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Find the solution of the following differential equations. dy=e^{x-y}dx
abreviatsjw 2021-12-28 Answered
Find the solution of the following differential equations.
dy={e}^{x-y}dx
dy={e}^{x-y}dx
dy={e}^{x}\cdot {e}^{-y}dx
\frac{dy}{{e}^{-y}}={e}^{x}dz
{e}^{y}dy={e}^{x}dx
Integrting both side
\int {e}^{y}dy=\int {e}^{x}dx
{e}^{y}={e}^{x}+c
In this tutorial we shall evaluate the simple differential equation of the form
\frac{dy}{dx}={e}^{x-y}
using the method of separating the variables.
The differential equation of the form is given as
\frac{dy}{dx}={e}^{x-y}
⇒\frac{dy}{dx}={e}^{x}{e}^{-y}
⇒\frac{dy}{dx}=\frac{{e}^{x}}{{e}^{y}}
Separating the variables, the given differential equation can be written as
{e}^{y}dy={e}^{x}dx
In the separating the variables technique we must keep the terms dy and dx in the numerators with their respective functions.
Now integrating both sides of the equation (i), we have
\int {e}^{y}dy=\int {e}^{x}dx
Using the formulas of integration
\int {e}^{x}dx={e}^{x}
{e}^{y}={e}^{x}+c
⇒y=\mathrm{ln}\left({e}^{x}+c\right)
\begin{array}{}dy={e}^{x-y}dx\\ dy={e}^{x}\cdot {e}^{-y}dx\\ \frac{dy}{{e}^{-y}}={e}^{x}dz\\ {e}^{y}dy={e}^{x}dx\\ \int {e}^{y}dy=\int {e}^{x}dx.\\ {e}^{y}={e}^{x}+c\end{array}
{L}^{-1}\left\{\frac{{e}^{-\pi s}}{{s}^{2}+1}\right\}
Show that any separable equation
M\left(x,y\right)+N\left(x,y\right){y}^{\prime }=0
{A}_{i}
of the differential equation are real, then real-valued solutions are generally preferable. Since non-real roots z then come in conjugate pairs, so do their corresponding basis functions
{x}^{k}{e}^{zx}
, and the desired result is obtained by replacing each pair with their real-valued linear combinations Re(y) and Im(y), where y is one of the pair.
Using only the Laplace transform table, obtain the Laplace transforms of the following function:
\text{cosh}\left(4t\right)-4t
where ”cosh” stands for hyperbolic cosine and cosh
x=\frac{{e}^{x}-{e}^{-x}}{2}
{L}^{-1}\left[F\left(s\right)\right]
for the given F.
F\left(s\right)=\frac{s-2}{{s}^{2}+2s+3}
\frac{dy}{dt}-y=z,\text{ }y\left(0\right)=0
x"+4x=\delta \left(t\right)+\delta \left(t-\pi \right)
x\left(0\right)={x}^{\prime }\left(0\right)=0
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Pls solve and answer the question in detail Q If x, y satisfy the equation yx=xy and x - Maths - Relations and Functions - 12676105 | Meritnation.com
Pls solve and answer the question in detail.
Q. If x, y satisfy the equation
{y}^{x}={x}^{y}
and x = 2 y then
\frac{{x}^{2}+{y}^{2}}{5}=?
\mathbit{G}\mathbit{i}\mathbit{v}\mathbit{e}\mathbit{n}\mathbf{:} {x}^{y}={y}^{x} and x=2y\phantom{\rule{0ex}{0ex}}So, {\left(2y\right)}^{y}={y}^{2y}\phantom{\rule{0ex}{0ex}}{2}^{y}\overline{)×{y}^{y}}={y}^{y}\overline{)×{y}^{y}} \left[\mathbf{U}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{ }\mathbf{p}\mathbf{r}\mathbf{o}\mathbf{p}\mathbf{e}\mathbf{r}\mathbf{t}\mathbf{y}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{e}\mathbf{x}\mathbf{p}\mathbf{o}\mathbf{n}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{s}\mathbf{:}\mathbf{ }{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{y}\mathbf{×}\mathbf{y}\right]\phantom{\rule{0ex}{0ex}}{2}^{y}={y}^{y}\phantom{\rule{0ex}{0ex}}On Compairing, we get\phantom{\rule{0ex}{0ex}}y=2\phantom{\rule{0ex}{0ex}}Substituting in x=2y, we get\phantom{\rule{0ex}{0ex}}x=2×2\phantom{\rule{0ex}{0ex}}x=4\phantom{\rule{0ex}{0ex}}Now, \frac{{x}^{2}+{y}^{2}}{5} = \frac{16+4}{5}= \frac{20}{5}= 4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}
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Trigonometric Parametrization of an Ellipse - Maple Help
Home : Support : Online Help : Tasks : Algebra : Trigonometric Parametrization of an Ellipse
Trigonometric Parametrization of an Ellipse
For an ellipse defined by a general quadratic equation in
x
y
, the trigonometric parametrization is obtained. The conic determined by this equation is graphed, and so is the parametrized version.
The graphs serve as validation of the parametrization - if the graph of the original quadratic and the graph of the parametric form coincide, then that is taken as evidence that the parametrization is correct.
Quadratic in
x
y
x\left(t\right)=
y\left(t\right)=
Student[Precalculus][ConicsTutor], geometry[ellipse], geometry[detail]
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List the various data abstraction concepts and the corresponding modeling
List the various data abstraction concepts and the corresponding modeling concepts in the EER model.
Here is a list of data abstraction concepts and their variants in EER model:
* Classification and instantiation - The EER variant of classification is represented by classifying entities into entity types based on their attributes and relationships. Relationship instances becomerelationship types. Forthe need offurther classification, we introduce superclass /subclass concepts and sometimes even categories. In EER model, there is no handling of class exceptions and no showing of examples (instances).
* Identification - EER model implements identification hy sticking to the rule of name uniqueness. Every entity type, relationship type, attribute, subclass, superclass and category has to have unique name that no other object in database has.
* Specialization and Generalization - The concept of specialization and generalization in the EER model is represented by subclasses. They are more specific objects than their respective superclasses and arerelated to them by IS-A-SUBCLASS-OF (or shortly IS-A) relationship.
* Aggregation and Association - EER schema uses aggregation abstraction in form of creating new compositeobjects. We can aggregate attributes to createa composite attribute, entity type or even a relationship type if there’s need to. Association in EER modeling is simply a relationship type between two or more independent entity types.
y=3.5x+2.8
\frac{\sqrt{7}}{\sqrt{28}}
To solve: The problem by modeling and solving a system of linear equations.
a) To write: A binomial that represents the height of the projection.
Given information: Height of wall is xft. and width of projection is
\left(x-3\right)ft
. Area is represented by
{x}^{2}-8x+15
b) To find: Perimeter of projection when height of the wall is 8ft.
Given information: Height of wall is
x=8ft
and width of projection is
\left(x-3\right)ft
. Height of projection is
\left(x-5\right)ft.
Modeling situations as functions is common. Describe whether each description represents a function. Explain.
a. The input is a day of the year. The output is the average temperature in Barcelona on that day.
b. The input is speed of a car. The output is the time it takes for a car moving constantly at that speed to travel 100 miles.
c. The input is a positive number. The output is a number whose absolute value is the input.
d. The input is a year. The output is the population of the United States during that year.
a) To calculate: A model of the form
N\left(t\right)=a+b\mathrm{ln}t
for the provided data.
b) To calculate: The sales in a 7 weeks.
c) The validity of the statement with explanation.
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Linear Programming Problem (LPP) Formulation with Numericals | Education Lessons
To formulate Linear Programming Problem (LPP) for given statement type problems(numerical) is easy if we go through its Mathematical Model.
General Mathematical Modelling of LP
As explained in the video, mathematical model of LP consists of the following:
i) Objective Function (denote with “Z”)
ii) Decision variables (represented in terms of “x”)
iii) Constraints
Note: To know more about above terms you can go through the notes of the Basics of Linear programming, link to the notes- click here
Now, let us generate the mathematical model of the LPP:
Decision variables be
x_1, x_2, x_3, x_4,…, x_n
with total ‘m’ constraints.
Now, LP can be formulate as:
\small \text {Optimize (Maximize or Minimize)} \space \space Z= c_1x_1 + c_2x_2 + c_3x_3 +\dotso+ c_nx_n \\ \begin{aligned} \text {Subject to linear constraints,} \quad a_{11}x_1 + a_{12}x_2 + a_{13}x_3 + \dotso + a_{1n}x_n \quad &(≤ , = , ≥) \quad b_1 \\ \quad a_{21}x_1 + a_{22}x_2 + a_{23}x_3 +\dotso+ a_{2n}x_n\quad &(≤ , = , ≥) \quad b_2 \\ \quad a_{31}x_1 + a_{32}x_2 + a_{33}x_3 +\dotso+ a_{3n}x_n \quad &(≤ , = , ≥) \quad b3\\ \quad \text upto \dotso a_{m1}x_1 + a_{m2}x_2 + a_{m3}x_3 + \dotso+ a_{1n}x_n \quad &(≤ , = , ≥) \quad b_m \\ \text and \quad x_1, x_2, x_3, x_4,\dotso, x_n ≥ 0 \end{aligned}
Thus, we will have any of the three conditions applied on constraints as per the provided problem
(≤ , = , ≥)
Steps for LPP Formulation
Identify the decision variables:
It is the most important step on LPP Formulation, because based on the decision variables only, the whole problem governs. We will learn about, how to identify decision variables from given numerical/problem in this note only. (Also you can go through video for the same. Link provided in cover page)
Identify Objective Function (Z) and express it as a linear function of decision variables:
As we have seen above in mathematical model, objective of any problem can be maximize or minimize. Based on that, we will have to define the objective function in terms of decision variables. (For eg: Maximize
Z= 2x_1 + 5x_2 + 9x_3
Identify all constraints and express it as linear inequalities or equalities in terms of decision variables:
As we know that, in the real world all the resources are limited and so in each particular problem you will find some limitation/constraint on the use of available resources.
Express decision variables as Feasible variables:
It means that, we have lastly define all the decision variables are greater than or equal to zero (For eg:
\ x_1, x_2, x_3 ≥ 0
Let’s move to numerical now,
Numerical 1: Two products ‘A’ and ‘B’ are to be manufactured. Single unit of ‘A’ requires 2.4 minutes of punch press time and 5 minutes of assembly time, while single unit of ‘B’ requires 3 minutes of punch press time and 2.5 minutes of welding time. The capacity of punch press department, assembly department and welding department are 1200 min/week, 800 min/week and 600 min/week respectively. The profit from ‘A’ is ₹60 and from ‘B’ is ₹70 per unit. Formulate LPP such that, profit is maximized.
Tip: Always relate/treat LPP with real life problems for better understanding and ease of solution
Here, the decision has to be taken for how much quantities of product ‘A’ and ‘B’ to be manufactured in order to maximize the profit. (Quantities of product is governing this problem)
Quantity of product ‘A’ manufactured =
x_1
Quantity of product ‘B’ manufactured =
x_2
Step 2: Identify Objective Function (Z)
Here, the objective is to maximize the profit from product ‘A’ and ‘B’. Each unit produced product ‘A’ gives profit of ₹60, while ‘B’ gives ₹70. So, the objective function can be defined in terms of decision variables as:
\text Maximize \quad Z = 60x_1 + 70x_2
Step 3: Identify all the constraints
Product ‘A’ manufactured by: Punch press and Assembly
Product ‘B’ manufactured by: Punch press and Welding
So, we have a total of three processes for the manufacturing of two products A and B.
Thus, there are total three constraints.
Now, we will bifurcate this problem based on processes to get a clear idea of constraints on resources, as follows:
Product ‘A’ requires 2.4 min and product ‘B’ requires 3 min of Punch press time.
Also, punch press time is limited to 1200 min/week.
\therefore 2.4x_1 + 3x_2≤ 1200
Product ‘A’ requires 5 min of Assembly time.
Also, assembly time is limited to 800 min/week.
\therefore 5x_1≤ 800
Product ‘B’ requires 2.5 min of Welding time.
Also, welding time is limited to 600 min/week.
\therefore 2.5x_1≤ 600
Step 4: Express decision variables as feasible variables
We assume that the decision variables has feasible solution, that implies - decision variables are greater than or equal to zero.
x_1, x_2 ≥ 0
So, complete LP can be written as,
\text {Maximize Z} = 60x_1 + 70x_2 \\ \begin{aligned} \text {Subject to } 2.4x_1 + 3x_2 &≤ 1200 \\ 5x1 &≤ 800 \\ 2.5x1 &≤ 600 \\ \text And \ \ x_1, x_2 &≥ 0 \end{aligned}
Note that, here the solution provided (in above numerical) to you with steps is just for explanation. You can always skip this step wise solution to LP problems and can use direct method as provided in video of this topic (click here to watch video).
Numerical 2: A tailor prepares two kinds of dresses. First kind of dress is having raw material cost of ₹150 and labour cost of ₹80, while the second kind of dress is raw material cost of ₹250 and labour cost of ₹100. He can sell the dresses of first and second kind at the rate of ₹300 and ₹500 respectively. First kind of dress takes 2 hours and second kind takes 2.5 hours of cutting. The total labour hours are restricted to 84 hours/week. Also, first kind of dress takes 1.5 hours of stitching and second requires 2 hours of stitching. The stitching hours are restricted to 60 hours/week. Formulate LPP, such that maximum profit can be earned by tailor.
The solution of this numerical will be such that, you can present this in your examination. This will save your time and gives complete idea of your explanation too in examination. All the explanation stuff will be provided in {"Explanation"} brackets for your understanding only, in the solution of this problem.
Number of first kind of dress =
x_1
x_2
{The decision variables selected here are the units of dresses of both kind so produced, as we are going to maximize the profit of tailor. Also, by understanding the whole problem you can easily make out that, to increase the profit, tailor needs to produce more goods.}
Now, Net Profit earned by selling a unit of first kind of dress
= ₹(300-150-80) = ₹70
Net Profit earned by selling a unit of first kind of dress
= ₹(500-250-100) = ₹150
{As you can understand that the problem provided here is based on revenue so generated by selling different kinds of dresses. Now, revenue will be the profit generated after excluding the raw material and manufacturing cost on each of these dresses. So, for example you can see above we have excluded raw material cost of first kind of dress ₹150 and labour cost ₹80 from total selling cost of ₹300.
Thus revenue/net profit generated by selling each unit of first kind of dress is ₹ (300-150-80) = ₹70 and the same for the second kind of dress.}
∴ The objective function can be defined as;
\text {Maximize} \quad Z = 70x_1 + 150x_2
Now, constraints are provided on labour hours for cutting and stitching. Thus, LP is subject to the following constraints;
\begin{aligned} 2x_1 + 2.5x_2 &≤ 84 \\ 1.5x_1 + 2x_2 &≤ 60 \\ \text {And} \ \ x_1, x_2 &≥ 0 \end{aligned}
{As we can find from problem provided, the cutting and stitching hour requirements for both kinds of dresses. Also total hours are restricted on cutting and stitching is 84 hours/week and 60 hours/week respectively.}
\text {Maximize Z}= 70x_1 + 150x_2 \\ \begin{aligned} \text {Subject to} \ \ 2x_1 + 2.5x_2 &≤ 84 \\ 1.5x_1 + 2x_2 &≤ 60 \\ \text {And} \quad x_1, x_2 &≥ 0 \end{aligned}
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2.1 Absolutely Continuous <==> Indefinite Integral
2.2 Apply Inequalities,Sum over n, and Use Hypothesis
3.2 Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem
3.3 Justification for Lebesgue Dominated Convergence Theorem
4 Solution 1c
6.1 Check Criteria for Lebesgue Dominated Convergence Theorem
6.1.1 g_n dominates hat{f}_n
6.1.2 g_n converges to g a.e.
6.1.3 integral of g_n converges to integral of g =
6.1.4 hat{f_n} converges to hat{f} a.e.
6.2 Apply LDCT
8.1 Show that g(x)=|x|^p is Lipschitz
8.2 Apply definitions to g(f(x))
10.1 f(x)= x^4sin^2(\frac{1}{x^2}) is Lipschitz (and then AC)
10.2 |f|^{1/2} is not of bounded variation (and then is not AC)
{\displaystyle \{f_{n}\}\!\,}
is a sequence of absolutely continuous functions defined on
{\displaystyle [0,1]\!\,}
{\displaystyle f_{n}(0)=0\!\,}
{\displaystyle n\!\,}
{\displaystyle \sum _{n=1}^{\infty }\int _{0}^{1}|f_{n}^{\prime }(x)|dx<+\infty \!\,}
{\displaystyle x\in [0,1]\!\,}
{\displaystyle \sum _{n=1}^{\infty }f_{n}(x)\!\,}
converges for each
{\displaystyle x\in [0,1]\!\,}
pointwise to a function
{\displaystyle f\!\,}
{\displaystyle f\!\,}
{\displaystyle [0,1]\!\,}
{\displaystyle f^{\prime }(x)=\sum _{n=1}^{\infty }f_{n}^{\prime }(x)\quad a.e.\,\,x\in [0,1]\!\,}
Absolutely Continuous <==> Indefinite IntegralEdit
{\displaystyle f_{n}(x)\!\,}
is absolutely continuous if and only if
{\displaystyle f_{n}(x)\!\,}
can be written as an indefinite integral i.e. for all
{\displaystyle x\in [0,1]\!\,}
{\displaystyle {\begin{aligned}f_{n}(x)-\underbrace {f_{n}(0)} _{0}&=\int _{0}^{x}f_{n}^{\prime }(t)dt\\f_{n}(x)&=\int _{0}^{x}f_{n}^{\prime }(t)dt\end{aligned}}}
Apply Inequalities,Sum over n, and Use HypothesisEdit
{\displaystyle x_{0}\in [0,1]\!\,}
be given. Then,
{\displaystyle {\begin{aligned}f_{n}(x_{0})&=\int _{0}^{x_{0}}f_{n}^{\prime }(t)dt\\&\leq \int _{0}^{x_{0}}|f_{n}^{\prime }(t)|dt\\&\leq \int _{0}^{1}|f_{n}^{\prime }(t)|dt\end{aligned}}}
{\displaystyle f_{n}(x_{0})\leq \int _{0}^{1}|f_{n}^{\prime }(t)|dt\!\,}
Summing both sides of the inequality over
{\displaystyle n\!\,}
and applying the hypothesis yields pointwise convergence of the series
{\displaystyle f_{n}\!\,}
{\displaystyle \sum _{n=1}^{\infty }f_{n}(x_{0})\leq \sum _{n=1}^{\infty }\int _{0}^{1}|f_{n}^{\prime }(t)|dt<+\infty \!\,}
{\displaystyle f(x)=\sum _{n=1}^{\infty }f_{n}(x)\!\,}
{\displaystyle f(x)=\int _{0}^{x}\sum _{n=1}^{\infty }f_{n}^{\prime }(t)dt\!\,}
Rewrite f(x) and Apply Lebesgue Dominated Convergence TheoremEdit
{\displaystyle {\begin{aligned}f(x)&=\sum _{n=1}^{\infty }f_{n}(x)\\&=\sum _{n=1}^{\infty }\int _{0}^{x}f_{n}^{\prime }(t)dt\quad {\mbox{ (since }}f_{n}{\mbox{ is absolutely continuous)}}\\&=\lim _{n\rightarrow \infty }\sum _{k=1}^{n}\int _{0}^{x}f_{k}^{\prime }(t)dt\\&=\lim _{n\rightarrow \infty }\int _{0}^{x}\sum _{k=1}^{n}f_{k}^{\prime }(t)dt\\&=\int _{0}^{x}\lim _{n\rightarrow \infty }\sum _{k=1}^{n}f_{k}^{\prime }(t)dt\quad {\mbox{ (by LDCT)}}\\&=\int _{0}^{x}\sum _{n=1}^{\infty }f_{n}^{\prime }(t)dt\end{aligned}}}
Justification for Lebesgue Dominated Convergence TheoremEdit
{\displaystyle {\begin{aligned}|\sum _{k=1}^{n}f_{k}^{\prime }(t)|&\leq \sum _{k=1}^{n}|f_{k}^{\prime }(t)|\\&\leq \underbrace {\sum _{n=1}^{\infty }|f_{n}^{\prime }(t)|} _{g(t){\mbox{ dominating function}}}\\\\\\\int _{0}^{1}\sum _{n=1}^{\infty }|f_{n}^{\prime }(x)|dx&=\sum _{n=1}^{\infty }\int _{0}^{1}|f_{n}^{\prime }(x)|dx\quad {\mbox{ (by Tonelli Theorem)}}\\&<\infty \quad {\mbox{ (by hypothesis) }}\end{aligned}}}
{\displaystyle g(t)\!\,}
The above inequality also implies
{\displaystyle \sum _{n=1}^{\infty }|f_{n}^{\prime }(x)|<\infty \!\,}
{\displaystyle [0,1]\!\,}
{\displaystyle |\sum _{k=1}^{n}f_{k}^{\prime }(t)|\rightarrow |\sum _{k=1}^{\infty }f_{k}^{\prime }(t)|\!\,}
{\displaystyle [0,1]\!\,}
to a finite value.
Solution 1cEdit
{\displaystyle f(x)=\int _{0}^{x}\sum _{n=1}^{\infty }f'_{n}(t)dt\!\,}
, by the Fundamental Theorem of Calculus
{\displaystyle f'(x)=\sum _{n=1}^{\infty }f'_{n}(x)\!\,}
{\displaystyle x\in [0,1]\!\,}
{\displaystyle \{f_{n}\}\!\,}
is a sequence of nonnegative integrable functions such that
{\displaystyle f_{n}\rightarrow f\!\,}
a.e., with
{\displaystyle f\!\,}
integrable, and
{\displaystyle \int _{R}f_{n}\rightarrow \int _{R}f\!\,}
{\displaystyle \int _{R}|f_{n}-f|\rightarrow 0\!\,}
Check Criteria for Lebesgue Dominated Convergence TheoremEdit
{\displaystyle {\hat {f}}_{n}=|f-f_{n}|\!\,}
{\displaystyle g_{n}=f+f_{n}\!\,}
g_n dominates hat{f}_nEdit
{\displaystyle f_{n}\!\,}
is positive, then so is
{\displaystyle f\!\,}
{\displaystyle |f_{n}|=f_{n}\!\,}
{\displaystyle |f|=f\!\,}
{\displaystyle |{\hat {f}}_{n}|=|f-f_{n}|\leq |f|+|f_{n}|=f+f_{n}=g_{n}\!\,}
g_n converges to g a.e.Edit
{\displaystyle g=2f\!\,}
{\displaystyle f_{n}\rightarrow f\!\,}
{\displaystyle f+f_{n}\rightarrow 2f\!\,}
{\displaystyle g_{n}\rightarrow g\!\,}
integral of g_n converges to integral of g =Edit
{\displaystyle {\begin{aligned}\int g&=\int (f+f)\\&=2\int f\\\\\\\lim _{n\rightarrow \infty }\int g_{n}&=\lim _{n\rightarrow \infty }\int (f+f_{n})\\&=\int f+\lim _{n\rightarrow \infty }\int f_{n}\\&=\int f+\int f{\mbox{ (from hypothesis) }}\\&=2\int f\end{aligned}}}
{\displaystyle \int g=\lim _{n\rightarrow \infty }\int g_{n}\!\,}
hat{f_n} converges to hat{f} a.e.Edit
{\displaystyle f_{n}\rightarrow f\!\,}
{\displaystyle |f_{n}-f|\rightarrow 0\!\,}
{\displaystyle {\hat {f_{n}}}\rightarrow 0={\hat {f}}\!\,}
Apply LDCTEdit
Since the criteria of the LDCT are fulfilled, we have that
{\displaystyle \lim _{n}\int {\hat {f_{n}}}=\int {\hat {f}}=0\!\,}
{\displaystyle \lim _{n}\int |f-f_{n}|=0\!\,}
Show that i{\displaystyle f\!\,}
{\displaystyle [0,1]\!\,}
{\displaystyle p>1\!\,}
{\displaystyle |f|^{p}\!\,}
{\displaystyle [0,1]\!\,}
Show that g(x)=|x|^p is LipschitzEdit
Consider some interval
{\displaystyle I=[\alpha ,\beta ]\!\,}
{\displaystyle x\!\,}
{\displaystyle y\!\,}
be two points in the interval
{\displaystyle I\!\,}
{\displaystyle K=\|g(x)\|_{\infty }\!\,}
{\displaystyle x\in I\!\,}
{\displaystyle {\begin{aligned}||x|^{p}-|y|^{p}|&=||x|-|y||(|x|^{p-1}+|x|^{p-2}|y|+\ldots +|x||y|^{p-2}+|y|^{p-1})\\&\leq |K^{p-1}+K^{p-2}K+\ldots +KK^{p-2}+K^{p-1}|||x|-|y||\\&=\underbrace {pK^{p-1}} _{M}||x|-|y||\\&\leq M|x-y|\end{aligned}}}
{\displaystyle g(x)\!\,}
is Lipschitz in the interval
{\displaystyle I\!\,}
Apply definitions to g(f(x))Edit
{\displaystyle f(x)\!\,}
{\displaystyle [0,1]\!\,}
{\displaystyle \epsilon >0\!\,}
{\displaystyle \delta >0\!\,}
{\displaystyle \{(x_{i},x_{i}^{'}\}\!\,}
is a finite collection of nonoverlapping intervals of
{\displaystyle [0,1]\!\,}
{\displaystyle \sum _{i=1}^{n}|x_{i}^{'}-x_{i}|<\delta \!\,}
{\displaystyle \sum _{i=1}^{n}|f(x_{i}^{'})-f(x_{i})|<\epsilon \!\,}
{\displaystyle g\circ f(x)=|f(x)|^{p}\!\,}
{\displaystyle g\!\,}
{\displaystyle {\begin{aligned}\sum _{i=1}^{n}|g(f(x_{i}^{'}))-g(f(x_{i}))|&\leq \sum _{i=1}^{n}M|f(x_{i}^{'})-f(x_{i})|\\&=M\underbrace {\sum _{i=1}^{n}|f(x_{i}^{'})-f(x_{i})|} _{<\epsilon }\\&<M\epsilon \end{aligned}}}
{\displaystyle g\circ f(x)=|f(x)|^{p}\!\,}
is absolutely continuous.
{\displaystyle 0<p<1\!\,}
. Give an example of an absolutely continuous function
{\displaystyle f\!\,}
{\displaystyle [0,1]\!\,}
{\displaystyle |f|^{p}\!\,}
is not absolutely continuous
f(x)= x^4sin^2(\frac{1}{x^2}) is Lipschitz (and then AC)Edit
{\displaystyle f(x)=x^{4}\sin ^{2}({\frac {1}{x^{2}}})\!\,}
. The derivate of f is given by
{\displaystyle f'(x)=4x^{3}\sin ^{2}({\frac {1}{x^{2}}})-2x\sin({\frac {2}{x^{2}}})\!\,}
The derivative is bounded (in fact, on any finite interval), so
{\displaystyle f\!\,}
is Lipschitz.
Hence, f is AC
|f|^{1/2} is not of bounded variation (and then is not AC)Edit
{\displaystyle |f(x)|^{1/2}=x^{2}\left|\sin \left({\frac {1}{x^{2}}}\right)\right|\!\,}
Consider the partition
{\displaystyle \left\{{\sqrt {\frac {2}{n\pi }}}\right\}\!\,}
{\displaystyle \left|f\left({\sqrt {\frac {2}{n\pi }}}\right)\right|^{1/2}={\frac {2}{n\pi }}\left|\sin \left({\frac {n\pi }{2}}\right)\right|\!\,}
Then, T(f) goes to
{\displaystyle \infty \!\,}
{\displaystyle n\!\,}
{\displaystyle \infty \!\,}
{\displaystyle |f|^{1/2}\!\,}
is not of bounded variation and then is not AC
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High School Calculus/Integration by Substitution - Wikibooks, open books for an open world
High School Calculus/Integration by Substitution
Integration by Substitution[edit | edit source]
There is a theorem that will help you with substitution for integration. It is called Change of Variables for Definite Integrals.
what the theorem looks like is this
{\displaystyle \int _{a}^{b}f(x)\operatorname {d} x=\int _{\alpha }^{\beta }f(g(u))g\prime (u)\operatorname {d} u}
{\displaystyle \alpha }
you must plug a into the function g and to get
{\displaystyle \beta }
you must plug b into the function g.
The tricky part is trying to identify what you want to make your u to be. Some times substitution will not be enough and you will have to use the rules for integration by parts. That will be covered in a different section
{\displaystyle \int _{0}^{2}x(x^{2}+1)^{2}\operatorname {d} x}
Instead of making this a big polynomial we will just use the substitution method.
Identify your u
{\displaystyle u=x^{2}+1}
{\displaystyle \operatorname {d} u}
{\displaystyle \operatorname {d} u=2x\operatorname {d} x}
Now we plug in our limits of integration to our u to find our new limits of integration
{\displaystyle x=0,u=0^{2}+1=1}
{\displaystyle x=2,u=2^{2}+1=5}
Now our integration problem looks something like this
{\displaystyle {\frac {1}{2}}\int _{0}^{5}(x^{2}+1)^{2}(2x)\operatorname {d} x}
write your new integration problem
When we plug in our u it looks like
{\displaystyle {\frac {1}{2}}\int _{0}^{5}(u)^{2}\operatorname {d} u}
{\displaystyle {\frac {1}{2}}\left[{\frac {1}{3}}u^{3}\right]_{0}^{5}}
{\displaystyle {\frac {1}{2}}\left[\left({\frac {1}{3}}*5^{3}\right)-\left({\frac {1}{3}}*0^{3}\right)\right]}
{\displaystyle {\frac {1}{2}}\left[{\frac {1}{3}}*125\right]}
{\displaystyle {\frac {1}{2}}\left[{\frac {125}{3}}\right]}
{\displaystyle {\frac {125}{6}}}
As you can see this all simplified fairly nice. Using substitution will be hard, for most people, at first. Once you get the hang of doing this it should come to you faster and faster each time.
I'll give you some other problems to work on as well.
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin(x)\cos(x)\operatorname {d} x}
{\displaystyle \int _{-1}^{2}{\sqrt {x^{2}+4}}(2x)\operatorname {d} x}
{\displaystyle u=\sin(x)}
{\displaystyle \operatorname {d} u=\cos(x)\operatorname {d} x}
{\displaystyle \sin(0)=0}
{\displaystyle x={\frac {\pi }{2}}}
{\displaystyle \sin \left({\frac {\pi }{2}}\right)=1}
{\displaystyle \int _{0}^{1}u\operatorname {d} u}
{\displaystyle \left[{\frac {1}{2}}u^{2}\right]_{0}^{1}}
{\displaystyle \left[{\frac {1}{2}}\sin ^{2}(1)\right]-\left[{\frac {1}{2}}\sin ^{2}(0)\right]}
{\displaystyle ={\frac {1}{2}}\sin ^{2}(1)}
{\displaystyle u=x^{2}+4}
{\displaystyle \operatorname {d} u=2x}
plug in our limits to get new limits
{\displaystyle (-1)^{2}+4=5}
and when x = 2
{\displaystyle 2^{2}+4=8}
Our new integration problem is
{\displaystyle \int _{5}^{8}(u)^{\frac {1}{2}}\operatorname {d} u}
{\displaystyle \left[{\frac {2}{3}}u^{\frac {3}{2}}\right]_{5}^{8}}
{\displaystyle =\left[{\frac {2}{3}}*(x^{2}+4)\right]_{5}^{8}}
{\displaystyle =\left[{\frac {2}{3}}*(8^{2}+4)\right]-\left[{\frac {2}{3}}*(5^{2}+4)\right]}
{\displaystyle =\left[{\frac {136}{3}}\right]-\left[{\frac {58}{3}}\right]}
{\displaystyle =26}
Retrieved from "https://en.wikibooks.org/w/index.php?title=High_School_Calculus/Integration_by_Substitution&oldid=2178799"
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Basics of Program Evaluation and Review Technique (PERT) | Education Lessons
PERT is the technique used to find project completion time of “variable activities”. In PERT, the time is combination of three different time estimations. Following are the three different time estimation:
The optimistic time estimate (to): The minimum time required for the completion of the activity as per the predetermined condition.
The pessimistic time estimate (tp): The maximum time that activity will take under worst condition.
The most likely time estimate (tm): The time an activity will take if executed under normal condition.
Important terms in PERT analysis:
Expected time or average time (te): Since there are three time values available in PERT, average time is to be calculate by following formula:
t_{e} = {t_o + 4t_m + t_p \over 6}
Variance (V): Variance is given by following formula:
V = \Big({t_p - t_o \over 6}\Big)^2
Standard Deviation (
\sigma
): Standard deviation is square root of summation of variance of critical activities. It is given by following formula:
\sigma = \sqrt{V_1 + V_2 + V_3}
Probability of completion of project (z): It is calculated in order to estimate that how many percentages are the chances of completion of project in certain time or given time (t). Let, tcp be the time of completion of project on critical path and t is any certain time or given time, then probability of completion of project in that given time t, is given by:
Z = {t - t_{cp} \over \sigma}
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Adaptive tests for periodic signal detection with applications to laser vibrometry
Fromont, Magalie ; Lévy-Leduc, Céline
Initially motivated by a practical issue in target detection via laser vibrometry, we are interested in the problem of periodic signal detection in a gaussian fixed design regression framework. Assuming that the signal belongs to some periodic Sobolev ball and that the variance of the noise is known, we first consider the problem from a minimax point of view: we evaluate the so-called minimax separation rate which corresponds to the minimal
{l}_{2}-
distance between the signal and zero so that the detection is possible with prescribed probabilities of error. Then, we propose a testing procedure which is available when the variance of the noise is unknown and which does not use any prior information about the smoothness degree or the period of the signal. We prove that it is adaptive in the sense that it achieves, up to a possible logarithmic factor, the minimax separation rate over various periodic Sobolev balls simultaneously. The originality of our approach as compared to related works on the topic of signal detection is that our testing procedure is sensitive to the periodicity assumption on the signal. A simulation study is performed in order to evaluate the effect of this prior assumption on the power of the test. We do observe the gains that we could expect from the theory. At last, we turn to the application to target detection by laser vibrometry that we had in view.
Mots clés : periodic signal detection, adaptive test, minimax separation rates, nonparametric regression
author = {Fromont, Magalie and L\'evy-Leduc, C\'eline},
title = {Adaptive tests for periodic signal detection with applications to laser vibrometry},
AU - Fromont, Magalie
AU - Lévy-Leduc, Céline
TI - Adaptive tests for periodic signal detection with applications to laser vibrometry
Fromont, Magalie; Lévy-Leduc, Céline. Adaptive tests for periodic signal detection with applications to laser vibrometry. ESAIM: Probability and Statistics, Tome 10 (2006), pp. 46-75. doi : 10.1051/ps:2006002. http://www.numdam.org/articles/10.1051/ps:2006002/
[1] Y. Baraud, Non-asymptotic minimax rates of testing in signal detection. Bernoulli 8 (2002) 577-606. | Zbl 1007.62042
[2] Y. Baraud, S. Huet, and B. Laurent, Adaptive tests of linear hypotheses by model selection. Ann. Statist. 31 (2003) 225-251. | Zbl 1018.62037
[3] L. Birgé, An alternative point of view on Lepski's method, in State of the Art in Probability and Statistics (Leiden, 1999), 113-133, IMS Lecture Notes Monogr. Ser. 36 (2000).
[4] P.J. Brockwell and R.A. Davis, Time series: theory and methods. Springer Series in Statistics. Springer-Verlag, New York, second edition (1991). | MR 1093459 | Zbl 0709.62080
[5] R. Eubank and J. Hart, Testing goodness-of-fit in regression via order selection criteria. Ann. Stat. 20 (1992) 1412-1425. | Zbl 0776.62045
[6] J. Fan and Q. Yao, Nonlinear Time series. Springer series in Statistics. Springer-Verlag, New York, Nonparametric and parametric methods (2003). | MR 1964455 | Zbl 1014.62103
[7] G. Gayraud and C. Pouet, Minimax testing composite null hypotheses in the discrete regression scheme. Math. Methods Stat. 10 (2001) 375-394. | Zbl 1005.62048
[8] P. Gregory and T. Loredo, A new method for the detection of a periodic signal of unknown shape and period. The Astrophysical J. 398 (1992) 146-168.
[9] W. Härdle and A. Kneip, Testing a regression model when we have smooth alternatives in mind. Scand. J. Stat. 26 (1999) 221-238. | Zbl 0934.62043
[10] J. Horowitz and V. Spokoiny, An adaptive, rate-optimal test of a parametric mean-regression model against a nonparametric alternative. Econometrica 69 (2001) 599-631. | Zbl 1017.62012
[11] Y. Ingster, Minimax nonparametric detection of signals in white Gaussian noise. Probl. Inf. Transm. 18 (1982) 130-140. | Zbl 0499.94002
[12] Y. Ingster, Asymptotically minimax testing for nonparametric alternatives I-II-III. Math. Methods Statist. 2 (1993) 85-114, 171-189, 249-268. | Zbl 0798.62059
[14] M. Lavielle and C. Lévy-Leduc, Semiparametric estimation of the frequency of unknown periodic functions and its application to laser vibrometry signals. IEEE Trans. Signal Proces. 53 (2005) 2306-2314.
[15] O. Lepski and V. Spokoiny, Minimax nonparametric hypothesis testing: The case of an inhomogeneous alternative. Bernoulli 5 (1999) 333-358. | Zbl 0946.62050
[16] O. Lepski and A. Tsybakov, Asymptotically exact nonparametric hypothesis testing in sup-norm and at a fixed point. Probab. Theory Relat. Fields 117 (2000) 17-48. | Zbl 0971.62022
[17] M. Prenat, Vibration modes and laser vibrometry performance in noise, in Proceedings of the Physics in Signal and Image Processing conference (PSIP'01), 23-24 janvier 2001, Marseille, France (2001).
[18] B.G. Quinn and E.J. Hannan, The estimation and tracking of frequency. Cambridge Series in Statistical and Probabilistic Mathematics. Cambridge University Press, Cambridge (2001). | MR 1813156 | Zbl 0969.62060
[19] V. Spokoiny, Adaptive hypothesis testing using wavelets. Ann. Stat. 24 (1996) 2477-2498. | Zbl 0898.62056
[20] V. Spokoiny, Adaptive and spatially adaptive testing of a nonparametric hypothesis. Math. Methods Stat. 7 (1998) 245-273. | Zbl 1103.62345
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The Discovery of the Number e | Brilliant Math & Science Wiki
The Discovery of the Number e
Andrew Ellinor, Patrick Corn, Geoff Pilling, and
Compound Interest and
e
e
^\text{th}
century was a time of rapid change. It was the era of the scientific revolution, the proliferation of colonialism, the emergence of mass literacy, and an explosion of international trade.The European age of exploration (and exploitation) brought disparate cultures of the world in contact, conflict, and business with each other to a degree that none of the large empires of old ever approached. Increasing the scale of commerce increased the demands for capital. Money lending began to play an increasingly large role in the prosperity of individuals, business ventures, and nations.
Given the growing presence of finance in the 17
^\text{th}
century, historians believe that the first person to calculate
was most likely a banker or trader exploring the properties of compound interest.
Interest is a fee charged by a lender on a borrower for the service of providing a loan. Interest fees offset the opportunity cost to the lender of not being able to do anything else with their assets while they are controlled by the borrower. These interest fees accumulate over time, depending on the interest rate. Borrowers think of interest as the cost of having debt, while lenders think of interest as the return on their loan as an investment.
Simple interest is where the interest accumulated is always the same proportion of the initial amount borrowed or invested (this initial amount is called the principal). For instance, if you borrow $1.00 at 5% interest per year, after one year, you will owe $1.05. After two years you will owe $1.10, etc. In effect, simple interest creates an arithmetic progression where a debt or investment grows at the same rate over all periods of time. Simple interest is rare and usually appears only in short term loans.
Compound interest is where interest accumulates on both the principal and the prior interest. Each time the interest rate is applied to the total accumulated debt or investment, it is referred to as compounding. If you invest $1 at 5% annual compounding interest, after the first year you would have $1.05. Unlike simple interest, in the second year it would increase by 5% of $1.05, yielding $1.1025. The general equation for compound interest, compounded once annually, is
S=P(1+r)^{t},
S
is the total accumulated debt or investment,
P
is the principal,
r
t
is the time elapsed in years. Under compound interest, debts and investments grow by a geometric progression.
Interest can be compounded multiple times within a given year or interest rate period. For instance, if you put one dollar into a bank account that returns 5% interest per year compounded biannually, then your account will grow by 2.5% twice in a given year. For multiple compounding intervals within each rate period, the formula for compound interest becomes
S=P \left( 1+\frac{r}{n} \right)^{nt},
n
is the number of times the interest is compounded in a given year and
nt
is therefore the total number of times it is compounded.
Compounding more times in a given time period causes your debt or investment to grow more often, but at a smaller rate each time it is compounded. As you can see, it would not take much imagination for an ambitious banker to wonder how much money could be made if the interest rate was really high and it was compounded as much as possible (daily, hourly, infinitely, etc.). If more compounding intervals make investments grow more often, would more compounding result in investments growing faster?
Historians believe that a businessman or banker likely beats the mathematicians to the thought experiment: if $1 is invested at 100% interest over one year, how much more money will be made if it is compounded often as opposed to only compounding it a few times in a year?
The equation for compound interest in this special case would simplify to
S= \left(1+\frac{1}{n} \right)^{n}.
It turns out that compounding weekly barely yields any more money than compounding monthly and at higher values of
n
, it gets closer and closer to what we recognize as the number
e
. As you can see, compounding more often does yield more money up to a point, but rapidly reaches an upper bound where growing more often does not yield faster growth. Mathematicians (led by Jacob Bernoulli, in 1683) would later go on to define
as the limit
\lim_{n \to \infty} \left( 1+1/n \right)^{n}=e.
n
going to infinity, growth occurs continuously, at every possible instant no matter how small the time interval. Simple interest reflects an arithmetic growth progression, compound interest reflects a discrete geometric growth progression, and infinitely compounding interest reflects a continuous growth progression (exponential growth). The number
e
is thought of as the base that represents the growth of processes or quantities that grow continuously in proportion to their current quantity. This is why
e
appears so often in modeling the exponential growth or decay of everything from bacteria to radioactivity.
Cite as: The Discovery of the Number e. Brilliant.org. Retrieved from https://brilliant.org/wiki/the-discovery-of-the-number-e/
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Robustness of Servo Controller for DC Motor - MATLAB & Simulink Example - MathWorks Italia
Data Structures for Uncertainty Modeling
DC Motor Example with Parameter Uncertainty and Unmodeled Dynamics
Electrical and Mechanical Equations
Uncertain Model of DC Motor
Robustness of Disturbance Rejection Characteristics
This example shows how to use uncertain objects in Robust Control Toolbox™ to model uncertain systems and assess robust stability and robust performance using the robustness analysis tools.
Robust Control Toolbox lets you create uncertain elements, such as physical parameters whose values are not known exactly, and combine these elements into uncertain models. You can then easily analyze the impact of uncertainty on the control system performance.
For example, consider a plant model
P\left(s\right)=\frac{\gamma }{\tau s+1}
where gamma can range in the interval [3,5] and tau has average value 0.5 with 30% variability. You can create an uncertain model of P(s) as in this example:
gamma = ureal('gamma',4,'range',[3 5]);
P = tf(gamma,[tau 1])
gamma: Uncertain real, nominal = 4, range = [3,5], 1 occurrences
tau: Uncertain real, nominal = 0.5, variability = [-30,30]%, 1 occurrences
Suppose you have designed an integral controller C for the nominal plant (gamma=4 and tau=0.5). To find out how variations of gamma and tau affect the plant and the closed-loop performance, form the closed-loop system CLP from C and P.
CLP = feedback(P*C,1)
CLP =
Type "CLP.NominalValue" to see the nominal value, "get(CLP)" to see all properties, and "CLP.Uncertainty" to interact with the uncertain elements.
Plot the step response of the plant and closed-loop system. The step command automatically generates 20 random samples of the uncertain parameters gamma and tau and plots the corresponding step responses.
subplot(2,1,1); step(P), title('Plant response (20 samples)')
subplot(2,1,2); step(CLP), title('Closed-loop response (20 samples)')
Figure 1: Step responses of the plant and closed-loop models
The bottom plot shows that the closed-loop system is reasonably robust despite significant fluctuations in the plant DC gain. This is a desirable and common characteristic of a properly designed feedback system.
This example builds on the example Reference Tracking of DC Motor with Parameter Variations by adding parameter uncertainty and unmodeled dynamics, to investigate the robustness of the servo controller to such uncertainty.
The nominal model of the DC motor is defined by the resistance R, the inductance L, the emf constant Kb, armature constant Km, the linear approximation of viscous friction Kf and the inertial load J. Each of these components varies within a specific range of values. The resistance and inductance constants range within ±40% of their nominal values. Use ureal to construct these uncertain parameters.
For physical reasons, the values of Kf and Kb are the same, even if they are uncertain. In this example, the nominal value is 0.015 with a range between 0.012 and 0.019.
Km = K;
Kb = K;
Viscous friction, Kf, has a nominal value of 0.2 with a 50% variation in its value.
The current in the electrical circuit, and the torque applied to the rotor can be expressed in terms of the applied voltage and the angular speed. Create the transfer function H relating these variables, and make AngularSpeed an output of H for later use.
H = [1;0;Km] * tf(1,[L R]) * [1 -Kb] + [0 0;0 1;0 -Kf];
H.InputName = {'AppliedVoltage';'AngularSpeed'};
H.OutputName = {'Current';'AngularSpeed';'RotorTorque'};
The motor typically drives an inertia, whose dynamic characteristics relate the applied torque to the rate-of-change of the angular speed. For a rigid body, this value is a constant. A more realistic, but uncertain, model might contain unknown damped resonances. Use the ultidyn object to model uncertain linear time-invariant dynamics. Set the nominal value of the rigid body inertia to 0.02 and we include 15% dynamic uncertainty in multiplicative form.
J = 0.02*(1 + ultidyn('Jlti',[1 1],'Type','GainBounded','Bound',0.15,...
'SampleStateDim',4));
It is a simple matter to relate the AngularSpeed input to the RotorTorque output through the uncertain inertia, J, using the lft command. The AngularSpeed input equals RotorTorque/(J*s). Therefore, use "positive" feedback from the third output to the second input of H to make the connection. This connection results in a system with one input (AppliedVoltage) and two outputs (Current and AngularSpeed).
Pall = lft(H,tf(1,[1 0])/J);
Select only the AngularSpeed output for the remainder of the control analysis.
P = Pall(2,:)
Jlti: Uncertain 1x1 LTI, peak gain = 0.15, 1 occurrences
K: Uncertain real, nominal = 0.015, range = [0.012,0.019], 2 occurrences
Kf: Uncertain real, nominal = 0.2, variability = [-50,50]%, 1 occurrences
L: Uncertain real, nominal = 0.5, variability = [-40,40]%, 1 occurrences
P is a single-input, single-output uncertain model of the DC motor. For analysis purposes, use the following controller.
Cont = tf(84*[.233 1],[.0357 1 0]);
First, compare the step response of the nominal DC motor with 15 samples of the uncertain model of the DC motor. Use usample to explicitly specify the number of random samples.
step(usample(P,15),P.NominalValue,3)
legend('Samples','Nominal')
Figure 2: Plant step response
Similarly, compare the Bode response of the nominal (red) and sampled (blue) uncertain models of the DC motor.
bode(usample(P,15),P.NominalValue);
Figure 3: Plant Bode response
In this section, analyze the robustness of the DC motor controller. A nominal analysis of the closed-loop system indicates the feedback loop is very robust with 22 dB gain margin and 66 deg of phase margin.
margin(P.NominalValue*Cont)
Figure 4: Closed-loop robustness analysis
The diskmargin function computes the disk-based gain and phase margins. By modeling gain and phase variations at all frequencies and in all feedback loops, disk margins tend to be more accurate estimates of robustness, especially in multi-loop control systems. Compute the disk-based margins for the DC motor loop.
DM = diskmargin(P.NominalValue*Cont)
While smaller than the classical gain and phase margins, the disk-based margins essentially confirm that the nominal feedback loop is very robust. Now, recall that the DC motor plant is uncertain. How does the modeled uncertainty affect these stability margins? For quick insight, plot the disk-based gain and phase margins for 20 samples of the uncertain open-loop response.
diskmarginplot(P*Cont,P.NominalValue*Cont)
Some combinations of plant uncertainty lead to smaller margins. The plot shows only a small sample. Use worst-case analysis to find out how bad the margins can really get. The wcdiskmargin function directly computes the worst-case gain and phase margins for the modeled uncertainty.
wcDM = wcdiskmargin(P*Cont,'siso')
wcDM = struct with fields:
mag2db(wcDM.GainMargin)
Here the worst-case margins are only 1.2 dB and 7.8 degrees, signaling that the closed loop is nearly unstable for some combinations of the uncertain elements.
The sensitivity function is a standard measure of closed-loop performance for the feedback system. Compute the uncertain sensitivity function S and compare the Bode magnitude plots for the nominal and sampled uncertain sensitivity functions.
S = feedback(1,P*Cont);
bodemag(S,S.Nominal)
Figure 5: Magnitude of sensitivity function S.
In the time domain, the sensitivity function indicates how well a step disturbance can be rejected. Plot its step response to see the variability in disturbance rejection characteristics (nominal appears in red).
step(S,S.Nominal)
Figure 6: Rejection of a step disturbance.
Use the wcgain function to compute the worst-case value of the peak gain of the sensitivity function.
[maxgain,worstuncertainty] = wcgain(S);
With the usubs function you can substitute the worst-case values of the uncertain elements into the uncertain sensitivity function S. This gives the worst-case sensitivity function Sworst over the entire uncertainty range. Note that the peak gain of Sworst matches the lower-bound computed by wcgain.
Sworst = usubs(S,worstuncertainty);
norm(Sworst,inf)
maxgain.LowerBound
Now compare the step responses of the nominal and worst-case sensitivity.
step(Sworst,S.NominalValue,6);
legend('Worst-case','Nominal')
Figure 7: Nominal and worst-case rejection of a step disturbance
Clearly some combinations of uncertain elements significantly degrade the ability of the controller to quickly reject disturbances. Finally, plot the magnitude of the nominal and worst-case values of the sensitivity function. Observe that the peak value of Sworst occurs at the frequency maxgain.CriticalFrequency:
bodemag(Sworst,S.NominalValue)
semilogx(maxgain.CriticalFrequency,20*log10(maxgain.LowerBound),'g*')
Figure 8: Magnitude of nominal and worst-case sensitivity
diskmargin | wcgain | uss | usubs
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Crashing Special Case - Indirect cost less than Crash Cost | Education Lessons
As we have understand the method of crashing in the video, let us move to the special case of Crashing. (If you have not watched the video yet, click on the button above and clear your doubts about the method now.)
In this case of crashing we will have indirect cost incurred in the project less than the minimum cost slope of the critical activity.
We will discuss here how to solve such problem here! And also, learn how to obtain the minimum project duration and optimum minimum cost.
Also, we will understand the whole concept by solving the numerical and comparing it with practical-real life problem to get clear idea.
Let us start by considering the following numerical;
Table below shows the normal duration & cost and crash duration & cost of the various activities in a project. Find the optimum duration and minimum project cost, assuming the indirect cost of the project as Rs. 300/day.
Normal Time(days)
Normal Cost(Rs.)
Crash Time(days)
Crash Cost(Rs.)
1-2 2 800 1 1400
1-3 5 1000 2 2000
1-4 5 1000 3 18000
Let us solve this numerical, starting with Critical Path method (CPM) to obtain project completion time.
If you don't know about Critical Path method (CPM), then check out - CPM Video.
Network Diagram using CPM
We can derive project completion time and critical path from above network diagram.
Also, we will derive project completion cost from the given data and network diagram.
\begin{aligned} \text{Critical Path} &: \ \ 1-3-4-5\\ \text{Project completion time} &= 12 \ \text{days}\\ \ \\ \text{Project completion cost} &= \text{Normal cost} + \text{Indirect cost for 12 days} \\ &= 8900 + (12\ \sdot\ 300)\\ &= \text{Rs.} \ 12,500 \end{aligned}
If you don't know above mentioned terms like - Project Cost, Normal Cost, Crash Cost, Indirect Cost, and also don't know the calculation, then check out - Crashing video.
Now, let us begin with crashing method by finding cost slope from the given data as follows:
In examination you can only copy acivites in the first column and add second column for the cost slope. For better understanding, we are providing whole table again with the addition of the last column of cost slope. Same type examination tactics are always illustrating in videos provided by Education Lessons.
Cost Slope =
{(C_c - C_n) \over (t_n - t_c)}
1-2 2 800 1 1400 600
1-3 5 1000 2 2000 333.33
1-4 5 1000 3 18000 8500
2-4 1 500 1 500 Not possible
2-5 5 1500 3 2100 300
3-4 4 2000 3 3000 1000
Here, in the above table, activites which are generating critical path are highlighted with bold format.
Let us now identify the activity having minimum cost slope from the above highlighted activites.
We found that, acitivity 1-3 is having the minimum cost slope of 333.33.
Cost Slope is the minimum cost one has to pay, if crash any particular acitivity by 1 day.
But wait, we have given indirect cost is Rs. 300/day. So, if we move to step-by-step carshing for that activity, and reduce 1 day (suppose), then also, we will have new cost of project as follows:
Step-by-step crashing:
It means that we have to crash the acitivity in parts. Suppose as we have given in this numerical, for activity 1-3 normal time as 5 days, whereas crash time is 2 days. So, we can crash this activity 3 times in parts by crashing for 1 day at a time. Also, if you crash this activity 3 times, then remaining days for activity completion will be equal to crash time, and so you will not able to crash that activity further.
We are using step-by-step crashing to understand this miscellaneous numerical very easily. You can use direct method as well, as shown in the Crashing video. After completion of this numerical, your mind will be trained to identify such kind of numerical.
