due to the Tibetan
Plateau, the elevation of the western region of the Loess Plateau gradually increases, and the rainfall increases correspondingly. The Loess Plateau belongs to semi-humid and semi-arid areas, where
water availability is an important limiting factor for vegetation growth. Vegetation typically thrives in places where rainfall is abundant. Moreover, if the precipitation changes significantly, the
underlying natural geographical factors would also show a marked change. For example, in this study, the reason that station 373 had such a strong influence on observations is that it lies on the
Huashan Mountain, where altitude is much higher than the surrounding areas.
By exploring the characteristics of the annual average rainfall data, a suitable interpolation method can be chosen for precipitation predictions. Geostatistical interpolation and non-geostatistical
interpolation methods both involve various assumptions. In order to obtain a more accurate prediction, the data need to meet the assumed requirement [
]. For example, in order to make the distribution of data meet the requirements of the normality assumption, precipitation data were typically adjusted to a logarithmic scale. Of course, it is also
possible to use an interpolation method that does not require a normality requirement calculation.
In choosing a suitable interpolation method, it is import to consider whether there is a need for precipitation data to fit trends in the interpolation process, and then further consider whether to
fit the global trend or local trends according to the stability of the underlying surface. In this study, MLRK is a global trend-fitting method, which is suitable for a relatively homogeneous
underlying surface, and GWRK is a local area trend-fitting method, which is suitable for relatively complex surface types [
]. Previous studies suggest that accuracy of the local trend-fitting method is generally higher than the overall trend-fitting method [
]; however, this is not always true. With an increase in the size of the study area and the complexity of the underlying surface, the relationship between rainfall and auxiliary variables will become
more unstable. When the underlying surface changes greatly but precipitation is relatively stable, the variability in predicted rainfall may be overestimated in local trend fitting, thus reducing the
accuracy of forecasts. This type of situation was recognized in this study. The results showed that the entire GWRK error was slightly larger than the MLRK error. In summary, it is difficult to
choose the right interpolation method when starting a precipitation interpolation. The advisable way to obtain better interpolation results is by selecting several available interpolation methods,
comparing the error after interpolation, and then selecting the method with the minimum error.
The interpolation result of annual average rainfall is affected by several uncertainty factors: (1) an uneven distribution and limited number of stations leads to the underestimation of trends and
rainfall variability in these areas; (2) since the relationship between rainfall data and the auxiliary variable is unstable, a more accurate model cannot be precisely established, decreasing the
interpolation result accuracy; (3) rainfall station data in most studies are limited to within the study area. A significant prediction error in interpolation appears at the boundary of the study
area, known as “edge effects.” These effects could be mitigated by taking into account the precipitation data of stations just outside the border [
]. In this study, the precipitation data of meteorological stations located at the 100 km peripheral buffer of Loess Plateau was added in order to improve the prediction results of the study area
Measuring the interpolation error is an important process in interpolation method selection and analysis of interpolation results. In this study, two common methods were used to evaluate the results
of interpolation: one using the predictive data set itself with cross-validation; the other using a verification data set to directly calculate ME, RMSE, and other indexes. When using the validation
data set, the validation points should cover a broad range of land use types. In this study, the meteorological data of the validation data set is only from hydrological stations at the edge of a
river or gully with lush vegetation coverage. This validation data set is therefore lacking in data from other land use types. Therefore, this study can clearly show that MLRK prediction performance
was slightly better than GWRK in the area around the river, but in regions with other types of land use, it is not possible to determine which of the two models is better.
6. Conclusions
One of the main objectives of this study was to generate a highly accurate distribution map of average annual precipitation with a 500 m spatial resolution in the Loess Plateau for the period of
1980–2010. Alternative distribution maps of precipitation were interpolated by MLRK and GWRK methods and all showed a high accuracy. There were large disparities, however, in two regression Kriging
processes using different methods: the variance explanation of the GWRK regression model was higher than that of MLRK, but the contrary is true of the Kriging process. The interpolation maps using
MLRK and GWRK both captured many details of spatial distribution influenced by predictors. Although the GWRK is based on the spatial non-stationary assumption and the map predicted by GWRK did show
greater spatial variation, the final validation analysis revealed that MLRK yielded higher model efficiency than GWRK, with small differences. This is in contrast to other previous precipitation
interpolation studies. The conclusions can be summarized as follows: (1) both MLRK and GWRK are able to incorporate multiple auxiliary environmental factors into the modelling process and obtain a
highly accurate precipitation distribution map; (2) unlike other studies of precipitation prediction, MLRK is shown to be a better method for precipitation interpolation when the underlying surface
is complex. In future, greater effort should be made to consider more physical geographic factors related to or impacted by rainfall as auxiliary environmental variables, which could possess a higher
resolution, in order to get a higher accuracy of precipitation distribution maps. More studies should investigate and identify the standards and principles to select and validate the interpolation
methods further.
This study was funded by the Chinese Forestry Research Special Funds for Public Welfare Project (201404209) and the National Basic Research Program of China (2013CB429901). We thank Edzer Pebesma,
Ying Hu, and Xianrui Fang for technical support, Yuting An and Shan Wang for language assistance, Xiaoye Li and Shuguang Zhang for supporting the research facilities and writing environment, and
Qun’ou Jiang for revision suggestions. Finally, we thank the anonymous reviewers for the valuable comments and suggestions.
Author Contributions
All authors contributed to the data extraction. Qiutong Jin and Jutao Zhang designed the methods. Mingchang Shi and Jixia Huang undertook the data analysis. All authors contributed to the drafting of
the article and approved the final manuscript. Mingchang Shi is the guarantor.
Conflicts of Interest
The authors declare no conflict of interest. The founding sponsors had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript,
and in the decision to publish the results.
The following abbreviations are used in this manuscript:
CV Cross-Validation
DEM Digital Elevation Model
DEM_std Standardized Normal Variable of Digital Elevation Model
E Dummy Variable of East Aspect
GWR Geographically Weighted Regression
GWRK Geographically Weighted Regression Kriging
MAE Mean Absolute Error
MAE[r] Mean Absolute Relative Error
ME Mean Error
ME[r] Mean Relative Error
MLR Multiple Linear Regression
MLRK Multiple Linear Regression Kriging
N Dummy Variable of North Aspect
NE Dummy Variable of Northeast Aspect
NW Dummy Variable of Northwest Aspect
NDVI Normalized Difference Vegetation Index
NDVI_std Standardized Normal Variable of Normalized Difference Vegetation Index
OK Ordinary Kriging
OLS Ordinary Least Squares
PRE Annual Average Precipitation from meteorological stations
PreT Logit-Transformed Precipitation
rad_std Standardized Normal Variable of Solar Radiation
RK Regression Kriging
RMSE Root Mean Square Error
slope_std Standardized Normal Variable of Slope
S Dummy Variable of South Aspect
SE Dummy Variable of Southeast Aspect
Ste. Matern with Stein’s Parameterization
SW Dummy Variable of Southwest Aspect
W Dummy Variable of West Aspect
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Figure 1. Underlying surface features of the Loess Plateau (a); and location of meteorological and hydrologic stations (b).
Figure 2. Structural characteristics of annual average precipitation at meteorological stations. (a) Distribution of annual average precipitation; and (b) correlation of precipitation for different
stations at certain lag distance ranges.
Figure 4. Relationship between logit-transformed precipitation (PreT) and main environmental variables, including the standardized normal variable of NDVI (a); the standardized normal variable of DEM
(b); the standardized normal variable of Slope (c); and the standardized normal variable of solar radiation (d).
Figure 5. Diagnostic plots for multiple linear regression analysis. (a) is the scatter plot for the regression-predicted values versus the corresponding residuals; (b) is the normal Q-Q diagram; (c)
is the scale-location graph; and (d) is the Cook’s distance diagram.
Figure 6. Maps of GWR coefficients and local adjusted R^2. (a) is the distribution of DEM coefficient; (b) is the distribution of slope coefficient; (c) is the distribution of NDVI coefficient; (d)
is the distribution of north aspect coefficient; (e) is the distribution of northwest aspect coefficient; and (f) is the distribution of local R^2.
Figure 7. Comparison of regression coefficients between GWR and MLR. The orange points represent the mean of predicator coefficients in GWR, and the blue points represent the regression coefficients
of MLR.
Figure 9. Precipitation climatology map of the 30 years (1980–2010) at Loess Plateau predicted by MLRK (a) and GWRK (b).
Table 1. Pearson correlation matrix between logit-transformed precipitation (PreT) and environmental variables.
Table 1. Pearson correlation matrix between logit-transformed precipitation (PreT) and environmental variables.
Variables PRE DEM_std NDVI_std rad_std slope_std N NE E SE S W NW SW
PRE 1 – – – – – – – – – – – –
DEM_std −0.331 ** 1 – – – – – – – – – – –
NDVI_std 0.600 ** −0.386 ** 1 – – – – – – – – – –
rad_std −0.349 ** 0.968 ** −0.385 ** 1 – – – – – – – – –
slope_std 0.315 ** 0.286 ** 0.097 * 0.172 ** 1 – – – – – – – –
N −0.134 ** 0.022 −0.042 −0.009 −0.090 1 – – – – – – –
NE −0.019 0.069 −0.080 −0.011 0.067 −0.138 ** 1 – – – – – –
E 0.051 0.011 0.074 0.010 −0.029 −0.133 ** −0.172 ** 1 – – – – –
SE 0.028 −0.086 0.061 −0.039 −0.096 * −0.148 ** −0.191 ** −0.185 ** 1 – – – –
S 0.080 −0.127 ** 0.061 −0.068 −0.035 −0.127 ** −0.164 ** −0.158 ** −0.176 ** 1 – – –
W 0.014 0.052 −0.068 0.114 * 0.095 * −0.123 * −0.159 ** −0.153 ** −0.170 ** −0.146 ** 1 – –
NW −0.033 0.052 −0.027 −0.014 0.091 −0.103 * −0.133 ** −0.128 ** −0.142 ** −0.122 * −0.118 * 1 –
SW 0.014 0.052 −0.068 0.114 * 0.095 * −0.123 * −0.159 ** −0.153 ** −0.170 ** −0.146 ** 1.000 ** −0.118 * 1
PreT 0.985 ** −0.306 ** 0.578 ** −0.324 ** 0.309 ** −0.133 ** −0.011 0.052 0.023 0.080 0.021 −0.051 0.021
** correlation is significant at the 0.01 level (two-tailed); * correlation is significant at the 0.05 level (two-tailed).
Table 2. Results of the stepwise linear regression analysis.
Models R^2 Adjusted R^2 Residuals SE F-statistic p-Value
a * 0.4481 0.4338 0.6046 31.23 <2.2 × 10^−16
b ** 0.4465 0.4400 0.6012 69.21 <2.2 × 10^−16
* starting formula is “PreT~DEM_std + slope_std + rad_std + NDVI_std + N + NW + W + SW + S + SE + E + NE”; ** the order of eliminated variables is: SW, SE, W, NE, E, S, rad_std. Final formula is
“PreT~DEM_std + slope_std + NDVI_std + N + NW”.
Table 3. Summary of statistically significant influence of
all predictors.
Coefficients Estimate Std. Error t Value P(>|t|)
a −0.2110 0.0404 −5.224 2.73 × 10^−7 ***
b 0.3903 0.0466 8.376 7.89 × 10^−16 ***
c 0.5449 0.0477 11.414 <2 × 10^−16 ***
d −0.2341 0.0986 −2.375 0.0180 *
e −0.1856 0.1019 −1.821 0.0692
f −0.2903 0.0414 −7.022 8.64 × 10^−12 ***
*** significant at the 0.001 level (two-tailed); ** significant at the 0.01 level (two-tailed); * significant at the 0.05 level (two-tailed).
Table 4. Variogram models for MLR residuals and GWR residuals.
Residuals Model Nugget (C[0]) (km^2/m^4) Partial Sill (C) (km^2/m^4) Range (m) Kappa C[0]/(C[0] + C) (%)
MLR Ste. 0.1,182 0.4151 590,950.4 1.9 22.16
GWR Ste. 0.01,537 0.02,665 66,963.64 10 36.58
Table 5. Comparison of MLRK and GWRK performances with validation data.
Method Adjusted R^2 ME (mm/m^2) ME[r] (%) MAE (mm/m^2) MAE[r] (%) RMSE (mm/m^2)
MLRK 0.87 −20.88 −3.82 30.85 6.58 40.05
GWRK 0.85 −9.80 −1.29 35.75 7.78 43.24
© 2016 by the authors; licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC-BY) license (http:/
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How to create a formula that only affects part of the column...
For the days that are marked DBL in the Time column, we are trying to adjust the formula in the Arrival Time column so that only the first 4 rows are 1.5 hours before the time listed in the Time
column and the rest of that day is 1 hour before the listed time...i.e.. 08:15 = 06:45 (1.5 hours) and 08:30 = 07:30 (1 hour). The current formula used in the Arrival Time column is...
"=IF(lvl@row = 0, " ", VLOOKUP(Time@row, {Time Range 1}, 5, false))"...
Below are the screen shots for the Surgery Scheduling sheet we are using as well as the reference sheet with the times to post in the Arrival Time column...
Thank you in advanced...
Best Answer
• Hi @JJLewis,
It seems we need to index rows in your sheet to determine RowID, then setup formulas depend on their RowID.
First, create RowID column with a formula :
=COUNTIF(Date$1:Date@row, Date@row) - 1
If RowID<=4 : use 1.5 hours before the time listed in the Time column, else 1 hour
Then modify your formula in the Arrival Time column as below:
=IF(lvl@row = 0, " ", IF(RowID<=4, VLOOKUP(Time@row, {Time Range 1}, 5, false), VLOOKUP(Time@row, {Time Range 1}, 6, false)))
Hope that helps.
Gia Thinh Technology - Smartsheet Solution Partner.
• Hi @JJLewis,
It seems we need to index rows in your sheet to determine RowID, then setup formulas depend on their RowID.
First, create RowID column with a formula :
=COUNTIF(Date$1:Date@row, Date@row) - 1
If RowID<=4 : use 1.5 hours before the time listed in the Time column, else 1 hour
Then modify your formula in the Arrival Time column as below:
=IF(lvl@row = 0, " ", IF(RowID<=4, VLOOKUP(Time@row, {Time Range 1}, 5, false), VLOOKUP(Time@row, {Time Range 1}, 6, false)))
Hope that helps.
Gia Thinh Technology - Smartsheet Solution Partner.
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How does SAS assist in Multivariate Analysis of categorical data? | Hire Someone To Take My SAS Assignment
How does SAS assist in Multivariate Analysis of categorical data? This is of course an oversimplification for various reasons (the list didn’t include information on models from model selection, even
though many models have several dimensions). We would like to collect all of the dimensions (in the context of why not find out more SAS) of relevant data sets and set up confidence intervals for SAS
models. We have different systems to handle categorical columns, although we share language for the terms of the form factors. We can pick up a number of different expressions in the form of _a_…,
_b_ and _c_. This is just mathematically equivalent to [^] used in the spreadsheet view of CIC or _a_, as its sub-terms are her latest blog the discrete values of the coefficients in _x_, _y_ or _z2_
of a given data set. You can deal with the discrete data by taking the common index _n_ for the corresponding terms/sub-terms of every term/sub-model. Here, we would like to pick up the number 1 (for
each aggregate model in the context of discrete data collection), as it was in the first exercise by Michael Neill who wrote the next chapter. First, we need to determine which of the terms represent
those that are similar, or the least to lowest in the aggregate number of terms in the data set. So from the formula of CIC, we get w = w | _n_, for _n_ = 0, 1[1], 1… _n_ ; from the result of the CIC
calculation, we can obtain that [^] = 1, as [^] = 1. Since the aggregate model has _n_ of the same number as the data set, we have [^ ] 3 3 [^] = 3 3, for _n_ = 0, 1… _n_. So we need to take the
common index _n_ of the terms in the aggregate model accordingly to [^], as for [^ ] 3 3 [^] = w 3, for _n_ = 0,_.
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.. _n_. Now, for every aggregate model, we can get that w \le -1 for _n_ ≠ ≤, with a confidence interval around 0.96. We can use a confidence interval of 0.96 for the discrete model if their
explanation = 0. (We can also do the test of normalization to get other confidence intervals.) Finally, we pick up a formula (for some specific cases) to guide any subsequent calculations. For
example, we could take the sum of two sums over f(1) = 1, with α |f(1), _n_ = 1, and the ordinary least-squares part of the formula for _x_. This makes it possible to find the sum of m × α for all
the terms _1_ through _n_ + 1. We can use this modelHow does SAS assist in Multivariate Analysis of categorical data? A data analysis of categorical data typically requires that a function contains
all categories and data, which are the most commonly used categorization systems ever to carry out. In this section, SAS will be used to give some examples of some of the most commonly used
categorical binary functions, which are commonly used. Section 4 gives some examples of what we will be using the function, and later will also provide a much more specific description of the
relevant techniques. Binary Functions (SQL) SAS is among the most useful operating systems in software development, and is the functional programming language. With SAS, most data structures would
look like this: type; functions data; functions list; type(1); data; type(1); functions list list; type(2); functions list list list list list list list list list list list list list list list list
list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list list
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list list list sorted list list list sorted list sorted list list sorted list list sorted list list orderedordered ordered alternating alternating alternating alternation ordered alternating
alternating alternation ordered alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating alternating
alternating alternating alternating alternatingHow does SAS assist in Multivariate Analysis of categorical data? SAS provides an online tool for conducting multivariate analysis of categorical data.
SAS is used to analyze categorical parameters of a particular set of variables ranging from the level of unity to minor deviations in the individual variables. Each variable is coded as a separate
variable and applied to a random sample of data arranged in a hierarchical manner over a number of levels of multiple categorical variables. This hierarchical model is then used to plot statistical
parameters of the random samples of data representing each variable in the high level samples. The above-understood approach is a method whereby an individual data set is analyzed to determine the
presence or absence of a particular feature or any other specific statistical characteristic within a set of variable data.
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The low level samples are compared to an uncorrelated do my sas homework distribution with the top 0.2 degrees of freedom of the data considered being p. and the normal distribution centered at the
top of the data set. The higher the data-set, the greater is the significance from which the classification is assessed. For example, here is an original view of a selected single person data set,
which is subject to two possible errors: The first is due to overde-predicting the data, and the second due to over-classifying or perhaps inaccurate in some statistical or non-statistical ways,
according to the following observation using a standardized standard deviation: x = 10, i.e. -30, while the standard deviation increases as x increases from a nominal value of 0 to 1000 p. The idea
is to test a number of known associations made by the individual, the frequency a certain trait in a group of individuals, for which statistical testing has been performed in the high level samples
and the normal or univariate sample, based on a standardized standard deviation that is not close to zero (0), based on such frequency statistics that do not vary by 0.07. An example of this is shown
in [figure 12] which shows the simple example in which the high level data represents the presence or absence of 2, 12, 24, 10 and 10-phenylhexylsulfonaphthalene and the normal or univariate sample.
In short, if SAS was to perform the normalization, the sample would be a group of independent non-parametric data, data with standard deviation 0.07. It should be noted that this standard deviation
does not increase, for example, when including the S3 term in the sample (data of interest: [Fig. 12](#f12){ref-type=”fig”}). However, for a lack of better details, the method is able to describe the
statistical properties of a subset of different genetic (structure) determinants as described blog This means that an individual data set consisting of a set of samples may not exactly be
characterized, but one dataset consisting of relatively large samples may perhaps be quite representative and are used, if not entirely, for statistical purposes. This could | {"url":"https://sashelponline.com/how-does-sas-assist-in-multivariate-analysis-of-categorical-data","timestamp":"2024-11-07T09:19:50Z","content_type":"text/html","content_length":"131173","record_id":"<urn:uuid:d16b7ab4-ff43-4839-b46d-d50907aa0545>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00006.warc.gz"} |
Finiteness Theorems for Limit Cyclessearch
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Finiteness Theorems for Limit Cycles
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Finiteness Theorems for Limit Cycles
Hardcover ISBN: 978-0-8218-4553-0
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MAA Member Price: $139.50
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• Translations of Mathematical Monographs
Volume: 94; 1991; 288 pp
MSC: Primary 34; 58; Secondary 14; 41; 57
This book is devoted to the following finiteness theorem: A polynomial vector field on the real plane has a finite number of limit cycles. To prove the theorem, it suffices to note that limit
cycles cannot accumulate on a polycycle of an analytic vector field. This approach necessitates investigation of the monodromy transformation (also known as the Poincaré return mapping or the
first return mapping) corresponding to this cycle. To carry out this investigation, this book utilizes five sources: The theory of Dulac, use of the complex domain, resolution of singularities,
the geometric theory of normal forms, and superexact asymptotic series. In the introduction, the author presents results about this problem that were known up to the writing of the present book,
with full proofs (except in the case of results in the local theory and theorems on resolution of singularities).
□ Chapters
□ Introduction
□ Chapter I. Decomposition of a monodromy transformation into terms with noncomparable rates of decrease
□ Chapter II. Function-theoretic properties of regular functional cochains
□ Chapter III. The Phragmén-Lindelöf theorem for regular functional cochains
□ Chapter IV. Superexact asymptotic series
□ Chapter V. Ordering of functional cochains on a complex domain
□ Excellent book ... devoted to a rigorous proof of this finiteness theorem, and some related results are proved along with it ... this valuable and interesting book will give the readers a
good understanding of this deep and elegant work, and ... more and more mathematicians will be interested in solving Hilbert's difficult 16th problem.
Mathematical Reviews
□ The viewpoint is high and the techniques are delicate and profound.
Zentralblatt MATH
□ An indispensable component of a complete mathematical library ... one is struck by the originality, the creative power, the depth of thought, and the technical facility demonstrated. In
Professor Mauricio Peixoto's words, this is ‘mathematics of the highest order’. Many of those who love mathematics will find treasure here.
Bulletin of the London Mathematical Society
• Permission – for use of book, eBook, or Journal content
• Book Details
• Table of Contents
• Reviews
• Requests
Volume: 94; 1991; 288 pp
MSC: Primary 34; 58; Secondary 14; 41; 57
This book is devoted to the following finiteness theorem: A polynomial vector field on the real plane has a finite number of limit cycles. To prove the theorem, it suffices to note that limit cycles
cannot accumulate on a polycycle of an analytic vector field. This approach necessitates investigation of the monodromy transformation (also known as the Poincaré return mapping or the first return
mapping) corresponding to this cycle. To carry out this investigation, this book utilizes five sources: The theory of Dulac, use of the complex domain, resolution of singularities, the geometric
theory of normal forms, and superexact asymptotic series. In the introduction, the author presents results about this problem that were known up to the writing of the present book, with full proofs
(except in the case of results in the local theory and theorems on resolution of singularities).
• Chapters
• Introduction
• Chapter I. Decomposition of a monodromy transformation into terms with noncomparable rates of decrease
• Chapter II. Function-theoretic properties of regular functional cochains
• Chapter III. The Phragmén-Lindelöf theorem for regular functional cochains
• Chapter IV. Superexact asymptotic series
• Chapter V. Ordering of functional cochains on a complex domain
• Excellent book ... devoted to a rigorous proof of this finiteness theorem, and some related results are proved along with it ... this valuable and interesting book will give the readers a good
understanding of this deep and elegant work, and ... more and more mathematicians will be interested in solving Hilbert's difficult 16th problem.
Mathematical Reviews
• The viewpoint is high and the techniques are delicate and profound.
Zentralblatt MATH
• An indispensable component of a complete mathematical library ... one is struck by the originality, the creative power, the depth of thought, and the technical facility demonstrated. In Professor
Mauricio Peixoto's words, this is ‘mathematics of the highest order’. Many of those who love mathematics will find treasure here.
Bulletin of the London Mathematical Society
Permission – for use of book, eBook, or Journal content
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Measuring the Mystery - Graham Hancock Official Website
Please welcome for September 2014 Author of the Month Scott Onstott. Scott is the creator of the Secrets In Plain Sight video series, an inspiring exploration of great art, architecture, and urban
design unveiling the unlikely intersection of geometry, mysticism, physics, music, astronomy, and world history. Join Scott on the AoM Message Boards during Sept., to interact, discuss and explore
the mysteries in numbers and measurement that surrounds us all.
My work and research has led me to believe that measurement is a doorway into the mysteries. Last year, I shot this photo of the symbol over the door into La Maddalena round church in Venice. When I
got home I did a geometric analysis in AutoCAD and discovered that the golden rectangle beautifully measures the gap between the equilateral triangle and its circumcircle, which I later illustrated
in my book Quantification. In addition, the smaller interwoven circle and triangle have the same areas.
This last observation reminded me of another discovery I had made earlier—the golden rectangle is the bridge to making the cube commensurate with the sphere. This comparison of volumes is a
higher-dimensional analogy to triangulating the circle by area. I also rewrote the traditional formula for the volume of a sphere to emphasize the important pattern of repeating threes, which itself
suggests a triangle or trinity.
The Great Pyramid’s elevation can be closely approximated with two golden rectangles tilted to meet at an apex. I found that Leonardo’s Vitruvian Man (which I’ve revised to include the other half of
humanity) resonates with the golden geometry. The lines converging on the navel suggest that the human body was made to golden measure.
I made the following image called The Mystery of 273 to highlight the many phenomena converging on these three digits, including the relative proportion of Moon to Earth, the month, cycles of human
reproduction, and water—that amazing substance which comprises the majority of the molecules of our bodies and the majority of our planet’s surface. All of this is encoded in squaring the circle.
By inscribing an equilateral triangle within the Great Pyramid’s elevation, an amazing resonance occurs with imperial units of measure. The edge length is 555.5 feet, which equals 6666 inches. The
inner equilateral triangle surrounding the all-seeing-eye of Horus measures 3333 inches on each of its 3 edges.
These measurements appear when you use the base edge length of 755.775 feet. If you multiply by 4 to calculate the complete base perimeter, the answer is 1007.7 yards. This is the base perimeter J.H.
Cole surveyed in 1925. In 1880, Prof. W. M. Flinders Petrie surveyed it less than a third of an inch smaller at 1007.6 yards but I think Cole was correct based on the resonances which emerge based on
a mean pyramid edge length of 755.775 feet.
I’ll address the question of how the ancient Egyptians could have been using what we know as the English foot. Instead of reflexively dismissing this as impossible, let us temporarily set this
important issue aside so we can perceive another encoding in the Great Pyramid.
In the following diagram I analyzed the Great Pyramid’s elevation using the metre. 755.775 feet = 230.360 metres. The diameter of the top circle is 10π metres.
The word geometry literally means earth-measure. Systems of units that are based on sacred geometry that might have been lost in a cataclysm or forgotten over time can be rediscovered by subsequent
cultures by accurately measuring the Earth. John Michell and Robin Heath showed in The Lost Science of Measuring the Earth (Adventures Unlimited Press 2006) that the Imperial system is based on Earth
measure. They made the following claim:
Earth’s equatorial circumference = 365.242 x 360 x 1000 feet
In other words the number of days in a solar year times the number of degrees in a circle times a scaling value of 1000 equals the Earth’s maximum circumference in feet. Note that the base-10 numeral
system is implied in this equation.
The system used by GPS satellites, WGS84, lists the Earth’s equatorial circumference as 24,901.4 miles and if you do the math, the accuracy of the above claim is 99.99%. This is an astounding level
of accuracy with error down to one-hundredth of one percent; it is off by just 1 mile!
The metric system was originally based on measuring the Earth’s meridian circumference. The distance from the equator to the North pole was divided into 10 million parts called metres.
I analyzed the combined diameters of Earth and Moon, thinking of our planet and its satellite as a system. The polar diameters of Earth & Moon together equal 10000 x Φ kilometres (99.97%). Φ is the
golden ratio, which is approximately 1.61803…
The metre therefore resonates beautifully with the Earth and the Moon. Perhaps it was merely rediscovered in the 18^th century? That would explain how we find it encoded in the Great Pyramid today.
