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Table 4 Changes in body composition from PRE to POST (
\overline{x}
Body Mass (kg)*
CON 78.2 ± 5.0 77.9 ± 5.0 -0.3 ± 0.5
EX 82.3 ± 4.1 82.0 ± 4.0 -0.3 ± 0.5
EXFS 84.7 ± 5.5 82.9 ± 5.2 -1.8 ± 1.0
Fat Mass (kg)**
EX 23.9 ± 2.2 22.8 ± 2.9 -1.1 ± 0.4a
EXFS 28.9 ± 2.3 26.2 ± 2.3 -2.7 ± 0.4a,b,c
% Body Fat**
Fat-Free Mass (kg)*
CON 55.6 ± 4.3 55.7 ± 4.4 0.10 ± 0.3
EX 58.4 ± 3.5 59.2 ± 3.7 0.80 ± 0.6
EXFS 55.8 ± 3.5 56.7 ± 3.4 0.90 ± 0.5
Muscle Mass (kg)**
EX 28.1 ± 2.2 28.4 ± 2.3 0.3 ± 0.2
CON = control; EX = exercise only; EXFS = exercise + food supplement. *Main effect for time (p ≤ 0.05); **Group-by-Time interaction (p ≤ 0.05). aDifferent from PRE (p ≤ 0.017); bDifferent from CON (p ≤ 0.05); cDifferent from EX (p ≤ 0.05).
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Butterworth filter prototype - MATLAB buttap - MathWorks India
Frequency Response of a Butterworth Analog Filter
[z,p,k] = buttap(n)
[z,p,k] = buttap(n) returns the poles and gain of an order n Butterworth analog lowpass filter prototype.
Design a 9th-order Butterworth analog lowpass filter. Display its magnitude and phase responses.
[z,p,k] = buttap(9); % Butterworth filter prototype
n — Order of Butterworth filter
Order of Butterworth filter, specified as a positive integer scalar.
Zeros of the system, returned as a matrix. z contains the numerator zeros in its columns. z is an empty matrix because there are no zeros.
Gains of the system, returned as a scalar. k contains the gains for each numerator transfer function.
The function buttap returns the poles in the length n column vector p and the gain in scalar k. z is an empty matrix because there are no zeros. The transfer function is
H\left(s\right)=\frac{z\left(s\right)}{p\left(s\right)}=\frac{k}{\left(s-p\left(1\right)\right)\left(s-p\left(2\right)\right)\cdots \left(s-p\left(n\right)\right)}
p = exp(sqrt(-1)*(pi*(1:2:2*n-1)/(2*n)+pi/2)).';
k = real(prod(-p));
The function buttap returns zeros, poles, and gain (z, p, and k) in MATLAB®. However, the generated C/C++ code for buttap returns only poles p and gain k since zeros z is always an empty matrix.
Butterworth filters are characterized by a magnitude response that is maximally flat in the passband and monotonic overall. In the lowpass case, the first 2n-1 derivatives of the squared magnitude response are zero at ω = 0. The squared magnitude response function is
{|H\left(\omega \right)|}^{2}=\frac{1}{1+{\left(\omega /{\omega }_{0}\right)}^{2n}}
corresponding to a transfer function with poles equally spaced around a circle in the left half plane. The magnitude response at the cutoff angular frequency ω0 is always
1/\sqrt{2}
regardless of the filter order. buttap sets ω0 to 1 for a normalized result.
[1] Parks, T. W., and C. S. Burrus. Digital Filter Design. New York: John Wiley & Sons, 1987.
besselap | butter | cheb1ap | cheb2ap | ellipap
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9 Apr 2017 • on keras localization
In the last few years, experts have turned to global average pooling (GAP) layers to minimize overfitting by reducing the total number of parameters in the model. Similar to max pooling layers, GAP layers are used to reduce the spatial dimensions of a three-dimensional tensor. However, GAP layers perform a more extreme type of dimensionality reduction, where a tensor with dimensions
h \times w \times d
is reduced in size to have dimensions
1 \times 1 \times d
. GAP layers reduce each
h \times w
feature map to a single number by simply taking the average of all
hw
In mid-2016, researchers at MIT demonstrated that CNNs with GAP layers (a.k.a. GAP-CNNs) that have been trained for a classification task can also be used for object localization. That is, a GAP-CNN not only tells us what object is contained in the image - it also tells us where the object is in the image, and through no additional work on our part! The localization is expressed as a heat map (referred to as a class activation map), where the color-coding scheme identifies regions that are relatively important for the GAP-CNN to perform the object identification task. Please check out the YouTube video below for an awesome demo!
We’ll begin with the Activation layer. This layer contains 2048 activation maps, each with dimensions
7\times7
f_k
k
-th activation map, where
k \in \{1, \ldots, 2048\}
The following AveragePooling2D GAP layer reduces the size of the preceding layer to
(1,1,2048)
by taking the average of each feature map. The next Flatten layer merely flattens the input, without resulting in any change to the information contained in the previous GAP layer.
The object category predicted by ResNet-50 corresponds to a single node in the final Dense layer; and, this single node is connected to every node in the preceding Flatten layer. Let
w_k
represent the weight connecting the
k
-th node in the Flatten layer to the output node corresponding to the predicted image category.
w_1 \cdot f_1 + w_2 \cdot f_2 + \ldots + w_{2048} \cdot f_{2048}
You can plot these class activation maps for any image of your choosing, to explore the localization ability of ResNet-50. Note that in order to permit comparison to the original image, bilinear upsampling is used to resize each activation map to
224 \times 224
. (This results in a class activation map with size
224 \times 224
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Predict responses using ensemble of bagged decision trees - MATLAB - MathWorks Switzerland
{\stackrel{^}{y}}_{\text{bag}}=\frac{1}{\sum _{t=1}^{T}{\alpha }_{t}I\left(t\in S\right)}\sum _{t=1}^{T}{\alpha }_{t}{\stackrel{^}{y}}_{t}I\left(t\in S\right).
{\stackrel{^}{y}}_{t}
I\left(t\in S\right)
{\stackrel{^}{P}}_{t}\left(c|x\right)
{\stackrel{^}{P}}_{\text{bag}}\left(c|x\right)=\frac{1}{\sum _{t=1}^{T}{\alpha }_{t}I\left(t\in S\right)}\sum _{t=1}^{T}{\alpha }_{t}{\stackrel{^}{P}}_{t}\left(c|x\right)I\left(t\in S\right).
{\stackrel{^}{y}}_{\text{bag}}=\underset{c\in C}{\mathrm{arg}\mathrm{max}}\left\{{\stackrel{^}{P}}_{\text{bag}}\left(c|x\right)\right\}.
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Table 5 Changes in upper- and lower-body strength from PRE to POST (
\overline{x}
Bench Press (kg)**
EX 47.7 ± 7.4 54.0 ± 5.7 6.3 ± 0.8a,b
EXFS 50.0 ± 6.4 58.5 ± 7.2 8.5 ± 1.0a,b
Squat (kg)**
EX 59.2 ± 7.2 70.8 ± 7.3 11.6 ± 1.4a,b
EXFS 61.0 ± 6.9 74.9 ± 7.2 13.9 ± 1.8a,b
Upper- and lower-body strength, as assessed by five-repetition maximum (5RM) bench press (upper) and squat (lower). CON = control; EX = exercise only; EXFS = exercise + food supplement. *Main effect for time (p ≤ 0.05); **Group-by-Time interaction (p ≤ 0.05). aDifferent from PRE (p ≤ 0.017); bDifferent from CON (p ≤ 0.05).
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Caesar Cipher - Maple Help
Home : Support : Online Help : Math Apps : Computer Science : Caesar Cipher
The Caesar Cipher is one of the simplest and most widely known encryption techniques. It is a form of substitution cipher in which each letter of the original text, known as the "plaintext", is replaced by a letter some fixed number of positions further down the alphabet. This technique is named after Julius Caesar, who used it with a left shift of 3 to protect messages of military significance.
Decryption occurs by performing a shift of the same magnitude but in the opposite direction on each letter of the encrypted text, known as the "ciphertext".
For example, with a right shift of 3: A would be replaced with D, B would be replaced with E, and so on until W is replaced with Z. Then, the beginning of the alphabet must wrap around so that X is replaced with A, Y is replaced with B, and finally, Z is replaced with C.
Encryption using a Caesar cipher can be represented using modular arithmetic by first transforming the letters of the alphabet into numbers with A = 0, B = 1, C = 2, ... , Y = 24, Z = 25.
Encryption of the letter x with a shift of n can be described as:
{E}_{n}\left(x\right) = \left(x + n\right) \mathbf{mod} 26
Decryption of the same letter x with a shift of n can then be described as:
{\mathrm{D}}_{n}\left(x\right) = \left(x - n\right) \mathbf{mod} 26
Like all types of monoalphabetic substitution, encryptions made using a Caesar cipher can be easily broken through the frequency analysis of letters and brute force attacks.
The black letters below illustrate the normal position of the letters of the alphabet, while the dark blue letters illustrate the shifted position of these letters in a Caesar cipher. Use the slider to adjust the magnitude of the shift, then type your message into the "Plaintext" text box and press the "Encrypt Text" button to transform it into the corresponding ciphertext.
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Assum T: R^m to R^n is a matrix transformation with matrix A. Prove that if the
Assum T: R^m to R^n is a matrix transformation with matrix A.Prove that if the column
Assum T:
{R}_{m}
{R}_{n}
is a matrix transformation with matrix A. Prove that if the columns of A are linearly independent, then T is one to one (i.e injective). (Hint: Remember that matrix transformations satisfy the linearity properties.
If A is a matrix, v and w are vectors and c is a scalar then
A0=0
A\left(cv\right)=cAv
A\left(v\text{ }+\text{ }w\right)=Av\text{ }+\text{ }Aw
Let T be defined as
T\left(v\right)=Av
v=\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\\ .\\ .\\ {v}_{m}\end{array}\right]\text{ }\in \text{ }{R}_{m}\text{ }and\text{ }A=\left[\begin{array}{ccccc}{a}_{11}& {a}_{12}& .& .& {a}_{1m}\\ {a}_{21}& {a}_{22}& .& .& {a}_{2m}\\ .& .& .& .& .\\ .& .& .& .& .\\ {a}_{n1}& {a}_{n2}& .& .& {a}_{nm}\end{array}\right].
Further obtain the result as follows:
T\left(v\right)=Av
=\left[\begin{array}{ccccc}{a}_{11}& {a}_{12}& .& .& {a}_{1m}\\ {a}_{21}& {a}_{22}& .& .& {a}_{2m}\\ .& .& .& .& .\\ .& .& .& .& .\\ {a}_{n1}& {a}_{n2}& .& .& {a}_{nm}\end{array}\right]\text{ }\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\\ .\\ .\\ {v}_{m}\end{array}\right]
\left[\begin{array}{ccccc}{a}_{11}{v}_{1}& +& {a}_{12}{v}_{2}& +\cdots \text{ }+& {a}_{1m}{v}_{m}\\ {a}_{21}{v}_{1}& +& {a}_{22}{v}_{2}& +\cdots \text{ }+& {a}_{2m}{v}_{m}\\ .& .& .& .& .\\ .& .& .& .& .\\ {a}_{n1}{v}_{1}& +& {a}_{n2}{v}_{2}& +\cdots \text{ }+& {a}_{nm}{v}_{m}\end{array}\right]
As the colums of A are linearly idependent, no rows Av will be equal. Thus, the vector obtained is different than one another in
{R}_{n}.
Hence, if the columns of A are linearly independent, then T is one to one.
y=-2\left(\frac{3}{2}-{e}^{3-x}\right)
Trying to go ABC is reflected across the X axis and then across the Y access which rotation is equivalent to this composition transformations transformations.
{R}_{2}
T(x, y) = (x + 4y, y)
\sum _{k=0}^{\mathrm{\infty }}\frac{10}{{k}^{2}+9}
The equation for inverse function
{g}^{-1}\left(x\right)\text{ }\text{for the function}\text{ }g\left(x\right)=\sqrt{x\text{ }+\text{ }2}
Mark each of the following statement true or false:
T:{R}_{n}\to R
defined by T(x) = | x | is a linear transformation.
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Find the Exact length of the curve. x=e^t+e^{-t} , y=5-2t
Find the Exact length of the curve. x=e^t+e^{-t} , y=5-2t between
x={e}^{t}+{e}^{-t},y=5-2t
0\le t\le 3
Find the inverse Laplace transforms of the functions given. Accurately sketch the time functions.
F\left(s\right)=\frac{3{e}^{-2s}}{s\left(s+3\right)}
F\left(s\right)=\frac{{e}^{-2s}}{s\left(s+1\right)}
F\left(s\right)=\frac{{e}^{-2s}-{e}^{-3s}}{2}
\frac{df}{dx}+\frac{df}{dy}=xy
Solve the differential equation using Laplace transform of
{y}^{″}-3{y}^{\prime }+2y={e}^{3t}
when y(0)=0 and y'(0)=0
t,1+t,{t}^{2},-t
y{}^{‴}+a\left(t\right)y{}^{″}+b\left(t\right)y{}^{″}+c\left(t\right)y=d\left(t\right)
L\left(t\right)=L\left(1+t\right)=L\left({t}^{2}\right)=L\left(-t\right)=d\left(t\right)
1,t,{t}^{2}
How to deal with two interdependent integrators?
I have two functions, f(t,x) and g(t,u), where
\frac{d}{dt}u=f\left(t,x\right)\text{ }\text{and}\text{ }\frac{d}{dt}x=g\left(t,u\right)
I am trying to discretize the integral of this system in order to track x and u. I have succeeded using Euler integration, which is quite simple, since x(t) and u(t) are both known at t:
u\left(t+h\right)=u\left(t\right)+hf\left(t,x\left(t\right)\right)
x\left(t+h\right)=x\left(t\right)+hg\left(t,u\left(t\right)\right)
However, I am now trying to implement mid-point integration to get more accurate results. (Eventually Runge-Kutta but I am stuck here for now.)
X\left(s\right)=\frac{s+1}{\left(s+2\right)\left({s}^{2}+2s+2\right)\left({s}^{2}+4\right)}
X\left(s\right)=\frac{1}{\left(2{s}^{2}+8s+20\right)\left({s}^{2}+2s+2\right)\left(s+8\right)}
X\left(s\right)=\frac{1}{{s}^{2}\left({s}^{2}+2s+5\right)\left(s+3\right)}
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Generate Random Numbers Using Uniform Distribution Inversion - MATLAB & Simulink - MathWorks Nordic
The inversion method relies on the principle that continuous cumulative distribution functions (cdfs) range uniformly over the open interval (0,1). If
u
is a uniform random number on (0,1), then
x={F}^{-1}\left(u\right)
x
from any continuous distribution with the specified cdf F.
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How do you simplify \frac{\sec x}{\tan x}?
\frac{\mathrm{sec}x}{\mathrm{tan}x}
\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}
\mathrm{sec}x=\frac{1}{\mathrm{cos}x}
\frac{\mathrm{sec}x}{\mathrm{tan}x}=\frac{1}{\mathrm{cos}x}\cdot \frac{\mathrm{cos}x}{\mathrm{sin}x}=\frac{1}{\mathrm{sin}x}=\mathrm{csc}x
Let's use the definitions of
\mathrm{sec}x
\mathrm{tan}x
to simplify this. We know
\mathrm{sec}x=\frac{1}{\mathrm{cos}x}
\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}
. Let's plug these values into our original expression. we get
\frac{\frac{1}{\mathrm{cos}x}}{\frac{\mathrm{sin}x}{\mathrm{cos}x}}=\frac{1}{\mathrm{sin}x}⇒\mathrm{csc}x
\frac{\mathrm{sec}x}{\mathrm{tan}x}=\mathrm{csc}x
{S}_{n}=\sum _{j=1}^{n}\mathrm{sin}\left(\sqrt{j\phantom{\rule{0.167em}{0ex}}}\pi \right)
{S}_{\left(2M{\right)}^{2}}<0
{S}_{\left(2M+1{\right)}^{2}}>0\phantom{\rule{1em}{0ex}}\left(M=1,2,3,...\right)
\mathrm{tan}28°=\mathrm{cot}
\mathrm{cos}\left(\mathrm{arcsin}x-\mathrm{arctan}2x\right)
\mathrm{tan}\left(\frac{\pi }{6}\right)
y=3\sqrt{x}+\frac{1}{\sqrt{x}}
0=-60°
\left(-1,-1\right)
\left(1,-\surd 3\right)
\left(-\surd 3,1\right)
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Minimal nutrition intervention with high-protein/low-carbohydrate and low-fat, nutrient-dense food supplement improves body composition and exercise benefits in overweight adults: A randomized controlled trial | Nutrition & Metabolism | Full Text
Table 6 Changes in cardiorespiratory fitness from PRE to POST (
\overline{x}
± SE).
From: Minimal nutrition intervention with high-protein/low-carbohydrate and low-fat, nutrient-dense food supplement improves body composition and exercise benefits in overweight adults: A randomized controlled trial
VO2max (ml·kg-1·min-1)**
CON 36.8 ± 2.1 36.9 ± 2.0 0.1 ± 0.5
EX 35.7 ± 2.7 37.5 ± 2.4 1.8 ± 0.6a
EXFS 32.9 ± 1.9 35.5 ± 1.8 2.6 ± 0.4a
VO2 (L·min-1)*
EX 2.87 ± 0.23 3.00 ± 0.82 0.13 ± 0.05
EXFS 2.76 ± 0.20 2.93 ± 0.21 0.17 ± 0.04a
Minute Ventilation (L·min-1)*
Maximum Heart Rate (bpm)*
CON 192.1 ± 2.7 190.9 ± 2.0 -1.2 ± 1.8
EX 190.6 ± 2.6 188.6 ± 2.8 -2.0 ± 1.3
EXFS 187.0 ± 6.2 184.6 ± 6.0 -2.4 ± 1.5
Time-to-Exhaustion (sec)**
CON 643.9 ± 44.8 653.0 ± 46.4 9.1 ± 13.4
EX 681.8 ± 62.0 748.6 ± 65.4 66.8 ± 13.8a,b
EXFS 559.9 ± 46.8 678.6 ± 50.1 118.7 ± 14.3a,b,c
Cardiorespiratory fitness, as assessed by graded treadmill test to exhaustion. CON = control; EX = exercise only; EXFS = exercise + food supplement. *Main effect for time (p ≤ 0.05); **Group-by-Time interaction (p ≤ 0.05). aDifferent from PRE (p ≤ 0.017). bDifferent from CON (p ≤ 0.05). cDifferent from EX (p ≤ 0.05).
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Using analytic geometry, find the coordinates of the point on
A\left(5,\text{ }8\right)
B\left(-3,\text{ }4\right)
trisanualb6
A\left(5,\text{ }8\right),\text{ }B\left(-3,\text{ }4\right)
Let, Equidistent point,
p\left(x,\text{ }0\right)
So, using distance formula:
\sqrt{{\left(x-3\right)}^{2}+{\left(-8\right)}^{2}}
\sqrt{\left(x+3{\right)}^{2}+\left(-4{\right)}^{2}}
\text{⧸}{x}^{2}+25-10x+64=\text{⧸}{x}^{2}+9+6x+16
25+64-9-16=10x+6x
16x=64
x=4
To find the distance between these two points we will use the distance formula
d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}
{d}_{1}=\sqrt{{\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}}
{d}_{2}=\sqrt{{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}}
{d}_{1}={d}_{2}
\sqrt{{\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}}
{\left(\sqrt{{\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}}\right)}^{2}={\left(\sqrt{{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}}\right)}^{2}
\sqrt{{\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}}
\sqrt{{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}}
with using rule:
\sqrt{a}={a}^{\frac{1}{2}}
\left(\left(\left(-5-x{\right)}^{2}+\left(2-0{\right)}^{2}{\right)}^{\frac{1}{2}}{\right)}^{2}=\left(\left(\left(2-x{\right)}^{2}+\left(3-0{\right)}^{2}{\right)}^{\frac{1}{2}}{\right)}^{2}
Apply exponenet rule:
{\left({a}^{b}\right)}^{c}={a}^{bc}
\left({\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}\right)6\left\{\frac{1}{2}×2\right\}={\left({\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}\right)}^{\frac{1}{2}×2}
Cancel th ecommon term into the exponent.
{\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}+{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}
Subtract 2-0 and 2-0
{\left(-5-x\right)}^{2}+{2}^{2}={\left(2-x\right)}^{2}+{3}^{2}
Find the area of the surface generaled by revolving the curve
x=\frac{{y}^{3}}{5},\text{ }0\le y\le 3
about the y-axis.
\frac{{\partial }^{2}}{\partial {x}^{2}}f+\frac{{\partial }^{2}}{\partial {y}^{2}}f=0
Write an equation for the circle that satisfies set of conditions: endpoints of a diameter at (11, 18) and (13, -19)
Micheal came up with a design of a rectangular patio that is 25 ft by 40 ft., and that is surrounded by a terraced strip of uniform width planted with small trees and shrubs. If the A of this terraced strip is
504f{t}^{2}
, find the width (x) of the strip.
y=\frac{x}{x-1}\left(x+3\right)
Point K is on line segment
\stackrel{―}{JL}
JK=2x-2,KL=x-9,\text{ }and\text{ }JL=2x+8,
determine the numerical length of
\stackrel{―}{KL}
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Compute the following derivatives \frac{d}{dx}[x^{4}\tan^{3}(x^{2})]
\frac{d}{dx}\left[{x}^{4}{\mathrm{tan}}^{3}\left({x}^{2}\right)\right]
y={x}^{4}{\mathrm{tan}}^{3}\left({x}^{2}\right)
{y}^{\prime }={x}^{4}×3{\mathrm{tan}}^{2}\left({x}^{2}\right)×{\mathrm{sec}}^{2}\left({x}^{2}\right)×2x+{\mathrm{tan}}^{3}\left({x}^{2}\right)×4{x}^{3}
Used chain rule which says derivative of
F(G(X))=F'(G'(X))
{y}^{\prime }=6{x}^{5}{\mathrm{tan}}^{2}\left({x}^{2}\right){\mathrm{sec}}^{2}\left({x}^{2}\right)+4{x}^{3}{\mathrm{tan}}^{3}\left({x}^{2}\right)
f\left(x\right)={\mathrm{cos}}^{3}{x}^{2}
How do you find the slope of the tangent line y=(x+1)(x-2) at x=1?
What is Lebniz Notation for the second derivative of y=f(x)?
How do you use the limit definition to find the derivative of y=-4x-5?
y=\frac{7{x}^{5}}{5}-2x+9{e}^{x}
\frac{dy}{dx}=
y=5x-2{x}^{3}
How do you find the points where the graph of the function
y=2x{\left(8-x\right)}^{5}
has horizontal tangents?
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Circles of Apollonius - Wikipedia
Any of several sets of circles associated with Apollonius of Perga
The circles of Apollonius are any of several sets of circles associated with Apollonius of Perga, a renowned Greek geometer. Most of these circles are found in planar Euclidean geometry, but analogs have been defined on other surfaces; for example, counterparts on the surface of a sphere can be defined through stereographic projection.
The main uses of this term are fivefold:
Apollonius showed that a circle can be defined as the set of points in a plane that have a specified ratio of distances to two fixed points, known as foci. This Apollonian circle is the basis of the Apollonius pursuit problem. It is a particular case of the first family described in #2.
The Apollonian circles are two families of mutually orthogonal circles. The first family consists of the circles with all possible distance ratios to two fixed foci (the same circles as in #1), whereas the second family consists of all possible circles that pass through both foci. These circles form the basis of bipolar coordinates.
The circles of Apollonius of a triangle are three circles, each of which passes through one vertex of the triangle and maintains a constant ratio of distances to the other two. The isodynamic points and Lemoine line of a triangle can be solved using these circles of Apollonius.
Apollonius' problem is to construct circles that are simultaneously tangent to three specified circles. The solutions to this problem are sometimes called the circles of Apollonius.
The Apollonian gasket—one of the first fractals ever described—is a set of mutually tangent circles, formed by solving Apollonius' problem iteratively.
1 Apollonius' definition of a circle
1.1 Proof using vectors in Euclidean spaces
1.2 Proof using the angle bisector theorem
1.3 Apollonius pursuit problem
2 Circles sharing a radical axis
3 Solutions to Apollonius' problem
4 Apollonian gasket
5 Isodynamic points of a triangle
Apollonius' definition of a circle[edit]
Main article: Circle § Circle of Apollonius
Figure 1. Apollonius' definition of a circle.
A circle is usually defined as the set of points P at a given distance r (the circle's radius) from a given point (the circle's center). However, there are other, equivalent definitions of a circle. Apollonius discovered that a circle could be defined as the set of points P that have a given ratio of distances k = d1/d2 to two given points (labeled A and B in Figure 1). These two points are sometimes called the foci.
Proof using vectors in Euclidean spaces[edit]
Let d1, d2 be non-equal positive real numbers. Let C be the internal division point of AB in the ratio d1 : d2 and D the external division point of AB in the same ratio, d1 : d2.
{\displaystyle {\overrightarrow {\mathrm {PC} }}={\frac {d_{2}{\overrightarrow {\mathrm {PA} }}+d_{1}{\overrightarrow {\mathrm {PB} }}}{d_{2}+d_{1}}},\ {\overrightarrow {\mathrm {PD} }}={\frac {d_{2}{\overrightarrow {\mathrm {PA} }}-d_{1}{\overrightarrow {\mathrm {PB} }}}{d_{2}-d_{1}}}.}
{\displaystyle \mathrm {PA} :\mathrm {PB} =d_{1}:d_{2}.}
{\displaystyle \Leftrightarrow d_{2}|{\overrightarrow {\mathrm {PA} }}|=d_{1}|{\overrightarrow {\mathrm {PB} }}|.}
{\displaystyle \Leftrightarrow d_{2}^{2}|{\overrightarrow {\mathrm {PA} }}|^{2}=d_{1}^{2}|{\overrightarrow {\mathrm {PB} }}|^{2}.}
{\displaystyle \Leftrightarrow (d_{2}{\overrightarrow {\mathrm {PA} }}+d_{1}{\overrightarrow {\mathrm {PB} }})\cdot (d_{2}{\overrightarrow {\mathrm {PA} }}-d_{1}{\overrightarrow {\mathrm {PB} }})=0.}
{\displaystyle \Leftrightarrow {\frac {d_{2}{\overrightarrow {\mathrm {PA} }}+d_{1}{\overrightarrow {\mathrm {PB} }}}{d_{2}+d_{1}}}\cdot {\frac {d_{2}{\overrightarrow {\mathrm {PA} }}-d_{1}{\overrightarrow {\mathrm {PB} }}}{d_{2}-d_{1}}}=0.}
{\displaystyle \Leftrightarrow {\overrightarrow {\mathrm {PC} }}\cdot {\overrightarrow {\mathrm {PD} }}=0.}
{\displaystyle \Leftrightarrow {\overrightarrow {\mathrm {PC} }}={\vec {0}}\vee {\overrightarrow {\mathrm {PD} }}={\vec {0}}\vee {\overrightarrow {\mathrm {PC} }}\perp {\overrightarrow {\mathrm {PD} }}.}
{\displaystyle \Leftrightarrow \mathrm {P} =\mathrm {C} \vee \mathrm {P} =\mathrm {D} \vee \angle {\mathrm {CPD} }=90^{\circ }.}
Therefore, the point P is on the circle which has the diameter CD.
Proof using the angle bisector theorem[edit]
Proof of Apollonius' definition of a circle
First consider the point
{\displaystyle C}
on the line segment between
{\displaystyle A}
{\displaystyle B}
, satisfying the ratio. By the definition
{\displaystyle {\frac {|AP|}{|BP|}}={\frac {|AC|}{|BC|}}}
and from the angle bisector theorem the angles
{\displaystyle \alpha =\angle APC}
{\displaystyle \beta =\angle CPB}
Next take the other point
{\displaystyle D}
on the extended line
{\displaystyle AB}
that satisfies the ratio. So
{\displaystyle {\frac {|AP|}{|BP|}}={\frac {|AD|}{|BD|}}.}
Also take some other point
{\displaystyle Q}
anywhere on the extended line
{\displaystyle AP}
. Also by the Angle bisector theorem the line
{\displaystyle PD}
bisects the exterior angle
{\displaystyle \angle QPB}
{\displaystyle \gamma =\angle BPD}
{\displaystyle \delta =\angle QPD}
{\displaystyle \beta +\gamma =90^{\circ }}
. Hence by Thales's theorem
{\displaystyle P}
lies on the circle which has
{\displaystyle CD}
as a diameter.