New project time & cost if crashed 1-3 by 1 day:
\begin{aligned} \text{Project completion time} &= 11 \ \text{days}\\ \ \\ \text{Project completion cost} &= \text{Cost obtained in last step} - \text{Indirect cost for 1 days} + \text{Cost slope for 1 day} \\ &= 12500 - (1\ \sdot\ 300) + (1 \ \sdot \ 333.33)\\ &= 12500 - 300 + 333.33 \\ &= 12,533.33\\ &> 12,500 \end{aligned}
\begin{aligned} \text{Project completion time} &= 11 \ \text{days}\\ \ \\ \text{Project completion cost} &= \text{Normal cost} + \text{Indirect cost for 11 days} + \text{Cost slope for 1 day} \\ &= 8900 + (11\ \sdot\ 300) + (1 \ \sdot \ 333.33)\\ &= 8900 + 3300 + 333.33 \\ &= 12,533.33\\ &> 12,500 \end{aligned}
So, we can observe that from above calculations, that the after crashing the critical acitivity, with minimum cost slope and by just 1 day (i.e., with least possible parameters); the total project cost will be increased only.
But, by crashing, our main aim is to reduce the project time with minimum cost. However, in this numerical if you crash with minimum parameters required for the crashing, then also project cost will be increased with reduction in duration of the project, and which is not profitable.
Any company or we can say company's project manager, will not plan the work with loss to the company.
So, optimum time and cost for this numerical will be the initially found values by CPM only.
\begin{aligned} \text{Optimum project completion time} &= 12 \ \text{days}\\ \ \\ \text{Optimum project completion cost} &= \text{Normal cost} + \text{Indirect cost for 12 days} \\ &= 8900 + (12\ \sdot\ 300)\\ &= \text{Rs.} \ 12,500 \end{aligned}
Here, we consider that optimum duartion is associated with the minimum cost of the project (relating the problem practically).
But, if we want to find optimum duration with compromising the project cost (i.e. cost will increase), then we can go for crashing, the same way explained in the Crashing video, until all the activities get converted to critical one (ignore stopping when cost increased than the last step).
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Function Arithmetic | Brilliant Math & Science Wiki
Ram Mohith, Beakal Tiliksew, Pi Han Goh, and
When performing arithmetic on functions, it is necessary to understand the rules that relate functions to each other. For the following examples, we'll use these functions:
a(x) = x^2
b(x) = x+3
c(x) = 4x
(f+g)(x) = f(x) + g(x)
, so for example,
(b+c)(x) = b(x) + c(x) = (x+3) + (4x) = 5x + 3 .
(f-g)(x) = f(x) - g(x)
(b-c)(x) = b(x) - c(x) = (x+3) - (4x) = -3x + 3.
( f\cdot g)(x) = f(x) \cdot g(x)
(a\cdot b)(x) = (x^2) \cdot (x+3) = x^3 + 3x^2.
\frac{f}{g}(x)=\frac{f(x)}{g(x)}
\frac{b}{c}(x)=\frac{b(x)}{c(x)}=\frac{x+3}{4x}.
Exponents:
f^n(x) = \big(f(x)\big)^n
n \in \mathbb N
a^5(x) = \big(a(x)\big)^5 = \big(x^2\big)^5 = x^{10}.
Square Root: It is a part of "exponents," but here
n
\sqrt{f} (x) = \sqrt{f(x)}
\sqrt{b} (x) = \sqrt{b(x)} = \sqrt{x + 3}.
Note: The domain of all these combinations is the intersection of the domains of
g
. That is, both functions must be defined at a point of the combination to be defined. Another requirement for division is that the denominator must be non-zero: in this case
g(x)\neq0
(f-g)(3)
f(x) = x^2 + 1
g(x) = 2x-2?
f(3) = 3^2 + 1 = 10
g(3) = 2(3) - 2 = 4
(f-g)(3) = f(3) - g(3) =10 - 4 = 6. \ _\square
(f+g)(4)
f(x) = 5x^{2} + 4
g(x) = 3x+6?
f(4) = 5\big(4^{2}\big)+4 = 84
g(4) = 3(4) +6 = 18
(f+g)(4) = f(4) + g(3) =84 +18 = 102. \ _\square
\begin{aligned} f(x)&=4{ x }^{ 3 }-5\\ g(x)&=6{ x }^{ 2 }+9\\ h(x)&=(f\cdot g)(x), \end{aligned}
h(3)?
h(3)=(f\cdot g)(3)=f(3)\cdot g(3).
f(3)=4(3)^{3}-5=103
g(3)=6(3)^{2}+9=63,
\begin{aligned} h(3)&=f(3)\cdot g(3)\\ &=103\times 63\\ &=6489.\ _\square \end{aligned}
\begin{aligned} f(x)&={ x }^{ 2 }+1\\ g(x)&=6x\\ h(x)&=3x+4, \end{aligned}
\big((f\cdot h)-g\big)(3)?
The expression can be rewritten as
\begin{aligned} \big((f\cdot h)-g\big)(3)&=(f\cdot h)(3)-g(3)\\ &=f(3)\cdot h(3)-g(3). \end{aligned}
f(3)=3^{2}+1=10
g(3)=6(3)=18
h(3)=3(3)+4=13,
f(3)\cdot h(3)-g(3)=10\cdot 13-18=112.\ _\square
In some questions, instead of giving the general form of the function along with its domain and co-domain, the function will be directly given mentioning for what value what output will come. In such a way, the value and its output will be enclosed and expressed as one pair in a pair of parenthesis "
( )
". So, every first term from each pair is taken as domain and the second term from each pair is taken as co-domain. The rules and algebra of functions are applicable here, too, but while applying any arithmetic operation between two functions, remember that the domain will not change and only co-domain will change. For example, if it is given that
f = {(x,a), (y,b), (z,c)},
it means that if we give input as
x
the output will be
a
, if we give input as
y
b,
and if we give input as
z
c,
so if the domain is
{x,y,z}
its co=domain is
{a,b,c}
. The below-given example gives the complete picture on how to do problems related to this way of expressing:
f = \big\{(4,5),(5,6),(6,-4) \big\}
g = \big\{ (4,-4),(6,5),(8,5) \big\},
then find the values of the following:
\ f + g
\ f - g
\ 2f + 4g
\ f + 4
\ fg
\ \frac{f}{g}
\ |f|
\ \sqrt{f}
\ f^2
\ f^3
f = A = \{4,5,6\}
and the domain of
g = B = \{4,6,8\},
and hence the domain of
f \pm g = A \cap B = {4,6}
f + g = \big\{(4,5 - 4),(6,-4 + 5)\big\} = \big\{(4,1), (6,1)\big\}
f - g = \big\{(4,5 + 4), (6, -4 - 5)\big\} = \big\{(4,9),(6,-9)\big\}
2f
4g
will not change but the co-domain will change. As we need
2f,
multiply the second term in each pair by two and do the same for
4g:
2f = \big\{(4,10),(5,12),(6,-8)\big\},\ 4g = \big\{(4,-16),(6,20),(8,20)\big\}.
2f + 4g
f + g = {4,6}:
2f + 4g = \big\{(4,10 - 16), (6, -8 + 20)\big\} = \big\{(4,-6),(6,12)\big\}.
f + 4 = \big\{(4, 5+4),(5,6+4),(6,-4+4)\big\} = \big\{(4,9),(5,10),(6,0)\big\}
f \cdot g = A \cap B = {4,6}
. So, we need only the pairs whose first term is 4 or 6 and only need to change their respective co-domains:
f \cdot g = \left\{\big(4, 5 \times (-4)\big),\big(6,(-4) \times (-5)\big)\right\} = \big\{(4,-20),(6,20)\big\}.
Here too the domain of
\frac{f}{g} = {4,6}
. So, we need to change only those two terms from the domain of each function:
\frac{f}{g} = \left\{\left(4,\dfrac{-5}{4} \right), \left(6,\dfrac{-4}{5} \right)\right\}.
|f| = \left\{\big(4,|5|\big),\big(5,|6|\big),\big(6,|-4|\big)\right\} = \big\{ (4,5),(5,6),(6,4) \big\}
\sqrt{f}
\{4,6\}
\sqrt{-4}
is not defined:
\sqrt{f} = \Big\{\big(4,\sqrt{5}\big),\big(5,\sqrt{6}\big)\Big\}.
f^2 = \Big\{\big(4,5^2\big),\big(5,6^2\big),\big(6,(-4)^2\big)\Big\} = \big\{(4,25),(5,36),(6,16)\big\}
f^3 = \Big\{\big(4,5^3\big),\big(5,6^3\big),\big(6,(-4)^3\big)\Big\} = \big\{(4,125),(5,216),(6,-64)\big\}
_\square
\frac{3}{5}
\frac{6}{5}
\sqrt { 6 }
\sqrt { 8 }
\frac { f\big( \sqrt { 6 } \big) -f\big( \sqrt { 8 } \big) }{ \sqrt { 6 } -\sqrt { 8 } }
f\left( x \right) =\frac { 3 }{ 5 } x+\sqrt { 3 },
f\left( x \right) +f\left( 2x \right) +f\left( 2-x \right) +f\left( 1+x \right) =x
x\in \mathbb R
f(0)
Cite as: Function Arithmetic. Brilliant.org. Retrieved from https://brilliant.org/wiki/functions-arithmetic/
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Why Gravitational Acceleration g is 9.8 m/s² ? [Calculation] | Education Lessons
All Notes / SCIENCE
Earth and other planets attract objects towards itself with some force which depends on the mass and the distance between them. In most of the cases, we say that the object is being attracted towards the planet because of the size and the mass of the object is negligible compared to that of the planet. Since the planet is a huge and a heavy body, it will not get attracted or move towards the object, until and unless the object is another heavy body with a considerable mass. This phenomenon of attraction is governed by Universal Law of Gravitation.
Now, let us understand and calculate with how much acceleration will the object move towards Earth under the effect of the Earth's gravitational force.
Example to understand the gravitational acceleration (g) of the Earth
A person standing on the top of a building drops a stone without applying any force as shown in the figure below. The stone will start moving in downward direction and will perform an accelerated motion towards the surface of the Earth.
Person dropping stone from top of a building
This is because according to Newton's law of gravitation, the force of attraction between any two particles or objects is given by
F = G \times {M \times m \over d^2} \quad ---- (1)
Gravitational force acting on two objects according to universal law of gravitation
Now according to the second law of motion, we know that the acceleration acting on any body is dependent on two things:
The total force acting on the object (in our example, only the gravitational force)
\begin{aligned} &F = m \times a \\ \\ &Where, \\ &F = \text{the force acting on the body} \\ &m = \text{the mass of the body} \\ &a = \text{acceleration with which the body the moving} \\ \end{aligned}
We know that, in our example the stone will move downwards with a force that the Earth attracts the stone with. And the acceleration will be the gravitational acceleration of the Earth, which we denote as g. Hence in the above equation, we can replace a the acceleration with g the gravitational acceleration. Which gives us:
F = m \times g \quad ---- (2)
In our example, this is the force that will act on the stone that was dropped from the building. And the variables will be
\begin{aligned} &M = \text{mass of the Earth} \\ &m = \text{mass of the stone} \\ &G = \text{gravitational constant} \\ &d = \text{the distance between the center of the stone and the center of the Earth} \end{aligned}
The distance between stone and the surface of the Earth is negligible compared to the radius of Earth
From equation (1) and equation (2), we can say that:
\begin{aligned} m \times g &= G \times {M \times m \over d^2} \\ \cancel{m} \times g &= G \times {M \times \cancel{m} \over d^2} \\ g &= G \times {M \over d^2} \end{aligned}
As mentioned earlier, d is the distance between the center of the stone and the center of the Earth. We can replace d with the radius of the Earth R because the distance from the center of the stone to the Earth is negligible compared to the radius of the Earth(as seen in the image above). Hence we get the equation:
g = G \times {M \over R^2} \quad ---- (3)
Now, to find the value of g, we just need to replace the values that we already know in the equation above.
Find the value of gravitational acceleration - step by step calculation
Also watch detailed step-by-step explanation for derivation of value of
g= 9.8 \ m/s^2
in our video.
Tip: The three values shown below are to be remembered along with the unit. This will also help you solve other numerical.
\begin{aligned} &M = \text{mass of the Earth} = 6 \times 10^{24} \thickspace kg \\ &G = \text{gravitational constant} = 6.7 \times 10^{-11} \thickspace N m^2 / kg^2 \\ &R = \text{radius of the Earth} = 6.4 \times 10^6 \thickspace m \end{aligned}
Putting all these values in equation (3):
\begin{aligned} g &= {6.7 \times 10^{-11} \times 6 \times 10^{24} \over (6.4 \times 10^6)^2} \\ \\ &= {6.7 \times 6 \times 10^{24-11} \over 6.4 \times 6.4 \times 10^{6} \times 10^{6}} \\ \\ &\text{removing decimals to simplify the equation} \\ &= {67 \times 6 \times 10^2 \times 10^{13} \over 64 \times 64 \times 10^1 \times 10^{12}} \\ \\ &= {67 \times \cancel{6} \times 10^2 \times 10^{13} \over \cancel{64} \times 64 \times 10^1 \times 10^{12}} \\ \\ &= {67 \times 3 \times 10^{2+13-1-12} \over 32 \times 64} \quad \text{(applying rule: \ } a^m \times a^n = a^{m+n} \text{\ and \ } {a^m \over a^n} = a^{m-n} ) \\ \\ &= {201 \times 10^{2} \over 2048} \\ \\ &= {20100 \over 2048} \\ \\ &= 9.8144 \\ \\ &\text{rounding the value} \\ &= 9.8 \quad m / s^2 \end{aligned}
Deriving the unit of gravitational acceleration
Tip: This is a bonus section just for extra information.
How do we know that the unit of g is
m / s^2
To understand that let us replace the units shown here to equation (3) rather than the values.
\begin{aligned} g &= G \times {M \over R^2} \\ \\ &= {N \cancel{m^2} \over kg^{\cancel{2}}} \times {\cancel{kg} \over \cancel{m^2}} \\ \\ &= {N \over kg} \quad ---- (4) \\ \end{aligned}
We know that, N (newton) is a unit of force. And according to the second law of motion
F = ma
. That means the unit of force can also be written as
kg \sdot m / s^2
. Replacing this in equation (4):
\begin{aligned} &= {N \over kg} \\ \\ &= {kg \sdot m / s^2 \over kg} \\ \\ &= {\cancel{kg} \times m \over s^2 \times \cancel{kg}} \\ \\ &= {m \over s^2} \end{aligned}
Hence we can say that the unit of gravitational acceleration is
m / s^2
Any object falling towards the Earth will move towards the Earth with an acceleration of
9.8 \space m / s^2
Gravitational Acceleration of Earth
Any object that is within the range of Earth's gravitational field will get attracted towards the Earth with a force. This force is called the force of gravitation of Earth(gravity). The acceleration with which the object moves towards Earth due to gravity is called Gravitational Acceleration.
Gravitational Acceleration is denoted by g (small g) and its value is
\thickspace 9.8 \ m/s^2
Can you answer this now ???
What will be the value of g (gravitational acceleration) on the moon?
Let us know your answers in the comment section of our YouTube video
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Lifting and data flow analysis.
Control flow analysis.
Type analysis.
Decompilation is the process of converting executable code(machine readable) into high level code(human readable).
A decompiler also referred to as a reverse compiler is the program that performs the inverse process of compiling, it takes a program compiled in a high level language and produces a high-level language program that performs the same functions as the executable. Its input is machine dependent and output is language dependent.
Note that some languages are easily decompiled and therefore produce code that is close in resemblance to the original e.g Java, in that, java stores high-level information in executables which is why it is portable.
Although being portable source code recovery is much easier since there are alot of details of how the application functions.
Decompilation involves four steps, namely,
disassembly - whereby machine code is transformed to assembly code.
Lifting and data flow analysis whereby we transform the assembly code into a high-level internal representation.
Control flow analysis where we recover the control flow structure information e.g if and while statements and their nesting levels.
Type analysis where we recover the types of the used variables, functions etc.
Since we have discussed disassembly in depth in a previous article, we shall go over a few details inorder to get the basic idea.
The mapping between assembly and machine code is a one-to-one correspondence and thereby the translation should be feasible. However as we have discussed it is difficult for the disassembler to distinguish between code and data.
Two strategies taken to solve this are as follows,
Disassembling sections filled with code then treat the rest as data.
Considering starting addresses provided in the binary's header and recursively disassembling reachable code from the address.
Disassembly has various pitfalls for example, programs get obfuscated and thus many steps are targeted to fool the disassembler and without correct disassembly, we cannot proceed with decompilation.
Assuming we have a correct disassembly, the structure cannot be analyzed. The decompiler will perform actions resembling a backwards form of instruction selection but they cannot tile sequences of assembly instruction with sequences of abstract instructions since different compilers will produce different assembly code for the same sequence of abstract instructions.
Additionally a single abstract instruction can expand into a long sequence of real instructions.
To deal with this, one approach would be to translate for example, a complex x86_64 into a simpler RISC instruction set.
Another approach would be to translate it into a semantics preserving complex instruction set which has cross platform methods for its analysis.
To translate x86_64 to a three-address code representation we perform reverse instruction selection.
Once we have this representation we eliminate machine specific details using data flow analysis so as to recover variables, expressions and statements.
We then replace registers with a temp and perform a function-call expansion step in reverse where we replace sequences of moves into arguments registers followed by a parameterized call.
For this we make passes over all functions so as to determine the number of arguments each take in order to deal with certain moves being optimized out.
We then perform a modified version of SSA analysis then an extended copy-propagation pass so as to collapse expressions.
At this stage we have obtained a control flow graph with real variables and can produce C code equivalent to the original machine code.
Although producing C code here is undesirable since few programs are written with as much abuse of the goto keyword similar to the code we obtain here.
Control flow graphs are generated by structured programs using if, for and while statements, only then is it desirable for a compiler to recover original structure so as to arrive at the original code.
Dominator nodes are a primary element of this analysis.
Given a start node a, a node b is said to dominate a node c if all paths from a to c go through b.
An immediate dominator of c is b such that for every node v, if d dominated c, then either d=b or d dominates b.
In this section we shall see how to structure three classes loops namely;
While loops where the starting node is a conditional and the latching an unconditional.
Repeat loops where the latching node is a conditional.
Endless loops where both the start and latching nodes are unconditional.
A latching node is a node with the back-edge to the start node.
To structure loops we consider intervals on a digraph.
If h is a node in G, interval l(h) is the maximal subgraph whereby h is the only entry node in which all closed paths contains h.
In theory there exists a set {h1,...,hk} of header nodes such that the set {l(h1,...,l(hk)} is a partition of the graph.
An algorithm exists to find such a partition.
Then we define derived graphs of G as follows,
{G}^{n+1}
- graph formed by contracting every interval of
{G}^{n}
into a single node.
Eventually we will reach a fixed point where the resulting graph is irreducible.
For any interval l(h) there is a loop rooted at h if there is a back-edge from h to a node z
\in
l(h).
To find such a node we perform a DFS algorithm on the interval.
To find nodes in the loop we define h as part of the loop and proceed by noting that a node k is in the loop iff its immediate dominator is in the loop and h is reachable from k.
To find loops, we compute the derived graphs of G until we arrive at a fixed point then find loops in each derived graph.
If a node is found to be a latching node for two loops, one loop is labeled with a goto instead.
An if statement is a 2-way conditional branch with a common end node.
The final node is referred to as the follow node and it is dominated by the header node.
To structure such, we perform a post-ordering and post-order traversal of the graph which will guarantee analysis of the internal nested ifs before the external ones.
To find if statements,
For every conditional node a, we find the set of nodes dominated by a.
Produce G' from G by reversing arrows and filter nodes from the above set the don't dominate a in G'.
Find the closest node to a in the resulting set by taking into consideration the one with the highest post-order number.
The resulting node is the follow node of a.
At this stage we determine what the types of variables are, each variable and function is assigned a type as part of the process of recovering the structure of the program.
Type analysis is an open problem meaning research is being done, however a model of a simple type analysis is as follows;
Multiplication, subtraction, division, bitwise operations such as xor, binary or are forced to be integers.
Dereferencing forces a parameter to be a pointer.
Return values of standard library functions are maintained.
A variable that is branched on becomes a boolean.
When two variables are added together and one is a pointer, the other will be an integer.
When two variables are added together and one is an integer, the other will be either a pointer or integer.
When two variables are compared with == or !=, they will have the same type.
When two variables are compared with <, >, >=, <=, they are both integers.
When a value is returned from main() function, it is an integer.
When the value of a variable is moved into another variable then they have the same type.
If the dereferenced value of a pointer has type T then the pointer is of type T*.
The sum of a pointer of type T* and an integer is a pointer, not necessarily of type T*.
Analysis is performed across function boundaries so as to to get high-quality types.
This analysis however cannot distinguish structures and arrays.
Decompilation has various uses for example, assuming there exists a legacy system whose source code is lost and we need to fix bugs and port into a modern architecture, decompilation will aid such a problem.
Decompilation can also be applied in malware analysis whereby we don't have the source code, we therefore decompile the executable to get an approximation of the original code.
A Structuring Algorithm for Decompilation
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Even and Odd Functions | Brilliant Math & Science Wiki
Hobart Pao, Pranshu Gaba, Azeem Hakim, and
Even and odd functions are functions that satisfy certain properties. This is a powerful concept; identifying even and odd functions can make some seemingly tough integration problems trivial.
Even functions are functions that satisfy
f(x) = f(-x)
x
. Even functions are symmetric about the line
x =0
Odd functions are functions that satisfy
f(x) = -f(-x)
x
. Odd functions exhibit point symmetry about the origin.
It is possible for a function to be neither odd nor even.
f(x) = \displaystyle \dfrac{x}{1-2^{x}}-\dfrac{x}{2}
\text{\_\_\_\_\_\_\_\_\_\_}.
\quad \text{A) }
even but not odd
\quad \text{B) }
odd but not even
\quad \text{C) }
both even and odd
\quad \text{D) }
neither even nor odd
f(-x)
\begin{aligned} f(-x)& = \dfrac{-x}{1-2^{-x}} - \dfrac{-x}{2} \\ &= \dfrac{-x \cdot 2^x}{2^x -1} + \dfrac{x}{2} \\ &= \dfrac{x}{1-2^x} - x + \dfrac{x}{2} \\ &= \dfrac{x}{1-2^x} - \dfrac{x}{2} \\ &= f(x). \end{aligned}
f(x)
is even and obviously not odd, so the answer is choice
\text{A)}
_\square
Even and odd functions can also be spotted by their graphs - specifically, even functions have the y-axis as a line of symmetry and odd functions have rotational symmetry about the origin. For example, take a point (a,b) on the coordinate plane such that the function f(x) is even, thus f(x) = f(-x) - so if (a,b) is on the graph, (-a,b) is as well. Odd functions can be rotated 180 degrees about the origin and then appear the same, and so are said to be symmetric about the origin.
Cite as: Even and Odd Functions. Brilliant.org. Retrieved from https://brilliant.org/wiki/even-and-odd-functions/
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Jacques_Touchard Knowpia
Jacques Touchard (1885–1968) was a French mathematician. In 1953, he proved that an odd perfect number must be of the form 12k + 1 or 36k + 9. In combinatorics and probability theory, he introduced the Touchard polynomials. He is also known for his solution to the ménage problem of counting seating arrangements in which men and women alternate and are not seated next to their spouses.
Touchard's Catalan identityEdit
The following algebraic identity involving the Catalan numbers
{\displaystyle C_{k}={1 \over {k+1}}{{2k} \choose {k}},\quad k\geq 0}
is apparently due to Touchard (according to Richard P. Stanley, who mentions it in his panorama article "Exercises on Catalan and Related Numbers" giving an overwhelming plenitude of different definitions for the Catalan numbers). For n ≥ 0 one has
{\displaystyle C_{n+1}=\sum _{k\,\leq \,n/2}2^{n-2k}{n \choose 2k}C_{k}.\,}
{\displaystyle C(t)=\sum _{n\geq 0}C_{n}t^{n}={{1-{\sqrt {1-4t}}} \over {2t}}}
it can be proved by algebraic manipulations of generating series that Touchard's identity is equivalent to the functional equation
{\displaystyle {t \over {1-2t}}C\left({t^{2} \over (1-2t)^{2}}\right)=C(t)-1}
satisfied by the Catalan generating series C(t).
Canadian Journal of Mathematics 1956, Vol 8, No 3.; Journal in French
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Trapezoidal rule (differential equations) - Wikipedia
Trapezoidal rule (differential equations)
In numerical analysis and scientific computing, the trapezoidal rule is a numerical method to solve ordinary differential equations derived from the trapezoidal rule for computing integrals. The trapezoidal rule is an implicit second-order method, which can be considered as both a Runge–Kutta method and a linear multistep method.
Suppose that we want to solve the differential equation
{\displaystyle y'=f(t,y).}
The trapezoidal rule is given by the formula
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{2}}h{\Big (}f(t_{n},y_{n})+f(t_{n+1},y_{n+1}){\Big )},}
{\displaystyle h=t_{n+1}-t_{n}}
is the step size.[1]
This is an implicit method: the value
{\displaystyle y_{n+1}}
appears on both sides of the equation, and to actually calculate it, we have to solve an equation which will usually be nonlinear. One possible method for solving this equation is Newton's method. We can use the Euler method to get a fairly good estimate for the solution, which can be used as the initial guess of Newton's method.[2] Cutting short, using only the guess from Eulers method is equivalent to performing Heun's method.
Integrating the differential equation from
{\displaystyle t_{n}}
{\displaystyle t_{n+1}}
{\displaystyle y(t_{n+1})-y(t_{n})=\int _{t_{n}}^{t_{n+1}}f(t,y(t))\,\mathrm {d} t.}
The trapezoidal rule states that the integral on the right-hand side can be approximated as
{\displaystyle \int _{t_{n}}^{t_{n+1}}f(t,y(t))\,\mathrm {d} t\approx {\tfrac {1}{2}}h{\Big (}f(t_{n},y(t_{n}))+f(t_{n+1},y(t_{n+1})){\Big )}.}
Now combine both formulas and use that
{\displaystyle y_{n}\approx y(t_{n})}
{\displaystyle y_{n+1}\approx y(t_{n+1})}
to get the trapezoidal rule for solving ordinary differential equations.[3]
Error analysis[edit]
It follows from the error analysis of the trapezoidal rule for quadrature that the local truncation error
{\displaystyle \tau _{n}}
of the trapezoidal rule for solving differential equations can be bounded as:
{\displaystyle |\tau _{n}|\leq {\tfrac {1}{12}}h^{3}\max _{t}|y'''(t)|.}
Thus, the trapezoidal rule is a second-order method.[citation needed] This result can be used to show that the global error is
{\displaystyle O(h^{2})}
as the step size
{\displaystyle h}
tends to zero (see big O notation for the meaning of this).[4]
The pink region is the stability region for the trapezoidal method.
The region of absolute stability for the trapezoidal rule is
{\displaystyle \{z\in \mathbb {C} \mid \operatorname {Re} (z)<0\}.}
This includes the left-half plane, so the trapezoidal rule is A-stable. The second Dahlquist barrier states that the trapezoidal rule is the most accurate amongst the A-stable linear multistep methods. More precisely, a linear multistep method that is A-stable has at most order two, and the error constant of a second-order A-stable linear multistep method cannot be better than the error constant of the trapezoidal rule.[5]
In fact, the region of absolute stability for the trapezoidal rule is precisely the left-half plane. This means that if the trapezoidal rule is applied to the linear test equation y' = λy, the numerical solution decays to zero if and only if the exact solution does.
^ Iserles 1996, p. 8; Süli & Mayers 2003, p. 324
Süli, Endre; Mayers, David (2003), An Introduction to Numerical Analysis, Cambridge University Press, ISBN 0521007941 .
Retrieved from "https://en.wikipedia.org/w/index.php?title=Trapezoidal_rule_(differential_equations)&oldid=1058820810"
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Dynamic Programming - Problem Solving Practice Problems Online | Brilliant
Suppose we are given two short DNA sequences seq1 and seq2.Suppose we are also given a set of operations:
Replace : Replace one character of a string with any other character.
Insert : Insert one character into a string.
Delete: Remove one character into another string.
Consider the two sequences below
seq1 = 'CTCCCAGGGTG'
seq2 = 'ACGTATAGCTTG'
What is the minimum number of operations required to convert one sequence into the other?
Brilli the ant sees a square
n \times n
matrix consisting of only
0
1
s. He is given the task of finding the largest
1
square space( the largest square submatrix consisting of only
1
's) . For example if he was given the below table, the shaded area would be the largest square area:
Unfortunately Brilli is given a matrix of much larger size. Being the competent programmer that he is, he divides the problem into subproblems and comes up with a brilliant dynamic programming solution.
His dynamic programming solution considers the subproblem
S[i,j]
S[i,j]
being the side length of the largest
1
square space whose bottom-right corner is at
i,j
Once he solves the problem, his table has the following values:
S[15][15] = 4
S[16][14] = 10
S[15][14] = 12
Given that there is a 1 at [16][15], what is the value of
S[16][15]
The longest common subsequence (LCS) problem is defined as follows:
Given two strings: string
S
n
and string
T
m
, the goal is to produce their longest common subsequence: the longest sequence of characters that appear left-to-right (but not necessarily in a contiguous block) in both strings.
S
= BZHAC
T
= HKABFT
Then the longest common subsequence has a length of
3
and is HAB.
Suppose you have the following snippets of two genes which you suspect to code for homologous proteins, but to have diverged long ago on the evolutionary scale. Find their LCS, what is its length?
sequence_1 = GGCAAGGTACTTCCGGTCTTAATGAATGGCCGGG
AAAGGTACGCACGCGGTATGGGGGGGTGAAGGGGCGAATAGACAGGC
TCCCCTCTCACTCGCTAGGAGGCAATTGTATAAGAATGCATACTGCA
TCGATACATAAAACGTCTCCATCGCTTGCCCAAGTTGTGAAGTGTCT
ATCACCCCTAGGCCCGTTTCCCGCA
sequence_2 = GGCTGGCGTTTTGAATCCTCGGTCCCCCTTGTCT
ATCCAGATTAATCCAATTCCCTCATTTAGGACCCTACCAAGTCAACA
TTGGTATATGAATGCGACCTCGAAGAGGCCGCCTAAAAATGACAGTG
GTTGGTGCTCTAAACTTCATTTGGTTAACTCGTGTATCAGCGCGATA
GGCTGTTAGAGGTTTAATATTGTAT
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Option price by Bates model using FFT and FRFT - MATLAB optByBatesFFT - MathWorks India
\mathrm{max}\left(St-K,0\right)
\mathrm{max}\left(K-St,0\right)
\begin{array}{l}d{S}_{t}=\left(r-q-{\lambda }_{p}{\mu }_{J}\right){S}_{t}dt+\sqrt{{v}_{t}}{S}_{t}d{W}_{t}+J{S}_{t}d{P}_{t}\\ d{v}_{t}=\kappa \left(\theta -{v}_{t}\right)dt+{\sigma }_{v}\sqrt{{v}_{t}}d{W}_{t}\\ \text{E}\left[d{W}_{t}d{W}_{t}^{v}\right]=pdt\\ \text{prob(}d{P}_{t}=1\right)={\lambda }_{p}dt\end{array}
\mathrm{ln}\left(1+{\mu }_{J}\right)-\frac{{\delta }^{2}}{2}
\frac{1}{\left(1+J\right)\delta \sqrt{2\pi }}\mathrm{exp}\left\{{\frac{-\left[\mathrm{ln}\left(1+J\right)-\left(\mathrm{ln}\left(1+{\mu }_{J}\right)-\frac{{\delta }^{2}}{2}\right]}{2{\delta }^{2}}}^{2}\right\}
{W}_{t}^{v}
{\lambda }_{p}
{\lambda }_{p}
{f}_{Bate{s}_{j}\left(\varphi \right)}
\begin{array}{l}{f}_{Bates\left(\varphi \right)}=\mathrm{exp}\left({C}_{j}+{D}_{j}{v}_{0}+i\varphi \mathrm{ln}{S}_{t}\right)\mathrm{exp}{\left({\lambda }_{p}\tau \left(1+{\mu }_{J}\right)}^{{m}_{j}+\frac{1}{2}}\left[{\left(1+{\mu }_{j}\right)}^{i\varphi }{e}^{{\delta }^{2}\left({m}_{j}i\varphi +\frac{{\left(i\varphi \right)}^{2}}{2}\right)}-1\right]-{\lambda }_{p}\tau {\mu }_{J}i\varphi \right)\\ {m}_{j}=\left\{\begin{array}{l}{m}_{1}=\frac{1}{2}\\ {m}_{2}=-\frac{1}{2}\end{array}\right\}\\ {C}_{j}=\left(r-q\right)i\varphi \tau +\frac{\kappa \theta }{{\sigma }_{v}{}^{2}}\left[\left({b}_{j}-p{\sigma }_{v}i\varphi +{d}_{j}\right)\tau -2\mathrm{ln}\left(\frac{1-{g}_{j}{e}^{{d}_{j}\tau }}{1-{g}_{j}}\right)\right]\\ Dj=\frac{{b}_{j}-p{\sigma }_{v}i\varphi +{d}_{j}}{{\sigma }_{v}^{2}}\left(\frac{1-{e}^{{d}_{j}\tau }}{1-{g}_{j}{e}^{{d}_{j}\tau }}\right)\\ {g}_{j}=\frac{{b}_{j}-p{\sigma }_{v}i\varphi +{d}_{j}}{{b}_{j}-p{\sigma }_{v}i\varphi -{d}_{j}}\\ {d}_{j}=\sqrt{{\left({b}_{j}-p{\sigma }_{v}i\varphi \right)}^{2}-{\sigma }_{v}^{2}\left(2{u}_{j}i\varphi -{\varphi }^{2}\right)}\\ \text{where for }j=1,2:\\ {u}_{1}=\frac{1}{2},{u}_{2}=-\frac{1}{2},{b}_{1}=\kappa +{\lambda }_{VolRisk}-p{\sigma }_{v},{b}_{2}=\kappa +{\lambda }_{VolRisk}\end{array}
\begin{array}{l}{C}_{j}=\left(r-q\right)i\varphi \tau +\frac{\kappa \theta }{{\sigma }_{v}{}^{2}}\left[\left({b}_{j}-p{\sigma }_{v}i\varphi -{d}_{j}\right)\tau -2\mathrm{ln}\left(\frac{1-{\epsilon }_{j}{e}^{-{d}_{j}\tau }}{1-{\epsilon }_{j}}\right)\right]\\ Dj=\frac{{b}_{j}-p{\sigma }_{v}i\varphi -{d}_{j}}{{\sigma }_{v}^{2}}\left(\frac{1-{e}^{-{d}_{j}\tau }}{1-{\epsilon }_{j}{e}^{-{d}_{j}\tau }}\right)\\ {\epsilon }_{j}=\frac{{b}_{j}-p{\sigma }_{v}i\varphi -{d}_{j}}{{b}_{j}-p{\sigma }_{v}i\varphi +{d}_{j}}\end{array}
\begin{array}{l}Call\left(k\right)=\frac{{e}^{-\alpha k}}{\pi }{\int }_{0}^{\infty }\mathrm{Re}\left[{e}^{-iuk}\psi \left(u\right)\right]du\\ \psi \left(u\right)=\frac{{e}^{-r\tau }{f}_{2}\left(\varphi =\left(u-\left(\alpha +1\right)i\right)\right)}{{\alpha }^{2}+\alpha -{u}^{2}+iu\left(2\alpha +1\right)}\\ Put\left(K\right)=Call\left(K\right)+K{e}^{-r\tau }-{S}_{t}{e}^{-q\tau }\end{array}
\mathrm{ln}\left({S}_{t}\right)-\frac{N}{2}\Delta k
\mathrm{ln}\left({S}_{t}\right)+\left(\frac{N}{2}-1\right)\Delta k
{S}_{t}\mathrm{exp}\left(-\frac{N}{2}\Delta k\right)
{S}_{t}\mathrm{exp}\left[\left(\frac{N}{2}-1\right)\Delta k\right]
Call\left({k}_{n}\right)=\Delta u\frac{{e}^{-\alpha {k}_{n}}}{\pi }\sum _{j=1}^{N}\mathrm{Re}\left[{e}^{-i\Delta k\Delta u\left(j-1\right)\left(n-1\right){e}^{i{u}_{j}}\left[\frac{N\Delta k}{2}-\mathrm{ln}\left({S}_{t}\right)\right]}\psi \left({u}_{j}\right)\right]{w}_{j}
\Delta k\Delta u=\left(\frac{2\pi }{N}\right)
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Open-loop transfer function at specified point using slLinearizer or slTuner interface - MATLAB getLoopTransfer - MathWorks India
Obtain Loop Transfer Function at Analysis Point
Obtain Negative-Feedback Loop Transfer Function at Analysis Point
Specify Temporary Loop Opening for Loop Transfer Function Calculation
Obtain Loop Transfer Function for Specific Parameter Combination
Obtain Offsets from Loop Transfer Function
Open-loop transfer function at specified point using slLinearizer or slTuner interface
linsys = getLoopTransfer(s,pt)
linsys = getLoopTransfer(s,pt,sign)
linsys = getLoopTransfer(s,pt,temp_opening)
linsys = getLoopTransfer(s,pt,temp_opening,sign)
linsys = getLoopTransfer(___,mdl_index)
[linsys,info] = getLoopTransfer(___)
linsys = getLoopTransfer(s,pt) returns the point-to-point open-loop transfer function at the specified analysis point for the model associated with the slLinearizer or slTuner interface, s.
The software enforces all the permanent loop openings specified for s when it calculates linsys. If you configured either s.Parameters, or s.OperatingPoints, or both, getLoopTransfer performs multiple linearizations and returns an array of loop transfer functions.
linsys = getLoopTransfer(s,pt,sign) specifies the feedback sign for computing the open-loop response. By default, linsys is the positive-feedback open-loop transfer function.
Set sign to -1 to compute the negative-feedback open-loop transfer function for applications that assume the negative-feedback definition of linsys. Many classical design and analysis techniques, such as the Nyquist or root locus design techniques, use the negative-feedback convention.
The closed-loop sensitivity at pt is equal to feedback(1,linsys,sign).
linsys = getLoopTransfer(s,pt,temp_opening) considers additional, temporary, openings at the point specified by temp_opening. Use an opening, for example, to calculate the loop transfer function of an inner loop, measured at the plant input, with the outer loop open.
linsys = getLoopTransfer(s,pt,temp_opening,sign) specifies temporary openings and the feedback sign.
linsys = getLoopTransfer(___,mdl_index) returns a subset of the batch linearization results. mdl_index specifies the index of the linearizations of interest, in addition to any of the input arguments in previous syntaxes.
Use this syntax for efficient linearization, when you want to obtain the loop transfer function for only a subset of the batch linearization results.
[linsys,info] = getLoopTransfer(___) returns additional linearization information.
Obtain the loop transfer function, calculated at e, for the ex_scd_simple_fdbk model.
To obtain the loop transfer function at e, add this point to sllin as an analysis point.
addPoint(sllin,'e');
Obtain the loop transfer function at e.
sys = getLoopTransfer(sllin,'e');
From input "e" to output "e":
The software adds a linearization output, breaks the loop, and adds a linearization input, de, at e.
sys is the transfer function from de to e. Because the software assumes positive-feedback, it returns sys as .
Obtain the negative-feedback loop transfer function, calculated at e, for the ex_scd_simple_fdbk model.
sys = getLoopTransfer(sllin,'e',-1);
sys is the transfer function from de to e. Because the third input argument indicates negative-feedback, the software returns sys as .
Obtain the loop transfer function for the inner loop, calculated at e2, for the scdcascade model.
To calculate the loop transfer function for the inner loop, use the e2 signal as the analysis point. To eliminate the effects of the outer loop, break the outer loop at y1m. Add these points to sllin.
addPoint(sllin,{'e2','y1m'});
Obtain the inner-loop loop transfer function at e2.
sys = getLoopTransfer(sllin,'e2','y1m');
Here, 'y1m', the third input argument, specifies a temporary loop opening. The software assumes positive-feedback when it calculates sys.
Suppose you batch linearize the scdcascade model for multiple transfer functions. For most linearizations, you vary the proportional (Kp2) and integral gain (Ki2) of the C2 controller, in the 10% range. For this example, calculate the loop transfer function for the inner loop at e2 for the maximum values of Kp2 and Ki2.
Obtain the inner-loop loop transfer function at e2, with the outer loop open.
sys = getLoopTransfer(sllin,'e2','y1m',-1,mdl_index);
The fourth input argument specifies negative-feedback for the loop transfer calculation.
Calculate the loop transfer function at the analysis point, and obtain the corresponding linearization offsets.
[sys,info] = getLoopTransfer(sllin,'watertank/Water-Tank System');
To calculate linsys, the software adds a linearization output, followed by a loop break, and then a linearization input at pt. Consider the following model:
Specify pt as 'u'.
If you specify pt as multiple signals, for example pt = {'u','y'}, the software adds a linearization output, loop break, and a linearization input at each point.
sign — Feedback sign
Feedback sign, specified as one of the following values:
+1 (default) — getLoopTransfer returns the positive-feedback open-loop transfer function.
-1 — getLoopTransfer returns the negative-feedback open-loop transfer function. The negative-feedback transfer function is -1 times the positive-feedback transfer function.
linsys — Point-to-point open-loop transfer function
state-space object
Point-to-point open-loop transfer function, returned as described in the following:
The loop transfer function at a point is the point-to-point open-loop transfer function from an additive disturbance at a point to a measurement at the same point.
To compute the loop transfer function at an analysis point, x, the software adds a linearization output, inserts a loop break, and adds a linearization input, dx. The software computes the transfer function from dx to x, which is equal to the loop transfer function at x.
How getLoopTransfer Interprets Analysis Point
For example, consider the following model where you compute the loop transfer function at e:
Here, at e, the software adds a linearization output, inserts a loop break, and adds a linearization input, de. The loop transfer function at e, L, is the transfer function from de to e. L is calculated as follows:
\begin{array}{l}e=-GKde\\ \therefore L=-GK\end{array}
To compute -KG, use u as the analysis point for getLoopTransfer.
The software does not modify the Simulink model when it computes the loop transfer function.
slLinearizer | slTuner | addPoint | addOpening | getIOTransfer | getSensitivity | getCompSensitivity
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Step-by-step solutions to hundreds of high school math problems!
High school math questions and answers
Recent questions in Secondary
Hyus Jsjs 2022-05-23
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dddtest test testtest test testtest test testtest test testtest test test
\sqrt{2}
question2answer 2022-05-23
\frac{5{x}^{2}+1}{3{x}^{3}-1}=8
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According to one survey in India, 75% of Instagram users love REELS. Suppose that 25
why is 4/3 πr^3 a monomial
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Write an equation of a line that goes through the points (0,9) and (4,11)
Shweta Goel 2022-05-22
Carefully describe the level surfaces of h(x, y, z) = x ^2− y^2 − z^2 − 4x + 3.
Your description must include the name and a general equation of the
family of level surfaces.
Gracie Tibbs 2022-05-22
calculating the correct sample size is a critical component of a study design. t/f. quizlet
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Explain FOUR (4) Second Partial Derivatives of the Partial Derivative of z with respect to x.
I found article in which it is said that it would be incorrect to say that g is not acceleration due to gravity but local gravitational field as there is no acceleration on a block placed on a table. Please can you explain this as I am in a school and I have read only that g is acceleration due to gravity and textbooks say this too.
For the electrical resistance of a conductor, we have
R=\rho \frac{l}{A}
Noting the structural similarity between the Hagen-Poiseuille law and Ohm's law, we can define a similar quantity for laminar flow through a long cylindrical pipe:
{R}_{V}=8\eta \frac{l}{A{r}^{2}}
So there's a structural difference of a factor of
{r}^{2}
between the two. What's the intuition behind this?
What is gamma in the damping equation?
{x}^{″}+\gamma {x}^{\prime }+{w}_{0}^{2}x=0
That is the general equation for damped harmonic motion. What is the term or name that describes gamma ?
Is it called the damping constant ? I know its the ration between the resistive coefficient (b) and mass of the system (m) but what do we actually call it ?
How much power is produced if a voltage of 12V is applied to a circuit with a resistance of
144\mathrm{\Omega }
The isentropic relations of ideal gases applicable for flows across:
(a) Normal shock waves
(b) Oblique shock waves.
(c) Prandtl-Meyer expansion waves.
Say there is a water bottle that is filled with 300 mL of water and has a circular hole with a radius of 2 mm. In this bottle, the water sits 7.8cm above the top of the hole (which has been drilled 1.5cm above the bottom of the bottle).
According to Bernoulli's law the velocity
v
of the water flowing out is equal to
\sqrt{2gh}
Using this, the flow rate can be calculated as
Q\text{ }=\text{ }Av\text{ }=\text{ }\pi \left(0.002\text{ }m{\right)}^{2}\ast 1.24\text{ }m/s=0.000016\text{ }{m}^{3}/s=16\text{ }mL/s
This doesn't seem accurate considering that the experimental flow rate is equal to 8 mL/s (40 mL over 5 seconds). However I understand that it ignores viscosity (and other things?)
I'm wondering a few things, firstly, does the theoretical math here apply to the situation I'm describing? The hole in the bottle isn't exactly a pipe and the only examples I've seen with water flow involve pipes.
Secondly, can Poiseuille's Law be used to determine the flow rate instead, with a more accurate result? (From what I understand Q=πPR^4/8nl, however I don't understand what P is, seeing as in Bernoulli's law pressure cancels and as aforementioned this isn't a typical pipe example.)
Thirdly, I assume the theoretical flow rate will still be different from the experimental flow rate, what factors cause this?
Let's consider rotating charged ring. Theoretically mass of this ring has no limit as rotation speed increases. So what about magnetic moment of the ring? Is it limited by the value of speed of light?
If the magnitude of the magnetic induction is less than that in a vaccum within a solid, so, this material called
a) Ferrimagnetic
b) Ferromagnetic
d) Paramagnetic
Acceleration due to gravity during its journey up and down
When we throw an object up into the air, ignoring air resistance, etc, we define acceleration to be -9.8 m/s^2. When it goes down after its journey up, like a parabola, do we define the acceleration as 9.8 m/S^2, OR still the same as -9.8m/s^2?
I know that acceleration due to gravity is always constant, but would it be wrong to say it like the question i have above?
Turning back to high school math can be essential to understand engineering tasks that you may encounter later. The high school math problems have all the basics that have good equations and answers, which will let you see things clearly. The list of high school math questions below will help you identify your weaknesses and find various solutions. Taking a look at high school math equations, you will see certain parts that can be applied to Physics. In either case, the best way is to learn by example, which is why high school math problems with answers will be essential.
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Biomolecules - Course Hero
Microbiology/Fundamental Building Blocks/Biomolecules
Biomolecules are organic compounds produced by living things, including microorganisms (microbes). By definition, organic molecules contain one or more carbon-hydrogen bonds. There are four families of biomolecules, all of which perform essential functions in cells: carbohydrates, lipids, proteins, and nucleic acids. These compounds can be described in part by how they interact with water. Compounds are often described as hydrophobic, hydrophilic, or amphipathic. Hydrophilic is the characteristic of having a strong affinity toward water. Hydrophobic is the characteristic of having a weak or no affinity to water, which occurs when the bonds that connect their atoms are nonpolar. Molecules become hydrophilic when the bonds that connect their atoms are polar, creating a partial charge. An amphipathic molecule has both hydrophobic and hydrophilic portions.
Carbohydrates are compounds consisting of only carbon, hydrogen, and oxygen that have structural and energy storage or transport roles.
A carbohydrate is an organic compound that contains carbon, hydrogen, and oxygen and provides energy to cells. Most carbohydrates exist as polymers, long, chainlike molecules or rings of five or six carbons, which can be bonded together. Three major groups of carbohydrates exist: monosaccharides, disaccharides, and polysaccharides. Monosaccharides consist of a single sugar molecule, such as glucose. Disaccharides, such as lactose, consist of two sugar molecules bonded together. Polysaccharides, such as cellulose, are composed of many sugar molecules. Common types of carbohydrates include sugars, starches, and cellulose. Sugars include glucose, the critical component of cellular respiration. Sugars provide both immediate energy and long-term energy storage for microbes. Carbohydrates also contribute to the physical framework of many biological structures. For example, the carbohydrate polymer peptidoglycan composes the cell walls of most bacteria. Cellulose is a primary component in rigid cells walls that support the vertical growth of many plants. Humans cannot digest cellulose, but it provides fiber needed for the health of the intestinal tract.
Carbohydrates, such as glucose and fructose, are organic molecules that provide energy for cellular function. A carbohydrate has carbon, oxygen, and hydrogen atoms. It usually forms long chains or rings.
A lipid is a long-chain hydrocarbon that is soluble in nonpolar solvents, that is, solvents without charge. Lipids are a class of hydrophobic compound that are involved in building cell membranes and storing energy. Lipids include fats, waxes, and sterols. Fats, which can be used to store energy, are a combination of fatty acids and the alcohol glycerol. Waxes are composed of fatty acid and a long-chain alcohol. They provide a water-repellent surface for the fur or feathers or some animals as well as the leaves and fruits of some plants. Sterols, such as cholesterol, consist of a hydroxyl group (
-\rm OH
) bonded to a fused four-ring structure.
Lipids include fats, waxes, and sterols. Lauric acid is a typical lipid known as a fatty acid, consisting of a long chain of carbon atoms, linked end to end. Attached to these are bonds where hydrogen can attach. Cholesterol is an example of a sterol, which includes a hydroxol group (
-\rm OH
) and four fused rings.
In all cells, whether prokaryotic (Archaea and Bacteria) or eukaryotic (all other organisms), the main structure of the plasma membrane is formed of phospholipids. A phospholipid is a molecule composed of a phosphate group, fatty acids, and glycerol. The thin plasma membrane surrounds cells and acts as a selective barrier that controls passage of ions and molecules into and out of cells. Phospholipids consist of glycerol molecule connected to two fatty acids and to a phosphate group. This combination makes phospholipids amphipathic. They are arranged in a bilayer in the membrane, with hydrophobic fatty acid tails pointing inward and hydrophilic phosphate heads pointing outward.
The cell membrane consists of a phospholipid bilayer that contains both hydrophobic (water-fearing) fatty acid tails and hydrophilic (water-loving) heads. The hydrophilic heads are found toward the more aqueous sides of the cell (inside and outside).
A protein is a large molecule composed of amino acids. Most enzymes are proteins that act as biological catalysts and speed up reactions. Transport proteins move materials in and out of cells. The amino acid chains of proteins fold into complex three-dimensional structures. For example, some proteins form cable-like shapes that provide scaffolding in the cell and others form pipe-like shapes that form channels in membranes allowing chemical transport. Other proteins form subcellular structures in membranes, cell walls, organelles, the cytoplasmic scaffolding skeleton, flagella, and pili. Proteins make up approximately 55 percent of the dry weight of prokaryotic cells.
The shape and function of a protein is determined by the specific sequence of amino acids that make up the protein. Different types of amino acids possess different chemical properties that help shape the final protein, such as having hydrophilic or hydrophobic sections, or differences in polarity. For example, cysteine is an important amino acid in structural proteins, as it is capable of forming very strong bonds with other cysteine molecules, called disulfide bridges. Different amino acids are more frequent in some groups of organisms than others, depending on their function. Twenty amino acids are used by all organisms and two additional amino acids are used by subsets of organisms. For example, the amino acid pyrrolysine is found only in some archaea species and one bacterium. Pyrrolysine is used by these species to produce an enzyme that correctly positions the methyl group of a compound called methylamine during a reaction that produces methane.
Proteins are composed of sequences of organic molecules called amino acids. Amino acids contain a carboxyl group and an amino group.
A nucleotide is an organic compound consisting of a sugar, a phosphate, and a nitrogenous base. It is the basic structural unit of genetic sequences. A nucleic acid is a molecule made of nucleotides. Nucleic acids exist in cells in two forms: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). DNA employs the nucleotides thymine (T), cytosine (C), adenine (A), and guanine (G). RNA employs the nucleotides uracil (U), cytosine (C), adenine (A), and guanine (G).
DNA and RNA are nucleic acids that encode and communicate the genetic information that sustains the structure and function of cells in living organisms. Though similar in their makeup, they do have some notable differences. DNA contains deoxyribose sugar, a 5-carbon sugar with a hydrogen attached to its fourth carbon atom. RNA contains ribose, a 5-carbon sugar with an
-\rm OH
group attached to its fourth carbon atom. Additionally, while DNA utilizes the nucleotide thymine (T), RNA utilizes uracil (U).
Together, DNA and RNA encode and communicate the genetic information necessary to sustain living things. Nucleic acids are the second most common component of prokaryotic cells, composing approximately 24 percent of cellular dry weight. In bacterial chromosomes, DNA is frequently circular, whereas DNA in chromosomes of eukaryotes is generally linear. Bacterial cells also contain plasmids, small circular pieces of DNA that can replicate on their own and transfer between cells. Unlike eukaryotic cells, where DNA is packed into nuclei, prokaryotic cells—which lack nuclei—store DNA only in the cytoplasm. The viral genome is made of DNA or RNA that can be double or single stranded.
DNA consists of two long chains of nucleotides bonded together in a double helix pattern. DNA in bacteria is found in cytoplasm and is typically circular.
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How do you find the zeroes for f(x)=x^{2}-9x-70?
Anderson Melton 2022-01-23 Answered
How do you find the zeroes for
f\left(x\right)={x}^{2}-9x-70
Bottisiooq
We need to think of two numbers, that, when I add them, sum up to -9, and when I multiply them, have a product of -70. Since the product is negative, we know the signs must be different.
Through some thought, we arrive at -14 and 5 as our two numbers, because
-14+5=-9
-14×5=-70
Thus, our equation is as follows:
\left(x-14\right)\left(x+5\right)=0
To find the zeroes, we take the opposite signs to get
x=14
x=-5
f\left(x\right)={x}^{2}-9x-70=0
To solve f(x), find 2 numbers (real roots) knowing the sum
\left(-b=9\right)
\left(c=-70\right)
. They are: - 5 and 14.
Note . This method avoids proceeding the lengthy factoring by grouping and solving the 2 binomials.
Would an equation in the form
ax+b=\frac{c}{x}
be considered quadratic?
For example is
2x+3=\frac{5}{x}
quadratic?