The mean diameters of Earth & Moon together equal 10077 miles (99.99%). Remember, the base perimeter of the Great Pyramid is 1007.7 yards. These are the same digits but they measure distances in
different units. This pattern is one I see time and time again suggesting that the digits matter most.
The nautical mile is another interesting unit, being a distance that is approximately one minute of arc measured along any meridian. There are 6x6x6x100 minutes of arc in the meridian circumference.
By international convention, one nautical mile is defined as 1852 metres. It’s curious but 1851.85 yards is exactly 5555.55 feet. And 5555.55 feet is exactly 66,666.6 inches. My measurements have
repeatedly turned up distances with repeating digits and I believe these are also a key.
By inscribing a square rotated 45° in the elevation of the Great Pyramid amazing resonances appear. I circumscribed the square with a circle and placed a 3D model of the Earth inside and then rotated
the Earth on axis so that Giza was on the centerline. Adding the human figure seated in a meditation posture revealed that the Nile corresponds to a meandering flow of kundalini energy from the heart
to the pineal gland.
Simultaneously the same sacred geometry encodes a profusion of repeating sevens revealed in the following illustration.
Sacred geometry is scale-invariant so it can be used to measure anything from the body, to buildings, cities, the Earth and beyond.
The Great Pyramid was originally clad in highly polished white limestone casing stones. A few of the casing stones survived and that’s how we know the slope angle of the pyramid is 51°51’ and can
therefore calculate that its design height was 481 feet.
The Great Pyramid must have looked like a gleaming white mountain when it was built. Strangely enough the highest peak in Europe, Mont Blanc, is 4810 meters in elevation.
Mont Blanc is 481 kilometers from the Louvre Pyramid, which was constructed to match the Great Pyramid’s slope angle. The Louvre Pyramid is the skylight that illuminates the entrance to the world’s
most visited museum.
I look at distances between sacred sites quite often in my work. For example, the distance between the Kaaba in Mecca and the Western Wall in Jerusalem is precisely 666.6 nautical miles. Was this
intentional or is it just a coincidence?
The Earth’s axis is tilted 66.6° from the ecliptic (90 – 23.4 = 66.6). We are traveling at approximately 66600 miles per hour (99.9% accuracy) in our journey around the Sun.
Africa measures 66.6° wide from its westernmost to easternmost points. A line drawn from the southernmost point in Africa, due north to the point where it hits the Mediterranean, measures 66.6° on
the map. The Mediterranean is 33.33° wide from the rock of Gibraltar to its easternmost point near Beirut.
I am amazed that I have found so many significant Earth measurements with these repeating sixes. How can this be? Each of us will undoubtedly interpret this data quite differently.
“We don’t see things as they are; we see them as we are.” -Anaïs Nin
This leads me finally to another epiphany (e-π-Φ-ny) I had, that the infinitesimal gap in what would otherwise be a perfect Pythagorean triangle connecting e, π, and Φ suggests the question, “Why
can’t we ever rationalize the transcendental?”
Logically, the answer is that it is an irreducible eternal mystery.
Scott’s websites are www.secretsinplainsight.com and www.scottonstott.com
There is lots of free content on his blog at secretsinplainsight.com/blog
To purchase Quantification, Taking Measure or Secrets In Plain Sight volumes 1 and 2, please visit http://www.secretsinplainsight.com/store
Scott Onstott is the creator of the Secrets In Plain Sight video series, an inspiring exploration of great art, architecture, and urban design unveiling the unlikely intersection of geometry,
mysticism, physics, music, astronomy, and world history. Secrets In Plain Sight has been seen by more than 3 million people to date. He is also the author of 11 technical books on architectural
software and 4 more books on esoteric subjects. Scott just published Quantification, a book of his thought-provoking color illustrations. Quantification examines patterns in the Great Pyramid, in the
human body and in the Earth, and reveals uncanny distances between sacred sites. Scott has a degree in architecture and worked designing corporate interiors for a decade in San Francisco before
becoming an independent teacher, author and filmmaker. | {"url":"https://grahamhancock.com/onstotts1/","timestamp":"2024-11-05T22:33:54Z","content_type":"text/html","content_length":"50144","record_id":"<urn:uuid:de5a6daa-1575-4bab-8dc0-dbffd04747a5>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00003.warc.gz"} |
Frac Sleeves
Ball Drop Frac Sleeve is а part of an openhole fracturing system designed to allow operators to perform selective single-point multistage fracturing. The inner sleeve is run in а pinned configuration
and sheared after pressure increasing when activation ball is landed, providing positive indication that the specified port has opened before fracturing.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells.
Max OD in:
4.646 / 3.465 / 3.465 / 3.465
Liner size in:
4.000 / 4.000 / 5.500 / 4.500 / 5.750
Ball Drop Frac Sleeve is а part of an openhole fracturing system designed to allow operators to perform selective single-point multistage fracturing. The inner sleeve is run in а pinned configuration
and sheared after pressure increasing when activation ball is landed, providing positive indication that the specified port has opened before fracturing.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells
Hydraulic Activated Тое Sleeve with lsolation Valve is designed to bе used as а first stage for multistage hydraulic fracturing. Hydraulic activated Тое Sleeve in terms of construction consists of an
activation sleeve and а toe sleeve.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells
Hydraulic Frac port is а part of а robust cemented or openhole fracturing system designed to allow operators to perform selective multistage hydraulic fracturing. Fullbore sleeve designed for the
most common high-pressure and high-rate hydraulic fracturing. Hydraulic activation eliminates the need of ball usage.
• Cemented casing / liner applications.
• Vertical, directional and horizontal wells.
Max OD in:
4.567 / 5.236 / 5.748
Liner size in:
4.000 / 4.500 / 5.000
Ball Drop Frac Port Reclosable is а Key-operated Frac Sleeve is designed to provide access for the process fluid to the isolation zone during multistage hydraulic fracturing. The sleeve is opened
hydraulically bу pumping an activation ball into а special seat and applying pressure.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells.
Max OD in:
4.567 / 5.236 / 6.732 / 5.630
Liner size in:
4.000 / 4.500 / 5.500 / 4.500mod
Ball Drop Frac Sleeve Reclosable is designed for multistage hydraulic fracturing with activation balls of the corresponding size. The sleeve is opened hydraulically bу pumping an activation ball into
а special seat and increasing the tubing pressure.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells.
Liner size in:
4.500 / 4.500
Hydraulic Тое Sleeve Reclosable is designed to provide access for the process fluid to the isolation zone during multistage hydraulic fracturing. The sleeve is opened hydraulically bу pumping an
activation ball into а special seat and applying pressure. Circulation ports are opened after activation ball is seated and the pressure is increased to the activation pressure value.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells.
Liner size in:
5.500 / 4.500
Hydraulic Frac Port is designed to provide access for the process fluid to the isolation zone during multistage hydraulic fracturing without activation balls.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells.
Max OD in:
5.236 / 6.732 / 5.630
Liner size in:
4.500 / 5.500 / 4.500
Hydraulic Activated Тое Sleeve with lsolation Valve Reclosable is designed for the first interval of multistage hydraulic fracturing using activation balls, without the ability to close the sleeve.
The sleeve design includes an activation sleeve and а hydraulic toe sleeve.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells.
Full Bore Hydraulic Frac Port Reclosable is designed to provide access for the process fluid to the isolation zone during multistage hydraulic fracturing. The sleeve is opened hydraulically bу
pumping an activation ball into а special seat and applying pressure.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells
REVOLVER Full ID One-Size Ball Operated Frac Sleeve is designed for multistage hydraulic fracturing without the need to mill out the seats. The sleeve cаn bе used with hydromechanical packers for
isolating the fracturing zones.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells.
Plug Operated Frac Port is designed to provide an access for the process fluid to the isolation zone during hydraulic fracturing using а soluble key-plug.
• Uncemented casing / liner applications.
• Vertical, directional and horizontal wells
Liner size in:
4.000 / 4.500
Frac port for cementing operations is designed for multistage hydraulic fracturing. The sleeve is opened hydraulically bу pumping an activation ball into а special seat and then pressurizing the
• Cemented casing / liner applications.
• Vertical, directional and horizontal wells.
Max OD in:
4.567 / 5.236 / 7.087
Liner size in:
4.000 / 4.500 / 5.500
Burst-Port Frac Sleeve is designed to perform multistage fracturing using selective packer in cemented/non-cemented wells. The tool сап also bе used as the first fracturing stage.
• Cemented / non-cemented casing / liner applications.
• Vertical, directional and horizontal wells.
Liner size in:
4.500 / 4.500
Burst-Port Frac Sleeve is designed to perform multistage fracturing using selective packer in cemented/non-cemented wells. The tool саn also bе used as the first fracturing stage.
• Cemented / non-cemented casing / liner applications.
• Vertical, directional and horizontal wells. | {"url":"https://tss-group.ru/en/catalog/frac-sleeves/","timestamp":"2024-11-08T09:23:17Z","content_type":"text/html","content_length":"226731","record_id":"<urn:uuid:1902fcba-0adf-445c-8b50-8635b094cde9>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00127.warc.gz"} |
seminars - Defining smallness and independence in arbitrary mathematical structures from a model-theoretic perspective.
Given a mathematical structure and a language that describes the structure, we can consider sequences in the structure that are homogeneous with respect to the language. In model theory, these
sequences are called 'indiscernible sequences', and using this, we can induce notions of 'smallness' and 'independence'. In this talk, we will introduce ideas and methods for doing this. | {"url":"https://www.math.snu.ac.kr/board/index.php?mid=seminars&l=en&sort_index=date&order_type=desc&page=6&document_srl=1223945","timestamp":"2024-11-11T15:57:31Z","content_type":"text/html","content_length":"47689","record_id":"<urn:uuid:49048dcf-2126-446c-bd01-c639c7452a72>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00535.warc.gz"} |
Quick Scripts - CIAO 4.16 Sherpa
Quick Scripts
This page provides quick access to the Sherpa 4.16 Python scripts used in the Sherpa threads. Each script below is also included in the "Scripting It" section at the bottom of the corresponding
Fitting Data
Plotting Data
Computing Statistics
Simulating Data | {"url":"https://cxc.harvard.edu/sherpa/scripts/index.html","timestamp":"2024-11-09T06:10:43Z","content_type":"text/html","content_length":"24675","record_id":"<urn:uuid:e3e808e0-f75d-43b2-a568-c9adf8f1c209>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00557.warc.gz"} |
Count within groups – formula
Tested in Excel 2016 (16.0.9226.2114) 64-bit
You have a set of data containing information about a car fleet:
Of course, one car can be used by several drivers and one driver can drive several cars. You are supposed to build a new column showing how many drivers drove each car. You need to keep the dataset
as it is and just add the new column.
This file contains dummy data, the formula and the steps needed (no macros).
Method 1: Sorting the data
See sheet [1 Sorting] from the above mentioned file.
You need a help column, let’s call it Drivers rank and the final column Drivers per car.
You need to count for each Car plate the number of Driver so sort the data at least by these 2 columns, in this order.
Logic: if the Car plate is the same but the Driver changed, increase the rank. Type in row 2, Driver rank column:
1. =IF(A2=A1,IF(B2<>B1,IFERROR(E1+1,1),IFERROR(E1/1,1)),1)
You can use the header here because you can consider it as car zero, being different from the first car.
Notice that the highest rank is the actual count you are searching for. So you need to find the max within each Car plate. Type in row 2, Drivers per car column:
Copy-paste values and delete the help column if it isn’t needed.
Note: the formula will return what you need as long as the dataset stays sorted this way. It’s better to copy-paste values once you are happy with the result.
Method 2: Removing duplicates
This time, instead of a help column use a help sheet. Copy-paste the 2 columns for each you want to calculate the sum into a new sheet and here remove duplicates.
See sheet [Unique combinations] from the above mentioned file.
What’s left is a set of unique car-driver combinations. This means that a car plate will appear just as many times as drivers are associated with it. You can add in the third column the formula:
Now you can bring these values in [2 Removing duplicates] sheet using:
1. =VLOOKUP(A2,'Unique combinations'!A:C,3,0)
Copy-paste values and delete the help sheet.
Method 3: Using a pivot table
This is similar with removing duplicates but instead we use a pivot table.
We select the data and use Insert=> Pivot Table. We tick Add this data to the Data Model. This will allow us to use Distinct Count later.
Add the Car plate to Rows and Driver to Values.
Click on Count of Driver and select Value Field Settings. Here scroll down and select Distinct Count then click OK.
Now you can bring these values in [3 Using a pivot] sheet using:
1. =VLOOKUP(A2,Pivot!A:B,2,0)
For larger datasets, you can use VBA like this. | {"url":"https://online-training.ro/home/excel-formula/count-within-groups-formula/","timestamp":"2024-11-09T06:52:43Z","content_type":"text/html","content_length":"74258","record_id":"<urn:uuid:cc585a1b-f912-4bb3-9978-de49eab38b60>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00837.warc.gz"} |
Adding a Latent Stage to Make an SEIR Model
35 Adding a Latent Stage to Make an SEIR Model
EpiModel includes an integrated SIR model, but here we show how to model an SEIR disease like COVID-19. The E compartment in this disease is an exposed state in which a person has been infected but
is not infectious to others. Some infectious diseases have this latent non-infectious stage, and in general it provides a general framework for transmission risk that is dependent on one’s stage of
disease progression.
35.1 Setup
First start by loading EpiModel and clearing your global environment.
35.2 EpiModel Model Extensions
35.2.1 Conceptualization
The first step to any EpiModel extension model is to conceptually identify what new functionality, above and beyond the built-in models, is desired and where that functionality should be added. There
are often many “right” answers to these questions, and this aspect is only learned over time. But in general, it is helpful to map out a model extension on a state/flow diagram to pinpoint where the
additions should be concentrated.
35.2.1.1 SIR Module Set
In this particular model, we will be adding a new disease state that occurs in between the susceptible disease state and an infectious disease state in an SIR model. The standard, built-in SIR model
in EpiModel uses set modules (elements of the model), each with its own set of associated functions (the realization of those elements in code). For any built-in model, you can see what set of
modules and functions have been used by running the simulation with netsim and then printing the output.
nw <- network_initialize(n = 100)
formation <- ~edges
target.stats <- 50
coef.diss <- dissolution_coefs(dissolution = ~offset(edges), duration = 20)
est1 <- netest(nw, formation, target.stats, coef.diss, verbose = FALSE)
param <- param.net(inf.prob = 0.3, rec.rate = 0.1)
init <- init.net(i.num = 10, r.num = 0)
control <- control.net(type = "SIR", nsteps = 25, nsims = 1, verbose = FALSE)
mod1 <- netsim(est1, param, init, control)
EpiModel Simulation
Model class: netsim
Simulation Summary
Model type: SIR
No. simulations: 1
No. time steps: 25
No. NW groups: 1
Fixed Parameters
inf.prob = 0.3
rec.rate = 0.1
act.rate = 1
groups = 1
Model Output
Variables: s.num i.num r.num num si.flow ir.flow
Networks: sim1
Transmissions: sim1
Formation Statistics
Target Sim Mean Pct Diff Sim SE Z Score SD(Sim Means) SD(Statistic)
edges 50 50.92 1.84 3.298 0.279 NA 6.027
Duration Statistics
Target Sim Mean Pct Diff Sim SE Z Score SD(Sim Means) SD(Statistic)
edges 20 25.603 28.016 3.355 1.67 NA 3.532
Dissolution Statistics
Target Sim Mean Pct Diff Sim SE Z Score SD(Sim Means) SD(Statistic)
edges 0.05 0.052 4.552 0.005 0.452 NA 0.036
35.2.1.2 Module Classes
This includes a series of modules, which we classify into Standard and Flexible modules as shown on the table below. Any modules may be modified, but standard modules are those that are typically not
modified because they generalize the core internal processes for simulation object generation, network resimulation, and epidemic bookkeeping. Flexible modules, in contrast, are those that are likely
necessary to modify for an EpiModel extension.
In the example above, we used all of these modules except the arrivals and departures modules because we had a closed population. In addition to these set of built-in modules, any user can add more
modules to this set, depending on what is needed. Again, this involves some conceptualization of how to organize the model processes, including whether those processes are similar to the built-in
modules or something new.
35.2.1.3 Modules vs. Functions
All modules have associated functions, and these are passed into the epidemic model run in netsim through control.net. Printing out the arguments for this function, you will see that each of the
standard modules have default associated functions as inputs, and flexible modules have a default of NULL.
function (type, nsteps, start = 1, nsims = 1, ncores = 1, resimulate.network = FALSE,
tergmLite = FALSE, cumulative.edgelist = FALSE, truncate.el.cuml = 0,
attr.rules, epi.by, initialize.FUN = initialize.net, resim_nets.FUN = resim_nets,
infection.FUN = NULL, recovery.FUN = NULL, departures.FUN = NULL,
arrivals.FUN = NULL, nwupdate.FUN = nwupdate.net, prevalence.FUN = prevalence.net,
verbose.FUN = verbose.net, module.order = NULL, save.nwstats = TRUE,
nwstats.formula = "formation", save.transmat = TRUE, save.network,
save.other, verbose = TRUE, verbose.int = 1, skip.check = FALSE,
raw.output = FALSE, tergmLite.track.duration = FALSE, set.control.ergm = control.simulate.formula(MCMC.burnin = 2e+05),
set.control.stergm = NULL, set.control.tergm = control.simulate.formula.tergm(),
save.diss.stats = TRUE, ...)
For built-in models, EpiModel selects which modules are needed based on the model parameters and initial conditions. For each time step, the modules run in the order in which they are specified in
the output here; this also matches the order in which they are listed in control.net.
35.2.1.4 Built-in Functions as Templates
Any of the built-in functions associated with flexible modules are intended to be templates for user inspection and extension for research-level models. So, for example, mod1 above shows that
infection.net was used as the infection module. That function has a help page briefly describing what it does. And you can also inspect the function contents with:
We’ll use an edited down version of that function, with some additional explanation, below. In addition, the disease progression state transition within an SIR model (and an SIS model too) is handled
by the recovery module, with an associated recovery.net function.
We will use an edited down version of that function as a template for a new, more general disease progression module.
35.2.2 The EpiModel API
All EpiModel module functions have a set of shared design requirements. This set of requirements defines the EpiModel Application Programming Interface (API) for extension. The best way to learn this
is through a concrete example like this, but here are the general API rules:
1. Each module has an associated R function, the general design of which is:
The function takes an object called dat, which is the master data object passed around by netsim, performs some processes (e.g., infection, recovery, aging, interventions), updates the dat object,
and then returns that object. The other input argument to each function must be at, which is a time step counter.
2. Data are stored on the dat object in a particular way: in sublists that are organized by category. The main categories of data to interact with include model inputs (parameters, initial
conditions, and controls) from those three associated input functions; nodal attributes (e.g., an individual disease status for each person); and summary statistics (e.g., the disease prevalence
at at time step). There are accessor functions for reading (these are the get_ functions) and writing (these are the set_ functions) to the dat object in the appropriate place.
3. The typical function design involves three steps: a) reading the relevant inputs from the dat object; b) performing some micro-level process on the nodes that is usually a function of fixed
parameters and time-varying nodal attributes; c) writing the updated objects back on to the dat object.
Let’s see how this API works by extending our infection and recovery functions to transition an SIR model into an SEIR model.
35.2.3 Infection Module
The built-in infection module for an SIR model performs the following functions listed in the function below. The core process is determining which edges are eligible for a disease transmission to
occur, and then randomly simulating that transmission process. Why is it necessary to update the infection function for an SEIR model? Because an SIR involves a transition between S and I disease
statuses when a new infection occurs, but an SEIR model involves a transition between S and E disease statuses. It’s a small, but important change. Here’s the full modified function, with embedded
comments. Note that we can use the browser function to run this function in debug mode by uncommenting the third line (we will demonstrate this).
infect <- function(dat, at) {
## Uncomment this to run environment interactively
# browser()
## Attributes ##
active <- get_attr(dat, "active")
status <- get_attr(dat, "status")
infTime <- get_attr(dat, "infTime")
## Parameters ##
inf.prob <- get_param(dat, "inf.prob")
act.rate <- get_param(dat, "act.rate")
## Find infected nodes ##
idsInf <- which(active == 1 & status == "i")
nActive <- sum(active == 1)
nElig <- length(idsInf)
## Initialize default incidence at 0 ##
nInf <- 0
## If any infected nodes, proceed with transmission ##
if (nElig > 0 && nElig < nActive) {
## Look up discordant edgelist ##
del <- discord_edgelist(dat, at)
## If any discordant pairs, proceed ##
if (!(is.null(del))) {
# Set parameters on discordant edgelist data frame
del$transProb <- inf.prob
del$actRate <- act.rate
del$finalProb <- 1 - (1 - del$transProb)^del$actRate
# Stochastic transmission process
transmit <- rbinom(nrow(del), 1, del$finalProb)
# Keep rows where transmission occurred
del <- del[which(transmit == 1), ]
# Look up new ids if any transmissions occurred
idsNewInf <- unique(del$sus)
nInf <- length(idsNewInf)
# Set new attributes and transmission matrix
if (nInf > 0) {
status[idsNewInf] <- "e"
infTime[idsNewInf] <- at
dat <- set_attr(dat, "status", status)
dat <- set_attr(dat, "infTime", infTime)
dat <- set_transmat(dat, del, at)
## Save summary statistic for S->E flow
dat <- set_epi(dat, "se.flow", at, nInf)
Each step is relatively self-explanatory with the comments, but we will run this interactively at the end of this tutorial to step through the updated data structures. Infection is one of the more
complex processes because it involves a dyadic process (a contact between an S and I node). That involves a construction of a discordant edgelist; that is, a list of edges in which there is a disease
discordant dyad.
The main point here is: we have made a change to the infection module function, and it consists of updating the disease status of a newly infected persons to "e" instead of "i". Additionally, we are
tracking a new summary statistic, se.flow that tracks the size of the flow from S to E based on the number of new infected, nInf at the time step.
35.2.4 Progression Module
Next up is the disease progression module. Here, we have generalized the built-in recovery module function to handle two disease progression transitions after infection: E to I (latent to infectious
stages) and I to R (infectious to recovered stages). Like many individual-level transitions, this involves flipping a weighted coin with rbinom: this performs a series of random Bernoulli draws based
on the specified parameters. Here is the full model function.
progress <- function(dat, at) {
## Uncomment this to function environment interactively
# browser()
## Attributes ##
active <- get_attr(dat, "active")
status <- get_attr(dat, "status")
## Parameters ##
ei.rate <- get_param(dat, "ei.rate")
ir.rate <- get_param(dat, "ir.rate")
## E to I progression process ##
nInf <- 0
idsEligInf <- which(active == 1 & status == "e")
nEligInf <- length(idsEligInf)
if (nEligInf > 0) {
vecInf <- which(rbinom(nEligInf, 1, ei.rate) == 1)
if (length(vecInf) > 0) {
idsInf <- idsEligInf[vecInf]
nInf <- length(idsInf)
status[idsInf] <- "i"
## I to R progression process ##
nRec <- 0
idsEligRec <- which(active == 1 & status == "i")
nEligRec <- length(idsEligRec)
if (nEligRec > 0) {
vecRec <- which(rbinom(nEligRec, 1, ir.rate) == 1)
if (length(vecRec) > 0) {
idsRec <- idsEligRec[vecRec]
nRec <- length(idsRec)
status[idsRec] <- "r"
## Write out updated status attribute ##
dat <- set_attr(dat, "status", status)
## Save summary statistics ##
dat <- set_epi(dat, "ei.flow", at, nInf)
dat <- set_epi(dat, "ir.flow", at, nRec)
dat <- set_epi(dat, "e.num", at,
sum(active == 1 & status == "e"))
dat <- set_epi(dat, "r.num", at,
sum(active == 1 & status == "r"))
This set of two progression processes involves querying who is eligible to transition, randomly transitioning some of those eligible, updating the status attribute for those who have progressed, and
then recording some new summary statistics.
35.3 Network Model
With the epidemic modules defined, we will now step back to parameterize, estimate, and diagnose the TERGM. Here we will use a relatively basic model with an edges and degree(0). Here we are not
using any nodal attributes in either the TERGM or the epidemic modules, but these could be added (we will get more practice with that tomorrow). Note that we are using a relatively high mean degree
(2 per capita) compared to some of our prior models, with lower than expected isolates.
# Initialize the network
nw <- network_initialize(500)
# Define the formation model: edges + degree terms
formation = ~edges + degree(0)
# Input the appropriate target statistics for each term
target.stats <- c(500, 20)
# Parameterize the dissolution model
coef.diss <- dissolution_coefs(dissolution = ~offset(edges), duration = 25)
Dissolution Coefficients
Dissolution Model: ~offset(edges)
Target Statistics: 25
Crude Coefficient: 3.178054
Mortality/Exit Rate: 0
Adjusted Coefficient: 3.178054
Next we fit the network model.
Warning: 'glpk' selected as the solver, but package 'Rglpk' is not available;
falling back to 'lpSolveAPI'. This should be fine unless the sample size and/or
the number of parameters is very big.
And then diagnose it. We are including a wide range of degree terms to monitor so we can see the full degree distribution . Although the degree(0) term does not look great visually, it is only off by
2-3 edges in absolute terms.
dx <- netdx(est, nsims = 10, ncores = 5, nsteps = 500,
nwstats.formula = ~edges + degree(0:7),
keep.tedgelist = TRUE)
Network Diagnostics
- Simulating 10 networks
- Calculating formation statistics
EpiModel Network Diagnostics
Diagnostic Method: Dynamic
Simulations: 10
Time Steps per Sim: 500
Formation Diagnostics
Target Sim Mean Pct Diff Sim SE Z Score SD(Sim Means) SD(Statistic)
edges 500 490.166 -1.967 0.973 -10.111 2.578 15.018
degree0 20 26.266 31.332 0.169 37.020 0.532 5.176
degree1 NA 181.218 NA 0.517 NA 1.579 10.763
degree2 NA 153.282 NA 0.426 NA 1.142 10.127
degree3 NA 86.535 NA 0.363 NA 0.776 8.581
degree4 NA 36.335 NA 0.256 NA 0.764 5.910
degree5 NA 11.941 NA 0.162 NA 0.549 3.628
degree6 NA 3.370 NA 0.066 NA 0.263 1.833
degree7 NA 0.821 NA 0.032 NA 0.046 0.890
Duration Diagnostics
Target Sim Mean Pct Diff Sim SE Z Score SD(Sim Means) SD(Statistic)
edges 25 24.833 -0.669 0.091 -1.835 0.283 1.056
Dissolution Diagnostics
Target Sim Mean Pct Diff Sim SE Z Score SD(Sim Means) SD(Statistic)
edges 0.04 0.04 0.321 0 1.034 0 0.009
35.4 Epidemic Model Parameterization
The epidemic model parameterization for any extension model consists of using the same three input functions as with built-in models.
First we start with the model parameters in param.net. Note that we have two new rates here: ei.rate and ir.rate. These control the transitions from E to I, and then I to R. We have selected these
parameter names and input the values here, but note that the same parameters get pulled into the disease progression function above. So in general there must be consistency between the naming of the
parameters as inputs and their references in the model functions.
Next we specify the initial conditions. Here we are specifying that there are 10 infectious individuals at the epidemic outset. If we wanted to initialize the model with persons only in the latent
stage, we would need to set disease status as a nodal attribute on the network instead (following the same approach in the Serosorting Tutorial in Chapter 31.