Apollonius pursuit problem[edit]
The Apollonius pursuit problem is one of finding whether a ship leaving from one point A at speed vA will intercept another ship leaving a different point B at speed vB. The minimum time in interception of the two ships is calculated by means of straight-line paths. If the ships' speeds are held constant, their speed ratio is defined by μ. If both ships collide or meet at a future point, I, then the distances of each are related by the equation:[1]
{\displaystyle a=\mu b}
Squaring both sides, we obtain:
{\displaystyle a^{2}=b^{2}\mu ^{2}}
{\displaystyle a^{2}=x^{2}+y^{2}}
{\displaystyle b^{2}=(d-x)^{2}+y^{2}}
{\displaystyle x^{2}+y^{2}=[(d-x)^{2}+y^{2}]\mu ^{2}}
{\displaystyle x^{2}+y^{2}=[d^{2}+x^{2}-2dx+y^{2}]\mu ^{2}}
{\displaystyle x^{2}+y^{2}=x^{2}\mu ^{2}+y^{2}\mu ^{2}+d^{2}\mu ^{2}-2dx\mu ^{2}}
Bringing to the left-hand side:
{\displaystyle x^{2}-x^{2}\mu ^{2}+y^{2}-y^{2}\mu ^{2}-d^{2}\mu ^{2}+2dx\mu ^{2}=0}
{\displaystyle x^{2}(1-\mu ^{2})+y^{2}(1-\mu ^{2})-d^{2}\mu ^{2}+2dx\mu ^{2}=0}
{\displaystyle 1-\mu ^{2}}
{\displaystyle x^{2}+y^{2}-{\frac {d^{2}\mu ^{2}}{1-\mu ^{2}}}+{\frac {2dx\mu ^{2}}{1-\mu ^{2}}}=0}
{\displaystyle \left(x+{\frac {d\mu ^{2}}{1-\mu ^{2}}}\right)^{2}-{\frac {d^{2}\mu ^{4}}{(1-\mu ^{2})^{2}}}-{\frac {d^{2}\mu ^{2}}{1-\mu ^{2}}}+y^{2}=0}
Bring non-squared terms to the right-hand side:
{\displaystyle {\begin{aligned}\left(x+{\frac {d\mu ^{2}}{1-\mu ^{2}}}\right)^{2}+y^{2}&={\frac {d^{2}\mu ^{4}}{(1-\mu ^{2})^{2}}}+{\frac {d^{2}\mu ^{2}}{1-\mu ^{2}}}\\&={\frac {d^{2}\mu ^{4}}{(1-\mu ^{2})^{2}}}+{\frac {d^{2}\mu ^{2}}{1-\mu ^{2}}}{\frac {(1-\mu ^{2})}{(1-\mu ^{2})}}\\&={\frac {d^{2}\mu ^{4}+d^{2}\mu ^{2}-d^{2}\mu ^{4}}{(1-\mu ^{2})^{2}}}\\&={\frac {d^{2}\mu ^{2}}{(1-\mu ^{2})^{2}}}\end{aligned}}}
{\displaystyle \left(x+{\frac {d\mu ^{2}}{1-\mu ^{2}}}\right)^{2}+y^{2}=\left({\frac {d\mu }{1-\mu ^{2}}}\right)^{2}}
Therefore, the point must lie on a circle as defined by Apollonius, with their starting points as the foci.
Circles sharing a radical axis[edit]
Main article: Apollonian circles
Figure 2. A set of Apollonian circles. Every blue circle intersects every red circle at a right angle, and vice versa. Every red circle passes through the two foci, which correspond to points A and B in Figure 1.
The circles defined by the Apollonian pursuit problem for the same two points A and B, but with varying ratios of the two speeds, are disjoint from each other and form a continuous family that cover the entire plane; this family of circles is known as a hyperbolic pencil. Another family of circles, the circles that pass through both A and B, are also called a pencil, or more specifically an elliptic pencil. These two pencils of Apollonian circles intersect each other at right angles and form the basis of the bipolar coordinate system. Within each pencil, any two circles have the same radical axis; the two radical axes of the two pencils are perpendicular, and the centers of the circles from one pencil lie on the radical axis of the other pencil.
Solutions to Apollonius' problem[edit]
Main article: Problem of Apollonius
Apollonius' problem may have up to eight solutions. The three given circles are shown in black, whereas the solution circles are colored.
In Euclidean plane geometry, Apollonius's problem is to construct circles that are tangent to three given circles in a plane.
Three given circles generically have eight different circles that are tangent to them and each solution circle encloses or excludes the three given circles in a different way: in each solution, a different subset of the three circles is enclosed.
Apollonian gasket[edit]
Main article: Apollonian gasket
Figure 4. A symmetrical Apollonian gasket, also called the Leibniz packing, after its inventor Gottfried Leibniz
By solving Apollonius' problem repeatedly to find the inscribed circle, the interstices between mutually tangential circles can be filled arbitrarily finely, forming an Apollonian gasket, also known as a Leibniz packing or an Apollonian packing.[2] This gasket is a fractal, being self-similar and having a dimension d that is not known exactly but is roughly 1.3,[3] which is higher than that of a regular (or rectifiable) curve (d = 1) but less than that of a plane (d = 2). The Apollonian gasket was first described by Gottfried Leibniz in the 17th century, and is a curved precursor of the 20th-century Sierpiński triangle.[4] The Apollonian gasket also has deep connections to other fields of mathematics; for example, it is the limit set of Kleinian groups;[5] see also Circle packing theorem.
Isodynamic points of a triangle[edit]
The circles of Apollonius may also denote three special circles
{\displaystyle {\mathcal {C}}_{1},{\mathcal {C}}_{2},{\mathcal {C}}_{3}}
defined by an arbitrary triangle
{\displaystyle \mathrm {A_{1}A_{2}A_{3}} }
{\displaystyle {\mathcal {C}}_{1}}
is defined as the unique circle passing through the triangle vertex
{\displaystyle \mathrm {A_{1}} }
that maintains a constant ratio of distances to the other two vertices
{\displaystyle \mathrm {A_{2}} }
{\displaystyle \mathrm {A_{3}} }
(cf. Apollonius' definition of the circle above). Similarly, the circle
{\displaystyle {\mathcal {C}}_{2}}
{\displaystyle \mathrm {A_{2}} }
{\displaystyle \mathrm {A_{1}} }
{\displaystyle \mathrm {A_{3}} }
, and so on for the circle
{\displaystyle {\mathcal {C}}_{3}}
All three circles intersect the circumcircle of the triangle orthogonally. All three circles pass through two points, which are known as the isodynamic points
{\displaystyle S}
{\displaystyle S^{\prime }}
of the triangle. The line connecting these common intersection points is the radical axis for all three circles. The two isodynamic points are inverses of each other relative to the circumcircle of the triangle.
The centers of these three circles fall on a single line (the Lemoine line). This line is perpendicular to the radical axis, which is the line determined by the isodynamic points.
Wikimedia Commons has media related to Circles of Apollonius.
^ Weintraub, Isaac; Garcia, Eloy; Pachter, Meir (2020). "Optimal guidance strategy for the defense of a non‐manoeuvrable target in 3‐dimensions". IET Control Theory & Applications. 14 (11): 1531–1538. doi:10.1049/iet-cta.2019.0541.
^ Kasner, E.; Supnick, F. (1943). "The Apollonian packing of circles". Proceedings of the National Academy of Sciences USA. 29 (11): 378–384. Bibcode:1943PNAS...29..378K. doi:10.1073/pnas.29.11.378. PMC 1078636. PMID 16588629.
^ Boyd, David W. (1973). "Improved Bounds for the Disk Packing Constants". Aequationes Mathematicae. 9: 99–106. doi:10.1007/BF01838194. S2CID 121089590.
Boyd, David W. (1973). "The Residual Set Dimension of the Apollonian Packing". Mathematika. 20 (2): 170–174. doi:10.1112/S0025579300004745.
McMullen, Curtis, T. (1998). "Hausdorff dimension and conformal dynamics III: Computation of dimension" (PDF). American Journal of Mathematics. 120 (4): 691–721. doi:10.1353/ajm.1998.0031. S2CID 15928775.
^ Mandelbrot, B. (1983). The Fractal Geometry of Nature. New York: W.H. Freeman. p. 170. ISBN 978-0-7167-1186-5.
Aste, T., and Weaire, D. (2008). The Pursuit of Perfect Packing (2nd ed.). New York: Taylor and Francis. pp. 131–138. ISBN 978-1-4200-6817-7. {{cite book}}: CS1 maint: multiple names: authors list (link)
^ Mumford, D., Series, C., and Wright, D. (2002). Indra's Pearls: The Vision of Felix Klein. Cambridge: Cambridge University Press. pp. 196–223. ISBN 0-521-35253-3. {{cite book}}: CS1 maint: multiple names: authors list (link)
Ogilvy, C.S. (1990) Excursions in Geometry, Dover. ISBN 0-486-26530-7.
Johnson, R.A. (1960) Advanced Euclidean Geometry, Dover.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Circles_of_Apollonius&oldid=1084577604"
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Monte Carlo Simulation of Conditional Variance Models - MATLAB & Simulink - MathWorks India
Conditional variance models specify the dynamic evolution of the variance of a process over time. Perform Monte Carlo simulation of conditional variance models by:
Specifying any required presample data (or use default presample data).
Generating the next conditional variance recursively using the specified conditional variance model.
Simulating the next innovation from the innovation distribution (Gaussian or Student’s t) using the current conditional variance.
For example, consider a GARCH(1,1) process without a mean offset,
{\epsilon }_{t}={\sigma }_{t}{z}_{t},
where zt either follows a standardized Gaussian or Student’s t distribution and
{\sigma }_{t}^{2}=\kappa +{\gamma }_{1}{\sigma }_{t-1}^{2}+{\alpha }_{1}{\epsilon }_{t-1}^{2}.
Suppose that the innovation distribution is Gaussian.
Given presample variance
{\sigma }_{0}^{2}
and presample innovation
{\epsilon }_{0},
realizations of the conditional variance and innovation process are recursively generated:
{\sigma }_{1}^{2}=\kappa +{\gamma }_{1}{\sigma }_{0}^{2}+{\alpha }_{1}{\epsilon }_{0}^{2}
{\epsilon }_{1}
from a Gaussian distribution with variance
{\sigma }_{1}^{2}
{\sigma }_{2}^{2}=\kappa +{\gamma }_{1}{\sigma }_{1}^{2}+{\alpha }_{1}{\epsilon }_{1}^{2}
{\epsilon }_{2}
{\sigma }_{2}^{2}
⋮
{\sigma }_{N}^{2}=\kappa +{\gamma }_{1}{\sigma }_{N-1}^{2}+{\alpha }_{1}{\epsilon }_{N-1}^{2}
{\epsilon }_{N}
{\sigma }_{N}^{2}
Random draws are generated from EGARCH and GJR models similarly, using the corresponding conditional variance equations.
\stackrel{^}{p}=\frac{#\text{\hspace{0.17em}}\text{\hspace{0.17em}}times\text{\hspace{0.17em}}\text{\hspace{0.17em}}event\text{\hspace{0.17em}}\text{\hspace{0.17em}}occurs\text{\hspace{0.17em}}\text{\hspace{0.17em}}in\text{\hspace{0.17em}}\text{\hspace{0.17em}}M\text{\hspace{0.17em}}\text{\hspace{0.17em}}draws}{M}.
se=\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{M}}.
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j=1..\mathrm{num}
\mathrm{with}\left(\mathrm{combstruct}\right):
u
\mathrm{eqns0}≔\mathrm{agfeqns}\left({T=\mathrm{Prod}\left(Z,\mathrm{Set}\left(T\right)\right)},{h\left(T\right)=\mathrm{Prod}\left(0,\mathrm{Set}\left(h\left(T\right)\right)+1\right)},\mathrm{labeled},z,[[u,h]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{eqns0}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{u}\right)}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{Z}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}]
\mathrm{eqns1}≔\mathrm{gfeqns}\left({E=\mathrm{Ε},T=\mathrm{Prod}\left(Z,\mathrm{Prod}\left(E,\mathrm{Set}\left(T\right)\right)\right)},\mathrm{labeled},z,[[u,E]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{eqns1}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{u}\right)}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{Z}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{z}]
u
u
\mathrm{agfmomentsolve}\left(\mathrm{eqns0},0\right)
{\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{Z}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{z}}
\mathrm{agfmomentsolve}\left(\mathrm{eqns0},1,\mathrm{new}\right)
{{\textcolor[rgb]{0,0,1}{T}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\frac{{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}}{\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{T}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}}{\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{Z}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{Z}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{z}}
\mathrm{new}
[{\textcolor[rgb]{0,0,1}{T}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{T}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{T}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{T}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{Z}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{Z}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{z}]
{z}^{n}
T\left(z\right)
\mathrm{coeff}\left(T\left(z\right),z,n\right)
nodes, and
\mathrm{coeff}\left({T}_{2}\left(z\right),z,n\right)
nodes for the number of internal nodes. The average number of internal nodes for a tree on
\frac{\mathrm{coeff}\left({T}_{2}\left(z\right),z,n\right)}{\mathrm{coeff}\left(T\left(z\right),z,n\right)}
\mathrm{sol}≔\mathrm{agfmomentsolve}\left(\mathrm{eqns0},2\right):
\mathrm{subs}\left(\mathrm{sol},T[2,2]\left(z\right)\right)
\frac{{\left({\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\right)}{{\left({\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\right)}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}{\left({\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{LambertW}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\right)}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}
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A triangular number is defined as a number which has the property of being expressed as a sum of consecutive - Maths - Real Numbers - 11405225 | Meritnation.com
A triangular number is defined as a number which has the property of being expressed as a sum of consecutive numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of two consecutive natural numbers?
We know " This sequence comes from a pattern of dots that form a triangle "
Triangular numbers : 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666 ...
So, nth triangular number =
\frac{\mathrm{n} \left(\mathrm{n} + 1 \right)}{2}
, So 44th triangular number =
\frac{44 \left( 44+ 1 \right)}{2} = 22 × 45
= 990 after that we get triangular number greater than 1000 . and we can see that there are same number of odd and even numbers in triangular number , So in 44 triangular number 22 odd number and 22 even numbers are there .
And we know difference of squares of two consecutive natural numbers : 3 , 5 , 7 , ... all odd number
So, from above condition we can say there are 21 triangular numbers that can satisfied the given condition ( Here we neglect number = 1 as that is not a difference of consecutive natural number ) . ( Ans )
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Applications of Percentages | Brilliant Math & Science Wiki
A percentage is a number that represents the fractional part out of 100 (per cent literally means per one hundred). Thus 87% means
\frac{87}{100}
0.87.
Because percentages are always out of 100, they are generally simpler to use than fractions when comparing quantities. Common uses of percentages include taxes, discounts, markups, profits, losses, and interest rates.
Tips, Taxes, and Discounts
When people tip others for their service, they generally use a percentage to calculate the amount to tip. For example, if you decide to tip 20% on a $50 meal, you would leave a
(0.20)(\$50) = \$10
tip.
Likewise, taxes and discounts are often percentage-based. For example, sales tax in a city might be 8% and a store might have a "40% OFF" sale.
We can calculate tips, taxes, and discounts by calculating percentages of quantities and then adding or subtracting as necessary.
Drew buys $70 worth of new clothing. If he must pay a 6% sales tax, what will his total be?
6% of $70 is
(0.06)(\$70) = \$4.20.
Therefore, Drew's total bill will be
\$70 + \$4.20 = \$74.20.
Anya gets a 30% sale discount and 10% club discount for the $200 jacket that she would like to purchase. How much does she have to pay for the jacket?
To begin, Anya gets a 30% discount, so she is paying 70% of the cost of the jacket, or
(0.7)(\$200) = \$140.
Next, Anya gets a 10% discount, so she is paying 90% of the remaining cost of the jacket, or
(0.9)(\$140) = \$126.
Department store Big mart Retail shop All the same
Magan is going to buy a new skirt. But the skirt that she wants to buy is offered at different prices and discount rates depending on the shops. The list prices and discount rates offered are shown in the table below. Choose the shop where she can buy it at the lowest price.
Percentage change is the measure of change in a quantity expressed as a percent.
Percentage change is
\frac {\text{amount of change}} {\text{ original value}} \times 100 \%.
The price of a sweatshirt changed from
\$20
\$30
. Find the percentage change in the price of the sweatshirt.
The change in price of the sweatshirt is
\$10,
and the original price of the sweatshirt was
\$20.
Therefore, the percentage change in the price of the sweatshirt is
\frac{10}{20} × 100 \% = 50\%.
Imagine a city with 10,000 people at the beginning of a year. Given that the population at the end of that year is 9,500, compute the percentage change of the population.
The population changed by
10000-9500 = 500
The percent change in the population is
\frac{500}{10000} \times 100 \% = 5\%.
The price of a lunch is
\$10
10
10% of $10 is $1, so the new price is
\$10 +\$1 = \$11.
1% 10% 9 % 2%
If a man's wages is increased by 10% and afterwards decreased by 10%, what is the total decrease of his wages?
Cite as: Applications of Percentages. Brilliant.org. Retrieved from https://brilliant.org/wiki/applications-of-percentages/
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FHSST Physics/Atom/Successive Ionization Energies - Wikibooks, open books for an open world
FHSST Physics/Atom/Successive Ionization Energies
Successive ionization energies to provide evidence for arrangement of electrons into core and valence[edit | edit source]
[Brink and Jones sections: de Broglie - matter shows particle and wave characteristics, proved by Davisson and Germer. Shroedinger and Heisenberg developed this model into quantum mechanics]
The nucleus (atomic nucleus) is the center of an atom. It is composed of one or more protons and usually some neutrons as well. The number of protons in an atom's nucleus is called the atomic number, and determines which element the atom is (for example hydrogen, carbon, oxygen, etc.).
The discovery of the electron was the first indication that the atom had internal structure. This structure was initially imagined according to the "raisin cookie" or "plum pudding" model, in which the small, negatively charged electrons were embedded in a large sphere containing all the positive charge. Ernest Rutherford and Marsden, however, discovered in 1911 that alpha particles from a radium source were sometimes scattered backwards from a gold foil, indicating that there must be a dense concentration of positive charge at the center of the atom. This discovery led to the acceptance of a planetary model, in which the electrons orbited a tiny nucleus in the same way that the planets orbit the sun.
Interesting Fact: The word atom is derived from the Greek atomos, indivisible, from a-, not, and tomos, a cut.
An atom is the smallest portion into which a chemical element can be divided while still retaining its properties. Atoms are the basic constituents of molecules and ordinary matter. Atoms are composed of subatomic particles.
Atoms are composed mostly of empty space, but also of smaller subatomic particles. At the center of the atom is a tiny positive nucleus composed of nucleons (protons and neutrons). The rest of the atom contains only the fairly flexible electron shells. Usually atoms are electrically neutral with as many electrons as protons.
Atoms are generally classified by their atomic number, which corresponds to the number of protons in the atom. For example, carbon atoms are those atoms containing 6 protons. All atoms with the same atomic number share a wide variety of physical properties and exhibit the same chemical behavior. The various kinds of atoms are listed in the Periodic table. Atoms having the same atomic number, but different atomic masses (due to their different numbers of neutrons), are called isotopes.
The simplest atom is the hydrogen atom, having atomic number 1 and consisting of one proton and one electron. It has been the subject of much interest in science, particularly in the early development of quantum theory.
The chemical behavior of atoms is largely due to interactions between the electrons. In particular the electrons in the outermost shell, called the valence electrons, have the greatest influence on chemical behavior. Core electrons (those not in the outer shell) play a role, but it is usually in terms of a secondary effect due to screening of the positive charge in the atomic nucleus.
There is a strong tendency for atoms to completely fill (or empty) the outer electron shell, which in hydrogen and helium has space for two electrons, and in all other atoms has space for eight. This is achieved either by sharing electrons with neighboring atoms or by completely removing electrons from other atoms. When electrons are shared a covalent bond is formed between the two atoms. Covalent bonds are the strongest type of atomic bond.
When one or more electrons are completely removed from one atom by another, ions are formed. Ions are atoms that possess a net charge due to an imbalance in the number of protons and electrons. The ion that stole the electron(s) is called an anion and is negatively charged. The atom that lost the electron(s) is called a cation and is positively charged. Cations and anions are attracted to each other due to coulombic forces between the positive and negative charges. This attraction is called ionic bonding and is weaker than covalent bonding.
As mentioned above covalent bonding implies a state in which electrons are shared equally between atoms, while ionic bonding implies that the electrons are completely confined to the anion. Except for a limited number of extreme cases, neither of these pictures is completely accurate. In most cases of covalent bonding, the electron is unequally shared, spending more time around the more electronegative atom, resulting in the covalent bond having some ionic character. Similarly, in ionic bonding the electrons often spend a small fraction of time around the more electropositive atom, resulting in some covalent character for the ionic bond.
Democritus' shaped-atom model (for want of a better name)
The Plum pudding model of the atom was made after the discovery of the electron but before the discovery of the proton or neutron. In it, the atom is envisioned as electrons surrounded by a soup of positive charge, like plums surrounded by pudding. This model was disproved by an experiment by Ernest Rutherford when he discovered the nucleus of the atom.
The Bohr Model is a physical model that depicts the atom as a small positively charged nucleus with electrons in orbit at different levels, similar in structure to the solar system. Because of its simplicity, the Bohr model is still commonly used and taught today.
In the early part of the 20th century, experiments by Ernest Rutherford and others had established that atoms consisted of a small dense positively charged nucleus surrounded by orbiting negatively charged electrons. However, theoretical predictions and experimental evidence quickly began to cast doubt on the classical orbital model.
The simplest possible atom is hydrogen, which consists of a positively charged proton and a negatively charged orbiting electron. According to the orbital model of the atom, since the nucleus (the proton) and the electron are oppositely charged they will attract one another by coulomb force, in the same way that the Sun attracts the Earth with the gravitational force. However, if the electron orbits the nucleus in a classical orbit, it is undergoing a central acceleration, and ought to emit electromagnetic radiation (light) according to well established theories of electromagnetism. If the orbiting electron radiates as it orbits, it must lose energy and therefore spiral into the nucleus in a very short time, which clearly does not occur.
Also, it was well known at the time that atoms emit light only at certain frequencies - the frequencies at which an atom can radiate are known as the spectra of the atom. It was clear that the emission of light by atoms was due to electrons radiating. However, if the electrons are simply orbiting the nucleus classically like planets, there is no reason for the spectra - one would expect the electrons to emit radiation in a continuous frequency spectrum. It soon became clear that something was very wrong with the orbital description of the atom.
These difficulties were resolved in 1913 by Niels Bohr who proposed that:
(1) The orbiting electrons existed in orbits that had discrete quantized energies. That is, not every orbit is possible but only certain specific ones. The exact energies of the allowed orbits depends on the atom in question.
(2) The laws of classical mechanics do not apply when electrons make the jump from one allowed orbit to another.
(3) When an electron makes a jump from one orbit to another the energy difference is carried off (or supplied) by a single quantum of light (called a photon) which has a frequency that directly depends on the energy difference between the two orbitals.
{\displaystyle f=E/h}
where f is the frequency of the photon, E the energy difference, and h is a constant of proportionality known as Planck's constant. Defining we can write
where ?? is the angular frequency of the photon.
(4) The allowed orbits depend on quantized (discrete) values of orbital angular momentum, L according to the equation
Where n = 1,2,3, and is called the angular momentum quantum number.
These assumptions explained many of the observations seen at the time, such as why spectra consist of discrete lines. Assumption 4) states that the lowest value of n is 1. This corresponds to a smallest possible radius (for the mathematics see Ohanian-principles of physics or any of the large, usually American, college introductory physics textbooks) of 0.0529 nm. This is known as the Bohr radius, and explains why atoms are stable. Once an electron is in the lowest orbit, it can go no further. It cannot emit any more light because it would need to go into a lower orbit, but it can't do that if it is already in the lowest allowed orbit.
The Bohr model is sometimes known as the semiclassical model because although it does include some ideas of quantum mechanics it is not a full quantum mechanical description of the atom. Assumption 2) states that the laws of classical mechanics don't apply during a quantum jump but doesn't state what laws should replace classical mechanics. Assumption 4) states that angular momentum is quantised but does not explain why.
In order to fully describe an atom we need to use the full theory of quantum mechanics, which was worked out by a number of people in the years following the Bohr model. This theory treats the electrons as waves, which create 3D standing wave patterns in the atom. (This is why quantum mechanics is sometimes called wave mechanics.) This theory considers that idea of electrons as being little billiard ball like particles that travel round in orbits as absurdly wrong instead electrons form probability clouds. You might find the electron here with a certain probability; you might find it over there with a different probability. However it is interesting to note that if you work out the average radius of an electron in the lowest possible energy state it turns out to be exactly equal to the Bohr radius (although it takes many more pages of mathematics to work it out).
The full quantum mechanics theory is a beautiful theory that has been experimentally tested and found to be incredibly accurate, however it is mathematically much more advanced, and often using the much simpler Bohr model will get you the results with much less hassle. The thing to remember is that it is only a model, an aid to understanding. Atoms are not really little solar systems.
See also: Hydrogen atom, quantum mechanics, Schrö dinger equation, Niels Bohr.
An interactive demonstration http://webphysics.davidson.edu/faculty/dmb/hydrogen/ of the probability clouds of electron in Hydrogen atom according to the full QM solution.
Retrieved from "https://en.wikibooks.org/w/index.php?title=FHSST_Physics/Atom/Successive_Ionization_Energies&oldid=3528358"
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Mahler_measure Knowpia
In mathematics, the Mahler measure
{\displaystyle M(p)}
{\displaystyle p(z)}
with complex coefficients is defined as
{\displaystyle M(p)=|a|\prod _{|\alpha _{i}|\geq 1}|\alpha _{i}|=|a|\prod _{i=1}^{n}\max\{1,|\alpha _{i}|\},}
{\displaystyle p(z)}
factorizes over the complex numbers
{\displaystyle \mathbb {C} }
{\displaystyle p(z)=a(z-\alpha _{1})(z-\alpha _{2})\cdots (z-\alpha _{n}).}
The Mahler measure can be viewed as a kind of height function. Using Jensen's formula, it can be proved that this measure is also equal to the geometric mean of
{\displaystyle |p(z)|}
{\displaystyle z}
on the unit circle (i.e.,
{\displaystyle |z|=1}
{\displaystyle M(p)=\exp \left(\int _{0}^{1}\ln(|p(e^{2\pi i\theta })|)\,d\theta \right).}
By extension, the Mahler measure of an algebraic number
{\displaystyle \alpha }
is defined as the Mahler measure of the minimal polynomial of
{\displaystyle \alpha }
{\displaystyle \mathbb {Q} }
{\displaystyle \alpha }
is a Pisot number or a Salem number, then its Mahler measure is simply
{\displaystyle \alpha }
The Mahler measure is named after the German-born Australian mathematician Kurt Mahler.
The Mahler measure is multiplicative:
{\displaystyle \forall p,q,\,\,M(p\cdot q)=M(p)\cdot M(q).}
{\textstyle M(p)=\lim _{\tau \to 0}\|p\|_{\tau }}
{\textstyle \,\|p\|_{\tau }=\left(\int _{0}^{1}|p(e^{2\pi i\theta })|^{\tau }d\theta \right)^{1/\tau }}
{\displaystyle L_{\tau }}
{\displaystyle p}
Kronecker's Theorem: If
{\displaystyle p}
is an irreducible monic integer polynomial with
{\displaystyle M(p)=1}
{\displaystyle p(z)=z,}
{\displaystyle p}
is a cyclotomic polynomial.
(Lehmer's conjecture) There is a constant
{\displaystyle \mu >1}
{\displaystyle p}
is an irreducible integer polynomial, then either
{\displaystyle M(p)=1}
{\displaystyle M(p)>\mu }
The Mahler measure of a monic integer polynomial is a Perron number.