On the one hand it has two solutions,
x=1\text{ }\text{and}\text{ }x=-\frac{5}{2}
which is the number of solutions we'd expect from the fundamental theorem of algebra but on the other hand any equation in the form
ax+b=\frac{c}{x}
would be undefined at
x=0
and every quadratic equation I'm familiar with is continuous over all real numbers. Is being continuous over all real numbers necessary for an equation to be called quadratic?
2{m}^{2}+2m-12=0
{x}^{2}+px+q
f\left(f\left(x\right)\right)=0
p\ge 0\text{ }\text{and}\text{ }q\ge 0
Local extrema of the function
f\left(x\right)={x}^{3}-3\alpha {x}^{2}+3\left({\alpha }^{2}-1\right)x+1
Find the value of the following expression in terms of a+bi:
\frac{\sqrt{-6}}{\sqrt{-3}\sqrt{-4}}
When factoring the quadratic
{x}^{2}+bx+c
, where b and c are integers, why do we not consider the case where u and v in the factored form
\left(x+u\right)\left(x+v\right)
are fractions?
2{x}^{-2}-{x}^{-1}=3
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Latency - Wikiversity
Latency from a general point of view is a time delay between the cause and the effect of some physical change in the system being observed, but, known within gaming circles as "lag", latency is a time interval between the input to a simulation and the visual or auditory response, often occurring because of network delay in online games.[1]
Latency is physically a consequence of the limited velocity which any physical interaction can propagate. The magnitude of this velocity is always less than or equal to the speed of light. Therefore, every physical system with any physical separation (distance) between cause and effect will experience some sort of latency, regardless of the nature of stimulation that it has been exposed to.
The precise definition of latency depends on the system being observed or the nature of the simulation. In communications, the lower limit of latency is determined by the medium being used to transfer information. In reliable two-way communication systems, latency limits the maximum rate that information can be transmitted, as there is often a limit on the amount of information that is "in-flight" at any one moment. In the field of human–machine interaction, perceptible latency has a strong effect on user satisfaction and usability.
2.3 Satellite transmission
4 Video latency
5 Operational latency
6 Mechanical latency
7 Computer hardware and operating system latency
8 In simulators
This learning resource is based on the Open Community Approach, so that the learning content is available on Wikiversity and the tools to explore the concept of Latency are Open Source Software to understand the underlying concept.
Latency Geogebra interactive example of trigonometric function.
(Online Music Jam) Analyse the application of online music jams when musician located at different geolocations and play together with an Open Source tool. Explain why latency is one of the main parameters, that defines if the online jam is possible.
(Mars Rover) The duration for the signal transmission from to a vehicle on the Mars lets to the fact that Mars rovers must operate as autonomous vehicles. Explain in dependency of the required response time for an operator act according to a transmitted signal of an events recorded from a Mars rover. Explore the latency of the signals transmitted from the Mars rover to earth.
(Communication Latency) Explain why Online games are sensitive to latency (or "lag"). Assess if low latency networks are an advantage or disadvantage for players according to response times or the time the player have to respond to new events occurring during a game session. Due to a delay in transmission of game events, a player with a high latency internet connection may have more time to respond to an event and in turn may show slow response to an event even if the response to event was within required appropriate reaction time. Access the advantages and disadvantages for players with low latency connections.
(Capital Markets) Most of relevant trading is driven by algorithms and trading actions within milliseconds. Explain why minimizing latency is a key to maximize the benefits of trading actions and a key in the capital markets,[2] particularly where algorithmic trading is used to process market updates and turn around orders within milliseconds. Low-latency trading occurs on the networks used by financial institutions to connect to stock exchanges and electronic communication networks (ECNs) to execute financial transactions.[3] Joel Hasbrouck and Gideon Saar (2011) measure latency based on three components: the time it takes for information to reach the trader, execution of the trader's algorithms to analyze the information and decide a course of action, and the generated action to reach the exchange and get implemented. Hasbrouck and Saar contrast this with the way in which latencies are measured by many trading venues who use much more narrow definitions, such as, the processing delay measured from the entry of the order (at the vendor's computer) to the transmission of an acknowledgement (from the vendor's computer).[4] Electronic trading now makes up 60% to 70% of the daily volume on the New York Stock Exchange and algorithmic trading close to 35%.[5] Trading using computers has developed to the point where millisecond improvements in network speeds offer a competitive advantage for financial institutions.[6]
(COVID-19) Explain why registration of infection, hospital admission and mortality of patients have a different specific latencies according to the event when a patient was infected. Describe, why integration of latency in Risk Management could be used to improve preparedness of the health system. What are the consequences for the Risk Management.
(Mathematics) Create a mathematical function
{\displaystyle f(x):=sin(3\cdot x)+2\cdot cos(x)}
in Geogebra and plot the function. Now plot the function
{\displaystyle g(x):=f(x-a)}
where a defines the latency, i.e. the delay in which the signal reaches the recipient of the signal. The unit 1 could represent a second, millisecond or minutes. Use the parameter
{\displaystyle a}
with the value slides in the interactive Geogebra visualization for the delay.
Packet-switched networks[edit | edit source]
Opus Bitrate and Latency - Comparision
Fiber optics[edit | edit source]
Satellite transmission[edit | edit source]
Audio latency[edit | edit source]
Main article: w:en:Latency (audio)
Video latency[edit | edit source]
Operational latency[edit | edit source]
Mechanical latency[edit | edit source]
Computer hardware and operating system latency[edit | edit source]
Further information: w:en:Access time
In simulators[edit | edit source]
↑ "Latency" Retrieved 2020-10-27.
↑ TABB (2009). High Frequency Trading Technology: a TABB Anthology. http://www.tabbgroup.com/PublicationDetail.aspx?PublicationID=498. Retrieved 2017-02-11.
↑ Mackenzie, Michael; Grant, Jeremy (2009). "The dash to flash" (PDF). Financial Times. Archived from the original (PDF) on 23 July 2011. Retrieved 18 July 2011. extracting tiny slices of profit from trading small numbers of shares in companies, often between different trading platforms, with success relying on minimal variations in speed - or "latency", in the trading vernacular.
↑ Hasbrouck, Joel; Saar, Gideon. "Low-Latency Trading" (PDF). p. 1. Archived from the original (PDF) on 11 November 2011. Retrieved 18 July 2011.
↑ Heires, Katherine. Code Green: Goldman Sachs & UBS Cases Heighten Need to Keep Valuable Digital Assets From Walking Out The Door. Millions in Trading Profits May Depend On It - Journal: Securities Industry News - Date: July 2009 URL: http://www.conatum.com/presscites/gogreen.pdf - (access date: 18 July 2011).
↑ "High-frequency trading: when milliseconds mean millions". The Telegraph. Retrieved 2018-03-25.
↑ "Don't misuse ping!". Retrieved 29 April 2015.
↑ Shane Chen (2005). "Network Protocols Discussion / Traffic Shaping Strategies". knowplace.org. Archived from the original on 2007-01-09.
↑ "Basic QoS part 1 – Traffic Policing and Shaping on Cisco IOS Router". The CCIE R&S. Retrieved 29 April 2015.
M. Brian Blake (December 2003). "Coordinating Multiple Agents for Workflow-Oriented Process Orchestration". Information Systems and E-Business Management Journal (Springer-Verlag). http://www.cs.georgetown.edu/~blakeb/pubs/blake_ISEB2003.pdf.
Latency_(engineering) https://en.wikipedia.org/wiki/Latency_(engineering)
Date: 3/1/2021 - Source History Wikipedia
Retrieved from "https://en.wikiversity.org/w/index.php?title=Latency&oldid=2263853"
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Two-axle vehicle body with translational and rotational motion - Simulink
\begin{array}{l}{\overline{F}}_{b}=\left[\begin{array}{c}{F}_{x}\\ {F}_{y}\\ {F}_{z}\end{array}\right]=m\left({\stackrel{˙}{\overline{V}}}_{b}+\overline{\omega }×{\overline{V}}_{b}\right)\\ \\ {\overline{M}}_{b}=\left[\begin{array}{c}L\\ M\\ N\end{array}\right]=I\stackrel{˙}{\overline{\omega }}+\overline{\omega }×\left(I\overline{\omega }\right)\\ \\ I=\left[\begin{array}{ccc}{I}_{xx}& -{I}_{xy}& -{I}_{xz}\\ -{I}_{yx}& {I}_{yy}& -{I}_{yz}\\ -{I}_{zx}& -{I}_{zy}& {I}_{zz}\end{array}\right]\end{array}
\left[\begin{array}{ccc}\stackrel{˙}{\varphi }\text{ }\text{\hspace{0.17em}}& \stackrel{˙}{\theta }\text{\hspace{0.17em}}\text{ }\text{ }& \stackrel{˙}{\psi }\end{array}{\right]}^{T}
\left[\begin{array}{c}p\\ q\\ r\end{array}\right]=\left[\begin{array}{c}\stackrel{˙}{\varphi }\\ 0\\ 0\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& \mathrm{cos}\varphi & \mathrm{sin}\varphi \\ 0& -\mathrm{sin}\varphi & \mathrm{cos}\varphi \end{array}\right]\left[\begin{array}{c}0\\ \stackrel{˙}{\theta }\\ 0\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& \mathrm{cos}\varphi & \mathrm{sin}\varphi \\ 0& -\mathrm{sin}\varphi & \mathrm{cos}\varphi \end{array}\right]\left[\begin{array}{ccc}\mathrm{cos}\theta & 0& -\mathrm{sin}\theta \\ 0& 1& 0\\ \mathrm{sin}\theta & 0& \mathrm{cos}\theta \end{array}\right]\left[\begin{array}{c}0\\ 0\\ \stackrel{˙}{\psi }\end{array}\right]\equiv {J}^{-1}\left[\begin{array}{c}\stackrel{˙}{\varphi }\\ \stackrel{˙}{\theta }\\ \stackrel{˙}{\psi }\end{array}\right]
\left[\begin{array}{c}\stackrel{˙}{\varphi }\\ \stackrel{˙}{\theta }\\ \stackrel{˙}{\psi }\end{array}\right]=J\left[\begin{array}{c}p\\ q\\ r\end{array}\right]\text{\hspace{0.17em}}=\left[\begin{array}{ccc}1& \left(\mathrm{sin}\varphi \mathrm{tan}\theta \right)& \left(\mathrm{cos}\varphi \mathrm{tan}\theta \right)\\ 0& \mathrm{cos}\varphi & -\mathrm{sin}\varphi \\ 0& \frac{\mathrm{sin}\varphi }{\mathrm{cos}\theta }& \frac{\mathrm{cos}\varphi }{\mathrm{cos}\theta }\end{array}\right]\left[\begin{array}{c}p\\ q\\ r\end{array}\right]
\begin{array}{l}{\overline{F}}_{b}=\left[\begin{array}{c}{F}_{x}\\ {F}_{y}\\ {F}_{z}\end{array}\right]=\left[\begin{array}{c}{F}_{d}{}_{{}_{x}}\\ {F}_{d}{}_{{}_{y}}\\ {F}_{d}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{F}_{g}{}_{{}_{x}}\\ {F}_{g}{}_{{}_{y}}\\ {F}_{g}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{F}_{ext}{}_{{}_{x}}\\ {F}_{ext}{}_{{}_{y}}\\ {F}_{ext}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{F}_{FL}{}_{{}_{x}}\\ {F}_{FL}{}_{{}_{y}}\\ {F}_{FL}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{F}_{FR}{}_{{}_{x}}\\ {F}_{FR}{}_{{}_{y}}\\ {F}_{FR}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{F}_{RL}{}_{{}_{x}}\\ {F}_{RL}{}_{{}_{y}}\\ {F}_{RL}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{F}_{RR}{}_{{}_{x}}\\ {F}_{RR}{}_{{}_{y}}\\ {F}_{RR}{}_{{}_{z}}\end{array}\right]\\ \\ {\overline{M}}_{b}=\left[\begin{array}{c}{M}_{x}\\ {M}_{y}\\ {M}_{z}\end{array}\right]=\left[\begin{array}{c}{M}_{d}{}_{{}_{x}}\\ {M}_{d}{}_{{}_{y}}\\ {M}_{d}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{M}_{ext}{}_{{}_{x}}\\ {M}_{ext}{}_{{}_{y}}\\ {M}_{ext}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{M}_{FL}{}_{{}_{x}}\\ {M}_{FL}{}_{{}_{y}}\\ {M}_{FL}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{M}_{FR}{}_{{}_{x}}\\ {M}_{FR}{}_{{}_{y}}\\ {M}_{FR}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{M}_{RL}{}_{{}_{x}}\\ {M}_{RL}{}_{{}_{y}}\\ {M}_{RL}{}_{{}_{z}}\end{array}\right]+\left[\begin{array}{c}{M}_{RR}{}_{{}_{x}}\\ {M}_{RR}{}_{{}_{y}}\\ {M}_{RR}{}_{{}_{z}}\end{array}\right]+{\overline{M}}_{F}\end{array}
{J}_{ij}={I}_{ij}+m\left({|R|}^{2}{\delta }_{ij}-{R}_{i}{R}_{j}\right)
\begin{array}{l}\overline{w}=\sqrt{{\left(\stackrel{˙}{x}-{w}_{x}\right)}^{2}+{\left(\stackrel{˙}{x}-{w}_{x}\right)}^{2}+{\left({w}_{z}\right)}^{2}}\\ \\ {F}_{dx}=-\frac{1}{2TR}{C}_{d}{A}_{f}{P}_{abs}{\left(}^{\overline{w}}\\ {F}_{dy}=-\frac{1}{2TR}{C}_{s}{A}_{f}{P}_{abs}{\left(}^{\overline{w}}\\ {F}_{dz}=-\frac{1}{2TR}{C}_{l}{A}_{f}{P}_{abs}{\left(}^{\overline{w}}\end{array}
\begin{array}{l}{M}_{dr}=-\frac{1}{2TR}{C}_{rm}{A}_{f}{P}_{abs}{\left(}^{\overline{w}}\left(a+b\right)\\ {M}_{dp}=-\frac{1}{2TR}{C}_{pm}{A}_{f}{P}_{abs}{\left(}^{\overline{w}}\left(a+b\right)\\ {M}_{dy}=-\frac{1}{2TR}{C}_{ym}{A}_{f}{P}_{abs}{\left(}^{\overline{w}}\left(a+b\right)\end{array}
x,\stackrel{˙}{x},\stackrel{¨}{x}
y,\stackrel{˙}{y},\stackrel{¨}{y}
z,\stackrel{˙}{z},\stackrel{¨}{z}
FSusp=\left[\begin{array}{cccc}{F}_{FLx}& {F}_{FRx}& {F}_{RLx}& {F}_{RRx}\\ {F}_{FLy}& {F}_{FRy}& {F}_{RLy}& {F}_{RRy}\\ {F}_{FLz}& {F}_{FRz}& {F}_{RLz}& {F}_{RRz}\end{array}\right]
MSusp=\left[\begin{array}{cccc}{M}_{FLx}& {M}_{FRx}& {M}_{RLx}& {M}_{RRx}\\ {M}_{FLy}& {M}_{FRy}& {M}_{RLy}& {M}_{RRy}\\ {M}_{FLz}& {M}_{FRz}& {M}_{RLz}& {M}_{RRz}\end{array}\right]
\text{FExt}={F}_{ext}=\left[\begin{array}{ccc}{F}_{ex{t}_{x}}& {F}_{ex{t}_{y}}& {F}_{ex{t}_{z}}\end{array}\right]or\left[\begin{array}{c}{F}_{ex{t}_{x}}\\ {F}_{ex{t}_{y}}\\ {F}_{ex{t}_{z}}\end{array}\right]
\text{MExt}={M}_{ext}=\left[\begin{array}{ccc}{M}_{ex{t}_{x}}& {M}_{ex{t}_{y}}& {M}_{ex{t}_{z}}\end{array}\right]or\left[\begin{array}{c}{M}_{ex{t}_{x}}\\ {M}_{ex{t}_{y}}\\ {M}_{ex{t}_{z}}\end{array}\right]
\beta =\frac{{V}_{y}}{{V}_{x}}
z1I=\left[\begin{array}{ccc}{I}_{xx}& {I}_{xy}& {I}_{xz}\\ {I}_{yx}& {I}_{yy}& {I}_{yz}\\ {I}_{zx}& {I}_{zy}& {I}_{zz}\end{array}\right]
z2I=\left[\begin{array}{ccc}{I}_{xx}& {I}_{xy}& {I}_{xz}\\ {I}_{yx}& {I}_{yy}& {I}_{yz}\\ {I}_{zx}& {I}_{zy}& {I}_{zz}\end{array}\right]
z3I=\left[\begin{array}{ccc}{I}_{xx}& {I}_{xy}& {I}_{xz}\\ {I}_{yx}& {I}_{yy}& {I}_{yz}\\ {I}_{zx}& {I}_{zy}& {I}_{zz}\end{array}\right]
z4I=\left[\begin{array}{ccc}{I}_{xx}& {I}_{xy}& {I}_{xz}\\ {I}_{yx}& {I}_{yy}& {I}_{yz}\\ {I}_{zx}& {I}_{zy}& {I}_{zz}\end{array}\right]
z5I=\left[\begin{array}{ccc}{I}_{xx}& {I}_{xy}& {I}_{xz}\\ {I}_{yx}& {I}_{yy}& {I}_{yz}\\ {I}_{zx}& {I}_{zy}& {I}_{zz}\end{array}\right]
z6I=\left[\begin{array}{ccc}{I}_{xx}& {I}_{xy}& {I}_{xz}\\ {I}_{yx}& {I}_{yy}& {I}_{yz}\\ {I}_{zx}& {I}_{zy}& {I}_{zz}\end{array}\right]
z7I=\left[\begin{array}{ccc}{I}_{xx}& {I}_{xy}& {I}_{xz}\\ {I}_{yx}& {I}_{yy}& {I}_{yz}\\ {I}_{zx}& {I}_{zy}& {I}_{zz}\end{array}\right]
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Rahul is younger than Radha by 10 yrs. If 5 yrs back their ages were in the ratio 1:2, how old is Radha?
Let Rahul age = x years
⇒ Radha's age = (x + 10)years
⇒ x - 5 /x + 10-5 = 1/ 2
Hence, Radha's age = (15 + 10)years = 25 years
A circular swimming pool is surrounded by a concrete wall 4m wide. If the area of the concrete wall surrounding the pool is 11 25 that of the pool, then the radius (in m) of the pool is
Of the 3 numbers,second is twice the first and is also thrice the third. If the average of the three numbers is 44,the largest number is
The biggest number should be 72.
First, let’s start off by defining the numbers in terms of x. If the second number (the largest number) is defined by x, the first number is 1/2x and the third number is 1/3x. This is because the first number is half of the second number while the third number is a third of the second number.
Now that we have defined the three numbers in terms of x, let’s focus on the average given in the question. To find the average, you add up all the values and divide by the number of values.
Average= (Number 1+Number 2+Number 3)/number of values.
Since there are three numbers and the average is given, we can say
44=(1/2x+x+1/3x)/3
Multiply both sides by three
132=1/2x+x+1/3x
Add up all the like terms
132=11/6x
One-third of two-fifth of three-seventh of a number is 10. What is 40% of that number?
A man buys 2 dozen bananas at Rs.16 per dozen. After selling 18 bananas at the rate of Rs.12 per dozen, the shopkeeper reduced the rate to Rs.4 per dozen. The perent loss is :
C.P. = Rs (16X 2) = 32
S.P. = Rs (12 X 1.5 + 4 X 0.5) = Rs(18 + 2) = Rs 20
∴ Loss% = ( 12 32 X 100)% = 37.5%
The ratio of the boys to that of the girls in a school is 6:5. The number of boys is more by 200. What is the total number of girls in the school?
and that of boys be(x + 200)
\frac{x + 200}{x}
\frac{6}{5}
⇒ 5x + 1000 = 6x
⇒ The number of girls is 1000
At what an amount is doubled at the rate of 10% per annum simple interest
Amount is doubled mean interest = Principal
⇒ P X 10 X T/ 100 = P
⇒ Time,T = 10 year
A wall of 100 mts can be built by 7 men or 10 women in 10 days. How many days will 14 men and 20 women take to build a wall of 600 mts?
If both the radius and height of a right circular cone are increased by 20%, its volume will be increased by :
I. √2304=a2304=a
II. b2 = 2304
a = √23042304
⇒ b = √23042304
Thus a = b
I. 12a2 - 7a + 1 = 0
II. 15b2 - 16b + 4 = 0
⇒ 12a2 - 4a - 3a + 1 = 0
⇒ 4a(3a - 1) -1 (3a - 1) = 0
⇒ (4a - 1) (3a - 1) = 0
⇒' 4a = 1 ⇒ a = 1414
3a = 1 ⇒ a = 1313
⇒ 15b2 - 10b - 6b + 4 = 0
⇒ 5b (3b - 2) - 2(3b - 2) = 0
⇒ (3b - 2)(5b - 2) = 0
⇒ 3b = 2 5b = 2
⇒ b = 2/3 ⇒ b = 2/5
Thus a < b
A survey conducted on 5800 villagers staying in various villagers and having variuos favourite fruits .
Mango is the favourite fruit of 50% of the people from Village C. People having their favourite fruit as Mango from Village C form approximately what percent of the people having their faviourite fruit as mango from all the villagers together ?
20% of the people from Village D have banana as their favourite fruit and 12% of the people from the same village have guava as their favourite fruit. How many people from the village like other fruits ?
People in villave D = 25% of 5800 = 1450
∴ Required no. of people
= {100 - (20 + 12)} % of 1450
= 68% of 1450 = 986
In the following table region-wise market share (in percentage) of Companies A, B, C for three products a, b and g is shown.
Note: N= Northern region, S= Southern region, E= Eastern region, W=Western region.
What is the total sales of products α,βα,β and γγ of Companay A?
Rs.1162.5crore
Rs.135.5crore
If another Company D also produce αα ,β,β and γγ and has the remaining market share of the three products in all the four region, then Company D has more than 50% market share.
Of only one product in only one of the regions.
Of only two products in one of the regions.
Of at least one product in all the four regions.
Of all the products in at least one region.
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Revision as of 12:24, 4 August 2013 by NikosA (talk | contribs) (→Deriving Physical Quantities: added band-specific radiometric parameters)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
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Newton's Second Law for Composite Systems | Brilliant Math & Science Wiki
Newton's Second Law for Composite Systems
Dale Gray and Jimin Khim contributed
Newton's three laws of motion form the basis for classical mechanics. Central to the modern discussion of mechanics is the concept of a particle, i.e. a body that is small enough that its motion is described sufficiently by giving its position as a function of time. Newton's first law defines how one recognizes the presence of a net force on a particle by describing how the particle behaves in the absence of a net force. The second law expresses the relation between a net force on a particle of mass,
m
, and the acceleration produced by that force. Using bold face letters for vectors and regular type for scalars, that relationship is expressed by the formula
\mathbf{F} = m \mathbf{a},
\mathbf{F}
is the net force on the particle of mass,
m
\mathbf{a}
is the particle's acceleration.
For extended bodies, where rotational motion becomes relevant, we must consider where the various forces that contribute to the net force are applied. This consideration leads to the concept of torque. Before considering torque, we will show that a system of particles obeys Newton's second law, for the net external force and the acceleration of the center of mass.
Torque on a System of Particles
An extended body will be treated as a collection of point particles. Since the particles in the system may exert forces on each other, the net force
\mathbf{F}_\textrm{i,net}
i^\text{th}
particle is the sum of the net external force
\mathbf{F}_\textrm{i,ext}
and the net internal force
\mathbf{F}_\textrm{i,int}
. It is shown below that, because of Newton's third law, the forces that the particles of a system exert on each other can be ignored when considering translational motion of the system as a whole:
\mathbf{F}_\textrm{i,net} = \mathbf{F}_\textrm{i,ext} + \mathbf{F}_\textrm{i,int}. \qquad (1)
The net force on the system is therefore
\mathbf{F}_\textrm{net} = \sum_i \mathbf{F}_\textrm{i,ext} + \sum_i \mathbf{F}_\textrm{i,int} . \qquad (2)
The first summation on the right side of (2) is the net external force on the system
\mathbf{F}_\textrm{net,ext}
, and the second sum will be seen to be zero when Newton's third law is used. Let
\mathbf{F}_\textrm{i,j}
be the force on the
i^\text{th}
particle due to the
j^\text{th}
particle. Newton's third law is
\mathbf{F}_\textrm{i,j} = - \mathbf{F}_\textrm{j,i}. \qquad (3)
The second term on the right side of (2) is
\sum_i \mathbf{F}_\textrm{i,int} = \sum_i \sum_j \mathbf{F}_\textrm{i,j} = 0, \qquad (4)
since the internal forces cancel in pairs. Therefore,
\mathbf{F}_\textrm{net} = \mathbf{F}_\textrm{net,ext}. \qquad (5)
It will now be shown, by introducing the concept of center of mass of the system of particles, that for translational motion, one can treat the entire system as a "particle" obeying Newton's second law, where the applied force is the net external force and the acceleration of the "particle" is the acceleration of the center of mass. This is a considerable simplification when considering a composite system.
The position vector
\mathbf{r}_\textrm{cm}
of the center of mass of a system of particles is a "weighted average" of the position vectors
\mathbf{r}_\textrm{i}
of all particles:
\mathbf{r}_\textrm{cm} = \sum_\textrm{i} \frac{m_i \mathbf{r}_i}{M}, \qquad (6)
M
M = \sum_i m_i
. The velocity and acceleration of the center of mass are, respectively,
\mathbf{v}_\textrm{cm} = \frac{d \mathbf{r}_\textrm{cm}}{dt},\quad \mathbf{a}_\textrm{cm} = \frac{d^2 \mathbf{r}_\textrm{cm}}{dt^2}. \qquad (7)
Apply Newton's second law to each particle:
\begin{aligned} \mathbf{F}_\textrm{i} &= m_\textrm{i} \mathbf{a}_\textrm{i} \\&= m_\textrm{i}\frac{d^2 \mathbf{r}_\textrm{i}}{dt^2}\\\\ \mathbf{F}_\textrm{net,ext} &= \frac{d^2}{dt^2} \sum_\textrm{i} m_\textrm{i} \mathbf{r}_\textrm{i} \\&= M \mathbf{a}_\textrm{cm}. \qquad (8) \end{aligned}
This means that the center of mass accelerates under the influence of the net external force just as a particle accelerates under the influence of the net force acting on it.
An artillery shell is fired at an angle of projection of
\theta_{0} = 30^\circ
v_{0} = 400\text{ m/s}
. At the apex of its trajectory the shell explodes into two pieces with the larger piece three times as massive as the smaller one. The smaller piece drops straight down below the point of the explosion. How far from the launch point does the larger piece land? Assume that there is no air resistance, that the ground is level, and that the acceleration of gravity is
g = 9.8\text{ m/s}^2
R
be the range of an identical shell which does not explode and is launched at the same angle of projection and with the same speed. Choose the origin at the launch point and let
m_1
be the mass of the smaller piece and
m_2
be the mass of the larger piece. We are given that
m_2 = 3m_1
x_1 = \frac{1}{2}R,
and are asked to find
x_2
The net external force on the system, after the shell is launched, is the weight of the shell. Therefore, the center of mass of the exploded shell follows the same trajectory as the unexploded shell and has, therefore, the same range. For
x_\text{cm}
after the pieces have landed, we have
x_\text{cm} = R
. Using the information we are given, we have
\begin{aligned} R &= \frac{0.5Rm_1 +3m_1 x_2}{4m_1}\\ &= \frac{0.5R + 3x_2}{4}\\ \Rightarrow x_2 &= \frac{7}{6}R. \qquad (9) \end{aligned}
The range of a projectile is
R = \frac{v_{0}^2 \sin (2\theta_0)}{g}. \qquad (10)
Calculation gives
x_2 = 19 \text{ km}
_\square
\mathbf{\tau}
about the origin produced by a force
\mathbf{F}
is defined in terms of the position vector
\mathbf{r}
of a particle and the force as
\mathbf{\tau} = \mathbf{r}\times \mathbf{F}. \qquad (11)
For a system of
N
m_\textrm{i}
\mathbf{r}_\textrm{i}
, the torque on the
i^\text{th}
particle produced by the net force on it,
\mathbf{F}_\textrm{i,net}
\mathbf{\tau}_\textrm{i} = \mathbf{r}_\textrm{i}\times \mathbf{F}_\textrm{i,net}. \qquad (12)
The net or total torque on the system about the chosen origin is
\mathbf{\tau}_\textrm{net} = \sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,net}. \qquad (13)
Using equation (1), the net torque on the system is
\mathbf{ \tau}_\textrm{net} = \sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,ext} + \sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,int}. \qquad (14)
The first term on the right side of the equation is the net external torque
\mathbf{ \tau}_\textrm{net,ext}
, and the second term is the sum of internal torques, which can be shown to be zero for purely mechanical forces. (In electromagnetic theory, this is not the case.)
The displacement vector from
m_i
m_j
\mathbf{r}_\textrm{i,j}
\mathbf{r}_j
\mathbf{r}_i
. (See the picture below.)
For mechanical forces one has, in addition to Newton's third law
\mathbf{F}_\textrm{i,j}
\mathbf{F}_\textrm{j,i}
, the so-called strong form of Newton's third law, i.e. that the internal forces
\mathbf{F}_\textrm{i,j}
\mathbf{r}_\textrm{i,j}
\sum_i \mathbf{F}_\textrm{i,int}
\sum_i \sum_j \mathbf{F}_\textrm{i,j}
, the sum of internal torque is
\sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,int} = \sum_i \sum_j \mathbf{r}_\textrm{i} \times \mathbf{F}_\textrm{i,j}. \qquad (15)
The summation on the right in equation (15) can be written as
\sum_i \sum_j \mathbf{r}_\textrm{i} \times \mathbf{F}_\textrm{i,j} = \frac{1}{2} \sum_i \sum_j \mathbf{r}_\textrm{i} \times \mathbf{F}_\textrm{i,j} +\frac{1}{2} \sum_j \sum_i \mathbf{r}_\textrm{j} \times \mathbf{F}_\textrm{j,i}. \qquad (16)
Collecting terms and using Newton's third law gives
\sum_i \sum_j \mathbf{r}_\textrm{i} \times \mathbf{F}_\textrm{i,j} = \frac{1}{2} \sum_i \sum_j ( \mathbf{r}_i - \mathbf{r}_j ) \times \mathbf{F}_\textrm{i,j} = \frac{1}{2} \sum_i \sum_j \mathbf{r}_{j.i} \times \mathbf{F}_\textrm{i,j}. \qquad (17)
Using the strong form of Newton's third law, the cross product in each term is seen to be zero. Therefore,
\sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,int} = 0, \qquad (18)
\mathbf{ \tau}_\textrm{net} = \sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,ext} = \mathbf{ \tau }_\textrm{ext}. \qquad (19)
Finally, we derive the relationship between torque and angular momentum and arrive at the rotational analogue of Newton's second law. The angular momentum of the
i^\text{th}
particle in the system is
\mathbf{L}_i
m_i
\mathbf{r}_i \times \mathbf{v}_i
. The total angular momentum for the system of particles is
\mathbf{L} = \sum_i \mathbf{L}_i = \sum_i m_i \mathbf{r}_i \times \mathbf{v}_i. \qquad (20)
\mathbf{L}
with respect to time gives
\frac{d \mathbf{L}}{dt} = \sum_i m_i \frac{d \mathbf{r}_i}{dt} \times \mathbf{v}_i + \sum_i m_i \mathbf{r}_i \times\frac{d \mathbf{v}_i}{dt}. \qquad (21)
The first sum on the right side of equation (21) is zero because of the cross product, and the second term is the net torque. Thus, we have
\mathbf{ \tau}_\textrm{net} = \frac{d \mathbf{L}}{dt}. \qquad (22)
APPLICATION: When riding a bicycle, turning the handle bars produces a torque on the turning front wheel. This forces the direction of the angular momentum vector to change. The change is either upward or downward, depending on whether a right turn or left turn, respectively, is made. This is the way a bicycle is kept upright by the rider.
We will now consider the rotation of a rigid body about an arbitrary fixed axis. The particles of the system rotate around the axis in fixed circles and maintain their positions relative to each other. The figure below shows a particle of the system, of mass
m_{i}
, rotating in a circle around an axis through the origin
O
. The axis is perpendicular to the plane of the circle and the direction of motion is counterclockwise as seen from above, as indicated by the curved arrow:
The velocity vector of the particle is the cross product
\mathbf{v}_{i} = \mathbf{ \omega}\times \mathbf{r}_{i}, \qquad (23)
and the angular momentum is
\mathbf{L}_{i} = m_i \mathbf{r}_{i} \times \mathbf{v}_{i} = m_i \mathbf{r}_{i} \times ( \mathbf{ \omega}\times \mathbf{r}_{i}). \qquad (24)
Applying an identity for the vector triple product, the so called BAC-CAB identity, gives
\mathbf{L}_{i} = m_{i} \big[ r_{i}^2 \mathbf{ \omega } - ( \mathbf{r}_{i} \centerdot \mathbf{ \omega }) \mathbf{r}_{i} \big]. \qquad (25)
The total angular momentum of the rigid body about the axis of rotation is, therefore,
\mathbf{L} = \sum_i \mathbf{L}_i = \sum_i m_{i} \big[ r_{i}^2 \mathbf{ \omega } - ( \mathbf{r}_{i} \centerdot \mathbf{ \omega }) \mathbf{r}_{i} \big]. \qquad (26)
Define the unit second rank tensor
\mathbf{ \iota }
such that, for each vector
\mathbf{w}
\mathbf{ \iota } \centerdot \mathbf{w} = \mathbf{w}. \qquad (27)
The moment of inertia tensor is defined by
\mathbf{I} = \sum_i m_{i} \big[ r_{i}^2 \mathbf{ \iota } - \mathbf{r}_{i} \otimes \mathbf{r}_{i}\big]. \qquad (28)
With this definition of the moment of inertia tensor, the angular momentum of the rigid body is
\mathbf{L} = \mathbf{I} \centerdot \mathbf{ \omega }. \qquad (29)
The components of the moment of inertia tensor produce a symmetric matrix, which consequently has three mutually orthogonal eigenvectors
\mathbf{a}_{k}\, (k = 1, 2, 3)
I_{k}
\mathbf{I} \centerdot \mathbf{a}_{k} = I_{k} \mathbf{a}_{k}. \qquad (30)
Axes of rotation through the center of mass, parallel to the eigenvectors of the moment of inertia tensor, are called principle axes of the body, and the eigenvalues are called principal moments of inertia. For a rigid body rotating around a principal axis, the angular momentum is
\mathbf{L}
I
\mathbf{\omega}
I
is the moment of inertia about the axis and
\mathbf{ \omega}
is the angular velocity vector. Since
I
is a constant, one has
\mathbf{ \tau}_\textrm{net} = \mathbf{ \tau }_\textrm{ext} = I\frac{d \mathbf{\omega}}{dt} = I \mathbf{\alpha}. \qquad (31)
This is known as Newton's second law for rotational motion.
Cite as: Newton's Second Law for Composite Systems. Brilliant.org. Retrieved from https://brilliant.org/wiki/newtons-second-law-for-composit-systems/
|
Stationary Distributions of Markov Chains | Brilliant Math & Science Wiki
Henry Maltby, Samir Khan, and Jimin Khim contributed
A stationary distribution of a Markov chain is a probability distribution that remains unchanged in the Markov chain as time progresses. Typically, it is represented as a row vector
\pi
whose entries are probabilities summing to
1
, and given transition matrix
\textbf{P}
\pi = \pi \textbf{P}.
\pi
is invariant by the matrix
\textbf{P}
Ergodic Markov chains have a unique stationary distribution, and absorbing Markov chains have stationary distributions with nonzero elements only in absorbing states. The stationary distribution gives information about the stability of a random process and, in certain cases, describes the limiting behavior of the Markov chain.
A sports broadcaster wishes to predict how many Michigan residents prefer University of Michigan teams (known more succinctly as "Michigan") and how many prefer Michigan State teams. She noticed that, year after year, most people stick with their preferred team; however, about
3\%
of Michigan fans switch to Michigan State, and about
5\%
of Michigan State fans switch to Michigan. However, there is no noticeable difference in the state's population of 10 million's preference at large; in other words, it seems Michigan sports fans have reached a stationary distribution. What might that be?
A reasonable way to approach this problem is to suppose there are
x
million Michigan fans and
y
million Michigan State fans. The state's population is
10
million, so
x + y = 10
. These numbers do not change each year. It follows that
\begin{aligned} x &= 0.97x + 0.05y \\ y &= 0.03x + 0.95y. \end{aligned}
Rearranging either equation,
x = \tfrac{5}{3} y
x + y = 10
y = \tfrac{3}{8} \cdot 10 = 3.75
x = 6.25
6.25
3.75
million Michigan state fans. In other words, the stationary distribution is
(0.625, \, 0.375)
_\square
Note that the limiting distribution does not depend on the number of fans in the state!
Relation to Limiting Distribution
Students of linear algebra may note that the equation
\pi \textbf{P} = \pi
looks very similar to the column vector equation
M v = \lambda v
for eigenvalues and eigenvectors, with
\lambda = 1
. In fact, by transposing the matrices,
\left(\pi \textbf{P}\right)^T = \pi^T \implies \textbf{P}^T \pi^T = \pi^T.
In other words, the transposed transition matrix
\textbf{P}^T
has eigenvectors with eigenvalue
1
that are stationary distributions expressed as column vectors. Therefore, if the eigenvectors of
\textbf{P}^T
are known, then so are the stationary distributions of the Markov chain with transition matrix
\textbf{P}
. In short, the stationary distribution is a left eigenvector (as opposed to the usual right eigenvectors) of the transition matrix.
When there are multiple eigenvectors associated to an eigenvalue of 1, each such eigenvector gives rise to an associated stationary distribution. However, this can only occur when the Markov chain is reducible, i.e. has multiple communicating classes.
In genetics, one method for identifying dominant traits is to pair a specimen with a known hybrid. Their offspring is once again paired with a known hybrid, and so on. In this way, the probability of a particular offspring being purely dominant, purely recessive, or hybrid for the trait is given by the table below.
States Child Dominant Child Hybrid Child Recessive
Parent Dominant
\hspace{1cm} 0.5
\hspace{1cm} 0.5
\hspace{1cm} 0
Parent Hybrid
\hspace{1cm} 0.25
\hspace{1cm} 0.5
\hspace{1cm} 0.25
Parent Recessive
\hspace{1cm} 0
\hspace{1cm} 0.5
\hspace{1cm} 0.5
What is a stationary distribution for this Markov chain?
The transition matrix is
\textbf{P} = \begin{pmatrix} 0.5 & 0.5 & 0 \\ 0.25 & 0.5 & 0.25 \\ 0 & 0.5 & 0.5 \end{pmatrix}.
The transpose of this matrix has eigenvalues satisfying the equation
\det \begin{pmatrix} 0.5 - \lambda & 0.25 & 0 \\ 0.5 & 0.5 - \lambda & 0.5 \\ 0 & 0.25 & 0.5 - \lambda \end{pmatrix} = 0.
(0.5 - \lambda)^3 - 0.125 (0.5 - \lambda) - 0.125 (0.5 - \lambda) = (0.5 - \lambda) (\lambda^2 - \lambda) = 0
. So the eigenvalues are
\lambda = 0
\lambda = 0.5
\lambda = 1
. The eigenvalue
\lambda = 0
gives rise to the eigenvector
(1,\, -2,\, 1)
, the eigenvalue
\lambda = 0.5
(-1,\, 0,\, 1)
, and the eigenvalue
\lambda = 1
(1,\, 2,\, 1)
. The only possible candidate for a stationary distribution is the final eigenvector, as all others include negative values.
Then, the stationary distribution must be
\tfrac{1}{1 + 2 + 1} \cdot (1,\, 2,\, 1) = \left(\tfrac{1}{4},\, \tfrac{1}{2},\, \tfrac{1}{4}\right)
_\square
\left(\frac{1}{2}, \ \frac{1}{2}\right)
\left(\frac{8}{15}, \ \frac{7}{15}\right)
\left(\frac{3}{5}, \ \frac{2}{5}\right)
(1, \ 0)
Find a stationary distribution for the 2-state Markov chain with stationary transition probabilities given by the following graph:
The limiting distribution of a Markov chain seeks to describe how the process behaves a long time after . For it to exist, the following limit must exist for any states
i
j
L_{i,j} = \lim_{n \to \infty} \mathbb{P}(X_n = j \mid X_0 = i).
Furthermore, for any state
i
, the following sum must be
1
\sum_{\text{states }j} \lim_{n \to \infty} \mathbb{P}(X_n = j \mid X_0 = i) = 1.
This ensures that the numbers obtained do, in fact, constitute a probability distribution. Provided these two conditions are met, then the limiting distribution of a Markov chain with
X_0 = i
is the probability distribution given by
\ell = \left(L_{i,j}\right)_{\text{states }j}
For any time-homogeneous Markov chain that is aperiodic and irreducible,
\lim_{n \to \infty} \textbf{P}^n
converges to a matrix with all rows identical and equal to
\pi
. Not all stationary distributions arise this way, however. Some stationary distributions (for instance, certain periodic ones) only satisfy the weaker condition that the average number
\tfrac{1}{n+1} \cdot H_i^{(n)}
of times the process is in state
in the first
n
steps approaches the corresponding value of the stationary distribution. That is, if
(x_1, \, x_2, \, \dots,\, x_m)
is the stationary distribution, then
\lim_{n \to \infty} \frac{H_i^{(n)}}{n+1} = x_i.
Not all stationary distributions are limiting distributions.
Consider the two-state Markov chain with transition matrix
\textbf{P} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.
n
increases, there is no limiting behavior to
\textbf{P}^n
. In fact, the expression simply alternates between evaluating to
P
I
, the identity matrix. However, the system has stationary distribution
\left(\tfrac{1}{2}, \, \tfrac{1}{2}\right)
\left(\tfrac{1}{2}, \, \tfrac{1}{2}\right) \cdot \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \left(\tfrac{1}{2}, \, \tfrac{1}{2}\right).
So, not all stationary distributions are limiting distributions. Sometimes no limiting distribution exists!
_\square
For time-homogeneous Markov chains, any limiting distribution is a stationary distribution.
Let the Markov chain have transition matrix
\textbf{P}
. Then, suppose
\lim_{n \to \infty} \textbf{P}^n = \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} \pi.
That is, the limit is an
n \times n
matrix with all rows equal to
\pi
. Then note that
\begin{aligned} \lim_{n \to \infty} \textbf{P}^n \cdot \textbf{P} &= \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} \pi \textbf{P} \\ \lim_{n \to \infty} \textbf{P}^{n+1} &= \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} \pi. \end{aligned}
Inspecting one row of the left matrix being multiplied on the right-hand side, it becomes clear that
\pi \textbf{P} = \pi
. Thus, the limiting distribution is also a stationary distribution.
_\square
Cite as: Stationary Distributions of Markov Chains. Brilliant.org. Retrieved from https://brilliant.org/wiki/stationary-distributions/
|
Find the area of the region enclosed by one loop
Find the area of the region enclosed by one loop of the curve. r=\sin 4 \thet
uneskovogl5 2021-11-16 Answered
Find the area of the region enclosed by one loop of the curve.
r=\mathrm{sin}4\theta
Annie Midgett
r=\mathrm{sin}\left(4\theta \right)
Area enclosed by one of the loops will be simply
{\int }_{a}^{b}\frac{{r}^{2}}{2}d\theta
We will find the limits of integration a, b, and then we will integrate the integral to find the area enclosed by one of the loops.
To find the limits of integration, we need to find two consecutive values of
\theta
for which r is zero.
y=\mathrm{sin}4x
We can see that 0 and
\frac{\pi }{4}
are two consecutive values for which
\mathrm{sin}4x
Therefore, we can integrate from 0 to
\frac{\pi }{4}
You can also integrate from
-\frac{\pi }{4}
The only condition is that you have to integrate between two consecutive values of
\theta
r=0
Therefore, the limit of integration is from 0 to
\frac{\pi }{4}
Area enclosed by one of the loops is
{\int }_{0}^{\frac{\pi }{4}}\frac{{r}^{2}}{2}d\theta
{\int }_{0}^{\frac{\pi }{4}}\frac{{\left(\mathrm{sin}\left(4\theta \right)\right)}^{2}}{2}d\theta
{\int }_{0}^{\frac{\pi }{4}}\frac{{\mathrm{sin}}^{2}\left(4\theta \right)}{2}d\theta
{\int }_{0}^{\frac{\pi }{4}}\left(\frac{1-\mathrm{cos}\left(8\theta \right)}{4}\le ft\right)d\theta
We used the formula
{\mathrm{sin}}^{2}\theta =\frac{1-\mathrm{cos}2\theta }{2}
{\left[\frac{\theta }{4}-\frac{\mathrm{sin}\left(8\theta \right)}{32}\right]}_{0}^{\frac{\pi }{4}}=\frac{\pi }{16}
Area enclosed by the loop is
\frac{\pi }{16}
r=\mathrm{sin}4\theta
Area of the curve enclosed in the first loop is,
A={\int }_{a}^{b}\frac{1}{2}{r}^{2}d\theta
Two consecutive value of theta for which
\mathrm{sin}4\theta
\mathrm{sin}4\theta =0
\theta =0,\text{ }\frac{\pi }{4}
Now integrate from 0 to
\frac{\pi }{4}
Area enclosed by one of the loop is,
A={\int }_{0}^{\frac{\pi }{4}}\frac{1}{2}{\left(\mathrm{sin}4\theta \right)}^{2}d\theta
=\frac{1}{2}{\int }_{0}^{\frac{\pi }{4}}\left(\frac{1-\mathrm{cos}8\theta }{2}\right)d\theta \left[\mathrm{cos}2\theta =1-2{\mathrm{sin}}^{2}\theta \right]
=\frac{1}{4}{\int }_{0}^{\frac{\pi }{4}}\left(1-\mathrm{cos}8\theta \right)d\theta
=\frac{1}{4}{\int }_{0}^{\frac{\pi }{4}}\left(1\right)d\theta -\frac{1}{4}{\int }_{0}^{\frac{\pi }{4}}\left(\mathrm{cos}8\theta \right)d\theta
=\frac{1}{4}\left({\left[\theta \right]}_{0}^{\frac{\pi }{4}}-{\left[\frac{\mathrm{sin}8\theta }{8}\right]}_{0}^{\frac{\pi }{4}}\right)
=\frac{1}{4}\left(\left(\frac{\pi }{4}-0\right)-\left(\frac{\mathrm{sin}2\pi -\mathrm{sin}0}{8}\right)\right)
=\frac{1}{4}\left(\frac{\pi }{4}-\left(\frac{0-0}{8}\right)\right)
=\frac{\pi }{16}
f\left(x\right)=\sqrt{4-x}
a=0
\sqrt{3.9}
\sqrt{3.99}
Cramer’s Rule to solve (if possible) the system of linear equations.
-8{x}_{1}+7{x}_{2}\mid -10{x}_{3}=-151
12{x}_{1}+3{x}_{2}-5{x}_{3}=86
15{x}_{1}-9{x}_{2}+2{x}_{3}=187
Do the equations
4x-3y=5
7y+2x=-8
I have been learning that the standard expression of a linear equation in two variables is of the form:
ax+by+c=0
a\ne 0
b\ne 0
. I want to understand the purpose of the constant 'c' in this equation and where it was derived from, whether it be graphically or algebraically.
I had one doubt in matrix form of linear equations
Say , we have a system of equations
Ax=b
such that A is an
n×n
matrix and b is a
n×1
matrix and so is x then, if we are told that
\text{rank}\left(A\right)=n
, then, do we need to check that
\text{rank}\left(A\mid b\right)=n
or can we say that since b is a linear combination of the component vectors in A then augmenting it in A won't increase the number of linearly independent column vectors and hence the
\text{rank}\left(A\mid b\right)=n
and can't ever be
n+1
|
Unusual microstructure of \(Zn_{45}Au_{30}Cu_{25}\) with satisfaction of the cofactor conditions – MURI: Managing the Mosaic of Microstructure
MURI: Managing the Mosaic of Microstructure
Sponsored by the MURI program. Managed by the Air Force Office of Scientific Research
High Temporal Resolution CT Reconstruction Using Interlaced View Sampling
Unusual microstructure of
Zn_{45}Au_{30}Cu_{25}
with satisfaction of the cofactor conditions
By admin in Research Highlights
The cofactor conditions, as the conditions of compatibility between phases consist of two main subconditions: (CC1)
\lambda_2 = 1
, which means the middle principle stretch of the transformation is unity, (CC2) (a.v2)(n.v2)=0 where the vector n is the normal of twinning plane, the vector a corresponds to the twinning shear and the vector v2 is the undistorted direction associated with the middle principle stretch, and a mild inequality condition (CC3) that is always true for Type I and Type II twin system of martensite. The microstructure of a material undergoing martensitic transformation has strongly dependence on the conditions of compatibility, and further influences the macroscopic properties of the material. Here we report an example of a phase-changing material
Zn_{45}Au_{30}Cu_{25}
with the satisfaction of the cofactor conditions for both Type I and Type II twins. Theoretically, as a result of the satisfaction of both Type I and Type II cofactor conditions, two martensite variants can form a triple junction with austenite with perfect fit at each of the interfaces , while they can also form a quad-junction with the other two variants through Type II twinning interfaces without paying additional elastic energy. Macroscopically, many compatible triple and quad junction with different volume fraction become building blocks for the microstructure of the material undergoing phase transformation. Experimentally, we have observed the riverine strip microstructures as well as zig-zag interfaces in this material by the optical microscopy, which is completely unreproducible for the consecutive transformation cycles while the hysteresis and the latent heat are stable for each of the cycles. To further quantify the orientation correspondences between the real microstructure observed under the optical microscopy and the calculated morphology by the theory, we use the electron backscatter diffraction technique to measure the orientation for each martensite variant, then verify the cofactor conditions by calculating the rotation matrix connecting the adjacent variants.
Protected: ICOTOM Crystallography Tutorial (11/5/17)
3D Visualization of Textures in Metals and Alloys
Electron Channeling Pattern Dictionary Indexing Manuscript
EBSD Data Files
3D Visualization of neo-Eulerian and other rotation representations
New Release of EMsoft Package
Software for Time Interlaced Model Based Iterative Reconstruction (TIMBIR)
July 2015 CMU Workshop for 3D Microstructural Studies
June 2015 MURI Program Review
Study mosaic of multi-variant martensites by Laue microdiffraction
3D Model-Based Iterative Reconstruction for Synchrotron Tomography
Construction of a new 3D projection dome at CMU
A predictive data mining approach for microstructure optimization
Model-Based Dynamic Sampling
© 2022 MURI: Managing the Mosaic of Microstructure.
|
Dynamic Snap-Through of a Shallow Arch Under a Moving Point Load | Journal of Vibration and Acoustics | ASME Digital Collection
e-mail: jschen@ccms.ntu.edu.tw
Contributed by the Technical Committee on Vibration and Sound for publication in the JOURNAL OF VIBRATION AND ACOUSTICS. Manuscript received October 2002; final revision, March 2004. Associate Editor: Chin An Tan.
Chen, J., and Lin, J. (December 21, 2004). "Dynamic Snap-Through of a Shallow Arch Under a Moving Point Load ." ASME. J. Vib. Acoust. October 2004; 126(4): 514–519. https://doi.org/10.1115/1.1804991
In this paper we study the dynamic behavior of a shallow arch under a point load Q traveling at a constant speed. Emphasis is placed on finding whether snap-through buckling will occur. In the quasi-static case when the moving speed is almost zero, there exists a critical load
Qcr
in the sense that no static snap-through will occur as long as Q is smaller than
Qcr.
In the dynamic case when the point load travels with a nonzero speed, the critical load
Qcrd
is, in general, smaller than the static one. When Q is greater than
Qcrd,
there exists a finite speed zone within which the arch runs the risk of dynamic snap-through either while the point load is still on the arch or after the point load leaves the arch. The boundary of this dangerous speed zone can be determined by a more conservative criterion, which employs the concept of total energy and critical energy barrier, to guarantee the safe passage of the point load. This criterion requires the numerical integration of the equations of motion only up to the instant when the point load reaches the other end of the arch.
buckling, bending, elastic deformation
Arches, Stress, Equilibrium (Physics), Buckling, Equations of motion, Bifurcation
Buckling of Flat Curved Bars and Slightly Curved Plates
The Energy Criterion and Snap Buckling of Arches
Collapse of Arches Under Repeated Loading
Buckling of Shallow Arches
Inelastic Buckling of Shallow Arches
Snapping of Low Pinned Arches on an Elastic Foundation
Stability of Structures With Small Imperfections
Simitses, G. J., 1990, Dynamic Stability of Suddenly Loaded Structures, Springer-Verlag, New York.
Dynamic Analysis of the Buckling of Laterally Loaded Flat Arches
On the Final States of Shallow Arches on Elastic Foundations Subjected to Dynamical Loads
Dynamic Stability of Shallow Arch With Elastic Supports—Application in the Dynamic Stability Analysis of Inner Winding of Transformer During Short Circuit
Effects of Prescribed End Motion on the Dynamic Stability of a Shallow Arch on an Elastic Foundation
On Dynamic Snap Buckling of Shallow Arches
The Snapping of a Shallow Sinusoidal Arch Under a Step Pressure Load
Dynamic Buckling of Shallow Arches
The Effect of Spatial Distribution on Dynamic Snap-Through
The Effect of Damping on Dynamic Snap-Through
Lin, J.-S., 2002, Dynamic Stability of a Shallow Arch Under Prescribed End Motion, Master Thesis, Department of Mechanical Engineering, National Taiwan University, Taipei, Taiwan.
Nayfeh, A. H., and Balachandran, B., 1995, Applied Nonlinear Dynamics, Wiley, New York.
|
Implement three-axis gyroscope - Simulink - MathWorks 日本
ωmeas
Second order dynamics
Natural frequency (rad/sec)
Scale factors and cross-coupling
G-sensitive bias
Update rate (sec)
Noise seeds
Lower and upper output limits
Implement three-axis gyroscope
Aerospace Blockset / GNC / Navigation
The Three-Axis Gyroscope block implements a gyroscope on each of the three axes. For more information on the measured body angular rates, see Algorithms.
Optionally, to apply discretizations to the block inputs and dynamics along with nonlinearizations of the measured body angular rates, use the Saturation block.
Anisoelastic bias and anisoinertial bias effects are not accounted for in this block.