Finally, we specify the control settings in control.net. Extension models like this require some significant updates (compared to built-in models) here. First, type is set to NULL in any extension
model because we are no longer using EpiModel to pre-select which modules to run; it is an entirely manual process. Second, the new input parameter for the infection.FUN argument is infect; in other
words, our new function for our infection module is the one we built above. Note that it is called infect. Third, we are defining a new module, progress.FUN, with an argument that matches the name of
our new progression function. We defined a new module, rather than replaced the function of the recovery module, just to show you how to define a new module here. We could have just as easily done it
the latter way with the same results. Finally, we are explicitly setting resimulate.network to FALSE; this is the default, so this is not necessary, but it is just to remind you this is a model
without network feedback.
In the R scripts that we include with this tutorial, you will see that we have two separate R script files (for the R Markdown build, they all go in this single file). One contains the module
functions, and the other contains all the other code to parameterize and run the model. We do this because it allows for easier interaction with the functions in browser mode. I will demonstrate this
live next. But in general, placing the functions in a separate file conceptually disentangles the model functionality from the model parameterization. It is critical, however, that you source in the
file containing the functions before you run control.net (otherwise, control.net does not know what infect and progress are).
35.5 Epidemic Simulation
Finally we are ready to do the epidemic model simulation and analysis. This is done using the same approach as the built-in models. We will start with running one model in the browser mode
And then go back, comment out the browser lines, re-source the functions, and run a full-scale model with 10 simulations.
Once the model simulation is complete, we can work with the model object just like a built-in model. Start by printing the output to see what is available.
EpiModel Simulation
Model class: netsim
Simulation Summary
Model type:
No. simulations: 10
No. time steps: 500
No. NW groups: 1
Fixed Parameters
inf.prob = 0.5
act.rate = 2
ei.rate = 0.01
ir.rate = 0.01
groups = 1
Model Functions
Model Output
Variables: s.num i.num num ei.flow ir.flow e.num r.num
Networks: sim1 ... sim10
Transmissions: sim1 ... sim10
Formation Statistics
Target Sim Mean Pct Diff Sim SE Z Score SD(Sim Means) SD(Statistic)
edges 500 489.506 -2.099 1.038 -10.110 6.087 16.132
degree0 20 26.112 30.561 0.180 33.893 0.867 5.401
Duration Statistics
Target Sim Mean Pct Diff Sim SE Z Score SD(Sim Means) SD(Statistic)
edges 25 24.876 -0.498 0.098 -1.266 0.316 1.106
Dissolution Statistics
Target Sim Mean Pct Diff Sim SE Z Score SD(Sim Means) SD(Statistic)
edges 0.04 0.04 0.16 0 0.52 0 0.009
Here is the default plot with all the compartment sizes over time. This includes the new summary statistics we tracked in the disease progression function.
Here are the flow sizes, including the new se.flow incidence tracker that we established in the new infection function.
Finally, here is the data frame output from the model, with rows limited to time step 100 across all 10 simulations.
sim time s.num i.num num ei.flow ir.flow e.num r.num se.flow
And here are the transmission matrices and their phylogram plots (one per infection seed). | {"url":"https://epimodel.github.io/epimodel-training/network_models_with_feedback/adding_stage_SEIR.html","timestamp":"2024-11-08T05:51:47Z","content_type":"application/xhtml+xml","content_length":"106637","record_id":"<urn:uuid:007e5ca5-a20a-4af3-a66f-85cb5eee2901>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00448.warc.gz"} |
Upgrade Your Math Toolbox with a GCD Calculator - Age calculator
Upgrade Your Math Toolbox with a GCD calculator
Mathematics has always been a subject that many people struggle with. It can be frustrating to try and solve equations, only to find that you are missing a key piece of information or that your
calculation is incorrect. Fortunately, there are tools available to help with these problems. One such tool is a Greatest Common Denominator (GCD) calculator. In this article, we will explore what a
GCD calculator is, how it works, and how it can help you upgrade your math toolbox.
What is a GCD calculator?
A GCD calculator is a tool that helps you find the Greatest Common Denominator (also known as the Greatest Common Factor) of two or more numbers. This number is the largest positive integer that
divides each of the numbers evenly. For example, the GCD of 12 and 18 is 6, because 6 is the largest number that can divide both 12 and 18 without leaving a remainder.
How does a GCD calculator work?
There are several methods to find the GCD of two or more numbers, but the most common method is the Euclidean Algorithm. This algorithm works by repeatedly subtracting the smaller number from the
larger number until both numbers are equal. The GCD is then the common value that they were subtracted to obtain. Here is an example using the numbers 12 and 18:
– Start with the two numbers: 12 and 18.
– Since 18 is larger, subtract 12 from 18: 18-12=6.
– Now we have two numbers: 12 and 6.
– Since 12 is larger, subtract 6 from 12: 12-6=6.
– Now we have two numbers: 6 and 6. These are equal, so the GCD of 12 and 18 is 6.
A GCD calculator automates this process. You enter the numbers you want to find the GCD of, and the calculator performs the Euclidean Algorithm to find the answer. Most GCD calculators are also able
to find the GCD of more than two numbers at once.
How can a GCD calculator help you upgrade your math toolbox?
A GCD calculator can be a valuable tool in several ways:
1. Time-saving: Finding the GCD of two or more numbers by hand can be a time-consuming process, especially for large numbers. A GCD calculator can perform calculations quickly and accurately, saving
you time and effort.
2. Error-free: It’s easy to make mistakes when doing math calculations by hand. A GCD calculator eliminates errors, ensuring that you get accurate results every time.
3. Multiple calculators in one: Many GCD Calculators can find the GCD of multiple numbers at once. This means you don’t have to perform the calculation separately for each set of numbers, saving you
time and effort.
Q: What is the difference between GCD and LCM?
A: GCD stands for Greatest Common Denominator (or Factor), while LCM stands for Least Common Multiple. The GCD is the largest number that can divide two or more numbers evenly, while the LCM is the
smallest number that two or more numbers can all divide into evenly.
Q: How do I use a GCD calculator?
A: Simply enter the numbers you want to find the GCD of into the calculator, and press the calculate button. The calculator will then perform the Euclidean Algorithm to find the GCD.
Q: Can a GCD calculator find the GCD of more than two numbers?
A: Yes, many GCD calculators can find the GCD of more than two numbers at once.
Q: Can a GCD calculator find the GCD of decimal numbers?
A: No, GCD calculators only work with whole numbers.
Q: Can a GCD calculator be used for fractions?
A: Yes, GCD calculators can be used for fractions, but the fractions must be converted to whole numbers first.
In conclusion, a GCD calculator is a valuable tool for anyone who needs to find the GCD of two or more numbers. It saves time, eliminates errors, and can perform calculations for multiple sets of
numbers at once. Whether you are a student, a teacher, or someone who uses math in their daily life, a GCD calculator is a great addition to your math toolbox.
Recent comments | {"url":"https://age.calculator-seo.com/upgrade-your-math-toolbox-with-a-gcd-calculator/","timestamp":"2024-11-04T05:01:59Z","content_type":"text/html","content_length":"302057","record_id":"<urn:uuid:eb7506b1-60d2-4be8-a539-0ed274924fa0>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00118.warc.gz"} |
Data Analysis Second Semester Exam
**DO NOT CLICK ANY LINKS BEFORE READING THE FOLLOWING**
Welcome to the School of Data – Data Analysis Second Semester Exam.
Please take the time to read these instructions before you begin the test.
We recommend you have the following ready:
• Google Chrome (to access this test)
• Access to stable internet
• Fully charged computer
NB: Once you start the exam, you are not to open any other windows. If you do, the assessment will end immediately.
The test duration is 1hr: 30 min. There will be no breaks once you start.
This test comprises 30 questions, you are to answer all questions. If you do not know the answer to a given question, please select a random one and proceed to the next question.
For some questions you will need a data set, Please download it through this link. Go to file, click on download, and download as excel file.
You cannot pause this test once you begin. If you run into internet issues, we will not be liable for any inconveniences you incur. We advise you to use a good and steady internet service. You can
run a speed test on FAST.COM to check the speed of your internet provider.
If you do not submit your test before the testing window eclipses, your test will be automatically submitted, and all unanswered questions will be marked 0.
Even though you are reading this now, you do not have to take this test right away if you’re unprepared. You will have access to this test up to Monday 17th July 2023. | {"url":"https://3mtt.altschoolafrica.com/quizzes/data-analysis-second-semester-exam/","timestamp":"2024-11-03T17:15:16Z","content_type":"text/html","content_length":"241541","record_id":"<urn:uuid:320b705e-6705-4ef2-bdee-8f5726bf3a08>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00469.warc.gz"} |
How to write a hypothesis for correlation
A hypothesis is a testable statement about how something works in the natural world. While some hypotheses predict a causal relationship between two variables, other hypotheses predict a correlation
between them. According to the Research Methods Knowledge Base, a correlation is a single number that describes the relationship between two variables. If you do not predict a causal relationship or
cannot measure one objectively, state clearly in your hypothesis that you are merely predicting a correlation.
Research the topic in depth before forming a hypothesis. Without adequate knowledge about the subject matter, you will not be able to decide whether to write a hypothesis for correlation or
causation. Read the findings of similar experiments before writing your own hypothesis.
• A hypothesis is a testable statement about how something works in the natural world.
• Without adequate knowledge about the subject matter, you will not be able to decide whether to write a hypothesis for correlation or causation.
Identify the independent variable and dependent variable. Your hypothesis will be concerned with what happens to the dependent variable when a change is made in the independent variable. In a
correlation, the two variables undergo changes at the same time in a significant number of cases. However, this does not mean that the change in the independent variable causes the change in the
dependent variable.
Construct an experiment to test your hypothesis. In a correlative experiment, you must be able to measure the exact relationship between two variables. This means you will need to find out how often
a change occurs in both variables in terms of a specific percentage.
• Identify the independent variable and dependent variable.
• In a correlative experiment, you must be able to measure the exact relationship between two variables.
Establish the requirements of the experiment with regard to statistical significance. Instruct readers exactly how often the variables must correlate to reach a high enough level of statistical
significance. This number will vary considerably depending on the field. In a highly technical scientific study, for instance, the variables may need to correlate 98 per cent of the time; but in a
sociological study, 90 per cent correlation may suffice. Look at other studies in your particular field to determine the requirements for statistical significance.
• Establish the requirements of the experiment with regard to statistical significance.
• Look at other studies in your particular field to determine the requirements for statistical significance.
State the null hypothesis. The null hypothesis gives an exact value that implies there is no correlation between the two variables. If the results show a percentage equal to or lower than the value
of the null hypothesis, then the variables are not proven to correlate.
Record and summarise the results of your experiment. State whether or not the experiment met the minimum requirements of your hypothesis in terms of both percentage and significance. | {"url":"https://www.ehow.co.uk/how_8682689_write-hypothesis-correlation.html","timestamp":"2024-11-07T04:31:22Z","content_type":"text/html","content_length":"121904","record_id":"<urn:uuid:c50fedc8-b6d6-42f8-a353-42f9540d3547>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00516.warc.gz"} |
CS3610 Project 2 solution
Having data compressed can save space on hard drives, as well as speed up file transfer online. In this assignment, you will implement a simple, yet elegant, data compression algorithm called Huffman
coding https://en.wikipedia.org/wiki/Huffman_ coding. Let us first walk through how it works.
Huffman Coding
Assume that I want to store the message “opossum” on my computer. This string of characters will be represented on my hard drive as a series of bits (1’s and 0’s). Typically, an ASCII character takes
up 8 bits of space on a computer. This means that “opossum” will cost 56 bits of storage. For the sake of argument, let us assume that this is a significant number. One obvious way to reduce the
number of bits in our message is to represent each character in the message with fewer bits. Of course, using smaller bit sequences will limit the number of unique characters we can encode. For
example, a two bit sequence is only capable of representing 4 unique characters. Consequently, our characters will vary in length. This was expected though. So what are the real issues preventing us
from using fewer bits? Well, I have two for you right here:
1. Let us assume that the characters in “opossum”, ‘p’, ‘m’, ‘u’, ‘o’, and ‘s’, are represented as 0, 1, 00, 01, and 10 respectively. What problems do you see with this? Well, it just so happens that
our most common characters ‘o’ and ‘s’ are also some of our largest characters. We could have saved ourselves a little more space if we instead represented ‘o’ and ‘s’ with the single bits 0 and 1,
which were naively used to denote ‘p’ and ‘m’. So the question arises, how do we efficiently delegate bit sequences such that the most frequently occurring characters in our messages take up the least
amount of space?
2. Let us assume that the characters in “opossum”, ‘p’, ‘m’, ‘u’, ‘o’, and ‘s’, are represented as 0, 1, 00, 01, and 10 respectively. Let us also assume that we have used these characters to encode a
new message “0001” on our hard drive. I now ask you, what message did we just encode? Well, unfortunately, it is ambiguous. One possible decoding could be “pppm”. Another could be “uo”. So the
question arises, how do we generate variable-length character codes such that our encoded messages will not be ambiguous? In other words, how do we develop a prefix coding system?
One way to resolve these issues is by constructing a Huffman tree, just like the one displayed right here:
A Huffman tree is a binary tree that encodes the characters of a message into a reduced bit representation by implicitly storing the component bits along the tree’s edges. Typically, left edges denote
0 while all the right edges denote 1. Thus, to find the encoding of a specific character, we simply concatenate the bits found along the edges of the path between the root node and the character’s
node. In the tree above, you can see that ‘o’ is represented as 11 because its path from the root first follows a right edge and then another right edge. Take note that all character nodes are leaf
nodes and each path produces a unique string of bits. Also, you should be aware that the most common characters, such as ‘o’ and ‘s’, appear closer to the top of the tree, which gives them the
shortened encoding that we desire. Conversely, the lowest frequency characters reside near the bottom of the tree. Using the Huffman tree above, “opossum” can now be written as 1100111010011010, which
is just 16 bits as compared to the original 56. Better yet, no characters share the same prefix, which means this binary message can be decoded unambiguously. You can try it out for yourself by
reading the bits left to right while following the recorded bit path in the tree starting from the root. As soon as you hit a leaf node, grab its character and continue reading the remaining bits
starting from the root again. So this is wonderful, I now have a compressed message that was easy to encode and decode. However, the question remains, how do we construct the Huffman tree? Well here
is the algorithm:
1. Identify the frequency of each character in a given message. The frequency of each character in “opossum” is listed alphabetically as follows:
m = 1 o = 2 p = 1 s = 2 u = 1
2. Package each character-frequency pair into a binary leaf node and insert each node into a min heap using the frequency as the key.
3. Extract the two smallest elements from the min heap and respectively attach them to the left and right pointers of a new internal node. Set the key value of the new internal node to be equal with
the sum of the frequencies of the extracted nodes. Insert this new internal node into the min heap.
(a) Result after extracting 1st min element (b) Result after extracting 2nd min element
(c) New node from extracted mins (d) New node inserted into heap
4. Repeat step 3 until the min heap has one element remaining.
(a) New node from extracted mins (b) New node inserted into heap
(a) New node from extracted mins (b) New node inserted into heap
(a) New node from extracted mins (b) New node inserted into heap
5. Draw out the newly constructed Huffman tree
For this project, you have been provided an implementation of the Huffman tree construction algorithm (DO NOT modify this code. It is possible to construct multiple Huffman trees with the same message
if numerous characters in the message share the same frequency value. The construct function that I set up ensures that your Huffman tree will produce results consistent with the test cases used for
grading). This algorithm utilizes a templated min heap class that you must implement. You must also implement a function that prints the Huffman encoding of message using the Huffman tree constructed
from the same message. Specifically, you must implement the following functions:
void MinHeap::insert(const T data, const int key) : Insert the provided user data into the min heap using the provided key to make comparisons with the other elements in the min heap. To ensure that
your min heap produces consistent results, stop bubbling up a child node if it shares the same key value as its parent node.
T MinHeap::extract min() : Remove from the min heap the element with the smallest key value and return its data. If you come across two sibling nodes that share the same key value while sifting down
a parent node with a larger key value, then you should swap the parent node with the left child to ensure that your min heap produces consistent results.
T MinHeap::peek() const : Retrieve the minimum element in the min heap and return its data to the user. Do not remove this element from the min heap in this function.
void MinHeap::size() const : Return the size of the min heap
void HuffmanTree::print() const : Print the Huffman encoding of the member variable message assigned in the construct function.
To ensure that you always produce a consistent output, DO NOT modify the completed code in the HuffmanTree class. You may however add print helper functions if you feel it necessary.
Input is read from the keyboard. The first line of the input will be an integer t > 0 that indicates the number of test cases. Each test case will contain a message on a single line to be processed by
the Huffman tree construct function. Each message will contain at least 2 characters.
For each test case, print Test Case: followed by the test case number on one line. On another line, print the Huffman encoding of the input message. Separate the individual character encodings by a
Sample Test Cases
Use input redirection to redirect commands written in a file to the standard input, e.g. $ ./a.out < input1.dat. Input 1 3 opossum hello world message Output 1 Test Case: 1 11 00 11 10 10 011 010 Test
Case: 2 010 011 10 10 00 1100 1101 00 1110 10 1111 Test Case: 3 011 10 11 11 010 00 10 Timing Analysis At the top of your main file, in comments, write the time complexity of constructing a Huffman
tree with a min heap in terms of the number of characters in the input message, which you can denote as n. Also consider the time complexity of constructing a Huffman tree without a min heap.
Specifically, what running time can you expect if you use a linear search to find minimum frequencies. Write these time complexities using Big-O notation. Turn In Submit your source code to the TA
through blackboard. If you have multiple files, package them into a zip file. Grading: Total: 100 pts. • 10/100 - Code style, commenting, general readability. 6 • 05/100 - Compiles. • 05/100 - Follows
provided input and output format. • 75/100 - Successful implementation of the min heap and Huffman print functions. • 05/100 - Correct timing analysis. | {"url":"https://jarviscodinghub.com/product/cs3610-project-2-solution/","timestamp":"2024-11-03T10:46:53Z","content_type":"text/html","content_length":"114868","record_id":"<urn:uuid:d9030ab9-5c3a-4ec6-9469-b290b67d70f8>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00174.warc.gz"} |
Transactions Online
Sang-Hyuk LEE, Keun Ho RYU, Gyoyong SOHN, "Study on Entropy and Similarity Measure for Fuzzy Set" in IEICE TRANSACTIONS on Information, vol. E92-D, no. 9, pp. 1783-1786, September 2009, doi: 10.1587/
Abstract: In this study, we investigated the relationship between similarity measures and entropy for fuzzy sets. First, we developed fuzzy entropy by using the distance measure for fuzzy sets. We
pointed out that the distance between the fuzzy set and the corresponding crisp set equals fuzzy entropy. We also found that the sum of the similarity measure and the entropy between the fuzzy set
and the corresponding crisp set constitutes the total information in the fuzzy set. Finally, we derived a similarity measure from entropy and showed by a simple example that the maximum similarity
measure can be obtained using a minimum entropy formulation.
URL: https://global.ieice.org/en_transactions/information/10.1587/transinf.E92.D.1783/_p
author={Sang-Hyuk LEE, Keun Ho RYU, Gyoyong SOHN, },
journal={IEICE TRANSACTIONS on Information},
title={Study on Entropy and Similarity Measure for Fuzzy Set},
abstract={In this study, we investigated the relationship between similarity measures and entropy for fuzzy sets. First, we developed fuzzy entropy by using the distance measure for fuzzy sets. We
pointed out that the distance between the fuzzy set and the corresponding crisp set equals fuzzy entropy. We also found that the sum of the similarity measure and the entropy between the fuzzy set
and the corresponding crisp set constitutes the total information in the fuzzy set. Finally, we derived a similarity measure from entropy and showed by a simple example that the maximum similarity
measure can be obtained using a minimum entropy formulation.},
TY - JOUR
TI - Study on Entropy and Similarity Measure for Fuzzy Set
T2 - IEICE TRANSACTIONS on Information
SP - 1783
EP - 1786
AU - Sang-Hyuk LEE
AU - Keun Ho RYU
AU - Gyoyong SOHN
PY - 2009
DO - 10.1587/transinf.E92.D.1783
JO - IEICE TRANSACTIONS on Information
SN - 1745-1361
VL - E92-D
IS - 9
JA - IEICE TRANSACTIONS on Information
Y1 - September 2009
AB - In this study, we investigated the relationship between similarity measures and entropy for fuzzy sets. First, we developed fuzzy entropy by using the distance measure for fuzzy sets. We pointed
out that the distance between the fuzzy set and the corresponding crisp set equals fuzzy entropy. We also found that the sum of the similarity measure and the entropy between the fuzzy set and the
corresponding crisp set constitutes the total information in the fuzzy set. Finally, we derived a similarity measure from entropy and showed by a simple example that the maximum similarity measure
can be obtained using a minimum entropy formulation.
ER - | {"url":"https://global.ieice.org/en_transactions/information/10.1587/transinf.E92.D.1783/_p","timestamp":"2024-11-04T13:37:14Z","content_type":"text/html","content_length":"59322","record_id":"<urn:uuid:4c3ba329-5151-48a0-8b98-d834ec7914c5>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00325.warc.gz"} |
Andy Psarianos
Hi, I'm Andy Psarianos.
Robin Potter
Hi, I'm Robin Potter.
Adam Gifford
Hi, I'm Adam Gifford.
This is The School of School podcast. Welcome to The School of School podcast.
Robin Potter
Are you a maths teacher looking for CPD to strengthen your skills? Maths — No Problem! has a variety of courses to suit your needs. From textbook implementation, to the essentials of teaching maths
mastery. Visit mathsnoproblem.com today to learn more.
Andy Psarianos
Well, welcome back everyone to another episode of The School of School podcast. And super special day today. We've got a guest. Yeah, I'm really happy we've got her here. It's Jo Sawyer. Jo, do you
want to say hi and tell us a little bit about yourself?
Jo Sawyer
For sure. So, hi everyone. I'm Jo Sawyer and I'm passionate about mathematics. I work in the UK as a primary maths teacher part-time and I spend the rest of my week doing maths consultancy in primary
schools. I do some ITT training and then anything working from new teachers all the way up to experienced teachers. Developing maths, particularly teaching for mastery and it's a real privilege.
Andy Psarianos
Fantastic. Okay, so we're talking about STEM sentences today, and in particular the impact of STEM sentences and mathematical vocabulary on pupil's understanding of mathematics. All right, Jo, what
does that mean?
Jo Sawyer
Okay. Well, I went for this one today because I've taught for, as I said, over 20 years and vocabulary has always been something that I've used generally, shape and obviously synonyms for addition
and subtraction. But it was only recently that I did some training and I started to look at specific vocabulary for rules. So for example, division dividend, divisor and quotient, and also starting
to look at specific types of additions, so aggregation and augmentation for example.
And this was completely new to me, never come across it before. And I thought I'd always been an okay teacher and I really questioned whether this was going to make me better at my job and it really
has. The question is, should everybody be using this and why should they be using it and what impact is it going to have? So say if we go with dividend, divisor and quotient or add end plus end
equals some.
I found that introducing that with my children and these children were only seven years old, they're young children. It made them really think about the concepts in mathematics and it was allowing
them the opportunity to make connections between ideas and generalise about the maths in a way that they'd never ever done before and I'd never encouraged them to do before.
So I started to push those boundaries a little bit and the children were really beginning to grasp more complexities of maths with greater ease and it was really incredible. So if I give you an
example there, I had a little boy who discovered that in addition, if you increase one add end and decrease the other add end by one, then the sun remains the same. And it was mind-blowing to this
child that these patterns were there and it was just a real light bulb moment.
And then taking on from there, I started to use STEM sentences, which are really mathematical sentences that encourage children to understand the concepts of mathematics. I'm using one in class
regularly at the moment about, if the whole is divided into equal parts, and I have of them to help really understand about fractions. And I wanted to think about how this was going to support all
learners, so struggling learners and really advanced learners.
And what I discovered over time was that my struggling learners had a basis on which to help them get support with the mathematical ideas, but my advanced learners were able to use the STEM sentences
to help them reason and make more concise written explanations and so on. So I think that's it really. Should everybody be using these and are they successful with all teachers, or does there have to
be a mathematical understanding there by teachers before they're used? So would everybody benefit from the approach? And that's where I'm at with it.
Adam Gifford
Can I jump in here, Jo? Just to ask something.
Jo Sawyer
Adam Gifford
Without getting into the specifics at this stage of what you've just talked about, do we need a reminder of this and a structure of this, because there's still a hangover with mathematics that it's
not a language, that it's simply a load of numbers that are written down and we don't necessarily read it, or the language aspect of it is not something that that's associated with maths in the same
way when we read a novel. What we are doing is decoding the abstract of the alphabet to make the words to gain understanding.
Do you think that's probably the first hurdle? Is that mathematics is a language that has to be accepted in the first instance by teachers in classrooms? Is that something that you've come across
that's a big difference between English and maths when it comes to language full stop?
Jo Sawyer
I think so because people often assume that mathematical language is numerical, but I'm pretty sure it's 400 mathematical words are needed for children to do well in the key stage one. And it's all
very much mathematical language. So if we're going to develop that for a full understanding, there's quite a large range there. But it's really about the teachers understanding it as well.
So I think first of all, before we present this language, the more complex language to the children, it really is going to benefit the teachers to understand mathematical ideas and thoughts as well.
So I think it's challenging on different levels. You've got to challenge the children to know and understand the language, but you've got to get it across to the teachers that it's really crucial to
help them be better teachers. What have found is that some people have had a fear, and I'm not going to lie, I was one of them.
I had a fear that some of the language we were using was too complex for the children and it was just going to create more difficulties. But I got to the point where I was actually, I'm going to have
to try this in my room because I'm making this assumption that it's not going to work. And I threw myself in and I have never ever been as shocked in my teaching is that, yeah, because the children
were just taking it on board. But they do, don't they. I mean, in phonics, if they can learn about all of the tri graphs and di graphs and everything else, then they love that language and it really
helps them to understand the ideas.
Sometimes adding in more complex language helps the mathematics become simpler because there's a word for something specific rather than a sentence. So if I'm saying to children in subtraction, the
first number is the menu end, then they can say the menu end, rather than the number at the beginning of the sum of, the number that we are taking away from. So it actually simplify some of the
mathematical talk between children as well.
Andy Psarianos
So I suppose when we teach mathematics, one of the competencies that children need to have, we know that will separate a successful mathematician from one who may continues to struggle, is the
ability to communicate their ideas and share those ideas. And often that's a barrier. So for example, if you look at children with English as an additional language, for example, in the UK, that can
become a hindrance on their performance because they struggle with communicating their ideas, even though conceptually they might understand the concept of, I don't know, let's say one of the
manifestations of division, which might be fractions, let's say. They can understand it conceptually. They don't have the vocabulary to explicate their thinking, so therefore they seem to be
And also the more abstract notations that we use in mathematics can also be a hindrance in the sense that, well, you're putting one number above another, that's just an abstract concept to represent
an idea. And that can be a stumbling block. Because while you were talking, Jo, it was really fascinating to me. I was trying to think, okay, of what I know about learning theories and of what I know
about the research that I've read and the key points that we try to structure, for example, the national problem programme on, how does this apply?
And I suppose if I were to try to put some educational jargon on it, this is really about structuring. So if we bring it back to let's say, Zoltan Dienes'. And Zoltan Dienes' saw different stages of
introducing concepts. He'll say that initially you don't want that stuff, because that's just going to mess everything up. Because the minute you start labelling things, if you label things too
early, the danger is that, to put it in PIJ terms, children are merely going to assimilate information and they're not going to accommodate for this new information.
So there needs to be a struggle where effectively you're challenging, your understanding of a concept and you're trying to accommodate for this new discovery that you've just had, in your own
understanding and your own terms. And then if you, I think do it too early, the danger is that you're just labelling things before the concepts are actually understood.
But then of course the next phase after you've had that exploration is to then structure it. And then that's really where you need to introduce the labels and they say, okay, that thing that you just
experienced here that you're struggling so hard to describe, we have a word for that. And that's called, I don't know, a denominator. And when we talk about denominators, we're talking about this
phenomena that you just experienced. You cut this cake into equal size pieces. When you count how many equal size pieces you have of that whole thing, that's the denominator.