Higher-dimensional Mahler measureEdit
The Mahler measure
{\displaystyle M(p)}
of a multi-variable polynomial
{\displaystyle p(x_{1},\ldots ,x_{n})\in \mathbb {C} [x_{1},\ldots ,x_{n}]}
is defined similarly by the formula[2]
{\displaystyle M(p)=\exp \left(\int _{0}^{1}\int _{0}^{1}\cdots \int _{0}^{1}\log {\Bigl (}{\bigl |}p(e^{2\pi i\theta _{1}},e^{2\pi i\theta _{2}},\ldots ,e^{2\pi i\theta _{n}}){\bigr |}{\Bigr )}\,d\theta _{1}\,d\theta _{2}\cdots d\theta _{n}\right).}
It inherits the above three properties of the Mahler measure for a one-variable polynomial.
The multi-variable Mahler measure has been shown, in some cases, to be related to special values of zeta-functions and
{\displaystyle L}
-functions. For example, in 1981, Smyth[3] proved the formulas
{\displaystyle m(1+x+y)={\frac {3{\sqrt {3}}}{4\pi }}L(\chi _{-3},2)}
{\displaystyle L(\chi _{-3},s)}
is the Dirichlet L-function, and
{\displaystyle m(1+x+y+z)={\frac {7}{2\pi ^{2}}}\zeta (3),}
{\displaystyle \zeta }
is the Riemann zeta function. Here
{\displaystyle m(P)=\log M(P)}
is called the logarithmic Mahler measure.
Some results by Lawton and BoydEdit
From the definition, the Mahler measure is viewed as the integrated values of polynomials over the torus (also see Lehmer's conjecture). If
{\displaystyle p}
vanishes on the torus
{\displaystyle (S^{1})^{n}}
, then the convergence of the integral defining
{\displaystyle M(p)}
is not obvious, but it is known that
{\displaystyle M(p)}
does converge and is equal to a limit of one-variable Mahler measures,[4] which had been conjectured by Boyd.[5][6]
This is formulated as follows: Let
{\displaystyle \mathbb {Z} }
denote the integers and define
{\displaystyle \mathbb {Z} _{+}^{N}=\{r=(r_{1},\dots ,r_{N})\in \mathbb {Z} ^{N}:r_{j}\geq 0\ {\text{for}}\ 1\leq j\leq N\}}
{\displaystyle Q(z_{1},\dots ,z_{N})}
{\displaystyle N}
{\displaystyle r=(r_{1},\dots ,r_{N})\in \mathbb {Z} _{+}^{N}}
define the polynomial
{\displaystyle Q_{r}(z)}
of one variable by
{\displaystyle Q_{r}(z):=Q(z^{r_{1}},\dots ,z^{r_{N}})}
{\displaystyle q(r)}
{\displaystyle q(r):=\min \left\{H(s):s=(s_{1},\dots ,s_{N})\in \mathbb {Z} ^{N},s\neq (0,\dots ,0)~{\text{and}}~\sum _{j=1}^{N}s_{j}r_{j}=0\right\}}
{\displaystyle H(s)=\max\{|s_{j}|:1\leq j\leq N\}}
Theorem (Lawton) — Let
{\displaystyle Q(z_{1},\dots ,z_{N})}
be a polynomial in N variables with complex coefficients. Then the following limit is valid (even if the condition that
{\displaystyle r_{i}\geq 0}
is relaxed):
{\displaystyle \lim _{q(r)\to \infty }M(Q_{r})=M(Q)}
Boyd's proposalEdit
Boyd provided more general statements than the above theorem. He pointed out that the classical Kronecker's theorem, which characterizes monic polynomials with integer coefficients all of whose roots are inside the unit disk, can be regarded as characterizing those polynomials of one variable whose measure is exactly 1, and that this result extends to polynomials in several variables.[6]
Define an extended cyclotomic polynomial to be a polynomial of the form
{\displaystyle \Psi (z)=z_{1}^{b_{1}}\dots z_{n}^{b_{n}}\Phi _{m}(z_{1}^{v_{1}}\dots z_{n}^{v_{n}}),}
{\displaystyle \Phi _{m}(z)}
is the m-th cyclotomic polynomial, the
{\displaystyle v_{i}}
are integers, and the
{\displaystyle b_{i}=\max(0,-v_{i}\deg \Phi _{m})}
are chosen minimally so that
{\displaystyle \Psi (z)}
is a polynomial in the
{\displaystyle z_{i}}
{\displaystyle K_{n}}
be the set of polynomials that are products of monomials
{\displaystyle \pm z_{1}^{c_{1}}\dots z_{n}^{c_{n}}}
and extended cyclotomic polynomials.
Theorem (Boyd) — Let
{\displaystyle F(z_{1},\dots ,z_{n})\in \mathbb {Z} [z_{1},\ldots ,z_{n}]}
be a polynomial with integer coefficients. Then
{\displaystyle M(F)=1}
{\displaystyle F}
{\displaystyle K_{n}}
This led Boyd to consider the set of values
{\displaystyle L_{n}:={\bigl \{}m(P(z_{1},\dots ,z_{n})):P\in \mathbb {Z} [z_{1},\dots ,z_{n}]{\bigr \}},}
and the union
{\textstyle {L}_{\infty }=\bigcup _{n=1}^{\infty }L_{n}}
. He made the far-reaching conjecture[5] that the set of
{\displaystyle {L}_{\infty }}
{\displaystyle \mathbb {R} }
. An immediate consequence of this conjecture would be the truth of Lehmer's conjecture, albeit without an explicit lower bound. As Smyth's result suggests that
{\displaystyle L_{1}\subsetneqq L_{2}}
, Boyd further conjectures that
{\displaystyle L_{1}\subsetneqq L_{2}\subsetneqq L_{3}\subsetneqq \ \cdots .}
Mahler measure and entropyEdit
An action
{\displaystyle \alpha _{M}}
{\displaystyle \mathbb {Z} ^{n}}
by automorphisms of a compact metrizable abelian group may be associated via duality to any countable module
{\displaystyle N}
{\displaystyle R=\mathbb {Z} [z_{1}^{\pm 1},\dots ,z_{n}^{\pm 1}]}
.[7] The topological entropy (which is equal to the measure-theoretic entropy) of this action,
{\displaystyle h(\alpha _{N})}
, is given by a Mahler measure (or is infinite).[8] In the case of a cyclic module
{\displaystyle M=R/\langle F\rangle }
for a non-zero polynomial
{\displaystyle F(z_{1},\dots ,z_{n})\in \mathbb {Z} [z_{1},\ldots ,z_{n}]}
the formula proved by Lind, Schmidt, and Ward gives
{\displaystyle h(\alpha _{N})=\log M(F)}
, the logarithmic Mahler measure of
{\displaystyle F}
. In the general case, the entropy of the action is expressed as a sum of logarithmic Mahler measures over the generators of the principal associated prime ideals of the module. As pointed out earlier by Lind in the case
{\displaystyle n=1}
of a single compact group automorphism, this means that the set of possible values of the entropy of such actions is either all of
{\displaystyle [0,\infty ]}
or a countable set depending on the solution to Lehmer's problem. Lind also showed that the infinite-dimensional torus
{\displaystyle \mathbb {T} ^{\infty }}
either has ergodic automorphisms of finite positive entropy or only has automorphisms of infinite entropy depending on the solution to Lehmer's problem.[9]
^ Although this is not a true norm for values of
{\displaystyle \tau <1}
^ Schinzel 2000, p. 224.
^ a b Boyd 1981a.
^ a b Boyd 1981b.
^ Kitchens, Bruce; Schmidt, Klaus (1989). "Automorphisms of compact groups". Ergodic Theory Dynam. Systems. 9 (4): 691–735. doi:10.1017/S0143385700005290.
^ Lind, Douglas; Schmidt, Klaus; Ward, Tom (1990). "Mahler measure and entropy for commuting automorphisms of compact groups". Invent. Math. 101: 593–629. doi:10.1007/BF01231517.
^ Lind, Douglas (1977). "The structure of skew products with ergodic group automorphisms". Israel J. Math. 28, no. 3: 205–248. doi:10.1007/BF02759810.
Borwein, Peter (2002). Computational Excursions in Analysis and Number Theory. CMS Books in Mathematics. Vol. 10. Springer. pp. 3, 15. ISBN 978-0-387-95444-8. Zbl 1020.12001.
Boyd, David (1981a). "Speculations concerning the range of Mahler's measure". Canadian Mathematical Bulletin. 24 (4): 453–469. doi:10.4153/cmb-1981-069-5.
Boyd, David (1981b). "Kronecker's Theorem and Lehmer's Problem for Polynomials in Several Variables". Journal of Number Theory. 13: 116–121. doi:10.1016/0022-314x(81)90033-0.
Boyd, David (2002a). "Mahler's measure and invariants of hyperbolic manifolds". In Bennett, M. A. (ed.). Number theory for the Millenium. A. K. Peters. pp. 127–143.
Boyd, David (2002b). "Mahler's measure, hyperbolic manifolds and the dilogarithm". Canadian Mathematical Society Notes. 34 (2): 3–4, 26–28.
Boyd, David; Rodriguez Villegas, F. (2002). "Mahler's measure and the dilogarithm, part 1". Canadian Journal of Mathematics. 54 (3): 468–492. doi:10.4153/cjm-2002-016-9. S2CID 10069657.
Brunault, François (2020). Many variations of Mahler measures : a lasting symphony. Cambridge, United Kingdom New York, NY: Cambridge University Press. ISBN 978-1-108-79445-9. OCLC 1155888228.
Everest, Graham and Ward, Thomas (1999). "Heights of polynomials and entropy in algebraic dynamics". Springer-Verlag London, Ltd., London. xii+211 pp. ISBN: 1-85233-125-9
"Mahler measure", Encyclopedia of Mathematics, EMS Press, 2001 [1994] .
Jensen, J.L. (1899). "Sur un nouvel et important théorème de la théorie des fonctions". Acta Mathematica. 22: 359–364. doi:10.1007/BF02417878. JFM 30.0364.02.
Knuth, Donald E. (1997). "4.6.2 Factorization of Polynomials". Seminumerical Algorithms. The Art of Computer Programming. Vol. 2 (3rd ed.). Addison-Wesley. pp. 439–461, 678–691. ISBN 978-0-201-89684-8.
Lawton, Wayne M. (1983). "A problem of Boyd concerning geometric means of polynomials". Journal of Number Theory. 16 (3): 356–362. doi:10.1016/0022-314X(83)90063-X. Zbl 0516.12018.
Mossinghoff, M.J. (1998). "Polynomials with Small Mahler Measure". Mathematics of Computation. 67 (224): 1697–1706. doi:10.1090/S0025-5718-98-01006-0. Zbl 0918.11056.
Schinzel, Andrzej (2000). Polynomials with special regard to reducibility. Encyclopedia of Mathematics and Its Applications. Vol. 77. Cambridge University Press. ISBN 978-0-521-66225-3. Zbl 0956.12001.
Smyth, Chris (2008). "The Mahler measure of algebraic numbers: a survey". In McKee, James; Smyth, Chris (eds.). Number Theory and Polynomials. London Mathematical Society Lecture Note Series. Vol. 352. Cambridge University Press. pp. 322–349. ISBN 978-0-521-71467-9. Zbl 1334.11081.
Mahler Measure on MathWorld
Jensen's Formula on MathWorld
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Digital Signal Processing/Z Transform - Wikibooks, open books for an open world
Digital Signal Processing/Z Transform
The Z Transform has a strong relationship to the DTFT, and is incredibly useful in transforming, analyzing, and manipulating discrete calculus equations. The Z transform is named such because the letter 'z' (a lower-case Z) is used as the transformation variable.
1 z Transform Definition
3 Equivalence to DTFT
6 Z-Plane
z Transform Definition[edit | edit source]
For a given sequence x[n], we can define the z-transform X(z) as such:
[Z Transform]
{\displaystyle {\mathcal {Z}}(x[n])=X(z)=\sum _{n=-\infty }^{\infty }x[n]z^{-n}}
it is important to note that z is a continuous complex variable defined as such:
{\displaystyle z=Re(z)+jIm(z)}
{\displaystyle j={\sqrt {-1}}}
There can be several sequences
{\displaystyle x[n]}
which will generate the same z-transform
{\displaystyle X(z)}
with the different functions being differentiated by the convergence region of
{\displaystyle z}
for which the summation in the z-transform will converge. These convergence regions are annuli centered at the orgin. In a given convergence region, only one
{\displaystyle x[n]}
will converge to a given
{\displaystyle X(z)}
{\displaystyle x[n]=\left\{{\begin{array}{ll}0&n<0\\a^{n}&n\geq 0\end{array}}\right.}
{\displaystyle X(z)=\sum _{n=-\infty }^{\infty }x[n]z^{-n}=\sum _{n=0}^{\infty }a^{n}z^{-n}={\frac {1}{1-az^{-1}}}}
{\displaystyle \left|z\right|>\left|a\right|}
{\displaystyle x[n]=\left\{{\begin{array}{ll}-a^{n}&n<0\\0&n\geq 0\end{array}}\right.}
{\displaystyle X(z)=\sum _{n=-\infty }^{\infty }x[n]z^{-n}=\sum _{n=-\infty }^{-1}-a^{n}z^{-n}=-\sum _{n=1}^{\infty }a^{-n}z^{n}=-\left({\frac {1}{1-za^{-1}}}-1\right)={\frac {1}{1-az^{-1}}}}
{\displaystyle \left|z\right|<\left|a\right|}
Note that both examples have the same function
{\displaystyle X(z)}
as their z-transforms but with different convergence regions needed for the respective infinite summations in their z-transforms to converge. Many textbooks on z-transforms are only concerned with so-called right-handed functions which is to say functions
{\displaystyle x[n]}
{\displaystyle x[n]=0}
{\displaystyle n}
less than some initial start point
{\displaystyle N_{i}}
{\displaystyle x[n]=0}
{\displaystyle n<N-i}
. So long as the function grows at most exponentially after the start point, the z-transform of these so-called right-handed functions will converge in an open annulus going to infinity,
{\displaystyle \left|z\right|>R}
for some positive real
{\displaystyle R}
It is important to note that the z-transform rarely needs to be computed manually, because many common results have already been tabulated extensively in tables, and control system software includes it (MatLab,Octave,SciLab).
The z-transform is actually a special case of the so-called Laurent series, which is a special case of the commonly used Taylor series.
The Inverse Z-Transform[edit | edit source]
The inverse z-transform can be defined as such:
[Inverse Z Transform]
{\displaystyle x[n]={\frac {1}{2\pi j}}\oint _{C}X(z)z^{n-1}dz}
Where C is a closed-contour that lies inside the unit circle on the z-plane, and encircles the point z = {0, 0}.
The inverse z-transform is mathematically very complicated, but luckily—like the z-transform itself—the results are extensively tabulated in tables.
Equivalence to DTFT[edit | edit source]
{\displaystyle z=e^{j\omega }}
{\displaystyle \omega }
is the frequency in radians per second, into the Z-transform, we get
{\displaystyle Z(e^{j\omega })=\sum _{n=-\infty }^{\infty }x[n]e^{-j\omega n}}
which is equivalent to the definition of the Discrete-Time Fourier Transform. In other words, to convert from the Z-transform to the DTFT, we need to evaluate the Z-transform around the unit circle.
Since the z-transform is equivalent to the DTFT, the z-transform has many of the same properties. Specifically, the z-transform has the property of duality, and it also has a version of the convolution theorem (discussed later).
The z-transform is a linear operator.
Convolution Theorem[edit | edit source]
Since the Z-transform is equivalent to the DTFT, it also has a convolution theorem that is worth stating explicitly:
Multiplication in the discrete-time domain becomes convolution in the z-domain. Multiplication in the z-domain becomes convolution in the discrete-time domain.
Y(s)=X(s).H(s)
Z-Plane[edit | edit source]
Since the variable z is a continuous, complex variable, we can map the z variable to a complex plane as such:
Let's say we have a system with an input/output relationship defined as such:
Y(z) = H(z)X(z)
We can define the transfer function of the system as being the term H(z). If we have a basic transfer function, we can break it down into parts:
{\displaystyle H(z)={\frac {N(z)}{D(z)}}}
Where H(z) is the transfer function, N(z) is the numerator of H(z) and D(z) is the denominator of H(z). If we set N(z)=0, the solutions to that equation are called the zeros of the transfer function. If we set D(z)=0, the solutions to that equation are called the poles of the transfer function.
The poles of the transfer function amplify the frequency response while the zero's attenuate it. This is important because when you design a filter you can place poles and zero's on the unit circle and quickly evaluate your filters frequency response.
{\displaystyle Y(z)={\frac {z^{-1}}{(z^{-1}+2)}}X(z)}
So by dividing through by X(z), we can show that the transfer function is defined as such:
{\displaystyle H(z)={\frac {z^{-1}}{(z^{-1}+2)}}}
We can also find the D(z) and N(z) equations as such:
{\displaystyle N(z)=z^{-1}}
{\displaystyle D(z)=(z^{-1}+2)}
And from those equations, we can find the poles and zeros:
z → -1/2
It can be shown that for any causal system with a transfer function H(z), all the poles of H(z) must lie within the unit-circle on the z-plane for the system to be stable. Zeros of the transfer function may lie inside or outside the circle. See Control Systems/Jurys Test.
Gain[edit | edit source]
Gain is the factor by which the output magnitude is different from the input magnitude. If the input magnitude is the same as the output magnitude at a given frequency, the filter is said to have "unity gain".
Here is a listing of the most common properties of the Z transform.
{\displaystyle x[n]={\mathcal {Z}}^{-1}\{X(z)\}}
{\displaystyle X(z)={\mathcal {Z}}\{x[n]\}}
{\displaystyle r_{2}<|z|<r_{1}\ }
{\displaystyle a_{1}x_{1}[n]+a_{2}x_{2}[n]\ }
{\displaystyle a_{1}X_{1}(z)+a_{2}X_{2}(z)\ }
{\displaystyle x[n-k]\ }
{\displaystyle z^{-k}X(z)\ }
{\displaystyle z=0\ }
{\displaystyle k>0\,}
{\displaystyle z=\infty }
{\displaystyle k<0\ }
{\displaystyle a^{n}x[n]\ }
{\displaystyle X(a^{-1}z)\ }
{\displaystyle |a|r_{2}<|z|<|a|r_{1}\ }
{\displaystyle x[-n]\ }
{\displaystyle X(z^{-1})\ }
{\displaystyle {\frac {1}{r_{2}}}<|z|<{\frac {1}{r_{1}}}\ }
{\displaystyle x^{*}[n]\ }
{\displaystyle X^{*}(z^{*})\ }
{\displaystyle \operatorname {Re} \{x[n]\}\ }
{\displaystyle {\frac {1}{2}}\left[X(z)+X^{*}(z^{*})\right]}
{\displaystyle \operatorname {Im} \{x[n]\}\ }
{\displaystyle {\frac {1}{2j}}\left[X(z)-X^{*}(z^{*})\right]}
{\displaystyle nx[n]\ }
{\displaystyle -z{\frac {\mathrm {d} X(z)}{\mathrm {d} z}}}
{\displaystyle x_{1}[n]*x_{2}[n]\ }
{\displaystyle X_{1}(z)X_{2}(z)\ }
{\displaystyle r_{x_{1},x_{2}}(l)=x_{1}[l]*x_{2}[-l]\ }
{\displaystyle R_{x_{1},x_{2}}(z)=X_{1}(z)X_{2}(z^{-1})\ }
{\displaystyle z^{-1}}
{\displaystyle x_{1}[n]x_{2}[n]\ }
{\displaystyle {\frac {1}{j2\pi }}\oint _{C}X_{1}(v)X_{2}({\frac {z}{v}})v^{-1}\mathrm {d} v\ }
{\displaystyle r_{1l}r_{2l}<|z|<r_{1u}r_{2u}\ }
{\displaystyle \sum ^{\infty }x_{1}[n]x_{2}^{*}[n]\ }
{\displaystyle {\frac {1}{j2\pi }}\oint _{C}X_{1}(v)X_{2}^{*}({\frac {1}{v^{*}}})v^{-1}\mathrm {d} v\ }
{\displaystyle x[0]=\lim _{z\rightarrow \infty }X(z)\ }
{\displaystyle x[n]\,}
{\displaystyle x[\infty ]=\lim _{z\rightarrow 1}(z-1)X(z)\ }
{\displaystyle (z-1)X(z)\ }
Phase Shift[edit | edit source]
Digital Signal Processing/Bilinear Transform
Control Systems/Z-transform
Retrieved from "https://en.wikibooks.org/w/index.php?title=Digital_Signal_Processing/Z_Transform&oldid=4017502"
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3Blue1Brown - Recommendations
If there’s one thing I hope 3blue1brown videos do, it’s to inspire people to spend more time playing with and learning about math. In a separate post, I lay out some recommendations for books on math which I've personally enjoyed. Here I thought I'd list some of the other places online you might enjoy exploring, should you find yourself with an itch for more math.
Passive consumption, say by watching videos or reading books, is certainly the lowest friction way to expose yourself to new math. But if you care about long-term retention and deeper understanding, there really is no substitute for practice. So before getting to other YouTube channels and books you might enjoy, here are a few places you can go to learn more actively.
The first is brilliant.org. I know a few of the people there, and I can vouch for the fact that they are extremely thoughtful about what makes for good learning. It shows in the product. Rather than putting lectures and videos front and center, problems and puzzles take precedence, all presented in a structured progression. The site also benefits from an active and helpful community of users.
Another good reservoir of problems with a helpful community surrounding them is the Art of Problem Solving. In particular, they have an extensive archive of contest math problems, ranging from more approachable ones like the AMC8 to extremely hard ones like the IMO, all with excellent surrounding discussion from the community.
The original Art of Problem Solving book by Richard Rusczyk, by the way, was one of the things that made me fall in love with creative math as a young student.
Mathigon is a very beautifully done set of interactive articles on all sorts of fun topics in math. The experience sits somewhere between reading and playing a game, with a really wonderful blend of artwork, technology and pedagogical clarity. One small feature, but which adds a ton in my opinion, is that you have to answer questions in order to continue reading. It's a nice way to force the reader to be honest with themselves about whether they're being passive or not.
For building up your fundamentals, say starting with algebra, take a look at Khan Academy. They are probably most famous for the videos Sal does, but arguably the site is most valuable as a source of problems you can drill on, starting from the very basics if necessary. I may be biased, though, given that I used to work there!
Mathologer, by Burkard Polster. Half of the appeal here comes from the delightful bits of math this channel covers, often with novel ways of presenting otherwise extremely complicated topics to make them easier. For example, he has a video showing why
\pi
is transcendental, which is a famously difficult topic, but which he skillfully boils down into an approachable step-by-step presentation. The other half of the appeal is watching Burkard delightedly laugh at his own jokes every 2 minutes or so.
Numberphile, by Brady Haran. I’m guessing if you’re here, you already know about Numberphile. If somehow you don’t, it’s a great channel full of interviews with mathematicians. As a result, you see a wide array of the types of topics which different mathematicians find exciting to showcase or teach, and a little more of the human element behind this subject. There is also an associated podcast, less about the math, and more about those human stories.
Standup Maths, by Matt Parker. This channel is more whimsical and comedic, but that’s not to say there isn’t a lot of serious math it covers.
Looking Glass Universe, by Mithuna Yoganathan. She covers ideas from physics and math explained simply and unusually honestly. Start with the one on spin, it’s great. She also places an emphasis on doing practice, often assigning “homework” at the end of each video.
Also, can we just step back to appreciate that the four channels above are all run by people either from or based in Australia? What’s going on over there? And that’s not even counting Tibees, Eddy Woo, James Tanton and countless others.
Welch Labs, by Stephen Welch. Perhaps best known for his series on imaginary numbers, this channel is full of high production quality and well-written videos. There are also several series on machine learning, the topics underlying self-driving cars, so this is certainly one of the more applied math channels out there.
Black pen red pen, by Steve Chow. If you like seeing how to work out integrals in all sorts of crazy ways, boy is this the channel for you. Steven is a math teacher, and the videos have a feeling of being in office hours, honestly working through problems with a very skillful and enthusiastic teacher. Many of the problems covered are the kind you might come across in school, and many or just for fun.
Patrick JMT, by… well… Patrick. If you want good, honest worked examples of practice problems, especially for calculus, it’s hard to find a deeper well of examples than what this channel provides.
Needless to say, there are countless others. But often less is more when it comes to recommendations, so let's leave things here for the time being.
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Short‐Term Hindcasts of Seismic Hazard in the Western Canada Sedimentary Basin Caused by Induced and Natural Earthquakes | Seismological Research Letters | GeoScienceWorld
Department of Earth Sciences, Western University, Biological & Geological Sciences Building, Room 1026, 1151 Richmond Street N., London, Ontario, Canada N6A 5B7, hghofra@uwo.ca, gatkins6@uwo.ca, kassatou@uwo.ca
Alberta Geological Survey, Alberta Energy Regulator, Suite 205, 4999 ‐ 98 Avenue, Edmonton, Alberta, Canada T6B 2X3, ryan.schultz@aer.ca
Hadi Ghofrani, Gail M. Atkinson, Ryan Schultz, Karen Assatourians; Short‐Term Hindcasts of Seismic Hazard in the Western Canada Sedimentary Basin Caused by Induced and Natural Earthquakes. Seismological Research Letters 2019;; 90 (3): 1420–1435. doi: https://doi.org/10.1785/0220180285
We produce year‐by‐year probabilistic seismic hazard maps based on the seismicity from 2011 to 2018 for the western Canada sedimentary basin (WCSB) and compare them with corresponding maps of the baseline historical hazard levels through 2010. Rates of earthquakes across the WCSB, of moment magnitude
(M)≥3.0
, have increased in the past decade because of oil and gas development activities, especially hydraulic fracturing in long horizontal wellbores. The increased activity rates have elevated the hazard on a short‐term basis, dramatically in some locations. For example, in the area west of Rocky Mountain House, the 1/2500 per annum ground motions (peak acceleration and high‐frequency response spectra) are greater than those in the 2015 national seismic hazard model by more than a factor of 4; the 2015 hazard model was based on seismicity to 2012 and attempted to exclude induced earthquakes.
The intent of this study is to provide 1 yr hindcasts of the impact of induced seismicity on regional hazard. Such retrospective forecasts show how the observed seismicity in each specific year would translate into a seismic hazard map. This is a useful first step in developing earthquake hazard forecasts for the WCSB, which might eventually forecast future hazard based on the relationship between operational and geological factors and seismicity response.
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Current meter - FIThydrowiki
Figure 1: Current meter: a) mechanic, b) electromagnetic, c) acoustic (Sources: SEBA hydrométrie, OTT, Sontek).
Figure 2: Flow cross-section with the verticals (in red) and the points (black dots) for velocity measurement with the current meter
Figure 3: Operator in the downstream migration channel performing measurements with a current meter.
The first devices to measure flow velocities appeared during the 1780s, with the first mechanical current meter which counts the rotations of a propeller and links it to the flow velocity (by proportionality principle).
Today there are 3 different techniques to assess the flow velocity: mechanically, electromagnetically, and acoustically (Figure 1). The acoustic current meters using the Doppler Effect are not presented here, as these tools (e.g ADCP, ADV) are presented separately.
In a mechanical flow meter, the rotation of the propeller produces an electrical impulse detected and recorded by a counter connected to the current meter. The electromagnetic flow meter uses the Faraday law. The sensor creates a magnetic field between two electrodes located at the end of the probe. The flow velocity is deduced from the measurement of the electromagnetic force generated by the passage of water through the magnetic field.
The measurement conditions of discharge through flow velocity fields are regulated by the European standard NF EN ISO 748. The velocities are measured along a cross section at verticals. The number of verticals (n) depends on the width of the channel (w):
W < 0.5 m, n = 5 or 6
W > 0.5 m and < 1 m, n = 6 or 7
W > 1 m and < 3 m, n = 7 or 12
W > 3 m and < 5 m, n = 13 to 16
W > 5 m, n ≥ 22
One method among others to measure the mean velocity on each vertical is to measure the velocity in one or several measurement points at different water levels on the vertical. The number of measurements points for each vertical depends on the water level. For example, only one point is needed for water level <20 cm at 60% of the water level. If there are only 2 measurement points, they should be at 20 and 80% of the flow depth, and the mean velocity corresponds to the average of the two measured velocities. In the following example, 6 measurements were selected (Figure 2). The example is linked to the velocity profile inside a downstream migration channel, which is not logarithmic.
To do the measurements, the operator needs to be equipped with waders and to be in the stream (Figure 3).