This block is not intended to model the internal dynamics of different forms of the instrument.
ω — Angular rates
Angular rates in the body-fixed axes, specified as a three-element vector, in radians per second.
G's — Accelerations
Accelerations in the body-fixed axes, specified as a three-element vector, in Gs.
ωmeas — Measured angular rates
Measured angular rates from the gyroscope, returned as a three-element vector, in radians per second.
Second order dynamics — Second-order dynamics
To apply second-order dynamics to gyroscope readings, select this check box.
Block Parameter: dtype_g
Natural frequency (rad/sec) — Natural frequency
Natural frequency of the gyroscope, specified as a double scalar, in radians per second.
Block Parameter: w_g
Damping ratio — Damping ratio
Damping ratio of the gyroscope, specified as a double scalar.
Block Parameter: z_g
Scale factors and cross-coupling — Scale factors and cross coupling
[1 0 0; 0 1 0; 0 0 1] (default) | 3-by-3 matrix
Scale factors and cross-coupling, specified as a 3-by-3 matrix, to skew the gyroscope from body axes and to scale accelerations along body axes.
Block Parameter: g_sf_cc
Values: 3-by-3 matrix
Default: '[1 0 0; 0 1 0; 0 0 1]'
Measurement bias — Measurement bias
Long-term biases along the gyroscope axes, specified as a three-element vector, in radians per second.
Block Parameter: g_bias
G-sensitive bias — Maximum change in rates
Maximum change in rates due to linear acceleration, specified as a three-element vector, in radians per second per g-unit.
Block Parameter: g_sen
Values: three-element vector
Update rate (sec) — Update rate
Update rate of the gyroscope, specified as a double scalar, in seconds. An update rate of 0 creates a continuous gyroscope. If the Noise on check box is selected and the update rate is 0, the block updates the noise at a rate of 0.1.
Update this parameter value to 0 (continuous)
Configure a fixed-step solver for the model
you must also select the Automatically handle rate transition for data transfer check box in the Solver pane. This check box enables the software to handle rate transitions correctly.
Block Parameter: g_Ts
Noise on — White noise
To apply white noise to the gyroscope readings, select this check box.
Block Parameter: g_rand
Noise seeds — Noise seeds
[23093 23094 23095] (default) | three-element vector
Scalar seeds for the Gaussian noise generator for each axis of the gyroscope, specified as a three-element vector.
To enable this parameter, select Noise on.
Block Parameter: g_seeds
Default: '[23093 23094 23095]'
Noise power — Noise power
Height of the power spectral density (PSD) of the white noise for each axis of the gyroscope, specified as a three-element vector, in (rad/s)2/Hz.
Block Parameter: g_pow
Default: '[0.0001 0.0001 0.0001]'
Lower and upper output limits — Minimum and maximum values of angular rates
[-inf -inf -inf inf inf inf] (default) | six-element vector
Three minimum values and three maximum values of angular rates in each of gyroscope axes, specified as a six-element vector, in radians per second.
Block Parameter: g_sat
Values: six-element vector
Default: '[-inf -inf -inf inf inf inf]'
The measured body angular rates
\left({\stackrel{¯}{\mathrm{Ï}}}_{meas}\right)
include the body angular rates
\left({\stackrel{¯}{\mathrm{Ï}}}_{b}\right)
, errors, and, optionally, the discretizations and nonlinearizations of the signals:
{\stackrel{¯}{\mathrm{Ï}}}_{meas}={\stackrel{¯}{\mathrm{Ï}}}_{b}Ã{\stackrel{¯}{\mathrm{Ï}}}_{SFCC}+{\stackrel{¯}{\mathrm{Ï}}}_{bias}+GsÃ{\stackrel{¯}{\mathrm{Ï}}}_{gsens}+noise
{\stackrel{¯}{\mathrm{Ï}}}_{SFCC}
is a 3-by-3 matrix of scaling factors on the diagonal and misalignment terms in the nondiagonal,
{\stackrel{¯}{\mathrm{Ï}}}_{bias}
are the biases, (Gs) are the Gs on the gyroscope, and
{\stackrel{¯}{\mathrm{Ï}}}_{gsens}
are the G-sensitive biases.
[1] Rogers, R. M., Applied Mathematics in Integrated Navigation Systems, AIAA Education Series, 2000.
Three-axis Accelerometer | Three-axis Inertial Measurement Unit | Saturation
|
Perform the indicated operation: 7[\cos(228^{\circ})+i\sin(228^{\circ})]\times8[\cos(232^{\circ})+i\sin(232^{\circ})]
Perform the indicated operation:
7\left[\mathrm{cos}\left({228}^{\circ }\right)+i\mathrm{sin}\left({228}^{\circ }\right)\right]×8\left[\mathrm{cos}\left({232}^{\circ }\right)+i\mathrm{sin}\left({232}^{\circ }\right)\right]
Topic - complex numbers
56\left[\mathrm{cos}\left(460\right)+i\mathrm{sin}\left(460\right)\right]
Multiplication of two complex numbers in polar coordinates: multiply the r components and add the
\theta
7\left[\mathrm{cos}\left({228}^{\circ }\right)+i\mathrm{sin}\left({228}^{\circ }\right)\right]×8\left[\mathrm{cos}\left({232}^{\circ }\right)+i\mathrm{sin}\left({232}^{\circ }\right)\right]
=7×8\left[\mathrm{cos}\left({228}^{\circ }+232\right)+i\mathrm{sin}\left({228}^{\circ }+{232}^{\circ }\right)\right]
=56\left[\mathrm{cos}\left({460}^{\circ }\right)+i\mathrm{sin}\left({460}^{\circ }\right)\right]
56\left[\mathrm{cos}\left({460}^{\circ }\right)+i\mathrm{sin}\left({460}^{\circ }\right)\right]
How do I find the trigonometric form of the complex number
\sqrt{3}-i
10{z}^{4}
into polar coordinates where
z=2+4i
How do you find the fourth root of
16\left(\mathrm{cos}\left(\frac{4\pi }{3}\right)+i\mathrm{sin}\left(\frac{4\pi }{4}\right)\right)?
-9-9i÷-1-8i
\left[3{\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)\right]}^{3}
How do you convert the complex the number into polar representation:
2\sqrt{3}+2i
|
(→Bandlimitation and timing: html audio api toy)
* '''Panel 8''': when selected, generate a synthetic 'squarewave' (this is not quite equivalent to a bandlimited analog squarewave; the harmonic amplitudes are a bit different) that when aligned with the sampling phase just right gives the appearance of having infinite rise and fall time. The slider allows us to shift the waveform sample alignment back and forth by +/- one sample to reveal that the underlying signal is still band-limited.
{\displaystyle \ x(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}x_{\mathrm {square} }(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
|
Flat top weighted window - MATLAB flattopwin - MathWorks Switzerland
w = flattopwin(L)
w = flattopwin(L,sflag)
w = flattopwin(L) returns an L-point symmetric flat top window
w = flattopwin(L,sflag) returns an L-point symmetric flat top window using the window sampling method specified by sflag.
Create a 64-point symmetric flat top window. View the result using wvtool.
w = flattopwin(N);
w — Flat top window
Flat top window, returned as a column vector.
Flat top windows are summations of cosines. The coefficients of a flat top window are computed from the following equation:
w\left(n\right)={a}_{0}-{a}_{1}\mathrm{cos}\left(\frac{2\pi n}{L-1}\right)+{a}_{2}\mathrm{cos}\left(\frac{4\pi n}{L-1}\right)-{a}_{3}\mathrm{cos}\left(\frac{6\pi n}{L-1}\right)+{a}_{4}\mathrm{cos}\left(\frac{8\pi n}{L-1}\right),
0\le n\le L-1
. The coefficient values are:
Flat top windows have very low passband ripple (< 0.01 dB) and are used primarily for calibration purposes. Their bandwidth is approximately 2.5 times wider than a Hann window.
[1] D’Antona, Gabriele, and A. Ferrero. Digital Signal Processing for Measurement Systems. New York: Springer Media, 2006, pp. 70–72.
[2] Gade, Svend, and Henrik Herlufsen. “Use of Weighting Functions in DFT/FFT Analysis (Part I).” Windows to FFT Analysis (Part I): Brüel & Kjær Technical Review. Vol. x, Number 3, 1987, pp. 1–28.
blackman | hamming | hann | WVTool
|
Investigation of Tribological Behavior of Al–Si Alloy Against Steel Lubricated With Ionic Liquids of 1-Diethylphosphonyl-n-propyl-3- Alkylimidazolium Tetrafluoroborate | J. Tribol. | ASME Digital Collection
Investigation of Tribological Behavior of Al–Si Alloy Against Steel Lubricated With Ionic Liquids of 1-Diethylphosphonyl-
n
-propyl-3- Alkylimidazolium Tetrafluoroborate
Zonggang Mu,
University of Jinnan
, Jinan, 250022, P.R.C.; State Key Laboratory of Solid Lubrication,
, Chinese Academy of Sciences, Lanzhou, 730000, P.R.C.
e-mail: chm̱mouzg@ujn.edu.cn
Xiaoxuan Wang,
, Jinan, 250022, P.R.C.
Shuxiang Zhang,
Yongmin Liang,
Yongmin Liang
Meng Bao,
Mu, Z., Wang, X., Zhang, S., Liang, Y., Bao, M., and Liu, W. (June 23, 2008). "Investigation of Tribological Behavior of Al–Si Alloy Against Steel Lubricated With Ionic Liquids of 1-Diethylphosphonyl-
n
-propyl-3- Alkylimidazolium Tetrafluoroborate." ASME. J. Tribol. July 2008; 130(3): 034501. https://doi.org/10.1115/1.2913553
A series of room temperature ionic liquids bearing with phosphonyl groups on the imidazolium cations, namely, 1-(
3′-O,O
-diethylphosphonyl-
n
-propyl)-3-alkylimidazolium tetrafluoroborate, were prepared and their physical properties were determined. They were also evaluated as promising lubricants for the contacts of aluminum on steel by using a SRV test rig. The tribological test results show that the synthetic ionic liquids exhibit better friction-reducing and antiwear abilities than the unsubstituted ionic liquid of 1-ethyl-3-hexylimidazolium tetrafluoroborate (coded as L206) and phosphazene (X-1P). Both the anions and the side substitutes attached to the imidazolium cations affect the tribological performance of lubricants. The scanning electron microscopy, energy-dispersive x-ray analysis, and x-ray photoelectron spectroscopy analyses of the worn surfaces show that complicated tribochemical reactions are involved in the sliding process. The anion decomposition and chemical adsorption of cation took place on the worn surface of aluminum alloy during the sliding process. As a result of the generation of boundary lubrication films which are composed of metal fluorides,
B2O3
, BN, nitrogen oxide, and
FePO4
help to effectively reduce the friction and wear of the contacts.
adsorption, aluminium alloys, decomposition, lubricants, lubrication, organic compounds, scanning electron microscopy, silicon alloys, sliding friction, steel, wear resistance, X-ray chemical analysis, X-ray photoelectron spectra, ionic liquid, 1-(3′-O,O-diethylphosphonyl-n-propyl)-3-alkylimidazolium tetrafluoroborate, lubricant, aluminum-on-steel system
Aluminum, Aluminum alloys, Friction, Lubricants, Steel, Tribology, Wear, Silicon alloys, Scanning electron microscopy, X-rays
Friction and Wear of Al-Si Alloys
Wear and Seizure of Binary Al-Si Alloys
Wear Characteristics of Al-Si Alloys
Room Temperature Ionic Liquids: A Kind of Novel Versatile Lubricant
Room Temperature Ionic Liquid 1-Ethyl-3-Hexylimidazolium-bis(Trifluoromethylisulfonyl)-Imde as Lubricant for Steel-Steel Contact
Tribological Behavior of the Ionic Liquid of 1-Methy-3-Butyl-Imidazolium Hexafluorophosphate as Lubricant
Friction and Wear Behaviors of Ionic Liquid of Alkylimidazolium Hexafluorophosphates as Lubricants for Steel/Steel Contact
Functional Room-Temperature Ionic Liquids as Lubricants for an Aluminum-on-Steel System
Effect of the Functional Groups in Ionic Liquid Molecules on the Friction and Wear Behavior of Aluminum Alloy in Lubricated Aluminum-on-Steel Contact
Tribological Performance of an Ionic Liquid as a Lubricant for Steel/Aluminum Contacts
Synth. Lubr.
Effect of Antiwear and Extreme Pressure Additives on the Wear of Aluminum Alloy in Lubricated Aluminum-on-Steel Contact
The Effects of Molecular Structure of Organo Chlorine on the Lubricity of Al 2024 Against Steel
The Effects of Alcohols and Ethers on the Wear Behavior of Aluminum
Tribological Properties of Alcohols as Lubricating Additives for Aluminum-on-Steel Contact
The Effect of Pentaerythritol Partial Ester on the Wear of Aluminum
Bidentate Organic Oxygen Compounds as Boundary Lubricants for Aluminum
Effects of Diol Compounds on the Friction and Wear of Aluminum Alloy in a Lubricated Aluminum-on-Steel Contact
Friction and Wear Behaviour of an Al-Si Alloy Against Steel Lubricated With N- and O-Containing Organic Compounds
|
Which set of ordered pairs could be generated by an exponential function? A. (-
Which set of ordered pairs could be generated by an exponential function? A. (-1,-1/2), (0, 0),(1,1/2), (2, 1) B. (–1, –1), (0, 0), (1, 1), (2, 8) C. (-1,1/2), (0, 1), (1, 2), (2, 4) D. (–1, 1), (0, 0), (1, 1), (2, 4)
\left(-1,-\frac{1}{2}\right)
\left(1,\frac{1}{2}\right)
\left(-1,\frac{1}{2}\right)
An exponential function has the property that the y-values are multiplied by some common factor a when x incresses by 1.
We note that ench set of ordered pairs contains -1, 0, 1, and 2 as x-values, which are increments of 1.
Let us next determine the ratio of each pair of consecutive y-values for each option.
\frac{0}{-\frac{1}{2}}=0
\left(\frac{1}{2}\right)0=
Does not: exist
\frac{1}{\frac{1}{2}}=2
\frac{0}{-1}=0
\frac{1}{0}
= Does not exist
\frac{8}{1}=8
\frac{1}{\frac{1}{2}}=2
\frac{2}{1}=2
\frac{4}{2}=2
Answer option Does
\frac{0}{1}=0
\frac{1}{0}
=Does no exist
\frac{4}{1}=4
We note that the three ratios are only constant for answer option C and thus answer option C could be generated by an exponential function.
\left(-1,\frac{1}{2}\right)
,(0,1),(1,2),(2,4)
Make a scatterplot of the data. Use the year on the horizontal scale and the number of ounces on the vertical scale.
Available Drink Sizes at a Convenience Store Year, Sizes Available (cm)
y=\text{ }\text{height (in centimeters) to}\text{ }x=\text{ }\text{age (in years).}
\text{After verifying that the conditions for the regression model were met, the nurse calculated the equation of the population regression line to be}\text{ }{\mu }_{0}=105\text{ }+\text{ }4.2x\text{ }\text{with}\text{ }\sigma =7\text{ }cm.
Which two of the four scatter plots have most positive correlation and closer to zero correlation
Given a scatterplot for a set of data, how can you draw an accurate trend line?
Two scatterplots are shown below.
A scatterplot has 14 points.
The horizontal axis is labeled "x" and has values from 30 to 110.
The vertical axis is labeled "y" and has values from 30 to 110.
The points are plotted from approximately (55, 60) up and right to approximately (95, 85).
The points are somewhat scattered.
The points are plotted from approximately (55, 55) steeply up and right to approximately (70, 90), and then steeply down and right to approximately (85, 60).
Explain why it makes sense to use the least-squares line to summarize the relationship between x and y for one of these data sets but not the other.
Scatterplot 1 seems to show a relationship between x and y, while Scatterplot 2 shows a relationship between the two variables. So it makes sense to use the least squares line to summarize the relationship between x and y for the data set in , but not for the data set in .
Make a scatterplot for the data.
Length (mi) and Water Flow (
1000\text{ }f\frac{{t}^{3}}{s}
) of rivers
Length: 2540, 1980, 1460, 1420, 1290, 1040, 886, 774, 724, 659
Flow: 76, 225, 41, 58, 56, 57, 68, 67, 67, 41
|
Server-aided access control for cloud computing
网络与信æ¯å®‰å…¨å¦æŠ¥ ›› 2016, Vol. 2 ›› Issue (10): 58-76.doi: 10.11959/j.issn.2096-109x.2016.00104
WENG Jian( ),WENG Jia-si,LIUJia-nan,HOU Lin
College of Information Science and Technology,Jinan University,Guangzhou 510632,China
作者简介:WENG Jian(1976-),born in Guangdong.In 2008,he received his Ph.D.degree in Computer Science and Engineering from Shanghai Jiaotong University.Currently,he is a professor and vice dean with the School of Information Technology,Jinan University.His research interests include cryptography and information security.|WENG Jiasi(1994-),born in Guangdong.She is a master in Jinan University.Her research interests include cryptography and information security.|LIU Jianan(1992-),born in Henan.He is a Ph.D student in Jinan University.His research interests include cryptography and information security.|HOU Lin (1991-),born in Hubei.She is a Ph.D student in Jinnan University.Her research interests include cryptography and information security.
With the massive diffusion of cloud computing,more and more sensitive data is being centralized into the cloud for sharing,which brings forth new challenges for the security and privacy of outsourced data.To address these challenges,the server-aided access control (SAAC) system was proposed.The SAAC system builds upon a variant of conditional proxy re-encryption (CPRE) named threshold conditional proxy re-encryption (TCPRE).In TCPRE,t out of n proxies can re-encrypt ciphertexts (satisfying some specified conditions) for the delegator (while up to t?1 proxies cannot),and the correctness of the re-encrypted ciphertexts can be publicly verified.Both features guarantee the trust and reliability on the proxies deployed in the SAAC system.The security models for TCPRE were formalized,several TCPRE constructions were proposed and that our final scheme was secure against chosen-ciphertext attacks was proved.
关键è¯: threshold conditional proxy re-encryption, server-aided access control, cloud computing, chosen-ciphertext attack
WENG Jian,WENG Jia-si,LIUJia-nan,HOU Lin. Server-aided access control for cloud computing[J]. 网络与信æ¯å®‰å…¨å¦æŠ¥, 2016, 2(10): 58-76.
Notations used in proposed system"
(pki ,ski ) i-th user’s public and secret key pair
w Condition
rk ij Re-encryption key from i-th user to j -th user
ij vk Verification key from i-th user to j -th user
θij Re-encryption share from i-th user to j -th user
?]v v-th share of a secret
L )(? Polynomial implicitly defined in our system
ct Original ciphertext
ct ′ Transformed ciphertext
k File encryption/decryption key
(n,t) Number of proxies and the threshold
Access control system architecture"
Data flow in server-aided access control system"
Computational overhead in SAAC"
Phase Computational overhead
System step sE G
File storage 3 P+
{E}_{{G}_{T}}
1| |T EG+G +ξ f
File access 5 G nE (file owner)
3 G E (access controller server)
2 ( 1) 2| |
{E}_{{G}_{T}}
tP+ t++ξ f (file user)
Time cost in simulation"
[1] BLAZE M , BLEUMER G , STRAUSS M . Divertible protocols and atomic proxy cryptography[C]// The International Conference on the Theory and Applications of Cryptographic Techniques. Berlin Heidelberg, 1998: 127-144.
[2] WENG J , DENG R H , DING X ,et al. Conditional proxy re-encryption secure against chosen-ciphertext attack[C]// The 4th International Symposium on Information,Computer,and Communications Security,ACM. 2009: 322-332.
[3] TANG Q , . Type-based proxy re-encryption and its construction[C]// The International Conference on Cryptology in India. 2008: 130-144.
[4] VRABLE M , SAVAGE S , VOELKER G M . Cumulus:filesystem backup to the cloud[J]. ACM Transactions on Storage (TOS), 2009,5(4):14.
[5] ATENIESE G , FU K , GREEN M ,et al. Improved proxy re-encryption schemes with applications to secure distributed storage[J]. ACM Transactions on Information and System Security (TISSEC), 2006,9(1): 1-30.
[6] LI J , CHEN X , LI J ,et al. Fine-grained access control system based on outsourced attribute-based encryption[C]// European Symposium on Research in Computer Security. 2013: 592-609.
[8] BONEH D , BOYEN X , GOH E J . Hierarchical identity based encryption with constant size ciphertext[C]// Annual International Conference on the Theory and Applications of Cryptographic Techniques. 2005: 440-456.
[9] Final report on main computational assumptions in cryptography[EB/OL]..2013
[10] CORON J S , . On the exact security of full domain hash[C]// Annual International Cryptology Conference. 2000: 229-235.
[11] Pairing-based cryptography library[EB/OL]..2013
[12] MAMBO M , OKAMOTO E . Proxy cryptosystems:delegation of the power to decrypt ciphertexts[J]. IEICE Transactions on Fundamentals of Electronics,Communications and Computer Sciences, 1997,80(1): 54-63.
[13] CANETTI R , HOHENBERGER S . Chosen-ciphertext secure proxy re-encryption[C]// The 14th ACM conference on Computer and Communications Security. 2007: 185-194.
[14] LIBERT B , VERGNAUD D . Unidirectional chosen-ciphertext secure proxy re-encryption[J]. IEEE Transactions on Information Theory, 2011,57(3): 1786-1802.
[15] LIBERT B , VERGNAUD D . Unidirectional chosen-ciphertext secure proxy re-encryption[C]// International Workshop on Public Key Cryptography. 2008: 360-379.
[16] WENG J , DENG R H , LIU S ,et al. Chosen-ciphertext secure bidirectional proxy re-encryption schemes without pairings[J]. Information Sciences, 2010,180(24): 5077-5089.
[17] DENG R H , WENG J , LIU S ,et al. Chosen-ciphertext secure proxy re-encryption without pairings[C]// International Conference on Cryptology and Network Security. 2008: 1-17.
[18] SHAO J , CAO Z . CCA-secure proxy re-encryption without pairings[C]// International Workshop on Public Key Cryptography. 2009: 357-376.
[19] CHOW S S M , WENG J , YANG Y ,et al. Efficient unidirectional proxy re-encryption[C]// International Conference on Cryptology in Africa. 2010: 316-332.
[20] WENG J , ZHAO Y , HANAOKA G . On the security of a bidirectional proxy re-encryption scheme from PKC 2010[C]// International Workshop on Public Key Cryptography. 2011: 284-295.
[21] HANAOKA G , KAWAI Y , KUNIHIRO N ,et al. Generic construction of chosen ciphertext secure proxy re-encryption[C]// Cryptographers’ Track at the RSA Conference. 2012: 349-364.
[22] WENG J , CHEN M , YANG Y ,et al. CCA-secure unidirectional proxy re-encryption in the adaptive corruption model without random oracles[J]. Science China Information Sciences, 2010,53(3): 593-606.
[23] MATSUDA T , NISHIMAKI R , TANAKA K . CCA proxy re-encryption without bilinear maps in the standard model[C]// International Workshop on Public Key Cryptography. 2010: 261-278.
[24] GREEN M , ATENIESE G . Identity-based proxy re-encryption[C]// Applied Cryptography and Network Security. 2007: 288-306.
[25] MATSUO T , . Proxy re-encryption systems for identity-based encryption[C]// International Conference on Pairing-Based Cryptography. 2007: 247-267.
[26] CHU C K , TZENG W G . Identity-based proxy re-encryption without random oracles[C]// International Conference on Information Security. 2007: 189-202.
[27] LIBERT B , VERGNAUD D . Tracing malicious proxies in proxy re-encryption[C]// The International Conference on Pairing-Based Cryptography. 2008: 332-353.
[28] ATENIESE G , BENSON K , HOHENBERGER S . Key-private proxy re-encryption[C]// Cryptographers’ Track at the RSA Conference. 2009: 279-294.
[29] CHANDRAN N , CHASE M , VAIKUNTANATHAN V . Functional re-encryption and collusion-resistant obfuscation[C]// Theory of Cryptography Conference. 2012: 404-421
|
BA*B3=57A - Maths - Playing with Numbers - 9940604 | Meritnation.com
BA*B3=57A
Given : BA
×
B3 = 57A
First we see that unit's place of the number we get when 3 is multiplied by A, must be A. This can happen in one case only and that is When A = 5 As :
×
3 = 15 , Here we see the unit's digit is same as 5. So A = 5.
Now , BA = ( 10B + A ) = ( 10 B + 5 ) and B 3 = ( 10 B + 3)
×
B3 = ( 10 B + 5 ) ( 10 B + 3) = 575 , So
100 B2 + 30 B + 50 B + 15 = 575
100 B2 + 80 B + 15 = 575
100 B2 + 80 B = 560
10 ( 10 B2 + 8 B ) = 560
10 B2 + 8 B = 56
Here we used hit and trial method and substitute B = 2 and get
10 ( 2 )2 + 8 ( 2 ) = 56
10 ( 4 ) + 8 ( 2 ) = 56
×
23 = 575 , A = 5 and B = 2 ( Ans )
Hope this information will clear your doubts about Playing with numbers.
|
Approximation of Square Roots | Brilliant Math & Science Wiki
Akshay Yadav, Kaustubh Miglani, and Abhiram Rao contributed
Square root is common function in mathematics. It has a wide range of applications from the field of mathematics to physics. Sometimes it gets hard to calculate square root of a number, especially the one which are not actually square of a number. Often the method we employ are to tedious work with decimals. Here is a guide to find square root or rather their approximates.
A formula for square root approximation
Conventional method of Long Division
n
be the number whose square root we need to calculate. Let
n
p+q
p
the largest perfect square less than
q
be any positive real number. Then,
\sqrt{n} = \sqrt{p+q} ≈ \sqrt{p} + \frac{q}{2\sqrt{p}+1}
Approximate the square root of 968.
Let us first find the perfect square less than
968
. To do this we would be comparing
968
with perfect squares that are easy to figure out like
30^2=900
900<968
30<\sqrt{968}
Now try the square of another number greater than
30
31^2=961
32^2=1024
961<968<1024
31<\sqrt{968}<32
Now applying my formula,
\sqrt{968} = \sqrt{961+7}≈ \sqrt{961} + \frac{7}{2\sqrt{961}+1}
\sqrt{968} ≈ 31+\frac{7}{2(31)+1}
\sqrt{968} ≈ 31+\frac{7}{63}
\sqrt{968} ≈ 31+\frac{1}{9}
\sqrt{968} ≈ \boxed{31.11}
You can cross check by squaring the answer
(31.11)^2=967.8321
That is indeed a good approximation.
Here is the proof of the formula mentioned above.
The proof of the formula lies inside the formula itself. The following formula is based on assumption that roots between two perfect squares are uniformly distributed.
To understand it better we can use a graph-
2\sqrt{p}+1
integers exist between a perfect square
p
(\sqrt{p}+1)²
n
be any real number that exists between
p
(\sqrt{p}+1)²
n-p=q
\sqrt{n}
will exist between
\sqrt{p}
\sqrt{p}+1
Clearly the difference between
\sqrt{p}
\sqrt{p}+1
1
, now
2\sqrt{p}+1
roots can be uniformly distributed between
\sqrt{p}
\sqrt{p}+1
, if each root occupies
\frac{1}{2\sqrt{p}+1}
q^{th}
number will occupy
\frac{q}{2\sqrt{p}+1}
space. Hence the formula turns out to be
\sqrt{n} = \sqrt{p+q} ≈ \sqrt{p} + \frac{q}{2\sqrt{p}+1}
This section deals with conventional method of square root-finding. This method is more accurate, however it is a tedious one too.
Cite as: Approximation of Square Roots. Brilliant.org. Retrieved from https://brilliant.org/wiki/approximation-of-square-roots/
|
m (→Bandlimitation and timing)
want to see is a system's behavior at a given isolated frequency. Now
let's look at something a bit more complex. What should we expect to
happen when I change the input to a [[WikiPedia:Square_wave|square wave]]...
Exactly what it should.
{\displaystyle \ x(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}x_{\mathrm {square} }(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
|
Assignment Model | Linear Programming Problem (LPP) | Introduction | Education Lessons
What is Assignment Model?
→ Assignment model is a special application of Linear Programming Problem (LPP), in which the main objective is to assign the work or task to a group of individuals such that;
i) There is only one assignment.
ii) All the assignments should be done in such a way that the overall cost is minimized (or profit is maximized, incase of maximization).
→ In assignment problem, the cost of performing each task by each individual is known.
→ It is desired to find out the best assignments, such that overall cost of assigning the work is minimized.
Suppose there are 'n' tasks, which are required to be performed using 'n' resources.
The cost of performing each task by each resource is also known (shown in cells of matrix)
In the above asignment problem, we have to provide assignments such that there is one to one assignments and the overall cost is minimized.
How Assignment Problem is related to LPP? OR Write mathematical formulation of Assignment Model.
→ Assignment Model is a special application of Linear Programming (LP).
→ The mathematical formulation for Assignment Model is given below:
→ Let,
\text {C}_{ij}
denotes the cost of resources 'i' to the task 'j'; such that
\begin{aligned} x_{ij} &= 1 \ ; \text{if} \ \ i^{th} \text{resource is original to } j^{th} \text{task}\\ x_{ij} &= 0 \ ; \text{if} \ \ i^{th} \text{resource is not original to } j^{th} \text{task} \end{aligned}
→ Now assignment problems are of the Minimization type. So, our objective function is to minimize the overall cost.
\therefore \text{Minimize Z}= \displaystyle\sum_{j=1}^n \ \sdot \ \displaystyle\sum_{i=1}^n \ C_{ij} \ \sdot \ x_{ij}
→ Subjected to constraint;
j^{th}
task, only one
i^{th}
resource is possible:
\therefore \displaystyle\sum_{j=1}^n \ x_{ij} = 1 \quad \text{(i=1,2,3, ...,n)}
(ii) For all
i^{th}
resource, there is only one
j^{th}
task possible;
\therefore \displaystyle\sum_{i=1}^n \ x_{ij} = 1 \quad \text{(j=1,2,3, ...,n)}
x_{ij}
is '0' or '1'.
Types of Assignment Problem:
(i) Balanced Assignment Problem
It consist of a suqare matrix (n x n).
Number of rows = Number of columns
(ii) Unbalanced Assignment Problem
It consist of a Non-square matrix.
\not=
Methods to solve Assignment Model:
(i) Integer Programming Method:
In assignment problem, either allocation is done to the cell or not.
So this can be formulated using 0 or 1 integer.
While using this method, we will have n x n decision varables, and n+n equalities.
So even for 4 x 4 matrix problem, it will have 16 decision variables and 8 equalities.
So this method becomes very lengthy and difficult to solve.
(ii) Transportation Methods:
As assignment problem is a special case of transportation problem, it can also be solved using transportation methods.
In transportation methods (NWCM, LCM & VAM), the total number of allocations will be (m+n-1) and the solution is known as non-degenerated. (For eg: for 3 x 3 matrix, there will be 3+3-1 = 5 allocations)
But, here in assignment problems, the matrix is a square matrix (m=n).
So total allocations should be (n+n-1), i.e. for 3 x 3 matrix, it should be (3+3-1) = 5
But, we know that in 3 x 3 assignment problem, maximum possible possible assignments are 3 only.
So, if are we will use transportation methoods, then the solution will be degenerated as it does not satisfy the condition of (m+n-1) allocations.
So, the method becomes lengthy and time consuming.
(iii) Enumeration Method:
It is a simple trail and error type method.
Consider a 3 x 3 assignment problem. Here the assignments are done randomly and the total cost is found out.
For 3 x 3 matrix, the total possible trails are 3!
So total 3! = 3 x 2 x 1 = 6 trails are possible.
The assignments which gives minimum cost is selected as optimal solution.
But, such trail and error becomes very difficult and lengthy.
If there are more number of rows and columns,
( For eg: For 6 x 6 matrix, there will be 6! trails. So 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 trails possible)
then such methods can't be applied for solving assignments problems.
(iv) Hungarian Method:
It was developed by two mathematicians of Hungary. So, it is known as Hungarian Method.
It is also know as Reduced matrix method or Flood's technique.
There are two main conditions for applying Hungarian Method:
(1) Square Matrix (n x n).
(2) Problem should be of minimization type.
|
Complex number in rectangular form What is (1+2j) + (1+3j)? Your answer should contain thr
Sum of complex numbers
\left({a}_{1}+{b}_{1}i\right)+\left({a}_{2}+{b}_{2}i\right)=\left({a}_{1}+{a}_{2}\right)+\left({b}_{1}+{b}_{2}\right)i=a+bi
\left(1+2i\right)+\left(1+3i\right)=\left(1+1\right)+\left(2+3\right)i=2+5i
this is the binomial form
Then find r and
\theta
r=\sqrt{{a}^{2}+{b}^{2}}=\sqrt{{2}^{2}+{5}^{2}}=\sqrt{29}
if a and b are possitive then
\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)={\mathrm{tan}}^{-1}\left(\frac{5}{2}\right)={68.2}^{\circ }
z=r\left(\mathrm{cos}\left(\theta \right)+\mathrm{sin}\left(\theta \right)i\right)
z=\sqrt{29}\left(\mathrm{cos}\left(68.2\right)+\mathrm{sin}\left(68.2\right)i\right)
Multiplicative Inverse of a Complex Number The multiplicative inverse of a complex number z is a complex number zm such that
z×zm=1
. Find the multiplicative inverse of complex number.
z=1+i
|z|\le Re\left(z\right)?
\left(7+9i\right)-\left(\frac{5}{7}+\frac{5i}{3}\right)
{z}_{1}{z}_{2}
\frac{{z}_{1}}{{z}_{2}}
. Express your answers in polar form.
{z}_{1}=\frac{7}{8}\left(\mathrm{cos}\left({35}^{\circ }\right)+i\mathrm{sin}\left({35}^{\circ }\right)\right),{z}_{2}=\frac{1}{8}\left(\mathrm{cos}\left({135}^{\circ }\right)+i\mathrm{sin}\left({135}^{\circ }\right)\right)
{z}_{1}{z}_{2}=?
\frac{{z}_{1}}{{z}_{2}}=?
f\left(z\right)=\frac{1}{{z}^{2}\left(1-z\right)}
\sum _{n=0}^{\mathrm{\infty }}{z}^{n}+\frac{1}{z}+\frac{1}{{z}^{2}}
1<|z|<\mathrm{\infty }
Find zw and
\frac{z}{w}
. Write each answer in polar form and in exponential form.
z=7\left(\mathrm{cos}\frac{\pi }{4}+i\mathrm{sin}\frac{\pi }{4}\right)
w=7\left(\mathrm{cos}\frac{\pi }{10}+i\mathrm{sin}\frac{\pi }{10}\right)
The product zw in polar form is ? and in exponential form is ?.
{e}^{2-i}
a+ib
|
Lines And Angles, Popular Questions: CBSE Class 7 MATH, Math - Meritnation
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Linearly Weighted Moving Average (LWMA)
What Is a Linearly Weighted Moving Average?
A linearly weighted moving average (LWMA) is a moving average calculation that more heavily weights recent price data. The most recent price has the highest weighting, and each prior price has progressively less weight. The weights drop in a linear fashion. LWMAs are quicker to react to price changes than simple moving averages (SMA) and exponential moving averages (EMA).
Use a linearly weighted moving average in the same way as an SMA or EMA.
Use a LWMA to more clearly define the price trend and reversals, provide trade signals based on crossovers, and indicate areas of potential support or resistance.
Traders who want a moving average with less lag than an SMA may wish to utilize a LWMA.
The Formula for the Linearly Weighted Moving Average (LWMA) Is:
\begin{aligned} &\text{LWMA}=\frac{\left(P_n*W_1\right)+\left(P_{n-1}*W_2\right)+\left(P_{n-2}*W_3\right)...}{\sum{W}}\\ &\textbf{where:}\\ &\text{P = Price for the period}\\ &\text{n = The most recent period, n-1 is the prior period,}\\ &\text{and n-2 is two periods prior}\\ &\text{W = The assigned weight to each period, with the}\\ &\text{highest weight going first and then descending linearly}\\ &\text{based on the number of periods being used}\\ \end{aligned}
LWMA=∑W(Pn∗W1)+(Pn−1∗W2)+(Pn−2∗W3)...where:P = Price for the periodn = The most recent period, n-1 is the prior period,and n-2 is two periods priorW = The assigned weight to each period, with thehighest weight going first and then descending linearlybased on the number of periods being used
How to Calculate the Linearly Weighted Moving Average (LWMA)
Choose a lookback period. This is how many n values will be calculated into the LWMA.
Calculate the linear weights for each period. This can be accomplished in a couple of ways. The easiest is to assign n as the weight for the first value. For example, if using a 100-period lookback, then the first value is multiplied by a weight of 100, the next value is multiplied by a weight of 99. A more complex way is to choose a different weight for the most recent value, such as 30. Now each value will need to drop by 30/100 so that when n-99 (100th period) is reached the weight is one.
Multiply the prices for each period by their respective weights, then get the sum total.
Divide the above by the sum of all the weights.
Let’s say we are interested in calculating the linearly weighted moving average of the closing price of a stock over the last five days.
Begin by multiplying today’s price by 5, yesterday’s by 4, and the price of the day before by 3. Continue multiplying each day’s price by its position in the data series until reaching the first price in the data series, which is multiplied by 1. Add these results together, divide by the sum of the weights, and you will have the linearly weighted moving average for this period.
((P5*5)+(P4*4)+(P3*3)+(P2*2)+(P1*1)) / (5+4+3+2+1)
Let’s say that the price of this stock fluctuates as so:
((90.90*5)+(90.36*4)+(90.28*3)+(90.83*2)+(90.91*1)) / (5+4+3+2+1) = 90.62
The LWMA of this stock over this time period is $90.62.
What Does the Linearly Weighted Moving Average (LWMA) Tell You?
The linearly weighted moving average is a method of calculating the average price of an asset over a given period of time. This method weights recent data more heavily than older data, and is used to analyze market trends.
Generally, when the price is above the LWMA, and the LWMA is rising, the price is above the weighted average which helps confirm an uptrend. If the price is below the LWMA, and the LWMA is pointed down, this helps confirm a downtrend in price.
When the price crosses the LWMA that could signal a trend change. For example, if the price is above the LWMA and then drops below it, that could indicate a shift from an uptrend to a downtrend.
When assessing trends, traders should be aware of the lookback period. The lookback period is how many periods are being calculated into the LWMA. A five-period LWMA will track price very closely and is useful for tracking small trends as the line will be easily breached by even minor price oscillations. A 100-period LWMA will not track the price as closely, meaning there will often be room between the LWMA and the price. This allows for the determination of longer-term trends and reversals.
Like other types of moving averages, the LWMA may sometime be used to indicate support and resistance areas. For example, in the past, the price bounced off the LWMA on multiple occasions and then moved higher. This indicates the line is acting as support. The line may continue to act as support in the future. Failure to do so could indicate the price trend has undergone a change. It could be reversing to the downside or may be starting a period where it moves more sideways.
What is the Difference Between a Linearly Weighted Moving Average (LWMA) and a Double Exponential Moving Average (DEMA)?
Both of these moving averages are designed to reduce the lag that is inherent in the SMA. The LWMA does this by applying greater weight to recent prices. The double exponential moving average (DEMA) does this through multiplying the EMA over a certain period by two, and then subtracting a smoothed EMA. Because the MAs are calculated differently they will provide different values on a price chart.
The Limitations of Using a Linearly Weighted Moving Average (LWMA)
All moving averages help to define trends when they are present, but provide little information when the price action is choppy or moving predominantly sideways. During such times the price will oscillate around the MA. The MA will not provide good crossover or support/resistance signals during such times.
A LWMA may not provide support or resistance. This is especially likely if it hasn't done so in the past.
Multiple false signals may also occur before a significant trend develops. A false signal is when the price crosses the LWMA but then fails to move in the direction expected, resulting in a poor trade.
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Parallelogram | Brilliant Math & Science Wiki
Niranjan Khanderia, Lawrence Chiou, Deleted User, and
A parallelogram is a quadrilateral whose opposite sides are parallel. A parallelogram, in its most general form, looks something like this: Note that the arrowheads are used to indicate which pair of sides are parallel.
Some Cooler Properties
The fundamental definition of a parallelogram is as follows:
In the diagram of a general parallelogram above,
AB || DC
AD || BC
. Several important properties then follow. First, it is clear that the opposite angles must be equal
( \angle A = \angle C
\angle B = \angle D)
since they make corresponding angles with the two sets of parallel lines. Meanwhile, consecutive angles are supplementary.
Property 1. The opposite angles of a parallelogram are equal.
Property 2. Consecutive angles of a parallelogram are supplementary.
One can also show that the opposite sides are equal
( AB = DC
AD = BC):
the two triangles formed by drawing in a diagonal of the parallelogram
(
i.e. either the segment
AC
BD
)
must be congruent by angle-side-angle, so the corresponding sides of the two triangles must be congruent as well.
Property 3. The lengths of the opposite sides of a parallelogram are equal.
Now that the lengths of all of the sides are known, it is easy to compute the perimeter of a parallelogram.
Property 4. A parallelogram whose side lengths are
and
b
2a + 2 b
Meanwhile, the area of the parallelogram can be found by computing the sum of the area of the two triangles formed.
Property 5. The area of a parallelogram with side lengths
and
b
, with the acute angle formed between them
\theta
ab \sin{\theta}
Drawing in both diagonals simultaneously produces four congruent triangles. Therefore, the intersection of the two diagonals must be the midpoint of each diagonal.
Property 6. The diagonals of a parallelogram bisect each other.
One last result is left as an exercise for the reader.
Property 7. Lines that connect midpoints of opposite sides with opposite vertices trisect the diagonal.
ABCD
AB
AD
E
AB
\angle DEC
\frac{1}{3}
\frac{3}{8}
\frac{1}{2}
\frac{3}{5}
In the given parallelogram, the midpoints of two adjacent sides of the parallelogram are joined and then connected to the opposite vertex to form a triangle.
What fraction of the total parallelogram is the shaded area?
Click here for more from this set.
A parallelogram has sides measuring 7 and 9. Its shorter diagonal has a length of 8.
Find the measure of the longer diagonal.
X
is drawn on the diagonal of a parallelogram
ABCD
. The lines parallel to the sides through
X
are constructed, and quadrilaterals
P
Q
are formed, as shown in the diagram.
Q
has base
12
and perpendicular height
4
and that the area of
ABCD
200
, find the sum of all possible values of the base of
P
Commonly encountered special cases of parallelograms include
rectangles, all of whose angles are equal;
rhombuses, all of whose sides are equal;
squares, all of whose sides and angles are equal.
Because opposite angles of a cyclic quadrilateral are supplementary, all cyclic parallelograms are rectangles. Furthermore, the only rectangles with an incircle are squares, so the only bicyclic parallelogram is a square.
Cite as: Parallelogram. Brilliant.org. Retrieved from https://brilliant.org/wiki/parallelogram/
|
Alexander Katz, Andy Hayes, Tejas Suresh, and
A certain fast-food restaurant gets an average of 3 visitors to the drive-through per minute. This is just an average, however. The actual amount can vary.
A Poisson distribution can be used to analyze the probability of various events regarding how many customers go through the drive-through. It can allow one to calculate the probability of a lull in activity (when there are 0 customers coming to the drive-through) as well as the probability of a flurry of activity (when there are 5 or more customers coming to the drive-through). This information can, in turn, help a manager plan for these events with staffing and scheduling.
In addition to its use for staffing and scheduling, the Poisson distribution also has applications in biology (especially mutation detection), finance, disaster readiness, and any other situation in which events are time-independent.
The Poisson distribution is applicable only when several conditions hold.
Of course, this situation isn't an absolute perfect theoretical fit for the Poisson distribution. For instance, the office certainly cannot receive a trillion calls during the time period, as there are less than a trillion people alive to be making calls. Practically speaking, the situation is close enough that the Poisson distribution does a good job of modeling the situation's behavior.
The following problem gives an idea of how the Poisson distribution was derived:
0
\frac{e^{-m}}{k!\ m}
\frac{e^{-k}}{m!\ k}
\frac{m^ke^{-m}}{k!}
\frac{m}{k! \ln{m}}
\frac{\ln{k}}{k! \ln{m}}
\frac{e^{-k}\ln{m}}{m!\ k}
Consider a binomial distribution of
X\sim B(n,p)
P(X=k)={n\choose k}p^k{(1-p)}^{n-k}
k=0,1,2,3,\ldots,n
Now, let's take the limit of the above using
n \to \infty
. Instead of having an infinitesimal
p
, let's assume that it is given that
np
, the mean of the probability distribution function, is some finite value
m
P(X=k)
m
k
for this new distribution, where
k=0,1,2,3,\ldots
, without looking anything up or reciting any formulas from memory.
Given that a situation follows a Poisson distribution, there is a formula which allows one to calculate the probability of observing
k
events over a time period for any non-negative integer value of
k
X
be the discrete random variable that represents the number of events observed over a given time period. Let
\lambda
be the expected value (average) of
X
X
follows a Poisson distribution, then the probability of observing
k
events over the time period is
P(X=k) = \frac{\lambda^ke^{-\lambda}}{k!},
where
In the World Cup, an average of 2.5 goals are scored each game. Modeling this situation with a Poisson distribution, what is the probability that
k
goals are scored in a game?
In this instance,
\lambda=2.5
. The above formula applies directly:
\begin{aligned} P(X=0) &= \frac{2.5^0e^{-2.5}}{0!} \approx 0.082\\\\ P(X=1) &= \frac{2.5^1e^{-2.5}}{1!} \approx 0.205\\\\ P(X=2) &= \frac{2.5^2e^{-2.5}}{2!} \approx 0.257\\\\ P(X=3) &= \frac{2.5^3e^{-2.5}}{3!} \approx 0.213\\\\ P(X=4) &= \frac{2.5^4e^{-2.5}}{4!} \approx 0.133\\\\ &\ \ \vdots \end{aligned}
The Poisson distribution with
\lambda=2.5
There is no upper limit on the value of
k
for this formula, though the probability rapidly approaches 0 as
k
_\square
A fast food restaurant gets an average of 2.8 customers approaching the register every minute.
Assuming the number of customers approaching the register per minute follows a Poisson distribution, what is the probability that 4 customers approach the register in the next minute?
The Poisson distribution can be used to calculate the probabilities of "less than" and "more than" using the rule of sum and complement probabilities.
A statistician records the number of cars that approach an intersection. He finds that an average of 1.6 cars approach the intersection every minute.
Assuming the number of cars that approach this intersection follows a Poisson distribution, what is the probability that 3 or more cars will approach the intersection within a minute?
For this problem,
\lambda=1.6.
The goal of this problem is to find
P(X \ge 3),
the probability that there are 3 or more cars approaching the intersection within a minute. Since there is no upper limit on the value of
k,
this probability cannot be computed directly. However, its complement,
P(X \le 2),
can be computed to give
P(X \ge 3):
\begin{aligned} P(X=0) &= \frac{1.6^0e^{-1.6}}{0!} \approx 0.202 \\\\ P(X=1) &= \frac{1.6^1e^{-1.6}}{1!} \approx 0.323 \\\\ P(X=2) &= \frac{1.6^2e^{-1.6}}{2!} \approx 0.258 \\\\ \Rightarrow P(X \le 2) &= P(X=0) + P(X=1) + P(X=2) \\ &\approx 0.783 \\ \\ \Rightarrow P(X \ge 3) &= 1-P(X \le 2) \\ &\approx 0.217. \end{aligned}
Therefore, the probability that there are 3 or more cars approaching the intersection within a minute is approximately
0.217.
_\square
\lambda = 0.2
There are other applications of the Poisson distribution that come from more open-ended problems. For example, it can be used to help determine the amount of staffing that is needed in a call center.
A call center receives an average of 4.5 calls every 5 minutes. Each agent can handle one of these calls over the 5 minute period. If a call is received, but no agent is available to take it, then that caller will be placed on hold. Assuming that the calls follow a Poisson distribution, what is the minimum number of agents needed on duty so that calls are placed on hold at most 10% of the time?
In order for all calls to be taken, the number of agents on duty should be greater than or equal to the number of calls received. If
X
is the number of calls received and
k
is the number of agents, then
k
should be set such that
P(X > k)\le 0.1,
P(X \le k) > 0.9.
The average number of calls is 4.5, so
\lambda=4.5:
\begin{array}{cl} P(X=0) = \frac{4.5^0 e^{-4.5}}{0!} \approx 0.011 & \\ P(X=1) = \frac{4.5^1 e^{-4.5}}{1!} \approx 0.050 &\implies P(X\le 1) \approx 0.061 \\ P(X=2) = \frac{4.5^2 e^{-4.5}}{2!} \approx 0.112 &\implies P(X\le 2) \approx 0.173 \\ P(X=3) = \frac{4.5^3 e^{-4.5}}{3!} \approx 0.169 &\implies P(X\le 3) \approx 0.342 \\ P(X=4) = \frac{4.5^4 e^{-4.5}}{4!} \approx 0.190 &\implies P(X\le 4) \approx 0.532 \\ P(X=5) = \frac{4.5^5 e^{-4.5}}{5!} \approx 0.171 &\implies P(X\le 5) \approx 0.703 \\ P(X=6) = \frac{4.5^6 e^{-4.5}}{6!} \approx 0.128 &\implies P(X\le 6) \approx 0.831 \\ P(X=7) = \frac{4.5^7 e^{-4.5}}{7!} \approx 0.082 &\implies P(X\le 7) \approx 0.913. \end{array}
If the goal is to make sure that less than 10% of calls are placed on hold, then
\boxed{7}
agents should be on duty.
_\square
The expected value of a Poisson distribution should come as no surprise, as each Poisson distribution is defined by its expected value.
Expected Value of Poisson Random Variable:
Given a discrete random variable
X
that follows a Poisson distribution with parameter
\lambda,
the expected value of this variable is
\text{E}[X]=\lambda.
\text{E}[X] = \sum_{x \in \text{Im}(X)}xP(X=x),
x \in \text{Im}(X)
simply means that
x
is one of the possible values of the random variable
X
. Applying this to the Poisson distribution,
\begin{aligned} \text{E}[X] &= \sum_{k = 0}^{\infty} k \cdot \frac{\lambda^ke^{-\lambda}}{k!} \\ &=\lambda e^{-\lambda}\sum_{k=1}^{\infty} \frac{\lambda^{k-1}}{(k-1)!} \\ &=\lambda e^{-\lambda}\sum_{j=0}^{\infty} \frac{\lambda^j}{j!} \\ &=\lambda e^{-\lambda}e^{\lambda} \\ &=\lambda, \end{aligned}
where the rescaling
j=k-1
and the Taylor series
e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}
_\square
The variance of the Poisson distribution is also conveniently simple.
Variance of Poisson Random Variable:
X
\lambda,
the variance of this variable is
\text{Var}[X]=\lambda.
The proof involves the routine (but computationally intensive) calculation that
E[X^2]=\lambda^2+\lambda
. Then using the formula for variance
\text{Var}[X] = E[X^2]-E[X]^2,
\text{Var}[X]=\lambda^2+\lambda-\lambda^2=\lambda
\sqrt{\lambda}
1+\lambda
1-\lambda^2
\lambda^{-2}
e^{\lambda z}e^{-\lambda}.
The mode is only slightly more complicated:
Mode of Poisson Random Variable:
\lambda
is not an integer, the mode of a Poisson distribution with parameter
\lambda
\lfloor \lambda \rfloor
. Otherwise, both
\lambda
\lambda-1
are modes.
The median of a Poisson distribution does not have a closed form, but its bounds are known:
Median of Poisson Random Variable:
\rho
of a Poisson distribution with parameter
\lambda
\lambda-\ln 2 \leq \rho \leq \lambda+\frac{1}{3}.
The sum of two independent Poisson random variables is a Poisson random variable.
X
Y
be Poisson random variables with parameters
\lambda_1
\lambda_2
X
Y
X+Y
is a Poisson random variable with parameter
\lambda_1+\lambda_2.
Its distribution can be described with the formula
P(X+Y=k)=\frac{(\lambda_1+\lambda_2)^k e^{-(\lambda_1+\lambda_2)}}{k!}.
Damon is working the evening shift at the register of his retail job. There are currently two registers open, but his coworker is about to go home for the day and close her register.
The number of customers approaching each register is an independent Poisson random variable. If each register was getting an average of 2 customers per minute, what is the probability that Damon will have more than 4 customers approaching his register in minute after his coworker goes home?
Additionally, the Poisson distribution can be thought of as the limiting case of the binomial distribution. If there are
n
independent trials,
p
is the probability of a successful trial, and
np
remains constant, then this binomial distribution will behave as a Poisson distribution as
n
Poisson Limit Theorem:
n
approaches infinity and
p
0
\lambda
is a constant with
\lambda=np,
the binomial distribution with parameters
n
p
is approximated by a Poisson distribution with parameter
\lambda
\binom{n}{k}p^k(1-p)^{n-k} \simeq \frac{\lambda^k e^{-\lambda}}{k!}.
This can be proved by considering the fact that convergence in moment generating functions implies convergence in distribution.
The classical example of the Poisson distribution is the number of Prussian soldiers accidentally killed by horse-kick, due to being the first example of the Poisson distribution's application to a real-world large data set. Ten army corps were observed over 20 years, for a total of 200 observations, and 122 soldiers were killed by horse-kick over that time period. The question is how many deaths would be expected over a period of a year, which turns out to be excellently modeled by the Poisson distribution
(
\lambda=0.61):
The interpretation of this data is important: since the Poisson distribution measures the frequency of events under the assumption of statistical randomness, the agreement of the expected distribution with the actual data suggests that the actual data was indeed due to randomness. If the actual data resulted in many more deaths than expected, an alternate explanation should be sought (e.g. inadequate training, a clever and subtle enemy plot, etc.).
\sum_{k=20}^{\infty}\frac{12^ke^{-12}}{k!} \approx 2.12\%,
In short, the list of applications is very long. A partial list[1] of recently studied phenomena that obey a Poisson distribution is below:
the number of mutations on a given strand of DNA per time unit
the number of bankruptcies that are filed in a month
the number of arrivals at a car wash in one hour
the number of network failures per day
the number of file server virus infection at a data center during a 24-hour period
the number of Airbus 330 aircraft engine shutdowns per 100,000 flight hours
the number of asthma patient arrivals in a given hour at a walk-in clinic
the number of hungry persons entering McDonald's restaurant per day
the number of work-related accidents over a given production time
the number of birth, deaths, marriages, divorces, suicides, and homicides over a given period of time
the number of customers who call to complain about a service problem per month
the number of visitors to a web site per minute
the number of calls to consumer hot line in a 5-minute period
the number of telephone calls per minute in a small business
the number of arrivals at a turnpike tollbooth per minute between 3 A.M. and 4 A.M. in January on the Kansas Turnpike.