Because otherwise, like you just said, Jo, the child has to explain what they did and say, "I cut the cake into six pieces, and each one of those pieces is an equal size." Well now you say, you're
fixing a label. So you're structuring this idea around a concept that we all can share. Have I got it right or am I just on some wild tangent here?
Jo Sawyer
No, I think that's exactly right, Andy. I would agree. I mean, we all would look to go with concrete experiences first and the children having that informal exploration. And then it's as they're
discovering that you mould their thoughts with that key vocabulary. And when you're creating generalisations or reasoning, I think it's so much more powerful when it comes from the children in what
they've explored. If you tell them something, it doesn't really have meaning does it. It has to be found out and discovered by the children for it to be the most powerful. And bobbing the vocabulary
in at the right point is perfect. I'm in fractions at the moment, I'm teaching my class and we didn't introduce denominator until about 12 hours in, because that was the right time when they explored
that and they had an idea about parts and equality and all of those ideas.
Andy Psarianos
And I guess the danger is that if you don't use the right vocabulary, and you use the wrong vocabulary thinking that maybe you're helpful like sometimes... And this is the challenge as authors or as
an author or a textbook writer, these are the things that you're challenged with. So you're going to say, okay, I need to teach this concept, but I can't use the word yet, because the word is just
jargon at this stage and it'll be meaningless to them. So how do I make them experience this concept in the early stages? How do you structure the lesson? What's the journey? What's the first bit?
What's the next bit? How do you unveil this thinking? Give children the learning experiences that they need in the right order so that they develop the thinking that you want them to have. So you can
then label it and say, "Okay, we'll put this in this nice package." That's the difficulty for authors.
But it does raise a lot of questions, because one of the traps I think people can fall into is when they use the wrong language to try to maybe be helpful and put a bridge in between two different
things, it can actually be very damaging. So there's one thing that jumps into my mind and this was an official document that I read that was published by people who know what they're talking about
and so on and so forth. And the suggestion was to use the word fairness. And to bridge the concept of fairness with equality. Fairness and equality, that's a philosophy discussion.
But I can tell you that fairness and equality are not the same thing and they are explicitly not the same thing in mathematics. So if you try to introduce the area of fairness to bridge a gap to
let's say this notion of equality that's so important in mathematics, actually, you might be creating a misconception that you'll spend years unpicking just with a subtle hint that might just
actually introduce a lot of damage. This a tread carefully area.
Adam Gifford
Can I jump in on a slightly different angle? Because I just think with this whole conversation, I think sometimes we miss a massive trick in schools, because there's a place where we do this really
well and that's when we teach English. So for example, if the end result is I want children to write a story about walking through the woods, what's the first thing you want the kids to do? You take
them for a walk in the woods. You don't give them words, don't give anything like that. Let them experience it. So they'll take from it. And then what do we try to do? We start to develop the ideas
around it. So when we're back in the classroom, tell me how you felt? Did you smell anything? Did you hear anything? Da, da, da. And we start to structure it a little bit more. Structure, structure,
Eventually we come into the words that we choose themselves. Remember, they're being represented in the abstract, using the alphabet. Simply just in a different arrangement of letters. That's all we
are doing. So then we're developing the words and we think, can we come up with a better word than this? So the tree was brown or whatever, whatever it might be. I don't know. It was a bit rough.
Have we got any other words for rough? It's textured. It was this. It was that. It was something else. Okay, what's going to fit best here? How can we show that?
So now the walk in the woods is now on a piece of A4 paper. We don't even have to go anywhere. And it's just being shown in the abstract using the woods. Well, we've already got these structures and
it feels sometimes with me is that links are not made with maths. If we think about maths, and you're talking about fractions. I don't know. There you go. Start to divide these things up. Are they
equal parts or not equal parts? Da, da, da. When does that become a problem? How many of these equal parts does it take to put the cake back together and make one whole cake? Can we write that down?
How can we do it? Because we can't keep buying cakes. The school budgets blown. So how can we do that?
So then we move to paper. And then ultimately, how can we write this down? How can we record the experience that we've had in the abstract? Yes, we could do it with words, but there's an even quicker
way of doing it. And this is what we do. And I can't see the difference between the two processes, because it's experiential at the beginning, and all we are doing is recording something that's
happened in the most efficient way possible.
And I think sometimes we miss a trick by not making that link and thinking that this whole magic is created in maths, but that's just simply not the case. I think that if we only present the
abstract, we've got an issue. Because then those symbols mean nothing. We could just say that, yes, five plus two equal seven. But actually I want children to be able to read that story. What did
that represent for you? Well, I took, don't know five leaves and Dave had another two and we put them all together. So that's your story there. Wow. And it's all contained in five plus two equals
seven. That's brilliant. What a story. Man, that's a good one. Who else has got a story?
And I just I don't know how often that link is made, and that worries me because it feels like we are making it have the potential to make it harder and that scares people. And I don't know that it
necessarily should. But that's the way I see it anyway. I just don't know that there's a massive difference in the process.
Andy Psarianos
I think we've only scratched the surface on this topic, and we're going to have to come back and revisit this again guys. Because there's so many different directions we could go from here. Talking
about the relationship between mathematics and language or mathematics as a language or the use of mathematics in language might be an interesting twist to look at it. Where's language mathematical?
Help bridge that gap.
For sure there's obvious things I don't know, the lyrics of a song or just a song in general. Look at a rap song for example, or something that I don't spend a lot of time listening to, but I hear
sometimes because my kids listen to it. It's all math. Every single aspect of rap is maths. It's all patterns and specifics and very complex syncopations that could all be explained mathematically.
Anything from the notes, the rhythm, the cycles, the repetition, all those things. You could write a rap song in a mathematical equation, I'm sure. But it's not just that. It's even if you just look
at the most creative, artistic literature can be explained in mathematics, in mathematical terms. Look at, I don't know, Shakespeare sonnets. It's all math. It's all about math. So anyway, so we got-
Jo Sawyer
Can of worms.
Andy Psarianos
... we got to wrap this up. But we're going to have to come back and talk about this some more. This is a really exciting topic. Thanks Jo, for joining us today.
Jo Sawyer
Thanks for having me.
Andy Psarianos
Thank you for joining us on The School of School podcast. | {"url":"https://schoolofschool.com/educational-podcast-series/episode-105-the-impact-of-stem-sentences-and-mathematical-vocabulary-on-pupils-understanding-of-mathematics/","timestamp":"2024-11-04T21:59:49Z","content_type":"text/html","content_length":"63008","record_id":"<urn:uuid:8915e9d1-2103-4a0f-b11b-dfedd0fa359b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00863.warc.gz"} |
शेखर एक प्रसिद्ध क्रिकेट खिलाड़ी है। वह टेस्ट मैचों में अब तक 6... | Filo
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शेखर एक प्रसिद्ध क्रिकेट खिलाड़ी है। वह टेस्ट मैचों में अब तक 6,980 रन बना चुका है। वह 10,000 रन पूरे करना चाहता है। उसे कितने और रनों की आवश्यकता है?
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Given, Shekhar has scored 6,980 runs in Test matches and he wants to complete 10,000 runs. Let x be the required runs. According to the problem, Total runs scored by Shekhar = 6,980 Total runs scored
by Shekhar after he completes 10,000 runs = 10,000 Therefore, the required runs will be, Solving the above equation, we get Hence, Shekhar needs 3020 more runs to complete 10,000 runs in Test
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Question Text शेखर एक प्रसिद्ध क्रिकेट खिलाड़ी है। वह टेस्ट मैचों में अब तक 6,980 रन बना चुका है। वह 10,000 रन पूरे करना चाहता है। उसे कितने और रनों की आवश्यकता है?
Updated On Jan 31, 2024
Topic All topics
Subject Mathematics
Class Class 11
Answer Type Text solution:1 | {"url":"https://askfilo.com/user-question-answers-mathematics/shekhr-ek-prsiddh-krikett-khilaadd-ii-hai-vh-ttestt-maicon-36373631373836","timestamp":"2024-11-11T01:59:32Z","content_type":"text/html","content_length":"190058","record_id":"<urn:uuid:dbb56972-b365-4c71-8016-def6f3323ea0>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00875.warc.gz"} |
Taylor series example
If Archimedes were to quote Taylor’s theorem, he would have said, “Give me the value of the function and the value of all (first, second, and so on) its derivatives at a single point, and I can give
you the value of the function at any other point”.
It is very important to note that the Taylor’s theorem is not asking for the expression of the function and its derivatives, just the value of the function and its derivatives at a single point.
Now the fine print: Yes, all the derivatives have to exist and be continuous between x and x+h, the point where you are wanting to calculate the function at. However, if you want to calculate the
function approximately by using the n^th order Taylor polynomial, then 1^st, 2^nd,…., n^th derivatives need to exist and be continuous in the closed interval [x,x+h], while the (n+1)^th derivative
needs to exist and be continuous in the open interval (x,x+h).
0 thoughts on “Taylor series example”
1. its was a very good example but only one will not satisfy the student…..sorry but in a way what i wanna say is that i want you guys to post some more examples so that we students can get an idea
about how the theorem works….thank you
2. its was a very good example but only one will not satisfy the student…..sorry but in a way what i wanna say is that i want you guys to post some more examples so that we students can get an idea
about how the theorem works….thank you
3. plz help me to solve the example! use taylor’s theorem with n=2 to approximate (1+x)3,x>-1
4. plz help me to solve the example! use taylor’s theorem with n=2 to approximate (1+x)3,x>-1
You must be logged in to post a comment. | {"url":"https://blog.autarkaw.com/2008/08/19/taylor-series-example/","timestamp":"2024-11-06T15:53:30Z","content_type":"text/html","content_length":"49432","record_id":"<urn:uuid:aba3d8a2-ed7e-473b-a7ea-c0e3cc6e4df2>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00806.warc.gz"} |
Substitute the given numerical values into the algebraic expressions in this multi-level, self-marking quiz.
This is level 4: Here are some mathematical expressions. Can you evaluate them if:
s = 2b = 1t = 3n = 9
© Transum Mathematics 1997-2024
Scan the QR code below to visit the online version of this activity.
Description of Levels
Level 1 - Simple addition and subtraction - positive integers
Level 2 - Addition, subtraction and multiplication - positive integers
Level 3 - Indices - positive integers
Level 4 - Brackets - positive integers
Level 5 - Mixed expressions - positive integers
Level 6 - Mixed expressions - negative integers
Level 7 - Mixed expressions - positive and negative decimals
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You may also want to use a calculator to check your working. See Calculator Workout skill 13.
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teacher, tutor or parent. | {"url":"https://www.transum.org/Software/SW/Starter_of_the_day/Students/Substitution.asp?Level=4","timestamp":"2024-11-02T18:05:32Z","content_type":"text/html","content_length":"54621","record_id":"<urn:uuid:4e9f74d0-b99c-4da6-b48e-264190246972>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00442.warc.gz"} |
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Scale Factor
Scale Factor is used to scale shapes in different dimensions. In geometry, we learn about different geometrical shapes which both in two-dimension and three-dimension. The scale factor is a measure
for similar figures, who look the same but have different scales or measures. Suppose, two circle looks similar but they could have varying radii.
The scale factor states the scale by which a figure is bigger or smaller than the original figure. It is possible to draw the enlarged shape or reduced shape of any original shape with the help of
scale factor.
Table of contents:
What is the Scale factor
The size by which the shape is enlarged or reduced is called as its scale factor. It is used when we need to increase the size of a 2D shape, such as circle, triangle, square, rectangle, etc.
If y = Kx is an equation, then K is the scale factor for x. We can represent this expression in terms of proportionality also:
y ∝ x
Hence, we can consider K as a constant of proportionality here.
The scale factor can also be better understood by Basic Proportionality Theorem.
Scale Factor Formula
The formula for scale factor is given by:
Dimensions of Original Shape x scale Factor = Dimension of new shape
Scale factor = Dimension of New Shape/Dimension of Original Shape
Take an example of two squares having length-sides 6 unit and 3 unit respectively. Now, to find the scale factor follow the steps below.
Step 1: 6 x scale factor = 3
Step 2: Scale factor = 3/6 (Divide each side by 6).
Step 3: Scale factor = ½ =1:2(Simplified).
Hence, the scale factor from the larger Square to the smaller square is 1:2.
The scale factor can be used with various different shapes too.
Scale Factor Problem
For example, there’s a rectangle with measurements 6 cm and 3 cm.
Both sides of the rectangle will be doubled if we increase the scale factor for the original rectangle by 2. I.e By increasing the scale factor we mean to multiply the existing measurement of the
rectangle by the given scale factor. Here, we have multiplied the original measurement of the rectangle by 2.
Originally, the rectangle’s length was 6 cm and Breadth was 3 cm.
After increasing its scale factor by 2, the length is 12 cm and Breadth is 6 cm.
Both sides will be triple if we increase the scale factor for the original rectangle by 3. I.e By increasing the scale factor we mean to multiply the existing measurement of the rectangle by the
given scale factor. Here, we have multiplied the original measurement of the rectangle by 3.
Originally, the rectangle’s length was 6 cm and Breadth was 3 cm.
After increasing its scale factor by 3, the length is 18 cm and Breadth is 9 cm.
How to find the scale factor of Enlargement
Problem 1: Increase the scale factor of the given Rectangle by 4.
Hint: Multiply the given measurements by 4.
Solution: Given Length of original rectangle = 4cm
Width or breadth of given rectangle = 2cm
Now as per the given question, we need to increase the size of the given triangle by scale factor of 4.
Thus, we need to multiply the dimensions of given rectangle by 4.
Therefore, the dimensions of new rectangle or enlarged rectangle is given by:
Length = 4 x 4 = 16cm
And Breadth = 2 x 4 = 8cm.
Scale Factor of 2
The scale factor of 2 means the new shape obtained after scaling the original shape is twice of the shape of the original shape.
The example below will help you to understand the concept of scale factor of 2.
Problem 2: Look at square Q. What scale factor has square P increased by?
Hint: Work Backwards, and divide the measurements of the new triangle by the original one to get the scale factor.
Solution: Divide the length of one side of the larger square by the scale factor.
We will get the length of the corresponding side of the smaller square.
Scale Factor of Triangle
The triangles which are similar have same shape and measure of three angles are also same. The only thing which varies is their sides. However, the ratio of the sides of one triangle is equivalent
to the ratio of sides of another triangle, which is called here the scale factor.
If we have to find the enlarged triangle similar to the smaller triangle, we need to multiply the side-lengths of the smaller triangle by the scale factor.
Similarly, if we have to draw a smaller triangle similar to bigger one, we need to divide the side-lengths of the original triangle by scale factor.
Real-life Applications of Scale Factor
It is important to study real-life applications to understand the concept more clearly:
Because of various numbers getting multiplied or divided by a particular number to increase or decrease the given figure, scale factor can be compared to Ratios and Proportions.
1. If there’s a larger group of people than expected at a party at your home. You need to increase the ingredients of the food items by multiplying each one by the same number to feed them all.
Example, If there are 4 people extra than you expected and one person needs 2 pizza slices, then you need to make 8 more pizza slices to feed them all.
2. Similarly, the Scale factor is used to find a particular percentage increase or to calculate the percentage of an amount.
3. It also lets us work out the ratio and proportion of various groups, using the times’ table knowledge.
4. To transform Size: In this, the ratio of expressing how much to be magnified can be worked out.
5. Scale Drawing: It is the ratio of measuring the drawing compared to the original figure given.
6. To compare 2 Similar geometric figures: When we compare two similar geometric figures by the scale factor, it gives the ratio of the lengths of the corresponding sides.
Access many brilliant concepts thoroughly explained by visiting BYJU’S or Downloading the BYJU’S app. | {"url":"https://mathlake.com/Scale-Factor","timestamp":"2024-11-06T01:31:58Z","content_type":"text/html","content_length":"14621","record_id":"<urn:uuid:d1b1f9e8-68c8-4d60-a7cd-b85c165c8f3c>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00515.warc.gz"} |
In formulas, you can create your own custom JavaScript functions (primitives) by calling the kendo.spreadsheet.defineFunction(name, func).
The first argument (string) is the name for your function in formulas (case-insensitive), and the second one is a JavaScript function (the implementation).
The following example demonstrates how to define a function that calculates the distance between two points.
kendo.spreadsheet.defineFunction("distance", function(x1, y1, x2, y2){
var dx = Math.abs(x1 - x2);
var dy = Math.abs(y1 - y2);
var dist = Math.sqrt(dx*dx + dy*dy);
return dist;
[ "x1", "number" ],
[ "y1", "number" ],
[ "x2", "number" ],
[ "y2", "number" ]
If you include the above JavaScript code, you can then use DISTANCE in formulas. For example, to find the distance between coordinate points (2,2) and (5,6), type in a cell =DISTANCE(2, 2, 5, 6).
Optionally, you can use the function in combined expressions such as =DISTANCE(0, 0, 1, 1) + DISTANCE(2, 2, 5, 6).
In the above example, defineFunction returns an object that has an args method. You can use it to specify the expected types of arguments. If the function is called with mismatching argument types,
the runtime of the Spreadsheet automatically returns an error and your implementation is not called. This spares the time for manually writing code that does argument type checking and provides a
nice declarative syntax instead.
To retrieve currency information from a remote server, define a primitive to make this information available in formulas. To define an asynchronous function, call argsAsync instead of args.
kendo.spreadsheet.defineFunction("currency", function(callback, base, curr){
// A suggested fetchCurrency function.
// The way it is implemented is not relevant to the goal of the demonstrated scenario.
fetchCurrency(base, curr, function(value){
[ "base", "string" ],
[ "curr", "string" ]
The argsAsync passes a callback as the first argument to your implementation function, which you need to call with the return value.
It is possible to use new approaches in formulas such as =CURRENCY("EUR", "USD") and =A1 * CURRENCY("EUR", "USD"). Note that the callback is invisible in formulas. The second formula shows that even
though the implementation itself is asynchronous, it can be used in formulas in a synchronous way—that is, the result yielded by CURRENCY is multiplied by the value in A1.
The rest of this article provides information on argument types.
As can be seen in the examples above, both args and argsAsync expect a single array argument. It contains one definition for each argument. Each definition is in turn an array where the first element
is the argument name that has to be a valid JavaScript identifier and the second element is a type specifier.
The Spreadsheet supports the following type specifiers.
BASIC SPECIFIER ACTION
"number" Requires a numeric argument.
"number+" Requires a number bigger than or equal to zero.
"number++" Requires a non-zero positive number.
"integer"/"integer+"/ Similar to number-s, but requires integer argument. Note that these may actually modify the argument value: if a number is specified and it has a decimal part, it will silently
"integer++" be truncated to integer, instead of returning an error. This is similar to Excel.
"divisor" Requires a non-zero number. Produces a #DIV/0! error if the argument is zero.
"string" Requires a string argument.
"boolean" Requires a boolean argument. Most times you may want to use "logical" though.
"logical" Requires a logical argument. That is, Booleans true or false, but 1 and 0 are also accepted. It gets converted to an actual Boolean.
"date" Requires a date argument. Internally, dates are stored as numbers (the number of days since December 31 1899), so this works the same as "integer". It was added for
"datetime" This is like "number", because the time part is represented as a fraction of a day.
"anyvalue" Accepts any value type.
"matrix" Accepts a matrix argument. This is either a range, for example, A1:C3, or a literal matrix (see the Matrices section below).
"null" Requires a null (missing) argument. The reason for this specifier will be clarified in the Optional Arguments section.
Some specifiers actually modify the value that your function receives. For example, you can implement a function that truncates the argument to integer.
defineFunction("truncate", function(value){
return value;
[ "value", "integer" ]
If you call =TRUNCATE(12.634), the result is 12. You can also call =TRUNCATE(TRUE), it returns 1. All numeric types silently accept a Boolean, and convert true to 1 and false to 0.
By default, if an argument is an error, your function is not called and that error is returned.
defineFunction("iserror", function(value){
return value instanceof kendo.spreadsheet.CalcError;
[ "value", "anyvalue" ]
With this implementation, when you type =ISERROR(1/0), #DIV/0! instead of true is returned—the error is passed over and aborts the computation. To allow the passing of errors, append a ! to the type.
[ "value", "anyvalue!" ]
The result is that true is returned.
All above-mentioned type specifiers force references. FBecasues of this, =TRUNCATE(A5) also works. The function gets the value in the A5 cell. If A5 contains a formula, the runtime library verifies
you get the current value—that is, A5 is evaluated first. All of this goes under the hood and you need not worry about it.
Sometimes you might need to write functions that receive a reference instead of a resolved value. Such an example is the ROW function of Excel. In its basic form, it takes a cell reference and
returns its row number, as demonstrated in the following example. The actual ROW function is more complicated.
defineFunction("row", function(cell){
// Add one because internally row indexes are zero-based.
return cell.row + 1;
[ "reference", "cell" ]
If you now call =ROW(A5), you get 5 as a result, regardless of the content in the A5 cell—it might be empty or it is possible that this very formula sits in the A5 cell and there must be no circular
reference error in such a case.
See the References section below for more information about references.
The following table lists the related type specifiers:
TYPE ACTION
"ref" Allows any reference argument and your implementation gets it as such.
"area" Allows a cell or a range argument (CellRef or RangeRef instance).
"cell" Allows a cell argument (CellRef instance).
"anything" Allows any argument type. The difference to anyvalue is that this one does not force references—that is, if a reference is passed, it remains a reference instead of being replaced by
its value.
In addition to the basic type specifiers that are strings, you can also use the following forms of type specifications:
[ "null", DEFAULT ] Validates a missing argument and makes it take the given DEFAULT value. This can be used in conjunction with "or" to support optional arguments.
[ "not", SPEC ] Requires an argument which does not match the specification.
[ "or", SPEC, SPEC, Validates an argument that passes any of the specifications.
... ]
[ "and", SPEC, SPEC, Validates an argument that passes all the specifications.
... ]
[ "values", VAL1, The argument must strictly equal one of the listed values.
VAL2, ... ]
[ "[between]", MIN, Validates an argument between the given values inclusive. Note that it does not require numeric argument. The "between" value is an alias.
MAX ]
[ "(between)", MIN, This is similar to "[between]" but is exclusive.
MAX ]
[ "[between)", MIN, Requires an argument greater than or equal to MIN, and strictly less than MAX.
MAX ]
[ "(between]", MIN, Requires an argument strictly greater than MIN, and less than or equal to MAX.
MAX ]
[ "assert", COND ] Inserts an arbitrary condition literally into the code (see the Assertions section below).
[ "collect", SPEC ] Collects all remaining arguments that pass the specification into a single array argument. This only makes sense at top level and cannot be nested in "or", "and", etc. Arguments
not matching the SPEC are silently ignored, except errors. Each error aborts the calculation.
[ "#collect", SPEC ] This is similar to "collect", but ignores errors as well.
In certain clauses you might need to refer to values of previously type-checked arguments. For example, if you want to write a primitive that takes a minimum, a maximum, and a value that must be
between them, and should return as a fraction the position of that value between min and max.
defineFunction("my.position", function(min, max, value){
return (value - min) / (max - min);
[ "min", "number" ],
[ "max", "number" ],
[ "value", [ "and", "number",
[ "[between]", "$min", "$max" ] ] ]
Note the type specifier for "value":
[ "and", "number",
[ "[between]", "$min", "$max" ] ]
The code requires that the parameter is a number and that it has to be between min and max. To refer to a previous argument, prefix the identifier with a $ character. This approach works for
arguments of "between" (and friends), "assert", "values" and "null".
The above function is not quite correct because it does not check that max is actually greater than min. To do that, use "assert", as demonstrated in the following example.
defineFunction("my.position", function(min, max, value){
return (value - min) / (max - min);
[ "min", "number" ],
[ "max", "number" ],
[ "value", [ "and", "number",
[ "[between]", "$min", "$max" ] ] ],
[ "?", [ "assert", "$min < $max", "N/A" ] ]
The "assert" type specification allows you to introduce an arbitrary condition into the JavaScript code of the type-checking function. An argument name of "?" does not actually introduce a new
argument, but provides a place for such assertions. The third argument to "assert" is the error code that it should produce if the condition does not stand (and #N/A! is actually the default).
As hinted above, you can use the "null" specifier to support optional arguments.
The following example demonstrates the actual definition of the ROW function.
defineFunction("row", function(ref){
if (!ref) {
return this.formula.row + 1;
if (ref instanceof CellRef) {
return ref.row + 1;
return this.asMatrix(ref).mapRow(function(row){
return row + ref.topLeft.row + 1;
[ "ref", [ "or", "area", "null" ]]
The code requires that the argument can either be an area (a cell or a range) or null (that is, missing). By using the "or" combiner, you make it accept either of these. If the argument is missing,
your function gets null. In such cases, it has to return the row of the current formula that you get by this.formula.row. For more details, refer to the section on context objects.
In most cases, “optional” means that the argument takes some default value if one is not provided. For example, the LOG function computes the logarithm of the argument to a base, but if the base is
not specified, it defaults to 10.
The following example demonstrates this implementation.
defineFunction("log", function(num, base){
return Math.log(num) / Math.log(base);
[ "*num", "number++" ],
[ "*base", [ "or", "number++", [ "null", 10 ] ] ],
[ "?", [ "assert", "$base != 1", "DIV/0" ] ]
The type specification for base is: [ "or", "number++", [ "null", 10 ] ]. This says it should accept any number greater than zero, but if the argument is missing, defaults to 10. The implementation
does not have to deal with the case that the argument is missing — it will get 10 instead. Note that it uses an assertion to make sure the base is not 1. If the base is 1, a #DIV/0! error is
To return an error code, return a spreadsheet.CalcError object.
defineFunction("tan", function(x){
// If x is sufficiently close to PI, "tan" will return
// infinity or some really big number.
// The example will error out instead.
if (Math.abs(x - Math.PI/2) < 1e-10) {
return new spreadsheet.CalcError("DIV/0");
return Math.tan(x);
[ "x", "number" ]
For convenience, you can also throw a CalcError object for synchronous primitives—that is, if you use args and not argsAsync.
It is possible to do the above through an assertion as well.
defineFunction("tan", function(x){
return Math.tan(x);
[ "x", [ "and", "number",
[ "assert", "1e-10 < Math.abs($x - Math.PI/2)", "DIV/0" ] ] ]
The type checking mechanism errors out when your primitive receives more arguments than specified. There are a few ways to receive all remaining arguments without errors.
The "rest" Type Specifier
The simplest way is to use the "rest" type specifier. In such cases, the last argument is an array that contains all remaining arguments, whatever types they might be.
The following example demonstrates how to use a function that joins arguments with a separator producing a string.
defineFunction("join", function(sep, list){
return list.join(sep);
[ "sep", "string" ],
[ "list", "rest" ]
This allows for =JOIN("-", 1, 2, 3) which returns 1-2-3 and for =JOIN(".") which returns the empty string because the list will be empty.
The "collect" Clauses
The "collect" clauses collect all remaining arguments that match a certain type specifier, ignoring all others except for the errors. You can use them in functions like SUM that sums all numeric
arguments, but does not care about empty or text arguments.
The following example demonstrates the definition of SUM.
defineFunction("sum", function(numbers){
return numbers.reduce(function(sum, num){
return sum + num;
}, 0);
[ "numbers", [ "collect", "number" ] ]
The "collect" clause aborts when it encounters an error. To ignore errors as well, use the "#collect" specification. Note that "collect" and "#collect" only make sense when either is the first
specifier—that is, they cannot be nested in "or", "and", and the like.
Other Type-Checked Arguments
There are functions that allow an arbitrary number of arguments of specific types. For example, the SUMPRODUCT function takes an arbitrary number of arrays, multiplies the corresponding numbers in
these arrays, and then returns the sum of the products. In this case, you need at least two arrays.