By the method of mean section, the mean velocity between the banks and the closer verticals can be calculated. The mean velocity on each vertical is calculated that way, in the case of 6 measurement points:
{\displaystyle V_{meansection}=0.1(V_{s}+2V_{0.8}+2V_{0.6}+2V_{0.4}+2V_{0.2}>+V_{pf})}
Then a mean velocity is calculated for each flow section between the verticals and the closer bank; by making an average for the sections without including the banks, and by applying the following formula:
{\displaystyle V_{meansection}={\frac {m}{(m+1)}}*V_{meancloservertical}}
where m is a parameter depicting the characteristics of the river bed or of the wall. m is generally comprised between 5 and 7, but can be equal to 10 if the walls are really smooth. If the river bed and the walls are rough, m=4.
The discharge per section can be calculated according to standard EN ISO 748:2007:
{\displaystyle Q_{section}=V_{meansection}*H_{watermean}*W}
{\displaystyle H_{watermean}}
is the mean water level and
{\displaystyle W}
is the width of the channel.
The total discharge is the sum of all the discharges per section.
The most recent devices have the possibility to directly treat the data and provide the total discharge.
Several companies commercialize these technologies.
At the French Test Cases an electromagnetic flow meter Marsh McBirney, FLO-MATE 2000 was used.
Aurélien Despax, 2016, Incertitude des mesures de débit des cours d’eau au courantomètre, Amélioration desméthodes analytiques et apports des essais interlaboratoires, Ingénierie de l’environnement, UniversitéGrenoble Alpes, 2016
Manon Dewitte, Dominique Courret, Fatma Lemkecher, Laurent David, 2019, Presentation of Las Rives Test Case, FIThydro
Retrieved from "https://www.fithydro.wiki/index.php?title=Current_meter&oldid=6062"
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Proca_action Knowpia
Proca action
In physics, specifically field theory and particle physics, the Proca action describes a massive spin-1 field of mass m in Minkowski spacetime. The corresponding equation is a relativistic wave equation called the Proca equation.[1] The Proca action and equation are named after Romanian physicist Alexandru Proca.
The Proca equation is involved in the Standard model and describes there the three massive vector bosons, i.e. the Z and W bosons.
This article uses the (+−−−) metric signature and tensor index notation in the language of 4-vectors.
Lagrangian densityEdit
The field involved is a complex 4-potential
{\displaystyle B^{\mu }=\left({\frac {\phi }{c}},\mathbf {A} \right)}
{\displaystyle \phi }
is a kind of generalized electric potential and
{\displaystyle \mathbf {A} }
is a generalized magnetic potential. The field
{\displaystyle B^{\mu }}
transforms like a complex four-vector.
The Lagrangian density is given by:[2]
{\displaystyle {\mathcal {L}}=-{\frac {1}{2}}(\partial _{\mu }B_{\nu }^{*}-\partial _{\nu }B_{\mu }^{*})(\partial ^{\mu }B^{\nu }-\partial ^{\nu }B^{\mu })+{\frac {m^{2}c^{2}}{\hbar ^{2}}}B_{\nu }^{*}B^{\nu }.}
{\displaystyle c}
{\displaystyle \hbar }
is the reduced Planck constant, and
{\displaystyle \partial _{\mu }}
is the 4-gradient.
The Euler–Lagrange equation of motion for this case, also called the Proca equation, is:
{\displaystyle \partial _{\mu }(\partial ^{\mu }B^{\nu }-\partial ^{\nu }B^{\mu })+\left({\frac {mc}{\hbar }}\right)^{2}B^{\nu }=0}
which is equivalent to the conjunction of[3]
{\displaystyle \left[\partial _{\mu }\partial ^{\mu }+\left({\frac {mc}{\hbar }}\right)^{2}\right]B^{\nu }=0}
with (in the massive case)
{\displaystyle \partial _{\mu }B^{\mu }=0\!}
which may be called a generalized Lorenz gauge condition. For non-zero sources, with all fundamental constants included, the field equation is:
{\displaystyle c{{\mu }_{0}}{{j}^{\nu }}=\left({{g}^{\mu \nu }}\left({{\partial }_{\sigma }}{{\partial }^{\sigma }}+{{m}^{2}}{{c}^{2}}/{{\hbar }^{2}}\right)-{{\partial }^{\nu }}{{\partial }^{\mu }}\right){{B}_{\mu }}}
{\displaystyle m=0}
, the source free equations reduce to Maxwell's equations without charge or current, and the above reduces to Maxwell's charge equation. This Proca field equation is closely related to the Klein–Gordon equation, because it is second order in space and time.
In the vector calculus notation, the source free equations are:
{\displaystyle \Box \phi -{\frac {\partial }{\partial t}}\left({\frac {1}{c^{2}}}{\frac {\partial \phi }{\partial t}}+\nabla \cdot \mathbf {A} \right)=-\left({\frac {mc}{\hbar }}\right)^{2}\phi \!}
{\displaystyle \Box \mathbf {A} +\nabla \left({\frac {1}{c^{2}}}{\frac {\partial \phi }{\partial t}}+\nabla \cdot \mathbf {A} \right)=-\left({\frac {mc}{\hbar }}\right)^{2}\mathbf {A} \!}
{\displaystyle \Box }
Gauge fixingEdit
The Proca action is the gauge-fixed version of the Stueckelberg action via the Higgs mechanism. Quantizing the Proca action requires the use of second class constraints.
{\displaystyle m\neq 0}
, they are not invariant under the gauge transformations of electromagnetism
{\displaystyle B^{\mu }\rightarrow B^{\mu }-\partial ^{\mu }f}
{\displaystyle f}
is an arbitrary function.
^ Particle Physics (2nd Edition), B.R. Martin, G. Shaw, Manchester Physics, John Wiley & Sons, 2008, ISBN 978-0-470-03294-7
^ W. Greiner, "Relativistic quantum mechanics", Springer, p. 359, ISBN 3-540-67457-8
Supersymmetry Demystified, P. Labelle, McGraw–Hill (USA), 2010, ISBN 978-0-07-163641-4
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Basic Mathematics Warmup: Level 2 Challenges Practice Problems Online | Brilliant
\Large{2^{3^2}}= \, ?
Consider a regular 5 pointed star (pentagram). What is the sum of the internal angles (in degrees) of the 5 points?
\Large \color{#3D99F6}{71^{70}} \quad \text{or} \quad \color{#D61F06}{ 70^{71} }
Which one of the two numbers above is greater?
{70^{71}}
{71^{70}}
by Guna Vikas
13, 13, 10
13, 13, 24\, ?
\large \color{#624F41}5^{55} + \color{#624F41}5^{55} + \color{#624F41}5^{55} + \color{#624F41}5^{55} +\color{#624F41} 5^{55} = \, ?
5^{56}
5^{275}
25^{56}
25^{275}
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Find the distance d between the points P_{1} and P_{2} P_{1}=(1.2,\
Find the distance d between the points
{P}_{1}
{P}_{2}
{P}_{1}=\left(1.2,\text{ }2.3\right)
{P}_{2}=\left(-0.3,\text{ }1.1\right)
The distance formula is given below so that we can calculate the distance between the two points then
d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}
now the points are given that
{P}_{1}
{P}_{2}
which are (1.2, 2.3) and (-0.3, 1.1) respectively. and
\left({x}_{1},\text{ }{y}_{1}\right)=\left(1.2,\text{ }2.3\right),\text{ }\left({x}_{2},\text{ }{y}_{2}\right)=\left(-0.3,\text{ }1.1\right)
{x}_{1}=1.2,\text{ }{y}_{1}=2.3,\text{ }{x}_{2}=-0.3,\text{ }{y}_{2}=1.1
now put the value into the distance formula
d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}
{x}_{1}=1.2,\text{ }{y}_{1}=2.3,\text{ }{x}_{2}=-0.3,\text{ }{y}_{2}=1.1
d=\sqrt{{\left(-0.3-1.2\right)}^{2}+{\left(1.1-2.3\right)}^{2}}
d=\sqrt{{\left(-1.5\right)}^{2}+{\left(-1.2\right)}^{2}}
d=\sqrt{{\left(1.5\right)}^{2}+{\left(1.2\right)}^{2}}
d=\sqrt{2.25+1.44}
d=\sqrt{3.69}
d=1.920937
The question is taken from the Analytical Geometry in which we have to find the distance between the two points. The given points are P1 and P2 which are (1.2, 2.3) and (-0.3, 1.1) respectively.
Given that distance between the two points is d
Now we have to solve the question by using the distance formula
We move to the next step -2 to solve the question
d=\sqrt{{\left(x2-x1\right)}^{2}+{\left(y2-y1\right)}^{2}}
now the points are given that P1 and P2 which are (1.2, 2.3) and (-0.3, 1.1) respectively.
\left(x1y1\right)=\left(1.2,2.3\right),\left(x2y2\right)=\left(-0.3,1.1\right)
x1=1.2,y1=2.3,x2=-0.3,y2=1.1
d=\sqrt{{\left(x2-x1\right)}^{2}+{\left(y2-y1\right)}^{2}}
x1=1.2,y1=2.3,x2=-0.3,y2=1.1
d=\sqrt{{\left(-0.3-1.2\right)}^{2}+{\left(1.1-2.3\right)}^{2}}
d=\sqrt{{\left(-1.5\right)}^{2}+{\left(-1.2\right)}^{2}}
d=\sqrt{{\left(1.5\right)}^{2}+{\left(1.2\right)}^{2}}
d=\sqrt{2.25+1.44}
d=\sqrt{3.69}
d=1.920937
the result is shown in the step-3
Result: From the above solution we got the distance between two points P1 and P2 are shown below
d=1.920937
d=\sqrt{\left({x}_{2}-{x}_{1}{\right)}^{2}+\left({y}_{2}-{y}_{1}{\right)}^{2}}
\left({x}_{1}{y}_{1}\right)=\left(1.2,2.3\right),\left({x}_{2}{y}_{2}\right)=\left(-0.3,1.1\right)
{x}_{1}=1.2,{y}_{1}=2.3,{x}_{2}=-0.3,{y}_{2}=1.1
\begin{array}{}d=\sqrt{\left(x2-x1{\right)}^{2}+\left(y2-y1{\right)}^{2}}\\ {x}_{1}=1.2,\text{ }{y}_{1}=2.3,\text{ }{x}_{2}=-0.3,\text{ }{y}_{2}=1.1\\ d=\sqrt{\left(-0.3-1.2{\right)}^{2}+\left(1.1-2.3{\right)}^{2}}\\ d=\sqrt{\left(-1.5{\right)}^{2}+\left(-1.2{\right)}^{2}}\\ d=\sqrt{\left(1.5{\right)}^{2}+\left(1.2{\right)}^{2}}\\ d=\sqrt{2.25+1.44}\\ d=\sqrt{3.69}\\ d=1.920937\end{array}
{r}^{2}=\mathrm{sin}\left(2\theta \right)
Write the equation of the circle with Center (-1, -8) and Radius:
\sqrt{7}
{x}^{2}+2x+{y}^{2}+16y+58=0
{x}^{2}+2x+1+{y}^{2}+16y+64=7
{x}^{2}+2x+{y}^{2}+16y+72=0
{x}^{2}-2x+{y}^{2}-16y+58=0
A bridge is built in the shape of a parabolic arch. The bridge has a span of 120 feet and a maximum height of 25 feet. Choose a suitable rectangular coordinate system and find the height of the arch at distances of 10, 30, and 50 feet from the center.
\pi \approx 3.14
\pi \approx 3.14
5x+2y=10
Ax+By=C
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GTON Unlocking Schedule - GTON Capital
What are the principles behind GTON distribution and its timeline? In this article, we explain the mechanics that underlie the emission of GTON in accordance with pre-defined allocations and vesting outline.
GTON is deflationary by design: its maximum supply is fixed at 21 million tokens, like Bitcoin. The generation of tokens, also called minting, occurs through gradual “unlocking” of the allocated part of supply to specific addresses. Unlocking takes place according to the established schedule represented by the emission curve, and all minted tokens are automatically assigned to a specific category of governance roles. This is why in the vocabulary of Graviton it is usually called farming — the issued tokens are deployed onto existing lists of accounts, depending on their role and activity.
There are several roles in the Graviton’s DAO:
early birds or early backers: those who made a deposit into the treasury at an early stage (early birds or early backers (EB))
backers: those who purchase tokens during the IDOs
strategic partners and investors (SPI): these are potential top-level industry players or strategic investors who enter the treasury in subsequent stages, which may include ecosystem-level actors, VC funds, and AMMs. This category of treasury investors does not undermine early backers’ share and will be negotiated individually case by case in a transparent manner;
contributors are the development team, signers of multisig and governers of the DAO;
operations, or, in other words, expenditure for operational needs, is an allocation aimed at working with the community, like bounty and airdrops, and market making;
LP, or liquidity providers, is the utility-oriented allocation intended for users participating in programs aimed to stimulate liquidity of wrapped assets and Graviton itself.
All these roles have a vesting period when unlocking GTON tokens. In the case of Operations and LP categories, the distribution parameters will be decided by the Graviton governance and are not predetermined. However, for the rest of allocations, e.g. EB (10% or 2.1 mln tokens), the vesting period is based on a special formula as described further.
At the end of the deposit period, all early-bird participants and their treasury balances in dollar equivalent are snapshot from the blockchain. For all backers, it is already predetermined how many tokens they will own (ownable allocation). At the end of the EB period, these tokens from ownable allocation will start getting gradually unlocked every hour, becoming a claimable allocation available for a withdrawal to the user’s wallet.
Initially, GTON will be available for withdrawal on the following networks: Ethereum, BSC and Fantom. The withdrawal in other integrated networks will be implemented in subsequent stages.
Unlocking of the ownable allocation into a claimable allocation follows the formula:
CA(g,t) = (C(g)*10⁶)/y(t),
Where C(g) — group (category) token allocation — g, y(t) — emission formula:
y = \frac{-a}{(x+{a}/{c}+o)}+c
where c = 2100000, a = 11099999999999, o is the starting block of farming, x is Ethereum's current block count.
For the EB group, the parameter C is 2.1 mln.
A simplified piecewise scheme shows that by the end of the first week, 5% (105k tokens) will be unlocked, 9% (189k) by the end of the 2nd week, 16% (336k) by the end of the month, 75% by the end of the year, 95% by the end of the next year, and so on.
A similar unlocking formula will be applied to all categories except LP and Ops. For Contributors, allocation will start in three months after the end of EB period; for SPI, two weeks after a finalization of the deal and transaction.
This emission formula provides certain advantages, as it allows for bringing in liquidity to the token at an early stage, gradually increasing its circulating supply without sudden leaps. At the same time, a part of allocations will remain unlocked even after two years from the project’s launch, which leaves all long-term shareholders interested in the future success of the project.
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APY Calculations - Oh! Finance Docs
Providing APY estimates is one of the most-requested and important pieces of information to our users. For the sake of transparency, this documentation outlays how our APYs are being calculated, and what an APY value means in our DeFi context.
Basic Assumption & Disclaimer
There are a myriad of inputs one could add into APY calculations (e.g., changing liquidity over time, entry and exits, intermittently called Finance and Rebalance functions, changes in underlying asset prices and platforms, gas fees, etc.).
In order to present the most honest data and avoid speculative inputs, only OH Virtual Price values are used, which are based on actual data from on-chain historical values.
Our APYs are calculated according to the formulas below. These are estimates based on previous and current information about OH, not future-looking predictive calculations. Stated simply: our APY values are backwards-looking estimates of historical performance on the platform. Past performance neither guarantees nor predicts the future performance of the platform.
Annual Percentage Return (APR) Calculations
The primary datapoint underpinning our APR calculations is the on-chain virtualPrice value of the OH Bank smart contract(s).
Notably, APR does not take into account recurring re-compounding of interest earned, unlike APY. APY is what we report, which is detailed in the next section, but APR is an input to our APY calculations.
The formula used to calculate APR is below. APR is defined as the growth of the virutalPirce over a window of time, expanded out to a 365-day year at constant similar performance. A sliding historical time window is used for APR calculation. Shorter time windows generally give a better picture of nearer-to-real-time current performance, whereas longer time windows are less noisy and generally give a better picture of longer-term performance.
APR_{n} = ((vp_{recent}/vp_{past})-1)*(n/365)
n : Number of Days in Sliding Window vp : Bank Contract virtualPrice at a given point in time, on either the recent front-end of the sliding window or the past back-end of the sliding window APR : APR calculated over n days, extrapolated to 365-days of returns assuming constant performance
Annual Percentage Yield (APY) is what we report to OH users as a more accurate representation of our returns. This is because we regularly (daily) compound any interest or rewards earned on our underlying platforms and re-invest them.
The formula used to calculate APY is below - at the time of writing, we report APY values over a 1-day, 7-day, and 30-day sliding time period. APY values are re-calculated approximately 24-times per day (every 1 hour).
APY_n = (1+\frac{APR_n}{p})^p -1
n : number of days in sliding APR time window APR_n : Annual Percentage Return calculated over an n-day time period p : Number of times re-compounding occurs during 1 year. OH compounds approximately once per day, so p=365.
APYs are presented on our platform as a percentage value, which multiplies the output of the above APY formula by 100.
Next - Community & Socials 📣
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Rates Module - Maker Protocol Technical Docs
The Maker Protocol's Rate Accumulation Mechanism
A fundamental feature of the MCD system is to accumulate stability fees on Vault debt balances, as well as interest on Dai Savings Rate (DSR) deposits.
R(t) \equiv R_0 \prod_{i=t_0 + 1}^{t} F_i = R_0 \cdot F_{t_0 + 1} \cdot F_{t_0 + 2} \cdots F_{t-1} \cdot F_t
D(t) \equiv A \cdot R(t) = D_0 \prod_{t=1}^{T} F_i
increases the system's total debt (i.e. issued Dai) by Art*rate_change.
Suppose at time 0, a Vault is opened and 20 Dai is drawn from it. Assume that rate is 1; this implies that the stored art in the Vault's Urn is also 20. Let the base and duty be set such that after 12 years, art*rate = 30 (this corresponds to an annual stability of roughly 3.4366%). Equivalently, rate = 1.5 after 12 years. Assuming that base + duty does not change, the growth of the effective debt can be graphed as follows:
Now suppose that at 12 years, an additional 10 Dai is drawn. The debt vs time graph would change to look like:
The art can be thought of as "debt at time 0", or "the amount of Dai that if drawn at time zero would result in the present total debt". The graph below demonstrates this visually; the length of the green bar extending upwards from t = 0 is the post-draw art value.
Vault owners wishing to draw Dai (if they don't call drip prior to drawing from their Vault, they will be charged fees on the drawn Dai going back to the last time drip was called—unless no one calls drip before they repay their Vault, see below)
MKR holders (they have a vested interest in seeing the system work well, and the collection of surplus in particular is critical to the ebb and flow of MKR in existence)
\text{rate} = 1 \text{ ; total fee} = f
\text{rate} = f^{28}
\text{totalfee} \xrightarrow{} g
\text{rate} = f^{28} g^{42}
\text{rate}_{ideal} = f^{56}g^{14}
Dai Savings Rate Accumulation
DSR accumulation is very similar to stability fee accumulation. It is implemented via the Pot, which interacts with the Vat (and again the Vow's address is used for accounting for the Dai created). The Pot tracks normalized deposits on a per-user basis (pie[usr]) and maintains a cumulative interest rate parameter (chi). A drip function analogous to that of Jug is called intermittently by economic actors to trigger savings accumulation.
\text{chi}(t) \equiv \text{chi}0 \prod{i=t_0 + 1}^{t} \text{dsr}_i
Suppose a user joins N Dai into the Pot at time t_0. Then, their internal savings Dai balance is set to:
\text{pie[usr]} = N / \text{chi}_0
The total Dai the user can withdraw from the Pot at time t is:
\text{pie[usr]} \cdot \text{chi}(t) = N \prod_{i=t_0 + 1}^{t} \text{dsr}_i
After updating chi, Pot.drip then calls Vat.suck with arguments such that the additional Dai created from this savings accumulation is credited to the Pot contract while the Vow's sin (unbacked debt) is increased by the same amount (the global debt and unbacked debt tallies are increased as well). To accomplish this efficiently, the Pot keeps track of a the total sum of all individual pie[usr] values in a variable called Pie.
There is no stored record of depositing or withdrawing Dai from the Pot
any user withdrawing Dai from the Pot (otherwise they lose money!)
any user putting Dai into the Pot—this is not economically rational, but is instead forced by smart contract logic that requires drip to be called in the same block as new Dai is added to the Pot (otherwise, an economic exploit that drains system surplus is possible)
r^N = 1.005
r = 1.000000000158153903837946258002097...
\text{dsr} = 1000000000158153903837946258
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The algorithm uses Todd-Coxeter coset enumeration, which is an inherently non-terminating process for infinite groups. Therefore, the algorithm will halt with an exception if too many cosets are generated during an attempt to enumerate cosets of a subgroup. The point at which the coset enumeration terminates is controlled by the environment variable _EnvMaxCosetsToddCoxeter, which has the default value
128000
\mathrm{with}\left(\mathrm{group}\right):
g≔\mathrm{grelgroup}\left({a,b,c,d},{[a,b,c,\frac{1}{d}],[b,c,d,\frac{1}{a}],[c,d,a,\frac{1}{b}],[d,a,b,\frac{1}{c}]}\right):
\mathrm{sg}≔\mathrm{subgrel}\left({x=[a,b],y=[a,c]},g\right):
\mathrm{pres}\left(\mathrm{sg}\right)
\textcolor[rgb]{0,0,1}{\mathrm{grelgroup}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}{[\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{y}}]\textcolor[rgb]{0,0,1}{,}[\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{y}}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}]}\right)
s≔\mathrm{subgrel}\left({x=[a,b,\frac{1}{a}]},\mathrm{grelgroup}\left({a,b},{[b,b],[a,a,a]}\right)\right):
\mathrm{pres}\left(s\right)
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Category density +> CalculatePlus
{\displaystyle \rho ={\frac {m}{V}}}
For a pure substance the density has the same numerical value as its mass concentration.Different materials usually have different densities, and density may be relevant to buoyancy, purity and packaging. Osmium and iridium are the densest known elements at standard conditions for temperature and pressure.
read more about on wikipedia.org
Latest from category density
0.8 kg/l to g/ml (kilogram/liter to gram/milliliter)
3 kg/l to avoirdupois pounds/quart (UK) (kilograms/liter to lb.av/qt.uk)
7 kg/l to kg/m3 (kilograms/liter to kilograms/cubic meter)
997 kg/m3 to kg/mm3 (kilograms/cubic meter to kilograms/cubic millimeter)
kg/l to kg/m3 (kilogram/liter to kilogram/cubic meter)
600 pounds (avoirdupois)/cubic meter to u/ft3.us (lb.av/m3 to atomic mass units/cubic foot (US))
1.12 g/cm3 to t/m3 (grams/cubic centimeter to tons/cubic meter)
200 mg/ml to g/cm3 (milligrams/milliliter to grams/cubic centimeter)
3.8 g/cm3 to atomic mass unit/cubic millimeter (grams/cubic centimeter to u/mm3)
5 kg/m3 to kg/mm3 (kilograms/cubic meter to kilograms/cubic millimeter)
50,000 g/cm3 to kg/m3 (grams/cubic centimeter to kilograms/cubic meter)
mp/ml to grams/milliliter (proton mass/milliliter to g/ml)
0.007 g/mm3 to kg/m3 (gram/cubic millimeter to kilogram/cubic meter)
50,000 kg/m3 to g/cm3 (kilograms/cubic meter to grams/cubic centimeter)
tonne/cubic meter to g/cm3 (t/m3 to gram/cubic centimeter)
g/cm3 to tons/cubic meter (gram/cubic centimeter to t/m3)
mg/m3 to kg/mm3 (milligram/cubic meter to kilogram/cubic millimeter)
1.293 kg/m3 to g/cm3 (kilograms/cubic meter to grams/cubic centimeter)
1.03 g/cm3 to kilograms/cubic centimeter (grams/cubic centimeter to kg/cm3)
7.8 g/cm3 to kg/mm3 (grams/cubic centimeter to kilograms/cubic millimeter)
360 g/l to kilogram/cubic meter (grams/liter to kg/m3)
0.5 mg/l to mg/ml (milligram/liter to milligram/milliliter)
3 kg/m3 to lb/cm3 (kilograms/cubic meter to avoirdupois pounds/cubic centimeter)
50 g/l to kg/m3 (grams/liter to kilograms/cubic meter)
10 g/l to gr/cm3 (grams/liter to grains/cubic centimeter)
3 lb/cm3 to kg/m3 (avoirdupois pounds/cubic centimeter to kilograms/cubic meter)
1,717.79 g/cm3 to kg/m3 (grams/cubic centimeter to kilograms/cubic meter)
8.41 g/ml to kg/l (grams/milliliter to kilograms/liter)
g/ft3 to lb/yd3 (gram/cubic foot to avoirdupois pound/cubic yard)
lb/yd3 to g/ft3 (avoirdupois pound/cubic yard to gram/cubic foot)
300 g/l to g/hl (grams/liter to grams/hectoliter)
18 g/cm3 to tonne/cubic meter (grams/cubic centimeter to t/m3)
1 g/ml to lb/cm3 (gram/milliliter to avoirdupois pound/cubic centimeter)
2 mg/ml to mg/l (milligrams/milliliter to milligrams/liter)
g/cm3 to grain/cubic yard (US) (gram/cubic centimeter to gr/yd3.us)
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nresults - Maple Help
Home : Support : Online Help : nresults
number of results expected to be returned from a procedure
Within a procedure, the special name _nresults has as its value the number of variables on the left-hand side of the expression in which the procedure was called. If the procedure call did not involve an assignment, then _nresults is given the value undefined. If the procedure call is nested inside another expression, the value of _nresults is also set to `undefined`. Any time the procedure call on the right-side of an assignment is part of an expression more complicated than a simple function call, _nresults will be set to `undefined`.
Do not use _nresults with option remember or option cache. Only the first result computed is stored in the remember table. Subsequent results with the same input and a different value for _nresults will not respect the current value of _nresults. The Cache package can be used to manually manipulate and simulate a remember table that works with _nresults.
A procedure to find both the minimum and maximum of an arbitrary sequence of numbers could be written as follows.
maxmin := proc () local max, min, i;
min := _passed[1];
elif _passed[i] < min then
min := _passed[i]
a≔\mathrm{maxmin}\left(1,2,3\right)
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{3}
a,b≔\mathrm{maxmin}\left(1,2,3\right)
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
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Compute or plot passivity index as function of frequency - MATLAB passiveplot - MathWorks Benelux
G=\left(s+2\right)/\left(s+1\right)
\left(G\left(s\right)+G{\left(s\right)}^{H}\right)/2
\left(G{\left(s\right)}^{-1}+G{\left(s\right)}^{-H}\right)/2
\text{G}\left(s\right)+\text{G}{\left(s\right)}^{H}>2\tau \left(\omega \right)\left(I+\text{G}{\left(s\right)}^{H}\text{G}\left(s\right)\right),
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Popular Science Monthly/Volume 9/July 1876/The Chromis Pater-Familias - Wikisource, the free online library
Popular Science Monthly/Volume 9/July 1876/The Chromis Pater-Familias
Ordeals and Oaths
The Chromis Pater-Familias by Louis Charles Émile Lortet
Bigotry in Scientific Controversy
599213Popular Science Monthly Volume 9 July 1876 — The Chromis Pater-Familias1876Louis Charles Émile Lortet
THE CHROMIS PATER-FAMILIAS.
By Dr. LORTET.