[1] Western New England University. Applications of the Poisson probability distribution. Retrieved February 9, 2016 from http://www.aabri.com/SA12Manuscripts/SA12083.pdf.
Cite as: Poisson Distribution. Brilliant.org. Retrieved from https://brilliant.org/wiki/poisson-distribution/
|
(→Bandlimitation and timing: original square wave)
==Bandlimitation and timing==
x_{\mathrm{square}}(t) = \frac{4}{\pi}\sin(\omega t) + \frac{4}{3\pi}\sin(3\omega t) + \frac{4}{5\pi}\sin(5\omega t) + \\
The rippling you see around sharp edges in a bandlimited signal is called the [[WikiPedia:/Gibbs_phenomenon|Gibbs effect]]. It happens whenever you slice off part of the frequency domain in the middle of nonzero energy.
{\displaystyle \ x(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}x_{\mathrm {square} }(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
|
Component: FAt
FAt=FA2iftime≥tAFA1otherwise
Component: FBt
FBt=FB2iftime≥tBFB1otherwise
Component: F0t
F0t=FAt+FBt
ddtimec15=p0V2c015-c15+p16c16+p17c17+pAA15pp-p15c19+p18+p19c19p20+c15+p29c15
ddtimec16=p0V2c016-c16+p15c15c19-p16c16+pAA16pp
ddtimec17=p0V2c017-c17+p1s1c1+p2s2c2+p3s3c3+p4s4c4+p5s5c5+p6s6c6+p7s7c7+p8s8c8+p9s9c9+p14s14c14+p18c15+p26c18-p17+p25c19+p28c17+pAA17pp
ddtimec18=p0V2c018-c18+p25c17c19-p26c18+pAA18pp
ddtimec19=p0V2c019-c19+p11s11c11+p12s12c12+p13c13s13+p10c10s10+p231-gtc17+p16c16+p26c18-p19c15p20+c15+p15c15+p25c17c19
ddtimec20=p0V2c020-c20+p19c15p20+c15c19
ddtimec21=p0V2c021-c21-p21c21
ddtimec22=p0V2c022-c22-p22c22
ddtimec23=p0V2c023-c23+p21c21+p22c22-p23c23+p31c15p20+c15c19
ddtimec24=p0V2c024-c24+2p30p23c23+p32c15p20+c15c19-p24c24
Component: pp
pp=p33iftime<t10otherwise
Component: gt
gt=0iftime<3p27otherwise
ddtimec1=p0V2c01-c1-p1c1+pAA1pp
ddtimec2=p0V2c02-c2-p2c2+pAA2pp
ddtimec3=p0V2c03-c3-p3c3+pAA3pp
ddtimec4=p0V2c04-c4-p4c4+pAA4pp
ddtimec5=p0V2c05-c5-p5c5+pAA5pp
ddtimec6=p0V2c06-c6-p6c6+pAA6pp
ddtimec7=p0V2c07-c7-p7c7+pAA7pp
ddtimec8=p0V2c08-c8-p8c8+pAA8pp
ddtimec9=p0V2c09-c9-p9c9+pAA9pp
ddtimec10=p0V2c010-c10-p10c10+pAA10pp
ddtimec11=p0V2c011-c11-p11c11+pAA11pp
ddtimec12=p0V2c012-c12-p12c12+pAA12pp
ddtimec13=p0V2c013-c13-p13c13+pAA13pp
ddtimec14=p0V2c014-c14-p14c14+pAA14pp
ddtimec01=FAtV1CA1+FBtV1CB1-F0tV1c01-p0V1c01-c1
ddtimec02=FAtV1CA2+FBtV1CB1-F0tV1c02-p0V1c02-c2
ddtimec03=FAtV1CA3+FBtV1CB1-F0tV1c03-p0V1c03-c3
ddtimec04=FAtV1CA4+FBtV1CB1-F0tV1c04-p0V1c04-c4
ddtimec05=FAtV1CA5+FBtV1CB1-F0tV1c05-p0V1c05-c5
ddtimec06=FAtV1CA6+FBtV1CB1-F0tV1c06-p0V1c06-c6
ddtimec07=FAtV1CA7+FBtV1CB1-F0tV1c07-p0V1c07-c7
ddtimec08=FAtV1CA8+FBtV1CB1-F0tV1c08-p0V1c08-c8
ddtimec09=FAtV1CA9+FBtV1CB1-F0tV1c09-p0V1c09-c9
ddtimec010=FAtV1CA10+FBtV1CB1-F0tV1c010-p0V1c010-c10
ddtimec011=FAtV1CA11+FBtV1CB1-F0tV1c011-p0V1c011-c11
ddtimec012=FAtV1CA12+FBtV1CB1-F0tV1c012-p0V1c012-c12
ddtimec013=FAtV1CA13+FBtV1CB1-F0tV1c013-p0V1c013-c13
ddtimec014=FAtV1CA14+FBtV1CB1-F0tV1c014-p0V1c014-c14
ddtimec015=FAtV1CA15+FBtV1CB15-F0tV1c015-p0V1c015-c15
ddtimec016=FAtV1CA16+FBtV1CB1-F0tV1c016-p0V1c016-c16
ddtimec017=FAtV1CA17+FBtV1CB1-F0tV1c017-p0V1c017-c17
ddtimec018=FAtV1CA18+FBtV1CB1-F0tV1c018-p0V1c018-c18
ddtimec019=FAtV1CA19+FBtV1CB1-F0tV1c019-p0V1c019-c19
ddtimec020=FAtV1CA20+FBtV1CB1-F0tV1c020-p0V1c020-c20
ddtimec021=FAtV1CA21+FBtV1CB1-F0tV1c021-p0V1c021-c21
ddtimec022=FAtV1CA22+FBtV1CB1-F0tV1c022-p0V1c022-c22
ddtimec023=FAtV1CA23+FBtV1CB1-F0tV1c023-p0V1c023-c23
ddtimec024=FAtV1CA24+FBtV1CB1-F0tV1c024-p0V1c024-c24
|
Please help me with the sum NO GUN NO ----------- HUNT ----------- Number should not be repeated - Maths - Playing with Numbers - 8481669 | Meritnation.com
Please help me with the sum
Number should not be repeated.
\mathrm{The} \mathrm{complete} \mathrm{solution} \mathrm{will} \mathrm{be},\phantom{\rule{0ex}{0ex}} \begin{array}{ccc}& 8& 7\\ 9& 0& 8\\ & 8& 7\end{array}\phantom{\rule{0ex}{0ex}}\overline{) 1 0 8 2 }
|
Make a scatterplot for the data in the table below.
Make a scatterplot for the data in the table below. Height (in.): 77 75 76 70 70 73 74 74 73 Weight (lb): 230 220 212 190 201 245 218 260 196
Height (in.): 77 75 76 70 70 73 74 74 73
Weight (lb): 230 220 212 190 201 245 218 260 196
Your scatterplot:
Sketch a scatterplot where the association is linear, but the correlation is close to r = 0.
Construct a scatterplot and identify the mathematical model that best fits the given data.
\left(-1,-\frac{1}{2}\right)
, (0, 0),
\left(1,\frac{1}{2}\right)
, (2, 1)
\left(-1,\frac{1}{2}\right)
, (0, 1), (1, 2), (2, 4)
D. (–1, 1), (0, 0), (1, 1), (2, 4)
What kind of plot is useful for deciding whether it is reasonable to find a regression to find a regression plane for a set of data points involving several predictor variables?
Height and Weight of Football Players
In this exercise, you will use the correlation and regression applet to create scatter plots with 10 points that have a correlation close to 0.7. The lesson here is that many models may have the same correlation. Always compile your data before trusting correlations. (a) Stop after adding the first two points. What is the value of correlation?
(Enter your answer, rounded to four decimal places).
Why does correlation matter? Two is the minimum number of data points required to calculate the correlation. This value is the default correlation.
Because two points define a line, correlation always matters.
The mean of these two values always has this value.
|
(→Bit-depth)
That's why seeing '[[WikiPedia:SPARS_code|D D D]]' on a [[WikiPedia:Compact_disk|Compact Disc]] used to be such a big,
high-end deal.
{\displaystyle \ squarewave(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}\ squarewave(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
|
Steric effects - Wikipedia
Geometric aspects of ions and molecules affecting their shape and reactivity
The parent cyclobutadiene (R = H) readily dimerizes but the R = tert-butyl derivative is robust.[1]
Steric effects arise from the spatial arrangement of atoms. When atoms come close together there is a rise in the energy of the molecule. Steric effects are nonbonding interactions that influence the shape (conformation) and reactivity of ions and molecules. Steric effects complement electronic effects, which dictate the shape and reactivity of molecules. Steric repulsive forces between overlapping electron clouds result in structured groupings of molecules stabilized by the way that opposites attract and like charges repel.
2 Measures of steric properties
2.2 A-values
2.3 Ceiling temperatures
2.4 Cone angles
Steric hindrance[edit]
Steric hindrance is a consequence of steric effects. Steric hindrance is the slowing of chemical reactions due to steric bulk. It is usually manifested in intermolecular reactions, whereas discussion of steric effects often focus on intramolecular interactions. Steric hindrance is often exploited to control selectivity, such as slowing unwanted side-reactions.
Steric hindrance between adjacent groups can also affect torsional bond angles. Steric hindrance is responsible for the observed shape of rotaxanes and the low rates of racemization of 2,2'-disubstituted biphenyl and binaphthyl derivatives.
Measures of steric properties[edit]
Because steric effects have profound impact on properties, the steric properties of substituents have been assessed by numerous methods.
Rate data[edit]
Relative rates of chemical reactions provide useful insights into the effects of the steric bulk of substituents. Under standard conditions methyl bromide solvolyzes 107 faster than does neopentyl bromide. The difference reflects the inhibition of attack on the compound with the sterically bulky (CH3)3C group.[3]
A-values[edit]
A values provide another measure of the bulk of substituents. A values are derived from equilibrium measurements of monosubstituted cyclohexanes.[4][5][6][7] The extent that a substituent favors the equatorial position gives a measure of its bulk.
The A-value for a methyl group is 1.74 as derived from the chemical equilibrium above. It costs 1.74 kcal/mol for the methyl group to adopt to the axial position compared to the equatorial position.
CH2CH3 1.75
Ceiling temperatures[edit]
Ceiling temperature (
{\displaystyle T_{c}}
) is a measure of the steric properties of the monomers that comprise a polymer.
{\displaystyle T_{c}}
is the temperature where the rate of polymerization and depolymerization are equal. Sterically hindered monomers give polymers with low
{\displaystyle T_{c}}
's, which are usually not useful.
Ceiling temperature (°C)[8]
ethylene 610 CH2=CH2
isobutylene 175 CH2=CMe2
1,3-butadiene 585 CH2=CHCH=CH2
isoprene 466 CH2=C(Me)CH=CH2
styrene 395 PhCH=CH2
α-methylstyrene 66 PhC(Me)=CH2
Cone angles[edit]
Ligand cone angle.
Cone angles of common phosphine ligands
P(OCH3)3 107
P(CH3)3 118
P(CH2CH3)3 132
P(C6H5)3 145
P(cyclo-C6H11)3 179
P(t-Bu)3 182
P(2,4,6-Me3C6H2)3 212
Ligand cone angles are measures of the size of ligands in coordination chemistry. It is defined as the solid angle formed with the metal at the vertex and the hydrogen atoms at the perimeter of the cone (see figure).[9]
Significance and applications[edit]
Steric effects are critical to chemistry, biochemistry, and pharmacology. In organic chemistry, steric effects are nearly universal and affect the rates and activation energies of most chemical reactions to varying degrees.
In biochemistry, steric effects are often exploited in naturally occurring molecules such as enzymes, where the catalytic site may be buried within a large protein structure. In pharmacology, steric effects determine how and at what rate a drug will interact with its target bio-molecules.
Prominent Sterically Hindered Compounds
Tris(2,4-di-tert-butylphenyl)phosphite, a widely used stabilizer in polymers.
Tricyclohexylphosphine, a bulky phosphine ligand used in homogeneous catalysis and, with B(C6F5)3, comprises the classic frustrated Lewis pair.[10]
2,6-Di-tert-butylphenol is used industrially as UV stabilizers and antioxidants for hydrocarbon-based products ranging from petrochemicals to plastics.[11]
Hindered amine light stabilizers are widely used in polymers.[12][13]
Titanium isopropoxide is a monomer, the corresponding titanium ethoxide is a tetramer.
An isolable selenenic acid owing to steric protection.[14]
Van der Waals strain, also known as steric strain
^ Günther Maier, Stephan Pfriem, Ulrich Schäfer, Rudolf Matusch (1978). "Tetra-tert-butyltetrahedrane". Angew. Chem. Int. Ed. Engl. 17 (7): 520–1. doi:10.1002/anie.197805201. {{cite journal}}: CS1 maint: uses authors parameter (link)
^ Gait, Michael (1984). Oligonucleotide synthesis: a practical approach. Oxford: IRL Press. ISBN 0-904147-74-6.
^ E.L. Eliel, S.H. Wilen and L.N. Mander, Stereochemistry of Organic Compounds, Wiley, New York (1994). ISBN 81-224-0570-3
^ Stevens, Malcolm P. (1999). "6". Polymer Chemistry an Introduction (3rd ed.). New York: Oxford University Press. pp. 193–194. ISBN 978-0-19-512444-6.
^ Tolman, Chadwick A. (1970-05-01). "Phosphorus ligand exchange equilibriums on zerovalent nickel. Dominant role for steric effects". J. Am. Chem. Soc. 92 (10): 2956–2965. doi:10.1021/ja00713a007.
^ Stephan, Douglas W. "Frustrated Lewis pairs": a concept for new reactivity and catalysis. Org. Biomol. Chem. 2008, 6, 1535-1539. doi:10.1039/b802575b
^ Helmut Fiege, Heinz-Werner Voges, Toshikazu Hamamoto, Sumio Umemura, Tadao Iwata, Hisaya Miki, Yasuhiro Fujita, Hans-Josef Buysch, Dorothea Garbe, Wilfried Paulus (2002). "Phenol Derivatives". Ullmann's Encyclopedia of Industrial Chemistry. Weinheim: Wiley-VCH. doi:10.1002/14356007.a19_313. ISBN 3527306730. {{cite encyclopedia}}: CS1 maint: uses authors parameter (link)
^ Pieter Gijsman (2010). "Photostabilisation of Polymer Materials". In Norman S. Allen (ed.). Photochemistry and Photophysics of Polymer Materials Photochemistry. Hoboken: John Wiley & Sons. pp. 627–679. doi:10.1002/9780470594179.ch17. ISBN 9780470594179. {{cite book}}: CS1 maint: uses authors parameter (link).
^ Klaus Köhler; Peter Simmendinger; Wolfgang Roelle; Wilfried Scholz; Andreas Valet; Mario Slongo (2010). "Paints and Coatings, 4. Pigments, Extenders, and Additives". Ullmann's Encyclopedia Of Industrial Chemistry. doi:10.1002/14356007.o18_o03. ISBN 978-3527306732.
^ Goto, Kei; Nagahama, Michiko; Mizushima, Tadashi; Shimada, Keiichi; Kawashima, Takayuki; Okazaki, Renji (2001). "The First Direct Oxidative Conversion of a Selenol to a Stable Selenenic Acid: Experimental Demonstration of Three Processes Included in the Catalytic Cycle of Glutathione Peroxidase". Organic Letters. 3 (22): 3569–3572. doi:10.1021/ol016682s. PMID 11678710.
Steric Effects (chem.swin.edu.au) at the Wayback Machine (archived July 25, 2008)
Steric: A Program to Calculate the Steric Size of Molecules (gh.wits.ac.za) at the Wayback Machine (archived December 22, 2017)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Steric_effects&oldid=1086221608"
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* '''Panel 6''': allows the user to play with the power of the dithering noise applied before quantizing the sine wave. Shaped or flat dither are available. The sine wave may also be modulated with a varying amplitude to highlight correlations between the input and the resulting quantization noise. The 'notch and gain' button applies a notch filter to the resulting signal, and boosts the gain of the remaining noise so that it's easily audible. The audio input from the audio device is discarded.
* '''Panel 7''': applies a sharper antialiasing (lowpass) filter than is likely to be built into the sound-card hardware (as there's generally no reason to use a filter quite this sharp in practice). The very sharp filter allows us to bandpass the demonstration squarewave without any harmonics landing in the transition band.
* '''Panel 7''': applies a sharper antialiasing (lowpass) filter than is likely to be built into the sound-card hardware (as there's generally no reason to use a filter quite this sharp in practice). The very sharp filter allows us to bandpass the demonstration square wave without any harmonics landing in the transition band. The input is read from the audio device, passed through this sharper filter, and then forwarded to the outputs.
* '''Panel 8''': when selected, generate a synthetic 'square wave' (this is not quite equivalent to a bandlimited analog square wave; the harmonic amplitudes are a bit different) that when aligned with the sampling phase just right gives the appearance of having infinite rise and fall time. The slider allows us to shift the waveform sample alignment back and forth by +/- one sample to reveal that the underlying signal is still band-limited.
* '''Panel 9''': as in panel 8, generate a 'perfect' synthetic 'square wave'. However, the slider now allows us to shift the sample alignment of the second channel with respect to the first, instead of shifting both channels. This allows us the trigger/lock the scope timing to the channel 1 waveform so we can see the fractional sample movement and alignment of the waveform on channel 2. The audio input from the audio device is discarded.
* '''Panel 10''': not used in the video; The audio device is configured to 24-bit input/output. The user may produce one of a range of test signals that are output to both the external applications and the audio device on the first channel. The input on the second channel is passed-through to the applications and audio device outputs unchanged. The first channel input is unused unless 'two input mode' is selected. When two input mode is selected, both input channels are read and the data sent to the external applications. Generated test signals are sent only to the audio hardware (on the first channel). This combination of test signals and input modes allows self-references frequency response, phase, noise, distortion and crosstalk testing of a given audio device.
{\displaystyle \ x(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}x_{\mathrm {square} }(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
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Decimal - New World Encyclopedia
Previous (Deciduous)
Next (Decision problem)
Counting rods Korean
Ionian/Greek
Roman Babylonian
3, 9, 12, 24, 30, 36, 60, more…
A decimal (or denary) system is a numeral system that has the number ten as its base. The term decimal is also used for a number written in this system, or for a fraction expressed using this system.
A number written in decimal notation involves the use of one or more of ten distinct symbols or fundamental units, called digits. The digits are often used with a decimal separator, which indicates the start of a fractional part. The decimal separator may be a dot, a period, or a comma.
7 Alternative numeral systems
The decimal system is the most widely used numeral system. It can be used to represent any number, no matter how large or small. In addition, it greatly simplifies arithmetical operations, a feature that is especially apparent when compared with the system of using Roman numerals. The decimal system forms the basis of the metric system of weights and measures, and it has been tapped to express the currencies of most nations of the world.
Evolution of Arabic numerals.
In the decimal system, the ten fundamental units that are currently in widespread use around the globe are: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. These ten symbols are called Arabic numerals by Europeans and Indian numerals by Arabs, the two groups' terms referring to the culture from which they learned the system. However, the symbols used in different areas are not identical. For instance, Western Arabic numerals (from which the European numerals are derived) differ from the forms used by other Arab cultures.
The decimal system is a positional numeral system; it has positions for units, tens, hundreds, and so forth. The position of each digit conveys the multiplier (a power of ten) to be used with that digit—each position has a value ten times that of the position to its right. For a mixed number (a number that is the sum of a whole number and a proper fraction), a decimal separator is used to separate the integer part from the fractional part. In English-speaking countries, a dot (·) or period (.) is used as the decimal separator; in many other languages, a comma is used. In addition, each number is preceded by one of the sign symbols, + or −, to indicate positive or negative sign, respectively.
The number ten is the count of the total number of fingers and thumbs on a person's two hands (or toes on two feet). In many languages, the word digit or its translation is also the anatomical term referring to fingers and toes. In English, the term decimal (from Latin decimus) means "tenth," decimate means "reduce by a tenth," and denary (Latin denarius) means "the unit of ten."
There were only two truly positional decimal systems in ancient civilization: the Chinese counting rods system and the Hindu-Arabic numeric system. Both required no more than ten symbols. Other numeric systems required more or fewer symbols.
A decimal fraction is a fraction in which the denominator is a power of ten.
Decimal fractions are commonly expressed without a denominator, the decimal separator being inserted into the numerator (with leading zeros added if needed), at the position from the right corresponding to the power of ten of the denominator. For example, 8/10, 83/100, 83/1000, and 8/10000 are expressed as: 0.8, 0.83, 0.083, and 0.0008, respectively.
The integer part (or integral part) of a decimal number is the part to the left of the decimal separator. The part to the right of the decimal separator is the fractional part; if considered as a separate number, a zero is often written in front. If the absolute value of a decimal number is less than one, it is usually expressed with a leading zero.
Trailing zeros after the decimal point are not necessary, but they may be retained in science, engineering, and statistics to indicate a required precision or to show a level of confidence in the accuracy of the number. For example, 0.080 and 0.08 are numerically equal, but in engineering, 0. 080 suggests a measurement with an error of up to 1 part in two thousand (±0.0005), while 0.08 suggests a measurement with an error of up to 1 in two hundred (±0.0005).
Any rational number that cannot be expressed as a decimal fraction has a unique infinite decimal expansion ending with recurring decimals.
Ten is the product of the first and third prime numbers (2x5 = 10); it is one greater than the square of the second prime number (3x3 + 1 = 10); and it is one less than the fifth prime number (11 - 1 = 10). This leads to a variety of simple decimal fractions, as follows:
1/3 = 0.333333… (with 3 repeating)
1/11 = 0.090909… (with 09 repeating)
1/12 = 0.083333… (with 3 repeating)
1/81 = 0.012345679012… (with 012345679 repeating)
Other prime factors in the denominator give longer recurring sequences. For instance, see 7 and 13.
That a rational number must have a finite or recurring decimal expansion can be seen as a consequence of the long division algorithm, in that there are only q-1 possible nonzero remainders on division by q, so that the recurring pattern will have a period less than q. For instance to find 3/7 by long division:
.4 2 8 5 7 1 4 ...
7 ) 3.0 0 0 0 0 0 0 0
2 8 30/7 = 4 r 2
1 4 20/7 = 2 r 6
5 6 60/7 = 8 r 4
3 5 40/7 = 5 r 5
4 9 50/7 = 7 r 1
7 10/7 = 1 r 3
2 8 30/7 = 4 r 2 (again)
The converse of this observation is that every recurring decimal represents a rational number p/q. This is a consequence of the fact that the recurring part of a decimal representation is, in fact, an infinite geometric series that will sum to a rational number. For instance,
{\displaystyle 0.0123123123\cdots ={\frac {123}{10000}}\sum _{k=0}^{\infty }0.001^{k}={\frac {123}{10000}}\ {\frac {1}{1-0.001}}={\frac {123}{9990}}={\frac {41}{3330}}}
Every real number has a (possibly infinite) decimal representation. That is, it can be written as
{\displaystyle x=\mathop {\rm {sign}} (x)\sum _{i\in \mathbb {Z} }a_{i}\,10^{i}}
sign() is the sign function,
ai ∈ { 0,1,…,9 } for all i ∈ Z, are its decimal digits, equal to zero for all i greater than some number (that number being the common logarithm of |x|).
Such a sum converges as i decreases, even if there are infinitely many nonzero ai.
Rational numbers (e.g. p/q) with prime factors in the denominator other than 2 and 5 (when reduced to simplest terms) have a unique recurring decimal representation.
Consider those rational numbers that have only the factors 2 and 5 in the denominator, that is, those that can be written as p/(2a5b). In this case, there is a terminating decimal representation. For instance, 1/1=1, 1/2=0.5, 3/5=0.6, 3/25=0.12, and 1306/1250=1.0448. Such numbers are the only real numbers that do not have a unique decimal representation, as they can also be written as a representation with a recurring 9; for instance, 1=0.99999…, 1/2=0.499999…, and so forth.
This leaves the irrational numbers. They also have unique infinite decimal representation, and can be characterized as the numbers whose decimal representations neither terminate nor recur.
So, in general, the decimal representation is unique if one excludes representations that end in a recurring 9.
The same trichotomy also holds for other base-n positional numeral systems:
and a version of this holds for irrational-base numeration systems as well, such as the golden mean base representation.
The following is a chronological list of writers and textual materials on decimals.
c. 3500 - 2500 B.C.E.: Elamites of Iran possibly used early forms of decimal system.[1]
c. 2900 B.C.E.: Egyptian hieroglyphs show counting in powers of 10 (for exampls, 1 million + 400,000 goats). (See Ifrah, below.)
c. 2600 B.C.E.: Indus Valley Civilization includes the earliest known physical use of decimal fractions in ancient weight system: 1/20, 1/10, 1/5, 1/2. See Ancient Indus Valley weights and measures
c. 1400 B.C.E.: Chinese writers show familiarity with the concept of decimals. For example, 547 is written 'Five hundred plus four decades plus seven of days' in some manuscripts.
c. 1200 B.C.E.: In ancient India, the Vedic text Yajur-Veda states the powers of 10, up to 1055.
c. 400 B.C.E.: Pingala develops the binary number system for Sanskrit prosody, with a clear mapping to the base-10 decimal system.
c. 250 B.C.E.: Archimedes writes the Sand Reckoner, which takes decimal calculation up to 1080,000,000,000,000,000.
c. 100–200 C.E.: The Satkhandagama is written in India—earliest use of decimal logarithms.
c. 476–550: Aryabhata uses an alphabetic cipher system for numbers that used zero.
c. 598–670: Brahmagupta explains the Hindu-Arabic numerals (modern number system) which uses decimal integers, negative integers, and zero.
c. 780–850: Muḥammad ibn Mūsā al-Ḵwārizmī is the first to expound on algorism outside India.
c. 920–980: Abu'l Hasan Ahmad ibn Ibrahim Al-Uqlidisi provides the earliest known direct mathematical treatment of decimal fractions.
c. 1300–1500: The Kerala School in South India uses decimal floating point numbers.
1548/1549–1620: Simon Stevin, author of De Thiende ('the tenth').
1561–1613: Bartholemaeus Pitiscus uses what appears to be decimal point notation.
1550–1617: John Napier uses decimal logarithms as a computational tool.
1765: Johann Heinrich Lambert discusses (with few proofs) patterns in decimal expansions of rational numbers and notes a connection with Fermat's little theorem in the case of prime denominators.
1800: Karl Friedrich Gauss uses number theory to systematically explain patterns in recurring decimal expansions of rational numbers (for example, the relation between period length of the recurring part and the denominator; fractions with the same denominator with recurring decimal parts that are shifts of each other, like 1/7 and 2/7). He also poses questions that remain open to this day.
1925: Louis Charles Karpinski publishes The History of Arithmetic.[2]
1959: Werner Buchholz writes Fingers or Fists? (The Choice of Decimal or Binary representation).[3]
1974: Hermann Schmid publishes Decimal Computation[4]
2000: Georges Ifrah's book, The Universal History of Numbers: From Prehistory to the Invention of the Computer, is published in English (translated from the French 1994 publication, Histoire Universelle des Chiffres.[5]
2003: Mike Cowlishaw presents Decimal Floating-Point: Algorism for Computers.[6].
A straightforward decimal system, in which 11 is expressed as ten-one and 23 as two-ten-three, is found in Chinese languages except Wu, and in Vietnamese with a few irregularities. Japanese, Korean, and Thai languages have imported the Chinese decimal system. Many other languages with a decimal system have special words for the numbers between 10 and 20, and decades.
Some psychologists suggest irregularities of numerals in a language may hinder children's counting ability[7].
Alternative numeral systems
Peoples of various cultures in history have used alternative numeral systems. Pre-Columbian Mesoamerican cultures, such as the Maya, used a vigesimal system (using all twenty fingers and toes); the Babylonians used a sexagesimal (base 60) system; and the Yuki tribe reportedly used an octal (base 8) system. Also, some Nigerians have used several duodecimal (base 12) systems.
Computer hardware and software systems commonly use a binary representation, internally. For external use by computer specialists, this binary representation is sometimes presented in the related octal or hexadecimal systems. For most purposes, however, binary values are converted to the equivalent decimal values for presentation to and manipulation by the public.
Both computer hardware and software also use internal representations that are effectively decimal for storing decimal values and doing arithmetic. Often, this arithmetic is done on data encoded using binary-coded decimal, but other decimal representations are also in use, especially in database implementations. Decimal arithmetic is used in computers so that decimal fractional results can be computed exactly, which is not possible using a binary fractional representation. This is usually important for financial and other calculations.[8]
↑ Robert Englund, 2001, The State of Decipherment of proto-Elamite. Max Planck Institute for the History of Science. Retrieved May 17, 2008.
↑ Louis Charles Karpinski, The History of Arithmetic, Rand McNally & Company, 1925.
↑ Werner Buchholz, Fingers or Fists? (The Choice of Decimal or Binary representation), Communications of the ACM, Vol. 2 #12, 3–11, ACM Press, December 1959.
↑ Hermann Schmid, Decimal Computation, John Wiley & Sons 1974 ISBN 047176180X; reprinted in 1983 by Robert E. Krieger Publishing Company ISBN 0898743184
↑ Georges Ifrah, and Robert Laffont, Histoire Universelle des Chiffres, 1994. (The Universal History of Numbers: From prehistory to the invention of the computer, Georges Ifrah, John Wiley and Sons Inc., 2000. Translated from the French by David Bellos, E.F. Harding, Sophie Wood and Ian Monk).
↑ Cowlishaw, M. F., Decimal Floating-Point: Algorism for Computers, Proceedings 16th IEEE Symposium on Computer Arithmetic, ISBN 076951894X, 104-111, IEEE Comp. Soc., June 2003
↑ Beth Azar, 1999, "English words may hinder math skills development", American Psychology Association Monitor 30(4) . Retrieved May 17, 2008.
↑ Decimal Arithmetic FAQ IBM. Retrieved May 17, 2008.
Long, Lynette. Delightful Decimals and Perfect Percents: Games and Activities That Make Math Easy and Fun. New York: J. Wiley, 2003. ISBN 978-0471210580
Mann, W. Wilberforce. A new system of measures, weights, and money; entitled the Linnbase decimal system; and designed for the adoption of all civilized nations, as the one common system. Ann Arbor, MI: Scholarly Publishing Office, University of Michigan Library, 2005. ISBN 978-1418192112
Nelson, Marvin N. Metric Handbook for Schools: Learning the Decimal System for Weights & Measures with Experiences and Problems. Minneapolis, MN: Burgess Pub. Co., 1980. ISBN 0808714481
Rasmussen, Steven, and Spreck Rosekrans. Decimal Concepts. Book 1. Berkeley, CA: Key Curriculum Project, 1985. ISBN 091368421X
Schmid, Hermann. Decimal Computation. New York: Wiley, 1974. ISBN 047176180X
Yates, James. Narrative Of The Origin And Formation Of The International Association For Obtaining A Uniform Decimal System Of Measures, Weights And Coins (1856). Whitefish, MT: Kessinger, 2008. ISBN 978-0548866672
Online Quiz: Decimal Place Value. – kwizNET Learning System.
Online Quiz: Decimal Sums. – kwizNET Learning System.
Online Quiz: Decimal to Fraction. – kwizNET Learning System.
Decimal Worksheets. – Maths Is Fun.
Convert Decimals to Fractions. – Maths Is Fun.
Decimal history
History of "Decimal"
Retrieved from https://www.newworldencyclopedia.org/p/index.php?title=Decimal&oldid=1037702
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Two Linear Equations in Two Unknowns Practice Problems Online | Brilliant
To begin our study of linear algebra, we’ll look at a simple case: a system of two linear equations with two unknowns. Though solving this system is quite easy, there are still quite a few interesting concepts and ideas that come up when we think deeply about the process. These concepts and ideas play a larger role when we look at more general systems of linear equations later.
So let’s start with the system
\begin{aligned} x-2y &= 6 \\ 3x+y &= 4. \end{aligned}
The first idea we might have is to solve for one of the variables. Let’s isolate
x
in the first equation and use it to solve
y
in the second. After isolating
x
in the first equation we get
x = 6+2y.
Now we can plug that into the second equation and solve:
\begin{aligned} 3(6+2y)+y &= 4 \\ 18+7y &= 4 \\ y &= -2. \end{aligned}
x
x=2,
so we have obtained a unique solution,
x=2,y=-2.
The next few problems are going to look at this exercise in several different ways: finding a different way to solve it, interpreting it geometrically, and imagining what might happen if we tweak some of the numbers.
Two Linear Equations in Two Unknowns
Another standard way to do the problem is similar, in that its goal is to eliminate a variable, but it achieves this goal slightly differently. That is, we can multiply both sides of one of the equations by a constant, and then add or subtract the two equations.
\begin{aligned} x-2y &= 6 \\ 3x+y &= 4. \end{aligned}
For instance, if we want to eliminate
x,
3
times the first equation minus the second equation:
\begin{aligned} 3x-6y &= 18 \\ 3x+y &= 4 \end{aligned}
\begin{aligned} (3x-6y)-(3x+y) &= 18-4 \\ -7y &= 14 \\ y &= -2 \end{aligned}
as before. (Also, substituting
y = -2
back into one of the original equations and solving gets
x = 2 .)
Just to make sure we’re paying attention, let’s say we wanted to eliminate
y
instead. Which combination of equations shown below would do that?
\begin{aligned} \text{Equation 1: }x-2y &= 6 \\ \text{Equation 2: }3x+y &= 4. \end{aligned}
1
plus equation
2
1
plus 2 times equation
2
1
minus equation
2
1
minus 2 times equation
2
Which tweaks to the original system of equations would change the number of solutions?
Remember, we started with
\begin{aligned} x-2y &= 6 \\ 3x+y &= 4. \end{aligned}
Let’s change the number “
3
” in the second equation to something else, and see what effect that has on the number of solutions. Remember that our original system has exactly one solution,
x=2,y=-2.
What number should we change the “
3
” to in order to create a system of equations that doesn’t have exactly one solution?
-2
-\frac12
0
It doesn’t matter what we change the “
3
” to, there is always exactly one solution There is more than one answer to this question
We’ll come back to algebra shortly, but let’s pause to think about what a pair of linear equations mean geometrically.
Each of the two equations involving
x
y
cuts out a line in the coordinate plane. For example, consider the system
x+y = 0 \ , \ y - \frac{1}{2} x = - 3.
\begin{aligned} y &= -x \\ y &= \frac{1}{2}x-3 \end{aligned}
The solution to this system (like the original) is the intersection of these two lines, which is the point
(2,-2).
Now think about all of the different ways that any two lines can intersect. Use that to answer the following question:
Suppose we have a system of two linear equations with real number coefficients in two real unknowns. Which one of the following options is impossible?
The system has no solutions The system has exactly one solution The system has exactly two solutions The system has infinitely many solutions
Now that we have some geometric intuition, let’s try to generalize. Say we have a system of two equations of the form
\begin{aligned} 2x+by &= m \\ 3x+dy &= n. \end{aligned}
What condition guarantees that the system will have exactly one solution?
Assume that it has at least one solution.
3m \ne 2n
3m=2n
3b \ne 2d
3b=2d
If we are even more general, we can consider the system of equations
\begin{aligned} ax+by &=m \\ cx+dy &= n. \end{aligned}
Expressed in slope-intercept form:
\begin{aligned} y&=-\frac{a}{b}x+\frac{m}{b} \\\\ y&=-\frac{c}{d}x+\frac{n}{d} \end{aligned}
A system of two linear equations has a unique solution if slopes are different:
\frac{a}{b}\neq \frac{c}{d} \implies ad-bc\neq 0.
The system has no solutions if slopes are equal
(ad-bc=0 ),
y
-intercepts are different:
\frac{m}{b}\neq\frac{n}{d} \implies dm-bn\neq 0
The system has infinite solutions if slopes are equal
(ad-bc=0 ),
y
-intercepts are equal
(dm-bn=0).
So there are at least two important takeaways: one is the useful fact that the “degeneration” of the system of equations can be boiled down to looking at a certain algebraic expression in the coefficients,
ad-bc
in this case, and deciding whether it is nonzero. This turns out to generalize to larger systems of equations.
The second takeaway is our intuition about the behavior of these systems. Because a “randomly” chosen real number is almost always nonzero, we should expect that a “random” system of two equations in two unknowns will have a unique solution. (Geometrically speaking, two “random” lines will almost always not be parallel, and hence will intersect in exactly one point.) This intuition will also follow us as we explore larger systems of equations.
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Plot worst-case gain of uncertain system - MATLAB wcsigmaplot - MathWorks India
wcsigmaplot
Plot Worst-Case Gain of Uncertain System
wcsigmaplot(usys)
wcsigmaplot(usys,w)
wcsigmaplot(___,opts)
wcsigmaplot(usys) plots the nominal and worst-case gains of the uncertain system usys as a function of frequency. For multi-input, multi-output (MIMO) systems, gain refers to the largest singular value of the frequency response matrix. (See sigma for more information about singular values.) The plot includes:
Nominal — Nominal gain of usys.
Worst perturbation — The response falling within the uncertainty of usys that has the highest peak gain. This curve corresponds to the wcu output argument of wcgain.
Worst-case gain (lower bound) — The lowest possible worst-case gain at each frequency.
Worst-case gain (upper bound) — The highest possible gain within the uncertainty at each frequency. This curve represents the envelope produced by finding the highest possible gain at each frequency.
Sampled Uncertainty — Responses randomly sampled from usys.
wcsigmaplot(usys,w) focuses the plot on the frequencies specified by w.
If w is a cell array of the form {wmin,wmax}, then wcsigmaplot plots the worst-case gains in the range {wmin,wmax}.
If w is an array of frequencies, then wcsigmaplot plots the worst-case gains at each frequency in the array.
wcsigmaplot(___,opts) specifies additional options for the computation. Use wcOptions to create opts.
Plot the worst-case gain of the following system:
sys=\frac{{s}^{2}+3s}{{s}^{2}+2s+a}.
The uncertain parameter a = 2
±
usys = tf([1 3 0],[1 2 a]);
The Worst perturbation curve identifies the single response within the uncertainty that yields the highest gain at any frequency. This perturbation corresponds to the wcu output of wcgain.
The Worst-case gain curves show the lower and upper bounds on the worst-case gain at each frequency. For any perturbation within the specified uncertainty range, the principal gains (singular values) of the perturbed system lie below the Worst-case gain (upper bound) curve. In other words, this curve is the envelope produced by finding the highest gain within the uncertainty at each frequency. For this system, the lower and upper bounds are close enough to appear identical on the plot. (See wcgain for more information about these bounds.)
w = {0.1 10};
Examine the effect on the worst-case response of increasing the uncertainty range. To do this without changing the uncertainty specified in usys, use the ULevel option of wcOptions. This option scales the normalized uncertainty by the factor you specify. For example, examine the worst-case response a 50% greater uncertainty range.
wcsigmaplot(usys,w,opts)
The plot shows that increasing the uncertainty range substantially increases the worst-case gain at low frequencies.
Dynamic system with uncertainty, specified as a uss, ufrd, genss, or genfrd model that contains uncertain elements.
For genss or genfrd models, wcsigmaplot uses the current value of any tunable blocks and folds them into the known (not uncertain) part of the model.
Frequencies at which to plot worst-case gains, specified as the cell array {wmin,wmax} or as a vector of frequency values.
If w is a cell array of the form {wmin,wmax}, then the function plots the worst-case gains at frequencies ranging between wmin and wmax.
If w is a vector of frequencies, then the function plots the worst-case gains at each specified frequency. For example, use logspace to generate a row vector with logarithmically spaced frequency values.
opts — Options for worst-case gain computation
Options for worst-case computation, specified as an object you create with wcOptions. Setting certain options for mussv can improve the results of the worst-case calculation. See wcOptions for more information.
Example: wcOptions('ULevel',2,'MussvOptions','m3')
wcsigmaplot uses wcgain to compute the worst-case gains. Use the opts argument to set options for the wcgain algorithm.
wcsigmaplot uses usample to compute the Sampled Uncertainty curves.
wcOptions | wcgain | sigma | uss
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When we convert a digital signal back to analog, the result is
''also'' smooth regardless of the [[WikiPedia:Audio_bit_depth|bit depth]]. 24 bits or 16 bits...
''also'' smooth regardless of the [[WikiPedia:Audio_bit_depth|bit depth]]. It doesn't matter if it's 24 bits or 16 bits or 8 bits.
or 8 bits... it doesn't matter.
So does that mean that the digital bit depth makes no difference at
all? Of course not.
Channel 2 here is the same sine wave input, but we quantize with
Channel 2 is the same sine wave input, but we quantize it with
[[WikiPedia:Dither|dither]] down to 8 bits.
On the scope, we still see a nice
smooth sine wave on channel 2. Look very close, and you'll also see a
bit more noise. That's a clue.
If we look at the spectrum of the signal... aha! Our sine wave is
If we look at the spectrum of the signal, our sine wave is
still there unaffected, but the noise level of the 8-bit signal on
the second channel is much higher!
the second channel is much higher. And that's the difference, the only difference, the number of bits makes.
When we digitize a signal, first we sample it. The
sampling step is perfect; it loses nothing. But then we [[WikiPedia:Quantization_(sound_processing)|quantize]] it,
and [[WikiPedia:Quantization_error|quantization adds noise]].
The number of bits determines how much noise and so the level of the
What does this dithered quantization noise sound like? Let's listen
What does this dithered quantization noise sound like?
to our 8-bit sine wave.
Those of you who have used analog recording equipment might think to yourselves, "My goodness! That sounds like tape hiss!"
That may have been hard to hear anything but the tone. Let's listen
to just the noise after we notch out the sine wave and then bring the
gain up a bit because the noise is quiet.
Those of you who have used analog recording equipment may have just
thought to yourselves, "My goodness! That sounds like tape hiss!"
Well, it doesn't just sound like tape hiss, it acts like it too, and
if we use a [[WikiPedia:Dither#Different_types|gaussian dither]] then it's
of [[WikiPedia:Magnetic_tape_sound_recording|magnetic audio tape]]
in [[WikiPedia:Shannon–Hartley_theorem#Examples|bits instead of decibels]], in order to put things in a
digital perspective. [[WikiPedia:Compact cassettes|Compact cassettes]] (for those of you who are old enough to remember them) could reach as
digital perspective. [[WikiPedia:Compact cassettes|Compact cassettes]], for those of you who are old enough to remember them, could reach as
deep as 9 bits in perfect conditions, though 5 to 6 bits was
deep as 9 bits in perfect conditions. 5 to 6 bits was
more typical, especially if it was a recording made on a
[[WikiPedia:Cassette_deck|tape deck]]. That's right... your mix tapes were only about 6 bits
[[WikiPedia:Cassette_deck|tape deck]]. That's right; your old mix tapes were only about 6 bits
deep... if you were lucky!
deep if you were lucky!
The very best professional [[WikiPedia:Reel-to-reel_audio_tape_recording|open reel tape]] used in studios could barely
hit... any guesses? 13 bits ''with'' [[WikiPedia:Reel-to-reel_audio_tape_recording#Noise_reduction|advanced noise reduction]]. And
hit 13 bits ''with'' [[WikiPedia:Reel-to-reel_audio_tape_recording#Noise_reduction|advanced noise reduction]].
{\displaystyle \ squarewave(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}\ squarewave(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
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Euclidean Geometry - Circles Problem Solving | Brilliant Math & Science Wiki
Euclidean Geometry - Circles Problem Solving
A Former Brilliant Member, A Former Brilliant Member, A Former Brilliant Member, and
This wiki is about problem solving on circles. You need to be familiar with some (if not all) theorems on circles.
Find the area of the shaded region in the above diagram.
The area of the shaded region is equal to the area of the semicircle with diameter
3
plus the area of the semicircle with diameter
4
plus the area of the right triangle minus the area of the semicircle with diameter
5
\begin{aligned} \text{Area}&=\dfrac{\pi}{4}\big(3^2\big)+\dfrac{\pi}{4}\big(4^2\big)+\dfrac{1}{2}(3)(4)-\dfrac{\pi}{4}(5)^2\\\\ &=\dfrac{9}{4}\pi + 4\pi + 6 - \dfrac{25}{4}\pi\\\\ &=6.\ _\square \end{aligned}
BC
is the diameter of a circle with center at
O
\angle ABO=30^\circ
\angle OAD=40^\circ
\angle COD
AC
. Then by Thales' theorem,
\triangle BAC
\triangle BOA
OB=OA=\text{radius}
\angle OAB=\angle ABO=30^\circ.
\angle CAD=90^\circ-30^\circ-40^\circ=20^\circ.
By the inscribed angle theorem,
\angle COD=2(\angle CAD)=2(20^\circ)=40^\circ.\ _\square
DC
is the diameter of the circle with center at
A.
BE
DC
DE=15
EC=9
BE.
\triangle DEB \sim \triangle DBC
\dfrac{BE}{DE}=\dfrac{BC}{BD}\implies \dfrac{BE}{15}=\dfrac{BC}{BD}\implies BE=15\dfrac{BC}{BD}\qquad (1)
\dfrac{BE}{BD}=\dfrac{BC}{DC}\implies \dfrac{BE}{BD}=\dfrac{BC}{24}\implies BE=\dfrac{(BD)(BC)}{24}. \qquad (2)
\begin{aligned} 15\dfrac{BC}{BD}&=\dfrac{(BD)(BC)}{24}\\ 15(24)&=(BD)^2\\ BD&=\sqrt{360}\\&=6\sqrt{10}. \end{aligned}
Using the Pythagorean theorem, we have
BC=\sqrt{DC^2-BD^2}=\sqrt{24^2-\big(6\sqrt{10}\big)^2}=\sqrt{216}=6\sqrt{6}.
BE=15\dfrac{BC}{BD}=15\left(\dfrac{6\sqrt{6}}{6\sqrt{10}}\right)=\dfrac{15\sqrt{6}}{\sqrt{10}}.
Rationalizing the denominator gives
BE=\dfrac{15\sqrt{6}}{\sqrt{10}} \cdot \dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{30}{10}\sqrt{15}=3\sqrt{15}.\ _\square
\sqrt{697}
\dfrac{1}{7}\pi+\sqrt{697}
\dfrac{5\sqrt{697}}{7}+\dfrac{2\sqrt{679}}{7}
\sqrt{\dfrac{17425}{49}}+\sqrt{\dfrac{27888}{49}}
BC
is tangent to both a circle with center at
A
and a circle with center at
D
. The area of the circle with center at
A
225\pi
and the area of the circle with center at
D
36\pi
BC=16
, find the distance between the centers of the two circles.
Cite as: Euclidean Geometry - Circles Problem Solving. Brilliant.org. Retrieved from https://brilliant.org/wiki/euclidean-geometry-circles-problem-solving/
|
Write the first six terms of the arithmetic sequence with
Write the first six terms of the arithmetic sequence with the first term,
{a}_{1}
, and common difference, d.
{a}_{1}=5
, d=3
A sequence is arithmetic if there is a constant difference between consecutive terms, called the common difference d. So, to find the next term, add the common difference to the previous term.
{a}_{1}=5
and d=3, the first six terms are:
{a}_{1}=5
{a}_{2}={a}_{1}+3=5+3=8
{a}_{3}={a}_{2}+3=8+3=11
{a}_{4}={a}_{3}+3=11+3=14
{a}_{5}={a}_{4}+3=14+3=17
{a}_{6}={a}_{5}+3=17+3=20
5,8,11,14,17, and 20
At a clothing store, 3 shirts and 8 hats cost $65. The cost for 2 shirts and 2 hats is $30. How much does each shirt and hat cost?
Consider the sequences
\left\{{a}_{n}\right\}=\left\{-2,5,12,19,\dots \right\}
\left\{{b}_{n}\right\}=\left\{3,6,12,24,48,\dots \right\}
(a) Find the next two terms of the sequence.
(b) Find a recurrence relation that generates the sequence.
(c) Find an explicit formula for the nth term of the sequence.
If a die is rolled 30 times, there are
{6}^{30}
different sequences possible. The following question asks how many of these sequences satisfy certain conditions.
What fraction of these sequences have exactly three 4s and three 5s?
2,4,6,8,10,..
2,4,8,16,32,\dots
a,a+2,a+4,a+6,a+8,\dots
Error Analysis On your homework, you write that the missing term in the arithmetic sequence 31, _, 41, ... is
35½
. Your friend says the missing term is 36. Who is correct? What mistake was made?
Suppose f :
\left(M,d\right)\to \left(N,p\right)
Prove or disprove that f is uniformly continuous
⇔
f maps a Cauchy seq. to Cauchy seq.
|
LMIs in Control/pages/Basic Matrix Theory - Wikibooks, open books for an open world
LMIs in Control/pages/Basic Matrix Theory
{\displaystyle A\in \mathbb {C} ^{n\times m}}
{\displaystyle A={\begin{bmatrix}a_{11}&\dots &a_{1m}\\\vdots &\ddots &\vdots \\a_{n1}&\dots &a_{nm}\end{bmatrix}}\in \mathbb {C} ^{n\times m}}
{\displaystyle A}
{\displaystyle A^{T}}
{\displaystyle A'}
{\displaystyle A^{T}={\begin{bmatrix}a_{11}&\dots &a_{n1}\\\vdots &\ddots &\vdots \\a_{1m}&\dots &a_{nm}\end{bmatrix}}\in \mathbb {C} ^{m\times n}.}
{\displaystyle A}
{\displaystyle A^{*}}
{\displaystyle A^{*}={\begin{bmatrix}a_{11}^{*}&\dots &a_{n1}^{*}\\\vdots &\ddots &\vdots \\a_{1m}^{*}&\dots &a_{nm}^{*}\end{bmatrix}}\in \mathbb {C} ^{m\times n}.}
{\displaystyle a_{nm}^{*}}
{\displaystyle a_{nm}}
{\displaystyle A\in \mathbb {R} ^{n\times m}}
{\displaystyle A^{*}=A^{T}}
{\displaystyle A\in \mathbb {C} ^{n\times n}}
{\displaystyle A=A^{*}}
{\displaystyle A\in \mathbb {R} ^{n\times n}}
{\displaystyle A\in \mathbb {C} ^{n\times n}}
{\displaystyle A^{*}=A^{-1}}
{\displaystyle A^{*}A=I}
Retrieved from "https://en.wikibooks.org/w/index.php?title=LMIs_in_Control/pages/Basic_Matrix_Theory&oldid=3609071"
|
Linear Diophantine Equations - One Equation Practice Problems Online | Brilliant
In a certain basketball game, you can only make 2 point shots and 3 point shots. If you scored exactly 20 points, how many of the following scenarios are possible?
1
2
3
4
Consider all ordered pairs of integers
(x,y)
\frac{5}{x}+\frac{7}{y} = \frac{12}{xy}.
The smallest positive integer value of
x
in these ordered pairs is
1,
x=y=1
satisfies the equation. What is the second smallest positive integer value of
x
in these ordered pairs?
2
3
4
5
6
7
8
9
Jimmy has the high score, 920, in his favorite video game. For each win, he gets 50 points, but for each loss, he loses 36 points. What is the minimum number of games that Jimmy could have played?
If you have infinite pennies ($0.01), nickels ($0.05), and dimes ($0.10), in how many different ways can you make change for $1.00?
Hint: Don't try to list all of the possibilities!
Gummy bears come in packs of 6 and 9. Is there a combination of these packs which gives a total of exactly 100 gummy bears?
|
Market dominance describes when a firm can control markets.[1] A dominant firm possesses the power to affect competition[2] and influence market price.[3] A firms' dominance is a measure of the power of a brand, product, service, or firm, relative to competitive offerings, whereby a dominant firm can behave independent of their competitors or consumers,[4] and without concern for resource allocation.[5] Dominant positioning is both a legal concept and an economic concept and the distinction between the two is important when determining whether a firm's market position is dominant.