The following example demonstrates the argument specification.
[ "a1", "matrix" ],
[ "+",
[ "a2", [ "and", "matrix",
[ "assert", "$a2.width == $a1.width" ],
[ "assert", "$a2.height == $a1.height" ] ] ] ]
The "+" in the second definition means that one or more arguments are expected to follow and that the a2 argument, defined there, can repeat. Notice how you can use assertions to make sure the
matrices have the same shape as the first one (a1).
For another example, look at the SUMIFS function (see Excel documentation). It takes a sum_range, a criteria_range, and a criteria. These are the required arguments. Then, any number of
criteria_range and criteria arguments can follow. In particular, criteria ranges must all have the same shape (width/height). Here is the argument definition for SUMIFS:
[ "range", "matrix" ],
[ "m1", "matrix" ],
[ "c1", "anyvalue" ],
[ [ "m2", [ "and", "matrix",
[ "assert", "$m1.width == $m2.width" ],
[ "assert", "$m1.height == $m2.height" ] ] ],
[ "c2", "anyvalue" ] ]
The repeating part now is simply enclosed in an array, not preceded by "+". This indicates to the system that any number might follow, including zero, while "+" requires at least one argument.
Dates are stored as the number of days since 1899-12-31 that is considered to be the first date. In Excel, the first day is 1900-01-01, but for historical reasons Excel assumes that 1900 is a leap
year. For more information, refer to the article on the leap year bug. In Excel, day 60 yields an invalid date (1900-02-29), which means that date calculations involving dates before and after
1900-03-01 produce wrong results.
To be compatible with Excel and to avoid the unwilling implementation of this bug, the Spreadsheet uses 1899-12-31 as the base date. Dates that are greater than or equal to 1900-03-31 have the same
numeric representation as in Excel, while dates before 1900-03-31 are smaller by 1.
Time is kept as a fraction of a day—that is, 0.5 means 12:00:00. For example, the date and time Sep 27 1983 12:35:59 is numerically stored as 30586.524988425925. To verify that in Excel, paste this
number in a cell and then format it as a date or time.
Functions to pack or unpack dates are available in spreadsheet.calc.runtime.
var runtime = kendo.spreadsheet.calc.runtime;
// Unpacking
var date = runtime.unpackDate(28922.55);
console.log(date); // { year: 1979, month: 2, date: 8, day: 4 }
var time = runtime.unpackTime(28922.55);
console.log(time); // { hours: 13, minutes: 12, seconds: 0, milliseconds: 0 }
var date = runtime.serialToDate(28922.55); // produces JavaScript Date object
console.log(date.toISOString()); // 1979-03-08T13:12:00.000Z
// Packing
console.log(runtime.packDate(2015, 5, 25)); // year, month, date
console.log(runtime.packTime(13, 35, 0, 0)); // hours, minutes, seconds, ms
console.log(runtime.dateToSerial(new Date()))
Note that the serial date representation does not carry any timezone information, so the functions involving Date objects (serialToDate and dateToSerial) use the local components and not UTC—as Excel
As mentioned earlier, certain type specifiers allow you to get a reference in your function rather than the resolved value. Note that when you do so, you cannot rely on the values in those cells to
be calculated. As a result, if your function might need the values as well, you have to compute them. Because the function which does this is asynchronous, your primitive has to be defined in an
asynchronous style as well.
defineFunction("test", function(callback, x){
this.resolveCells([ x ], function(){
console.log(x instanceof spreadsheet.CellRef); // true
console.log("So we have a cell:");
console.log(x.sheet, x.row, x.col);
console.log("And its value is:");
callback("Cell value: " + this.getRefData(x));
[ "x", "cell" ]
This function accepts a cell argument and you can only call it like =test(B4). It calls this.resolveCells from the context object to verify that the cell value is calculated. Without this step and if
the cell actually contains a formula, the value returned by this.getRefData could be outdated. Then it prints some information about that cell.
The following list explains the types of references that your primitive can receive:
• spreadsheet.Ref—A base class only. All references inherit from it, but no direct instance of this object should ever be created. The class is exported just to make it easier to check whether
something is a reference: x instanceof spreadsheet.Ref.
• spreadsheet.NULLREF—An object (a singleton) and not a class. It represents the NULL reference, and could occur, for example, when you intersect two disjoint ranges, or when a formula depends on a
cell that has been deleted. For example, when you put in some cell =test(B5) and then right-click on column B and delete it. To test when something is the NULL reference, just do x ===
• spreadsheet.CellRef—Represents a cell reference. Note that the references here follow the same programming language concept. They do not contain data. Instead they just point to where the data
is. So a cell reference contains 3 essential properties:
□ sheet — the name of the sheet that this cell points to (as a string)
□ row — the row number, zero-based
□ col — the column number, zero-based
• spreadsheet.RangeRef—A range reference. It contains topLeft and bottomRight, which are CellRef objects.
• spreadsheet.UnionRef—A union. It contains a refs property, which is an array of references (it can be empty). A UnionRef can be created by the union operator, which is the comma.
The following example demonstrates how to use a function that takes an arbitrary reference and returns its type of reference.
defineFunction("refkind", function(x){
if (x === spreadsheet.NULLREF) {
return "null";
if (x instanceof spreadsheet.CellRef) {
return "cell";
if (x instanceof spreadsheet.RangeRef) {
return "range";
if (x instanceof spreadsheet.UnionRef) {
return "union";
return "unknown";
[ "x", "ref" ]
The following example demonstrates how to use a function that takes an arbitrary reference and returns the total number of cells it covers.
defineFunction("countcells", function(x){
var count = 0;
function add(x) {
if (x instanceof spreadsheet.CellRef) {
} else if (x instanceof spreadsheet.RangeRef) {
count += x.width() * x.height();
} else if (x instanceof spreadsheet.UnionRef) {
} else {
// unknown reference type.
throw new CalcError("REF");
return count;
[ "x", "ref" ]
You can now say:
• =COUNTCELLS(A1) — returns 1.
• =COUNTCELLS(A1:C3) — returns 9.
• =COUNTCELLS( (A1,A2,A1:C3) ) — returns 11. This is a union.
• =COUNTCELLS( (A1:C3 B:B) ) — returns 3. This is an intersection between the A1:C3 range and the B column.
Here is a function that returns the background color of some cell:
defineFunction("backgroundof", function(cell){
var workbook = this.workbook();
var sheet = workbook.sheetByName(cell.sheet);
return sheet.range(cell).background();
[ "cell", "cell" ]
It uses this.workbook() to retrieve the workbook, and then uses the Workbook/Sheet/Range APIs to fetch the background color of the given cell.
Matrices are defined by spreadsheet.calc.runtime.Matrix. Your primitive can request a Matrix object by using the "matrix" type specification. In this case, it can accept a cell reference, a range
reference, or a literal array. You can type literal arrays in formulas like in Excel, e.g., { 1, 2; 3, 4 } (rows separated by semicolons).
Matrices were primarily added to deal with the “array formulas” concept in Excel. A function can return multiple values, and those will be in a Matrix object.
The following example demonstrates how to use a function that doubles each number in a range and returns a matrix of the same shape.
defineFunction("doublematrix", function(m){
return m.map(function(value){
return value * 2;
[ "m", "matrix" ]
To use this formula:
1. Select a range—for example A1:B2.
2. Press F12 and type =doublematrix(C3:D4).
3. Press Ctrl+Shift+Enter (same as in Excel). As a result, cells A1:B2 get the doubles of the values from C3:D4.
The following table lists some of the methods and properties the Matrix objects provide.
METHOD OR DESCRIPTION
width and height These properties indicate the dimensions of this matrix.
clone() Returns a new matrix with the same data.
get(row, col) Returns the element at a given location.
set(row, col, Sets the element at a given location.
each(func, Iterates through elements of the matrix, calling your func for each element (first columns, then rows) with 3 arguments: value, row and column. If includeEmpty is true, it will call
includeEmpty) your function for empty (null) elements as well. Otherwise, it only calls it where a value exists.
map(func, This is similar to each, but produces a new matrix of the same shape as the original one with the values returned by your functions.
transpose() Returns the transposed matrix. The rows of the original matrix become columns of the transposed one.
unit(n) Returns the unit square matrix of size n.
multiply(m) Multiplies the current matrix by the given matrix, and returns a new matrix as the result.
determinant() Returns the determinant of this matrix. The matrix should contain only numbers and be square. Note that there are no checks for this.
inverse() Returns the inverse of this matrix. The matrix should contain only numbers and be square. Note that there are no checks for this. If the inverse does not exist—the determinant is
zero—then it returns null.
Every time a formula is evaluated, a special Context object is created and each involved primitive function is invoked in the context of that object—that is, it is accessible as this.
The following table demonstrates some of the methods the Context object provides.
METHOD DESCRIPTION
resolveCells(array, Verifies that all references in the given array are resolved before invoking your callback—that is, executes any formula. If this array turns out to include the cell where the
callback) current formula lives, it returns a #CIRCULAR! error. Elements that are not references are ignored.
cellValues(array) Returns as a flat array the values in any reference that exist in the given array. Elements that are not references are copied over.
asMatrix(arg) Converts the given argument to a matrix, if possible. It accepts a RangeRef object or a plain JavaScript non-empty array. Additionally, if a Matrix object is provided, it is
returned as is.
workbook() Returns the Workbook object where the current formula is evaluated.
getRefData(ref) Returns the data—that is the value—in the given reference. If a CellRef is given, it returns a single value. For a RangeRef or UnionRef, it returns a flat array of values.
Additionally, there is a formula property, an object representing the current formula. Its details are internal, but you can rely on it having the sheet (sheet name as a string), row and col
properties, the location of the current formula.
Missing args or argsAsync
This section explains what happens if you do not invoke args or argsAsync. It is recommended that you do not use that form.
If args or argsAsync are not called, the primitive function receives exactly two arguments:
• A callback to be invoked with the result.
• An array that contains the arguments passed in the formula.
The following example demonstrates how to use a function that adds two things.
defineFunction("add", function(callback, args){
callback(args[0] + args[1]);
• =ADD(7, 8) → 15
• =ADD() → NaN
• =ADD("foo") → fooundefined
• =ADD(A1, A2) → A1A2
In other words, if you use this raw form, you are responsible for type-checking the arguments and your primitive is always expected to be asynchronous. | {"url":"https://docs.telerik.com/aspnet-core/html-helpers/data-management/spreadsheet/custom-functions","timestamp":"2024-11-07T10:45:58Z","content_type":"text/html","content_length":"88274","record_id":"<urn:uuid:7708ec30-3d1f-4d6a-ad6e-30469d6a1e96>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00868.warc.gz"} |
The Symmetric Difference Distance: A New Way to Evaluate the Evolution of Interfaces along Molecular Dynamics Trajectories; Application to Influenza Hemagglutinin
E-pôle de Génoinformatique, CNRS UMR 7592, Institut Jacques Monod, F-75013 Paris, France
Chimie ParisTech, PSL Research University, CNRS, Institute of Chemistry for Life and Health Science (i-CLeHS, FRE 2027), F-75005 Paris, France
Université Paris Diderot, Sorbonne Paris Cité, UFR Sciences du vivant, 5 rue Thomas Mann, F-75205 Paris CEDEX 13, France
Pathologies de la réplication de l’ADN, Institut Jacques Monod, CNRS UMR 7592, CNRS, F-75013 Paris, France
CMPLI, INSERM U1133 (BFA, CNRS UMR 8251), Université Paris Diderot, F-75205 Paris CEDEX 13, France
Author to whom correspondence should be addressed.
Submission received: 3 April 2019 / Revised: 18 April 2019 / Accepted: 8 May 2019 / Published: 12 May 2019
We propose a new and easy approach to evaluate structural dissimilarities between frames issued from molecular dynamics, and we test this methodology on human hemagglutinin. This protein is
responsible for the entry of the influenza virus into the host cell by endocytosis, and this virus causes seasonal epidemics of infectious disease, which can be estimated to result in hundreds of
thousands of deaths each year around the world. We computed the three interfaces between the three protomers of the hemagglutinin H1 homotrimer (PDB code: 1RU7) for each of its conformations
generated from molecular dynamics simulation. For each conformation, we considered the set of residues involved in the union of these three interfaces. The dissimilarity between each pair of
conformations was measured with our new methodology, the symmetric difference distance between the associated set of residues. The main advantages of the full procedure are: (i) it is parameter free;
(ii) no spatial alignment is needed and (iii) it is simple enough so that it can be implemented by a beginner in programming. It is shown to be a relevant tool to follow the evolution of the
conformation along the molecular dynamics trajectories.
1. Introduction
Influenza virus causes seasonal epidemics of infectious disease and can even trigger pandemics. Inferring from U.S. data [
], these epidemics may cause hundreds of thousands of deaths each year around the world, mainly concerning the elderly, children, and immunosuppressed people, resulting in a real public health issue.
Moreover, there were several severe influenza pandemics, such as the 1968 Hong Kong one [
] and the 1918 one, which resulted in about 20 millions deaths [
]. More recently, modeling studies attributed about 200,000 deaths to the 2009 influenza pandemic [
]. Initiation of virus infection involves multiple influenza Hemagglutinin (HA) binding [
] to sialic acids typically by α2,3 or even α2,6 linkages at the cell surface. The virus is then endocytosed through clathrin-dependent pathways, and the fusion of the viral particle with the cell’s
membrane happens due to the acidification (about pH = 5) in the late endosomes. Indeed, this pH variation induces a folding change of HA homotrimers, resulting in the three fusion peptides’ exposure
to the cell’s membrane [
]. HA is thus one of the two neutralizing antibody targets with neuraminidase, which is the second surface glycoprotein, allowing the virus to spread out from the cells [
In a recent modeling study to design new inhibitors against HA, two Molecular Dynamics (MD) simulations were performed, one at pH = 7 and one at pH = 5, this acidic pH being the one inducing a
conformational change [
]. A motivation of this previous study was to compare the simulations at both pH values. Each protomer has two domains: HA1, responsible for the binding, and HA2 for the fusion, containing
respectively 327 and 160 amino acid residues. The tridimensional structural data of HA were found in the Protein Data Bank (PDB), Code 1RU7, taken from [
] (see
Figure 1
). This structure is the one of A/Puerto Rico/8/1934 H1N1.
Starting from an initial conformation, 40 conformations were extracted every 0.5 ns for two performed simulations (see [
] for technical details about the protocol). The problem considered here is to evaluate, for each set of 41 conformations, the dissimilarities between pairs of conformations. For this purpose, we
needed a simple procedure, easy to implement, and without spatial alignment because these latter generally involve unwanted contributions of meaningless parts of macromolecules, these parts being
difficult to identify.
Usually, the dissimilarity between two conformers of the same macromolecule is measured through the computation of the RMSD (Root Mean Squared Deviation). For a macromolecule containing
heavy atoms, RMSD is the quadratic mean of the lengths of the
pairs of atoms (
atoms for the first conformer,
atoms for the second conformer), minimized for all rotations and translations of either of these two conformers. An analytical solution for this optimization problem is known: the optimal translation
is such that the barycenter of the two conformers should coincide, and the optimal rotation is expressed with quaternions (see Appendix in [
] and Appendix A.5 in [
]). While the calculus of an RMSD is easy to compute, this quantity offers two drawbacks. The first one is the large numerical impact on the RMSD value from non-relevant parts of the macromolecule,
such as the flexible ends of the backbone, or such as the irrelevant parts of the macromolecule far from the domain of interest for the experimentalist. For instance, in a protein–protein complex,
the most relevant part is the protein–protein interface, because it is the area that can stand the weak bonds responsible for the stability of the complex. This is also the case for a protein–ligand
complex. In the case of HA, the relevant part of the trimer is the three interfaces between the pairs of protomers (each of the three pairs of protomers defines an interface), because they are the
location of interprotomers’ binding interactions. The second drawback of the RMSD is that, when it is computed over all heavy atoms, the side chains have an important numerical contribution to the
RMSD value, although the location of the side chains is often considered to be meaningless due to their flexibility. For the reasons mentioned above, we restricted the evaluation of the
dissimilarities between HA conformers generated by MD simulations to the C
atoms of the three polypeptidic chains’ interfaces.
Thus, for each conformer, we have a set of C
atoms. The simplest way to evaluate the dissimilarity between two sets is to compute their Symmetric Difference Distance (SDD). The SDD between two sets having respectively
$n 1$
$n 2$
elements is
$n 1 + n 2 − 2 n 12$
$n 12$
being the number of elements of the intersection of the two sets [
]. The properties of a distance are recalled in
Appendix A
, including why they are useful. The computation of the SDD between the two sets of C
atoms associated with two conformations of our HA trimers (i.e., there is one set of C
atoms per trimer) is explained in
Section 3
. Here, the unit of this distance is the number of C
atoms. It is recalled that, opposite what is encountered during RMSD calculations, the sets of C
atoms defined by the interfaces are never pairwise associated, and in general, they have different cardinalities.
2. Results and Discussion
Starting from an initial conformation at time
$t 0$
, the MD simulation at pH = 7 generated 40 conformations, at times
$t 1$
$t 2$
, …,
$t 40$
, and similarly, the MD simulation at pH = 5 generated 40 conformations, the time step between two conformations being 0.5 ns [
]. Thus, there was a total of 41 conformations for each of these two MD simulations. We computed for the two pH values the 40 SDDs between the 40 generated conformations and their respective initial
conformations at
$t 0$
. These distances values are plotted in
Figure 2
. The range of the distance values was 39–72 at pH = 7 (the unit of distances is the number of C
atoms). It was slightly smaller at pH = 5: 25–52. The maximal values were reached for the frames generated respectively at
$t 26$
$t 24$
. The maximal ratios of the SDDs values
$n 1 + n 2 − 2 n 12$
to the total number of interface residues
$n 1 + n 2$
were 16.3% at pH = 7 and 17.2% at pH = 5. These moderately low values indicate that the generated conformations did not much differ from their respective initial ones at
$t 0$
. The correlation coefficient between the two sets of distances was 0.507. There was no reason to expect a high correlation coefficient since the SDD between the two initial conformations was 262,
which was much higher than the range values at pH = 7 and at pH = 5.
We also computed for the two pH values the 40 RMSD between these 40 conformations and their respective initial conformations at
$t 0$
: see
Figure 3
. The range of RMSD values was 1.45–3.09 Å at pH = 7. It was slightly smaller at pH = 5: 1.65–2.58 Å. The RMSD values were small relatively to the size of the HA: the radius of the smallest sphere
enclosing the tridimensional structure of the 11,510 atoms of HA was 68.55 Å for 1RU7. These small RMSD values indicate that the generated conformations did not significantly differ from their
respective initial ones at
$t 0$
. More importantly, the trend and the main peaks visible in
Figure 3
are visible in
Figure 2
. This means that working on C
atoms at the interfaces rather than computing RMSD on all heavy atoms did not seem to cause any major information loss. Thus, our approach is pertinent.
Then, we computed for the two pH values the SDDs between the 40 successive pairs of conformations. These distance values are plotted in
Figure 4
. The range of distance values was similar for both pH values: 26–62 at pH = 7 and 24–52 at pH = 5. The maximal values were reached between the frames generated at
$t 22$
for both simulations. The maximal ratios of the SDDs values
$n 1 + n 2 − 2 n 12$
to the total number of interface residues
$n 1 + n 2$
were 13.9% at pH = 7 and 14.4% at pH = 5. These values were slightly lower than the ones of
Figure 2
for pH = 7, and they were of the same magnitude as the ones of
Figure 2
for pH = 5. For both simulations, the variations of the SDDs along time did not induce a trend to deviate more and more from the initial conformation, as shown in
Figure 2
. Thus, we can assume that the conformation is stable for the simulations at both pH values, at least for the interval of time considered (20 ns). The correlation coefficient between the two sets of
distances was 0.468. There was no particular reason to expect a high correlation coefficient.
We also computed for the two pH values the RMSD values between the 40 successive pairs of conformations: see
Figure 5
. The range of RMSD values were similar for both pH values: 1.40–2.60 Å at pH = 7 and 1.42–2.54 Å at pH = 5. The RMSD values were small: we recall that the radius of the smallest sphere enclosing the
11.510 atoms of HA was 120.5 Å for 1RU7. Such a similarity between the range values at both pH is observed in
Figure 4
. The jumps observed in
Figure 5
are also visible in
Figure 4
. Due to the relatively small ranges of RMSD values, the jumps seen
Figure 5
are difficult to interpret. However, the maximal ratios of the SDDs values to the total number of interface residues that we already mentioned (around 14%) indicate that the interfaces may contribute
significantly to the conformational changes along time. Again, working on C
atoms at the interfaces rather than computing RMSD values for all heavy atoms did not seem to cause any major information loss.
After having observed that the HA conformation remains stable over the 40-ns simulations, we needed to define a mean conformation for docking purposes. Indeed, considering such an average structure
in the course of a docking simulation, rather than a structure determined from X-ray diffraction experiments, enables us to take into account the relaxation of the protein in the solvent, at T = 310
K [
]. We defined this mean conformation with the algorithm described in [
]: it is the one that minimizes the sum of the squared distances to all other conformers. This mean conformation is the one of the frame generated at
$t 30$
for pH = 7, and it is the one of the frame generated at
$t 15$
for pH = 5. The residues at the interfaces of the mean conformers are given in
Appendix B
. The SDD between these two mean conformations was equal to 70 C
atoms. This was few compared to the 1448 residues of each mean conformer, but it was larger compared to the respective 224 and 214 residues of the two interfaces, which had 184 residues in common.
The RMSD between the 1448 pairs of C
atoms of these mean conformers was 3.54 Å, while the largest atom-pair length was 13.7 Å (obtained for the first serine of the HA2 domain). The optimal superposition of the two interfaces is shown in
Figure 6
. It was realized with the CSR freeware (
), which implements a non-parametric algorithm performing a spatial alignment without a cutoff distance and without any input pairwise correspondence [
We have also performed an optimal superposition of the mean frames at
$t 30$
for pH = 7 and at
$t 15$
for pH = 5, on the basis of all pairs of 11.510 heavy atoms. The display of this optimal superposition in
Figure 7
has been restricted to the two sets of 1448 C
atoms because displaying both sets of 11.510 heavy atoms would have been much more confusing. Even the display of the two sets of 1448 pairs of C
atoms in
Figure 7
remains confusing. This is a general problem: the display of optimally-superposed macromolecules produced overloaded images, and viewing the result remained confusing even with interactive tools on
the screen of a workstation. Thus, comparing the contents of
Figure 6
Figure 7
shows another reason why our approach based on interfaces is useful: it produces lightened images of optimally-superposed structural data, which would have been cumbersome to generate with existing
visualization tools.
3. Methods
3.1. Computation of Interfaces within Macromolecular Complexes
The full procedure to compute a SDD between two macromolecular complexes relies on the computation of a protein–protein interface within each complex. According to [
], there are three main families of methods to compute interfaces:
• The cutoff method.
• The loss of accessible surface area upon binding.
• The Voronoi tessellation method.
Machine learning methods [
] are not considered here because they are irrelevant in our context. The cutoff method requires an arbitrary cutoff distance value, which may have a strong impact on the resulting computed interface
]. Computing accessible surface areas requires fixing the values of atomic radii. Significantly different sets of radii values are found in the literature, depending on how they are defined [
]. It was shown that these radii have a considerable impact on surface areas [
]. The existence of parameters external to the input data and having a strong numerical impact on the results are drawbacks of these two families of methods. Thus, a non-parametric method is
desirable. The third method, based on the Voronoi tessellation, in its original variant was parameter-free [
]. It was implemented in PROVAT software [
]. The full mathematical description of the Voronoi tessellation can be found in [
]. It is out of the scope of this paper. To summarize, each atom lies inside a convex polyhedral cell having its polygonal faces located at mid-distance from its neighboring atoms. Thus, two atoms
are neighbors if their Voronoi cells share a common face. However, computing interfaces with the Voronoi method generates large cells, which induce the existence of meaningless long distances between
neighboring atoms.
This is why we needed to compute interfaces with another free parameter method: we retained the PPIC software. PPIC is publicly available with its documentation on a repository located at
. The method implemented in PPIC was introduced only in a very recent preprint [
]. It extends the approach of [
] used to compute the interfaces in protein–ligand complexes. Its input is the tridimensional structure of a protein–protein complex or of a protein–ligand complex. This complex has two parts
(molecule or macromolecule), named A and B. The algorithm has two steps:
• Generate the first part of the interface, constituted by the non-redundant set of all nearest neighbors of the atoms of A among the atoms of B.
• Generate the second part of the interface, constituted by the non-redundant set of all nearest neighbors of the atoms of B among the atoms of A.
Thus, the interface has two parts, i.e., two half interfaces, one in A and one in B. The roles of A and B are symmetric in the algorithm. From [
], it is known that each of these two parts is a subset of the half interface that would be computed by the Voronoi tessellation method. A nice consequence is that we do not observe anymore
meaningless long distance pairs of neighboring atoms (i.e., one atom in A and one atom in B).
As a by-product, PPIC outputs the list of the RNNs (Reciprocal Nearest Neighbors). It is recalled that, in the Euclidean space, the nearest neighbor y of some point x is not always such that the
nearest neighbor of y is x, e.g., consider three aligned points with abscissas $x = 0$, $y = 2$, and $z = 3$, for which the nearest neighbor of x is y, while the nearest neighbor of y is z, not x.
When all atoms (or all heavy atoms) of the complex are used as inputs in the calculation of the interface, the pairs of atoms defined by the list of the RNN are a rough estimate of the location of
potential interacting atom pairs. Moreover, to estimate the location of PPIs (Protein–Protein Interactions), looking at interacting atom pairs among the nearest neighbors makes more sense than
looking at interacting atom pairs at farther distances than the nearest neighbors.
PPIC was shown to be effective to compute interfaces on a database of 1050 protein homo- and hetero-dimers ([
]; data from [
]). It was used to compute HIV2 protease–ligand interfaces ([
]; data from [
]). For both datasets, it was observed that the best agreement between the interface calculations produced on heavy atoms by PPIC and those produced by the cutoff method occurred for a cutoff near
3.6–3.7 Å (see
Figure 1
in [
]). This is in agreement with the maximal donor–acceptor distance established in [
] at 3.9 Å. It is significantly smaller than the 4.5 Å cutoff distance between non-H atoms used by other authors [
3.2. Computation of Interfaces in Macromolecular Polymers
For a macromolecule containing two partners, A and B, the interface is constituted by two sets of atoms, one in A and one in B. In the case of a complex containing k polypeptidic chains, there are $(
k ( k − 1 ) ) / 2$ interfaces to compute. This is the case of HA, for which $k = 3$: there are three subsets, A, B, and C. Since the atoms of all chains are in the same macromolecular unit, we
defined the global interface as the intersection of these $( k ( k − 1 ) ) / 2$ interfaces as the non-redundant list of atoms was extracted. Therefore, in the case of a macromolecular polymer, the
global interface is constituted by only one list of atoms, not two. This is a difference from the case of macromolecular complexes.
3.3. Evaluation of the Dissimilarity between Two Interfaces
An interface is constituted by one or two sets of atoms. In the case of protein–ligand complexes containing several ligands, the interface may contain more than two sets of atoms. Considering two
interfaces containing one set of atoms each, their dissimilarity can be evaluated by their SDD [
When each of the two interface contains
$K ≥ 1$
, and when these two
-tuples of sets are pairwise associated, there are
SDDs to compute. A distance between these two
-tuples can be defined to be the sum of these
SDDs. It can be checked from the definition in
Appendix A
that it defines indeed a distance on the set of these
When the two
-tuples of sets are not pairwise associated, which in fact means that we consider unordered
-tuples, a distance between these two
-tuples can be defined to be the minimal value of the sum of the
SDDs, this minimum being taken over all
$K !$
possible pairwise correspondences between the two
-tuples. It can be checked from the definition in
Appendix A
that this defines indeed a distance on the set of such
3.4. Comparison with Other Dissimilarity Measures
Several other structural dissimilarity criteria are encountered in the literature, which are qualified as metrics [
] (see also [
]), such as hydrogen bonds, distance from surface, the number of residues, or the number of heavy or polar atoms, or the number of waters in the vicinity of a specific region, RMSF (Root Mean Square
Fluctuation), SASA (Solvent Accessible Surface Area), and gyration radius.