UP to the present we know but a small number of fishes which hatch their eggs and bring up their young in the cavity of the mouth or among the gills. Agassiz, during his voyage on the Amazonas, discovered one species. Afterward there was brought from China the macropod, the singular habits of which are now known to all the world. All these species belong to the great group of the Labyrinthobranchiata; and Agassiz supposes that the fishes of this order only can hatch their eggs in so abnormal a manner, thanks to the branchial pockets which allow of the eggs being easily kept in place. But the Chromis, of which we give a faithful representation, proves the assertion of Agassiz to be erroneous. The Chromis pater-familias has the gills disposed in simple laminæ; it is unprovided with any special apparatus for retaining the eggs or the young ones, and yet it brings up about 200 young in the mouth and gills. It is always the male that performs these functions of incubation. After the female has deposited the eggs in a depression of the sand or between the tufts of reeds, the male approaches and takes them by inhalation into the cavity of the mouth. From there some movement, the mechanism of which we have not been able to observe, sends them between the leaflets of the gills. The pressure exerted on the eggs by the branchial laminæ suffices to keep them in place. There, in the midst of the organs of respiration, the eggs undergo all their metamorphoses. The young ones grow rapidly, and soon appear much inconvenienced in their narrow prison. They leave it, not by the gills but through the opening by which the bronchial cavity communicates with the mouth. Here they remain in great number, pressed against one another like the seeds of a pomegranate. The animal's mouth becomes so distended by the presence of this numerous progeny that actually the jaws cannot meet. The cheeks are swollen and the animal
Chromis Pater-familias of Lake Tiberias.
presents the strangest aspect. Some of the young, arrived at the perfect state, continue to live in the gills. All have the head directed toward the buccal opening of the father, the protecting cavity of which we have not seen them leave even for a moment. Though so numerous, they hold their ground very firmly, yet how they do so we have not discovered. Neither can we understand how the nursing father avoids swallowing his progeny; we are also ignorant at what period of their life the young ones leave the paternal mouth to live independently.
The Chromis pater-familias is 7 inches long by 1 3/4 inch thick. The teeth are very fine and sharp, disposed in several rows. The snout is obtuse, conical, the upper profile oblique. The nasal prominence is very conspicuous. The caudal fin is almost truncated. The soft rays of the dorsal reach to the beginning of the caudal. The length of the body, including the tail, is 4 1/2 times its thickness. The snout is in length twice the diameter of the orbit. The mouth is slightly oblique, large, as wide as it is long. The teeth are slightly recurved, disposed in three or four rows, tinged with deep yellow at the free end. The first row presents 26 on each side of the upper maxillary. The fins show the number of rays following:
Dorsal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
{\displaystyle +}
Anal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
{\displaystyle +}
Caudal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
Pectoral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Ventral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
{\displaystyle +}
The lateral line comprises 32 scales disposed 20+12. The scales are cycloidal, their length greater than their breadth; three-fourths of their surface is covered by the succeeding scales. Color, olive-green on the back, barred with blue. The belly has a silvery lustre, with green and blue spots.
I caught this interesting species, with a net, on the 29th of April, 1875, in shallow water full of reeds, on the margin of Lake Tiberias, at a place called Ain-Tin, the ancient Capernaum. There are numerous warm springs there which unite to form a rather considerable stream. It is in these warm waters that the Chromis lives.—La Nature.
Retrieved from "https://en.wikisource.org/w/index.php?title=Popular_Science_Monthly/Volume_9/July_1876/The_Chromis_Pater-Familias&oldid=8851926"
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Is square a rectangle? Or is rectangle a square?
For example, a quadrilateral is a rectangle if all four internal angles are 90° and all four sides are equal in measure.
However, note that the first condition for a square is the same as the only condition for a rectangle, and so all squares are rectangles. However, there is no condition which requires a rectangle to have four equal sides, and that means not all rectangles are squares.
eskalopit
All squares are rectangles. However, not all rectangles are squares.
{P}_{1}\left({\rho }_{1},\text{ }{\theta }_{1},\text{ }{\varphi }_{1}\right)
{P}_{2}\left({\rho }_{2},\text{ }{\theta }_{2},\text{ }{\varphi }_{2}\right)
{L}_{1}:\frac{x-2}{1}=\frac{y-3}{-2}=\frac{z-1}{-3}
{r}^{2}=\mathrm{sin}\left(2\theta \right)
{r}^{2}=\mathrm{sin}\left(2\theta \right)
\mathrm{cos}2\theta =\frac{28}{53}
\mathrm{sin}\theta
{0}^{\circ }<\theta <{90}^{\circ }
\mathrm{tan}\theta
\mathrm{sin}\theta
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Show that the series converges. \sum_{n=1}^\infty\frac{n!^n}{n^{n^2}}
Show that the series converges.
\sum _{n=1}^{\mathrm{\infty }}\frac{n{!}^{n}}{{n}^{{n}^{2}}}
Since by the Stirling approximation
n\mathrm{ln}n!-{n}^{2}\mathrm{ln}\in n\left(n\mathrm{ln}-n+o\left(n\right)\right)-{n}^{2}\mathrm{ln}n=-{n}^{2}\left(1-o\left(1\right)\right)
we have a superexponentially decaying upper bound on the n-th term, so the series converges.
alkaholikd9
By Stirlings
If you don't want to use Stirling the simpler is to go for AM-GM inequality followed by rough majoration
\left(n!{\right)}^{\frac{1}{n}}\le \frac{1}{n}×\frac{n\left(n+1\right)}{2}=\frac{n+1}{2}\le \frac{2n}{3}
for n large enough.
The term of your series is then to be compared to
\left(\frac{2}{3}{\right)}^{{n}^{2}}
\sum _{k=1}^{\mathrm{\infty }}\frac{{x}^{2k}}{k}
\frac{1}{200}\sum _{n=1}^{399}\frac{{5}^{200}}{{5}^{n}+{5}^{200}}=\frac{a}{b}
|a-b|
Is it possible to write this in closed form:
\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\mathrm{log}\left(\left(\begin{array}{c}n\\ k\end{array}\right)\right)
S\left(n\right)=\sum _{k\ge 1}\frac{n!}{k\left(n-k\right)!{n}^{k}}=\sum _{k\ge 1}\frac{{n}^{\underset{―}{k}}}{k{n}^{k}}\approx \frac{1}{2}\mathrm{log}\left(n\right)
How to prove the following equality?
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\prod _{k=1}^{m}\left(n+k\right)}=\frac{1}{\left(m-1\right)m!}
S=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}-{a}^{2}}
a\in \left(0,1\right)
Use the Root Test to determine the convergence or divergence of the series.
\sum _{n=1}^{\mathrm{\infty }}\left(\frac{3n+2}{n+3}{\right)}^{n}
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3Blue1Brown - Darts in Higher Dimensions
Darts in Higher Dimensions
There is a lovely puzzle hidden in the midst of darts, higher dimensional geometry, statistics, and some of the most famous numbers in math. We're going to play a game of darts where we lose the game when the dart misses the bullseye. But this bullseye is too small, so to begin let's make it the entire size of the dartboard.
Each time we hit the bullseye, it will shrink depending on where the dart lands. This makes the rounds get harder and harder as we continue. A line will be drawn from the center of the dartboard to the landing point, we'll call this line
h
. Then we'll draw the chord perpendicular to
h
, whose length will define the diameter of the new bullseye.
Where does the dart have to land in order for the circle to remain unchanged?
Exactly in the center of the circle. Then the line
h
will have a distance of
0
and the chord's length will be the current diameter, so the circle does not shrink.
This game rewards good shots because the length of the chord will be close to the old diameter when the dart lands close to the center. When the dart lands near the edge, the new diameter will be dramatically smaller.
When a dart is thrown and it doesn't land in the bullseye, the game is over. The score is the number of darts thrown, so even if the first dart misses, the minimum score is still
1
. Try playing the game for yourself by clicking below.
Instead of having a human playing this game, we'll create a robot who throws the darts randomly. Imagine a square that surrounds the dartboard, this robot's throws will be uniformly distributed on the square. This is not exactly a realistic distribution, which would be rotationally symmetric.
To make things simpler the dartboard is the unit circle, so each side of the square has a length of
2
x
y
coordinates of the dart's landing are chosen uniformly:
x\in [-1,1] \\ y\in [-1,1]
The square does not shrink with the bullseye, it always stays with the dartboard. The question to be answered is what is the expected score for this robot? As the robot plays more and more games, what is the average score?
We can calculate the expected score
by adding together all of the scores multiplied by the probability that the game ends on that score.
\begin{align*} E[S] &= P(S=1)\cdot 1 + P(S=2)\cdot 2 + P(S=3)\cdot 3 + \cdots \\ &= \sum_{i=1} P(S=i)\cdot i \end{align*}
We'll need to find what the probabilities are for each score. In theory, it is possible to have a perfect game. An expert player could hit the origin with every dart, never shrinking the bullseye. But the probability of such a game occurring is zero. A paradoxical fact about statistics is that there exist events that are possible, but their probability is zero.
When faced with a difficult problem, we can try to get a foothold by breaking it down into smaller sub-problems. The first question we might ask is what is the probability of making the first shot? The probability
P(S>1)
can be found by dividing the area of the circle
A=\pi r^2=\pi
by the area of the square
2\cdot 2=4
The probability of making the first shot is
\frac{\pi}{4}
, so what is the probability of making the second shot? This is where things get more interesting because the probability of the second shot depends on where the first shot landed. The next radius
r_1
forms a right triangle with the current radius and
h_0
, so we can use the Pythagorean theorem to calculate it.
We can build a table of the radiuses and hit distances to keep track of the information.
\begin{array}{c|c} r & h \\ \hline \\ r_0 = 1 & h_0=\sqrt{x_0^2+y_0^2} \\\ \\ r_1 = \sqrt{r_0^2-h_0^2} & h_1=\sqrt{x_1^2+y_1^2} \end{array}
The probability the second shot goes in
P(S>2)
now depends on four numbers. It may be helpful to write out what the actual requirement is:
h_1<r_1
. We can also rewrite it by taking the squares, because it doesn't change the truth value of the statement.
\begin{align*} h_1^2 &< r_1^2 \\ x_1^2+y_1^2 &< r_0^2-h_0^2 \\ x_1^2+y_1^2 &< 1 - x_0^2 - y_0^2 \\ x_0^2 + y_0^2 + x_1^2+y_1^2 &< 1 \end{align*}
We're asking what is the probability that the sum of squares of four random numbers chosen in the range
[-1,1]
1
This question is a bit tricky because we could guarantee if all the numbers are less than
0.5
then the sum of their squares would be less than
1
, but if one was
0.9
that doesn't necessarily throw it out because the others could be very small. We used geometric intuition for the first shot, thinking of it as a point inside or outside of a circle. Rather than thinking of the probabilistic event
P(S>2)
as two separate throws, we can also think about them as a single point in 4D space:
(x_0,y_0,x_1,y_1)\in\mathbb{R}^4
Just like how choosing two random numbers and seeing if the sum of their squares is less than
1
brought us to the area of a circle, doing the same with 4 numbers brings us to the volume of a 4D ball.
\text{Volume of a 4D ball: } \frac{\pi^2}2 R^4 \\ \quad \\ \text{Volume of a 4D cube: } 2\cdot2\cdot2\cdot2 = 2^4 = 16 \\ \quad \\ P(S>2) = \frac{\pi^2}{2^4\cdot2}
Why did we skip to 4 dimensions and not 3?
For every dart throw, we are picking an additional 2 random numbers (the x and y coordinates). This means we never consider the odd numbered dimensions.
The pattern of dividing the volume of higher dimensional balls by the volume of higher dimensional cubes continues when calculating the probability of larger scores.
P(S>3)
\frac{\text{Vol}(\text{6D Ball})}{\text{Vol}(\text{6D Cube})}
because the question is analogous to randomly choosing six numbers and having the sum of their squares be less than
1
x_0^2+y_0^2+x_1^2+y_1^2+x_2^2+y_2^2 < 1
The next problem we face is how to calculate the volume of a higher dimensional ball. Luckily mathematicians have calculated them and you could look at Wikipedia's list and pick them out.
\begin{matrix} 1D & 2R & \quad & 2D & \pi R^2 \\ \quad \\ 3D & \frac43\pi R^3 & \quad & 4D & \frac12\pi^2 R^4 \\ \quad \\ 5D & \frac8{15}\pi^2 R^5 & \quad & 6D & \frac16\pi^3 R^6 \\ \quad \\ 7D & \frac{16}{105}\pi^3 R^7 & \quad & 8D & \frac1{24}\pi^4 R^8 \\ \quad \\ 9D & \frac{32}{945}\pi^4 R^9 & \quad & 10D & \frac1{120}\pi^5 R^{10} \\ \quad \\ 11D & \frac{64}{10,395}\pi^5 R^{11} & \quad & 12D & \frac1{720}\pi^6 R^{12} \end{matrix}
What we'd like is a general formula for the volume of an even dimensional ball:
\text{Vol}(2n\text{D Ball}) = \frac{\pi^n}{n!} R^{2n}
What is the volume of a 14-dimensional unit ball?
\frac{\pi^{14}}{8\times 10^{10}}
\frac{\pi^7}{8\times 10^{10}}
\frac{\pi^7}{5040}
\frac{\pi^{14}}{5040}
The expected value of our dart throwing robot was defined as:
E[S] = \sum_{i=1} P(S=i)\cdot i
However we've been calculating probabilities in the form of greater than
P(S>i)
instead of equal to
P(S=i)
. That is completely fine because
P(S=2)
P(S>1)-P(S>2)
P(S=2)
\frac{\pi^2}{32}
\frac{\pi}{4}
\frac{\pi^2}{32}-\frac{\pi}{4}
\frac{\pi}{4}-\frac{\pi^2}{32}
Now we can expand things out and let them collapse down.
\begin{align*} E[S] =\ &P(S=1)\cdot 1 + P(S=2)\cdot 2 + P(S=3)\cdot 3 + \cdots \\ E[S] =\ & \ 1 \cdot \left(P(S>0)-P(S>1)\right) \\ +&\ 2 \cdot \left(P(S>1)-P(S>2)\right) \\ +&\ 3 \cdot \left(P(S>2)-P(S>3)\right) \\ +&\ \cdots \end{align*}
Notice that we're subtracting off 2 of
P(S>2)
but in the next term we're adding back 3 of them. Each of these terms collapse down to just the sum of each probability:
E[S] = P(S>0)+P(S>1)+P(S>2)+P(S>3)+\cdots
We can begin substituting the values we have calculated so far. Since the minimum score is 1, we will always have a score above 0, so
P(S>0)=1
. From dividing the area of a circle by the area of a square, we know
P(S>1)=\frac{\pi}4
. We had previous calculated
P(S>2) = \frac{\pi^2}{2^4\cdot2}
, which can also be written as
P(S>2) = \left( \frac{\pi}4 \right) ^2 \cdot \frac12
. By reformatting our solutions and adding them up, we can get the formula:
E[S] = 1+\frac{\pi}4+\left( \frac{\pi}4 \right) ^2 \cdot\frac1{2!}+\left( \frac{\pi}4 \right) ^3 \cdot\frac1{3!}+\cdots
This sequence of numbers is rather similar to another famous sequence: the Taylor series for powers of Euler's number is
e^x=1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots
A Taylor series is evaluated as an infinite polynomial where each term gets added together and it approaches the correct value in the limit. This particular series is the definition of
e^x
and I would argue it is the healthy way to view
e^x
Notice how our expected value is very similar to this Taylor polynomial, and the value of
x
\frac{\pi}4
E[S]=e^{0.25\pi}\approx 2.1932...
If our dart throwing robot played a large number of games, the average of all of its scores should be around this 2.1932 figure.
One of the reasons I enjoy this puzzle so much is that we are playing around with higher dimensional geometry naturally. In the middle of the puzzle we were asking what is the volume of a six dimensional ball without suggesting the universe has more dimensions or something similar. That isn't necessarily why mathematicians care about higher dimensions.
What happened is that we had six numbers and we were encoding a property of those numbers in something we like to describe geometrically. Someone else was tackling this puzzle and described how they were working through intensely difficult integrals. What they were really doing was rediscovering the volume of a 4D ball. Thinking about these higher dimensional shapes came about from a 2D plane, nothing about the dartboard was four dimensional.
Mathematicians will often describe manifolds in four dimensions or how the Poincaré conjecture has been solved for everything but four dimensions. But a common misconception people will have is that mathematicians don't actually care about moving in four dimensions, it is about encoding quadruplets of points.
Another reason why this puzzle is beautiful is it helps builds a healthier relationship with
e
. The Taylor series is more important than the number itself, it helps explain why
e^x
is its own derivative, why
e^{\pi i}=-1
, and many other abstract concepts that go beyond simple repeated multiplication.
Another nice puzzle which distills down the essence of
e
without involving circles and complicating the answer with
\pi
is described below.
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Wick rotation - Wikipedia
(Redirected from Wick rotated)
Mathematical trick using imaginary numbers to simplify certain formulas in physics
In physics, Wick rotation, named after Italian physicist Gian Carlo Wick, is a method of finding a solution to a mathematical problem in Minkowski space from a solution to a related problem in Euclidean space by means of a transformation that substitutes an imaginary-number variable for a real-number variable. This transformation is also used to find solutions to problems in quantum mechanics and other areas.
6 Interpretation and rigorous proof
Wick rotation is motivated by the observation that the Minkowski metric in natural units (with metric signature (−1, +1, +1, +1) convention)
{\displaystyle ds^{2}=-\left(dt^{2}\right)+dx^{2}+dy^{2}+dz^{2}}
{\displaystyle ds^{2}=d\tau ^{2}+dx^{2}+dy^{2}+dz^{2}}
are equivalent if one permits the coordinate t to take on imaginary values. The Minkowski metric becomes Euclidean when t is restricted to the imaginary axis, and vice versa. Taking a problem expressed in Minkowski space with coordinates x, y, z, t, and substituting t = −iτ sometimes yields a problem in real Euclidean coordinates x, y, z, τ which is easier to solve. This solution may then, under reverse substitution, yield a solution to the original problem.
Statistical and quantum mechanics[edit]
Wick rotation connects statistical mechanics to quantum mechanics by replacing inverse temperature
{\displaystyle 1/(k_{\text{B}}T)}
with imaginary time
{\displaystyle it/\hbar }
. Consider a large collection of harmonic oscillators at temperature T. The relative probability of finding any given oscillator with energy E is
{\displaystyle \exp(-E/k_{\text{B}}T)}
, where kB is Boltzmann's constant. The average value of an observable Q is, up to a normalizing constant,
{\displaystyle \sum _{j}Q_{j}e^{-{\frac {E_{j}}{k_{\text{B}}T}}},}
where the j runs over all states,
{\displaystyle Q_{j}}
is the value of Q in the j-th state, and
{\displaystyle E_{j}}
is the energy of the j-th state. Now consider a single quantum harmonic oscillator in a superposition of basis states, evolving for a time t under a Hamiltonian H. The relative phase change of the basis state with energy E is
{\displaystyle \exp(-Eit/\hbar ),}
{\displaystyle \hbar }
is reduced Planck's constant. The probability amplitude that a uniform (equally weighted) superposition of states
{\displaystyle |\psi \rangle =\sum _{j}|j\rangle }
{\displaystyle |Q\rangle =\sum _{j}Q_{j}|j\rangle }
is, up to a normalizing constant,
{\displaystyle \left\langle Q\left|e^{-{\frac {iHt}{\hbar }}}\right|\psi \right\rangle =\sum _{j}Q_{j}e^{-{\frac {E_{j}it}{\hbar }}}\langle j|j\rangle =\sum _{j}Q_{j}e^{-{\frac {E_{j}it}{\hbar }}}.}
Statics and dynamics[edit]
Wick rotation relates statics problems in n dimensions to dynamics problems in n − 1 dimensions, trading one dimension of space for one dimension of time. A simple example where n = 2 is a hanging spring with fixed endpoints in a gravitational field. The shape of the spring is a curve y(x). The spring is in equilibrium when the energy associated with this curve is at a critical point (an extremum); this critical point is typically a minimum, so this idea is usually called "the principle of least energy". To compute the energy, we integrate the energy spatial density over space:
{\displaystyle E=\int _{x}\left[k\left({\frac {dy(x)}{dx}}\right)^{2}+V{\big (}y(x){\big )}\right]dx,}
where k is the spring constant, and V(y(x)) is the gravitational potential.
The corresponding dynamics problem is that of a rock thrown upwards. The path the rock follows is that which extremalizes the action; as before, this extremum is typically a minimum, so this is called the "principle of least action". Action is the time integral of the Lagrangian:
{\displaystyle S=\int _{t}\left[m\left({\frac {dy(t)}{dt}}\right)^{2}-V{\big (}y(t){\big )}\right]dt.}
{\displaystyle iS=\int _{t}\left[m\left({\frac {dy(it)}{dt}}\right)^{2}+V{\big (}y(it){\big )}\right]dt=i\int _{t}\left[m\left({\frac {dy(it)}{dit}}\right)^{2}-V{\big (}y(it){\big )}\right]d(it).}
Both thermal/quantum and static/dynamic[edit]
The Schrödinger equation and the heat equation are also related by Wick rotation. However, there is a slight difference. Statistical-mechanical n-point functions satisfy positivity, whereas Wick-rotated quantum field theories satisfy reflection positivity.[further explanation needed]
Wick rotation also relates a quantum field theory at a finite inverse temperature β to a statistical-mechanical model over the "tube" R3 × S1 with the imaginary time coordinate τ being periodic with period β.
Note, however, that the Wick rotation cannot be viewed as a rotation on a complex vector space that is equipped with the conventional norm and metric induced by the inner product, as in this case the rotation would cancel out and have no effect.
Interpretation and rigorous proof[edit]
Wick rotations can be seen a useful trick that holds due to the similarity between the equations of two seemingly distinct fields of physics. Quantum Field Theory in a Nutshell by Anthony Zee discusses Wick rotations, saying that[1]
Surely you would hit it big with mystical types if you were to tell them that temperature is equivalent to cyclic imaginary time. At the arithmetic level this connection comes merely from the fact that the central objects in quantum physics exp(−iH T) and in thermal physics exp(βH) are formally related by analytic continuation. Some physicists, myself included, feel that there may be something profound here that we have not quite understood.
It has been proven that a more rigorous link between Euclidean and quantum field theory can be constructed using the Osterwalder–Schrader theorem.[2]
^ Zee, A. (2010-02-01). Quantum Field Theory in a Nutshell: Second Edition. Princeton University Press. ISBN 978-1-4008-3532-4.
^ Schlingemann, Dirk (1999-10-01). "From euclidean field theory to quantum field theory". Reviews in Mathematical Physics. 11 (9): 1151–1178. arXiv:hep-th/9802035. doi:10.1142/S0129055X99000362. ISSN 0129-055X. S2CID 9851483.
Wick, G. C. (1954). "Properties of Bethe-Salpeter Wave Functions". Physical Review. 96 (4): 1124–1134. Bibcode:1954PhRv...96.1124W. doi:10.1103/PhysRev.96.1124.
Wikiquote has quotations related to Wick rotation.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Wick_rotation&oldid=1088940880"
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Electron Configurations - Course Hero
General Chemistry/Electron Behavior and Periodic Properties of Elements/Electron Configurations
Electrons have specific arrangements in atoms, called the electron configuration. Electrons fill orbitals, the area of an atom in which an electron has the greatest probability of being located, in order of increasing energy, and electrons must fill all orbitals in a subshell singly before doubly occupying any other orbital with the same energy.
Every atom has an electron configuration, which is the orbital filling of electrons in an atom based on their quantum numbers in increasing energies. A quantum number is one of four numbers,
n
\ell
m
s
, that together describe the orbital state of subatomic electrons. The shell in which the electron is found is described by
n
. The subshell an electron is found in is described by
\ell=0
for the s subshell,
\ell=1
for the p subshell,
\ell=2
for the d subshell, and
\ell=3
for the f subshell. The
m
quantum number gives the orientation of the electron. There are
2\ell+1
possible values for any
m
\ell
m
can be –1, 0, or 1. The spin of the electron,
s
, is either
+1\rm{/}2
-1\rm{/}2
. Hydrogen has one electron, and the quantum numbers of that electron are
n=1
\ell=0
m=0
s=+1\rm{/}2
When writing the notation for an element, the sequence of subshells is written, with the number of electrons in each subshell identified with a superscript. Thus, for hydrogen, the notation is 1s1. This notation is given in accordance with the Aufbau principle, which states that electrons fill orbitals in order of increasing energy. This order is given as follows:
\begin{gathered} 1s\lt2s\lt2p\lt3s\lt3p\lt4s\lt3d\lt4p\lt5s\lt4d\\\lt5p\lt6s\lt4f\lt5d\lt6p\lt7s\lt5f\lt6d\lt7p\end{gathered}
This order can be visualized in a grid according to Madelung's rule.
To visualize the order in which electrons fill subshells, follow the increasing energy down each arrow. Electrons fill subshells in this order of increasing energy, according to the Aufbau principle. The quantum number
n
represents what shell the electron is in, and s, p, d, and f represent the subshell, quantum number
\ell
Thus, the electron configuration notation for bromine, which has 35 electrons, is 1s22s22p63s23p64s23d104p5. Scientists often use an abbreviated form called the noble gas notation. This notation indicates in brackets the highest noble gas that fills the same subshells as the given element, followed by additional electrons above that noble gas. Noble gases have filled orbitals, so that part of the abbreviated form represents filled orbitals. Noble gas notation for bromine is thus [Ar]4s23d104p5. Argon is the closest noble gas before bromine on the periodic table. A degenerate orbital is one of two or more orbitals with the same energy. When filling degenerate orbitals, electrons must first singly occupy all the empty orbitals in the subshell before pairing within the same orbital, a process known as Hund's rule. Thus, electrons singly fill orbitals before they pair. The filling of orbitals can be denoted using an electron box diagram that shows the spin (up or down) of each electron in each orbital. In the orbital box diagram for oxygen, the electrons in the 2p degenerate orbital fill out as singletons before doubling up with an electron of opposite spin. The direction of the electron's spin is denoted by the direction of the arrows drawn in the orbital box diagram.
According to Hund's rule, when filling orbitals of the same energy, electrons occupy empty orbitals before doubly occupying the same orbital.
<The Bohr Atom>Quantum Theory
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Integration | Brilliant Math & Science Wiki
Pi Han Goh, q !, Andrew Ellinor, and
Integration is the process of evaluating integrals. It is one of the two central ideas of calculus and is the inverse of the other central idea of calculus, differentiation. Generally, we can speak of integration in two different contexts: the indefinite integral, which is the anti-derivative of a given function; and the definite integral, which we use to calculate the area under a curve.
Note that many integration problems will require the use of integration techniques.
Notations in Indefinite Integrals
f(x)
g(x)
be functions related in such a way that
\int f(x)\ dx = g(x) + C.
The integral sign
\int
denotes integration.
f(x)\ dx
is called the indefinite integral of
f(x)
x
f(x)
is called the integrand.
dx
is part of the integrand; it identifies the variable used in integration.
g(x) + C
is called the general anti-derivative of
f(x)
C
is called the constant of integration; it is an arbitrary constant.
C
has a specific value from given conditions,
g(x) + C
is called a particular antiderivative.
What is the anti-derivative of
\left(x^2-1\right)^3?
Expanding the expression gives
x^6 - 3x^4 + 3x^2 - 1
. Applying the reverse power rule
\int x^n \, dx = \frac1{n+1} x^{n+1} + C
n\ne-1,
\begin{aligned} \int \left(x^2-1\right)^3 \, dx &= \int \left(x^6 - 3x^4 + 3x^2 - 1\right) \, dx \\ \displaystyle &= \int x^6 \, dx - 3 \int x^4 \, dx + 3 \int x^2 \, dx - \int \, dx \\ \displaystyle &= \frac17 x^7 - \frac35x^5 + x^3 - x+ C, \end{aligned}
C
(
\int \, dx = \int 1 \, dx = x + C.)
_\square
What is the indefinite integral of
e^{2x}?
y=e^{2x}
, then differentiating
y
x
\frac{dy}{dx} = 2e^{2x} = 2y
dx = \frac{dy}{2y}
\int e^{2x} \, dx = \int y \cdot \frac{dy}{2y} = \frac12 \int dy = \frac12 y + C = \frac12 e^{2x} + C,
C
_\square
x \ln x ?
x = e^y
y = \ln \, x
\frac{d \, (\ln \, x)}{dx} = \frac{1}{x}. \frac{d \, (\ln \, e^y)}{dx} = \frac{1}{e^y}.
Back to the function
x \ln \, x
\frac{dy}{dx} = \frac{1}{e^y} \implies dy = \frac{dx}{e^y} \implies dx = e^y \, dy.
\int e^y \, y \cdot e^y \, dy = \int y e^{2y} \, dy.
Integrating by parts, we let
u = y \implies du = dy,
dv = e^{2y} \, dy \implies v = \frac12 e^{2y}.
\begin{aligned} \int y e^{2y} \, dy &= \int u \, dv \\ \displaystyle &= uv - \int v \, du \\ \displaystyle &= \frac12 y e^{2y} - \frac12 \int e^{2y} \, dy \\ \displaystyle &= \frac12 y e^{2y} - \frac12 \cdot \frac12 e^{2y} + C \\ \displaystyle &= \frac12 x^2 \ln x - \frac14 x^2 + C, \end{aligned}
C
_\square
\displaystyle \int_{\ln 2}^{\ln 5} e^{2x} \, dx.