Sources of Market DominanceEdit
Economies of scale.[6]
First-Mover AdvantagesEdit
Many dominant firms are the first "important" competitor in their industry.[7] These firms can achieve short- or long-term advantages over their competitors when they are the first offering in a new industry. First-movers can set a benchmark for competitors and consumers regarding expectations of product and service offering, technology, convenience, quality, or price.[8] These firms are representative of their industry and their brand can become synonymous with the product category itself, such as the company Band-Aid. First-mover advantage is a limited source of market dominance if a firm becomes complacent or fails to keep up innovation by competitors.[9]
Measuring Market DominanceEdit
Impact on competitorsEdit
Herfindahl IndexEdit
{\displaystyle D=\sum _{i=1}^{n-1}(s_{i}-s_{i+1})^{2}}
{\displaystyle s_{1}\geq ...\geq s_{i}\geq s_{i+1}\geq ...\geq s_{n}}
{\displaystyle ID=\sum _{i=1}^{n}h_{i}^{2}}
{\displaystyle h_{i}=\left.s_{i}^{2}\right/HHI.}
{\displaystyle AI=\left.\sum _{i=1}^{n}\left(s_{i}-{1 \over n}\right)^{2}\right/n.}
Customer PowerEdit
|
\dots \mathrm{f1}{!}^{i}\mathrm{f2}{!}^{j}\mathrm{f3}{!}^{k}\dots
i,j,k
\mathrm{f1}
\mathrm{f2}
\mathrm{f3}
\sqrt{\mathrm{\pi }}=\mathrm{\Gamma }\left(\frac{1}{2}\right)
0<i
j,k<0
\mathrm{f1}-\mathrm{f2}-\mathrm{f3}=n
\frac{\left(\genfrac{}{}{0}{}{\mathrm{f1}}{\mathrm{f2}}\right)c\mathrm{f2}!\mathrm{f3}!}{\mathrm{f1}!}
c
is a correction factor depending on
\mathrm{f3}
i,0<j
k<0
\mathrm{f3}-\mathrm{f1}-\mathrm{f2}=n
\frac{c\mathrm{f3}!}{\mathrm{f1}!\mathrm{f2}!\left(\genfrac{}{}{0}{}{\mathrm{f2}}{\mathrm{f1}}\right)}
\dots \mathrm{f1}{!}^{i}\mathrm{f2}{!}^{j}\dots
i,j
\frac{\mathrm{f1}}{\mathrm{f2}}
r
1<|r|
\frac{\mathrm{f2}!\left(\genfrac{}{}{0}{}{\mathrm{f1}}{\mathrm{f2}}\right)\left(\mathrm{f1}-\mathrm{f2}\right)!}{\mathrm{f1}!}
|r|<1
\frac{\mathrm{f2}!}{\mathrm{f1}!\left(\genfrac{}{}{0}{}{\mathrm{f2}}{\mathrm{f1}}\right)\left(\mathrm{f2}-\mathrm{f1}\right)!}
a≔\frac{n!}{k!\left(n-k\right)!}
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{!}}{\textcolor[rgb]{0,0,1}{k}\textcolor[rgb]{0,0,1}{!}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{k}\right)\textcolor[rgb]{0,0,1}{!}}
\mathrm{convert}\left(a,\mathrm{binomial}\right)
\left(\genfrac{}{}{0}{}{\textcolor[rgb]{0,0,1}{n}}{\textcolor[rgb]{0,0,1}{k}}\right)
a≔\frac{n\left({n}^{2}+m-k+2\right)\left({n}^{2}+m\right)!}{k!\left({n}^{2}+m-k+2\right)!}
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{n}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{k}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{n}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{m}\right)\textcolor[rgb]{0,0,1}{!}}{\textcolor[rgb]{0,0,1}{k}\textcolor[rgb]{0,0,1}{!}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{n}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{k}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{!}}
\mathrm{convert}\left(a,\mathrm{binomial}\right)
\frac{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{}\left(\genfrac{}{}{0}{}{{\textcolor[rgb]{0,0,1}{n}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{m}}{\textcolor[rgb]{0,0,1}{k}}\right)}{{\textcolor[rgb]{0,0,1}{n}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{k}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}
a≔\frac{{m!}^{3}}{\left(3m\right)!}
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\frac{{\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{!}}^{\textcolor[rgb]{0,0,1}{3}}}{\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{m}\right)\textcolor[rgb]{0,0,1}{!}}
\mathrm{convert}\left(a,\mathrm{binomial}\right)
\frac{\textcolor[rgb]{0,0,1}{1}}{\left(\genfrac{}{}{0}{}{\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{m}}{\textcolor[rgb]{0,0,1}{m}}\right)\textcolor[rgb]{0,0,1}{}\left(\genfrac{}{}{0}{}{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{m}}{\textcolor[rgb]{0,0,1}{m}}\right)}
a≔\frac{\mathrm{\Gamma }\left(m+\frac{3}{2}\right)}{\mathrm{sqrt}\left(\mathrm{\pi }\right)\mathrm{\Gamma }\left(m\right)}
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{\mathrm{\Gamma }}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{2}}\right)}{\sqrt{\textcolor[rgb]{0,0,1}{\mathrm{\pi }}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\Gamma }}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{m}\right)}
\mathrm{convert}\left(a,\mathrm{binomial}\right)
\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left(\genfrac{}{}{0}{}{\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}}\right)
|
How do you solve by factoring and using the principle
How do you solve by factoring and using the principle of zero products
{x}^{2}+7x+6=0
=-6\text{ }\text{or}\text{ }x=-1
using the a-c method
the factors of +6 which sum to +7 are +6 and +1
⇒\left(x+6\right)\left(x+1\right)=0
equate each factor to zero and solve for x
x+6=0⇒x=-6
x+1=0⇒x=-1
Find the complex zeros of the following polynomial function. Write f in factored form.
f\left(x\right)={x}^{3}-8
Cannot solve quadratic equation through product sum product when the the first product is fraction
\frac{16}{3}{x}^{2}-2x-45=0
Find all the values of w such that
\left(w+\frac{21}{17}{\right)}^{2}=\frac{2645}{441}
Graph the quadratic function
f\left(x\right)={\left(x+4\right)}^{2}
. Give the (a) vertex, (b) axis, (c) domain, and (d) range. Then determine (e) the largest open interval of the domain over which the function is increasing and (f) the largest open interval over which the function is decreasing.
(a) The vertex is (-4,0).
(b) The axis is
Use the vertex and tercopts ic eketen the Aen of the quadrant function. Give the equation for the parabolas
{z}^{2}+12=-z
Quadratic equation solve
x2+2x-8=0
|
Over-Damped System (ξ>1) | Education Lessons
Fig-1 (Over damped system)
We know that the characteristic equation of the damped free vibration system is,
mS^2 + cS + K = 0
This is a quadratic equation having two roots
S_1
S_2
S_{1,2} = {-c \over 2m} \pm \sqrt{\bigg({-c \over 2m}\bigg)^2 - {K \over m}}
In order to convert the whole equation in the form of
\xi
, we will use two parameters, critical damping coefficient '
c_c
' and damping factor '
\xi
'. So the roots
S_1
S_2
can be written as follows;
\begin{aligned} \xi= {c \over c_c} \quad \text {OR} \quad \xi &= {c \over 2m\omega_0} \quad (\text {as} \quad c_c=2m\omega_n)\\ \therefore {c \over 2m} &= \xi \omega_n \end{aligned}
\omega_n
= natural frequency of undamped free vibration =
\sqrt{K \over m}
\therefore \omega_n^2 = {K \over m}
So we can write roots
S_1
S_2
and as;
\begin{aligned} S_{1,2} &= -\xi \omega_n \pm \sqrt{(\xi \omega_n)^2 - \omega_n^2}\\ \therefore S_{1,2} &= [-\xi \pm \sqrt{\xi^2 - 1}]\omega_n\\ \therefore \ \ S_{1} &= [-\xi + \sqrt{\xi^2 - 1}]\omega_n\\ \text{And, }\\ S_{2} &= [-\xi - \sqrt{\xi^2 - 1}]\omega_n \end{aligned}
Overdamped system (ξ>1)
If the damping factor ‘
\xi
’ is greater than one or the damping coefficient ‘
c
’ is greater than critical damping coefficient ‘
c_c
’, then the system is said to be over-damped.
\xi > 1 \quad \text{Or} \quad {c \over c_c}\quad \text{Or} \quad c > c_c
In overdamped system, the roots are given by;
\begin{aligned} S_{1} &= [-\xi + \sqrt{\xi^2 - 1}]\omega_n\\ \text{And, }\\ S_{2} &= [-\xi - \sqrt{\xi^2 - 1}]\omega_n \end{aligned}
\xi > 1
S_1
S_2
as real and negative so we get,
\begin{aligned} x &= Ae^{S_1t} + Be^{S_2t}\\ \therefore x &= Ae^{[-\xi + \sqrt{\xi^2-1}]\omega_nt} + Be^{[-\xi - \sqrt{\xi^2-1}]\omega_nt} \quad \dots \dots \text{(1)} \end{aligned}
Now differentiating equation (1) with respect to ‘t’, we get;
\begin{aligned} \mathring x = &Ae^{[-\xi + \sqrt{\xi^2-1}]\omega_nt}[-\xi + \sqrt{\xi^2-1}]\omega_n \ + \\ &Be^{[-\xi - \sqrt{\xi^2-1}]\omega_nt} [-\xi - \sqrt{\xi^2-1}]\omega_n\quad \dots \dots \dots \text{(2)} \end{aligned}
t = 0
x = X_0
t = 0
\mathring x =0
Substituting this value in equation (1) we get;
X_0 = A+B\quad \dots \dots \dots \text{(3)}
0= A[-\xi + \sqrt{\xi^2-1}]\omega_n + B[-\xi - \sqrt{\xi^2-1}]\omega_n \quad \dots \dots \text{(4)}
B= X_0 - A
and putting value of B in (4);
\begin{alignedat}{2} &\therefore \qquad 0 &=& \ A[-\xi + \sqrt{\xi^2-1}]\omega_n + (X_0 - A)[-\xi - \sqrt{\xi^2-1}]\omega_n\\ &\therefore \qquad 0 &=& \ -A\xi + A\sqrt{\xi^2-1} + X_0 [-\xi - \sqrt{\xi^2-1}]+A\xi+A\sqrt{\xi^2-1}\\ &\therefore \qquad 0 &=& \ 2A\sqrt{\xi^2-1} + X_0 [-\xi - \sqrt{\xi^2-1}]\\ &\therefore 2A\sqrt{\xi^2-1} \ &=& \ X_0 [\ \xi + \sqrt{\xi^2-1}]\\ &\therefore \qquad A &=& \ {X_0 [\xi + \sqrt{\xi^2-1}] \over 2\sqrt{\xi^2-1} }\quad \dots \dots \text{(5)} \end{alignedat}
A=X_0-B
, in equation (4);
\begin{alignedat}{2} &\therefore \qquad 0 &=& \ (X_0-B)[-\xi + \sqrt{\xi^2-1}]\omega_n + B[-\xi - \sqrt{\xi^2-1}]\omega_n\\ &\therefore \qquad 0 &=& \ X_0[-\xi + \sqrt{\xi^2-1}] + B\xi - B\sqrt{\xi^2-1} - B\xi - B\sqrt{\xi^2-1}\\ &\therefore \qquad 0 &=& \ X_0[-\xi + \sqrt{\xi^2-1}] - 2B\sqrt{\xi^2-1}\\ &\therefore 2B\sqrt{\xi^2-1} &=& \ X_0[-\xi + \sqrt{\xi^2-1}]\\ &\therefore \qquad B &=& \ {X_0[-\xi + \sqrt{\xi^2-1}] \over 2\sqrt{\xi^2-1}} \quad \dots \dots \text{(6)} \end{alignedat}
Now putting equation (5) and (6) in equation (1), we get;
\begin{aligned} x = &{X_0 [\xi + \sqrt{\xi^2-1}] \over 2\sqrt{\xi^2-1} } e^{[-\xi + \sqrt{\xi^2-1}]\omega_nt} +\\ &{X_0[-\xi + \sqrt{\xi^2-1}] \over 2\sqrt{\xi^2-1}} e^{[-\xi - \sqrt{\xi^2-1}]\omega_nt}\\ x = &{X_0 \over 2\sqrt{\xi^2-1}} \bigg [\bigg (\xi + \sqrt{\xi^2-1} \bigg) e^{[-\xi + \sqrt{\xi^2-1}]\omega_nt} + \\ &\bigg (-\xi + \sqrt{\xi^2-1} \bigg ) e^{[-\xi - \sqrt{\xi^2-1}]\omega_nt} \bigg ] \end{aligned}
Above equation represents the equation of motion for overdamped system.
The motion obtained by above equation is aperiodic (aperiodic motion motions are those motions in which the motion does not repeats after a regular interval of time i.e non periodic motion) and so the system does not shows vibrations.
This type of system does not show much damping, so this systems are used very rarely.
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Principles of radiation astronomy is a course of forty-eight lectures, sixteen mini-lectures for quiz sections, three hourly quizzes that are timed at an hour, a mid-term that covers the first half of the course, and a final which covers everything in the course. This is the first of three hourlies. It covers the first sixteen lectures, the first five mini-lectures, problem sets, lessons, and laboratories.
To improve your score, read and study the lectures and the rest, the links contained within, listed under See also, External links and in the {{principles of radiation astronomy}} template. This should give you adequate background to get 100 %.
1 What is a pfu?
4 Yes or No, The spin carried by quarks is sufficient to account for the total spin of protons.
5 Spin-charge separation has which characteristics?
6 Which of the following are characteristic of interstellar extinction?
What is the blue-radiation source in the image at right?
9 True or False, A theory may not be right but it should be testable.
{\displaystyle a={\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}}
{\displaystyle S}
{\displaystyle B}
{\displaystyle S=\iint _{\mathrm {source} }B(\Omega )\mathrm {d} \Omega =\iint _{\mathrm {source} }B(\theta ,\phi )\sin \theta \,d\theta \,d\varphi }
36 True or False, A calculation of energy is not possible unless a mass is involved.
43 In which of the following constellations does the ecliptic and the Galaxy or the galactic plane occur?
44 Which of the following may be the first astronomical source of the Earth?
the former protoplanetary disc
57 Which of the following are astronomical observatory phenomena associated with the Sun?
59 Phenomena associated with Kepler-36b?
has a gaseous surface
about 30% of its mass is iron
about 4.5 times the mass of Earth
has a rocky surface
discovered by the Kepler spacecraft
about 1.5 times as large as the Earth
62 Observations of Io have benefited greatly from what phenomenon?
66 Which of the following is not a studied characteristic of a planet?
an orbit around a star
a hydrostatic equilibrium (nearly round) shape
has cleared the neighbourhood around its orbit
an initially fractionated elemental composition
67 Which of the following is not a first astronomical source of rotation?
71 The Sun as a planet has what property?
When imaged in visible light Venus appears like a gas
77 What may be the first astronomical source referred to by the term "God"?
78 Which of the following is not an idea?
80 True or False, In the exosphere, temperature rises from around 1,500°C to 105 K.
83 The heliosphere might be the first astronomical source of?
virtually all the material emanating from the Sun itself
a termination shock
90 An X-ray source has been located at B1950 right ascension 01h 00m 06.91s and declination +62° 05' 45.5", which of the following constellations does this source occur in?
94 What may be the first astronomical source of the rocky-object Mercury?
100 Four scores in time is eighty
101 Which of the following are characteristic of a volume?
Essays such as for laboratories could be handled much like cases in Upper Limb Orthotics
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Convert zero-pole-gain filter parameters to second-order sections form - MATLAB zp2sos
Second-Order Sections from Zero-Pole-Gain Parameters
Convert zero-pole-gain filter parameters to second-order sections form
[sos,g] = zp2sos(z,p,k)
[sos,g] = zp2sos(z,p,k,order)
[sos,g] = zp2sos(z,p,k,order,scale)
[sos,g] = zp2sos(z,p,k,order,scale,zeroflag)
sos = zp2sos(___)
[sos,g] = zp2sos(z,p,k) finds a second-order section matrix sos with gain g that is equivalent to the transfer function H(z) whose n zeros, m poles, and scalar gain are specified in z, p, and k:
H\left(z\right)=k\frac{\left(z-{z}_{1}\right)\left(z-{z}_{2}\right)\cdots \left(z-{z}_{n}\right)}{\left(z-{p}_{1}\right)\left(z-{p}_{2}\right)\cdots \left(z-{p}_{m}\right)}.
[sos,g] = zp2sos(z,p,k,order) specifies the order of the rows in sos.
[sos,g] = zp2sos(z,p,k,order,scale) specifies the scaling of the gain and numerator coefficients of all second-order sections.
[sos,g] = zp2sos(z,p,k,order,scale,zeroflag) specifies the handling of real zeros that are negatives of each other.
sos = zp2sos(___) embeds the overall system gain in the first section.
Design a 5th-order Butterworth lowpass filter using the function butter with output expressed in zero-pole-gain form. Specify the cutoff frequency to be one-fifth of the Nyquist frequency. Convert the result to second-order sections. Visualize the magnitude response.
sos = zp2sos(z,p,k)
order — Row order
Row order, specified as one of the following:
'up' — Order the sections so the first row of sos contains the poles farthest from the unit circle.
'down' — Order the sections so the first row of sos contains the poles closest to the unit circle.
scale — Scaling of gain and numerator coefficients
'none' (default) | 'inf' | 'two'
Scaling of gain and numerator coefficients, specified as one of the following:
'none' — Apply no scaling.
'inf' — Apply infinity-norm scaling.
'two' — Apply 2-norm scaling.
Using infinity-norm scaling with 'up'-ordering minimizes the probability of overflow in the realization. Using 2-norm scaling with 'down'-ordering minimizes the peak round-off noise.
Infinity-norm and 2-norm scaling are appropriate only for direct-form II implementations.
zeroflag — Ordering of real zeros
Ordering of real zeros that are negatives of each other, specified as a logical scalar.
If you specify zeroflag as false, the function orders those zeros according to proximity to poles.
If you specify zeroflag as true, the function keeps those zeros together. This option results in a numerator with a middle coefficient equal to zero.
Second-order section representation, returned as a matrix. sos is an L-by-6 matrix
\text{sos}=\left[\begin{array}{cccccc}{b}_{01}& {b}_{11}& {b}_{21}& 1& {a}_{11}& {a}_{21}\\ {b}_{02}& {b}_{12}& {b}_{22}& 1& {a}_{12}& {a}_{22}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ {b}_{0L}& {b}_{1L}& {b}_{2L}& 1& {a}_{1L}& {a}_{2L}\end{array}\right]
H\left(z\right)=g\prod _{k=1}^{L}{H}_{k}\left(z\right)=g\prod _{k=1}^{L}\frac{{b}_{0k}+{b}_{1k}{z}^{-1}+{b}_{2k}{z}^{-2}}{1+{a}_{1k}{z}^{-1}+{a}_{2k}{z}^{-2}}.
If the transfer function has n zeros and m poles, then L is the closest integer greater than or equal to max(n/2,m/2).
Overall system gain, returned as a real scalar.
If you call zp2sos with one output argument, the function embeds the overall system gain in the first section, H1(z), so that
H\left(z\right)=\prod _{k=1}^{L}{H}_{k}\left(z\right).
Embedding the gain in the first section when scaling a direct-form II structure is not recommended and can result in erratic scaling. To avoid embedding the gain, use zp2sos with two outputs.
zp2sos uses a four-step algorithm to determine the second-order section representation for an input zero-pole-gain system:
It groups the zeros and poles into complex conjugate pairs using the cplxpair function.
It forms the second-order section by matching the pole and zero pairs according to the following rules:
Match the poles closest to the unit circle with the zeros closest to those poles.
Match the poles next closest to the unit circle with the zeros closest to those poles.
Continue until all of the poles and zeros are matched.
zp2sos groups real poles into sections with the real poles closest to them in absolute value. The same rule holds for real zeros.
It orders the sections according to the proximity of the pole pairs to the unit circle. zp2sos normally orders the sections with poles closest to the unit circle last in the cascade. You can tell zp2sos to order the sections in the reverse order using the order argument.
zp2sos scales the sections by the norm specified in scale. For arbitrary H(ω), the scaling is defined by
{‖H‖}_{p}={\left[\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}{|H\left(\omega \right)|}^{p}d\omega \right]}^{1/p}
where p can be either infinity or 2. This scaling is an attempt to minimize overflow or peak round-off noise in fixed-point filter implementations.
[2] Mitra, Sanjit Kumar. Digital Signal Processing: A Computer-Based Approach. 3rd ed. New York: McGraw-Hill Higher Education, 2006.
[3] Vaidyanathan, P. P. "Robust Digital Filter Structures." Handbook for Digital Signal Processing (S. K. Mitra and J. F. Kaiser, eds.). New York: John Wiley & Sons, 1993.
Any character or string input must be a constant at compile time.
cplxpair | filternorm | sos2zp | ss2sos | tf2sos | zp2ss | zp2tf
|
Diffusion-rate model component - MATLAB - MathWorks India
Create a diffusion Object
Diffusion-rate model component
The diffusion object specifies the diffusion-rate component of continuous-time stochastic differential equations (SDEs).
The diffusion-rate specification supports the simulation of sample paths of NVars state variables driven by NBrowns Brownian motion sources of risk over NPeriods consecutive observation periods, approximating continuous-time stochastic processes.
The diffusion-rate specification can be any NVars-by-NBrowns matrix-valued function G of the general form:
G\left(t,{X}_{t}\right)=D\left(t,{X}_{t}^{\alpha \left(t\right)}\right)V\left(t\right)
And a diffusion-rate specification is associated with a vector-valued SDE of the form:
d{X}_{t}=F\left(t,{X}_{t}\right)dt+G\left(t,{X}_{t}\right)d{W}_{t}
D is an NVars-by-NVars diagonal matrix, in which each element along the main diagonal is the corresponding element of the state vector raised to the corresponding power of α.
The diffusion-rate specification is flexible, and provides direct parametric support for static volatilities and state vector exponents. It is also extensible, and provides indirect support for dynamic/nonlinear models via an interface. This enables you to specify virtually any diffusion-rate specification.
DiffusionRate = diffusion(Alpha,Sigma)
DiffusionRate = diffusion(Alpha,Sigma) creates default DiffusionRate model component.
The diffusion object that you create encapsulates the composite drift-rate specification and returns the following displayed parameters:
Rate — The diffusion-rate function, G. Rate is the diffusion-rate calculation engine. It accepts the current time t and an NVars-by-1 state vector Xt as inputs, and returns an NVars-by-1 diffusion-rate vector.
Alpha — Access function for the input argument Alpha.
Sigma — Access function for the input argument Sigma.
Alpha — Return represents the parameter D
If you specify Sigma as an array, it must be an NVars-by-NBrowns two-dimensional matrix of instantaneous volatility rates. In this case, each row of Sigma corresponds to a particular state variable. Each column corresponds to a particular Brownian source of uncertainty, and associates the magnitude of the exposure of state variables with sources of uncertainty.
Although diffusion enforces no restrictions on the signs of these volatility parameters, each parameter is specified as a positive value.
Rate — Composite diffusion-rate function
value stored from diffusion-rate function (default) | function accessible by (t, Xt)
Composite diffusion-rate function, specified as: G(t,Xt)). The function stored in Rate fully encapsulates the combined effect of Alpha and Sigma where:
Alpha is the state vector exponent, which determines the format of D(t,Xt) of G(t,Xt).
Sigma is the volatility rate, V(t,Xt), of G(t,Xt).
Create a diffusion-rate function G:
The diffusion object displays like a MATLAB® structure and contains supplemental information, namely, the object's class and a brief description. However, in contrast to the SDE representation, a summary of the dimensionality of the model does not appear, because the diffusion class creates a model component rather than a model. G does not contain enough information to characterize the dimensionality of a problem.
When you specify the input arguments Alpha and Sigma as MATLAB arrays, they are associated with a specific parametric form. By contrast, when you specify either Alpha or Sigma as a function, you can customize virtually any diffusion-rate specification.
Accessing the output diffusion-rate parameters Alpha and Sigma with no inputs simply returns the original input specification. Thus, when you invoke diffusion-rate parameters with no inputs, they behave like simple properties and allow you to test the data type (double vs. function, or equivalently, static vs. dynamic) of the original input specification. This is useful for validating and designing methods.
When you invoke diffusion-rate parameters with inputs, they behave like functions, giving the impression of dynamic behavior. The parameters Alpha and Sigma accept the observation time t and a state vector Xt, and return an array of appropriate dimension. Specifically, parameters Alpha and Sigma evaluate the corresponding diffusion-rate component. Even if you originally specified an input as an array, diffusion treats it as a static function of time and state, by that means guaranteeing that all parameters are accessible by the same interface.
drift | sdeddo
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The interesting and non-obvious bit is that[[WikiPedia:Nyquist%E2%80%93Shannon_sampling_theorem|there's only one
A signal that differs even minutely from the original includes frequency content at or beyond Nyquist, breaks the bandlimiting requirement and isn't a valid solution.
Before you say, "I can draw a different signal that passes through those points", if it differs even
minutely from the original, it includes frequency content at or beyond Nyquist, breaks the bandlimiting requirement and isn't a valid solution.
{\displaystyle \ squarewave(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}\ squarewave(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
|
a
b,
c
\frac{\left(\frac{a}{b}\right)}{c} = \frac{a}{\left(\frac{b}{c}\right)}.
a
b,
c
\color{#D61F06} {\textbf{false}}
\frac{\left(\frac{a}{b}\right)}{c} \neq \frac{a}{\left(\frac{b}{c}\right)}
\frac{\left(\frac{36}{6}\right)}{2} = \frac{36}{\left(\frac{6}{2}\right)}
\frac{\left(\frac{36}{6}\right)}{2} = \frac{6}{2} = 3.
\frac{36}{\left(\frac{6}{2}\right)} = \frac{36}{3} =12.
3 \neq 12
_\square
\frac{\left(\frac{a}{b}\right)}{c} = \frac{a}{\left(\frac{b}{c}\right)}
a
b,
c
\frac{\left(\frac{a}{b}\right)}{c} = \frac{a}{\left(\frac{b}{c}\right)}
\begin{aligned} \frac{a}{b} \times \frac{b}{c} &= a \times c\\ \frac{a \times b}{b \times c} &= a \times c\\ \frac{a}{c} &= a \times c\\ \frac{1}{c} &= c. \end{aligned}
\frac1c = c
c=1
c=-1
c = 1 \text{ or } -1
c
\frac{\left(\frac{36}{6}\right)}{-1} = \frac{36}{\left(\frac{6}{-1}\right)}
\frac{\left(\frac{36}{6}\right)}{-1} = \frac{6}{-1} = -6.
\frac{36}{\left(\frac{6}{-1}\right)} = \frac{36}{-6} =-6.
(a,b,c)=(36,6,-1)
\square
36 \div 6 \div 3 =\, ?
36 \div 6 \div 3 = (36 \div 6) \div 3 = 6 \div 3 = 2.
\frac{\left(\frac{a}{b}\right)}{c}
\frac{a}{\left(\frac{b}{c}\right)}.
a
b,
c
and we want to know what's left of them at the end of evaluating these fractions in each of the two given situations
\frac{\left(\frac{a}{b}\right)}{c}\ \text{ and }\ \frac{a}{\left(\frac{b}{c}\right)}.
\frac{\left(\frac{a}{b}\right)}{c}
b
a
a
comes and attacks
as well, cutting our number down even further.
\frac{a}{\left(\frac{b}{c}\right)}
c
b
a
c,
a
c
a
c
A
B
\large A = \frac{1}{\left(\frac{0.5}{0.25}\right)} \hspace{5mm} \text{or} \hspace{5mm} B = \frac{\left(\frac{1}{0.5}\right)}{0.25}
|
Piecewise or exponential diode - MATLAB - MathWorks í•œêµ
\begin{array}{l}I=ISâ
\left({e}^{\frac{qV}{Nk{T}_{m1}}}â1\right)\text{}\text{}\text{}V>âBV\\ I=âISâ
\left({e}^{\frac{âq\left(V+Vz\right)}{k{T}_{m1}}}â{e}^{\frac{qV}{Nk{T}_{m1}}}\right)\text{}\text{}Vâ¤âBV\end{array}
{e}^{\frac{qV}{Nk{T}_{m1}}}
{e}^{\frac{qV}{Nk{T}_{m1}}}
\text{N}=\left(\left({V}_{1}â{V}_{2}\right)/{V}_{t}\right)/\left(\mathrm{log}\left({I}_{1}\right)â\mathrm{log}\left({I}_{2}\right)\right)
\text{IS}=\left({I}_{1}/\left(\mathrm{exp}\left({V}_{1}/\left(\text{N}{V}_{t}\right)\right)â1\right)+{I}_{2}/\left(\mathrm{exp}\left({V}_{2}/\left(\text{N}{V}_{t}\right)\right)â1\right)\right)/2
N={V}_{1}/\left({V}_{t}\mathrm{log}\left(\frac{{I}_{1}}{IS}+1\right)\right)
IS={I}_{1}/\left(\mathrm{exp}\left({V}_{1}/\left(N{V}_{t}\right)â1\right)\right)
CJ0={C}_{1}{\left(\left({V}_{R2}â{V}_{R1}\right)/\left({V}_{R2}â{V}_{R1}{\left({C}_{2}/{C}_{1}\right)}^{â1/M}\right)\right)}^{M}
VJ=â\left(â{V}_{R2}{\left({C}_{1}/{C}_{2}\right)}^{â1/M}+{V}_{R1}\right)/\left(1â{\left({C}_{1}/{C}_{2}\right)}^{â1/M}\right)
M=\mathrm{log}\left({C}_{3}/{C}_{2}\right)/\mathrm{log}\left({V}_{R2}/{V}_{R3}\right)
{Q}_{j}=CJ0â
\left(VJ/\left(Mâ1\right)\right)â
\left({\left(1âV/VJ\right)}^{1âM}â1\right)
{Q}_{j}=CJ0â
{F}_{1}+\left(CJ0/{F}_{2}\right)â
\left({F}_{3}â
\left(VâFCâ
VJ\right)+0.5\left(M/VJ\right)â
\left({V}^{2}â{\left(FCâ
VJ\right)}^{2}\right)\right)
{F}_{1}=\left(VJ/\left(1âM\right)\right)â
\left(1â{\left(1âFC\right)}^{1âM}\right)\right)
{F}_{2}={\left(1âFC\right)}^{1+M}\right)\right)
{F}_{3}=1âFCâ
\left(1+M\right)
i=\frac{{q}_{E}â{q}_{M}}{{T}_{M}}
\frac{d{q}_{M}}{dt}+\frac{{q}_{M}}{\mathrm{Ï}}â\frac{{q}_{E}â{q}_{M}}{{T}_{M}}=0
{q}_{E}=\left(\mathrm{Ï}+{T}_{M}\right)i
Ï„ is the carrier lifetime.
The block calculates transit time TM and carrier lifetime Ï„ based on the values you enter for the Charge Dynamics parameters. The block uses TM and Ï„ to solve the charge dynamics equations 1, 2, and 3.
{i}_{F}+at=\frac{{q}_{E}â{q}_{M}}{{T}_{M}}
\frac{d{q}_{M}}{dt}+\frac{{q}_{M}}{\mathrm{Ï}}={i}_{F}+at
{q}_{M}={i}_{F}\mathrm{Ï}âa{\mathrm{Ï}}^{2}+\frac{k}{\mathrm{exp}\left(\frac{t}{\mathrm{Ï}}\right)}+a\mathrm{Ï}t,
When t is zero, i = iF and qM = Ï„iF because the system is in steady state.
Substituting these relationships into equation 6 and solving the equation gives k = aτ2.
{q}_{M}={i}_{F}\mathrm{Ï}+a{\mathrm{Ï}}^{2}\left(\frac{1}{\mathrm{exp}\left(\frac{t}{\mathrm{Ï}}\right)}â1\right)+a\mathrm{Ï}t.
{i}_{RM}=\frac{â{q}_{M}}{{T}_{M}}
â{T}_{M}{i}_{RM}={i}_{F}\mathrm{Ï}+a{\mathrm{Ï}}^{2}\left(\frac{1}{\mathrm{exp}\left(\frac{{t}_{s}}{\mathrm{Ï}}\right)}â1\right)+a\mathrm{Ï}{t}_{s}
{t}_{s}=\frac{{i}_{RM}â{i}_{F}}{a}
i={i}_{RM}\text{exp[}\frac{â\left(tâ{t}_{s}\right)}{{\mathrm{Ï}}_{rr}}\right],
\frac{1}{{\mathrm{Ï}}_{rr}}=\frac{1}{\mathrm{Ï}}+\frac{1}{{T}_{M}}.
t=\frac{{i}_{RM}}{a}+{t}_{rr},
\frac{{i}_{RM}}{10}.
\mathrm{exp}\left(â\frac{tâ{t}_{s}}{{\mathrm{Ï}}_{rr}}\right)=0.1
{t}_{rr}={\mathrm{Ï}}_{rr}\mathrm{log}\left(10\right)+\frac{{i}_{RM}}{a}.
The block uses equations 9 and 14 to calculate values for TM and Ï„. The calculation uses an iterative scheme because of the exponential term in Equation 9.
Reverse recovery time stretch factor λ
The relationship between reverse recovery time stretch factor λ and trr is expressed by the equation
\mathrm{λ}=\frac{{t}_{rr}a}{{i}_{RM}}.
\frac{{i}_{RM}}{a}
3\left(\frac{{i}_{RM}}{a}\right).
Therefore, a typical value for λ is 3. λ must be greater than 1.
{Q}_{s}=\frac{1}{2}\left(â{i}_{RM}\right)\frac{{i}_{RM}}{a}.
{\mathrm{Ï}}_{rr}{i}_{RM}.
{Q}_{rr}=â\frac{{i}_{RM}^{2}}{2a}+{\mathrm{Ï}}_{rr}{i}_{RM}.
Rearranging equation 16 to solve for Ï„rr and substituting the result into equation 14 gives an equation that expresses trr in terms of Qrr:
{t}_{rr}=\left(\frac{{Q}_{rr}}{{i}_{RM}}+\frac{{i}_{RM}}{2a}\right)\mathrm{log}\left(10\right)+\frac{{i}_{RM}}{a}.
Alternatively, the block calculates Ï„rr by using the reverse recovery energy, Erec. This equation defines the diode voltage curve:
{v}_{d}={v}_{R}â\left({v}_{R}â{v}_{RM}\right)\mathrm{exp}\left(â\left(tâ{t}_{1}\right)/{\mathrm{Ï}}_{rr}\right),
{v}_{RM}={v}_{R}â{i}_{RM}\left(Râ\frac{L}{{\mathrm{Ï}}_{rr}}\right)
\frac{L}{{\mathrm{Ï}}_{rr}}â«R
{v}_{RM}={v}_{R}+{i}_{RM}\frac{L}{{\mathrm{Ï}}_{rr}}.
{i}_{F}+didt\text{â}·\text{â}\text{t}\text{â}\text{=}\text{â}\frac{{q}_{E}â{q}_{M}}{{T}_{M}}.
\frac{d{q}_{M}}{d}+\frac{{q}_{M}}{\mathrm{Ï}}={i}_{F}+didt\text{â}·\text{â}t.
{q}_{M}={i}_{F}\mathrm{Ï}+didt\text{â}·\text{â}{\mathrm{Ï}}^{2}\left(\frac{1}{\mathrm{exp}\left(\frac{t}{\mathrm{Ï}}\right)}â1\right)+didt\text{â}·\text{â}\mathrm{Ï}t,
t={t}_{1}=\frac{{i}_{RM}â{i}_{F}}{didt}
{i}_{RM}=\frac{â{q}_{M}}{{T}_{M}}.
\begin{array}{l}â{i}_{RM}{T}_{M}={i}_{F}\mathrm{Ï}+\text{â}didt\text{â}·\text{â}{\mathrm{Ï}}^{2}\left(\mathrm{exp}\left(â\frac{{i}_{RM}â{i}_{F}}{didt\text{â}·\text{â}\mathrm{Ï}}\right)â1\right)+\mathrm{Ï}\left({i}_{RM}â{i}_{F}\right)\\ â\left({T}_{M}+\mathrm{Ï}\right){i}_{RM}=didt\text{â}·\text{â}{\mathrm{Ï}}^{2}\left(\mathrm{exp}\left(â\frac{{i}_{RM}â{i}_{F}}{didt\text{â}·\text{â}\mathrm{Ï}}\right)â1\right)\\ {i}_{RM}=F\left(didt,{i}_{F},{T}_{M},\mathrm{Ï}\right).\end{array}
\begin{array}{l}{E}_{rec}={â«}_{{t}_{1}}^{{t}_{2}}{i}_{d}{v}_{d}dt\\ {E}_{rec}=\frac{{\mathrm{Ï}}_{rr}{i}_{RM}\left({v}_{R}+{v}_{RM}\right)}{2}\\ {E}_{rec}=\frac{{i}_{RM}\left(2{v}_{R}{\mathrm{Ï}}_{rr}+{i}_{RM}L\right)}{2}.\end{array}
{q}_{E}=\left(\mathrm{Ï}+{T}_{M}\right){i}_{diode},
\begin{array}{l}\frac{d{q}_{M}}{dt}+\frac{{q}_{M}}{\mathrm{Ï}}â\frac{{q}_{E}â{q}_{M}}{{T}_{M}}=0\\ i=\frac{{q}_{E}â{q}_{M}}{{T}_{M}}{Q}_{scale}â\left({Q}_{scale}â1\right){i}_{diode}\end{array}
I{S}_{Ts}=I{S}_{Tm1}â
{\left({T}_{s}/{T}_{m1}\right)}^{XTI/N}â
\mathrm{exp}\left(â\frac{EG}{Nk{T}_{s}}\left(1â{T}_{s}/{T}_{m1}\right)\right)
BVTs = BVTm1 – TCV· (Ts – Tm1) (24)
Use peak reverse current and stretch factor — Model charge dynamics by providing values for peak reverse current iRM and stretch factor λ plus information on the initial forward current and rate of change of current used in the test circuit when measuring iRM and trr.
Use peak reverse current and reverse recovery time — Model charge dynamics by providing values for peak reverse current iRM and reverse recovery time trr plus information on the initial forward current and rate of change of current used in the test circuit when measuring iRM and trr. Use this option if the manufacturer datasheet does not provide values for transit time TT and carrier lifetime Ï„.
Use transit time and carrier lifetime — Model charge dynamics by providing values for transit time TT and carrier lifetime Ï„.
-750 A/μs (default) | negative scalar
1500 s*μA (default)
â\frac{{i}^{2}{}_{RM}}{2a},
[3] Lauritzen, P.O. and C.L. Ma. “A Simple Diode Model with Reverse Recovery.†IEEE® Transactions on Power Electronics. Vol. 6, No. 2, April 1991, pp. 188–191.
|
[[Image:Xiph_ep02_test.png|400px|right]]
”Hi, I'm Monty Montgomery from [http://www.redhat.com/ Red Hat] and [http://xiph.org/ Xiph.Org].
“Hi, I'm Monty Montgomery from [http://www.redhat.com/ Red Hat] and [http://xiph.org/ Xiph.Org].
”A few months ago, I wrote
“A few months ago, I wrote
[http://people.xiph.org/~xiphmont/demo/neil-young.html an article on digital audio and why 24bit/192kHz music downloads don't make sense].
In the article, I
[[WikiPedia:Digital-to-analog_converter|convert from digital back to analog]].
”Of everything in the entire article, '''that''' was the number one thing
“Of everything in the entire article, '''that''' was the number one thing
people wrote about. In fact, more than half the mail I got was questions and
comments about basic digital signal behavior. Since there's interest, let's
take a little time to play with some ''simple'' digital signals. “
take a little time to play with some ''simple'' digital signals. ”
==Veritas ex machina==
{\displaystyle \ x(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}x_{\mathrm {square} }(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
|
Van der Pol oscillator - Wikipedia
Phase portrait of the unforced Van der Pol oscillator, showing a limit cycle and the direction field
Evolution of the limit cycle in the phase plane. The limit cycle begins as circle and, with varying μ, become increasingly sharp. An example of a Relaxation oscillator.
In dynamics, the Van der Pol oscillator is a non-conservative oscillator with non-linear damping. It evolves in time according to the second-order differential equation:
{\displaystyle {d^{2}x \over dt^{2}}-\mu (1-x^{2}){dx \over dt}+x=0,}
where x is the position coordinate—which is a function of the time t, and μ is a scalar parameter indicating the nonlinearity and the strength of the damping.
2 Two-dimensional form
3 Results for the unforced oscillator
4 Hamiltonian for Van der Pol oscillator
5 Forced Van der Pol oscillator
The Van der Pol oscillator was originally proposed by the Dutch electrical engineer and physicist Balthasar van der Pol while he was working at Philips.[1] Van der Pol found stable oscillations,[2] which he subsequently called relaxation-oscillations[3] and are now known as a type of limit cycle in electrical circuits employing vacuum tubes. When these circuits were driven near the limit cycle, they become entrained, i.e. the driving signal pulls the current along with it. Van der Pol and his colleague, van der Mark, reported in the September 1927 issue of Nature that at certain drive frequencies an irregular noise was heard,[4] which was later found to be the result of deterministic chaos.[5]
The Van der Pol equation has a long history of being used in both the physical and biological sciences. For instance, in biology, Fitzhugh[6] and Nagumo[7] extended the equation in a planar field as a model for action potentials of neurons. The equation has also been utilised in seismology to model the two plates in a geological fault,[8] and in studies of phonation to model the right and left vocal fold oscillators.[9]
Two-dimensional form[edit]
Liénard's theorem can be used to prove that the system has a limit cycle. Applying the Liénard transformation
{\displaystyle y=x-x^{3}/3-{\dot {x}}/\mu }
, where the dot indicates the time derivative, the Van der Pol oscillator can be written in its two-dimensional form:[10]
{\displaystyle {\dot {x}}=\mu \left(x-{\tfrac {1}{3}}x^{3}-y\right)}
{\displaystyle {\dot {y}}={\frac {1}{\mu }}x}
Another commonly used form based on the transformation
{\displaystyle y={\dot {x}}}
{\displaystyle {\dot {x}}=y}
{\displaystyle {\dot {y}}=\mu (1-x^{2})y-x}
Results for the unforced oscillator[edit]
Relaxation oscillation in the Van der Pol oscillator without external forcing. The nonlinear damping parameter is equal to μ = 5.
Two interesting regimes for the characteristics of the unforced oscillator are:[11]
When μ = 0, i.e. there is no damping function, the equation becomes:
{\displaystyle {d^{2}x \over dt^{2}}+x=0.}
This is a form of the simple harmonic oscillator, and there is always conservation of energy.
When μ > 0, the system will enter a limit cycle. Near the origin x = dx/dt = 0, the system is unstable, and far from the origin, the system is damped.
The Van der Pol oscillator does not have an exact, analytic solution.[12] However, such a solution does exist for the limit cycle if f(x) in the Lienard equation is a constant piece-wise function.
Hamiltonian for Van der Pol oscillator[edit]
Randomly chosen initial conditions are attracted to a stable orbit.
One can also write a time-independent Hamiltonian formalism for the Van der Pol oscillator by augmenting it to a four-dimensional autonomous dynamical system using an auxiliary second-order nonlinear differential equation as follows:
{\displaystyle {\ddot {x}}-\mu (1-x^{2}){\dot {x}}+x=0,}
{\displaystyle {\ddot {y}}+\mu (1-x^{2}){\dot {y}}+y=0.}
Note that the dynamics of the original Van der Pol oscillator is not affected due to the one-way coupling between the time-evolutions of x and y variables. A Hamiltonian H for this system of equations can be shown to be[13]
{\displaystyle H(x,y,p_{x},p_{y})=p_{x}p_{y}+xy-\mu (1-x^{2})yp_{y},}
{\displaystyle p_{x}={\dot {y}}+\mu (1-x^{2})y}
{\displaystyle p_{y}={\dot {x}}}
are the conjugate momenta corresponding to x and y, respectively. This may, in principle, lead to quantization of the Van der Pol oscillator. Such a Hamiltonian also connects[14] the geometric phase of the limit cycle system having time dependent parameters with the Hannay angle of the corresponding Hamiltonian system.
Forced Van der Pol oscillator[edit]
Chaotic behaviour in the Van der Pol oscillator with sinusoidal forcing. The nonlinear damping parameter is equal to μ = 8.53, while the forcing has amplitude A = 1.2 and angular frequency ω = 2π / 10.
The forced, or driven, Van der Pol oscillator takes the 'original' function and adds a driving function Asin(ωt) to give a differential equation of the form:
{\displaystyle {d^{2}x \over dt^{2}}-\mu (1-x^{2}){dx \over dt}+x-A\sin(\omega t)=0,}
where A is the amplitude, or displacement, of the wave function and ω is its angular velocity.
Electrical circuit involving a triode, resulting in a forced Van der Pol oscillator.[15] The circuit contains: a triode, a resistor R, a capacitor C, a coupled inductor-set with self inductance L and mutual inductance M. In the serial RLC circuit there is a current i, and towards the triode anode ("plate") a current ia, while there is a voltage ug on the triode control grid. The Van der Pol oscillator is forced by an AC voltage source Es.
Author James Gleick described a vacuum tube Van der Pol oscillator in his book from 1987 Chaos: Making a New Science.[16] According to a New York Times article,[17] Gleick received a modern electronic Van der Pol oscillator from a reader in 1988.
Mary Cartwright, British mathematician, one of the first to study the theory of deterministic chaos, particularly as applied to this oscillator.[18]
The quantum van der Pol oscillator, which is the quantum version of the classical van der Pol oscillator, has been proposed using a Lindblad equation to study its quantum dynamics and quantum synchronization.[19] Note the above Hamiltonian approach with an auxiliary second-order equation produces unbounded phase-space trajectories and hence cannot be used to quantize the van der Pol oscillator. In the limit of weak nonlinearity (i.e. μ→0) the van der Pol oscillator reduces to the Stuart–Landau equation. The Stuart–Landau equation in fact describes an entire class of limit-cycle oscillators in the weakly-nonlinear limit. The form of the classical Stuart–Landau equation is much simpler, and perhaps not surprisingly, can be quantized by a Lindblad equation which is also simpler than the Lindblad equation for the van der Pol oscillator. The quantum Stuart–Landau model has played an important role in the study of quantum synchronisation[20][21] (where it has often been called a van der Pol oscillator although it cannot be uniquely associated with the van der Pol oscillator). The relationship between the classical Stuart–Landau model (μ→0) and more general limit-cycle oscillators (arbitrary μ) has also been demonstrated numerically in the corresponding quantum models.[19]
^ Cartwright, M. L. (1960). "Balthazar Van Der Pol". Journal of the London Mathematical Society. Wiley. s1-35 (3): 367–376. doi:10.1112/jlms/s1-35.3.367. ISSN 0024-6107.
^ B. van der Pol: "A theory of the amplitude of free and forced triode vibrations", Radio Review (later Wireless World) 1 701–710 (1920)
^ van der Pol, Balth. (1926). "On "relaxation-oscillations"". The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. Informa UK Limited. 2 (11): 978–992. doi:10.1080/14786442608564127. ISSN 1941-5982.
^ VAN DER POL, BALTH; VAN DER MARK, J. (1927). "Frequency Demultiplication". Nature. Springer Science and Business Media LLC. 120 (3019): 363–364. doi:10.1038/120363a0. ISSN 0028-0836.
^ Kanamaru, T., "Van der Pol oscillator", Scholarpedia, 2(1), 2202, (2007).
^ FitzHugh, Richard (1961). "Impulses and Physiological States in Theoretical Models of Nerve Membrane". Biophysical Journal. Elsevier BV. 1 (6): 445–466. doi:10.1016/s0006-3495(61)86902-6. ISSN 0006-3495. PMC 1366333.
^ Nagumo, J.; Arimoto, S.; Yoshizawa, S. (1962). "An Active Pulse Transmission Line Simulating Nerve Axon". Proceedings of the IRE. Institute of Electrical and Electronics Engineers (IEEE). 50 (10): 2061–2070. doi:10.1109/jrproc.1962.288235. ISSN 0096-8390.
^ Cartwright, Julyan H. E.; Eguíluz, Víctor M.; Hernández-García, Emilio; Piro, Oreste (1999). "Dynamics of Elastic Excitable Media". International Journal of Bifurcation and Chaos. World Scientific Pub Co Pte Lt. 09 (11): 2197–2202. arXiv:chao-dyn/9905035. doi:10.1142/s0218127499001620. ISSN 0218-1274.
^ Lucero, Jorge C.; Schoentgen, Jean (2013). Modeling vocal fold asymmetries with coupled van der Pol oscillators. Proceedings of Meetings on Acoustics. Vol. 19. p. 060165. doi:10.1121/1.4798467. ISSN 1939-800X.
^ Kaplan, D. and Glass, L., Understanding Nonlinear Dynamics, Springer, 240–244, (1995).
^ Grimshaw, R., Nonlinear ordinary differential equations, CRC Press, 153–163, (1993), ISBN 0-8493-8607-1.
^ Panayotounakos, D.E.; Panayotounakou, N.D.; Vakakis, A.F. (2003-09-01). "On the lack of analytic solutions of the Van der Pol oscillator". ZAMM. Wiley. 83 (9): 611–615. doi:10.1002/zamm.200310040. ISSN 0044-2267.
^ Shah, Tirth; Chattopadhyay, Rohitashwa; Vaidya, Kedar; Chakraborty, Sagar (2015). "Conservative perturbation theory for nonconservative systems". Physical Review E. 92 (6): 062927. arXiv:1512.06758. Bibcode:2015PhRvE..92f2927S. doi:10.1103/physreve.92.062927. PMID 26764794. S2CID 14930486.
^ Chattopadhyay, Rohitashwa; Shah, Tirth; Chakraborty, Sagar (2018). "Finding the Hannay angle in dissipative oscillatory systems via conservative perturbation theory". Physical Review E. 97 (6): 062209. arXiv:1610.05218. doi:10.1103/PhysRevE.97.062209. PMID 30011548. S2CID 51635019.
^ K. Tomita (1986): "Periodically forced nonlinear oscillators". In: Chaos, Ed. Arun V. Holden. Manchester University Press, ISBN 0719018110, pp. 213–214.
^ Gleick, James (1987). Chaos: Making a New Science. New York: Penguin Books. pp. 41–43. ISBN 0-14-009250-1.
^ Colman, David (11 July 2011). "There's No Quiet Without Noise". New York Times. Retrieved 11 July 2011.
^ Cartwright, M. L.; Littlewood, J. E. (1945). "On Non-Linear Differential Equations of the Second Order: I. the Equation y¨ − k (1-y 2 )y˙ + y = b λk cos(λl + α), k Large". Journal of the London Mathematical Society. Wiley. s1-20 (3): 180–189. doi:10.1112/jlms/s1-20.3.180. ISSN 0024-6107.
^ a b Chia, A.; Kwek, L. C.; Noh, C. (2020-10-16). "Relaxation oscillations and frequency entrainment in quantum mechanics". Physical Review E. American Physical Society (APS). 102 (4): 042213. doi:10.1103/physreve.102.042213. ISSN 2470-0045.
^ Walter, Stefan; Nunnenkamp, Andreas; Bruder, Christoph (2014-03-06). "Quantum Synchronization of a Driven Self-Sustained Oscillator". Physical Review Letters. American Physical Society (APS). 112 (9): 094102. doi:10.1103/physrevlett.112.094102. ISSN 0031-9007.
^ Lee, Tony E.; Sadeghpour, H. R. (2013-12-04). "Quantum Synchronization of Quantum van der Pol Oscillators with Trapped Ions". Physical Review Letters. American Physical Society (APS). 111 (23): 234101. doi:10.1103/physrevlett.111.234101. ISSN 0031-9007.
"Van der Pol equation", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
Van der Pol oscillator on Scholarpedia
Van Der Pol Oscillator Interactive Demonstrations
Retrieved from "https://en.wikipedia.org/w/index.php?title=Van_der_Pol_oscillator&oldid=1085958541"
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[[Image:Dsat_011.png|360px|right]]
I keep saying that I'm quantizing with [[Wikipedia:dither|dither]], so what is dither
We've been quantizing with [[Wikipedia:dither|dither]]. What is dither
exactly and, more importantly, what does it do?
The simple way to quantize a signal is to choose the digital
The simplest way to quantize a signal is to choose the digital
amplitude value closest to the original analog amplitude. [[WikiPedia:Rounding|Obvious]],
amplitude value [[WikiPedia:Rounding|closest to the original analog amplitude]].
right? Unfortunately, the exact noise you get from this simple
Unfortunately, the exact noise that results from this simple
quantization scheme depends somewhat on the input signal,
quantization scheme depends somewhat on the input signal.
It may be inconsistent, cause distortion, or be
</div> </center>
quantization noise, it actually replaces it with noise characteristics
quantization noise, it replaces it with noise characteristics
of our choosing that aren't influenced by the input.
Let's ''watch'' what dither does. The signal generator has too much noise for this test so we'll produce a mathematically perfect sine wave with the ThinkPad and quantize it to 8 bits with dithering.
The signal generator has too much noise for this test so we produce a mathematically perfect sine wave with the ThinkPad and quantize it to 8 bits with dithering.
We see the sine wave on waveform display and output scope, and
We see a nice sine wave on the waveform display and output scope and, once the analog spectrum analyzer catches up...
a clean frequency peak with a uniform noise floor on both spectral displays
just like before. Again, this is with dither.
The quantization noise, that dither had spread out into a nice, flat noise
The quantization noise that dither had spread out into a nice, flat noise
floor, piles up into harmonic distortion peaks. The noise floor is
lower, but the level of distortion becomes nonzero, and the distortion
At 8 bits this effect is exaggerated. At 16 bits,
even without dither, harmonic distortion is going to be so low as to
be completely inaudible.
be completely inaudible. Still, we can use dither to eliminate it completely if we so choose.
Turning the dither off again for a moment, you'll notice that the
In a sense, everything quantizing to zero is just 100% distortion!
Dither eliminates this distortion too. We reenable dither
Dither eliminates this distortion too. When we reenable dither, we clearly see our signal at 1/4 bit with a nice flat noise floor.
and ... there's our signal back at 1/4 bit, with our nice flat noise floor.
The noise floor doesn't have to be flat. Dither is noise of our
choosing, so let's choose a noise as [http://www.acoustics.salford.ac.uk/res/cox/sound_quality/?content=subjective inoffensive] and
choosing, so it makes sense to choose a noise as [http://www.acoustics.salford.ac.uk/res/cox/sound_quality/?content=subjective inoffensive] and
[[WikiPedia:Absolute_threshold_of_hearing|difficult to notice]]
Our hearing is most sensitive in the midrange from 2kHz to 4kHz,
so that's where background noise is going to be the most obvious.
16-bit dithering noise is normally much too quiet to hear at all, but
let's listen to our noise shaping example, again with the gain
brought way up...
Lastly, dithered quantization noise ''is'' higher [[WikiPedia:Sound_power|power]] overall
than undithered quantization noise even when it sounds quieter, and
than undithered quantization noise, even though it often sounds quieter, and
you can see that on a [[WikiPedia:VU_meter|VU meter]] during passages of near-silence. But
you can see that on a [[WikiPedia:VU_meter|VU meter]] during passages of near-silence. However,
dither isn't only an on or off choice. We can reduce the dither's
power to balance less noise against a bit of distortion to minimize
We'll also [[WikiPedia:Amplitude_modulation|modulate the input signal]] like this to show how a varying input affects the quantization noise. At
For the next test, we also [[WikiPedia:Amplitude_modulation|modulate the input signal]] like this to show how a varying input affects the quantization noise. At
full dithering power, the noise is uniform, constant, and featureless
just like we expect:
just like we expect.
As we reduce the dither's power, the input increasingly
affects the amplitude and the character of the quantization noise.
Shaped dither behaves similarly, but noise shaping lends one more nice
advantage. To make a long story short, it can use a somewhat lower
advantage; it can use a somewhat lower
dither power before the input has as much effect on the output.