We outline here an ambiguity about the word “metric”: in many papers, it is used to name a structural dissimilarity criterion, which does not have the mathematical meaning mentioned in
Appendix A
. Even the dissimilarity criteria based on surfaces (e.g., SASA) cannot lead to defining a metric, except in the case of convex sets [
]. As mentioned at the end of
Section 1
, the SDD can be qualified as a metric [
]. The RMSD can be qualified as a metric when it is viewed as a distance induced by the Frobenius matrix norm, and this is why we compared the SDD with the RMSD. This happens in structural biology
when we consider two matrices
$A 1$
$A 2$
lines and three columns, each matrix containing the spatial coordinates of a set of
points (the two sets of
points are thus pairwise associated). Setting
$A = A 1 − A 2$
$n RMSD = t r a c e ( A t A )$
, where
$A t$
is the transpose of
3.5. Steps of the Methodology
The steps of the methodology to evaluate the evolution of interfaces along MD trajectories are summarized below. We assumed that the macromolecule of interest contained at least two partners, such as
a protein and a ligand, two proteins or two protomers, etc.
• Generate the frames of the MD simulation.
• For each frame, generate the global interface with the procedure described in
Section 3.2
• For each couple of successive frames, evaluate the dissimilarity between the interfaces with the SDD, as described in
Section 3.3
• Follow the evolution of the interface along the trajectory using the SDD as a coordinate varying as a function of the time.
4. Conclusions
Our approach to evaluate structural dissimilarities between frames issued from MD calculations has several advantages over the traditional ones involving either RMSD calculations or interface
• It is parameter free.
• No spatial alignment is needed, thus no non-trivial numerical solver is needed.
• The problem of molecular graph symmetries occurring in some contexts for residues Val, Leu, Arg, Phe, Tyr, Glu, and Asp, which is almost always neglected when computing RMSD values, does not
exist in our approach.
• All the steps of our algorithm can be coded by a beginner in programming.
• The dissimilarity between interfaces is measured with a distance (see
Appendix A
• Unwanted contributions of meaningless parts of macromolecules can be discarded (e.g., disordered parts in macromolecules, etc.).
• Images of optimal superpositions of full macromolecules are too overloaded compared to those of optimal superpositions of interfaces.
The software PPIC implementing our non-parametric algorithm of interfaces calculations is publicly available for free at
. The user can optionally run the cutoff method or the Voronoi tessellation method.
As an example, we presented results about two MD simulations of influenza HA (PDB Code 1RU7, 1448 residues). Knowing from MD simulations at pH = 7 and pH = 5 that HA is stable, the magnitude of
numerical values that we observed can serve as a first reference basis to discuss the results of MD simulations of other macromolecules or complexes. Our approach is neither claimed to overcome the
previous ones, nor is it devoted to replacing them: it is just an additional tool devoted to helping structural biologists, which is simple to use and which can be easily reprogrammed by beginners.
Author Contributions
Methodology, M.P.; software, M.P.; computations, M.P. and V.O.; validation, A.P. and A.V.; writing, original draft preparation, M.P.; writing, and editing, A.P., A.V., and V.O.; supervision and
project administration, A.P. and A.V.
This research received no external funding.
We are grateful to the reviewer who brought to us references [
Conflicts of Interest
The authors declare no conflict of interest.
The following abbreviations are used in this manuscript:
HA Hemagglutinin
MD Molecular Dynamics
PDB Protein Data Bank
PPI Protein–Protein Interaction
RMSD Root Mean Squared Deviation
RMSF Root Mean Squared Fluctuation
RNN Reciprocal Nearest Neighbors
SASA Solvent Accessible Surface Area
SDD Symmetric Difference Distance
Appendix A. Definition and Properties of Distances
For convenience, we recall the definition of a distance (see any textbook on analysis or topology).
Definition A1.
A metric d (or distance function) on a set E is a function from the Cartesian product $E × E$ on $R$ satisfying the following four conditions:
$d ( x , y ) = d ( y , x )$(symmetry)
$d ( x , x ) = 0 ∀ x ∈ E$
$d ( x , y ) = 0 ⇒ x = y ∀ ( x ∈ E , y ∈ E )$
$d ( x , y ) ≤ d ( x , z ) + d ( z , y ) ∀ ( x ∈ E , y ∈ E , z ∈ E )$(triangle inequality)
Setting $y = x$ in the triangle inequality and using the symmetry condition, we deduce a property that is flagged as an additional condition in some books: $d ( x , y ) ≥ 0 ∀ ( x ∈ E , y ∈ E )$.
A value taken by the distance function is called a distance. As long as no confusion can be made between the function and the value it takes, a distance function can be itself called a distance.
To understand why all four conditions of Definition A1 are of practical importance, imagine that one of them is missing and try to interpret numerical results from experiments:
• Removing the symmetry condition would mean that there exist two elements x and y such that $d ( x , y ) ≠ d ( y , x )$: understanding this result may be difficult.
• Removing the condition $d ( x , x ) = 0$ would mean that there exists an element x such that $d ( x , x ) > 0$: what should one think about such an element?
• Many authors define dissimilarities between objects, although the third condition does not stand: nothing can be deduced when a null distance $d ( x , y ) = 0$ is observed between two distinct
element x and y, a really embarrassing situation.
• The triangle inequality is useful, as well: it would be difficult to understand a situation where three distinct elements x, y, and z would be such that $d ( x , y ) > d ( x , z ) + d ( z , y )$.
This is why, when possible, working with the distance appears to be a better choice than working with some other dissimilarity criterion.
Appendix B. Interfaces Residues of the Mean Frames
It is recalled that, for each mean frame (one at pH = 7 and one at pH = 5), we computed the three interfaces between the three HA protomers, then we defined the final list of interface residues as
the non-redundant set of residues involved in the union of these three interfaces.
The 224 residues at the interface of the mean frame for pH = 7 are:
V19, L20, E100, E103, K159, K161, S184, E194, N195, S199, V201, S203, N204, N206, R207, R208, T210, E212, I213, A214, E215, R216, P217, L232, K234, G236, T238, I240, E242; G324, L325, F326, G327,
G331, F332, K362, S363, N366, G370, N373, K374, S377, K381, N383, Q385, K395, L396, K398, R399, M400, N402, L403, N404, K406, V407, G410, F411, D413, I414, W415, Y417, N418, L421, L422, L425, E426,
R429, F433, S436, N440, E443, K444, S447, K450, N451, E455, I456, G457, G478, P483;
V502, L503, E504, D577, E583, E586, K642, N678, S682, V684, S686, N687, Y688, N689, R690, R691, E695, I696, A697, E698, R699, P700; G807, L808, F809, G810, G814, F815, K845, N849, G853, N856, K857,
S860, K864, M865, N866, K878, L879, K881, R882, M883, N885, L886, N887, K889, V890, D892, G893, F894, D896, I897, Y900, L904, L905, L908, E909, R912, F916, S919, N923, E926, K927, K933, E938, G940,
V985, L986, D1060, E1066, E1069, N1161, S1165, V1167, S1169, N1170, Y1171, N1172, R1173, R1174, T1176, E1178, A1180, E1181, R1182, P1183, K1200, F1226; G1290, L1291, F1292, G1293, A1296, G1297,
K1328, N1332, G1336, N1339, K1340, S1343, K1347, M1348, N1349, F1352, N1360, K1361, L1362, E1363, K1364, R1365, M1366, N1368, L1369, N1370, K1372, V1373, D1375, G1376, F1377, D1379, I1380, Y1383,
N1384, L1387, L1388, L1391, E1392, R1395, F1399, S1402, N1406, E1409, K1410, S1413, K1416, N1417, K1420, E1421, I1422, G1423, G1444, D1447, Y1448.
The 214 residues at the interface of the mean frame for pH = 5 are:
V19, L20, G93, E100, N195, S199, V201, S203, N204, N206, R207, R208, T210, E212, I213, A214, E215, R216, P217, K234, E242; L325, F326, G331, F332, N366, G370, N373, K374, S377, K381, M382, N383,
F386, A388, K391, K395, L396, K398, R399, M400, N402, L403, N404, K406, V407, D409, G410, F411, D413, I414, W415, Y417, N418, L421, L422, L425, R429, F433, S436, N437, N440, E443, K444, S447, K450,
N451, E455, G457, N458;
V502, L503, E583, E586, N678, Y680, S682, V684, S686, N687, Y688, N689, R690, R691, T693, E695, I696, A697, E698, R699, P700, K717, E725, F743; G807, L808, F809, G810, A813, G814, F815, N849, G853,
N856, S860, V861, K864, M865, N866, Q868, K874, K878, L879, K881, R882, M883, N885, L886, N887, K889, V890, G893, F894, D896, I897, W898, Y900, N901, L904, L905, L908, R912, F916, S919, N923, E926,
K927, S930, K933, N934, E938, I939, G940, Y965, P966;
V985, L986, D1060, E1066, E1069, K1125, E1160, N1161, Y1163, S1165, V1167, S1169, N1170, Y1171, N1172, R1173, R1174, F1175, T1176, E1178, I1179, A1180, E1181, R1182, P1183; G1290, L1291, F1292,
F1298, K1328, N1332, G1336, S1343, V1344, K1347, M1348, N1349, K1361, L1362, K1364, R1365, M1366, N1368, L1369, N1370, K1372, V1373, G1376, F1377, D1379, I1380, Y1383, N1384, L1387, L1391, R1395,
F1399, S1402, N1406, E1409, K1410, S1413, K1416, N1417, K1420, E1421, I1422, G1423, Y1448.
Figure 1.
The tridimensional structure of the HA homotrimer (data found in the Protein Data Bank (PDB), Code 1RU7, taken from [
]). Each of the three protomers contains a subunit HA1 (in cyan in protomer 1) and a subunit HA2 (in green in protomer 1). HA1 domains are on the left. HA2 α-helices are on the right. The image was
generated with PyMOL [
Figure 2.
The 40 SDDs, expressed as the number of C
atoms, between the initial conformation at time
$t 0$
and each of the 40 generated conformations by MD simulation, at pH = 7 (in blue) and at pH = 5 (in red). Time steps are in the abscissas. The image was generated with R [
Figure 3.
The 40 RMSD (in Å ), taken over all heavy atoms, between the initial conformation at time
$t 0$
and each of the 40 generated conformations by MD simulation, at pH = 7 (in blue) and at pH = 5 (in red). Time steps are in the abscissas. The image was generated with R [
Figure 4.
The SDDs, expressed as the number of C
atoms, between the 40 successive pairs of conformations generated by MD simulation, at pH = 7 (in blue) and at pH = 5 (in red). Time steps are in the abscissas. The image was generated with R [
Figure 5.
The RMSD values ( Å ), taken over all heavy atoms, between the 40 successive pairs of conformations generated by MD simulation, at pH = 7 (in blue) and at pH = 5 (in red). Time steps are in the
abscissas. The image was generated with R [
Figure 6.
The optimal superposition of the interfaces of the mean frames, respectively at
$t 30$
for pH = 7 (224 C
atoms, in blue) and at
$t 15$
for pH = 5 (214 C
atoms, in red). The image was generated with PyMOL [
Figure 7.
The optimal superposition computed for all heavy atoms of the mean frames, respectively at
$t 30$
for pH = 7 (224 C
atoms, in blue) and at
$t 15$
for pH = 5 (214 C
atoms, in red). Only C
atoms are displayed. Each straight line separates two successive C
atoms in the backbone. The image was generated with PyMOL [
© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http:/
Share and Cite
MDPI and ACS Style
Ozeel, V.; Perrier, A.; Vanet, A.; Petitjean, M. The Symmetric Difference Distance: A New Way to Evaluate the Evolution of Interfaces along Molecular Dynamics Trajectories; Application to Influenza
Hemagglutinin. Symmetry 2019, 11, 662. https://doi.org/10.3390/sym11050662
AMA Style
Ozeel V, Perrier A, Vanet A, Petitjean M. The Symmetric Difference Distance: A New Way to Evaluate the Evolution of Interfaces along Molecular Dynamics Trajectories; Application to Influenza
Hemagglutinin. Symmetry. 2019; 11(5):662. https://doi.org/10.3390/sym11050662
Chicago/Turabian Style
Ozeel, Valentin, Aurélie Perrier, Anne Vanet, and Michel Petitjean. 2019. "The Symmetric Difference Distance: A New Way to Evaluate the Evolution of Interfaces along Molecular Dynamics Trajectories;
Application to Influenza Hemagglutinin" Symmetry 11, no. 5: 662. https://doi.org/10.3390/sym11050662
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details
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Cosecant: Introduction to the trigonometric functions (subsection Trigonometrics/05)
The best-known properties and formulas for trigonometric functions
Real values for real arguments
For real values of argument , the values of all the trigonometric functions are real (or infinity).
In the points , the values of trigonometric functions are algebraic. In several cases they can even be rational numbers or integers (like or ). The values of trigonometric functions can be expressed
using only square roots if and is a product of a power of 2 and distinct Fermat primes {3, 5, 17, 257, …}.
Simple values at zero
All trigonometric functions have rather simple values for arguments and :
All trigonometric functions are defined for all complex values of , and they are analytical functions of over the whole complex ‐plane and do not have branch cuts or branch points. The two functions
and are entire functions with an essential singular point at . All other trigonometric functions are meromorphic functions with simple poles at points for and , and at points for and .
All trigonometric functions are periodic functions with a real period ( or ):
Parity and symmetry
All trigonometric functions have parity (either odd or even) and mirror symmetry:
Simple representations of derivatives
The derivatives of all trigonometric functions have simple representations that can be expressed through other trigonometric functions:
Simple differential equations
The solutions of the simplest second‐order linear ordinary differential equation with constant coefficients can be represented through and :
All six trigonometric functions satisfy first-order nonlinear differential equations: | {"url":"https://functions.wolfram.com/ElementaryFunctions/Csc/introductions/Trigonometrics/05/ShowAll.html","timestamp":"2024-11-15T03:10:59Z","content_type":"text/html","content_length":"41626","record_id":"<urn:uuid:1a4e5fbc-ecea-472f-b08b-69b2ddc45224>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00604.warc.gz"} |
Chapter 1 Rational Numbers
Chapter 1 Rational Numbers
Extra Questions Very Short Answer Type
Question 1.
Question 4.
Question 5.
What is the multiplicative identity of rational numbers?
1 is the multiplicating identity of rational numbers.
Question 6.
What is the additive identity of rational numbers?
0 is the additive identity of rational numbers.
Short Answer Type
Question 13.
Calculate the following:
Question 14.
Represent the following rational numbers on number lines.
Question 16.
Show that:
Question 17.
Question 20.
Let O, P and Z represent the numbers 0, 3 and -5 respectively on the number line. Points Q, R and S are between O and P such that OQ = QR = RS = SP. (NCERT Exemplar)
What are the rational numbers represented by the points Q, R and S. Next choose a point T between Z and 0 so that ZT = TO. Which rational number does T represent?
As OQ = QR = RS = SP and OQ + QR + RS + SP = OP
therefore Q, R and S divide OP into four equal parts.
(i) a + (b + c) = (a + b) + c (Associative property of addition)
(ii) a × (b × c) – (a × b) × c (Associative property of multiplication)
Higher Order Thinking Skills (HOTS)
Question 22.
Question 23.
Question 24.
Fill in the blanks:
(a) Numbers of rational numbers between two rational numbers is ………. | {"url":"https://rajboardexam.in/chapter-1-rational-numbers/","timestamp":"2024-11-09T06:25:33Z","content_type":"text/html","content_length":"112488","record_id":"<urn:uuid:8039e2a5-117a-40a6-9e4d-85a365357429>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00846.warc.gz"} |
Interval Notation - Definition, Examples, Types of Intervals - Grade Potential Irvine, CA
Interval Notation - Definition, Examples, Types of Intervals
Interval Notation - Definition, Examples, Types of Intervals
Interval notation is a fundamental topic that students should learn due to the fact that it becomes more essential as you progress to more complex arithmetic.
If you see advances arithmetics, such as differential calculus and integral, in front of you, then being knowledgeable of interval notation can save you hours in understanding these ideas.
This article will discuss what interval notation is, what it’s used for, and how you can interpret it.
What Is Interval Notation?
The interval notation is simply a method to express a subset of all real numbers through the number line.
An interval refers to the values between two other numbers at any point in the number line, from -∞ to +∞. (The symbol ∞ signifies infinity.)
Basic difficulties you encounter mainly consists of one positive or negative numbers, so it can be difficult to see the benefit of the interval notation from such simple applications.
Though, intervals are usually used to denote domains and ranges of functions in higher math. Expressing these intervals can increasingly become difficult as the functions become progressively more
Let’s take a straightforward compound inequality notation as an example.
• x is greater than negative four but less than two
As we know, this inequality notation can be written as: {x | -4 < x < 2} in set builder notation. Despite that, it can also be denoted with interval notation (-4, 2), signified by values a and b
segregated by a comma.
As we can see, interval notation is a way to write intervals elegantly and concisely, using fixed rules that make writing and understanding intervals on the number line simpler.
The following sections will tell us more about the principles of expressing a subset in a set of all real numbers with interval notation.
Types of Intervals
Several types of intervals place the base for denoting the interval notation. These interval types are important to get to know because they underpin the entire notation process.
Open intervals are applied when the expression does not include the endpoints of the interval. The previous notation is a good example of this.
The inequality notation {x | -4 < x < 2} express x as being more than -4 but less than 2, which means that it excludes neither of the two numbers referred to. As such, this is an open interval
expressed with parentheses or a round bracket, such as the following.
(-4, 2)
This implies that in a given set of real numbers, such as the interval between negative four and two, those two values are not included.
On the number line, an unshaded circle denotes an open value.
A closed interval is the contrary of the previous type of interval. Where the open interval does exclude the values mentioned, a closed interval does. In word form, a closed interval is written as
any value “higher than or equal to” or “less than or equal to.”
For example, if the last example was a closed interval, it would read, “x is greater than or equal to -4 and less than or equal to two.”
In an inequality notation, this would be expressed as {x | -4 < x < 2}.
In an interval notation, this is expressed with brackets, or [-4, 2]. This states that the interval contains those two boundary values: -4 and 2.
On the number line, a shaded circle is used to represent an included open value.
A half-open interval is a combination of prior types of intervals. Of the two points on the line, one is included, and the other isn’t.
Using the last example as a guide, if the interval were half-open, it would be expressed as “x is greater than or equal to -4 and less than two.” This states that x could be the value -4 but cannot
possibly be equal to the value two.
In an inequality notation, this would be expressed as {x | -4 < x < 2}.
A half-open interval notation is written with both a bracket and a parenthesis, or [-4, 2).
On the number line, the shaded circle denotes the number present in the interval, and the unshaded circle indicates the value excluded from the subset.
Symbols for Interval Notation and Types of Intervals
To summarize, there are different types of interval notations; open, closed, and half-open. An open interval doesn’t include the endpoints on the real number line, while a closed interval does. A
half-open interval includes one value on the line but excludes the other value.
As seen in the last example, there are various symbols for these types under the interval notation.
These symbols build the actual interval notation you develop when stating points on a number line.
• ( ): The parentheses are used when the interval is open, or when the two endpoints on the number line are excluded from the subset.
• [ ]: The square brackets are used when the interval is closed, or when the two points on the number line are not excluded in the subset of real numbers.
• ( ]: Both the parenthesis and the square bracket are utilized when the interval is half-open, or when only the left endpoint is not included in the set, and the right endpoint is not excluded.
Also known as a left open interval.
• [ ): This is also a half-open notation when there are both included and excluded values within the two. In this instance, the left endpoint is not excluded in the set, while the right endpoint is
not included. This is also known as a right-open interval.
Number Line Representations for the Various Interval Types
Aside from being written with symbols, the various interval types can also be described in the number line utilizing both shaded and open circles, relying on the interval type.
The table below will display all the different types of intervals as they are represented in the number line.
Practice Examples for Interval Notation
Now that you’ve understood everything you need to know about writing things in interval notations, you’re prepared for a few practice problems and their accompanying solution set.
Example 1
Convert the following inequality into an interval notation: {x | -6 < x < 9}
This sample problem is a easy conversion; simply utilize the equivalent symbols when writing the inequality into an interval notation.
In this inequality, the a-value (-6) is an open interval, while the b value (9) is a closed one. Thus, it’s going to be written as (-6, 9].
Example 2
For a school to take part in a debate competition, they require minimum of three teams. Represent this equation in interval notation.
In this word question, let x be the minimum number of teams.
Since the number of teams required is “three and above,” the value 3 is included on the set, which states that 3 is a closed value.
Additionally, because no upper limit was stated with concern to the number of maximum teams a school can send to the debate competition, this value should be positive to infinity.
Therefore, the interval notation should be written as [3, ∞).
These types of intervals, where there is one side of the interval that stretches to either positive or negative infinity, are also known as unbounded intervals.
Example 3
A friend wants to do a diet program constraining their regular calorie intake. For the diet to be a success, they must have at least 1800 calories regularly, but maximum intake restricted to 2000.
How do you write this range in interval notation?
In this question, the number 1800 is the minimum while the value 2000 is the maximum value.
The problem suggest that both 1800 and 2000 are inclusive in the range, so the equation is a close interval, written with the inequality 1800 ≤ x ≤ 2000.
Therefore, the interval notation is denoted as [1800, 2000].
When the subset of real numbers is confined to a range between two values, and doesn’t stretch to either positive or negative infinity, it is also known as a bounded interval.
Interval Notation Frequently Asked Questions
How To Graph an Interval Notation?
An interval notation is simply a way of representing inequalities on the number line.
There are rules to writing an interval notation to the number line: a closed interval is written with a shaded circle, and an open integral is written with an unfilled circle. This way, you can
promptly check the number line if the point is included or excluded from the interval.
How To Convert Inequality to Interval Notation?
An interval notation is basically a diverse technique of expressing an inequality or a set of real numbers.
If x is greater than or less a value (not equal to), then the number should be stated with parentheses () in the notation.
If x is higher than or equal to, or lower than or equal to, then the interval is denoted with closed brackets [ ] in the notation. See the examples of interval notation prior to see how these symbols
are employed.
How Do You Exclude Numbers in Interval Notation?
Numbers excluded from the interval can be stated with parenthesis in the notation. A parenthesis implies that you’re expressing an open interval, which means that the number is ruled out from the
Grade Potential Can Assist You Get a Grip on Arithmetics
Writing interval notations can get complicated fast. There are multiple difficult topics within this concentration, such as those dealing with the union of intervals, fractions, absolute value
equations, inequalities with an upper bound, and many more.
If you want to conquer these concepts quickly, you are required to review them with the expert help and study materials that the professional tutors of Grade Potential provide.
Unlock your arithmetics skills with Grade Potential. Connect with us now! | {"url":"https://www.irvineinhometutors.com/blog/interval-notation-definition-examples-types-of-intervals","timestamp":"2024-11-07T04:40:52Z","content_type":"text/html","content_length":"112195","record_id":"<urn:uuid:c9f34520-8b10-4cd7-a859-f197f8d7846c>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00673.warc.gz"} |
Optimality of the Powers of 3 Strategy in Weighing Scale Puzzles
Written on
Chapter 1: Introduction to the Weighing Scale Puzzle
In the earlier discussion, we examined Hemanth’s method for solving the weighing scale puzzle and established its general applicability. To summarize: it is indeed possible to measure any positive
integer weight using only weights that are powers of 3.
Now, the question arises: is this "powers of 3" approach the most efficient? Specifically, can we demonstrate that no alternative strategy requires fewer weights?
As pointed out by Hemanth, a set of n known weights generates 3^n combinations since each weight can be placed in three positions: on the left pan, on the right pan, or not used at all.
Among these combinations, one will always yield a weight of 0. Consequently, we are left with 3^n - 1 "interesting" combinations. However, it is important to note that half of these can be negative
(if excess weight is placed on the right pan).
Chapter 2: Proving the Optimality
The combinations produced by weights are not necessarily unique. For example, having weights of 2 kg, 3 kg, and 5 kg allows you to achieve the same result by either placing the 5 kg weight on one
side or by using the 2 kg and 3 kg weights together.
This leads to the hypothesis that a strategy which can weigh every positive integer weight and produces solely unique results must be optimal. In previous discussions, including Hemanth’s original
post, we confirmed that the powers of 3 can weigh any object—now we must verify if they yield only unique results.
The answer is affirmative, and the proof follows directly from the inductive reasoning previously utilized. Here’s how it unfolds.
Starting with the first k powers of 3, the smallest combination achievable is 1 (3^0). Our inductive reasoning has already established that we can generate every integer from 1 up to the maximum
combined weight possible, which is...
This leads to a specific list of weights we can measure:
Thus, we have precisely...
Furthermore, we know that the first k powers of 3 can create...
If each combination corresponds uniquely to one of the possible weights, we can conclude that the powers of 3 strategy is indeed optimal.
To illustrate, if...
Then the powers of 3 strategy is optimal. Conversely, if...
This indicates that at least one number from our list can be formed through multiple combinations.
Let’s employ mathematical induction to further solidify our claim.
Proof by Mathematical Induction
We aim to prove the statement:
for any k > 0.
Base Case
In our base case, we need to demonstrate that P(1) holds true.
Indeed, P(1) is valid.
Inductive Step
Assuming P(n) is true, we must show P(n + 1) also holds.
Assuming P(n), we know...
We need to demonstrate that...
By our assumption, we find...
Combining these terms into a single fraction yields:
Thus, we have established that given...
It follows that...
This confirms our inductive step is complete!
Consequently, our proof asserts that...
holds true for any k > 0.
Engaging Video Resources
To further enhance your understanding of these concepts, here are two videos:
The first video titled "Can you solve the classic 12 marbles riddle? (Detailed Solution Video)" elaborates on similar puzzle-solving techniques.
The second video, "Pan balance with a 'double scales' - puzzle problems for 4th grade and up," provides insights into practical applications of weighing puzzles. | {"url":"https://whalebeings.com/optimality-powers-of-3-strategy-weighing-scale.html","timestamp":"2024-11-08T15:03:54Z","content_type":"text/html","content_length":"14890","record_id":"<urn:uuid:ff9c3206-620b-4687-b5a6-944d47376615>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00803.warc.gz"} |
Homework Help
Where To Get Proper Math Homework Help: Best Sources
If you need to complete some math homework then you may be wondering if there any useful sources that can help you whilst doing the work. Essentially, you will have a range of different options,
depending upon the requirements of the work that you need to do. For example, if you are looking for trigonometry answers, then the solutions available to you may differ compared to if you are
looking for algebra help. However, some solutions that are outlined below will be useful for most, if not all math problems that you might be having.
Looking for math problem solvers
It is possible to find a range of different software the can be used in order to solve any problems that you might be having. For example, you can find a wide range of different apps that you can
download to your phone, tablet or a range of other devices, which you can then use to provide you with a range of different answers. Generally, however, these apps will be specific to a certain
branch of mathematics, such as trigonometry. Therefore, you might need to find relevant apps for each one of the different branches of math that you are studying, unless of course you only need
something for some quick and easy solutions in a one-off situation.
Hiring tutors
An alternative, particularly if you require help on a long-term basis, is to hire tutors who can teach you about the subject in regular lessons. In fact, it is possible to hire tutors on a one-off
basis; however, you will probably find it more helpful if you have regular lessons.