From the previous example, we have
\displaystyle\int e^{2x} \, dx = \frac12 e^{2x} + C .
Setting the upper and lower limits to be
\ln 5
\ln 2
, respectively, the integral is evaluates to
\left[ \frac12 e^{2x} \right]_{\ln 2}^{\ln 5} = \frac12 \cdot \left(5^2-2^2\right) = \frac{21}2 = 10.5. \ _\square
\displaystyle \int_{-1}^1 \left(x^2-1\right)^3 \, dx.
From the earlier example, the indefinite integral of
\left(x^2-1\right)^3
\frac17 x^7-\frac35x^5 + x^3 - x + C.
Applying the limits, we have
\left [ \frac17 x^7\right]_{-1}^1 - \left[\frac35 x^5\right]_{-1}^1 + \left[x^3\right]_{-1}^1 - [x]_{-1}^1 =-\frac{32}{35}. \ _\square
Main Article: Integration by Parts and Integration with Partial Fractions
Find the anti-derivative of
x \sin x
We should integrate by parts.
Using the LIATE (Logarithm, Inverse trigonometric, Algebraic, Trigonometric, Exponential) rules, we have
u = x, dv = \sin x \, dx \implies du = dx, v = -\cos x.
\begin{aligned} \int x \sin x \, dx & = \int u \, dv \\ &= uv - \int v du \\ &= -x \cos x + \int \cos x \, dx \\ &= -x \cos x + \sin x + C, \end{aligned}
C
_\square
\frac1{x(x+1)}.
By partial fraction decomposition, we have
\frac1{x(x+1)} = \frac1x - \frac1{x+1} .
Recall that the anti-derivative of
\frac1x
\ln|x|
, so the anti-derivative of the expression in question is
\ln|x| - \ln|x+1| + C = \ln\left| \frac x{x+1} \right| + C,
C
_\square
\left\lfloor \frac{\displaystyle{ \int_{0}^{\infty} e^{-x^2}\, dx}}{\displaystyle{\int_{0}^{\infty} e^{-x^2} \cos 2x \, dx}}\right\rfloor= \ ?
\displaystyle\int_{-2014}^{2014}\underbrace{\sin(\sin(\sin(\ldots\sin(\sin x)\ldots)))}_{\mbox{2014 times}}\, dx = \ ?
Verify by integration that the area of a circle with radius
r
\pi r^2
By coordinate geometry, let the circle of radius
r
have its center at the origin, then the circle satisfies the equation
x^2 + y^2 =r^2
y = \sqrt{r^2-x^2}
. We want to integrate from
-r
r
to get the area of the semicircle:
\int_{-r}^r \sqrt{r^2-x^2} \, dx.
x = r \sin\theta
\frac{dx}{d\theta} = r\cos\theta.
\theta
to ranges from
-\frac{\pi}{2}
\frac{\pi}{2}.
Then, the integral becomes
\int_{-\frac\pi2}^{\frac\pi2} \sqrt{r^2-r^2\sin^2\theta} \cdot r\cos\theta \, dr = r^2 \int_{-\frac\pi2}^{\frac\pi2} \cos^2 \theta \,dr = r^2 \cdot \frac{\pi}{2}.
\cos^2\theta
can be found via the half-angle substitution. Thus, the area of the circle is
2\cdot r^2 \cdot \frac{\pi}{2} = \pi r^2.\ _\square
The figure at right shows a unit semicircle (in blue) and the graph of
f(x)=x+1
(in green).
Let the purple line segment
AB
be the longest vertical line segment that joins the blue semicircle and green line in the domain
[-1,0].
R
is the area of the black shaded region, what is
\left\lfloor 100R \right\rfloor ?
\int _{-2}^{2} \min\big\{x-\lfloor x \rfloor,-x - \lfloor-x \rfloor\big\} \, dx = \, ?
\displaystyle{\int _{ 0 }^{ \frac{\pi}{4}}{ \ln { (\tan { x+1)\, dx } } } .}
The answer must be a decimal in terms of
\pi
. For example, if the calculation result is
2.654\pi,
then the answer should be
2.654
\int_{0}^{\infty} \dfrac{\sin(1729x)}{x} \, dx = \, ?
Try more integral problems here.
Cite as: Integration. Brilliant.org. Retrieved from https://brilliant.org/wiki/integration/
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(Redirected from Basicity)
Find sources: "Base" chemistry – news · newspapers · books · scholar · JSTOR (September 2010) (Learn how and when to remove this template message)
2 Reactions between bases and water
5.1 Superbases
8 Solid bases
9 Bases as catalysts
10 Uses of bases
11 Monoprotic and polyprotic bases
11.1 Monoacidic bases
11.2 Diacidic bases
11.3 Triacidic bases
12 Etymology of the term
{\displaystyle NH_{3}(aq)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq)+OH^{-}(aq)}
Retrieved from "https://en.wikipedia.org/w/index.php?title=Base_(chemistry)&oldid=1078301418"
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Venn diagram — Wikipedia Republished // WIKI 2
A Venn diagram is a widely used diagram style that shows the logical relation between sets, popularized by John Venn in the 1880s. The diagrams are used to teach elementary set theory, and to illustrate simple set relationships in probability, logic, statistics, linguistics and computer science. A Venn diagram uses simple closed curves drawn on a plane to represent sets. Very often, these curves are circles or ellipses.
Similar ideas had been proposed before Venn. Christian Weise in 1712 (Nucleus Logicoe Wiesianoe) and Leonhard Euler (Letters to a German Princess) in 1768, for instance, came up with similar ideas. The idea was popularised by Venn in Symbolic Logic, Chapter V "Diagrammatic Representation", 1881.
5 Extensions to higher numbers of sets
5.1 Edwards–Venn diagrams
5.2 Other diagrams
In Venn diagrams, the curves are overlapped in every possible way, showing all possible relations between the sets. They are thus a special case of Euler diagrams, which do not necessarily show all relations. Venn diagrams were conceived around 1880 by John Venn. They are used to teach elementary set theory, as well as illustrate simple set relationships in probability, logic, statistics, linguistics, and computer science.
Stained-glass window with Venn diagram in Gonville and Caius College, Cambridge
Venn diagrams were introduced in 1880 by John Venn in a paper entitled "On the Diagrammatic and Mechanical Representation of Propositions and Reasonings"[3] in the Philosophical Magazine and Journal of Science,[4] about the different ways to represent propositions by diagrams.[5][6][7] The use of these types of diagrams in formal logic, according to Frank Ruskey and Mark Weston, is "not an easy history to trace, but it is certain that the diagrams that are popularly associated with Venn, in fact, originated much earlier. They are rightly associated with Venn, however, because he comprehensively surveyed and formalized their usage, and was the first to generalize them".[8]
Venn himself did not use the term "Venn diagram" and referred to his invention as "Eulerian Circles".[7] For example, in the opening sentence of his 1880 article Venn writes, "Schemes of diagrammatic representation have been so familiarly introduced into logical treatises during the last century or so, that many readers, even those who have made no professional study of logic, may be supposed to be acquainted with the general nature and object of such devices. Of these schemes one only, viz. that commonly called 'Eulerian circles,' has met with any general acceptance..."[5][6] Lewis Carroll (Charles L. Dodgson) includes "Venn's Method of Diagrams" as well as "Euler's Method of Diagrams" in an "Appendix, Addressed to Teachers" of his book Symbolic Logic (4th edition published in 1896). The term "Venn diagram" was later used by Clarence Irving Lewis in 1918, in his book A Survey of Symbolic Logic.[8][9]
Venn diagrams are very similar to Euler diagrams, which were invented by Leonhard Euler in the 18th century.[note 1][10][11] Margaret Baron has noted that Leibniz (1646–1716) produced similar diagrams before Euler in the 17th century, but much of it was unpublished.[12] She also observes even earlier Euler-like diagrams by Ramon Llull in the 13th Century.[13]
In the 20th century, Venn diagrams were further developed. David Wilson Henderson showed, in 1963, that the existence of an n-Venn diagram with n-fold rotational symmetry implied that n was a prime number.[14] He also showed that such symmetric Venn diagrams exist when n is five or seven. In 2002, Peter Hamburger found symmetric Venn diagrams for n = 11 and in 2003, Griggs, Killian, and Savage showed that symmetric Venn diagrams exist for all other primes. These combined results show that rotationally symmetric Venn diagrams exist, if and only if n is a prime number.[15]
Venn diagrams and Euler diagrams were incorporated as part of instruction in set theory, as part of the new math movement in the 1960s. Since then, they have also been adopted in the curriculum of other fields such as reading.[16]
See also: Set (mathematics) § Basic operations
{\displaystyle ~A\cap B}
{\displaystyle ~A\cup B}
Symmetric difference of two sets
{\displaystyle A~\triangle ~B}
Relative complement of A (left) in B (right)
{\displaystyle A^{c}\cap B~=~B\setminus A}
Absolute complement of A in U
{\displaystyle A^{c}~=~U\setminus A}
{\displaystyle A=\{1,\,2,\,5\}}
{\displaystyle B=\{1,\,6\}}
{\displaystyle C=\{4,\,7\}}
Extensions to higher numbers of sets
Non-example: This Euler diagram is not a Venn diagram for four sets as it has only 13 regions (excluding the outside); there is no region where only the yellow and blue, or only the red and green circles meet.
Five-set Venn diagram using congruent ellipses in a five-fold rotationally symmetrical arrangement devised by Branko Grünbaum. Labels have been simplified for greater readability; for example, A denotes A ∩ Bc ∩ Cc ∩ Dc ∩ Ec, while BCE denotes Ac ∩ B ∩ C ∩ Dc ∩ E.
Edwards–Venn diagrams
Anthony William Fairbank Edwards constructed a series of Venn diagrams for higher numbers of sets by segmenting the surface of a sphere, which became known as Edwards–Venn diagrams.[19] For example, three sets can be easily represented by taking three hemispheres of the sphere at right angles (x = 0, y = 0 and z = 0). A fourth set can be added to the representation, by taking a curve similar to the seam on a tennis ball, which winds up and down around the equator, and so on. The resulting sets can then be projected back to a plane, to give cogwheel diagrams with increasing numbers of teeth—as shown here. These diagrams were devised while designing a stained-glass window in memory of Venn.[19]
Edwards–Venn diagrams are topologically equivalent to diagrams devised by Branko Grünbaum, which were based around intersecting polygons with increasing numbers of sides. They are also two-dimensional representations of hypercubes.
Henry John Stephen Smith devised similar n-set diagrams using sine curves[19] with the series of equations
{\displaystyle y_{i}={\frac {\sin \left(2^{i}x\right)}{2^{i}}}{\text{ where }}0\leq i\leq n-2{\text{ and }}i\in \mathbb {N} .}
Charles Lutwidge Dodgson (also known as Lewis Carroll) devised a five-set diagram known as Carroll's square. Joaquin and Boyles, on the other hand, proposed supplemental rules for the standard Venn diagram, in order to account for certain problem cases. For instance, regarding the issue of representing singular statements, they suggest to consider the Venn diagram circle as a representation of a set of things, and use first-order logic and set theory to treat categorical statements as statements about sets. Additionally, they propose to treat singular statements as statements about set membership. So, for example, to represent the statement "a is F" in this retooled Venn diagram, a small letter "a" may be placed inside the circle that represents the set F.[20]
Venn diagrams correspond to truth tables for the propositions
{\displaystyle x\in A}
{\displaystyle x\in B}
Existential graph (by Charles Sanders Peirce)
Information diagram
Marquand diagram (and as further derivation Veitch chart and Karnaugh map)
Spherical octahedron – A stereographic projection of a regular octahedron makes a three-set Venn diagram, as three orthogonal great circles, each dividing space into two halves.
Three circles model
Vesica piscis
^ In Euler's Lettres à une princesse d'Allemagne sur divers sujets de physique et de philosophie [Letters to a German Princess on various physical and philosophical subjects] (Saint Petersburg, Russia: l'Academie Impériale des Sciences, 1768), volume 2, pages 95-126. In Venn's article, however, he suggests that the diagrammatic idea predates Euler, and is attributable to Christian Weise or Johann Christian Lange (in Lange's book Nucleus Logicae Weisianae (1712)).
^ "Intersection of Sets". web.mnstate.edu. Retrieved 2020-09-05.
^ a b "Sets and Venn Diagrams". www.mathsisfun.com. Retrieved 2020-09-05.
^ Venn, John. "On the Diagrammatic and Mechanical Representation of Propositions and Reasonings" (PDF). Penn Engineering. {{cite web}}: CS1 maint: url-status (link)
^ "The Philosophical Magazine: A Journal of Theoretical Experimental and Applied Physics". Taylor & Francis. Retrieved 2021-08-06.
^ a b Venn, John (July 1880). "I. On the Diagrammatic and Mechanical Representation of Propositions and Reasonings" (PDF). The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. 5. 10 (59): 1–18. doi:10.1080/14786448008626877. Archived (PDF) from the original on 2017-05-16. [1] [2]
^ a b Venn, John (1880). "On the employment of geometrical diagrams for the sensible representations of logical propositions". Proceedings of the Cambridge Philosophical Society. 4: 47–59.
^ a b Sandifer, Ed (2003). "How Euler Did It" (PDF). MAA Online. The Mathematical Association of America (MAA). Retrieved 2009-10-26.
^ a b Ruskey, Frank; Weston, Mark (2005-06-18). "A Survey of Venn Diagrams". The Electronic Journal of Combinatorics.
^ a b c Lewis, Clarence Irving (1918). A Survey of Symbolic Logic. Berkeley: University of California Press.
^ a b Venn, John (1881). Symbolic logic. Macmillan. p. 108. Retrieved 2013-04-09.
^ Mac Queen, Gailand (October 1967). The Logic Diagram (PDF) (Thesis). McMaster University. Archived from the original (PDF) on 2017-04-14. Retrieved 2017-04-14. (NB. Has a detailed history of the evolution of logic diagrams including but not limited to the Venn diagram.)
^ Leibniz, Gottfried Wilhelm (1903) [ca. 1690]. "De Formae Logicae per linearum ductus". In Couturat, Louis (ed.). Opuscules et fragmentes inedits de Leibniz (in Latin). pp. 292–321.
^ Henderson, David Wilson (April 1963). "Venn diagrams for more than four classes". American Mathematical Monthly. 70 (4): 424–426. doi:10.2307/2311865. JSTOR 2311865.
^ Ruskey, Frank; Savage, Carla D.; Wagon, Stan (December 2006). "The Search for Simple Symmetric Venn Diagrams" (PDF). Notices of the AMS. 53 (11): 1304–1311.
^ "Strategies for Reading Comprehension Venn Diagrams". Archived from the original on 2009-04-29. Retrieved 2009-06-20.
^ Weisstein, Eric W. "Venn Diagram". mathworld.wolfram.com. Retrieved 2020-09-05.
^ "Euler Diagrams 2004: Brighton, UK: September 22–23". Reasoning with Diagrams project, University of Kent. 2004. Retrieved 2008-08-13.
^ a b c Edwards, Anthony William Fairbank (2004). Cogwheels of the Mind: The Story of Venn Diagrams. Baltimore, Maryland, USA: Johns Hopkins University Press. p. 65. ISBN 978-0-8018-7434-5. .
^ Joaquin, Jeremiah Joven; Boyles, Robert James M. (June 2017). "Teaching Syllogistic Logic via a Retooled Venn Diagrammatical Technique". Teaching Philosophy. 40 (2): 161–180. doi:10.5840/teachphil201771767. Archived from the original on 2018-11-21. Retrieved 2020-05-12.
^ Grimaldi, Ralph P. (2004). Discrete and combinatorial mathematics. Boston: Addison-Wesley. p. 143. ISBN 978-0-201-72634-3.
^ Johnson, David L. (2001). "3.3 Laws". Elements of logic via numbers and sets. Springer Undergraduate Mathematics Series. Berlin, Germany: Springer-Verlag. p. 62. ISBN 978-3-540-76123-5.
Mahmoodian, Ebadollah S.; Rezaie, M.; Vatan, F. (March 1987). "Generalization of Venn Diagram" (PDF). Eighteenth Annual Iranian Mathematics Conference. Tehran and Isfahan, Iran. Archived from the original (PDF) on 2017-05-01. Retrieved 2017-05-01.
Edwards, Anthony William Fairbank (1989-01-07). "Venn diagrams for many sets". New Scientist. 121 (1646): 51–56.
Watkinson, John (1990). "4.10. Hamming distance". Coding for Digital Recording. Stoneham, MA, USA: Focal Press. pp. 94–99, foldout in backsleeve. ISBN 978-0-240-51293-8. (NB. The book comes with a 3-page foldout of a seven-bit cylindrical Venn diagram.)
Stewart, Ian (June 2003) [1992]. "Chapter 4. Cogwheels of the Mind". Another Fine Math You've Got Me Into (reprint of 1st ed.). Mineola, New York, USA: Dover Publications, Inc. (W. H. Freeman). pp. 51–64. ISBN 978-0-486-43181-9.
Glassner, Andrew (2004). "Venn and Now". Morphs, Mallards, and Montages: Computer-Aided Imagination. Wellesley, MA, USA: A. K. Peters. pp. 161–184. ISBN 978-1568812311.
Mamakani, Khalegh; Ruskey, Frank (2012-07-27). "A New Rose: The First Simple Symmetric 11-Venn Diagram". p. 6452. arXiv:1207.6452. Bibcode:2012arXiv1207.6452M. Archived from the original on 2017-05-01. Retrieved 2017-05-01.
Wikimedia Commons has media related to Venn diagrams.
"Venn diagram", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
Lewis Carroll's Logic Game – Venn vs. Euler at Cut-the-knot
Six sets Venn diagrams made from triangles
Interactive seven sets Venn diagram
VBVenn a free open source program for calculating and graphing quantitative two-circle Venn diagrams
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Practical Electronics/Resistors - Wikibooks, open books for an open world
Practical Electronics/Resistors
Resistors are passive devices, meaning that they cannot provide any power gain (amplification). They have a linear IV characteristic, meaning that the voltage across the resistor is directly proportional to the current through the resistor. The coefficient of proportionality is called the resistance (R). This is Ohm's Law:
{\displaystyle V=IR}
1 Resistor's Characteristics
1.3 Resistor's Construction
3.3 Series and Parallel
4 Uses of Resistors
5 2 Port Resistor Network
Resistor's Characteristics[edit | edit source]
ISO standard resistor symbol
Old-fashioned, but common, American resistor symbol
Resistors have the symbol on the right. Resistance is defined as the voltage developed across the resistor per ampere of current through the resistor. Resistance has a symbol of R measured in unit called Ohm which has a symbol Ω.
{\displaystyle R={\frac {V}{I}}}
Conductance[edit | edit source]
Conductance is the reciprocal of resistance (current per unit voltage), and is given the symbol G and the units siemens with symbol S.
{\displaystyle G={\frac {1}{R}}={\frac {I}{V}}}
Resistor's Construction[edit | edit source]
If we assume a resistor to be made of a prismatic (invariant cross-sectional area A along its length) conductor with length l, and conductivity ρ, we can express the conductance as follows:
{\displaystyle G=\rho {\frac {A}{l}}}
Form above, conductivity of materials can be calculated by
{\displaystyle \rho ={\frac {I}{V}}{\frac {l}{A}}}
When a resistor is connected with a DC source voltage, the resistance is calculated by Ohm's law:
{\displaystyle R={\frac {V}{I}}}
When a resistor is connected with an AC voltage source, voltage and current in the resistor have zero phase difference. The impedance of a resistor is calculated in the same way as in the DC case:
{\displaystyle Z_{R}={\frac {V}{I}}=R}
Carbon and ceramic resistors of different power ratings alongside a scale. The scale smallest division is millimetres (mm) and the largest (numbered) is centimetres (cm).
Resistors used in practical electronics range from one ohm to several million ohms
1 Ohm Ω
1,000 Kilo-ohm KΩ
1,000,000 Megaohm MΩ
Often a shorthand is used which means that the Ω symbol, which is usually not easily accessed on a computer, is not needed. The shorthand also eliminates the need for decimal points which are sometimes lost or missed off when documents are copied. The shorthand works by replacing a decimal point with the prefix of the resistance (e.g. K for kilo-ohms) or, for resistances in just ohms, R:
1 Ω 1R
1.2 Ω 1R2
1.2 KΩ 1K2
This notation is preferred, and will be used in this book. please note that the resistances are still said the same, so the value of a 1K2 resistor is pronounced "one point two kilo-ohms". The shorthand is not used when we are not talking about an individual resistor, for example, when measuring the resistance of a combination of resistors, we then express the value in ohms.
Resistors are not made perfect, and so they each have a tolerance. This is the maximum that a resistor can deviate from its specified value. It is expressed in percent, and the standard tolerance is 10%, which is more than adequate for most of our needs.
Preferred Values[edit | edit source]
To prevent thousands of different values of resistors clogging everything up, resistors only come in specified values. These values are spread across multiples of ten so that each is a constant multiple of the one beneath. The number of divisions per multiple of ten (decade) depends on the tolerance of the resistors. Standard (10%) resistors belong to the E12 series. The values are spread according to the rounded results of the following rule:
{\displaystyle R_{n}=R_{n-1}\times {\sqrt[{12}]{10}}}
After that, the cycle repeats, but a power of ten higher. Therefore, the first three decades are as follows:
1R 1R2 1R5 1R8 2R2 2R7 3R3 3R9 4R7 5R6 6R8 8R2
10R 12R 15R 18R 22R 27R 33R 39R 47R 56R 68R 82R
This pattern continues right up to the top of the resistor values that are available. For more accurate resistors, there exists an E24 (5%), E96 (1%) and an E192 (0.5%) series. There is also an E6 series for 20% resistors. To see a list of all values in these series, see this appendix.
Another way to understand the choice of resistor values in a given series (based on tolerance of the resistors) is as follows: Say there is a 10K resistor (nominal value) of tolerance 10%, then the actual value can be anything between 9K and 11K (approximately). So specifying another resistor value within this range is nonsense. The next available value is actually 12K (nominal), since its actual value can spread from 11K to 13.5K (approx). The following nominal value is 15K (with actual stagger within 13.5K and 16.5K) and so on. Hence the values available in this series are 10K, 12K, 15K, 18K, 22K, 27K, 33K, 39K, 47K, 56K, 68K, 82K and 100K (which starts the next series).
68K (range (68K - 6K8) to (68K + 6K8) or between 61K to 75K)
82K (range (82K -8K2) to (82K + 8K2) or between 74K to 90K) etc.
Colour Codes[edit | edit source]
For more in-depth info , read up on this chapter Practical Electronics/Finding component information/Resistors
Power dissipation[edit | edit source]
Besides resistance, the resistors available in the market also vary according to the power they can dissipate. In general terms, the greater the power dissipation, the larger the resistor. Thus size can be taken as a rough measure of power dissipation.
Resistors in Series and Parallel[edit | edit source]
Resistors can be connected in series to increase resistance or in parallel to decrease resistance
When resistors are wired in series, the total resistance is the sum of all the individual resistances. For an explanation of this, see this proof. So for the resistor network to the right, the total resistance, Rtot, is:
{\displaystyle R_{tot}=R_{1}+R_{2}+\ldots +R_{n}}
If n identical resistors are in series, then the total resistance is n times the resistance of those identical resistors. This is useful when you want a simple multiple of the resistance.
When a resistance is added in series with others, the total resistance always increases. The total resistance will therefore be more than the greatest value of resistor present.
When resistors are wired in parallel, the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances. For an explanation of this, see this proof. So for the resistor network to the right, the total resistance, Rtot, is:
{\displaystyle {\frac {1}{R_{tot}}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+\ldots +{\frac {1}{R_{n}}}}
If n identical resistors are in parallel, then the total resistance is 1/n the resistance of those identical resistors. This is useful when you want a simple fraction of the resistance.
When a resistance is added in parallel with others, the total resistance always decreases. The total resistance will therefore be less than the smallest value of resistor present.
Series and Parallel[edit | edit source]
Resistors can also be combined in a combination of series and parallel. To calculate the total resistance in these cases, simply break it into smaller parts that are basic series/parallel combinations and treat each one as one resistor. Consider the arrangement to the right. This is made up of two resistors in parallel, in series with another resistor.
The resistance of the two in parallel is:
{\displaystyle {\frac {1}{R_{p}}}={\frac {1}{100}}+{\frac {1}{10}}}
{\displaystyle R_{p}={\frac {100}{11}}}
When combined in series with the other resistance,
{\displaystyle R_{tot}=R_{p}+3.3\,}
{\displaystyle R_{tot}={\frac {100}{11}}+3.3\,}
{\displaystyle R_{tot}={\frac {1363}{110}}}
{\displaystyle R_{tot}=12.4\Omega \,}
Uses of Resistors[edit | edit source]
As we saw in a previous chapter, the current flowing in a simple circuit depends only on the voltage and resistance. This means that given a fixed voltage, by changing the resistance, the current can be tuned. The same applies when the voltage is what we want to fix, and the current is constant (or cannot be changed). By altering the resistance, we can set the voltage.
Because they resist flow of charge, resistors are also used to limit the current flowing through sensitive components. For example, a component that has a low resistance but cannot tolerate too much current should be used in series with a resistor, which will limit the current. The exact value of the resistor depends on the specifications of the device being used. A common application like this is when an LED is used. LEDs generally need 2V at 20mA to work properly. Say we have a 9V circuit, we need to lose 7V to operate the LED. Putting this into Ohm's Law, we get:
{\displaystyle R={\frac {V}{I}}}
{\displaystyle R={\frac {7}{20\times 10^{-3}}}}
{\displaystyle R=350\,}
Therefore, the resistor should be 350R to give the LED 2V at 20mA. However, this is not in the E12 series, so we find the next best, which is 390R. 330R is closer, but may exceed the LEDs current capacity, damaging it.
2 Port Resistor Network[edit | edit source]
Wikipedia has related information at Resistor
Circuit Theory page on resistive circuit analysis
Electronics page on resistors
Retrieved from "https://en.wikibooks.org/w/index.php?title=Practical_Electronics/Resistors&oldid=3819776"
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Represent the plane curve by a vector-valued function. y =
Sherry Becker 2021-11-28 Answered
y=x+5
Witheyesse47
To Determine: represent the plane curve by a vector function .
Given: we have a function
y=x+5
Explanation: we have a function
y=x+5
now let us consider
x=t
y=t+5
so we know that the vector valued function can be expressed by
r\left(t\right)=x\left(t\right)\stackrel{^}{i}+y\left(t\right)\stackrel{^}{j}
now putting the value in the function then we have
r\left(t\right)=ti+\left(t+5\right)j
=⟨t,\text{ }t+5⟩
\left(x,\text{ }y\right)=
Differentiate the following function.
r\left(t\right)=⟨4,\text{ }3\mathrm{cos}2t,\text{ }2\mathrm{sin}3t⟩
Critical Values:
z0.005=2.575,z0.01=2.325,z0.025=1.96,z0.05=1.645,z0.1=1.282
d.f.=31:t0.005=2.744,t0.01=2.453,t0.025=2.040,t0.05=1.696t0.1=1.309
In a random sample of soldiers who fought in the Battle of Preston, 774 soldiers who fought in the Battle of Preston, 774 soldiers were from the New Model Army, and 226 were from the New Model Army, and 226 were from the Royalist Army. Use a 0.05 significance level to test the claim that fewer than one quarter of the soldiers were Royalist
Find parametric equations the line (Use the parameter t.)
The line through the origin and the point
\left(9,\text{ }3,\text{ }-1\right)
\left(x\left(t\right),\text{ }y\left(t\right),\text{ }z\left(t\right)\right)=?
Find the symmetric equations.
\frac{x}{9}=\frac{y}{3}=z
x+9=y+3=z-1
x-9=y-3=z+1
\frac{x}{9}=\frac{y}{3}=-z
\frac{x}{3}=\frac{y}{9}=-z
r\left(t\right)=-2{t}^{5}i-6{t}^{4}j
{r}^{\prime }\left(t\right)×r{}^{″}\left(t\right)
\left(-2,\text{ }4,\text{ }7\right)
-x+2y+z=4
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3Blue1Brown - Linear combinations, span, and basis vectors
Chapter 2Linear combinations, span, and basis vectors
“Mathematics requires a small dose, not of genius, but of an imaginative freedom which, in a larger dose, would be insanity.”