Despite all the time I just spent on dither, we're talking about
Despite all this text spent on dither, the differences exist 100 decibels or more below [[WikiPedia:Full_scale|full scale]]. If the CD had been
differences that start 100 decibels and more below [[WikiPedia:Full_scale|full scale]]. Maybe
if the CD had been
[http://www.research.philips.com/technologies/projects/cd/index.html 14 bits as originally designed],
dither ''might'' be
perhaps dither ''might'' be
more important. Maybe. At 16 bits, really, it's mostly a wash. You
more important. At 16 bits it's mostly a wash. It's reasonable to treat
can think of dither as an insurance policy that gives several extra
dither as an insurance policy that gives several extra
decibels of dynamic range just in case. The simple fact is, though, no
decibels of dynamic range just in case. That said no
one ever ruined a great recording by not dithering the final master.
{\displaystyle \ squarewave(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}\ squarewave(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
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Wavy Curve Method | Brilliant Math & Science Wiki
Ashish Menon, Andres Gonzalez, A Former Brilliant Member, and
Prem Chebrolu
The wavy curve method (also called the method of intervals) is a strategy used to solve inequalities of the form
\frac{f(x)}{g(x)} > 0
\left(<0, \, \geq 0, \, \text{or} \, \leq 0\right).
The method uses the fact that
\frac{f(x)}{g(x)}
can only change sign at its zeroes and vertical asymptotes, so we can use the roots of
f(x)
g(x)
to sketch a graph of the function over different intervals.
\frac{3x - x^2}{{(x + 4)}^2} \geq 0.
Factor the polynomials:
\begin{aligned} \dfrac{x(3 - x)}{{(x + 4)}^2} & \geq 0 \end{aligned}.
\begin{aligned} \dfrac{-x(x - 3)}{{(x + 4)}^2} & \geq 0 \end{aligned}.
Multiply/divide both sides of the inequality by -1 to remove the minus sign (remember that in doing so the inequality would reverse):
\begin{aligned} \dfrac{x(x - 3)}{{(x + 4)}^2} & \leq 0 \end{aligned}.
Find the roots and asymptotes of the inequality by equating each factor to 0:
\begin{aligned} x & = 0\\ x - 3 = 0 \implies x & = 3\\ x + 4 = 0 \implies x & = -4. \end{aligned}
Plot the points on the number line. Now, start with the largest factor, i.e. 3. Initially, a curve from the positive region of the number line should intersect that point (here 3). Now, look at the power of the respective factors. If it is odd, then we have to change the path of the curve from their respective roots. If it is even, continue in the same region. Here, the curve would change its path at 0 and 3 because their factors are odd powers. However, at 4, it would not change its direction since its factor has an even power.
\geq
\leq
x
x \in [0,3]. \ _\square
\{-5\}\cup\left(-1;0\right]\cup\left[3;4\right)\cup\left(4;6\right]
(-\infty,-1) \cup [0,2)\cup (2,3] \cup [6,\infty)
\mathbb R \setminus\{6\}
(-\infty,-5) \cup [2,5) \cup (3,7] \cup [8,\infty)
\dfrac{x{(x+5)}^{2016}{(x-3)}^{2017}{(6-x)}^{1231}}{{(x-2)}^{10000}{(x+1)}^{2015}{(4-x)}^{242}} \geq 0
x
\cup
[1 , 3)
(-\infty , 4]
(-2 , -1.5)
[4 , \infty)
\dfrac{(8x^3 + 36x^2 + 54x + 27){(x-1)}^2}{(x^3 + 6x^2 + 12x + 8){(x+1)}^2} < 0
x
Cite as: Wavy Curve Method. Brilliant.org. Retrieved from https://brilliant.org/wiki/wavy-curve-method/
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Hi, I'm Monty Montgomery from [http://www.redhat.com/ Red Hat] and [http://xiph.org/ Xiph.Org].
A few months ago, I wrote
Of everything in the entire article, '''that''' was the number one thing
take a little time to play with some ''simple'' digital signals.
{\displaystyle \ x(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}x_{\mathrm {square} }(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
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Instant centre of rotation - Calculations & Formula
Instant centre of rotation (5392 views - Calculations (Mech&Elec))
The instant centre of rotation, also called instantaneous velocity center, or also instantaneous centre or instant centre, is the point fixed to a body undergoing planar movement that has zero velocity at a particular instant of time. At this instant, the velocity vectors of the trajectories of other points in the body generate a circular field around this point which is identical to what is generated by a pure rotation. Planar movement of a body is often described using a plane figure moving in a two-dimensional plane. The instant centre is the point in the moving plane around which all other points are rotating at a specific instant of time. The continuous movement of a plane has an instant centre for every value of the time parameter. This generates a curve called the moving centrode. The points in the fixed plane corresponding to these instant centres form the fixed centrode. The generalization of this concept to 3-dimensional space is that of a twist around a screw. The screw has an axis which is a line in 3D space (not necessarily through the origin, and the screw also has a finite pitch (a fixed translation along its axis corresponding to a rotation about the screw axis.
Licensed under Creative Commons Attribution-Share Alike 3.0 (VanBuren).
The instant centre of rotation, also called instantaneous velocity center,[1] or also instantaneous centre or instant centre, is the point fixed to a body undergoing planar movement that has zero velocity at a particular instant of time. At this instant, the velocity vectors of the trajectories of other points in the body generate a circular field around this point which is identical to what is generated by a pure rotation.
Planar movement of a body is often described using a plane figure moving in a two-dimensional plane. The instant centre is the point in the moving plane around which all other points are rotating at a specific instant of time.
The continuous movement of a plane has an instant centre for every value of the time parameter. This generates a curve called the moving centrode. The points in the fixed plane corresponding to these instant centres form the fixed centrode.
The generalization of this concept to 3-dimensional space is that of a twist around a screw. The screw has an axis which is a line in 3D space (not necessarily through the origin, and the screw also has a finite pitch (a fixed translation along its axis corresponding to a rotation about the screw axis.
1 Pole of a planar displacement
1.1 Pure translation
2 Instant centre of a wheel rolling without slipping
3 Relative centre of rotation for two contacting planar bodies
4 Instant centre of rotation and mechanisms
Pole of a planar displacement
The instant centre can be considered the limiting case of the pole of a planar displacement.
The planar displacement of a body from position 1 to position 2 is defined by the combination of a planar rotation and planar translation. For any planar displacement there is a point in the moving body that is in the same place before and after the displacement. This point is the pole of the planar displacement, and the displacement can be viewed as a rotation around this pole.
Construction for the pole of a planar displacement: First, select two points A and B in the moving body and locate the corresponding points in the two positions; see the illustration. Construct the perpendicular bisectors to the two segments A1A2 and B1B2. The intersection P of these two bisectors is the pole of the planar displacement. Notice that A1 and A2 lie on a circle around P. This is true for the corresponding positions of every point in the body.
If the two positions of a body are separated by an instant of time in a planar movement, then the pole of a displacement becomes the instant centre. In this case, the segments constructed between the instantaneous positions of the points A and B become the velocity vectors VA and VB. The lines perpendicular to these velocity vectors intersect in the instant centre.
The algebraic construction of the Cartesian coordinates
{\displaystyle (P_{x},P_{y})}
can be arranged as follows: The midpoint between
{\displaystyle A^{1}}
{\displaystyle A^{2}}
has the Cartesian coordinates
{\displaystyle A_{x}^{m}={\frac {1}{2}}(A_{x}^{1}+A_{x}^{2});\quad A_{y}^{m}={\frac {1}{2}}(A_{y}^{1}+A_{y}^{2}),}
and the midpoint between
{\displaystyle B^{1}}
{\displaystyle B^{2}}
{\displaystyle B_{x}^{m}={\frac {1}{2}}(B_{x}^{1}+B_{x}^{2});\quad B_{y}^{m}={\frac {1}{2}}(B_{y}^{1}+B_{y}^{2}).}
The two angles from
{\displaystyle A^{1}}
{\displaystyle A^{2}}
{\displaystyle B^{1}}
{\displaystyle B^{2}}
measured counter-clockwise relative to the horizontal are determined by
{\displaystyle \tan \tau _{A}={\frac {A_{y}^{2}-A_{y}^{1}}{A_{x}^{2}-A_{x}^{1}}},\quad \tan \tau _{B}={\frac {B_{y}^{2}-B_{y}^{1}}{B_{x}^{2}-B_{x}^{1}}},}
taking the correct branches of the tangent. Let the center
{\displaystyle (P_{x},P_{y})}
of the rotation have distances
{\displaystyle d_{A}}
{\displaystyle d_{B}}
to the two midpoints. Assuming clockwise rotation (otherwise switch the sign of
{\displaystyle \pi /2}
{\displaystyle {\begin{aligned}P_{x}&=A_{x}^{m}+d_{A}\cos(\tau _{A}-\pi /2);\\P_{x}&=B_{x}^{m}+d_{B}\cos(\tau _{B}-\pi /2);\\P_{y}&=A_{y}^{m}+d_{A}\sin(\tau _{A}-\pi /2);\\P_{y}&=B_{y}^{m}+d_{B}\sin(\tau _{B}-\pi /2).\end{aligned}}}
Rewrite this as a
{\displaystyle 4\times 4}
inhomogeneous system of linear equations with 4 unknowns (the two distances
{\displaystyle d}
and the two coordinates
{\displaystyle P}
of the center):
{\displaystyle \left({\begin{array}{cccc}1&0&-\sin \tau _{A}&0\\1&0&0&-\sin \tau _{B}\\0&1&\cos \tau _{A}&0\\0&1&0&\cos \tau _{B}\end{array}}\right)\cdot \left({\begin{array}{c}P_{x}\\P_{y}\\d_{A}\\d_{B}\end{array}}\right)=\left({\begin{array}{c}A_{x}^{m}\\B_{x}^{m}\\A_{y}^{m}\\B_{y}^{m}\end{array}}\right).}
The coordinates of the center of the rotation are the first two components of the solution vector
{\displaystyle \left({\begin{array}{c}P_{x}\\P_{y}\\d_{A}\\d_{B}\end{array}}\right)={\frac {1}{\sin(\tau _{A}-\tau _{B})}}\left({\begin{array}{c}\sin \tau _{A}\sin \tau _{B}(B_{y}^{m}-A_{y}^{m})+B_{x}^{m}\cos \tau _{B}\sin \tau _{A}-A_{x}^{m}\sin \tau _{B}\cos \tau _{A}\\\cos \tau _{A}\cos \tau _{B}(A_{x}^{m}-B_{x}^{m})+A_{y}^{m}\cos \tau _{B}\sin \tau _{A}-B_{y}^{m}\sin \tau _{B}\cos \tau _{A}\\(B_{x}^{m}-A_{x}^{m})\cos \tau _{B}+(B_{y}^{m}-A_{y}^{m})\sin \tau _{B}\\(B_{x}^{m}-A_{x}^{m})\cos \tau _{A}+(B_{y}^{m}-A_{y}^{m})\sin \tau _{A}\\\end{array}}\right).}
If the displacement between two positions is a pure translation, then the perpendicular bisectors of the segments A1B1 and A2B2 form parallel lines. These lines are considered to intersect at a point on the line at infinity, thus the pole of this planar displacement is said to "lie at infinity" in the direction of the perpendicular bisectors.
In the limit, pure translation becomes planar movement with point velocity vectors that are parallel. In this case, the instant centre is said to lie at infinity in the direction perpendicular to the velocity vectors.
Instant centre of a wheel rolling without slipping
Consider the planar movement of a circular wheel rolling without slipping on a linear road; see sketch 3. The wheel rotates around its axis M, which translates in a direction parallel to the road. The point of contact P of the wheel with road does not slip, which means the point P has zero velocity with respect to the road. Thus, at the instant the point P on the wheel comes in contact with the road it becomes an instant centre.
The set of points of the moving wheel that become instant centres is the circle itself, which defines the moving centrode. The points in the fixed plane that correspond to these instant centres is the line of the road, which defines the fixed centrode.
The velocity vector of a point A in the wheel is perpendicular to the segment AP and is proportional to the length of this segment. In particular, the velocities of points in the wheel are determined by the angular velocity of the wheel in rotation around P. The velocity vectors of a number of points are illustrated in sketch 3.
The further a point in the wheel is from the instant centre P, the proportionally larger its speed. Therefore, the point at the top of the wheel moves in the same direction as the centre M of the wheel, but twice as fast, since it is twice the distance away from P. All points that are a distance equal to the radius of the wheel 'r' from point P move at the same speed as the point M but in different directions. This is shown for a point on the wheel that has the same speed as M but moves in the direction tangent to the circle around P.
Relative centre of rotation for two contacting planar bodies
If two planar rigid bodies are in contact, and each body has its own distinct centre of rotation, then the relative centre of rotation between the bodies has to lie somewhere on the line connecting the two centres. As a result, since pure rolling can only exist when the centre of rotation is at the point of contact (as seen above with the wheel on the road), it is only when the point of contact goes through the line connecting the two rotation centres that pure rolling can be achieved. This is known in involute gear design as the pitch point, where there is no relative sliding between the gears. In fact, the gearing ratio between the two rotating parts is found by the ratio of the two distances to the relative centre. In the example in Sketch 4 the gearing ratio is
{\displaystyle \gamma ={\frac {AD}{BD}}}
Instant centre of rotation and mechanisms
Sketch 1 above shows a four-bar linkage where a number of instant centres of rotation are illustrated. The rigid body noted by the letters BAC is connected with links P1-A and P2-B to a base or frame.
The three moving parts of this mechanism (the base is not moving) are: link P1-A, link P2-B, and body BAC. For each of these three parts an instant centre of rotation may be determined.
Considering first link P1-A: all points on this link, including point A, rotate around point P1. Since P1 is the only point not moving in the given plane it may be called the instant centre of rotation for this link. Point A, at distance P1-A from P1, moves in a circular motion in a direction perpendicular to the link P1-A, as indicated by vector VA.
The same applies to link P2-B: point P2 is the instant centre of rotation for this link and point B moves in the direction as indicated by vector VB.
For determining the instant centre of rotation of the third element of the linkage, the body BAC, the two points A and B are used because its moving characteristics are known, as derived from the information about the links P1-A and P2-B.
The direction of speed of point A is indicated by vector VA. Its instant centre of rotation must be perpendicular to this vector (as VA is tangentially located on the circumference of a circle). The only line that fills the requirement is a line colinear with link P1-A. Somewhere on this line there is a point P, the instant centre of rotation for the body BAC.
What applies to point A also applies to point B, therefore this instant centre of rotation P is located on a line perpendicular to vector VB, a line colinear with link P2-B. Therefore, the instant centre of rotation P of body BAC is the point where the lines through P1-A and P2-B cross.
Since this instant centre of rotation P is the centre for all points on the body BAC for any random point, say point C, the speed and direction of movement may be determined: connect P to C. The direction of movement of point C is perpendicular to this connection. The speed is proportional to the distance to point P.
Continuing this approach with the two links P1-A and P2-B rotating around their own instant centres of rotation the centrode for instant centre of rotation P may be determined. From this the path of movement for C or any other point on body BAC may be determined.
In biomechanical research the instant centre of rotation is observed for the functioning of the joints in the upper and lower extremities.[2] For example, in analysing the knee,[3][4][5] ankle,[6] or shoulder joints.[7][8] Such knowledge assists in developing artificial joints and prosthesis, such as elbow [9] or finger joints.[10]
Study of the joints of horses: "...velocity vectors determined from the instant centres of rotation indicated that the joint surfaces slide on each other.".[11]
Studies on turning a vessel moving through water.[12]
The braking characteristics of a car may be improved by varying the design of a brake pedal mechanism.[13]
Designing the suspension of a bicycle,[14] or of a car.[15]
In the case of the coupler link in a four-bar linkage, such as a double wishbone suspension in front view, the perpendiculars to the velocity lie along the links joining the grounded link to the coupler link. This construction is used to establish the kinematic Roll centre of the suspension.
^ Illustrated Dictionary of Mechanical Engineering: English, German, French, Dutch, Russian (Springer Science & Business Media, 17 Apr. 2013 - 422 pages)
^ "Muscle Physiology — Joint Moment Arm".
^ Knee joint motion description and measurement
^ Moorehead JD, Montgomery SC, Harvey DM (Sep 2003). "Instant centre of rotation estimation using the Reuleaux technique and a Lateral Extrapolation technique". J Biomech. 36 (9): 1301–7. PMID 12893038. doi:10.1016/S0021-9290(03)00156-8.
^ Hollman JH, Deusinger RH, Van Dillen LR, Matava MJ (Aug 2003). "Gender differences in surface rolling and gliding kinematics of the knee". Clin Orthop Relat Res. 413 (413): 208–21. PMID 12897612. doi:10.1097/01.blo.0000072902.36018.fe.
^ Maganaris CN, Baltzopoulos V, Sargeant AJ (Aug 1998). "Changes in Achilles tendon moment arm from rest to maximum isometric plantarflexion: in vivo observations in man". J Physiology. 510 (Pt 3): 977–85. PMC 2231068 . PMID 9660906. doi:10.1111/j.1469-7793.1998.977bj.x. [dead link]
^ Biomechanics of shoulder
^ Poppen NK, Walker PS (Mar 1976). "Normal and abnormal motion of the shoulder". J Bone Joint Surg Am. 58 (2): 195–201. PMID 1254624.
^ US 5030237 Elbow prosthesis
^ Pyrocarbon Finger Joint Implant
^ Colahan P, Piotrowski G, Poulos P (Sep 1988). "Kinematic analysis of the instant centres of rotation of the equine metacarpophalangeal joint". Am J Vet Res. 49 (9): 1560–5. PMID 3223666.
^ PART VI Vessel Navigation and Manoeuvering
^ GB 1443270 Variable Mechanical Ratio Brake Pedal Mounts - General Motors, 1976
^ US 7100930 Bicycle rear suspension system
^ Reza N. Jazar (2008). Vehicle Dynamics: Theory and Application. Berlin: Springer. ISBN 0-387-74243-3.
Rotor (electric)Axle
This article uses material from the Wikipedia article "Instant centre of rotation", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia
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Binance DEX Trading - Binance Chain Docs
Binance Chain Docs Binance DEX Trading
Binance DEX Trading Binance DEX Trading Table of contents
Trading Pairs Info
Trading View and Market Depth
Trade history and Orderbook
Your Trading Activities
Binance DEX Trading Interface
Binance DEX trading page is designed to provide a user-friendly trading interface to everyone. To get started, go to the trading page first at https://www.binance.org/trade. The trading page is composed of the following part:
Tradig Pairs info
Trading Video on Youtube
At the left-right part of the trading page, there is the section of “Trading Pairs”. The trading pairs are categorized into four markets: - BNB - BTC - ALT - FIATS
Click on tabs to switch between markets. Start Icon will show your favorite trading pairs. You can click at the star icon to add this trading pair to your list.
Trading pairs can be ordered by two “Volume” or “Price Change”. The list is in descending order by default. You can also change to ascending by clicking the header.
At the center of the trading page, there is the TradingView chart. The TradingView displays information about your selected trading pair. The chart allows you to display indicators such as the RSI, moving average, Bollinger Bands and many more. The chart also allows you to draw on the chart using the tools within the TradingView charts. More information on how to use TradingView charts can be found on the TradingView Wiki. Click on Depth button to switch to Depth View. Market depth is the market's ability to sustain relatively large market orders without impacting the price of the security. Market depth considers the overall level and breadth of open orders and usually refers to trading within individual trading pairs. The number of levels can be zoomed in and out. The key indicator of market depth is the spread. The spread is the gap between the bid and the ask prices. If the bid price for a stock is $19 and the ask price for the same stock is $20, then the spread for the stock in question is 1 divided by the lowest ask price) to yield a bid-ask spread of 5% (
1 divided by the lowest ask price) to yield a bid-ask spread of 5% (
1 / $20 x 100). Market with thinner spread is much liquid than others.
The order book shows a list of open orders on the exchange. The top half is the asking side of the order book, where you will see sell orders. The bottom half is the bidding side of the order book where you can see the buy orders for that trading pair. You can switch to different display mode: buy order only or sell order only to see more levels.
You can see the trading history from all users that have occurred on the trading pair that you currently have selected. The color of trades indicates taker/maker information. * Green: buyer is taker * Red: Seller is taker
In the bottom left-hand side of the interface, you can locate Open Orders, Order History, Trade History, and Balances. - Open Orders: This section shows which orders are not filled in their entirety and are still currently open. You can view all of the orders on the blockchain by looking at their TxHash. - Order History: This section shows the orders you have placed, filled, unfilled or canceled. You can view all of the orders on the blockchain by looking at their TxHash. - Trade History: This section shows your executed trades on the exchange. - Balances: This section displays the current balance of your Binance Chain Wallet.
Binance DEX currently only supports Limit orders. The Limit order section allows you to define the bid/ask price and the quantity you wish to trade. Please note that all order’s parameters’ constrains: * Price: Order price must be changed by multiple of tick size. Tick size stands for the smallest unit on price change * Quantity: Order quantity must be multiple of lot size. Lot size stands for the smallest quantity change * Timeinforce: It is Good Till Expire (GTE). The order would stay effective until expire time. The order may expire in the UTC midnight after more than 259, 200 blocks, which is 72 hours in terms of blocking time.
Previous Create Address
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Inner Product | Brilliant Math & Science Wiki
Orthogonal Inner Products
Symplectic Inner Products
Orthogonalization Algorithm
Consider some vectors in a vector space
V
\mathbb{F}
. We would like to define some concepts such as length of a vector or angle between two vectors, both of which are not intrinsic properties of the space itself. We notice that scalar multiplying a vector
\vec{v}
c
multiplies its length by
c
, and forming the sum
\vec{v} + \vec{w}
causes the norm to change from
\sqrt{\sum_i v_i^2}
\sqrt{\sum_i v_i^2 + \sum_i w_i^2 + 2 \sum_i v_i w_i}
. Let us abstract the concept. Such a function
L: V \times W \to \mathbb{F}
taking in two inputs has the property of being linear with respect to each input; that is,
L(c \vec{v} + \vec{a}, d \vec{w} + \vec{b} ) = cdL(\vec{v}, \vec{w} ) + cL(\vec{v}, \vec{b}) + dL(\vec{a}, \vec{w}) + L(\vec{b}, \vec{w})
. For a function of
n
inputs linear in each input, this function is called
n
linear. For
n =2, V = W
, such a function is called an inner product. Inner products will be used to develop the ideas of the magnitude of a vector and the angles between two vectors in Euclidean spaces as well as some other more abstract ideas.
g: V \to V \to \mathbb{F}
be an inner product. Choose the basis
\mathfrak{B} = \{ \vec{e}_{i=1, ..., n} \}
V
and define the matrix
G = G(i, j) = g(\vec{e}_i, \vec{e}_j), 1 \leqslant i, j \leqslant n
G
is called the Gram matrix of the inner product. Then to compute the inner product of two vectors
\vec{v}, \vec{w}
, first compute the coordinates
[\vec{v}]_{\mathfrak{B} }
[\vec{w}]_{\mathfrak{B}}
g( [\vec{v}]_{\mathfrak{B} } , [\vec{w}]_{\mathfrak{B}} ) = [\vec{v}]_{\mathfrak{B} }^t G [\vec{w}]_{\mathfrak{B}}
. This notation also makes the idea of multilinearity given in the motivation more natural. Moreover, if a basis
\mathfrak{B}
n
vectors is chosen and
G
n \times n
matrix over
\mathbb{F}
g([\vec{v}]_{\mathfrak{B}}, [\vec{w}]_{\mathfrak{B}} ) = [\vec{v}]_{\mathfrak{B}}^t G[\vec{w}]_{\mathfrak{B}}
defines an inner product on the space spanned by
\mathfrak{B}
(the 2-linearity is obvious by the definition of matrix multiplication). It follows that there is a bijection between inner products on a this space of
\dim n
n \times n
Let us now discuss how
G
is affected by a change of basis. Let
S
be the change of basis matrix so that
[\vec{v}]_{\mathfrak{B}} = S [\vec{v}]_{\mathfrak{A}}
. Then the following is obtained:
g([\vec{v}]_{\mathfrak{B}}, [\vec{w}]_{\mathfrak{B}}) = [S\vec{v}]_{\mathfrak{A}}^t G [S\vec{w}]_{\mathfrak{A}} = [\vec{v}]_{\mathfrak{A}}^t (S^t G S) [\vec{w}]_{\mathfrak{A}}
so that the Gram matrix in the basis
\mathfrak{A}
S^t G S
Exercise. Prove that
g^t (\vec{v}, \vec{w} ) = g (\vec{w}, \vec{v} )
Kostrikin, A., and Manin, Y. Linear Algebra and Geometry.
Cite as: Inner Product. Brilliant.org. Retrieved from https://brilliant.org/wiki/inner-product/
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(→GTK-Bounce: source boxes gray)
(→Cairo Animations: source box gray)
*Source for the Cairo animations is found at:
{\displaystyle \ x(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}x_{\mathrm {square} }(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
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As suggested by Sandeep Saurav sir, I am asking the below question again ​ Please solve - Maths - Playing with Numbers - 12478961 | Meritnation.com
As suggested by Sandeep Saurav sir, I am asking the below question again.
Please solve the below question as soon as possible... Please Don't say that it is not of our syllabus... I know that experts can solve this using simple understanding language too:
Q) While writing the amount in the Bank Cheque for withdrawal, Ryan by mistake wrote the last two digits in place of the first two digits and first two digits in place of last two digits. While he was coming back from the bank after withdrawing, he bought a 5 rupees packet with those withdrawn money. Then.... after that, He saw and realised that whatever money {(actual/formal money amount that he should write in the cheque)}, now that is double of that. Now question is that, how much money did he withdraw?
Let us assume that the amount Ryan had to withdraw is a 4-digit number.\phantom{\rule{0ex}{0ex}}Also, let a be the digit in the thousands place, b in the hundreds place, c in the tens place and d in the ones place. So, the amount he had to withdraw = 1000a + 100b + 10c + d.\phantom{\rule{0ex}{0ex}}Now, Ryan wrote the last two digits in place of the first two digits, and first two digits in place of the last two digits in the cheque.\phantom{\rule{0ex}{0ex}}In other words, he wrote c in the thousands place, d in the hundreds place, a in the tens place and b in the ones place. So, the amount he actually withdrew = 1000c + 100d + 10a + b.\phantom{\rule{0ex}{0ex}}Now, Ryan brought a 5 rupees pouch with the withdrawn money. So, the amount now left with him = 1000c + 100d + 10a + b - 5.\phantom{\rule{0ex}{0ex}}Ryan now realised that the actual amount he should have written in the cheque is twice the amount left with him.\phantom{\rule{0ex}{0ex}}\therefore 1000a + 100b + 10c + d = 2\left(1000c + 100d + 10a + b - 5\right)\phantom{\rule{0ex}{0ex}}⇒ 1000a + 100b + 10c + d = 2000c + 200d + 20a + 2b - 10\phantom{\rule{0ex}{0ex}}⇒ 980a + 98b - 1990c - 199d + 10 = 0\phantom{\rule{0ex}{0ex}}⇒ 98\left(10a + b\right) - 199\left(10c + d\right) + 10 = 0\phantom{\rule{0ex}{0ex}}The above equation is satisfied when a = 7, b = 3, c = 3, and d = 6.\phantom{\rule{0ex}{0ex}}So, the amount Ryan withdrew = 1000c + 100d + 10a + b = 3000 + 600 + 70 + 3 = Rs. 3673\phantom{\rule{0ex}{0ex}}NOTE: The above amount is not unique. There can be more solutions to this question, where the amount withdrawn could be a 5-digit, 6-digit, or a number with greater number of digits - it must satisfy the given conditions.
If you have any more doubts, please ask here on the forum and our experts will try and help you out as soon as possible.
V.sivagami Prabakar answered this
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Krull_dimension Knowpia
We say that a chain of prime ideals of the form
{\displaystyle {\mathfrak {p}}_{0}\subsetneq {\mathfrak {p}}_{1}\subsetneq \ldots \subsetneq {\mathfrak {p}}_{n}}
has length n. That is, the length is the number of strict inclusions, not the number of primes; these differ by 1. We define the Krull dimension of
{\displaystyle R}
to be the supremum of the lengths of all chains of prime ideals in
{\displaystyle R}
Given a prime
{\displaystyle {\mathfrak {p}}}
in R, we define the height of
{\displaystyle {\mathfrak {p}}}
{\displaystyle \operatorname {ht} ({\mathfrak {p}})}
, to be the supremum of the lengths of all chains of prime ideals contained in
{\displaystyle {\mathfrak {p}}}
{\displaystyle {\mathfrak {p}}_{0}\subsetneq {\mathfrak {p}}_{1}\subsetneq \ldots \subsetneq {\mathfrak {p}}_{n}={\mathfrak {p}}}
.[1] In other words, the height of
{\displaystyle {\mathfrak {p}}}
is the Krull dimension of the localization of R at
{\displaystyle {\mathfrak {p}}}
. A prime ideal has height zero if and only if it is a minimal prime ideal. The Krull dimension of a ring is the supremum of the heights of all maximal ideals, or those of all prime ideals. The height is also sometimes called the codimension, rank, or altitude of a prime ideal.
In a Noetherian ring, every prime ideal has finite height. Nonetheless, Nagata gave an example of a Noetherian ring of infinite Krull dimension.[2] A ring is called catenary if any inclusion
{\displaystyle {\mathfrak {p}}\subset {\mathfrak {q}}}
of prime ideals can be extended to a maximal chain of prime ideals between
{\displaystyle {\mathfrak {p}}}
{\displaystyle {\mathfrak {q}}}
, and any two maximal chains between
{\displaystyle {\mathfrak {p}}}
{\displaystyle {\mathfrak {q}}}
have the same length. A ring is called universally catenary if any finitely generated algebra over it is catenary. Nagata gave an example of a Noetherian ring which is not catenary.[3]
More generally, the height of an ideal I is the infimum of the heights of all prime ideals containing I. In the language of algebraic geometry, this is the codimension of the subvariety of Spec(
{\displaystyle R}
) corresponding to I.[6]
SchemesEdit
It follows readily from the definition of the spectrum of a ring Spec(R), the space of prime ideals of R equipped with the Zariski topology, that the Krull dimension of R is equal to the dimension of its spectrum as a topological space, meaning the supremum of the lengths of all chains of irreducible closed subsets. This follows immediately from the Galois connection between ideals of R and closed subsets of Spec(R) and the observation that, by the definition of Spec(R), each prime ideal
{\displaystyle {\mathfrak {p}}}
of R corresponds to a generic point of the closed subset associated to
{\displaystyle {\mathfrak {p}}}
by the Galois connection.
The dimension of a polynomial ring over a field k[x1, ..., xn] is the number of variables n. In the language of algebraic geometry, this says that the affine space of dimension n over a field has dimension n, as expected. In general, if R is a Noetherian ring of dimension n, then the dimension of R[x] is n + 1. If the Noetherian hypothesis is dropped, then R[x] can have dimension anywhere between n + 1 and 2n + 1.
For example, the ideal
{\displaystyle {\mathfrak {p}}=(y^{2}-x,y)\subset \mathbb {C} [x,y]}
has height 2 since we can form the maximal ascending chain of prime ideals
{\displaystyle (0)={\mathfrak {p}}_{0}\subsetneq (y^{2}-x)={\mathfrak {p}}_{1}\subsetneq (y^{2}-x,y)={\mathfrak {p}}_{2}={\mathfrak {p}}}
Given an irreducible polynomial
{\displaystyle f\in \mathbb {C} [x,y,z]}
{\displaystyle I=(f^{3})}
is not prime (since
{\displaystyle f\cdot f^{2}\in I}
, but neither of the factors are), but we can easily compute the height since the smallest prime ideal containing
{\displaystyle I}
{\displaystyle (f)}
The ring of integers Z has dimension 1. More generally, any principal ideal domain that is not a field has dimension 1.
An integral domain is a field if and only if its Krull dimension is zero. Dedekind domains that are not fields (for example, discrete valuation rings) have dimension one.
The Krull dimension of the zero ring is typically defined to be either
{\displaystyle -\infty }
{\displaystyle -1}
. The zero ring is the only ring with a negative dimension.
A ring is Artinian if and only if it is Noetherian and its Krull dimension is ≤0.
An integral extension of a ring has the same dimension as the ring does.
Let R be an algebra over a field k that is an integral domain. Then the Krull dimension of R is less than or equal to the transcendence degree of the field of fractions of R over k.[7] The equality holds if R is finitely generated as algebra (for instance by the noether normalization lemma).
Let R be a Noetherian ring, I an ideal and
{\displaystyle \operatorname {gr} _{I}(R)=\oplus _{0}^{\infty }I^{k}/I^{k+1}}
be the associated graded ring (geometers call it the ring of the normal cone of I.) Then
{\displaystyle \operatorname {dim} \operatorname {gr} _{I}(R)}
is the supremum of the heights of maximal ideals of R containing I.[8]
A commutative Noetherian ring of Krull dimension zero is a direct product of a finite number (possibly one) of local rings of Krull dimension zero.
A Noetherian local ring is called a Cohen–Macaulay ring if its dimension is equal to its depth. A regular local ring is an example of such a ring.
A Noetherian integral domain is a unique factorization domain if and only if every height 1 prime ideal is principal.[9]
For a commutative Noetherian ring the three following conditions are equivalent: being a reduced ring of Krull dimension zero, being a field or a direct product of fields, being von Neumann regular.
Of a moduleEdit
{\displaystyle \dim _{R}M:=\dim(R/\operatorname {Ann} _{R}(M))}
For non-commutative ringsEdit
Gelfand–Kirillov dimension
Krull's principal ideal theorem
^ Matsumura, Hideyuki: "Commutative Ring Theory", page 30–31, 1989
^ Eisenbud, D. Commutative Algebra (1995). Springer, Berlin. Exercise 9.6.
^ Matsumura, H. Commutative Algebra (1970). Benjamin, New York. Example 14.E.
^ Serre 2000, Ch. III, § B.2, Theorem 1, Corollary 4.
^ Eisenbud, Corollary 10.3. harvnb error: no target: CITEREFEisenbud (help)
^ Krull dimension less or equal than transcendence degree?
^ Eisenbud 2004, Exercise 13.8 harvnb error: no target: CITEREFEisenbud2004 (help)
^ Hartshorne,Robin:"Algebraic Geometry", page 7,1977
^ McConnell, J.C. and Robson, J.C. Noncommutative Noetherian Rings (2001). Amer. Math. Soc., Providence. Corollary 6.4.8.
L.A. Bokhut'; I.V. L'vov; V.K. Kharchenko (1991). "I. Noncommuative rings". In Kostrikin, A.I.; Shafarevich, I.R. (eds.). Algebra II. Encyclopaedia of Mathematical Sciences. Vol. 18. Springer-Verlag. ISBN 3-540-18177-6. Sect.4.7.
Eisenbud, David (1995), Commutative algebra with a view toward algebraic geometry, Graduate Texts in Mathematics, vol. 150, Berlin, New York: Springer-Verlag, ISBN 978-0-387-94268-1, MR 1322960
Serre, Jean-Pierre (2000). Local Algebra. Springer Monographs in Mathematics (in German). doi:10.1007/978-3-662-04203-8. ISBN 978-3-662-04203-8. OCLC 864077388.
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(→Dither)
Human hearing is most sensitive in the midrange from 2kHz to 4kHz; that's where background noise is going to be the most obvious.
Human hearing is [[WikiPedia:Equal-loudness_contour|most sensitive in the midrange from 2kHz to 4kHz]]; that's where background noise is going to be the most obvious.
We can [[WikiPedia:Noise_shaping|shape dithering noise]] away from sensitive frequencies to where
hearing is less sensitive, usually the highest frequencies.
{\displaystyle \ squarewave(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}\ squarewave(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
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On the Brunk-Chung type strong law of large numbers for sequences of blockwise $m$-dependent random variables
On the Brunk-Chung type strong law of large numbers for sequences of blockwise
m
-dependent random variables
Thanh, Le Van
For a sequence of blockwise
m
\left\{{X}_{n},n\ge 1\right\}
, conditions are provided under which
{lim}_{n\to \infty }\left({\sum }_{i=1}^{n}{X}_{i}\right)/{b}_{n}=0
almost surely where
\left\{{b}_{n},n\ge 1\right\}
is a sequence of positive constants. The results are new even when
{b}_{n}\equiv {n}^{r},r>0
. As special case, the Brunk-Chung strong law of large numbers is obtained for sequences of independent random variables. The current work also extends results of Móricz [Proc. Amer. Math. Soc. 101 (1987) 709-715], and Gaposhkin [Teor. Veroyatnost. i Primenen. 39 (1994) 804-812]. The sharpness of the results is illustrated by examples.
Mots clés : strong law of large numbers, almost sure convergence, blockwise
m
author = {Thanh, Le Van},
title = {On the {Brunk-Chung} type strong law of large numbers for sequences of blockwise $m$-dependent random variables},
AU - Thanh, Le Van
TI - On the Brunk-Chung type strong law of large numbers for sequences of blockwise $m$-dependent random variables
Thanh, Le Van. On the Brunk-Chung type strong law of large numbers for sequences of blockwise $m$-dependent random variables. ESAIM: Probability and Statistics, Tome 10 (2006), pp. 258-268. doi : 10.1051/ps:2006010. http://www.numdam.org/articles/10.1051/ps:2006010/
[1] S. Chobanyan, S. Levental and V. Mandrekar, Prokhorov blocks and strong law of large numbers under rearrangements. J. Theoret. Probab. 17 (2004) 647-672. | Zbl 1062.60028
[2] Y.S. Chow and H. Teicher, Probability Theory: Independence, Interchangeability, Martingales. 3rd ed. Springer-Verlag, New York (1997). | MR 1476912 | Zbl 0891.60002
[3] V.F. Gaposhkin, On the strong law of large numbers for blockwise independent and blockwise orthogonal random variables. Teor. Veroyatnost. i Primenen. 39 (1994) 804-812 (in Russian). English translation in Theory Probab. Appl. 39 (1994) 667-684 (1995). | Zbl 0847.60022
[4] M. Loève, Probability Theory I. 4th ed. Springer-Verlag, New York (1977). | MR 651017 | Zbl 0359.60001
[5] F. Móricz, Strong limit theorems for blockwise
m
-dependent and blockwise quasiorthogonal sequences of random variables. Proc. Amer. Math. Soc. 101 (1987) 709-715. | Zbl 0632.60025
|
Spectroscopic and Magnetic Properties - Course Hero
General Chemistry/Transition Metals and Coordination Chemistry/Spectroscopic and Magnetic Properties
Crystal field theory is a model used to predict bonding in octahedral and tetrahedral complexes considering a central cation surrounded by anionic (ligand) charge points in 3D space using simple VSEPR concepts.
Main-group elements form ionic or covalent compounds in order to satisfy the octet rule. However, this is not the case for transition metals. The bonding rules that govern coordination compounds are explained by an electrostatic model called crystal field theory that is used to describe and predict differences between
d orbital sets in transition metal complexes. This describes the structures of coordination compounds as well as their colors and magnetic properties.
In crystal field theory the focus is on the nonbonding electrons rather than the bonds between the central metal ion and the ligands. The ligands and the metal are considered as infinitesimally small point charges.
Because electrons all carry a negative charge, the electrons on the ligands repel those found on the central metal ion. In the tetrahedral configuration, the ligands form a triangular pyramid with the cation at the center of the pyramid. In the octahedral configuration, the ligand bonds are found along the Cartesian axes (x, y, z). The five d orbitals have lobes that lie along (x2-y2 and z2) and between (xz, yz, xy) on the Cartesian axes, respectively.
Orientation of d Orbitals
In the octahedral configuration, the central metal ion (M) lies at the intersections of the x, y, and z axes. Ligands are bonded along these axes. The
{d_{{z^2}}}
d_{x^2-y^2}
orbitals are called the
e_g
orbitals. The
e_g
orbitals have set lobes along x, y, and z axes. The
{d_{xy}}
{d_{yz}}
{d_{xz}}
{t_{2g}}
orbitals. Note that the torus of the z2 orbital lies in the xy plane.
When the metal ion is isolated (no ligands are attached to it; gas phase), the five orbitals,
d_{{z}^2}
d_{x^2-y^2}
d_{xy}
d_{yz}
d_{xz}
, all have the same energies. However, when ligands bond along the axes, this is no longer the case. The
{d_{{z^2}}}
{d_{{x^2}-{y^2}}}
orbitals, collectively known as the
e_g
orbitals, lie along the axes. The
{d_{xy}}
{d_{yz}}
{d_{xz}}
{t_{2g}}
orbitals, lie between the axes. This means that electrons in the
e_g
orbitals experience greater repulsion than those in the
{t_{2g}}
orbitals since the electrons of the ligands are points directly at the electrons of the metal. Thus, the energy of electrons in the
e_g
orbitals are higher than those in the
{t_{2g}}
orbitals, that is, more energy is required to fill the
e_g
orbitals than to fill the
{t_{2g}}
orbitals. The difference between these energies is known as crystal field splitting energy (
\Delta
{t_{2g}}
e_g
change places when the coordination complex has tetrahedral geometry. The magnitude of
\Delta
depends on the charge of the central metal ion, the orbitals the metal is using, and the nature of the ligands. The spectrochemical series is a list of ligands ordered according to their ability to induce crystal field splitting for a common metal center. Ligands on the left side of the series give small crystal field splittings while those on the right give greater energy separations between the
{t_{2g}}
e_g
orbital sets. A sample spectrochemical series can be formed using the following common ligands.
\begin{gathered}{\rm{O}_2}^{2-}\lt\rm{I}^-\lt\rm{S}^{2-}\lt\rm{SCN}^-\rm{(S\rm{-}bonded)}\lt\rm{Cl}^-\lt\rm{OH}^-\lt\rm{C}_2{\rm{O}_4}^{2-}\lt\rm{H}_2\rm{O}\\\lt\rm{NCS}^-\rm{(N\rm{-}bonded)}\lt\rm{CH}_3\rm{CN}\lt\rm{en}{\lt\rm{NO}_2}^-\lt\rm{CN}^-\lt\rm{CO}\end{gathered}
In this spectrochemical series, the ligand with the lowest
\Delta
is peroxide ion (O22−), and the ligand with the highest
\Delta
is carbon monoxide (CO). Ligands with low
\Delta
(those on the left side of the series) are known as weak-field ligands, and ligands with high
\Delta
(those on the right side of the series) are known as strong-field ligands. Metal ions can also be arranged according to their
\Delta
\rm{Mn}^{2+}\lt\rm{Ni}^{2+}\lt\rm{Co}^{2+}\lt\rm{Fe}^{2+}\lt\rm{V}^{2+}\lt\rm{Cu}^{2+}\lt\rm{Fe}^{3+}\lt\rm{Cr}^{3+}\lt\rm{V}^{3+}\lt\rm{Co}^{3+}
Similarly, in this spectrochemical series, the metal ion with the lowest
\Delta
is manganese ion (Mn2+), and the one with the highest
\Delta
is cobalt(III) ion (Co3+).
Generally, it is not possible to predict whether a particular ligand will have a strong or weak effect on a particular metal ion. However,
\Delta
tends to increase with increasing oxidation state, and
\Delta
tends to increase down a group on the periodic table.
The energy required for two electrons to occupy a single orbital is known as pairing energy (P). Because electrons in a single orbital repel each other, electrons will occupy orbitals singly, in the lowest energy levels possible, before they pair, according to Hund's rule. Orbitals that have the same energy level are degenerate orbitals. Electrons occupy each degenerate orbital singly before they pair.
For example, consider two different sets of ligands pairing with iron(II) ion (Fe2+ or FeII). When no ligands are bonded to the metal, the electrons occupy the lowest available orbital. When the weak-field ligand water (H2O) binds,
\Delta
is less than P, so electrons fill the
e_g
orbitals singly. A high-spin complex is a complex in which electrons fill higher orbitals singly and the pairing energy is large relative to the crystal field splitting energy. However, when a strong-field ligand, such as cyanide (CN–), binds, P is less than
\Delta
, so electrons will pair in the
{t_{2g}}
orbitals. A low-spin complex is a complex in which electrons preferentially pair in lower energy orbitals and the pairing energy is small relative to the crystal field splitting energy.
Filling Orbitals According to Energy
Fe2+ without ligands fills degenerate orbitals because this is the lowest energy state. H2O is a weak-field ligand, with
\Delta
small relative to P. When this ligand binds, electrons fill the
e_g
degenerate orbitals because the energy required to fill the higher orbital is less than the energy required to pair electrons. In contrast, CN- is a strong-field ligand, with
\Delta
large relative to P. When this ligand binds, electrons pair in the
t_{2g}
orbitals because the energy required to pair electrons is less than the energy required to fill the higher orbitals.
The magnetic properties of a coordination complex are affected by the arrangement of electrons, ligand field strength, electron-electron repulsion and pairing energies.
Molecules that contain unpaired electrons are paramagnetic, that is, attracted to magnetic fields. Molecules that contain no unpaired electrons are diamagnetic, that is, repelled by magnetic fields. Because it is possible to predict the way in which electrons will fill orbitals based on the energies required to pair electrons, it is possible to predict the magnetic moments, the strength and orientation of a magnetic field, of coordination complexes. The greater the number of unpaired electrons in an ion, complex, or molecule, the larger the magnetic moment. The magnetic moment is directly related to the number of unpaired electrons present. Therefore, high-spin complexes tend to be paramagnetic, and low-spin complexes tend to be diamagnetic. For example, high-spin and cationic [FeII(OH2)6]2+ complexes have four unpaired electrons (a
{t_{2g}}^4{e_g}^2
electronic configuration), so it is paramagnetic. On the other hand, low-spin [FeII(CN)6]4- anions have no unpaired electrons (a
{t_{2g}}^6
electronic configuration), so they are diamagnetic. Recall that water (aqua ligands) are weak-field (small
\Delta
) while cyanides (CN–) are strong-field ligands (big
\Delta
The arrangement of electrons in a coordination complex or transition metal ion and their interaction with light results in the bright colors associated with them.
The ways in which electrons are distributed in orbitals is related to the energy of those electrons. Light energy is absorbed or reflected differently depending on the arrangement of electrons in the complex. For this reason, the distribution of electrons also affects the color of the coordination complex.
Most main-group elements are not brightly colored, because they absorb visible wavelengths of light and reflect wavelengths that are outside the range of visible light (e.g., ultraviolet region of the electromagnetic spectrum). In comparison, many transition complexes absorb visible light wavelengths, as electronic transitions between d orbitals are at lower energies—that is, in the visual range of our eyes. When one or a few wavelengths of visible light are absorbed and the rest are reflected, the eye perceives color based on the reflected light. In other words, an object appears red if light of other wavelengths is all absorbed and red frequency wavelengths are reflected to reach a person's eye. Transition metals often reflect some of the wavelengths of visible light, resulting in bright colors, whether the metal is in its elemental form or bonded to ligands in a coordination complex.
Coordination complexes fill orbitals according to the lowest energy requirements; that is, orbitals are filled such that the electrons are at their lowest energy state. When light strikes these electrons, it excites them to a higher state. Electrons move from the
{t_{2g}}
orbital to the
e_g
orbital, absorbing the energy from a specific wavelength of light (or a small range of wavelengths). Because some wavelengths of light have been absorbed, the complex appears brightly colored to human eyes. For example, [Fe(H2O)6]2+ has four unpaired electrons, two of which occupy the
{t_{2g}}
orbitals. When white light strikes the complex, a photon excites one of these electrons to the
e_g
orbital. The amount of energy required to do so is equal to
\Delta
for the ligand. In this case, the complex absorbs red light and appears green.
Furthermore, different oxidation states of the metal ion can result in different colors. For example, vanadium is purple at +2, green at +3, blue at +4, and yellow at +5.
<Coordination Chemistry of Transition Metals>Suggested Reading
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Making Better Swirl Brakes Using Computational Fluid Dynamics: Performance Enhancement From Geometry Variation | J. Eng. Gas Turbines Power | ASME Digital Collection
J. Mike Walker '66 Department of Mechanical Engineering, Turbomachinery Laboratory, Texas A&M University
e-mail: yangjing@tamu.edu
e-mail: lsanandres@tamu.edu
Jing Yang Mem. ASME
Yang, J., and San Andrés, L. (January 13, 2022). "Making Better Swirl Brakes Using Computational Fluid Dynamics: Performance Enhancement From Geometry Variation." ASME. J. Eng. Gas Turbines Power. February 2022; 144(2): 021027. https://doi.org/10.1115/1.4051962
A fluid with a large swirl (circumferential) velocity entering an annular pressure seal influences the seal cross-coupled dynamic stiffness coefficients, and hence it affects system stability. Typically comprising a large number of angled vanes around the seal circumference, a swirl brake (SB) is a mechanical element installed to reduce (even reverse) the swirl velocity entering an annular seal. SB design guidelines are not readily available, and existing configurations appear to reproduce a single source. By using a computational fluid dynamics (CFD) model, the paper details a process to engineer a SB upstream of a 16-tooth labyrinth seal (LS) with tip clearance Cr = 0.203 mm. The process begins with a known nominal SB* geometry and considers variations in vane length (
LV*
= 3.25 mm) and width (
WV*
= 1.02 mm), and stagger angle (θ* = 0 deg). The vane number
NV*
= 72 and vane height
HV*
= 2.01 mm remain unchanged. The SB–LS operates with air supplied at pressure PS = 70 bar, a pressure ratio PR = exit pressure Pa/PS = 0.5, and rotor speed Ω = 10.2 krpm (surface speed ΩR = 61 m/s). Just before the SB, the preswirl velocity ratio = average circumferential velocity U/shaft surface speed (ΩR) equals α = 0.5. For the given conditions, an increase in LV allows more space for the development of vortexes between two adjacent vanes. These are significant to the dissipation of fluid kinetic energy and thus control the reduction of α. A 42% increase in vane length (LV = 4.6 mm) produces a ∼43% drop in swirl ratio at the entrance of the LS (exit of the SB), from αE = 0.23 to 0.13. Based on the SB with LV = 4.6 mm, the stagger angle θ varies from 0 deg to 50 deg. The growth in angle amplifies a vortex at ∼70% of the vane height, while it weakens a vortex at 30% of HV. For θ = 40 deg, the influence of the two vortexes on the flow produces the smallest swirl ratio at the LS entrance, αE = −0.03. For a SB with LV = 4.6 mm and θ = 40 deg, the vane width WV varies from 0.51 mm to 1.52 mm (±50% of
WV*
). A reduction in WV provides more space for the strengthening of the vortex between adjacent vanes. Therefore, a SB with greater spacing of vanes also reduces the inlet circumferential velocity. For WV = 0.51 mm, αE further decreases to −0.07. Besides the design condition (α = 0.5), the engineered SB having LV = 4.6 mm, θ = 40 deg, and WV = 0.51 mm effectively reduces the circumferential velocity at the LS entrance for other inlet preswirl ratios equaling α = 0 and 1.3. Rather than relying on extensive experiments, the CFD analysis proves effective to quickly engineer a best SB configuration from the quantification of performance while varying the SB geometry and inlet swirl condition.
Brakes, Computational fluid dynamics, Pressure, Rotors, Geometry, Flow (Dynamics), Fluids
Turbomachinery Rotordynamics With Case Studies
Minter Spring Publishing
Flow Induced Spring Coefficients of Labyrinth Seals for Application in Rotor Dynamics
Proceedings of the Workshop on Rotordynamic Instability Problems in High-Performance Turbomachinery
, Texas A&M University, College Station, TX, May 12–14, pp.
Improving the Stability ofLabyrinth Gas Seals
ASME Paper No. 98-GT-328.
Experimental and Theoretical Comparison of Two Swirl Brakes Designs
Optimization of Swirl Brake Design and Assessment of Its Stabilizing Effect on Compressor Rotordynamic Performance
Proceedings of the 43rd Turbomachinery and Pump Symposium
, Turbomachinery Laboratory, Texas A&M University, College Station, TX, Sept. 23–25, pp.
Numerical Investigation of the Effect of Geometric Design Parameters on Swirl Brake Performance
Hayrapetiau
Design and Implementation of Swirl Brakes for Enhanced Rotordynamic Stability in an Off-Shore Centrifugal Compressor
Proceedings of the Asia Turbomachinery and Pump Symposium
, Turbomachinery Laboratory, Texas A&M University, College Station, TX, Singapore, Mar. 12–15, pp.
, “ANSYS CFX User's Guide 19.1,”
|
I’m torn about talking about floating point numbers out in the open. I feel like this is the sort of thing that should be hidden away from polite society, so as not to scare the children. It’s just indecent!
A few years ago, I picked up the book Handbook of Floating Point Arithmetic, by Muller et al, which is fantastic. Chapter 7 alone is worth the price of admission for me, so I definitely recommend this one if you’re interested in this sort of thing. In the first chapter, the authors present a beautiful example originally due to the mathematician Jean-Michel Muller (one of the book’s coauthors), almost as a throwaway. I was previously unaware of this example, and as much as I like the book, it doesn’t really go into any detail for this deeply interesting, somewhat amazing example. So I will present an explanation here.
Frankly, I’ll take any excuse to use my advanced math degree for something other than kindling.
We’re going to define a sequence of numbers:
x_n = 111 - \frac{1130}{x_{n-1}} + \frac{3000}{x_{n-1}x_{n-2}}
n > 1
I’m going to state without proof that this sequence of numbers converges to a finite real number…an integer even! So what does it converge to? Well, this looks kind of hard, so let’s fire up the old computing machine and see what it has to say. In R, we might try computing, say, the n=90’th element of this sequence as follows:
for (i in 1L:n){
tmp <- 111 - 1130/x1 + 3000/x0/x1
x1 <- tmp
Which might suggest that the limit is 100. To help further convince ourselves, here are the first 25 values of the sequence computed as above:
And indeed, if we try higher values of n, we will always get 100 back. And that’s all perfectly well and good, except that the limit isn’t 100. Not even close; in fact, the limit is 6.
A Closed Form for the Recurrence
An ancient trick for this kind of thing at least as old as your mom is to let
x_n = \frac{y_{n+1}}{y_n}
. Then from the definition of
x_n
and some high school algebra, we have
So the characteristic polynomial of this (now linear) recurrence relation is
Therefore, we know that the closed form to the original recurrence is just:
\alpha, \beta, \gamma
depend on the initial values and are not all zero. We may choose
\gamma=0
, and so to satisfy our initial values, we could take
\alpha=4
\beta=-3
. So the above simply reduces to:
If staring at the closed form and taking the limit doesn’t do it for you, we can evaluate the closed form for n=90:
(4*5^(n+1) - 3*6^(n+1)) / (4*5^n - 3*6^n)
and we do indeed get 6. Not 100. What’s going on here?