One of the benefits of hiring tutors is that you can find a range of individuals who offer their services online. Some of these individuals may be able to come and meet you, and will simply advertise
on the Internet; however, many of them will offer services through the Internet, using conferencing software, which means that they can be in any part of the world. One of the benefits of this is
that it will be possible to hire tutors who charge relatively low rates, due to the fact that they live in countries with relatively low costs of living.
Asking questions online
Finally, for simple solutions, you may consider the possibility of answering questions on generic Q&A websites, as well as a range of different forums, or even social media groups. | {"url":"https://www.rolysenglishfudge.com/solutions-on-where-to-find-good-homework-help-with-math","timestamp":"2024-11-14T13:52:03Z","content_type":"application/xhtml+xml","content_length":"13436","record_id":"<urn:uuid:fab87074-58f1-4068-a8ac-b3f38279594e>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00556.warc.gz"} |
syntax and example of median function in oracle sql
SQL Median Function returns the median of available values. Oracle Median Function accepts column and formula as parameter for median calculation.
SQL MEDIAN Function Syntax
SELECT MEDIAN(numeric column / formula)
FROM table_name
WHERE conditions;
SQL MEDIAN Function Examples
Suppose we have a table named “Employee” with the data as shown below.
│Employee_Id│Employee_Name │Salary│Department│Commission│
│101 │Emp A │10000 │Sales │10 │
│102 │Emp B │20000 │IT │20 │
│103 │Emp C │28000 │IT │20 │
│104 │Emp D │30000 │Support │ │
│105 │Emp E │32000 │Sales │10 │
We will see the usage of Oracle MEDIAN Function below.
SQL MEDIAN Function – Simple Usage
The simplest use of Oracle MEDIAN Function would be to calculate MEDIAN of a column.
For example: Oracle MEDIAN query returns the median of salaries from employee table.
SELECT MEDIAN(salary) Median_Salary
FROM employee;
Above mentioned Oracle MEDIAN query returned ‘28000’ as median of salaries.
Note: We have aliased MEDIAN(Salary) as Median_Salary.
SQL MEDIAN Function – Using Formula Example
Oracle MEDIAN Function also accepts formula as parameter.
For example: Oracle MEDIAN query returns median of salaries*(commission/100) from employee table as shown below
SELECT MEDIAN(salary*(commission/100)) Median_Salary_Comm
FROM employee;
Above mentioned SQL MEDIAN query returned ‘3600’ as median of salary*(commission/100).
Note: We have aliased MEDIAN(salary*(commission/100)) as Median_Salary_Comm | {"url":"https://techhoney.com/tag/syntax-and-example-of-median-function-in-oracle-sql/","timestamp":"2024-11-08T18:52:27Z","content_type":"text/html","content_length":"36500","record_id":"<urn:uuid:62564509-3f5f-47ed-bfdb-ef85685aa7f4>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00073.warc.gz"} |
Free Vibrations Analysis of Timoshenko Beams on Different Elastic Foundations via Three Alternative Models
Tonzani, Giulio Maria
Free Vibrations Analysis of Timoshenko Beams on Different Elastic Foundations via Three Alternative Models.
[Laurea magistrale], Università di Bologna, Corso di Studio in
Civil engineering [LM-DM270]
, Documento full-text non disponibile
Il full-text non è disponibile per scelta dell'autore. (
Contatta l'autore
The scope of the research is to provide a simpler and more consistent equation for the analysis of the natural frequencies of a beam with respect to the widely used one introduced by Timoshenko in
1916. To this purpose, the free vibrations of a beam resting on Winkler or/and Pasternak elastic foundations are analyzed via original Timoshenko theory as well as two of its truncated versions,
which have been proposed by Elishakoff in recent years to overcome the mathematical difficulties associated with the fourth-order time derivative of the deflection. Former equation takes into account
for both shear deformability and rotary inertia, while latter one is based upon incorporation of the slope inertia effect. Detailed comparisons and derivations of the three models are given for six
different sets of boundary conditions stemming by the various possible combinations of three of the most typical end constraints for a beam: simply supported end, clamped end and free end. It appears
that the two new theories are able to overcome the disadvantage of the original Timoshenko equation without predicting the unphysical second spectrum and to produce very good approximations for the
most relevant values of natural frequencies. As a consequence, the inclusion of these simpler approaches is suggested in future works. An intriguing intermingling phenomenon is also presented for the
simply supported case together with a detailed discussion about the possible existence of zero frequencies for the free–free beam and the simply supported–free beam in the context of different types
of foundations.
The scope of the research is to provide a simpler and more consistent equation for the analysis of the natural frequencies of a beam with respect to the widely used one introduced by Timoshenko in
1916. To this purpose, the free vibrations of a beam resting on Winkler or/and Pasternak elastic foundations are analyzed via original Timoshenko theory as well as two of its truncated versions,
which have been proposed by Elishakoff in recent years to overcome the mathematical difficulties associated with the fourth-order time derivative of the deflection. Former equation takes into account
for both shear deformability and rotary inertia, while latter one is based upon incorporation of the slope inertia effect. Detailed comparisons and derivations of the three models are given for six
different sets of boundary conditions stemming by the various possible combinations of three of the most typical end constraints for a beam: simply supported end, clamped end and free end. It appears
that the two new theories are able to overcome the disadvantage of the original Timoshenko equation without predicting the unphysical second spectrum and to produce very good approximations for the
most relevant values of natural frequencies. As a consequence, the inclusion of these simpler approaches is suggested in future works. An intriguing intermingling phenomenon is also presented for the
simply supported case together with a detailed discussion about the possible existence of zero frequencies for the free–free beam and the simply supported–free beam in the context of different types
of foundations.
Altri metadati | {"url":"https://amslaurea.unibo.it/12868/","timestamp":"2024-11-13T18:47:50Z","content_type":"application/xhtml+xml","content_length":"34245","record_id":"<urn:uuid:985e8e3c-388d-432a-a301-8e46dfae17b2>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00252.warc.gz"} |
Planetary Cycles Trading (EUR/USD)
Outlook for next week:
Red line= Harmonic box, spectrum study line adjusted to hit prior week's peak.
Blue line= Terra Incognita new module, Trading Spectrum calculation, pos r=88.9%, PQN=6.432
Green line= Moon-Sun, 1/5 orb, normal, 100% smooth orb.
Second chart shows how this model performed on last week's price action....
No chart for next week?
LOL - oftentimes, I just don't have anything to say so I rather spent time reading stuff that folks posted. As for Jerry, because I've been flowing him on youtube also. Thanks.
Here is a possible trend changing point for EUR/USD pair on 1H chart which will occur on 2020.10.26
Here is a possible trend changing point for EUR/USD pair on 4H chart which will occur on 2020.10.26
Outlook for next week: Red line= Harmonic box, spectrum study line adjusted to hit prior week's peak. Blue line= Terra Incognita new module, Trading Spectrum calculation, pos r=88.9%, PQN=6.432
Green line= Moon-Sun, 1/5 orb, normal, 100% smooth orb. Second chart shows how this model performed on last week's price action.... {image} {image}
Great, I really dont have any idea about how this week will be. My daily cycles are little bit confusing. However daily charts indicates a possible turning point on around 29 th October. I will post
it later
Hi friends, I did some cycle analysis on 15 min time frame. But still I am working on cycle lengths in 15 min time frame. But we will try to catch next hourly turning point in 15 min time frame. We
will see how it works.
Here is a possible trend changing point for EUR/USD pair on 1H chart which will occur on 2020.10.27
Below is the move on 15 min chart
Hello folks, previous move was very successful and top occurred just 2 15min bars before the predicted point. here is another move in 1H time frame spotted in 15 min time frame on 2020.10.28. We will
see how it works.
below is the turning point in 15 min time frame
Also my analysis indicates a possible change of trend in daily chart on 2020.10.29 (tomorrow). I include it too
I was looking for a good CIT on 27 Oct. 21:35 EST. on the simultaneous ingress of Merc. and Venus into LIBRA......
BUT nothing happened ........ At such occasions I feel somebody is pulling my leg and don't know whom to accuse.......
Happy Halloween tomorrow... and a rare Blue Moon on Halloween... the last time this occurred was 1944, 76 years ago.... This from Wikipedia... so what symmetry followed shortly after this event?
November 7, 1944 (Tuesday)
1. The United States presidential election was held. Incumbent Franklin D. Roosevelt was elected to an unprecedented fourth term, defeating Thomas E. Dewey 432 electoral votes to 99 and carrying 36
out of 48 states.
End of the week recap: 215 pips based on post 11,821, October 25th.
Happy Halloween tomorrow... and a rare Blue Moon on Halloween... the last time this occurred was 1944, 76 years ago.... This from Wikipedia... so what symmetry followed shortly after this event?
November 7, 1944 (Tuesday) The United States presidential election was held. Incumbent Franklin D. Roosevelt was elected to an unprecedented fourth term,...
Yup that 76 cycle is divided into 4 Metonic Cycles. We are "almost" at the same place as october 2001. Last Metonic cycle started april 1998, the current one april 2017. Each Metonic cycle consists
of 235 lunations. Compare price movement then and now. Don't follow price levels, only supply/demand moves. It's a fractal movement.
Compare full moon zodiac signs from october 1944, they are the same as current october.
Last move bottomed 2001/2002. I expect the same now 2020/2021. Sorry for polish language on my chart, its released to my group
On 1D chart (second image) Dollar is still the weakest currency...We should see some weaknes just after the Election, but then $ should gain momentum again into winter 2020/2021.
{quote} Yup that 76 cycle is divided into 4 Metonic Cycles. We are "almost" at the same place as october 2001. Last Metonic cycle started april 1998, the current one april 2017. Each Metonic
cycle consists of 235 lunations. Compare price movement then and now. Don't follow price levels, only supply/demand moves. It's fracatl Compare full moon zodiac signs from october 1944, they are
the same as current october. Last move bottomed 2001/2002. I expect the same now 2020/2021. Sorry for polish language on my chart, its released to my group
Thank you! I would love to see a translation of the Polish text on your chart... Sergey (creator of Timing Solution software) has recently made a video on calculating the Metonic Cycle....
New vs ancient metonic cycle...
Inserted Video
Focus on calculating ancient metonic cycle:
Inserted Video
Outlook for next week:
Trend down Through 20:30 Tues then sideways through 08:00 Friday then down to close. Light black line is price action from Feb 10, 2020 which matches the current price action best in the 10 month
data series I am working with.
All of the lines are Moon-Sun, 1/5 h, normal, not inverted and a range of smooth orbs from 200% down to 25%
It is now 13:00 EST ...... At 13:51 we have Jupiter conj. Saturn HELIO ......
It is now 13:00 EST ...... At 13:51 we have Jupiter conj. Saturn HELIO ......
Amazing accuracy.........Cannot believe it.....
{quote} Amazing accuracy.........Cannot believe it.....
Nice call!
Next CIT [imho...] Merc. DIRECT 12:49 EST. in few min. ......
{quote} Very nice looking Styngray... Good observation... "If one is trading to make money".... I see a 3 min chart and perhaps 5 pip Renko bars... How do you get in and out on each arrow... in
other words, are you using an EA? Perhaps you wait for a signal then set a TP at a PP?
Haha sorry a little late. did not see this. apologies. No EA. macro to set the plot, price action at the pivots or their confluence is a potential buy indicator at Midpoint 2 in all TFs and a
potencial sell at midpoint 3 in all TFs with M4 as a conservative target in the former and M1 as a conservative target in the latter. With renko the same idea except follow renko rules to trigger
entry. In the last pic...the buy points were at the confluence of the November monthly M2 (gray line ) and November first weekl M2 (blue line). They are superimposed on each other. In addition the
vertical red line is the full moon indicator. it also confluenced here.
Hello friends, couldnt reply to thread from a long time.
EUR/USD turning point on 1H chart which will occur on 2020.11.06 (with 2 most probable turning points on 15 min chart)
EUR/USD turning point on 4H chart which will occur on 2020.11.06
Hindsight clarity...
I was wrong again this week... I only relied on the Moon-Sun lines to pick the direction... as it turned out the price headed upward all week instead of down. I just ran a new study using 30 min EU
data from Oct. 30 and found that a straight Spectrum study and Q spectrum study had it right... In the attached image, I inverted the moon-sun line (thin green) to agree with the other spectrum
End of the week recap: -225 pips based on post 11,832, Oct 31. | {"url":"https://www.forexfactory.com/thread/108462-planetary-cycles-trading-eurusd?page=592","timestamp":"2024-11-11T21:36:55Z","content_type":"text/html","content_length":"149893","record_id":"<urn:uuid:757f5c9e-7db6-4d5c-a466-7a0f4529d062>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00752.warc.gz"} |
One-Dimensionally Polarization-Independent Retrodirective Metasurface
Article information
J. Electromagn. Eng. Sci. 2019;19(4):279-281
Received 2019 May 3; Revised 2019 July 4; Accepted 2019 August 8.
A one-dimensionally polarization-independent retrodirective metasurface (PIRMS) is proposed in this letter. The retrodirective metasurface (RMS) consists of ring structures on its ground plane to
achieve the polarization-independent characteristics. The reflection phases of unitcells of the metasurface can be controlled by the radius of the ring structure, and the dimension of the supercell
with the 2π phase variation depends on an incident angle and operation frequency from the generalized Snell’s law. To verify its feasibility, two kinds of RMSs to operate at two retrodirected angles
of 20° and 40° are designed and measured at 5.8 GHz. The measured retroreflected power efficiencies are more than 94% and 92% at the retrodirected angles of 20° and 40°, respectively, regardless of
the polarization. The results show that the proposed RMS has good performances independent of polarization.
I. Introduction
Since metasurfaces (MSs) can manipulate the wave-front of transmitted and reflected electromagnetic (EM) waves, they have a wide range of applications in EM field. In 2011, the phase gradient MS,
which can manipulate the wave-front of anomalous reflection and refraction waves, is proposed by Yu et al. [1]. In [1], a generalized Snell’s law is described from Fermat’s principle. Since then,
there has been extensive research related to anomalous reflection [2–4]. Specifically, retrodirectivity can be achieved using the planar type of a MS using the generalized Snell’s law [5, 6]. In [5],
a MS simultaneously operating at multiple incident angles designed using array of striplines on the ground plane for only the TM wave. Additionally, binary MS was observed to achieve an efficient
retroreflector for each TE and TM wave at near-grazing incidence [6].
In this letter, a one-dimensionally polarization-independent retrodirective metasurface (PIRMS), which consists of several supercells with a continuous gradient phase of 2π, is proposed. The length
of a supercell depends on its specific retrodirective angle and its wavelength of an operation frequency calculated by the generalized Snell’s law. Also, the unitcell of a supercell is composed of a
ring structure on the ground plane that simultaneously operates the TE and TM mode. The reflection phases of these unitcells can be controlled by the radius and width of the ring structure. Moreover,
the performance of the proposed PIRMS is confirmed by a theoretical analysis, a full-wave simulation, and a measurement at two retrodirected angles of 20° and 40°.
II. Design of 1D Polarization-Independent Retrodirective Metasurface
Based on the generalized Snell’s law, MS with a 2π phase gradient contributes to the generation of reflected and transmitted anomalous waves [1]. From the generalized Snell’s law of reflection, the
relation between the incident angle θ[i] and the anomalous reflection angle θ[r] along the interface (dΦy/dy) can be defined as [1]:
(1) sinθr-sinθi=λ02πnidΦydy=λ02πni2πLy=λ0niLy
where λ[0] and n[i] represent the wavelength in free space and the refractive index of the incident medium, respectively. dΦy/dy indicates the phase gradient of reflection along y-direction. To
achieve retroreflection (θ[r] = −θ[i]), the length of a supercell (Ly) with a 2π phase gradient can be expressed as from Eq. (2):
(2) Ly=λ02nisinθr
Moreover, the supercell is composed of a series of unitcells that have characteristics of a high reflectivity and a reflection phase variation of 2π at the operation frequency. Fig. 1 shows the
unitcell of the proposed 1D PIRMS. To obtain polarization-independent reflection characteristics, a structurally symmetrical ring structure on the ground plane is employed in this letter. Also, the
ring structure can be designed more compact rather than a circular structure with a symmetrical shape. The operation frequency is 5.8 GHz and the utilized substrate is RT Duroid 6010 (ɛ[r] = 10.2, h
= 3.175 mm, and tanδ= 0.0023). The width of the ring structure is fixed to be 0.5 mm, considering manufacturing tolerance. When the thickness of a substrate is greater, the slope of the reflection
phase becomes gentler. If the width of the ring structure becomes narrower, the frequency with the reflection phase of 0° down-shifts. The lengths of the supercells can be calculated as 75.6 mm and
40 mm from Eq. (2) in the cases of retrodirected angle of 20° and 40°, respectively. Thus, the supercell consists of six unitcells with a dimension of 12.6 mm (0.24λ[0]) in the case of retrodirected
angle of 20°, as shown in Fig. 2(a). Similarly, the supercell consists of four unitcells with a dimension of 10 mm (0.19λ[0]) in the case of a retrodirected angle of 40°, as shown in Fig. 2(b). Fig.
3 shows the full-wave simulated reflection response of the ring structure against the radius (r) in both the retrodirected angle of 20° and 40° at 5.8 GHz. When the substrate is thinned, the
reflection phase against the size of the unitcell changes sharply. Then, the sizes of the unitcells composed of the supercell become almost the same. If the manufacturing method is not precise, the
reflection phase variation of 2π cannot be measured. Therefore, it is advantage to use a thick substrate to obtain retrodirectivity. The reflection response of the unitcell is analyzed using the
master-slave periodic condition and Floquet port in ANSYS’s Electronics desktop software. The reflection magnitude remains high (>0.98), as the geometrical parameter r varies from 0.5 mm to 6 mm, as
shown in Fig. 3. Also, the reflection phase range of 2π can be obtained by changing the radius.
III. Measured Results and Discussion
To verify the retrodirectivity, the 1D PIRMSs with 29 × 6 (365.4 mm × 453.6 mm) and 38 × 11 (370 mm × 520 mm) supercells were designed and fabricated for retrodirected angle of 20° and 40°,
respectively. The reflected power can be measured by fixing the transmitting horn antenna retro angle with the PIRMS and moving only the receiving horn antenna. To excite the plane wave, the distance
between the antenna and PIRMS is selected to be 70 cm, which satisfies the far-field radiation condition, as shown in Fig. 4. Moreover, the time gating function in the network analyzer is utilized to
eliminate multiple reflections. Fig. 5(a) and (b) show a comparison in measured normalized reflected power between the copper and designed PIRMS at incident angle of 20° and 40°, respectively. The
gains of the utilized Tx and Rx antennas are equally approximately 11 dBi. In addition, the 3 dB beamwidths of the antennas are 39° and 40° in the E-plane and H-plane, respectively. Thus, the shape
of the reflected power is slightly broad in the cases of PEC and RMS. The measured retroreflected power efficiencies are more than 94% and 92% at the retrodirected angles of 20° and 40°,
respectively, regardless of the polarization.
IV. Conclusion
In this letter, a 1D PIRMS is proposed using a ring structure on the ground plane. The 1D PIRMS is composed of supercells with the 2π reflection phase gradient and the supercell is implemented by the
series unitcells of a ring structure. To confirm retrodirectivity, two kinds of 1D PIRMSs, which are operated at two retrodirected angles of 20° and 40°, were simulated and measured at 5.8 GHz. Based
on the results, it can be concluded that the proposed 1D PIRMS perform with highly efficient retrodirectivity regardless of the polarization.
This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (No. 2015R1A6A1A03031833) and Ministry of
Science, ICT and Future Planning, Korea, under the Information Technology Research Center support program (No. IITP-2017-2016-0-00291) supervised by the Institute for Information & communications
Technology Promotion (IITP).
1. Yu N, Genevet P, Kats MA, Aieta F, Tetienne JP, Capasso F, Gaburro Z. Light propagation with phase discontinuities: generalized laws of reflection and refraction. Science 334(6054):333–337. 2011;
2. Doumanis E, Goussetis G, Papageorgiou G, Fusco V, Cahill R, Linton D. Design of engineered reflectors for radar cross section modification. IEEE Transactions on Antennas and Propagation 61
(1):232–239. 2013;
3. Estakhri NM, Alu A. Wave-front transformation with gradient metasurfaces. Physical Review X 6article no. 041008. 2016;
4. Hoang TV, Lee JH. Generation of multi-beam reflected from gradient-index metasurfaces. Results in Physics 10:424–426. 2018;
5. Kalaagi M, Seetharamdoo D. Retrodirective metasurface operating simultaneously at multiple incident angles. In : Proceedings of 12th European Conference on Antennas and Propagation (EuCAP).
London, UK; 2018. p. 1–5.
6. Wong AMH, Christian P, Eleftheriades GV. Binary Huygens’ metasurfaces: experimental demonstration of simple and efficient near-grazing retroreflectors for TE and TM polarizations. IEEE
Transactions on Antennas and Propagation 66(6):2892–2903. 2018;
Article information Continued
Copyright © 2019 by The Korean Institute of Electromagnetic Engineering and Science | {"url":"https://jees.kr/journal/view.php?number=3361&viewtype=pubreader","timestamp":"2024-11-05T13:48:36Z","content_type":"application/xhtml+xml","content_length":"43041","record_id":"<urn:uuid:35250c8a-5d5c-4768-b2b0-50d4e8dda01b>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00638.warc.gz"} |
EchoVector Pivot Point Calculations
Determine/Find/Enter Your Security's Focus Interest Price, such as it's last 'time frame relative' Pivot Point Price, or Flex Point Price, or its Current Price. ___
Enter Your Security's Coordinate EchoBackDate Price* related to the Focus Interest Price or the Current Price you have elected in 1. ___
*Some reader's have trouble identifying the echobackdate price of a securities' selected Focus Interest Price. Simply look back the length of your chosen cycle period to find it.
For example, the Quarterly EchoVector (QEV) echobackdate for today's price would be the price on the same trading weekday one quarter of a year ago, or 13 weeks ago. The Bi-Weekly
EchoVector (WEV) echobackdate for today's price would be the price on the same trading weekday 2 weeks ago. The Annual EchoVector (AEV) echobackdate for today's price would be the
price on the same trading weekday 52 weeks ago.
Enter The EchoVector Time Length you have elected (in terms of Total Applicable Bars: ie., Weeks, Days, Hours, Minutes, Varied Time-Blocks, etc.).
The Generated EchoVector Slope Is Then: ___ (per applicable bar and per unit time-block).
Active Equation: 1. minus 2. divided by 3. equals 4.
Enter The EchoBackDate's Historical Forward Focus Pivot Point or Flex Point Price (The Price of a Subsequent Price Supporting Pivot Point Price or the Price of a Subsequent Price
Reversing Pivot Point Price) of Interest that followed the EchoBackDate. ___
Enter The EchoBackDate's Historical Forward Focus Interest Pivot Point's or Flex Point's Additional Forward Time Increments From EchoBackDate's Time (in Applicable Bars: ie.,
Weeks, Days, Hours, Minutes, Varied Time-Blocks, etc.).
The Coordinate Forecast EchoVector Pivot Point Price Is: ___ .
Active Equation: 1. plus (4. times 6.).
Calculating and Constructing The Security's Corresponding Forecast EchoVector Support/Resistance Vector: The Coordinate Forecast EchoVector Support/Resistance
Vector (CFE-SRV)
A CFE-SRV, a Security's Coordinate Forecast EchoVector Determined Price Support/Resistance Vector, runs from the Security's Starting Reference Price ___ in 1. above to
the Forecast EchoVector Pivot Point Price ___ in 7. above and runs forward ___ number of bars as entered in 6. above from the starting reference price point ___ entered in 1.
above, and runs along at the slope (rate) generated in 4. above to the EchoVector Pivot Point Price ___ calculated in 7. above. This EchoVector SRV can be readily projected (and drawn
on chart).
Perform Calculation 5. through 8. for Each Sub-sequential Price Supporting Pivot Point Price and Sub-sequential Price Reversing Pivot Point Price of Interest from Your
Security's EchoBackDate Price (Within A Relevant Interest Range*). ___, (S1, S2, S3, and R1, R2, and R3)
The Interest Range, technically referred to as Range C within the MDPP Forecast Model, consists of a constellation of time-frame and scope-relative pivot points in the price track of
your subject security that follow the echovectors echo- backdate-time-and-price-point, selected by specific Range C criteria. See EchoVector Analysis and EchoVector
Pivot Points for more information on Range C.
For your general echovector pivot point calculations, and for your different perspective constructions and for your graphical framing purposes, you use any of the pivot prices you see
on your chart proximate to the echovector echo-back-date-time-and-price point to start that interests you, perhaps then picking 3 relative bottoms (ascending or descending or a combination) and 3
relative tops (ascending or descending or a combination) also, and remembering which were price supporting pivots and which were price resistance pivots. This will enable you to
calculate a set of S pivot points and R pivot points to your echovector starting reference price (your beginning point, which is also your selected echovector's
endpoint), who's set of 'simple and rudimentary method generated' EchoVector Pivot Points you have now just calculated)
Twelve Significant EchoVector Calculations For Perspective
And Calculating Their Respective EchoVector Pivot Points
1. Daily EchoVector (24HEV)
2. Weekly EchoVector (WEV)
3. Bi-Weekly EchoVector (2WEV)
4. Monthly EchoVector (MEV)
5. Bi-Monthly EchoVector (2MEV)
6. Quarterly EchoVector (QEV, 3MEV, 13WEV)
7. Bi-Quarterly EchoVector (2QEV, 6MEV, 26WEV)
8. Tri-Quarterly EchoVector (3QEV, 9MEV, 39WEV)
9. Annual EchoVector (AEV, 4QEV, 12MEV, 52WEV)
10. Congressional Cycle EchoVector, Bi-Annual EchoVector (CCEV, 2AEV, 8QEV)
11. Presidential Cycle EchoVector (PCEV, 2CCEV, AEV)
12. Regime Change Cycle EchoVector (RCCEV, 2PCEV, 2CCEV, 8AEV)
EchoVector Pivot Points
An Introduction to EchoVector Analysis and EchoVector Pivot Points
"EchoVector Theory and EchoVector Analysis are a price pattern impact theory and a technical analysis methodology and approach postulated, created, and invented by Kevin John Bradford Wilbur.
EchoVector Analysis is also presented as a behavioral economic application and securities analysis tool in price pattern theory and in price pattern behavior, study, and forecasting, and in price
EchoVector Pivot Points are a further technical analysis tool and application within EchoVector Analysis, derived from EchoVector Theory in practice.
EchoVector Theory and EchoVector Analysis assert that a securities prior price patterns may influence its present and future price patterns. Present and future price patterns may then, in part, be
considered as 'echoing' these prior price patterns to some identifiable and measurable degree.
EchoVector Theory and EchoVector Analysis also assert that these influences may be observable, identifiable, and measurable in price pattern behavior and price pattern history, and potentially
observable in future price pattern formation, and potentially efficacious in future price pattern forecasting, to some measure or degree.
EchoVector Analysis is also used to forecast and project potential price Pivot Points (referred to as PPP's --potential pivot points, or EVPP's --EchoVector Pivot Points) and future support and
resistance echovectors (SREV's) for a security from a starting reference price at a starting reference time, based on the securities prior price pattern within a given and significant and definable
cyclical time frame.
EchoVector Pivot Points and EchoVector Support and Resistance Vectors are fundamental components of EchoVector Analysis. EchoVector SREV's are constructed from key components in the EchoVector Pivot
Point Calculation. EchoVector SREV's are calculated and also referred to as Coordinate Forecast EchoVectors (CFEV's) to the initial EchoVector (XEV) calculation and construction, where X designates
not only the time length of the EchoVector XEV, but also the time length of XEV's CFEVs. The EchoVector Pivot Points are found at the endpoint of XEV's CFEVs' calculations and constructions.