\qquad
— Angus K. Rodgers
In the last chapter, along with the ideas of vector addition and scalar multiplication, I described vector coordinates, where there’s this back-and-forth between pairs of numbers and two-dimensional vectors.
Now I imagine that vector coordinates were already familiar to many of you, but there’s another interesting way to think about these coordinates, which is central to linear algebra. When you have a pair of numbers meant to describe a vector, like
(3, -2)
, I want you to think of each coordinate as a scalar, meaning think about how each one stretches or squishes vectors.
xy
-coordinate system, there are two special vectors. The one pointing to the right with length
1
, commonly called “i hat”
\hat i
or “the unit vector in the x-direction”. The other one is pointing straight up with length
1
, commonly called “j hat”
\hat j
or “the unit vector in the y-direction”. Now, think of the x-coordinate as a scalar that scales
\hat i
, stretching it by a factor of
3
, and the y-coordinate as a scalar that scales
\hat j
, flipping it and stretching it by a factor of
2
In this sense, the vector that these coordinates describe is the sum of two scaled vectors. This idea of adding together two scaled vectors is a surprisingly important concept. Those two vectors
\hat i
\hat j
have a special name: Together they are called the “basis” of the coordinate system. What this means is that when you think about coordinates as scalars, the basis vectors are what those scalars actually scale.
There’s also a more technical definition of basis, but we’ll get to that later. Framing our familiar coordinate system in terms of these two special basis vectors raises an interesting and subtle point: We could choose a different pair of basis vectors and get a perfectly reasonable new coordinate system.
Choosing Different Basis Vectors
For example, take some vector pointing up and to the right, along with a vector pointing down and to the right.
\overrightarrow{\mathbf{v}}=\begin{bmatrix}1\\ 2\end{bmatrix}
\overrightarrow{\mathbf{w}}=\begin{bmatrix}3\\ -1\end{bmatrix}
What values of the scalars
\alpha
\beta
satisfy the following equation?
\alpha\vec{\mathbf{v}}+\beta\vec{\mathbf{w}}=5\hat i -\frac12\hat j
\alpha=1\ \beta=0.5
\alpha=0.5\ \beta=1.5
\alpha=-1\ \beta=-1.5
\alpha=0.5\ \beta=-1.5
We have a new pair of basis vectors
\overrightarrow{\mathbf{v}}
\overrightarrow{\mathbf{w}}
. Take a moment to think about all the different vectors you can get by choosing two scalars, using each to scale one of the vectors, then adding them. Which two-dimensional vectors can you reach by altering your choice of scalars?
The answer is that you can describe every possible two-dimensional vector this way, and I think it’s a good puzzle to contemplate why. A new pair of basis vectors like this still gives you a way to go back and forth between pairs of numbers and two-dimensional vectors, but the association is definitely different from the version you get using the standard basis of
\hat i
\hat j
I’ll go into much more detail on this point in a later chapter, describing the relationship between different coordinate systems, but for now I just want you to appreciate that any way to describe vectors numerically depends on your choice of basis vectors.
Any time you’re scaling two vectors and adding them like this, it’s called a “linear combination” of those two vectors. Where does the word “linear” come from here? What does this have to do with lines? Well, when you multiply a scalar by a vector, it changes the magnitude of that vector. Multiplying every real number by the vector produces an infinite line that passes through the origin and the point defined by the vector.
So a linear combination of two vectors is a method of combining these two lines. For most pairs of vectors, if you let both scalars range freely and consider every possible vector you could get, you will be able to reach every possible point on the plane. Every two-dimensional vector is within your grasp.
However, if your two original vectors happen to line up, the lines produced by the scalar multiplication will be the same line, so adding them together can't yield a vector outside of that line. There’s a third possibility too: Both your vectors could be the 0 vector, in which case you’ll just be stuck at the origin.
The set of all possible vectors you can reach with linear combinations of a given pair of vectors is called the “span” of those two vectors. Restating what we just saw in this lingo, the span of most pairs of 2D vectors is all vectors in 2D space, but when they line up, their span is all vectors whose tip sit on a certain line.
Remember how I said linear algebra revolves around vector addition and scalar multiplication? The span of two vectors is basically a way of asking what are all the possible vectors you can reach using these two by only using those fundamental operations of vector addition and scalar multiplication.
What is the span of
\begin{bmatrix}3\\ -2\end{bmatrix}
\begin{bmatrix}-6\\ 4\end{bmatrix}
\mathbf{0}
\mathbb{R}
\mathbb{R}^2
Vectors vs Points
This is a good time to talk about how people commonly think about vectors as points. It gets very crowded to think about a whole collection of vectors sitting on a line, and more crowded still to think about all two-dimensional vectors all at once, filling up the plane.
So when dealing with collections of vectors like this, it’s common to represent each one just with a point in space, the point at the tip of this vector. That way, if you want to think about every possible vector whose tip sits on a certain line, just think about that line itself.
Likewise, to think about all possible two-dimensional vectors, conceptualize each one as the point where its tip sits. Then to think about all of them all at once, you can just think about the infinite flat sheet that is two-dimensional space, leaving the arrows out of it.
In general, if you’re thinking of a vector on its own, think of it as an arrow, and if you’re thinking of a collection, it’s convenient to think of them as points.
Span in 3D
The idea of span gets more interesting if we start thinking about vectors in three-dimensional space. For example, if you take two vectors in three-dimensional space that are not pointing in the same direction, what does it mean to take their span?
Well, their span is the collection of all possible linear combinations of those two vectors, meaning all possible vectors you get by scaling each of the two you start with in some way, then adding them together. You can imagine turning two knobs to change the two scalars defining the linear combination, adding the scaled vectors and following the tip of the resulting vector. That tip traces out some kind of flat sheet cutting through the origin of three-dimensional space.
\color{green}\overrightarrow{\mathbf{x}} \color{black}= \color{red}a\overrightarrow{\mathbf{v}} \color{black}+ \color{purple}b\overrightarrow{\mathbf{w}}
Which vectors in 3D space are not in this span?
The set of all possible vectors whose tips sit on this flat sheet is the span of your two vectors. Any vector which does not lie on the plane is not in the span.
So, what happens if you add on a third vector, and consider the span of all three of those guys? A linear combination of three vectors is defined pretty much the same way as for two: Choose three scalars, use them to scale each of your vectors, then add them all together. And again, the span of these vectors is the set of all possible linear combinations.
\color{green}\overrightarrow{\mathbf{x}} \color{black}= \color{red}a\overrightarrow{\mathbf{v}} \color{black}+ \color{purple}b\overrightarrow{\mathbf{w}} \color{black}+ \color{blue}c\overrightarrow{\mathbf{u}}
Two things could happen. The first possibility is if your third vector happens to be sitting on the span of the first two. Then the span doesn’t change, you’re sort of trapped on that same flat sheet. In other words, adding a scaled version of the third vector to linear combinations of the first two doesn’t give you access to any new vectors. This means the third vector can also be expressed as a linear combination of the other two:
\color{blue}\overrightarrow{\mathbf{u}} \color{black}= \color{red}a\overrightarrow{\mathbf{v}} \color{black}+ \color{purple}b\overrightarrow{\mathbf{w}}
There is another possibility though, if you just randomly choose a third vector, it’s almost certainly not sitting on the span of the first. Since it’s then pointing in a separate direction, it unlocks access to every possible three-dimensional vector! The way I like to think about this is that as you scale the new third vector, it moves around the span of the first two to sweep it through all of space.
It’s kind of like you’re making full use of the three freely-changing scalars that you have at your disposal to access the full three dimensions of space.
In the case where the third vector was sitting on the span of the first two, or the case where two vectors happen to line up, we want some terminology to describe the fact that at least one of these vectors is redundant, not adding anything to our span. Whenever this happens, where you have multiple vectors, and you could remove one without reducing their span, the relevant terminology is to say they are “linearly dependent”.
Another way of phrasing this would be to say that one of the vectors can be expressed as a linear combination of the others. That is, it’s already in the span of the other two. On the other hand, if each vector really does add another dimension to the span, they are said to be “linearly independent”.
\begin{align*} \text{Linearly Dependent: }\quad \color{blue}\overrightarrow{\mathbf{u}} \color{black}&= \color{red}a\overrightarrow{\mathbf{v}} \color{black}+ \color{purple}b\overrightarrow{\mathbf{w}} \quad\color{black}\text{ for some }\color{red}a\color{black}\text{ and }\color{purple}b \\ \text{Linearly Independent: }\quad \color{blue}\overrightarrow{\mathbf{u}} \color{black}&\neq \color{red}a\overrightarrow{\mathbf{v}} \color{black}+ \color{purple}b\overrightarrow{\mathbf{w}} \quad\color{black}\text{ for all }\color{red}a\color{black}\text{ and }\color{purple}b \end{align*}
The technical definition for the “basis” of a space is a set of linearly independent vectors that span that space, given how I described a basis earlier, and given your understanding of the words “span” and “linearly independent”, why does this definition make sense?
Earlier we learned that any pair of vectors could form a new basis as long as they didn't line up. If pair of vectors are linearly independent, their linear combination can span the entire 2D plane, meaning they can form the basis for the plane.
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Power of a Wave | Brilliant Math & Science Wiki
Matt DeCross, A Former Brilliant Member, A Former Brilliant Member, and
Abhi Kumbale
Waves are oscillatory disturbances in physical quantities, like light waves, sound waves, or transverse oscillations of a string. These disturbances take energy to create and propagate, in order to move the constituent particles or change the electric/magnetic fields. The power of a wave is therefore energy transported per unit time by the oscillations of a particular wave. The derivation of a formula for the power depends on the medium -- for light waves, the power is measured by the poynting vector, whereas for oscillations on a string, the power can be computed directly by balancing forces using Newton's second law. However, for all types of waves, the formula and physical meaning of the power takes similar forms, typically depending on the square amplitude of the waves among other factors.
Deriving the Power In String Oscillations
Energy in a Wavelength of Oscilation
To derive the power carried by the oscillations of a string, first note that a general solution to the wave equation for the amplitude of transverse oscillations of the string is:
y = A\sin { (\omega t-kx+\phi ) }
where the constants above are amplitude
A
, angular frequency
\omega
, wavenumber
k
, and phase shift
\phi
From considering the forces acting on a small element of the string, one can find the power carried by that element and from there the power carried by the oscillations throughout the string. See below diagram:
A small piece of an oscillating string is accelerated by tension forces due to the displacements of nearby parts of the string.
The power carried by a particle simply obeys the formula
P = \vec{F} \cdot \vec{v}
. Since the force acting on a small part of the string is just the tension
T
due to a nearby piece of the string, this power is:
P = Tv \cos (90 - \theta) = Tv \sin \theta,
\theta
is the angle between the tension force and the horizontal as in the above diagram. For small oscillations that satisfy the wave equation,
\sin \theta \approx \theta \approx \tan \theta
by the small-angle approximation. So the power can be rewritten in terms of the displacement
y
of the small piece of string as:
P = -T\frac { \partial y }{ \partial t } \frac { \partial y }{ \partial x }.
Using the general solution to the wave equation written above, this is:
P = -T\left( \omega A\cos { (\omega t-kx+\phi ) } \right) \left( -kA\cos { (\omega t-kx+\phi ) } \right) = Tk{ A }^{ 2 }\omega \cos ^{ 2 }{ (\omega t-kx+\phi ) } .
Time-averaging over a period of oscillation, one finds the average power transmitted to be:
\langle P \rangle = Tk{ A }^{ 2 }\omega \langle \cos ^{ 2 }{ (\omega t-kx+\phi ) } \rangle = \frac { Tk{ A }^{ 2 }\omega }{ 2 },
since the average of cosine squared over an oscillation is
\frac12
This formula can also be written with different variables, using the fact that wave velocity is given in terms of mass density
\mu
T
v = \sqrt{\frac{T}{\mu}}
and that by definition
k = \frac{\omega}{v}
. In this case the power is given by:
\langle P \rangle = \frac12 \mu v \omega^2 A^2.
\frac{2P}{9}
\frac{P}{3}
\frac{2P}{3}
\frac{P}{4}
An oscillating string carries a power
P
per unit time. If the length of the string is doubled (while keeping the mass of the string fixed) and the oscillations of the string are twice as rapid (while keeping the wave velocity constant), but the amplitude of oscillation is cut by a factor of 3, what is the new power carried by the oscillating string?
When a wave travels along a string, energy is transmitted along the direction of propagation of the wave, in the form of potential energy and kinetic energy of the string oscillation. The total energy transferred in a given time is given in terms of the average power as an integral:
E(t) = \int_0^t \langle P \rangle dt.
Show that the energy transferred in one period of an oscillating wave on a string is equal to the energy associated with one wavelength of oscillation of a string wave:
E_{\lambda} = \frac12m\omega^2 A^2 = \frac { \mu \lambda { \omega }^{ 2 }{ A }^{ 2 } }{ 2 },
\mu
is the mass per unit length. Note that the wave velocity is given in terms of
\mu
T
v = \sqrt{\frac{T}{\mu}}
k = \frac{\omega}{v}
Computing the integral above with
T_0
the period of oscillation,
T_0 = \frac{2\pi}{\omega}
E(T_0) = \int_0^{T_0} \langle P \rangle dt = \int_0^{T_0} \frac12 T k A^2 \omega dt = \frac12 T k A^2 \omega T_0.
Substituting in for
T
T_0
k
E(T_0) = \frac12 \mu v^2 \frac{\omega}{v} A^2 \omega \frac{2\pi}{\omega} = \frac12 \mu v \omega A^2 (2\pi) = \frac12 \mu \left(\frac{\lambda}{2\pi}\right) \omega^2 A^2 (2\pi) = \frac12 \mu \lambda \omega^2 A^2.
Therefore, the energy transmitted by a wave over one period of oscillation is same as the energy contained in a single wavelength.
This formula can also be proved using dimensional analysis. Note that the energy transmitted by the wave depends on the source of the waves. The source determines the amplitude of the wave as the source oscillates. The frequency, i.e., how rapid the oscillations are transmitted, also depends on the source. The mass density of the string determines the amount of inertia of the string; a heavier string oscillating at a particular frequency carries more energy because it takes more energy to move something with more inertia. Properties of the string likes its tension just characterize how massive the string is; the role of the tension is that it is the force internally transmitting the wave, so a greater tension corresponds to greater mass at a fixed velocity, consistent with the above formulae.
From the analysis, one can conclude that there are three fundamental quantities that fix the energy carried by a wave per unit wavelength. They are the:
Mass per unit length of the string
Thus, the energy per unit wavelength
u
is proportional to:
u \propto \mu^{a} A^b \omega^c,
for some exponents
a
b
c
. By requiring that the right-hand side have units of energy density, these exponents can be fixed to be
a=1
b=2
c=2
u \propto \mu \omega^2 A^2
, which is the correct energy per unit wavelength up to the prefactor of one-half.
1.2
0.7
0.3
0.02
20 \text{ g}/\text{cm}
2 \text{ cm}
50 \text{ Hz}
12 \text{ J}
Cite as: Power of a Wave. Brilliant.org. Retrieved from https://brilliant.org/wiki/power-of-a-wave/
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The following table gives a two-way classification of all basketball players at
The following table gives a two-way classification of all basketball players at a state university who began their college careers between 2004 and 20
The following table gives a two-way classification of all basketball players at a state university who began their college careers between 2004 and 2008, based on gender and whether or not they graduated.
\begin{array}{|ccc|}\hline & \text{Graduated}& \text{Did not Graduate}\\ \text{Male}& 129& 51\\ \text{Female}& 134& 36\\ \hline\end{array}\phantom{\rule{0ex}{0ex}}
If one of these players is selected at random, find the following probability.
Round your answer to four decimal places.
P\left(\text{graduated or male}\right)=
Enter your answer in accordance to the question statement
we have given that the following table of two-way classification of all basketball players at a state university who began their college careers between 2004 and 2008, based on gender and whether or not they graduated.
\begin{array}{|cccc|}\hline & \text{Graduated}& \text{Did not Graduate}& \text{Total}\\ \text{Male}& 129& 51& 180\\ \text{Female}& 134& 367& 170\\ \text{Total}& 263& 87& 350\\ \hline\end{array}\phantom{\rule{0ex}{0ex}}
We want to find the probability that randomly selected player is graduated or male
P\left(\text{graduated or male}\right)=?
P\left(\text{graduated or male}\right)=P\left(\text{graduated}\right)+P\left(\text{Male}\right)-P\left(\text{Graduated and Male}\right)
=\left(263/350\right)+\left(180/350\right)-\left(129/350\right)
=\frac{314}{350}
The probability that randomly selected player is graduated or male = 0.8971
\begin{array}{|ccc|}\hline & Like\text{ }Aerobic& Exercise\\ Like\text{ }Weight\text{ }Lifting& Yes& No& Total\\ Yes& 7& 14& 21\\ No& 12& 7& 19\\ Total& 29& 21& 40\\ \hline\end{array}
\begin{array}{|ccc|}\hline \text{Education}& \text{Use of vitamins takes}& \text{Does not take}\\ \text{No High School Diploma}& 0.03& 0.07\\ \text{High School Diploma}& 0.11& 0.39\\ \text{Undergraduate Degree}& 0.09& 0.27\\ \text{Graduate Degree}& 0.02& 0.02\\ \hline\end{array}
You randomly survey students at your school about what type of books they like to read. The two-way table shows your results. Find and interpret the marginal frequencies. Fiction LikesDislikesNon FictionLikes2622 Dislikes202
Researchers carried our a survey of fourth-, fifth-, and sixth-grade students in Michigan. Students were asked if good grades, athletic ability, or being popular was most important to them. The two-way table summarizes the survey data.
Create a two-way table that shows the joint and marginal relative frequencies using the table below.
\begin{array}{lccc}& Married& Notmarried& Total\\ Own& 172& 82& 254\\ Rent& 40& 54& 94\\ Total& 212& 136& 348\end{array}
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Find the indefinite integral. \int \sqrt[3]{\tan x}\sec^{2}xdx
\int \sqrt[3]{\mathrm{tan}x}{\mathrm{sec}}^{2}xdx
We have to solve the indefinite integral:
\int \sqrt[3]{\mathrm{tan}x}{\mathrm{sec}}^{2}xdx
Solving the integral by substitution method.
t=\mathrm{tan}x
\frac{dt}{dx}=\frac{d\mathrm{tan}x}{dx}
={\mathrm{sec}}^{2}x
dt={\mathrm{sec}}^{2}xdx
Substituting above values in the given integral, we get
\int \sqrt[3]{\mathrm{tan}x}{\mathrm{sec}}^{2}xdx=\int \sqrt[3]{t}dt
=\int {t}^{\frac{1}{3}}dt
=\frac{{t}^{\frac{1}{3}+1}}{\frac{1}{3}+1}+C
\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C
=\frac{{t}^{\frac{1+3}{3}}}{\frac{1+3}{3}}+C
=\frac{{t}^{\frac{4}{3}}}{\frac{4}{3}}+C
=\frac{3}{4}{t}^{\frac{4}{3}}+c
=\frac{3}{4}{\left(\mathrm{tan}x\right)}^{\frac{4}{3}}+C
Hence, value of given indefinite integration is
\frac{3}{4}{\left(\mathrm{tan}x\right)}^{\frac{4}{3}}+C
\int {\mathrm{sec}}^{2}\left(x\right)\sqrt{3}\left\{\mathrm{tan}\left(x\right)\right\}dx
u=\mathrm{tan}\left(x\right)⇒\frac{du}{dx}={\mathrm{sec}}^{2}\left(x\right)
=\int \sqrt{3}\left\{u\right\}du
\int {u}^{n}du=\frac{{u}^{n+1}}{n+1}
n=\frac{1}{3}:
=\frac{3{u}^{\frac{4}{3}}}{4}
u=\mathrm{tan}\left(x\right):
=\frac{3{\mathrm{tan}}^{\frac{4}{3}}\left(x\right)}{4}
\int {\mathrm{sec}}^{2}\left(x\right)\sqrt{3}\left\{\mathrm{tan}\left(x\right)\right\}dx
=\frac{3{\mathrm{tan}}^{\frac{4}{3}}\left(x\right)}{4}+C
A straight vertical wire carries a current of 1.20 A downward in a region b/t the poles of a large superconducting electromagnet,where the magnetic filed has magnitude B = 0.588 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is a) east? b)south? c) 30 degrees south of west?
According to the solutions, the force stays the same for parts
a,b,c:F=I\cdot l\cdot B=7.06×{10}^{-3}
but for part c,why isn't the 30 degree angle used to calculate the force?
A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.200 kg, and its center of gravity is located tits geometrical center. On the tray is a 1.00 kg plate of food and a 0.190 kg cup of coffee. Obtain the force T exerted by the thumb and the force F exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground.
Assume the temperature of the exhaust in an exhaust pipe can be approximated by
T={T}_{0}\left(1+a{e}^{-bx}\right)\left[1+c\mathrm{cos}\left(\omega t\right)\right]
{T}_{0}={100}^{\circ }rm\left\{C\right\},a=3,b=0.03rm{\left\{m\right\}}^{-1},c=0.05,\text{ }\text{ and }\text{ }\omega =100ra\frac{d}{s}.
If the exhaust speed is a constant 3 m/s, determine the time rate of change of temperature of the fluid particles at x = 0 and x = 4 m when t = 0.
Which transition in the hydrogen atom produces emitted light with the longest wavelength?
n=4\to n=3,
n=2\to n=1
n=3\to n=2
How many tiles 4 in. on a side should be used to cover a portion of a wall 48 in. long by 36 in. high?
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For f(t)=e^{-t}-1 , find the Laplace transform of \{\frac{f(t)}{t}\}.For g(t)=e^{-t}-2
For f(t)=e^{-t}-1 , find the Laplace transform of \{\frac{f(t)}{t}\}.For g(t)=e^{-t}-2 ,examine if the Laplace transform of \{\frac{g(t)}{t}\} exists
f\left(t\right)={e}^{-t}-1
, find the Laplace transform of
\left\{\frac{f\left(t\right)}{t}\right\}
. Then , for
g\left(t\right)={e}^{-t}-2
, examine if the Laplace transform of
\left\{\frac{g\left(t\right)}{t}\right\}
f\left(t\right)={e}^{-t}-1
Now , we know laplace transformation of
f\left(t\right)
L\left(f\left(t\right)\right)
L\left\{f\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt
Also we know , property of unilafenal Linear transformation is
L\left[\frac{f\left(t\right)}{t}\right]={\int }_{0}^{\mathrm{\infty }}
F(w)dw : where F(w) is the Laplace transformation of f(t)
Now , Laplace transformation of
f\left(t\right)={e}^{-t}-1
F\left(s\right)=L\left\{f\left(t\right)\right\}=L\left\{{e}^{-t}-1\right\}={\int }_{0}^{\mathrm{\infty }}\left({e}^{-t}-1\right){e}^{-st}dt
={\int }_{0}^{\mathrm{\infty }}\left({e}^{-\left(s+1\right)t}-{e}^{-st}\right)dt
={\left[\frac{{e}^{-\left(s+1\right)t}}{-\left(s+1\right)}+\frac{{e}^{-st}}{s}\right]}_{0}^{\mathrm{\infty }}
=\frac{1}{s+1}-\frac{1}{s}.\left(\text{since }\text{ }{e}^{-\mathrm{\infty }}=0\right)
Now , linear transformation of
\left\{\frac{f\left(t\right)}{t}\right\}
L\left\{\frac{f\left(t\right)}{t}\right\}=L\left\{\frac{{e}^{-t}-1}{t}\right\}={\int }_{s}^{\mathrm{\infty }}F\left(w\right)dw
={\int }_{s}^{\mathrm{\infty }}\left[\frac{1}{w+1}-\frac{1}{w}\right]dw
={\left[\mathrm{ln}\left(w+1\right)-\mathrm{ln}\left(w\right)\right]}_{s}^{\mathrm{\infty }}
={\left[\mathrm{ln}\left(\frac{w+1}{w}\right)\right]}_{s}^{\mathrm{\infty }}
=-\mathrm{ln}\left(\frac{s+1}{s}\right).⟨\underset{w\to \mathrm{\infty }}{lim}\mathrm{ln}\left(\frac{w+1}{w}\right)=0⟩
=\mathrm{ln}\left(\frac{s}{s+1}\right)
g\left(t\right)={e}^{-t}-2
G\left(s\right)=L\left\{g\left(t\right)\right\}=\frac{1}{s+1}-\frac{2}{s}
. Simply as previous
L\left\{\frac{g\left(t\right)}{t}\right\}=L\left\{\frac{{e}^{-t}-2}{t}\right\}={\int }_{s}^{\mathrm{\infty }}G\left(w\right)dw
y\prime \prime +3y\prime +2y={e}^{-t},y\left(0\right)=0y\prime \left(0\right)=0
{L}^{-1}\left\{\frac{3}{p\left(p+3\right)}\right\}
As part of trying to solve a differential equation using Laplace transforms, I have the fraction
\frac{-10s}{\left({s}^{2}+2\right)\left({s}^{2}+1\right)}
which I am trying to perform partial fraction decomposition on so that I can do a inverse Laplace transform. When I try to work out this fraction, I get that
-10=0\left(A+B\right)
{s}^{1}
does not show up in the fraction. How does partial fraction decomposition work for a fraction like this?
y"+4y=g\left(t\right)
y\left(0\right)=-1
{y}^{\prime }\left(0\right)=0,\text{ where }g\left(t\right)=\left\{\begin{array}{ll}t& ,t\le 2\\ 5& ,t>2\end{array}
y"-y=g\left(t\right)
y\left(0\right)=1
{y}^{\prime }\left(0\right)=2,\text{ where }g\left(t\right)=\left\{\begin{array}{ll}1& ,t\le 3\\ t& ,t>3\end{array}
The laplace transform for ramp function with a time delay can be expressed as follows
F\left(s\right)=\frac{1}{s\left({s}^{2}+2s+2\right)}
{L}^{-1}\left\{\frac{{e}^{-\pi s}}{{s}^{2}+1}\right\}
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Rivers Casino did an audit of all of their blackjack
Rivers Casino did an audit of all of their blackjack tables and found that the amount of money.What is the probability
Rivers Casino did an audit of all of their blackjack tables and found that the amount of money collected per night by a certain table has a mean of $6,000, a standard deviation of $200, and is approximately normally distributed.
What is the probability that a table will collect less than $6000 tomorrow night?
Determine whether this table represents a probability distribution.
\begin{array}{|cc|}\hline x& P\left(x\right)\\ 0& 0.05\\ 1& 0.3\\ 2& 0.25\\ 3& 0.4\\ \hline\end{array}
A. Yes, it is a probability distribution
B. No, it is not a probability distribution
The table shows the results of a survey of 100 authors by a publishing company.
\begin{array}{|cccc|}\hline & \text{New Authors}& \text{Established Authors}& Total\\ Successful& 15& 30& 45\\ Unsuccessful& 20& 35& 55\\ Total& 35& 65& 100\\ \hline\end{array}
Compute the following conditional probability. (Enter your probability as a fraction.)
An author is established, given that he is unsuccessful.
The following probability table is a breakdown of managers in four setors of economy:
\begin{array}{|ccccc|}\hline & Banking& Insurance& Real\text{ }Estate& Manufacturing\\ Male& 0.12& 0.14& 0.13& 0.21\\ Female& 0.08& 0.06& 0.17& 0.09\\ \hline\end{array}
a) What is the probability that the selected manager is female?
b) What is the probability that the selected manager is female or in insurane business?
c) What is the probability that the selected manager is neither female nor in insurance business?