The reason for the discrepancy is due to rounding in floating point calculations. This is the kind of stuff that a numerical analysis nerd has been saving up for ages and can’t wait to throw in your face, like how you shouldn’t fit a linear model numerically by solving the normal equations because it squares the condition number. The reason the behavior is so drastic in this particular case is interesting, and we’ll save for last. But for the moment, it’s worth talking a bit about precision. Now, as I plugged at the beginning, there are lengthy books written about this sort of thing. If this simple exercise isn’t enough for you, then I would recommend you check out that book.
In R, we were using the “numeric” type, which is just ordinary, standards-conforming double precision. A double stores 15-ish significant decimal digits. There are lots of great reasons to use floating point arithmetic, not the least of which is our modern processors are very good at doing that sort of thing quickly. But one of the many uncomfortable facts of life is that if you use the finite collection of double precision numbers as a proxy for all real numbers, you’re going to be doing some rounding when you do arithmetic. And all that rounding can cause problems, as we have seen. I emphasize this point because there are, quite literally, more total real numbers that we can’t represent in finite precision than there are rational numbers that we can’t represent in finite precision. So put that in your pipe and smoke it, pragmatists.
A fairly typical example demonstrating the annoyances of floating point arithmetic goes something like:
Take away 0.1 from 1.0 a total of 10 times. What is left?
This simple grade school problem obviously leaves us with the nice round figure of
for (i in 1:10) x <- x - 0.1
Returning to our original problem, we don’t have to restrict ourselves to double precision. For example, using the closed form version with 100 decimal digits of precision, the n=90’th element of the sequence is:
Which is pretty close to 6 (and would be closer if we increased n). But using the recurrence relation is a very different story. The n=90’th element was computed as 100.000000000000000 as a double. In fact, you aren’t likely to get anything different until you have well past 50 decimal digits of precision. With 110 decimal digits of precision using bc on this particular machine with the recurrence relation version, we get:
Rounding error or not, this is a pretty weird example, right? Throwing your hands up in the air and saying “floats are broken, the world is doomed” doesn’t really answer the question as far as I’m concerned. We can compute some things quite reliably in floating point arithmetic; why not this?
Notice that the zeros of the characteristic polynomial above are fixed points for the operator
F <- function(u, v) 111 - 1130/u + 3000/u/v
So why look at this operator? Because
And we can learn something very interesting by studying the eigenvalues of the Jacobian of this operator applied to our fixed points. The Jacobian is easily shown to be:
And so for our fixed points, we have:
J <- function(u, v) matrix(c(1130/u/u - 3000/u/u/v, 1, -3000/u/v/v, 0), 2, 2)
## [,1] [,2]
## [1,] 21.2 -24
## [2,] 1.0 0
eigen(J(5, 5))$values
## [1] 20.0 1.2
## [,1] [,2]
## [1,] 17.5 -13.88889
## [2,] 1.0 0.00000
## [1] 16.6666667 0.8333333
## [,1] [,2]
## [1,] 0.11 -0.003
## [2,] 1.00 0.000
eigen(J(100, 100))$values
There’s a well-known theorem in dynamical systems (holy crap, I can’t believe I get to reference this!) that if all of the eigenvalues of the Jacobian evaluated at a fixed point of the dynamical system are (in modulus) below 1, then that point is an attractor (which is exactly what it sounds like), but if one or more of the eigenvalues is (in modulus) above 1, then that point is a repeller (which is also exactly what it sounds like). This means that if a point is “close” to 100, then the sequence will very quickly converge to 100, as we have seen. And while 6 is a fixed point, it is repelling, so you have to sneak up on it (approach it along one particular path) or the sequence quickly runs away from it. The floating point errors exploit this behavior.
So if you’re having float problems, I feel bad for you, son. I got 99 problems but a float ain’t 1.00000000000000022204
|
{\displaystyle V_{B}}
{\displaystyle W_{D}}
{\displaystyle {\hat {H}}={\rm {Rep}}_{{\hat {B}},{\hat {D}}}(h)}
{\displaystyle {\hat {B}}}
{\displaystyle {\hat {D}}}
{\displaystyle {\rm {Rep}}_{{\hat {B}},B}({\mbox{id}})}
{\displaystyle H={\rm {Rep}}_{B,D}(h)}
{\displaystyle {\rm {Rep}}_{D,{\hat {D}}}({\mbox{id}})}
{\displaystyle {\hat {H}}={\rm {Rep}}_{D,{\hat {D}}}({\mbox{id}})\cdot H\cdot {\rm {Rep}}_{{\hat {B}},B}({\mbox{id}})\qquad \qquad (*)}
{\displaystyle T={\begin{pmatrix}\cos(\pi /6)&-\sin(\pi /6)\\\sin(\pi /6)&\cos(\pi /6)\end{pmatrix}}={\begin{pmatrix}{\sqrt {3}}/2&-1/2\\1/2&{\sqrt {3}}/2\end{pmatrix}}}
{\displaystyle {\mathcal {E}}_{2},{\mathcal {E}}_{2}}
{\displaystyle t:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}
{\displaystyle \pi /6}
{\displaystyle {\mathcal {E}}_{2},{\mathcal {E}}_{2}}
{\displaystyle {\hat {B}}=\langle {\begin{pmatrix}1\\1\end{pmatrix}}{\begin{pmatrix}0\\2\end{pmatrix}}\rangle \qquad {\hat {D}}=\langle {\begin{pmatrix}-1\\0\end{pmatrix}}{\begin{pmatrix}2\\3\end{pmatrix}}\rangle }
{\displaystyle *}
{\displaystyle {\hat {T}}={\rm {Rep}}_{{\mathcal {E}}_{2},{\hat {D}}}({\mbox{id}})\cdot T\cdot {\rm {Rep}}_{{\hat {B}},{\mathcal {E}}_{2}}({\mbox{id}})}
{\displaystyle {\rm {Rep}}_{{\mathcal {E}}_{2},{\hat {D}}}({\mbox{id}})}
{\displaystyle {\rm {Rep}}_{{\hat {D}},{\mathcal {E}}_{2}}({\mbox{id}})}
{\displaystyle {\begin{array}{rl}{\rm {Rep}}_{{\hat {B}},{\hat {D}}}(t)&={\begin{pmatrix}-1&2\\0&3\end{pmatrix}}^{-1}{\begin{pmatrix}{\sqrt {3}}/2&-1/2\\1/2&{\sqrt {3}}/2\end{pmatrix}}{\begin{pmatrix}1&0\\1&2\end{pmatrix}}\\&={\begin{pmatrix}(5-{\sqrt {3}})/6&(3+2{\sqrt {3}})/3\\(1+{\sqrt {3}})/6&{\sqrt {3}}/3\end{pmatrix}}\end{array}}}
{\displaystyle t}
{\displaystyle {\hat {B}}}
{\displaystyle {\rm {Rep}}_{\hat {B}}({\begin{pmatrix}1\\3\end{pmatrix}})={\begin{pmatrix}1\\1\end{pmatrix}}_{\hat {B}}}
{\displaystyle {\hat {T}}}
{\displaystyle {\begin{pmatrix}(5-{\sqrt {3}})/6&(3+2{\sqrt {3}})/3\\(1+{\sqrt {3}})/6&{\sqrt {3}}/3\end{pmatrix}}_{{\hat {B}},{\hat {D}}}{\begin{pmatrix}1\\1\end{pmatrix}}_{\hat {B}}={\begin{pmatrix}(11+3{\sqrt {3}})/6\\(1+3{\sqrt {3}})/6\end{pmatrix}}_{\hat {D}}}
{\displaystyle {\hat {D}}}
{\displaystyle {\frac {11+3{\sqrt {3}}}{6}}\cdot {\begin{pmatrix}-1\\0\end{pmatrix}}+{\frac {1+3{\sqrt {3}}}{6}}\cdot {\begin{pmatrix}2\\3\end{pmatrix}}={\begin{pmatrix}(-3+{\sqrt {3}})/2\\(1+3{\sqrt {3}})/2\end{pmatrix}}}
{\displaystyle \mathbb {R} ^{3}}
{\displaystyle {\begin{pmatrix}x\\y\\z\end{pmatrix}}{\stackrel {t}{\longmapsto }}{\begin{pmatrix}y+z\\x+z\\x+y\end{pmatrix}}}
{\displaystyle {\rm {Rep}}_{{\mathcal {E}}_{3},{\mathcal {E}}_{3}}(t)={\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}}}
{\displaystyle B=\langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}1\\1\\-2\end{pmatrix}},{\begin{pmatrix}1\\1\\1\end{pmatrix}}\rangle }
{\displaystyle {\rm {Rep}}_{B,B}(t)={\begin{pmatrix}-1&0&0\\0&-1&0\\0&0&2\end{pmatrix}}}
{\displaystyle *}
{\displaystyle H}
{\displaystyle {\hat {H}}}
{\displaystyle P}
{\displaystyle Q}
{\displaystyle {\hat {H}}=PHQ}
{\displaystyle H}
{\displaystyle {\hat {H}}}
{\displaystyle {\hat {H}}=PHQ}
{\displaystyle P}
{\displaystyle Q}
{\displaystyle P}
{\displaystyle Q}
{\displaystyle Q}
{\displaystyle {\begin{pmatrix}1&0\\0&0\end{pmatrix}}\qquad {\begin{pmatrix}1&1\\0&0\end{pmatrix}}}
{\displaystyle -1}
{\displaystyle m\!\times \!n}
matrix of ran{\displaystyle k}
{\displaystyle m\!\times \!n}
{\displaystyle k}
{\displaystyle {\begin{pmatrix}1&0&\ldots &0&0&\ldots &0\\0&1&\ldots &0&0&\ldots &0\\&\vdots \\0&0&\ldots &1&0&\ldots &0\\0&0&\ldots &0&0&\ldots &0\\&\vdots \\0&0&\ldots &0&0&\ldots &0\end{pmatrix}}}
{\displaystyle \left({\begin{array}{c|c}I&Z\\\hline Z&Z\end{array}}\right)}
{\displaystyle P}
{\displaystyle Q}
{\displaystyle P}
{\displaystyle Q}
{\displaystyle {\begin{pmatrix}1&2&1&-1\\0&0&1&-1\\2&4&2&-2\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}1&-1&0\\0&1&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}1&0&0\\0&1&0\\-2&0&1\end{pmatrix}}{\begin{pmatrix}1&2&1&-1\\0&0&1&-1\\2&4&2&-2\end{pmatrix}}={\begin{pmatrix}1&2&0&0\\0&0&1&-1\\0&0&0&0\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}1&2&0&0\\0&0&1&-1\\0&0&0&0\end{pmatrix}}{\begin{pmatrix}1&-2&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}}{\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&1\\0&0&0&1\end{pmatrix}}={\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&0&0&0\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&0&0&0\end{pmatrix}}{\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}}={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\end{pmatrix}}}
{\displaystyle P}
{\displaystyle Q}
{\displaystyle PHQ}
{\displaystyle {\begin{pmatrix}1&-1&0\\0&1&0\\-2&0&1\end{pmatrix}}{\begin{pmatrix}1&2&1&-1\\0&0&1&-1\\2&4&2&-2\end{pmatrix}}{\begin{pmatrix}1&0&-2&0\\0&0&1&0\\0&1&0&1\\0&0&0&1\end{pmatrix}}={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\end{pmatrix}}}
{\displaystyle 2\!\times \!2}
{\displaystyle 2\!\times \!2}
{\displaystyle 2\!\times \!2}
{\displaystyle B=\langle {\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\rangle }
{\displaystyle D=\langle {\vec {\delta }}_{1},\dots ,{\vec {\delta }}_{m}\rangle }
{\displaystyle c_{1}{\vec {\beta }}_{1}+\dots +c_{k}{\vec {\beta }}_{k}+c_{k+1}{\vec {\beta }}_{k+1}+\dots +c_{n}{\vec {\beta }}_{n}\;\longmapsto \;c_{1}{\vec {\delta }}_{1}+\dots +c_{k}{\vec {\delta }}_{k}+{\vec {0}}+\dots +{\vec {0}}}
{\displaystyle k}
{\displaystyle {\begin{pmatrix}c_{1}\\\vdots \\c_{k}\\c_{k+1}\\\vdots \\c_{n}\end{pmatrix}}_{B}\;\mapsto \;{\begin{pmatrix}c_{1}\\\vdots \\c_{k}\\0\\\vdots \\0\end{pmatrix}}_{D}}
{\displaystyle B}
{\displaystyle D}
{\displaystyle {\begin{pmatrix}1&3&0\\2&3&0\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}2&2&1\\0&5&-1\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}0&3\\1&1\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}4&0\\0&5\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}1&3\\2&6\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}1&3\\2&-6\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}2&1&0\\4&2&0\end{pmatrix}}}
{\displaystyle {\begin{pmatrix}0&1&0&2\\1&1&0&4\\3&3&3&-1\end{pmatrix}}}
{\displaystyle B={\mathcal {E}}_{2}\qquad D=\langle {\begin{pmatrix}1\\1\end{pmatrix}},{\begin{pmatrix}1\\-1\end{pmatrix}}\rangle }
{\displaystyle t:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}
{\displaystyle {\begin{pmatrix}1&2\\3&4\end{pmatrix}}}
Use change of basis matrices to represen{\displaystyle t}
{\displaystyle {\hat {B}}=\langle {\begin{pmatrix}0\\1\end{pmatrix}},{\begin{pmatrix}1\\1\end{pmatrix}}\rangle }
{\displaystyle {\hat {D}}=\langle {\begin{pmatrix}-1\\0\end{pmatrix}},{\begin{pmatrix}2\\1\end{pmatrix}}\rangle }
{\displaystyle {\hat {B}}=\langle {\begin{pmatrix}1\\2\end{pmatrix}},{\begin{pmatrix}1\\0\end{pmatrix}}\rangle }
{\displaystyle {\hat {D}}=\langle {\begin{pmatrix}1\\2\end{pmatrix}},{\begin{pmatrix}2\\1\end{pmatrix}}\rangle }
{\displaystyle P}
{\displaystyle Q}
{\displaystyle {\hat {H}}=PHQ}
{\displaystyle A}
{\displaystyle P}
{\displaystyle Q}
{\displaystyle PAQ=I}
{\displaystyle QP=A^{-1}}
{\displaystyle k=0}
{\displaystyle \mathbb {R} ^{1}}
{\displaystyle \mathbb {R} ^{3}}
{\displaystyle t:\mathbb {R} ^{n}\to \mathbb {R} ^{n}}
{\displaystyle T}
{\displaystyle {\mathcal {E}}_{n},{\mathcal {E}}_{n}}
{\displaystyle {\rm {Rep}}_{B,B}(t)}
{\displaystyle T={\begin{pmatrix}1&1\\3&-1\end{pmatrix}}\qquad B=\langle {\begin{pmatrix}1\\2\end{pmatrix}},{\begin{pmatrix}-1\\-1\end{pmatrix}}\rangle }
{\displaystyle {\rm {Rep}}_{B,B}(t)}
{\displaystyle B=\langle {\vec {\beta }}_{1},\ldots ,{\vec {\beta }}_{n}\rangle }
{\displaystyle V}
{\displaystyle B_{1}}
{\displaystyle B_{2}}
{\displaystyle W}
{\displaystyle D}
{\displaystyle h:V\to W}
{\displaystyle {\rm {Rep}}_{B_{2},D}(h)}
{\displaystyle {\rm {Rep}}_{B_{1},D}(h)}
{\displaystyle V}
{\displaystyle W}
{\displaystyle {\rm {Rep}}_{B_{1},B_{1}}(t)}
{\displaystyle {\rm {Rep}}_{B_{2},B_{2}}(t)}
{\displaystyle T}
{\displaystyle {\hat {T}}}
{\displaystyle T^{2}}
{\displaystyle {\hat {T}}^{2}}
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Yield Curve | Brilliant Math & Science Wiki
Mohamed Hazim and Jimin Khim contributed
A yield curve is a line that plots the interest rates, at a set point in time, of bonds having equal credit quality but differing maturity dates. The most frequently reported yield curve compares the three-month, two-year, five-year and 30-year U.S. Treasury debt. This yield curve is used as a benchmark for other debt in the market, such as mortgage rates or bank lending rates. The curve is also used to predict changes in economic output and growth.
\text{Yield Curve}
Cite as: Yield Curve. Brilliant.org. Retrieved from https://brilliant.org/wiki/yield-curve/
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High School Calculus/Techniques of Differentiation - Wikibooks, open books for an open world
High School Calculus/Techniques of Differentiation
Techniques of Differentiation[edit | edit source]
The definition of a derivative,
{\displaystyle f'(x)=\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}}
is not the only method of finding a derivative. This section provides three different techniques to help find derivatives. These techniques are often quicker than the definition of a derivative, depending on the given equation. The last part of the section presents how to calculate higher order derivatives.
The Power Rule[edit | edit source]
The first technique allows you to find a derivative of a summation of multiple terms.
This is known as The Power Rule.
Power Rule only applies in certain conditions.
Theorem[edit | edit source]
If n is a rational number, then the function
{\displaystyle f(x)=x^{n}}
{\displaystyle {\frac {d}{dx}}[x^{n}]=nx^{n-1}.}
For f to be differentiable at x=0, n must be a number such that
{\displaystyle x^{n-1}}
is defined on an interval containing 0.
{\displaystyle {\frac {d}{dx}}[x^{n}]=\lim _{{\Delta }x\rightarrow 0}{\frac {(x+{\Delta }x)^{n}-x^{n}}{{\Delta }x}}}
{\displaystyle =\lim _{{\Delta }x\rightarrow 0}{\frac {x^{n}+nx^{n-1}({\Delta }x)+{\frac {n(n-1)x^{n-2}}{2}}({\Delta }x)^{2}+.....+({\Delta }x)^{n}-x^{n}}{{\Delta }x}}}
{\displaystyle =\lim _{{\Delta }x\rightarrow 0}\left[nx^{n-1}+{\frac {n(n-1)x^{n-2}}{2}}({\Delta }x)+.....+({\Delta }x)^{n-1}\right]}
{\displaystyle =nx^{n-1}+0+.....+0}
{\displaystyle =nx^{n-1}{\square }.}
The following is an example of the Power Rule:
{\displaystyle f(x)=3x^{2}+2x+1}
Normally you would work through the definition of a derivative,
{\displaystyle f'(x)=\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}}
but instead you can apply the power rule to find your derivative.
First, take a look at the first term of the equation,
{\displaystyle 3x^{2}.}
{\displaystyle f'(x)}
first drop the power on the variable of this term down to the front of the equation,
{\displaystyle (2)*(3x^{2}).}
Next, subtract one from the power that was dropped down in front of the equation,
{\displaystyle (2)*(3x^{1}).}
{\displaystyle (2)(3)x=6x.}
So now we have the first part of the derivative,
{\displaystyle 6x.}
Now, perform this operation to the second and the third terms of the equation.
For the second part,
{\displaystyle 2x^{1}}
{\displaystyle (1)*(2x^{1})}
{\displaystyle (1)(2)(x^{0})=2.}
This shows that the second part is
{\displaystyle 2.}
Now for the third part we notice that there is no variable,
{\displaystyle x.}
There are two ways to look at this last part, either: one, there is no variable so the part disappears, or two, since there is no variable, we multiply the whole piece by 0, since that would be the power of the variable.
Either way, there is no longer a third piece.
Now, if we put the two parts together, we have our answer
{\displaystyle f'(x)=6x+2.}
Here is another example of this method in action
{\displaystyle f(x)=3x^{3}-4x^{2}+3x-4}
{\displaystyle f'(x)=(3)(3x^{3-1})-(2)(4x^{2-1})+3x^{1-1}}
{\displaystyle f'(x)=9x^{2}-8x+3.}
{\displaystyle f(x)=x^{2}-7x+13}
{\displaystyle h(x)=2x^{3}+2x-3}
{\displaystyle g(x)=4x^{4}-7x^{3}+2x^{2}+5x-4}
The Product Rule[edit | edit source]
The next technique is called The Product Rule.
As previously stated, not all derivatives can be found through the power rule.
While it is a nice way to find many derivatives, other methods exist depending on the equation you are currently solving.
The Power Rule can be seen as the sum of two separate functions, and their respective derivative is the sum of the individual derivatives.
The product rule introduces how to solve the product of two functions.
The product of two differentiable functions
{\displaystyle f(x)}
{\displaystyle g(x)}
is itself differentiable.
{\displaystyle f(x)g(x)}
{\displaystyle f(x)*g'(x)+g(x)*f'(x).}
{\displaystyle {\frac {d}{dx}}[f(x)g(x)]=\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)*g(x+{\Delta }x)-f(x)g(x)}{{\Delta }x}}}
{\displaystyle =\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)*g(x+{\Delta }x)-f(x+{\Delta }x)g(x)+f(x+{\Delta }x)g(x)-f(x)g(x)}{{\Delta }x}}}
{\displaystyle =\lim _{{\Delta }x\rightarrow 0}\left[f(x+{\Delta }x)*{\frac {g(x+{\Delta }x)-g(x)}{{\Delta }x}}+g(x)*{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}\right]}
{\displaystyle =\lim _{{\Delta }x\rightarrow 0}\left[f(x+{\Delta }x)*{\frac {g(x+{\Delta }x)-g(x)}{{\Delta }x}}\right]+\lim _{{\Delta }x\rightarrow 0}\left[g(x)*{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}\right]}
{\displaystyle =\lim _{{\Delta }x\rightarrow 0}f(x+{\Delta }x)*\lim _{{\Delta }x\rightarrow 0}{\frac {g(x+{\Delta }x)-g(x)}{{\Delta }x}}+\lim _{{\Delta }x\rightarrow 0}g(x)*\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}}
{\displaystyle =f(x)g'(x)+g(x)f'(x){\square }.}
For example, take the equation
{\displaystyle h(x)=(x^{2}+1)(x^{2}-2x-1).}
To start, think of the derivative of each part. Let
{\displaystyle f(x)=x^{2}+1}
{\displaystyle g(x)=x^{2}-2x-1.}
{\displaystyle f'(x)=2x}
{\displaystyle g'(x)=2x-2.}
Now apply each part to the previous theorem.
{\displaystyle h'(x)=f(x)*g'(x)+g(x)*f'(x)}
{\displaystyle h'(x)=(x^{2}+1)*(2x-2)+(x^{2}-2x-1)*(2x).}
Now all we need to do is simplify.
{\displaystyle h'(x)=(2x^{3}+2x-2x^{2}-2)+(2x^{3}-4x^{2}-2x)}
{\displaystyle h'(x)=(2x^{3}-2x^{2}+2x-2)+(2x^{3}-4x^{2}-2x)}
{\displaystyle h'(x)=4x^{3}-6x^{2}-2.}
Now, if product rule doesn't seem to appeal to you, when possible, you can just simplify the original equation.
If you multiple the parts by each other and then find the derivative, you are simply doing the Power Rule.
Sometimes this may be easier, but in some cases you may be unable to multiply the parts together.
Here is an example of changing a product rule.
{\displaystyle h(x)=(x^{2}+1)(x^{2}-2x-1)}
{\displaystyle h(x)=x^{4}+x^{2}-2x^{3}-2x-x^{2}-1}
{\displaystyle h(x)=x^{4}-2x^{3}-2x-1.}
Then using the power rule,
{\displaystyle h'(x)=4x^{3}-6x^{2}-2.}
As you can see, both methods yield the same answer.
Also, product rule can be expanded beyond two pieces.
{\displaystyle {\frac {d}{dx}}[f(x)g(x)h(x)]=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x).}
For further work, here are some practice problems.
{\displaystyle h(x)=(x^{2}-2)(3x+1)}
{\displaystyle w(x)=(3x+2)(2x^{2}+4x)}
{\displaystyle p(y)=(y^{2}-3y)(5y-4)(2y^{2}+1)}
The Quotient Rule[edit | edit source]
The next technique is called The Quotient Rule .
Before we learned what to do with the product of two functions, quotient rule teaches us what to do with a function divided by another function.
The quotient of two differentiable functions,
{\displaystyle {\frac {f(x)}{g(x)}}}
is itself differentiable for all
{\displaystyle g(x)\neq 0.}
This derivative is given by the bottom function multiplied by the derivative of the top function, minus the top function multiplied by the derivative of the bottom function, all divided by the bottom function squared.
{\displaystyle {\frac {d}{dx}}\left[{\frac {f(x)}{g(x)}}\right]={\frac {g(x)f'(x)-f(x)g'(x)}{[g(x)]^{2}}}.}
{\displaystyle {\frac {d}{dx}}\left[{\frac {f(x)}{g(x)}}\right]=\lim _{{\Delta }x\rightarrow 0}{\frac {{\frac {f(x+{\Delta }x)}{g(x+{\Delta }x)}}-{\frac {f(x)}{g(x)}}}{{\Delta }x}}}
{\displaystyle =\lim _{{\Delta }x\rightarrow 0}{\frac {g(x)f(x+{\Delta }x)-f(x)g(x+{\Delta }x)}{{\Delta }xg(x)g(x+{\Delta }x)}}}
{\displaystyle =\lim _{{\Delta }x\rightarrow 0}{\frac {g(x)f(x+{\Delta }x)-f(x)g(x)+f(x)g(x)-f(x)g(x+{\Delta }x)}{{\Delta }xg(x)g(x+{\Delta }x)}}}
{\displaystyle ={\frac {\lim _{{\Delta }x\rightarrow 0}{\frac {g(x)[f(x+{\Delta }x)-f(x)]}{{\Delta }x}}-\lim _{{\Delta }x\rightarrow 0}{\frac {f(x)[g(x+{\Delta }x)-g(x)]}{{\Delta }x}}}{\lim _{{\Delta }x\rightarrow 0}[g(x)g(x+{\Delta }x)]}}}
{\displaystyle ={\frac {g(x)[\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}]-f(x)[\lim _{{\Delta }x\rightarrow 0}{\frac {g(x+{\Delta }x)-g(x)}{{\Delta }x}}]}{\lim _{{\Delta }x\rightarrow 0}[g(x)g(x+{\Delta }x)]}}}
{\displaystyle ={\frac {g(x)f'(x)-f(x)g'(x)}{[g(x)]^{2}}}{\square }.}
{\displaystyle h(x)={\frac {3x-2}{x^{2}+1}}}
{\displaystyle f(x)=3x-2}
{\displaystyle g(x)=x^{2}+1.}
Start by finding the derivatives of each individual piece,
{\displaystyle f'(x)=3}
{\displaystyle g'(x)=2x.}
Now all we need do is plug each piece into the equation, so we have
{\displaystyle h'(x)={\frac {(x^{2}+1)(3)-(3x-2)(2x)}{(x^{2}+1)^{2}}}}
{\displaystyle ={\frac {3x^{2}+3-(6x^{2}-4x)}{x^{4}+2x^{2}+1}}}
{\displaystyle ={\frac {-3x^{2}+4x+3}{x^{4}+2x^{2}+1}}.}
In many cases, the derivative will not be able to be simplified further due to the division.
Sometimes, you may be able to perform some sort of division to turn a quotient rule into a product rule.
If not, you may be able to set the bottom function to the power of (-1) and then perform a product rule.
The only problem with this is that, in order to deal with the negative power, the Chain Rule must be used, which is covered in another section.
Even so, if quotient rule doesn't appeal to you, there are ways to change it.
Here are some practice problems for quotient rule:
{\displaystyle h(x)={\frac {x^{2}+2}{x}}}
{\displaystyle z(x)={\frac {3x-1}{x^{2}+2x+1}}}
{\displaystyle r(x)={\frac {2x^{2}-2x+3}{3}}}
Higher Order Derivatives[edit | edit source]
The last thing to mention is Higher Order Derivatives .
It is possible to find a second, even third and fourth, derivative of a function.
It is just finding the derivative of a derivative.
You denote higher order derivatives with two or more prime symbols, for example,
{\displaystyle f''(x)}
or even,
{\displaystyle g'''(x).}
Since the number of prime symbols can get out of hand quickly, many mathematicians stop at three, in favor of using small numbers,
{\displaystyle f^{4}(x)}
or any number higher,
{\displaystyle f^{n}(x).}
An example of a higher order derivative is:
{\displaystyle f(x)=x^{4}+x^{3}+2x^{2}-3x+1}
{\displaystyle f'(x)=4x^{3}+3x^{2}+4x-3}
{\displaystyle f''(x)=12x^{2}+6x+4}
{\displaystyle f'''(x)=24x+6}
{\displaystyle f^{4}(x)=24.}
Some practice problems are:
{\displaystyle f(x)=3x^{3}+2x^{2}-3}
{\displaystyle g(x)=(3x^{2}-1)(6x+2)}
{\displaystyle r(x)={\frac {2x+1}{3x}}}
Retrieved from "https://en.wikibooks.org/w/index.php?title=High_School_Calculus/Techniques_of_Differentiation&oldid=3292621"
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Lagrange's Zero-One Blocks (LZOB) | Brilliant Math & Science Wiki
Lagrange's Zero-One Blocks (LZOB)
Nihar Mahajan contributed
Lagrange's Zero-One Blocks (LZOB) is somewhat a sub-part of Lagrange Interpolation which especially aims at constructing a general formula of a certain sequence on basis of known few terms of that sequence.
Construction of Zero-One Blocks
Constructing a formula for any sequence of numbers. This technique was developed by a great mathematician-astronomer, Joseph Louis Lagrange.
No matter how many first consecutive terms of a sequence are given , they do not force on us any specific pattern. It is always possible to use a suitable number of zero-one blocks and construct a formula which will agree with the first given
k
terms and the next term of sequence can be assigned any number of our choice.
As an explicit example , suppose the sequence is
2,5,10,17
, then what is the next term? If you go by seeking pattern you would find that the differences are
3,5,7
between the two consecutive terms. So , you would get the next term as
17+9=\boxed{26}
. But what if you want next term as
1729
\sqrt{\pi}
? This is where LZOB proves useful. With the help of LZOB , we can assign any number as the next term in the sequence , in fact we can find a general formula for sequence! So let's find it!
We have to find general formula for the sequence
2,5,10,17
n=1
in the formula , it will give you
a_1=2
n=2
a_2=5
n=3
a_3=10
n=4
a_4=17
n=5
a_5=1729
First let us define
B_1(n)=\dfrac{(n-2)(n-3)(n-4)(n-5)}{(1-2)(1-3)(1-4)(1-5)}
Note: When we define
B_k(n)
, we exclude
(n-k)
term in the numerator and the terms in the denominator are of the form
(k-m)
m\neq k
Now we shall compute
B_1(n)
n=1,2,3,4,5
Indeed ,
B_1(2)=B_1(3)=B_1(4)=B_1(5)=0
because if we put either
n=2
3
4
5
, the numerator is anyway going to be equal to zero. Now what about
B_1(1)
n=1
, we see that the numerator does look the same as the denominator and we have
B_1(1)=1
\boxed{B_1(1)=1 \ , \ B_1(2)=B_1(3)=B_1(4)=B_1(5)=0}
Do you see now , why is
B_1(n)
called a 'zero-one' block?
Now we move on to compute second block , that is
B_2(n)=\dfrac{(n-1)(n-3)(n-4)(n-5)}{(2-1)(2-3)(2-4)(2-5)}
Following the same method as done for first block , we have
\boxed{B_2(2)=1 \ , \ B_2(1)=B_2(3)=B_2(4)=B_2(5)=0}
Similarly ,
B_3(n)=\dfrac{(n-1)(n-2)(n-4)(n-5)}{(3-1)(3-2)(3-4)(3-5)}
\boxed{B_3(3)=1 \ , \ B_3(1)=B_3(2)=B_3(4)=B_3(5)=0}
B_4(n)=\dfrac{(n-1)(n-2)(n-3)(n-5)}{(4-1)(4-2)(4-3)(4-5)}
\boxed{B_4(4)=1 \ , \ B_4(1)=B_4(2)=B_4(3)=B_4(5)=0}
B_5(n)=\dfrac{(n-1)(n-2)(n-3)(n-4)}{(5-1)(5-2)(5-3)(5-4)}
\boxed{B_5(5)=1 \ , \ B_5(1)=B_5(2)=B_5(3)=B_5(4)=0}
Since we want the fifth term
a_5=1729
, we make 5 blocks.
Now we are ready to construct a formula for
a_n
. Indeed the formula is :
\boxed{a_n=B_1(n)a_1 + B_2(n)a_2 + B_3(n)a_3 + B_4(n)a_4+B_5(n)a_5}
You may compute
a_1,a_2,a_3,a_4,a_5
to confirm the above claim.
Suppose you are given first
k
terms of sequence
<a_i>
and we need to find the formula for
n^{th}
k
terms are given of sequence , we need to make
k
zero-one blocks.
Each block is in the form
\large{B_i(n)=\frac{\displaystyle\prod_{j=1}^k (n-j)}{(n-i)\displaystyle\prod_{j=1 \ , \ j\neq i}^k (i-j)}}
1\leq i,j \leq k
For every block
B_i(i) = 1
and all other
B_i(j)=0
Now we have general formula for nth term in sequence as:
\Large \boxed{a_n=\displaystyle\sum_{i=1}^k B_i(n)a_i}
Cite as: Lagrange's Zero-One Blocks (LZOB). Brilliant.org. Retrieved from https://brilliant.org/wiki/lagranges-zero-one-blocks-lzob/
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Frequency - 3D CAD Models & 2D Drawings
Frequency (6568 views - Mechanical Engineering)
Frequency is the number of occurrences of a repeating event per unit of time. It is also referred to as temporal frequency, which emphasizes the contrast to spatial frequency and angular frequency. The period is the duration of time of one cycle in a repeating event, so the period is the reciprocal of the frequency. For example, if a newborn baby's heart beats at a frequency of 120 times a minute, its period—the time interval between beats—is half a second (that is, 60 seconds divided by 120 beats). Frequency is an important parameter used in science and engineering to specify the rate of oscillatory and vibratory phenomena, such as mechanical vibrations, audio signals (sound), radio waves, and light.
For a broader coverage of this topic, see Temporal rate.
Frequency is the number of occurrences of a repeating event per unit of time.[1] It is also referred to as temporal frequency, which emphasizes the contrast to spatial frequency and angular frequency. The period is the duration of time of one cycle in a repeating event, so the period is the reciprocal of the frequency.[2] For example, if a newborn baby's heart beats at a frequency of 120 times a minute, its period—the time interval between beats—is half a second (that is, 60 seconds divided by 120 beats). Frequency is an important parameter used in science and engineering to specify the rate of oscillatory and vibratory phenomena, such as mechanical vibrations, audio signals (sound), radio waves, and light.
4 Related types of frequency
5 In wave propagation
6.4 Heterodyne methods
7.3 Line current
For cyclical processes, such as rotation, oscillations, or waves, frequency is defined as a number of cycles per unit time. In physics and engineering disciplines, such as optics, acoustics, and radio, frequency is usually denoted by a Latin letter f or by the Greek letter
{\displaystyle \nu }
or ν (nu) (see e.g. Planck's formula).
The relation between the frequency and the period
{\displaystyle T}
of a repeating event or oscillation is given by
{\displaystyle f={\frac {1}{T}}.}
The SI derived unit of frequency is the hertz (Hz), named after the German physicist Heinrich Hertz. One hertz means that an event repeats once per second. A previous name for this unit was cycles per second (cps). The SI unit for period is the second.
A traditional unit of measure used with rotating mechanical devices is revolutions per minute, abbreviated r/min or rpm. 60 rpm equals one hertz.[3]
Angular frequency, usually denoted by the Greek letter ω (omega), is defined as the rate of change of angular displacement, θ, (during rotation), or the rate of change of the phase of a sinusoidal waveform (notably in oscillations and waves), or as the rate of change of the argument to the sine function:
{\displaystyle y(t)=\sin \left(\theta (t)\right)=\sin(\omega t)=\sin(2\mathrm {\pi } ft)}
{\displaystyle {\frac {\mathrm {d} \theta }{\mathrm {d} t}}=\omega =2\mathrm {\pi } f}
Angular frequency is commonly measured in radians per second (rad/s) but, for discrete-time signals, can also be expressed as radians per sampling interval, which is a dimensionless quantity. Angular frequency (in radians) is larger than regular frequency (in Hz) by a factor of 2π.
{\displaystyle y(t)=\sin \left(\theta (t,x)\right)=\sin(\omega t+kx)}
{\displaystyle {\frac {\mathrm {d} \theta }{\mathrm {d} x}}=k}
Wavenumber, k, is the spatial frequency analogue of angular temporal frequency and is measured in radians per meter. In the case of more than one spatial dimension, wavenumber is a vector quantity.
Further information: Wave propagation
{\displaystyle f={\frac {v}{\lambda }}.}
{\displaystyle f={\frac {c}{\lambda }}.}
When waves from a monochrome source travel from one medium to another, their frequency remains the same—only their wavelength and speed change.
See also: Frequency meter
Measurement of frequency can done in the following ways,
{\displaystyle f={\frac {71}{15\,{\text{s}}}}\approx 4.73\,{\text{Hz}}}
If the number of counts is not very large, it is more accurate to measure the time interval for a predetermined number of occurrences, rather than the number of occurrences within a specified time.[4] The latter method introduces a random error into the count of between zero and one count, so on average half a count. This is called gating error and causes an average error in the calculated frequency of
{\displaystyle \Delta f={\frac {1}{2T_{m}}}}
, or a fractional error of
{\displaystyle {\frac {\Delta f}{f}}={\frac {1}{2fT_{m}}}}
{\displaystyle T_{m}}
is the timing interval and
{\displaystyle f}
is the measured frequency. This error decreases with frequency, so it is generally a problem at low frequencies where the number of counts N is small.
An older method of measuring the frequency of rotating or vibrating objects is to use a stroboscope. This is an intense repetitively flashing light (strobe light) whose frequency can be adjusted with a calibrated timing circuit. The strobe light is pointed at the rotating object and the frequency adjusted up and down. When the frequency of the strobe equals the frequency of the rotating or vibrating object, the object completes one cycle of oscillation and returns to its original position between the flashes of light, so when illuminated by the strobe the object appears stationary. Then the frequency can be read from the calibrated readout on the stroboscope. A downside of this method is that an object rotating at an integral multiple of the strobing frequency will also appear stationary.
Main article: Frequency counter
Modern frequency counter
Above the range of frequency counters, frequencies of electromagnetic signals are often measured indirectly by means of heterodyning (frequency conversion). A reference signal of a known frequency near the unknown frequency is mixed with the unknown frequency in a nonlinear mixing device such as a diode. This creates a heterodyne or "beat" signal at the difference between the two frequencies. If the two signals are close together in frequency the heterodyne is low enough to be measured by a frequency counter. This process only measures the difference between the unknown frequency and the reference frequency. To reach higher frequencies, several stages of heterodyning can be used. Current research is extending this method to infrared and light frequencies (optical heterodyne detection).
Visible light is an electromagnetic wave, consisting of oscillating electric and magnetic fields traveling through space. The frequency of the wave determines its color: 7014400000000000000♠4×1014 Hz is red light, 7014800000000000000♠8×1014 Hz is violet light, and between these (in the range 4-7014800000000000000♠8×1014 Hz) are all the other colors of the visible spectrum. An electromagnetic wave can have a frequency less than 7014400000000000000♠4×1014 Hz, but it will be invisible to the human eye; such waves are called infrared (IR) radiation. At even lower frequency, the wave is called a microwave, and at still lower frequencies it is called a radio wave. Likewise, an electromagnetic wave can have a frequency higher than 7014800000000000000♠8×1014 Hz, but it will be invisible to the human eye; such waves are called ultraviolet (UV) radiation. Even higher-frequency waves are called X-rays, and higher still are gamma rays.
{\displaystyle \displaystyle c=f\lambda }
where c is the speed of light (c in a vacuum, or less in other media), f is the frequency and λ is the wavelength.
Main article: Audio frequency
Sound propagates as mechanical vibration waves of pressure and displacement, in air or other substances.[5]. In general, frequency components of a sound determine its "color", its timbre. When speaking about the frequency (in singular) of a sound, it means the property that most determines pitch.[6]
The frequencies an ear can hear are limited to a specific range of frequencies. The audible frequency range for humans is typically given as being between about 20 Hz and 20,000 Hz (20 kHz), though the high frequency limit usually reduces with age. Other species have different hearing ranges. For example, some dog breeds can perceive vibrations up to 60,000 Hz.[7]
Main article: Utility frequency
See also: Frequency (disambiguation) and Category:Units of frequency
Mechanical engineeringMechanical equilibriumOscillationPendulumPendulum clockRestoring forceTransmission (mechanics)PhysicsMotor-generatorRadio wave
This article uses material from the Wikipedia article "Frequency", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia
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Engineering Tables/Normal Distribution/Tail Function Graph - Wikibooks, open books for an open world
Engineering Tables/Normal Distribution/Tail Function Graph
The tail function is the area under the normal distribution N(0,1) between x and +∞:
{\displaystyle {\mbox{T}}(x)=\int \limits _{x}^{\infty }{{1 \over {\sqrt {2\pi }}}\exp \left({-{{y^{2}} \over 2}}\right)dy}}
For x > 6.5, the tail function can be approximated as:
{\displaystyle {\mbox{T}}(x)={1 \over {x{\sqrt {2\pi }}}}\exp \left({-{{x^{2}} \over 2}}\right)}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Tables/Normal_Distribution/Tail_Function_Graph&oldid=3232802"
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Write and solve a compound inequality that represents the possible
Write and solve a compound inequality that represents the possible temperatures
C=\frac{5}{9}\left(F-32\right)
C=-{20}^{\circ }
F=\frac{9}{5}\left(-20\right)+32
F=-36+32
F=-{4}^{\circ }
C=-{15}^{\circ }
F=\frac{9}{5}\left(-15\right)+32
F=-27+32
F={5}^{\circ }
The value of temperature at
C=-{20}^{\circ }
F=-{4}^{\circ }
C=-{15}^{\circ }
F=5\text{ }F={5}^{\circ }
Hence, the compound inequality of the temperature is found
-{4}^{\circ }\le F\le {5}^{\circ }
y=3.5x+2.8
Find the symbolic representation of
{f}^{-1}\left(x\right)
f\left(x\right)=\frac{2}{\sqrt{x}}
The total sales T(x) of Northwest Creations, to the nearest dollar using the model where the distribution of x thousand catalogues is approximated by
T\left(x\right)=5000\mathrm{log}\left(x+1\right)
Write an inequality to represent how many more people can enter the pool
Whether the given function is an appropriate model for extended years and explain the reason.
Given: The polynomial function is
f\left(x\right)=-0.87{x}^{3}+0.35{x}^{2}+81.62x+7684.94
Here, x represents the number of years after 1987 and f(x) represents the number of thefts in that respective year.
A cheetah that is running 90 feet per second is 120 feet behind an antelope that is running 60 feet per second. How long will it take cheetah to catch up to the antelope?
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Programming Blackjack | Brilliant Math & Science Wiki
One direction where we can take our programming skills is game development. Here, we'll build a text based Blackjack engine that allows us to play against a dealer, who follows conventional house rules. The logic of blackjack is simple, but is sufficiently complex that we can gain valuable experience for making more complicated games later on. As we said, this engine has several simplifications as compared to a commercial Blackjack platform:
It will be entirely text based.
It will have just a single player, and the dealer.
It will have no gambling system.
Each of these are things we can add in a future post, by using a GUI system such as Tk, by folding the game logic into an object oriented game design, and building out an accounting system to keep track of chip counts, tabs, and bets.
For those who'd like to follow along, here is the complete code.
To remind ourselves, the rules of blackjack are as follows:
The player and the dealer receive two cards from a shuffled deck. In our case, we'll use a single deck, though casinos usually use a 'shoe' consisting of six decks.
After the first two cards are dealt to dealer and player, the player is asked if they'd like another card (called 'hitting'), or if they are happy with the cards they have already (called 'staying'). The object is to make the sum of your card values as close to 21, without going over. If we make 21 exactly, we have blackjack, which can't be beat. If we go over 21, we 'bust' and we lose the round. The player is allowed to stop hitting at any point.
The number cards (2 through 10) are worth the number displayed, face cards are worth 10, and an Ace can be worth either 1 or 11. For example, if our first two cards are a Jack and an Ace, we'd want to count the Ace as 11 since 10 + 11 = 21 and we'd have blackjack, but, if we had already had a hand worth 18, decided to hit, and got an Ace, we'd want to count it as 1, since counting it as 11 would put us at 29 and we'd bust.
Once our hand is finished, the dealer tries to do the same. The dealer must keep hitting until they get to 17. If they get above 17 without busting, they can stay.
Scoring the game of Blackjack
Settling the winner
Sample game play
Finally, the game is settled by simple rules
If the player has blackjack, they win, unless the dealer also has blackjack, in which case the game is a tie.
If the dealer busts and the player doesn't, the player wins.
If the player busts, the dealer wins.
If the player and the dealer both don't bust, whoever is closest to 21 wins.
The first thing we need to play Blackjack is a shuffled deck of cards. First we will write code that accomplishes this.
# define the card ranks, and suits
ranks = [_ for _ in range(2, 11)] + ['JACK', 'QUEEN', 'KING', 'ACE']
suits = ['SPADE', 'HEART ', 'DIAMOND', 'CLUB']
"""Return a new deck of cards."""
return [[rank, suit] for rank in ranks for suit in suits]
# get a deck of cards, and randomly shuffle it
We laid out the ranks and suits of the cards with lists, and then wrote a simple function get_deck() that constructs the deck using a list comprehension.
Finally, we used Python's random library, which has various functions used in generating randomness. In particular, we employ shuffle, which takes any list and returns it in random order, to shuffle our deck of cards.
With this functionality in place, we can deal the player and dealer's first two cards. We use the destructive pop operation which returns the last element from a list and removes it from the list as a side effect. We also define a boolean variable to keep track of whether the player is still actively betting (that they haven't busted, and haven't asked to 'stay' with their current hand)
# boolean variable that indicates whether player has gone bust yet
player_in = True
# issue the player and dealer their first two cards
player_hand = [deck.pop(), deck.pop()]
dealer_hand = [deck.pop(), deck.pop()]
To start the game of Blackjack, players are dealt two cards at random from a shuffled deck.
You write the following code to simulate the act of dealing an initial hand. To test the code, you deal a hand
10^6
times and record the number of times the player makes Blackjack on their first two cards. If the code is written correctly, what do you expect to find for
\hat{f}_\text{blackjack}
, the fraction of initial hands that are Blackjack?
A two card hand is said to be "blackjack" if it consists of an Ace and any card worth 10 (i.e. a ten, or a face card).
Next, we write a function that accepts a list containing the cards of the hand as tuples in the form (rank, suit). The logic is quite simple. First we write a helper function that takes a single card, and returns its value according to the scheme we outlined above. By default, we count each ACE as 11. We map this function over the hand and store the sum of values as tmp_value.
The second major piece of logic is to count up the total number of aces in the hand. We then check to see that the hand value is less than or equal to 21. If it isn't, and there was an ace in the hand, we subtract 11. If the hand is then less than 21, we return the hand. Otherwise, if there is a second ace in the hand, we again subtract 11 and check again. We continue to do this until we've exhausted all the hand's aces.
The final step is to return the value of the hand. If we have a score below 21, we return a two entry list containing a string representation of the score, and the integer value of the score. As a convention, we count busts as 100.
"""Returns the integer value of a single card."""
rank = card[0]
elif rank is 'ACE':
def hand_value(hand):
"""Returns the integer value of a set of cards."""
# Naively sum up the cards in the deck.
tmp_value = sum(card_value(_) for _ in hand)
# Count the number of Aces in the hand.
num_aces = len([_ for _ in hand if _[0] is 'ACE'])
# Aces can count for 1, or 11. If it is possible to bring the value of
#the hand under 21 by making 11 -> 1 substitutions, do so.
while num_aces > 0:
if tmp_value > 21 and 'ACE' in ranks:
tmp_value -= 10
num_aces -= 1
# Return a string and an integer representing the value of the hand. If
# the hand is bust, return 100.
if tmp_value < 21:
return [str(tmp_value), tmp_value]
elif tmp_value == 21:
return ['Blackjack!', 21]
return ['Bust!', 100]
The next thing we code is the logic of gameplay. As we outlined above, we have to ask the player whether they'd like to hit or stay, and continue to ask them until they bust, or they decide to stay. One piece of information that's crucial to the player's decision is their current tally, therefore, we print the player's hand and current tally each time before asking for their response.
As long as the player's hand isn't a bust, we ask the question using Python's raw_input function, which can take keyboard input. For convenience, we take 1 to symbolize "hit me", and 0 to symbolize "stay." If they ask to hit, we again pop the deck, append the new card to our current hand, and immediately print the newly drawn card to the screen. If they ask to stay, we change the value of player_in to False and move on to the dealer.
# As long as the player remains in the game, ask them if they'd like to hit
# for another card, or stay with their current hand.
while player_in:
# Display the player's current hand, as well as its value.
current_score_str = '''\nYou are currently at %s\nwith the hand %s\n'''
print current_score_str % (hand_value(player_hand)[0], player_hand)
# If the player's hand is bust, don't ask them for a decision.
if hand_value(player_hand)[1] == 100:
if player_in:
response = int(raw_input('Hit or stay? (Hit = 1, Stay = 0)'))
# If the player asks to be hit, take the first card from the top of
# deck and add it to their hand. If they ask to stay, change
# player_in to false, and move on to the dealer's hand.
new_player_card = deck.pop()
player_hand.append(new_player_card)
print 'You draw %s' % new_player_card
player_in = False
At this point, we calculate the player's score, and the dealer's current score (on his first two cards). If the player's hand isn't a bust, we print the dealer's tally and current hand. Then, while the dealer's hand is worth less than 17, the dealer is made to hit. After each dealer hit, we print their new card. By design, this loop halts when the dealer exceeds 17.
player_score_label, player_score = hand_value(player_hand)
dealer_score_label, dealer_score = hand_value(dealer_hand)
dealer_hand_string = '''\nDealer is at %s\nwith the hand %s\n'''
print dealer_hand_string % (dealer_score_label, dealer_hand)
print "Dealer wins."
while hand_value(dealer_hand)[1] < 17:
new_dealer_card = deck.pop()
dealer_hand.append(new_dealer_card)
print 'Dealer draws %s' % new_dealer_card
In a pack of 52 cards there are four equal suits, two of black and two of red. If we remove two random cards, what is the probability that the two cards are of the same colour?
At this point, the game is nearly finished. All that remains is to check the player's score and the dealer's score against the list of scoring rules we outlined above. To start, we obtain the label, and number for the score of the dealer's hand. Next, we simply enumerate the possible end states, and ask which of them our game satisfies. Based on the outcome, we print to the screen declaring the victor.
if player_score < 100 and dealer_score == 100:
print 'You beat the dealer!'
elif player_score == dealer_score:
print 'You tied the dealer, nobody wins.'
elif player_score < dealer_score:
print "Dealer wins!"
Consider the two sample games below.
In the first, we're dealt an A
\heartsuit
and a 4
\heartsuit
, which gives us 15. We then proceed to draw three cards (Q
\spadesuit
\heartsuit
\heartsuit
) that keep us under 20. Drawing another ace is unlikely, so we stay.
In the process, our A
\heartsuit
is downvalued in value from 11 to 1 to keep from going bust.
The dealer is forced to hit on his starting 13, draws a 9
\spadesuit
, and goes bust.
python blackjack_no_class.py
You are at currently at 15
with the hand [['ACE', 'HEART '], [4, 'HEART ']]
Hit or stay? (Hit = 1, Stay = 0)1
You draw ['QUEEN', 'SPADE']
with the hand [['ACE', 'HEART '], [4, 'HEART '], ['QUEEN', 'SPADE']]
You draw [2, 'HEART ']
with the hand [['ACE', 'HEART '], [4, 'HEART '], ['QUEEN', 'SPADE'], [2, 'HEART ']]
with the hand [['ACE', 'HEART '], [4, 'HEART '], ['QUEEN', 'SPADE'], [2, 'HEART '], [3, 'HEART ']]
Dealer is at 13
with the hand [[10, 'CLUB'], [3, 'SPADE']]
Dealer draws [9, 'SPADE']
You beat the dealer!
In a second case, we're the ones to go bust and the program tells us we've done so. In this case, the dealer wins.
with the hand [[8, 'SPADE'], [4, 'DIAMOND']]
You draw ['QUEEN', 'DIAMOND']
You are at currently at Bust!
with the hand [[8, 'SPADE'], [4, 'DIAMOND'], ['QUEEN', 'DIAMOND']]
Dealer wins!
Cite as: Programming Blackjack. Brilliant.org. Retrieved from https://brilliant.org/wiki/programming-blackjack/
|
==== Automatising Sharpening & RGB Composition ====
An ''experimental'' bash script to automatise the sharpening/fusion and RGB composition process is demonstrated here. The script uses the Spectral Reflectance values (double precision values), gained from previous steps (see above), and produces Pan-sharpened images by applying all of the three methods that {{cmd|i.pansharpen}} offers. In addition, it attempts to produce True-Color composite images, without and with re-balancing the color tables by using the {{cmd|i.landsat.rgb}} module.
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L_{\lambda {\text{Pixel, Band}}}={\frac {K_{\text{Band}}*q_{\text{Pixel, Band}}}{\Delta \lambda _{\text{Band}}}}}
{\displaystyle L_{\lambda {\text{Pixel,Band}}}}
{\displaystyle K_{\text{Band}}}
{\displaystyle q_{\text{Pixel,Band}}}
{\displaystyle \Delta _{\lambda _{\text{Band}}}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
|
m (→Credits: hr under the credits.)
(→Use The Source Luke: Gromit-mpx link)
== Use The Source Luke ==
As stated in the Epilogue, everything that appears in the video demos is driven by open source software, which means the source is both available for inspection and freely usable by the community. The Thinkpad that appears in the video was running Fedora 17 and Gnome Shell (Gnome 3). The demonstration software does not require Fedora specifically, but it does require Gnu/Linux to run in its current form. In all, the video involved just under 50,000 lines of new and custom-purpose code (including contributions to non-Xiph projects such as Cinelerra and [http://sourceforge.net/projects/gromit-mpx/ Gromit]).
=== The Spectrum and Waveform Viewer ===
{\displaystyle \ squarewave(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}\ squarewave(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
|
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