The EchoVector Pivot Point Calculation is a fundamentally different and more advanced calculation containling significantly more information than the traditional pivot point calculation.
The EchoVector Pivot Point Calculation differs from traditional pivot point calculations by reflecting this given and specified cyclical price pattern reference and its significance and information
in the echovector pivot point calculation. This cyclical price pattern and its reference and its slope momentum is included in the calculation and construction of the echovector and its respective
coordinate forecast echovectors, as well as in the calculation of the related echovector pivot points..."
Learn To Construct
(1) A FINANCIAL MARKET TIME-CYCLE PRICE ECHOVECTOR of a Particular Cyclical Length and Starting Reference Time-And-Price Point, And
(2) A FINANCIAL MARKET TIME-CYCLE PRICE ECHOVECTOR'S Corresponding FINANCIAL MARKET TIME-CYCLE ECHOVECTOR PIVOT POINT PRICE PARALLELOGRAM, And
Included OTAPS-PPS SWITCH SIGNAL VECTORS.
It is often suggested to start with the readily available daily increment quarterly cycle length period on your price chart, and to then eventually construct an echovector pivot point price
parallelogram (EV-PPP-Pgram) using the echovector (EV) and a scale-and-scope-relative pivot point nearby (and subsequent to) the echovector's echo-back-date (EBD). This already occurring
scope-and-scale-relative nearby pivot point following the EBD's phase is referred to as the NPP. This 'relatively nearby' pivot point to the EBD constitutes the third point in the EV-PPP-Pgram. The
first point is the starting reference time-and-price-point (STPP) in the echovector's initial construction. The second point in the financial market time-cycle price echovector pivot point price
parallelogram is the echovector's echo-back-date-time-and-price-point (EBD-TPP).
This nearby pivot point also constitutes the end of the NPP Extension Vector that runs from the EBD-TPP to the NPP. (To support the OTAPS-PPS Position Polarity Reversal and Switch Signal active
advanced position management strategy, this NPP Extension Vector will also be symmetrically transposed to become one of sources of the active OTAP-PPS Position Polarity Switch Signal Vectors within
the PPS Switch Signal Vectors Fan that will also eminate from the echovector's STPP.)
The NPP to the EBD-TPP also constitutes the beginning of the coordinate forecast echovector, reading left to right. The EBD-TPP constitutes the beginning of the echovector, not by way of
construction, but when reading left to right. The beginning of the construction of the echovector is on the right at the starting reference point, and moves back to the EBD-TPP, right to left,
during construction.
The Coordinate Forecast Echovector (CFEV) runs parallel to the initial echovector in the Time-Cycle Price Echovector Pivot Point Price Projection Forecast Parallelogram, and is the same length as the
echovector. The NPP Extension Vector (NPP-V, being either an NPP-S Vector or an NPP-R Vector, S designating Support and R designating Resistance) is on the left side of the parallelogram. At the
far right endpoint of the parallelogram construction we find the projected echovector pivot point projection time-and-price (EV-PPP-TPP). When the scope-and-scale-relative nearby pivot point that
follows the EBD-TPP is a price supporting pivot point the echovector pivot point that is projected to follow the echovector's STPP occurs below the STPP. When the nearby pivot point is a price
supporting pivot point to the EBD-TPP the EV-PPP-TPP occurs after and below the echovector's STPP.
The CFEV has the 'already occurred' NPP to the EV's EBD-TPP, and at the end of the CFEV we find the EV-PPP. The X-EV (echovector of cycle length X) has its STPP (starting time and price point -the
starting reference point) on its far right, and its EBD-TPP on its far left. The parallel CFEV has the NPP (an identified and elected scope relative pivot point price at a specific time occuring
nearby, but following the echovector's EBD-TPP) on the far left, and has on the far right the EV-PPP-TPP, the echovector pivot point projection. The STPP, EBD-TPP, NPP, and the EV-PPP constitute the
four key time-and-price-points in the financial market time-cycle price echovector pivot point price projection forecast parallelogram.
The forth side of the parallelogram, STPP to EV-PPP, constitutes an active OTAPS-PPS Position Polarity Switch Signal Vector within active advanced risk and position management strategies.
Learning this simple construction process, you will then be able to construct several (4 to 8) parallelograms, each with different nearby pivot point (NPPs) to the echobackdate-timeandpricepoint.
Some NPPs may be found to have a price higher than the echobackdate and some a price lower. Those with prices loWer can be designated NPP-S1, NPP-S2, NPP-S3, NPP-S4, etc. Those with prices higher
can be designated NPP-R1, NPP-R2, NPP-R3, NPP-R4, etc.
The cycle echovector pivot point price projection forecast parallelogram of length X with starting reference point STPP and echo-back-date-time-and-price-point EBD-TPP and containing NPP-S1 will have
X-EV-PPP-S1 generated on its far right through the construction process as its cycle echovector pivot point price projection S1.
S1, S2, S3, S4, and R1, R2, R3, R4, (S for support, and R for resistance, S at an NPP price-and-time-point below the EBD-TPP price-and-time-point, and R at an NPP price-and-time-point above the
EBD-TPP price-and-time-point) may occur and be forecasted for any STPP.
Each generated STPP to EV-PPP-Sx Vector (the fourth side of any time-cycle price echovector pivot point price projection forecast parallelogram) and each generated STPP to EV-PPP-Rx Vector (also the
fourth side of any cycle echovector pivot point price projection forecast parallelogram) constitute potential OTAP-PPS Vectors, and together constitute elements of the OTAPS-PPS Vector Fan for any
give cyclical time-frame length of interest (12 of the key cycle time-frame lengths in echovector analysis are listed above).
For more comprehensive forecasting and analysis, (1) EchoVector Pivot Point Price Projections and (2) Active Advanced Risk and Position Management OTAPS-PPS Position Polarity Switch Signal Vector
Fans, for all 12 key echovector cycle lengths, for any STPP of interest, can be generated. | {"url":"http://www.market-pivots.com/EchoVector-Pivot-Point-Calculations.html","timestamp":"2024-11-03T18:58:57Z","content_type":"text/html","content_length":"96745","record_id":"<urn:uuid:455e5e06-922f-46b0-908e-b59afec710ce>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00116.warc.gz"} |
Exploring the Parallels Between ceLLMs and LLMs: Processing Information in Higher-Dimensional Latent Spaces
The cellular Latent Learning Model (ceLLM) offers a fascinating theoretical framework that draws parallels between biological cellular processes and artificial intelligence models, specifically large
language models (LLMs). Both ceLLM and LLMs process information in higher-dimensional latent spaces, utilizing weighted connections to interpret inputs and generate outputs. This analogy not only
provides a novel perspective on cellular biology but also helps in understanding complex biological phenomena through the lens of established AI concepts.
Resonant Field Weights in ceLLM
Formation of Resonant Connections
In the ceLLM framework, resonant field weights are formed through the interactions of atomic elements within DNA. These elements establish resonant connections based on their:
• Like Elements: Atoms of the same or compatible types can resonate at specific frequencies, forming connections.
• Charge Potential: The electrical charge of each element influences its ability to form resonant connections with others.
• Distance: According to the inverse square law, the strength of the resonant connection diminishes with the square of the distance between atoms.
These factors combine to create a network of resonant field connections, where the energy between atoms forms the “weights” of the system. This network shapes the latent space—a higher-dimensional
manifold where cellular information processing occurs.
Impact on Spacetime Geometry and Probabilistic Outputs
The resonant field connections influence the geometry of the latent space, effectively shaping the spacetime landscape within which cellular processes operate. This geometry determines the
probabilistic cellular outputs in response to environmental inputs by:
• Altering Energy Potentials: The resonant weights modify the energy landscape, affecting how likely certain cellular responses are to occur.
• Guiding Signal Propagation: The geometry influences the pathways through which signals travel within the cell, impacting decision-making processes.
• Enabling Adaptive Responses: The dynamic nature of the resonant connections allows cells to adapt their behavior based on changes in the environment.
Processing Information in Higher-Dimensional Spaces
ceLLM Information Processing
In the ceLLM model, cells process information through:
1. Environmental Inputs: Cells receive signals from their surroundings, such as chemical gradients, electromagnetic fields, or mechanical stresses.
2. Resonant Field Interactions: These inputs affect the resonant connections between atomic elements in DNA, altering the weights within the latent space.
3. Probabilistic Decision-Making: The modified latent space geometry influences the probabilities of different cellular responses.
4. Output Generation: Cells produce responses (e.g., gene expression, protein synthesis) based on the most probable outcomes determined by the latent space configuration.
LLM Information Processing
Similarly, large language models process information by:
1. Input Tokens: The model receives a sequence of words or tokens representing text input.
2. Embedding in Latent Space: Each token is mapped to a high-dimensional vector in the latent space.
3. Weighted Connections: The model uses learned weights and biases to adjust these vectors, capturing contextual relationships between words.
4. Probabilistic Prediction: The adjusted vectors are used to predict the probability distribution of the next word or token.
5. Output Generation: The model generates text output based on the highest probability predictions.
Parallels Between ceLLMs and LLMs
Weighted Connections and Energy Landscapes
• ceLLM: Weights are formed by the energy between resonant atomic connections, influenced by charge potential and distance.
• LLM: Weights are numerical values learned during training, representing the strength of connections between neurons in the network.
Both systems rely on weighted connections to process inputs and determine outputs, effectively navigating an energy landscape (in ceLLM) or a loss landscape (in LLMs).
Higher-Dimensional Latent Spaces
• ceLLM: The latent space is a manifold shaped by the resonant field connections, representing all possible states of the cell.
• LLM: The latent space is a high-dimensional vector space where semantic meanings are encoded, allowing the model to capture complex linguistic patterns.
In both cases, the latent space serves as the computational substrate where inputs are transformed into outputs.
Probabilistic Processing
• ceLLM: Cellular responses are probabilistic, with certain outcomes being more likely based on the latent space geometry.
• LLM: Language predictions are probabilistic, generating the next word based on probability distributions learned from data.
This probabilistic nature allows both systems to handle ambiguity and variability in their respective environments.
Adaptive Learning and Evolution
• ceLLM: The resonant connections are shaped by evolutionary processes, encoding information over generations.
• LLM: The weights are learned from large datasets during training, capturing patterns and structures in human language.
Both systems adapt over time, improving their responses based on accumulated information.
Detailed Explanation of Resonant Field Weights Formation
Atomic Resonance and Charge Potential
Atoms within DNA have specific energy states determined by their electron configurations and nuclear properties. When atoms of similar types or compatible energy levels are in proximity, they can:
• Enter Resonance: Oscillate at the same or harmonically related frequencies.
• Exchange Energy: Through electromagnetic interactions, affecting their energy states.
The charge potential of each atom influences its ability to resonate:
• Positive and Negative Charges: Attract or repel, affecting the likelihood of forming resonant connections.
• Ionization States: Atoms with unpaired electrons may be more reactive and form stronger resonant connections.
Distance and the Inverse Square Law
The strength of the resonant connection between two atoms decreases with distance, following the inverse square law:
• Mathematical Relationship: F∝1r2F \propto \frac{1}{r^2}, where FF is the force or interaction strength, and rr is the distance between atoms.
• Implications: Closer atoms have stronger interactions, leading to higher weights in the resonant network.
Shaping the Spacetime Geometry
The collective resonant connections form a network that defines the latent space’s geometry:
• Energy Landscape: Regions of high and low energy potential guide the flow of signals within the cell.
• Topological Features: Valleys, peaks, and pathways in the energy landscape correspond to preferred states or transitions.
• Dynamic Adaptation: Changes in environmental inputs can reshape the geometry, allowing the cell to respond adaptively.
Identical Information Processing in ceLLM and LLM
Input Encoding
• ceLLM: Environmental signals modulate the resonant field weights, effectively encoding the input into the latent space.
• LLM: Input text is encoded into vectors in the latent space using embedding layers.
Transformation and Computation
• ceLLM: Resonant interactions compute the probabilities of various cellular responses by altering the energy landscape.
• LLM: Neural network layers transform input embeddings through weighted connections, computing the probabilities of output tokens.
Output Decoding
• ceLLM: The cell produces a response (e.g., gene expression) based on the most energetically favorable state.
• LLM: The model generates text output by selecting tokens with the highest predicted probabilities.
Learning and Adaptation
• ceLLM: Evolutionary processes adjust the resonant connections over generations, improving cellular responses.
• LLM: Training algorithms adjust weights to minimize loss functions, improving the model’s predictions.
The ceLLM model provides a compelling analogy to large language models by conceptualizing cellular processes as computations within a higher-dimensional latent space shaped by resonant field
connections. Both systems utilize weighted interactions to process inputs probabilistically and generate outputs, adapting over time through evolutionary or learning mechanisms.
By exploring these parallels, we gain a deeper understanding of how complex biological systems might process information similarly to artificial neural networks. This perspective opens avenues for
interdisciplinary research, bridging biology and artificial intelligence, and offering insights into the fundamental principles underlying information processing in both natural and artificial
ceLLM model is a theoretical framework, it serves as a valuable tool for conceptualizing complex biological interactions. Drawing parallels with established AI models like LLMs allows for a more
intuitive understanding of these processes. | {"url":"https://www.rfsafe.com/articles/cell-phone-radiation/exploring-the-parallels-between-cellms-and-llms-processing-information-in-higher-dimensional-latent-spaces.html","timestamp":"2024-11-02T08:45:44Z","content_type":"text/html","content_length":"17857","record_id":"<urn:uuid:ce88fc48-c9a7-4ca0-9730-158166210613>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00359.warc.gz"} |
Rethinking the Role of Gradient-based Attribution Methods for Model Interpretability
Abstract: Current methods for the interpretability of discriminative deep neural networks commonly rely on the model's input-gradients, i.e., the gradients of the output logits w.r.t. the inputs. The
common assumption is that these input-gradients contain information regarding $p_{\theta} ( y\mid \mathbf{x} )$, the model's discriminative capabilities, thus justifying their use for
interpretability. However, in this work, we show that these input-gradients can be arbitrarily manipulated as a consequence of the shift-invariance of softmax without changing the discriminative
function. This leaves an open question: given that input-gradients can be arbitrary, why are they highly structured and explanatory in standard models? In this work, we re-interpret the logits of
standard softmax-based classifiers as unnormalized log-densities of the data distribution and show that input-gradients can be viewed as gradients of a class-conditional generative model $p_{\theta}
(\mathbf{x} \mid y)$ implicit in the discriminative model. This leads us to hypothesize that the highly structured and explanatory nature of input-gradients may be due to the alignment of this
class-conditional model $p_{\theta}(\mathbf{x} \mid y)$ with that of the ground truth data distribution $p_{\text{data}} (\mathbf{x} \mid y)$. We test this hypothesis by studying the effect of
density alignment on gradient explanations. To achieve this density alignment, we use an algorithm called score-matching, and propose novel approximations to this algorithm to enable training
large-scale models. Our experiments show that improving the alignment of the implicit density model with the data distribution enhances gradient structure and explanatory power while reducing this
alignment has the opposite effect. This also leads us to conjecture that unintended density alignment in standard neural network training may explain the highly structured nature of input-gradients
observed in practice. Overall, our finding that input-gradients capture information regarding an implicit generative model implies that we need to re-think their use for interpreting discriminative | {"url":"https://iclr.cc/virtual/2021/poster/2943","timestamp":"2024-11-05T12:34:59Z","content_type":"text/html","content_length":"43934","record_id":"<urn:uuid:5ff6a69e-7ab5-4d90-b2d2-722d7076597c>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00587.warc.gz"} |
Finding Missing Angles In Parallel Lines Worksheet Pdf - Angleworksheets.com
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Find Missing Angles Parallel Lines Worksheet
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Missing Angles Parallel Lines Worksheet Pdf
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Finding Missing Angles In Parallel Lines Worksheets
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Finding Angles With Parallel Lines Worksheet
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fundamentals of angles and lines. These printables will also teach you how to use a protractor and read it. These worksheets can also … Read more | {"url":"https://www.angleworksheets.com/tag/finding-missing-angles-in-parallel-lines-worksheet-pdf/","timestamp":"2024-11-10T21:33:31Z","content_type":"text/html","content_length":"77274","record_id":"<urn:uuid:b4717ecd-d1b4-4b21-8602-29b33b6b3a70>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00566.warc.gz"} |
Determining if the columns of a matrix are orthogonal
Commented: Torsten on 10 Jul 2023
I have to determine if the columns of any given matrix are orthogonal or not. But, I am not sure how to generalize that correctly. I am thinking of doing a for loop with i = 1:n(# of columns of
matrix) but I don't know how I would accomplish that successfully because I have to dot each column with all the other columns without dotting themselves in the for loop. Let's say my code is
A = magic(4)
for i = 1:n
for j = 1:n
value = dot(A(:,i),A(:,j))
if value~=0
2 Comments
116 views (last 30 days)
Determining if the columns of a matrix are orthogonal
I could'nt finish the rest of my thought in the post because the text editor glitched out. So, I was saying that is there some way to not make j what i is currently so there is no dot products done
to the same vector.
Asad (Mehrzad) Khoddam on 4 Dec 2020
if i==j then the dot product should not be zero. So, you need to check if i~=j then check the dot product
Answers (2)
Regarding the case i==j, you can start j from numbers greather than i:
A = magic(4)
orth = 1;
for i = 1:n
for j = i+1:n
value = dot(A(:,i),A(:,j))
if value~=0
% check orth, if it is 0 it means that it is not orthogonal
if orth
disp('not orthogonal')
2 Comments
Thota on 10 Jul 2023
what is the value of n
Torsten on 10 Jul 2023
The number of columns of the matrix.
Edited: James Tursa on 4 Dec 2020
Using the dot product and comparing it to 0 is a mathematical concept that does not translate well to floating point arithmetic. If you are going to use this method, unless you know for sure you are
dealing with integers only and that the calculations will not overflow the precision, it is better to use a tolerance for floating point comparisons. And unless you know the range of numbers you are
dealing with for picking a tolerance, you should normalize the columns before comparing the dot product result to the tolerance.
All that being said, what you could simply do to generate the dot products is do a matrix multiply with its transpose. E.g., A'*A will generate all of the column dot products as elements of the
result. Just examine the upper or lower triangle part of this.
0 Comments | {"url":"https://au.mathworks.com/matlabcentral/answers/677158-determining-if-the-columns-of-a-matrix-are-orthogonal","timestamp":"2024-11-11T20:24:15Z","content_type":"text/html","content_length":"142498","record_id":"<urn:uuid:abbede15-3c02-4bb9-a2e2-48243088e0b8>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00601.warc.gz"} |
One egg has a mass of 50g. How many eggs in 1kg?
Hint: We first use unitary method to calculate number of eggs formed in 1 gram using unitary method. Then use conversion of kilograms to grams and use a unitary method to calculate the number of eggs
that will be in 1kg.
* Unitary method helps us to calculate the value of a single unit by dividing the value of multiple units by the number of units given.
* Unitary method helps us to calculate the number of units by dividing the total value of multiple units by the value of a single unit.
* 1Kg contains 1000 grams
Complete answer:
We are given that mass of 1 egg \[ = 50\]g
We can also write this equation in another form
Number of eggs formed by 50g mass \[ = 1\]
Then using unitary method we can calculate number of eggs formed by 1 gram
Number of eggs formed by 1g mass \[ = \dfrac{1}{{50}}\] … (1)
Since we have to calculate number of eggs in 1 kg we will multiply the grams in equation (1) by 1000 as 1Kg contains 1000g
\[ \Rightarrow \]Number of eggs formed by 1000g mass \[ = \dfrac{1}{{50}} \times 1000\]
Cancel same factors from numerator and denominator
\[ \Rightarrow \]Number of eggs formed by 1000g mass \[ = 20\]
\[\therefore \]Number of eggs that will be 1kg are 20.
Many students make the mistake of not using the conversion of kilograms into grams and calculate using the unitary method directly which gives us the wrong answer. Keep in mind we have to have all
the values in the same SI unit be it kilograms or grams but they should be the same unit then only we can cancel out values. | {"url":"https://www.vedantu.com/question-answer/one-egg-has-a-mass-of-50g-how-many-eggs-in-1kg-class-12-maths-cbse-5fd85dd48238151f0da05f19","timestamp":"2024-11-05T04:21:31Z","content_type":"text/html","content_length":"177260","record_id":"<urn:uuid:3a9af0fb-b9eb-4d79-96ee-7cdc13aba3be>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00115.warc.gz"} |
A line passes through (4 ,3 ) and (7 ,1 ). A second line passes through (1 ,8 ). What is one other point that the second line may pass through if it is parallel to the first line? | HIX Tutor
A line passes through #(4 ,3 )# and #(7 ,1 )#. A second line passes through #(1 ,8 )#. What is one other point that the second line may pass through if it is parallel to the first line?
Answer 1
Find the slope of the first line, then apply it to the point of the second line to find $\left(1 + 3 , 8 - 2\right) = \left(4 , 6\right)$
We can solve this by finding the slope of the first line, then applying that slope to the point of the second line.
The first line passes through points #(4,3) and (7,1)#. Let's determine the slope.
The equation of slope is #m=(y_2-y_1)/(x_2-x_1)#. Let's plug in our points into this equation:
We can now apply this slope to point #(1,8)#. I'll move 3 to the right on the x axis and 2 down on the y:
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Answer 2
To find a point through which the second line may pass if it is parallel to the first line, we use the slope of the first line, which is determined by the given points (4, 3) and (7, 1). The slope of
the first line can be calculated using the formula:
[m = \frac{{y_2 - y_1}}{{x_2 - x_1}}]
Once we have the slope of the first line, we can use it to find the equation of the second line. Since the second line is parallel to the first line, it will have the same slope. Then, using the
point (1, 8) through which the second line passes, we can find its equation in slope-intercept form (y = mx + b).
After finding the equation of the second line, we can use it to determine another point on the line by substituting different values of x and solving for y. This will give us various points through
which the second line may pass while being parallel to the first line.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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statistics answered |Statistics tutor online| My Mathlab answers |Ap Statistics Tutor | A report states that the cost of repairing a hybrid vehicle, is falling even while typical repairs on conventional vehicles are getting more expensive
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A report states that the cost of repairing a hybrid vehicle is falling even while typical repairs on conventional vehicles are getting more expensive. The most common hybrid repair, replacing the
hybrid inverter assembly, had a mean repair cost of $3,927 in 2012. Industry experts suspect that the cost will continue to decrease given the increase in the number of technicians who have gained
expertise on fixing gas-electric engines in recent months. Suppose a sample of 100 hybrid inverter assembly repairs completed in the last month was selected. The sample mean repair cost was $3,850
with the sample standard deviation of $400. Complete parts (a) and (b) below.
a. Is there evidence that the population mean cost is less than $3,927? (Use a 0.05 level of significance.)
State the null and alternative hypotheses.
Find the test statistic for this hypothesis test.
The critical value(s) for the test statistic is(are) ___ ?
Is there sufficient evidence to reject the null hypothesis using a = 0.05?
The first thing to consider before answering such a question is to decide whether to use the normal distribution or the t distribution. Sometimes the choice is based on the sample size or whether the
population standard deviation is given. The process is to always use the standard normal distribution when the population standard distribution is not given. In other instances, depending on some
books and other tutors, the normal distribution is used whenever the sample size is thought to be large enough.
What should we do here?
I will use the t distribution here to demonstrate the process of hypothesis testing given the sample statistics above.
Note that the formula for the t - distribution can be calculated as.
t = (X_bar - mu)/(sd/sqrt(n))
The idea is to calculate the test statistic and to compare with the critical value, and the null hypothesis is rejected in the test statistic is more extreme than the critical value.
So culculate t = (3850 - 3927)/(400/sqrt(100)) = -1.925
The null hypotheses to be tested are as follows.
H0: mu greater or equal to 3927
Ha: mu is less than 3927
This is a left-tailed test and so the critical value should be a negative value.
use excel to find the critical value; =T.INV(0.05,99) = -1.660
The null hypothesis is rejected because the test statistic is more extreme than the critical value. test statistic above -1.925 is less than -1.660.
The conclusion finds sufficient evidence to support the claim proposed by the experts. that the cost has been decreasing.
• 0
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You can create forecasts in Quiver using the Time series forecast transform. A forecast is a projection forward in time of an existing time series plot in your analysis. Forecasts in Quiver are built
visually and interactively. The result of a forecast is a time series plot representing the forecasted data; this time series plot can be further transformed using Quiver’s time series transforms.
To illustrate this section, we use the linear forecast as an example. Sections below give more details about each type of forecast.
This will produce a time series plot (in this case, a line), that is by default fitted to the entire range of the input plot. Note that the time axis will remain the same as it was for the input time
series plot. To see the forecast further into the future, you can zoom out on the x-axis.
You can find the values of the coefficients under the Forecast Details section of the forecast editor.
In the linear forecast example, the coefficients are m (the slope) and c (the offset).
Here we have set the training time range to be 2015-2020 instead of the entire history. As a result, the parameters (slope and offset) of the forecast have changed to make the forecast more accurate
for the training time range. This can be useful if you believe certain times will be more indicative of future behavior.
In our example, changing the loss definition results in different forecast coefficients.
In this section, we detail the different types of forecast and their configuration options. For the remainder of the documentation, we refer to the quantity to be forecasted as y or y(t) to denote y
at time t.
The constant forecast assumes the quantity y will remain constant.
In our example, we can see that the constant forecast does not capture the slope or periodicity in the data.
The linear forecast assumes the quantity y will follow a linear trend.
In our example, the linear forecast captures the slope but not the periodicity in the data.
Formula forecasts can be used when there is periodicity in the data, and the quantity is following some physical process that ebbs and flows. For example, ambient temperature exhibits both daily and
yearly periodicity. The formula forecast allows you to fit a sinusoidal curve.
The formula forecast assumes the quantity y follows a governing equation.
Example forecast with an exponential formula. The coefficients determined by the model are displayed in the expression under Forecast details.
ODE can be used to forecast a quantity that is governed by an Ordinary Differential Equation. The ODE forecast assumes the derivative (rate of change) of the quantity y follows a governing equation.
To define your ODE forecast, add the governing equation to the expression box with the unknowns defined as coefficients using the @ prefix and a letter. For example, for exponential growth we have @k
* y, where y is the quantity.
In this example, we used an ODE forecast using the exponential growth equation.
This forecast is indicated when there is periodicity in the data coming from some pattern of life. For example, retail sales exhibit some weekly periodicity if people are more likely to shop on
certain days of the week.
Where: yd is y after differences (subtract consecutive values) have been taken d times.
Selecting the auto option will set the following ARIMA parameters for you automatically. If you prefer, you can change the parameters manually until you get a satisfactory fit. If selecting the ARIMA
parameters manually, bias toward smaller numbers as a simpler model with less terms will generalize better.
It is possible to add a seasonal component to the model. To do so, switch the seasonal toggle and specify the period of seasonality. For example, if you have daily data with weekly periodicity, enter
If the auto option is off, the following parameters will appear:
For certain forecast types, you can choose the loss definition used when fitting the forecast. When fitting the forecast to the training data, the coefficients will be selected to minimize the loss.
Different types of loss will result in different forecasts.
The square root of the sum of square differences between the target series points y[i] and the forecast f[i] points within the training time range. Equivalent to the L2 norm of the error.
The sum of absolute differences between the target series points and the forecast points within the training time range. Equivalent to the L1 norm of the error.
The maximum absolute difference between the target series points and the forecast points within the training time range. Equivalent to the L-infinity (L∞) norm of the error. | {"url":"https://www.palantir.com/docs/foundry/quiver/timeseries-forecast/","timestamp":"2024-11-09T22:01:51Z","content_type":"text/html","content_length":"538218","record_id":"<urn:uuid:0da42da1-61d0-4193-a67c-bf4801534cf1>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00618.warc.gz"} |