Consider the following discrete probability distribution. What is the probability that x equals 7?
\begin{array}{|ccccc|}\hline x& 3& 4& 7& 9\\ P\left(X=x\right)& 0.22& 0.17& ?& 0.18\\ \hline\end{array}
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Evaluate the indefinite integral. \int \frac{e^{t}dt}{e^{2t}+2e^{t}+1}
Patricia Crane 2021-12-12 Answered
\int \frac{{e}^{t}dt}{{e}^{2t}+2{e}^{t}+1}
I=\int \frac{{e}^{t}dt}{{e}^{2t}+2{e}^{t}+1}
for evaluating given integral we substitute
{e}^{t}=x
now differentiating equation (2) with respect to t
\frac{d}{dt}\left({e}^{t}\right)=\frac{d}{dt}\left(x\right)\text{ }\text{ }\text{ }\left(\because \frac{d}{dx}\left({e}^{x}\right)={e}^{x}\right)
{e}^{t}=\frac{dx}{dt}
{e}^{t}dt=dx
now replacing
{e}^{t}dt\text{ }ask\text{ }with\text{ }dx\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{e}^{t}
with x in equation (1)
I=\int \frac{dx}{{x}^{2}+2x+1}\text{ }\text{ }\text{ }\left(\because {a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}\right)
=\int \frac{dx}{{x}^{2}+2\left(x\right)\left(1\right)+{1}^{2}}
=\int \frac{dx}{{\left(x+1\right)}^{2}}
=\int {\left(x+1\right)}^{-2}dx\text{ }\text{ }\text{ }\left(\because \int {\left(x+a\right)}^{n}dx=\frac{{\left(x+a\right)}^{n+1}}{n+1}+c\right)
=\frac{{\left(x+1\right)}^{2+1}}{-2+1}+c
=\frac{{\left(x+1\right)}^{-1}}{-1}+c
=-\frac{1}{x+1}+c
now replacing x with et in equation (3)
I=-\frac{1}{{e}^{t}+1}+c
-\frac{1}{{e}^{t}+1}+c
\int \frac{{e}^{t}}{{e}^{2t}+2{e}^{t}+1}dt
\int \frac{1}{{u}^{2}+2u+1}du
\int \frac{1}{{\left(u+1\right)}^{2}}du
\int \frac{1}{{v}^{2}}dv
-\frac{1}{v}
-\frac{1}{u+1}
-\frac{1}{{e}^{t}+1}
-\frac{1}{{e}^{t}+1}+C
\int {x}^{2}{\mathrm{tan}}^{-1}xdx
\left({x}^{2}\right)\left({e}^{y}\right)\frac{dy}{dx}=4
\int \left(2x+1\right){e}^{{x}^{2}+x}dx
{y}^{\prime }+{x}^{2}y={x}^{2}
\int {e}^{x}\mathrm{tan}\left({e}^{x}\right)dx
What is a solution to the differential equation xy'=y?
Find the following indefinite integral:
\int 12x{\left({x}^{2}-9\right)}^{-3}
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Train support vector machine (SVM) classifier for one-class and binary classification - MATLAB fitcsvm - MathWorks Benelux
G\left({x}_{j},{x}_{k}\right)=\mathrm{exp}\left(-{‖{x}_{j}-{x}_{k}‖}^{2}\right)
G\left({x}_{j},{x}_{k}\right)={x}_{j}\prime {x}_{k}
G\left({x}_{j},{x}_{k}\right)={\left(1+{x}_{j}\prime {x}_{k}\right)}^{q}
\left\{\begin{array}{l}{\alpha }_{j}\left[{y}_{j}f\left({x}_{j}\right)-1+{\xi }_{j}\right]=0\\ {\xi }_{j}\left(C-{\alpha }_{j}\right)=0\end{array}
f\left({x}_{j}\right)=\varphi \left({x}_{j}\right)\prime \beta +b,
0.5\sum _{jk}{\alpha }_{j}{\alpha }_{k}G\left({x}_{j},{x}_{k}\right)
{\alpha }_{1},...,{\alpha }_{n}
\sum {\alpha }_{j}=n\nu
0\le {\alpha }_{j}\le 1
f\left(x\right)=x\prime \beta +b,
2/‖\beta ‖.
‖\beta ‖
0.5{‖\beta ‖}^{2}+C\sum {\xi }_{j}
{y}_{j}f\left({x}_{j}\right)\ge 1-{\xi }_{j}
{\xi }_{j}\ge 0
0.5\sum _{j=1}^{n}\sum _{k=1}^{n}{\alpha }_{j}{\alpha }_{k}{y}_{j}{y}_{k}{x}_{j}\prime {x}_{k}-\sum _{j=1}^{n}{\alpha }_{j}
\sum {\alpha }_{j}{y}_{j}=0
0\le {\alpha }_{j}\le C
\stackrel{^}{f}\left(x\right)=\sum _{j=1}^{n}{\stackrel{^}{\alpha }}_{j}{y}_{j}x\prime {x}_{j}+\stackrel{^}{b}.
\stackrel{^}{b}
{\stackrel{^}{\alpha }}_{j}
\stackrel{^}{\alpha }
\text{sign}\left(\stackrel{^}{f}\left(z\right)\right).
0.5\sum _{j=1}^{n}\sum _{k=1}^{n}{\alpha }_{j}{\alpha }_{k}{y}_{j}{y}_{k}G\left({x}_{j},{x}_{k}\right)-\sum _{j=1}^{n}{\alpha }_{j}
\sum {\alpha }_{j}{y}_{j}=0
0\le {\alpha }_{j}\le C
\stackrel{^}{f}\left(x\right)=\sum _{j=1}^{n}{\stackrel{^}{\alpha }}_{j}{y}_{j}G\left(x,{x}_{j}\right)+\stackrel{^}{b}.
{C}_{j}=n{C}_{0}{w}_{j}^{\ast },
{x}_{j}^{\ast }=\frac{{x}_{j}-{\mu }_{j}^{\ast }}{{\sigma }_{j}^{\ast }},
\begin{array}{c}{\mu }_{j}^{\ast }=\frac{1}{\sum _{k}{w}_{k}^{*}}\sum _{k}{w}_{k}^{*}{x}_{jk},\\ {\left({\sigma }_{j}^{\ast }\right)}^{2}=\frac{{v}_{1}}{{v}_{1}^{2}-{v}_{2}}\sum _{k}{w}_{k}^{*}{\left({x}_{jk}-{\mu }_{j}^{\ast }\right)}^{2},\\ {v}_{1}=\sum _{j}{w}_{j}^{*},\\ {v}_{2}=\sum _{j}{\left({w}_{j}^{*}\right)}^{2}.\end{array}
\sum _{j=1}^{n}{\alpha }_{j}=n\nu .
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Traveler's dilemma - Wikipedia
In game theory, the traveler's dilemma (sometimes abbreviated TD) is a non-zero-sum game in which each player proposes a payoff. The lower of the two proposals wins; the lowball player receives the lowball payoff plus a small bonus, and the highball player receives the same lowball payoff, minus a small penalty. Surprisingly, the Nash equilibrium is for both players to aggressively lowball. The traveler's dilemma is notable in that naive play appears to outperform the Nash equilibrium; this apparent paradox also appears in the centipede game and the finitely-iterated prisoner's dilemma.
5 Payoff matrix
The original game scenario was formulated in 1994 by Kaushik Basu and goes as follows:[1][2]
"An airline loses two suitcases belonging to two different travelers. Both suitcases happen to be identical and contain identical antiques. An airline manager tasked to settle the claims of both travelers explains that the airline is liable for a maximum of $100 per suitcase—he is unable to find out directly the price of the antiques."
"To determine an honest appraised value of the antiques, the manager separates both travelers so they can't confer, and asks them to write down the amount of their value at no less than $2 and no larger than $100. He also tells them that if both write down the same number, he will treat that number as the true dollar value of both suitcases and reimburse both travelers that amount. However, if one writes down a smaller number than the other, this smaller number will be taken as the true dollar value, and both travelers will receive that amount along with a bonus/malus: $2 extra will be paid to the traveler who wrote down the lower value and a $2 deduction will be taken from the person who wrote down the higher amount. The challenge is: what strategy should both travelers follow to decide the value they should write down?"
The two players attempt to maximize their own payoff, without any concern for the other player's payoff.
One might expect a traveler's optimum choice to be $100; that is, the traveler values the antiques at the airline manager's maximum allowed price. Remarkably, and, to many, counter-intuitively, the Nash equilibrium solution is in fact just $2; that is, the traveler values the antiques at the airline manager's minimum allowed price.
For an understanding of why $2 is the Nash equilibrium consider the following proof:
Alice, having lost her antiques, is asked their value. Alice's first thought is to quote $100, the maximum permissible value.
On reflection, though, she realizes that her fellow traveler, Bob, might also quote $100. And so Alice changes her mind, and decides to quote $99, which, if Bob quotes $100, will pay $101.
But Bob, being in an identical position to Alice, might also think of quoting $99. And so Alice changes her mind, and decides to quote $98, which, if Bob quotes $99, will pay $100. This is greater than the $99 Alice would receive if both she and Bob quoted $99.
This cycle of thought continues, until Alice finally decides to quote just $2—the minimum permissible price.
Another proof goes as follows:
If Alice only wants to maximize her own payoff, choosing $99 trumps choosing $100. If Bob chooses any dollar value 2–98 inclusive, $99 and $100 give equal payoffs; if Bob chooses $99 or $100, choosing $99 nets Alice an extra dollar.
A similar line of reasoning shows that choosing $98 is always better for Alice than choosing $99. The only situation where choosing $99 would give a higher payoff than choosing $98 is if Bob chooses $100—but if Bob is only seeking to maximize his own profit, he will always choose $99 instead of $100.
This line of reasoning can be applied to all of Alice's whole-dollar options until she finally reaches $2, the lowest price.
The ($2, $2) outcome in this instance is the Nash equilibrium of the game. By definition this means that if your opponent chooses this Nash equilibrium value then your best choice is that Nash equilibrium value of $2. This will not be the optimum choice if there is a chance of your opponent choosing a higher value than $2.[3] When the game is played experimentally, most participants select a value higher than the Nash equilibrium and closer to $100 (corresponding to the Pareto optimal solution). More precisely, the Nash equilibrium strategy solution proved to be a bad predictor of people's behavior in a traveler's dilemma with small bonus/malus and a rather good predictor if the bonus/malus parameter was big.[4]
Furthermore, the travelers are rewarded by deviating strongly from the Nash equilibrium in the game and obtain much higher rewards than would be realized with the purely rational strategy. These experiments (and others, such as focal points) show that the majority of people do not use purely rational strategies, but the strategies they do use are demonstrably optimal. This paradox could reduce the value of pure game theory analysis, but could also point to the benefit of an expanded reasoning that understands how it can be quite rational to make non-rational choices, at least in the context of games that have players that can be counted on to not play "rationally." For instance, Capraro has proposed a model where humans do not act a priori as single agents but they forecast how the game would be played if they formed coalitions and then they act so as to maximize the forecast. His model fits the experimental data on the Traveler's dilemma and similar games quite well.[5] Recently, the traveler's dilemma was tested with decision undertaken in groups rather than individually, in order to test the assumption that groups decisions are more rational, delivering the message that, usually, two heads are better than one.[6] Experimental findings show that groups are always more rational – i.e. their claims are closer to the Nash equilibrium - and more sensitive to the size of the bonus/malus.[7]
Some players appear to pursue a Bayesian Nash equilibrium.[8][9]
The traveler's dilemma can be framed as a finitely repeated prisoner's dilemma.[8][9] Similar paradoxes are attributed to the centipede game and to the p-beauty contest game[7] (or more specifically, "Guess 2/3 of the average"). One variation of the original traveler's dilemma in which both travelers are offered only two integer choices, $2 or $3, is identical mathematically to the standard non-iterated Prisoner's dilemma and thus the traveler's dilemma can be viewed as an extension of prisoner's dilemma. (The minimum guaranteed payout is $1, and each dollar beyond that may be considered equivalent to a year removed from a three-year prison sentence.) These games tend to involve deep iterative deletion of dominated strategies in order to demonstrate the Nash equilibrium, and tend to lead to experimental results that deviate markedly from classical game-theoretical predictions.
Payoff matrix[edit]
The canonical payoff matrix is shown below (if only integer inputs are taken into account):
Canonical TD payoff matrix
100, 100 97, 101 96, 100 95, 99 ⋯ 1, 5 0, 4
101, 97 99, 99 96, 100 95, 99 ⋯ 1, 5 0, 4
100, 96 100, 96 98, 98 95, 99 ⋯ 1, 5 0, 4
99, 95 99, 95 99, 95 97, 97 ⋯ 1, 5 0, 4
⋮ ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ ⋮
5, 1 5, 1 5, 1 5, 1 ⋯ 3, 3 0, 4
{\displaystyle S=\{2,...,100\}}
the set of strategies available to both players and by
{\displaystyle F:S\times S\rightarrow \mathbb {R} }
the payoff function of one of them we can write
{\displaystyle F(x,y)=\min(x,y)+2\cdot \operatorname {sgn}(y-x)}
(Note that the other player receives
{\displaystyle F(y,x)}
since the game is quantitatively symmetric).
^ Kaushik Basu, "The Traveler's Dilemma: Paradoxes of Rationality in Game Theory"; American Economic Review, Vol. 84, No. 2, pp. 391–395; May 1994.
^ Kaushik Basu,"The Traveler's Dilemma"; Scientific American, June 2007
^ Wolpert, D (2009). "Schelling Formalized: Strategic Choices of Non-Rational Personas". SSRN 1172602. {{cite journal}}: Cite journal requires |journal= (help)
^ Capra, C. Monica; Goeree, Jacob K.; Gomez, Rosario; Holt, Charles A. (1999-01-01). "Anomalous Behavior in a Traveler's Dilemma?". The American Economic Review. 89 (3): 678–690. doi:10.1257/aer.89.3.678. JSTOR 117040.
^ Capraro, V (2013). "A Model of Human Cooperation in Social Dilemmas". PLOS ONE. 8 (8): e72427. arXiv:1307.4228. doi:10.1371/journal.pone.0072427. PMC 3756993. PMID 24009679.
^ Cooper, David J; Kagel, John H (2005-06-01). "Are Two Heads Better Than One? Team versus Individual Play in Signaling Games" (PDF). American Economic Review. 95 (3): 477–509. doi:10.1257/0002828054201431. ISSN 0002-8282.
^ a b Morone, A.; Morone, P.; Germani, A. R. (2014-04-01). "Individual and group behaviour in the traveler's dilemma: An experimental study". Journal of Behavioral and Experimental Economics. 49: 1–7. doi:10.1016/j.socec.2014.02.001.
^ a b Becker, T., Carter, M., & Naeve, J. (2005). Experts Playing the Traveler's Dilemma (No. 252/2005). Department of Economics, University of Hohenheim, Germany.
^ a b Baader, Malte; Vostroknutov, Alexander (October 2017). "Interaction of reasoning ability and distributional preferences in a social dilemma". Journal of Economic Behavior & Organization. 142: 79–91. doi:10.1016/j.jebo.2017.07.025.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Traveler%27s_dilemma&oldid=1045850476"
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Conservative vs. non-conservative forces Practice Problems Online | Brilliant
As shown in the figure above, a spring is attached horizontally to a vertical wall, and an object lies on a frictionless horizontal floor, attached to the other end of the spring. A man pulls the object so that the spring doubles in length, and then let the object go. Which of the following statements is/are true concerning this situation?
I. The force exerted by the spring on the object is a conservative force.
II. The potential energy in the spring is constant.
III. If no other force is applied after the man lets it go, the object will keep on moving back and forth.
A conservative force is
\text{\_\_\_\_\_\_\_\_\_}.
a force that conserves the total energy of an object a force that conserves the form of an object a force that conserves the kinetic energy of an object a force that conserves the mechanical energy of an object
A ball
A
2\text{ kg}
moves in a straight line at
5\text{ m/s}
on a frictionless horizontal floor. It collides with another
2\text{ kg}
B
that is at rest. After the collision, ball
A
comes to a complete stop, and ball
B
starts to move at a speed of
5\text{ m/s}.
Which of the following is NOT correct concerning this situation?
The total linear momentum is conserved. The force exerted on
B
A
is a conservative force. The collision is perfectly elastic. The force exerted on
A
B
is a non-conservative force.
Which of the following choices correctly fills the blanks of the sentence below, in the right order?
"Gravity is a
\text{\_\_\_\_\_\_\_\_\_}
force and friction is a
\text{\_\_\_\_\_\_\_\_\_}
force."
conservative, conservative conservative, non-conservative non-conservative, conservative non-conservative, non-conservative
Consider a car is traveling down a hill. Which of the following is a conservative force that acts on the car?
The gravitational force that pulls the car down the hill. The force generated by the engine of the car. The air resistance that acts on the car. The kinetic friction between the wheels of the car and the road.
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Generalized 3-circular projections for unitary congruence invariant norms
July 2016 Generalized
3
-circular projections for unitary congruence invariant norms
Abdullah Bin Abu Baker
{P}_{0}
on a complex Banach space is generalized
3
- circular if its linear combination with two projections
{P}_{1}
{P}_{2}
having coefficients
{\lambda }_{1}
{\lambda }_{2}
, respectively, is a surjective isometry, where
{\lambda }_{1}
{\lambda }_{2}
are distinct unit modulus complex numbers different from
1
{P}_{0}\oplus {P}_{1}\oplus {P}_{2}=I
. Such projections are always contractive. In this paper, we prove structure theorems for generalized
3
-circular projections acting on the spaces of all
n×n
symmetric and skew-symmetric matrices over
\mathbb{C}
when these spaces are equipped with unitary congruence invariant norms.
Abdullah Bin Abu Baker. "Generalized
3
-circular projections for unitary congruence invariant norms." Banach J. Math. Anal. 10 (3) 451 - 465, July 2016. https://doi.org/10.1215/17358787-3599609
Received: 16 March 2015; Accepted: 17 August 2015; Published: July 2016
Keywords: generalized 3-circular projection , isometry , spectral theorem , unitary congruence invariant norm
Abdullah Bin Abu Baker "Generalized
3
-circular projections for unitary congruence invariant norms," Banach Journal of Mathematical Analysis, Banach J. Math. Anal. 10(3), 451-465, (July 2016)
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Train support vector machine (SVM) classifier for one-class and binary classification - MATLAB fitcsvm - MathWorks Nordic
G\left({x}_{j},{x}_{k}\right)=\mathrm{exp}\left(-{‖{x}_{j}-{x}_{k}‖}^{2}\right)
G\left({x}_{j},{x}_{k}\right)={x}_{j}\prime {x}_{k}
G\left({x}_{j},{x}_{k}\right)={\left(1+{x}_{j}\prime {x}_{k}\right)}^{q}
\left\{\begin{array}{l}{\alpha }_{j}\left[{y}_{j}f\left({x}_{j}\right)-1+{\xi }_{j}\right]=0\\ {\xi }_{j}\left(C-{\alpha }_{j}\right)=0\end{array}
f\left({x}_{j}\right)=\varphi \left({x}_{j}\right)\prime \beta +b,
0.5\sum _{jk}{\alpha }_{j}{\alpha }_{k}G\left({x}_{j},{x}_{k}\right)
{\alpha }_{1},...,{\alpha }_{n}
\sum {\alpha }_{j}=n\nu
0\le {\alpha }_{j}\le 1
f\left(x\right)=x\prime \beta +b,
2/‖\beta ‖.
‖\beta ‖
0.5{‖\beta ‖}^{2}+C\sum {\xi }_{j}
{y}_{j}f\left({x}_{j}\right)\ge 1-{\xi }_{j}
{\xi }_{j}\ge 0
0.5\sum _{j=1}^{n}\sum _{k=1}^{n}{\alpha }_{j}{\alpha }_{k}{y}_{j}{y}_{k}{x}_{j}\prime {x}_{k}-\sum _{j=1}^{n}{\alpha }_{j}
\sum {\alpha }_{j}{y}_{j}=0
0\le {\alpha }_{j}\le C
\stackrel{^}{f}\left(x\right)=\sum _{j=1}^{n}{\stackrel{^}{\alpha }}_{j}{y}_{j}x\prime {x}_{j}+\stackrel{^}{b}.
\stackrel{^}{b}
{\stackrel{^}{\alpha }}_{j}
\stackrel{^}{\alpha }
\text{sign}\left(\stackrel{^}{f}\left(z\right)\right).
0.5\sum _{j=1}^{n}\sum _{k=1}^{n}{\alpha }_{j}{\alpha }_{k}{y}_{j}{y}_{k}G\left({x}_{j},{x}_{k}\right)-\sum _{j=1}^{n}{\alpha }_{j}
\sum {\alpha }_{j}{y}_{j}=0
0\le {\alpha }_{j}\le C
\stackrel{^}{f}\left(x\right)=\sum _{j=1}^{n}{\stackrel{^}{\alpha }}_{j}{y}_{j}G\left(x,{x}_{j}\right)+\stackrel{^}{b}.
{C}_{j}=n{C}_{0}{w}_{j}^{\ast },
{x}_{j}^{\ast }=\frac{{x}_{j}-{\mu }_{j}^{\ast }}{{\sigma }_{j}^{\ast }},
\begin{array}{c}{\mu }_{j}^{\ast }=\frac{1}{\sum _{k}{w}_{k}^{*}}\sum _{k}{w}_{k}^{*}{x}_{jk},\\ {\left({\sigma }_{j}^{\ast }\right)}^{2}=\frac{{v}_{1}}{{v}_{1}^{2}-{v}_{2}}\sum _{k}{w}_{k}^{*}{\left({x}_{jk}-{\mu }_{j}^{\ast }\right)}^{2},\\ {v}_{1}=\sum _{j}{w}_{j}^{*},\\ {v}_{2}=\sum _{j}{\left({w}_{j}^{*}\right)}^{2}.\end{array}
\sum _{j=1}^{n}{\alpha }_{j}=n\nu .
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Conditional Probability Warmup Practice Problems Online | Brilliant
You spin a spinner with equal probability of landing on each of the numbers 1 through 10. Your friend covers up the number it lands on, but you can tell that it only has one digit (and therefore can't be 10). What is the probability it landed on a 5?
\frac{1}{8}
\frac{1}{9}
\frac{1}{10}
\frac{1}{11}
Suppose that Phil is a compulsive liar. Every statement he says has a 75% probability of being a lie. He flips a fair coin and tells you it's a head. Is the coin flip more likely to have been a head or a tail?
A Head A Tail
Phil is still a compulsive liar whose every statement has a 75% probability of being a lie. However, we know that he is aware of the winning number in a lottery that consists of choosing a single integer from 1 to a million. He says the winning number is 123. If you were to enter the lottery, which number should you pick to maximize the probability of winning?
Note: if Phil decides to lie about the winning number, he will pick any incorrect but plausible number with equal probability.
123 Any number but 123
Suppose that in a given population of people everyone is either good or bad at jumping, and good or bad at running. These are independent of each other (so knowing if someone is good at jumping tells you nothing about how they are at running). An athletic training camp only accepts people who are either good at running or good at jumping (possibly good at both).
You meet a random person who went to the camp, and you find out they are good at running. Are they more, less, or equally likely to be good at jumping than someone random at the camp whose running ability you don't know?
Note: Assume that there is at least one person good at only running and one person good at only jumping at the camp.
More Equally Less
You flip two coins and your friend tells you that both landed on the same side. Conditional on this information, what is the probability that at least one of the coins landed on heads?
\frac{1}{4}
\frac{1}{2}
\frac{3}{4}
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Electro Dynamo Theory of Gravity (g)
Industrial Tests, Inc., Rocklin, CA, USA
In this paper, flux transfer events from the Sun to the Earth are represented as an electrical frequency. The Earth’s inner core is modeled as a motor armature and the mechanical speed is calculated from the frequency of flux transfer events. The speed of the inner core creates a centripetal acceleration at the center of the Earth that is much higher than ever thought possible. The synchronous speed is the speed of the electromagnetic field that orbits or rotates around the Earth. All electrical machines have a rotating electromagnetic field that establishes the synchronous speed. Gravity is a centripetal acceleration derived from the synchronous speed of the electromagnetic field. The mechanical speed is somewhat less based on slip, which ranges from 1% to 10% for the electrical machine we call Earth. Flux transfer event from the Sun to the Earth is what powers the machine. By adjusting for the altitude distance, from the inner core to the surface of the Earth, a value of 9.806 m/s2 is calculated. The very nature of gravity has been discovered and explained using electrical equations and classical physics. There are only three forces of nature: electromagnetic, strong nuclear and weak nuclear.
Centripetal Acceleration, Dynamo, Flux Transfer Event, Gravity (g), Gyroscope, Synchronous Speed
N=\frac{120f}{P}
v=\frac{r2\pi \text{RPM}}{60}
v=r\times \text{RPM}\times 0.10472
V=\sqrt{\frac{GM}{r}}
\text{Newton}’\text{sConstant}=G=6.674\times {10}^{-11}
\text{MassoftheEarth}=M=5.972\times {10}^{24}\text{kg}
\text{Radiusofinnercore}=r=1.22\times {10}^{6}\text{m}
a=\frac{{v}^{2}}{r}
{g}_{alt}=g{\left(\frac{{r}_{e}}{{r}_{e}}+h\right)}^{2}
{g}_{alt}
{r}_{e}
\text{d}B=\frac{{\mu }_{0}I\overrightarrow{\text{d}L}\times \overrightarrow{\widehat{r}}}{4\pi {R}^{2}}=\frac{{\mu }_{0}I\text{d}L\mathrm{sin}\theta }{4\pi {R}^{2}}
B=\frac{{\mu }_{0}I}{4\pi {R}^{2}}\oint \text{d}L=\frac{{\mu }_{0}I}{4\pi {R}^{2}}2\pi R=\frac{{\mu }_{0}I}{2R}
{\mu }_{0}=4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}
F=qvB\mathrm{sin}\theta
F=q\frac{L}{t}B\mathrm{sin}\theta
F=\frac{q}{t}LB\mathrm{sin}\theta
F=ILB\mathrm{sin}\theta
F=ILB\mathrm{sin}\theta
F=ILB\mathrm{sin}90
F=ILB
\text{Torque}=\tau =rF\mathrm{sin}\theta
{K}_{t}=\frac{\text{Torque}}{\text{Current}}
{K}_{t}=\frac{2.0217\times {10}^{24}}{1.373\times {10}^{10}}
{K}_{t}=4.73\times {10}^{9}
{K}_{s}=\frac{1}{4.73\times {10}^{9}}
{K}_{s}=6.79\times {10}^{-11}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\text{Newton}’\text{s}\text{\hspace{0.17em}}\text{Constant}
emf=vBL\mathrm{sin}\theta
emf=vBL
\text{Generated}\text{\hspace{0.17em}}\text{voltage}=emf=\text{Volcity}\times B\text{\hspace{0.17em}}\text{field}\times \text{Length}
{k}_{v}=\frac{\text{Voltage}}{\omega \left(\text{rad}/\text{s}\right)}
V=2.22\times {10}^{8}\text{V}
\omega =18075\text{\hspace{0.17em}}\text{m}/\text{s}=0.0148155\left(\text{rad}/\text{s}\right)
{k}_{v}=1.5\times {10}^{10}
{k}_{s}=\frac{1}{{k}_{v}}=\frac{1}{1.5}\times {10}^{10}
{k}_{s}=6.68\times {10}^{-11}
{K}_{s}=\frac{1}{{K}_{v}}=\frac{1}{\frac{220309330}{0.01466\text{\hspace{0.17em}}\text{rad}/\text{s}}}=6.65\times {10}^{-11}\left(\text{Newton}’\text{s}\text{\hspace{0.17em}}\text{Constant}\right)
\mu =r{v}^{2}
r=1.22\times {10}^{6}\text{\hspace{0.17em}}\text{m}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{InnerCoreRadius}\right)
\mu =3.986\times {10}^{14}{\text{m}}^{\text{3}}/{\text{s}}^{\text{2}}
v=18075\text{\hspace{0.17em}}\text{m}/\text{s}
a=\frac{{v}^{2}}{r}
a=268\text{\hspace{0.17em}}\text{m}/{\text{s}}^{\text{2}}
g=268\times {\left(\frac{1.22\times {10}^{6}}{6.376\times {10}^{6}}\right)}^{2}
g=9.806\text{\hspace{0.17em}}\text{m}/{\text{s}}^{\text{2}}
B2\pi r=\mu NI
B=\frac{\mu NI}{2\pi r}
\text{Magneticfield}=\text{permeability}\times \text{turndensity}\times \text{current}
a=\frac{{v}^{2}}{r}
\text{slip}=\frac{{n}_{s}-n}{{n}_{s}}\times 100
{n}_{s}
Poole, G. (2019) Electro Dynamo Theory of Gravity (g). Journal of High Energy Physics, Gravitation and Cosmology, 5, 773-789. https://doi.org/10.4236/jhepgc.2019.53040
1. Elsasser, W. (1958) The Earth as a Dynamo. Scientific American, 198, 44-48. https://doi.org/10.1038/scientificamerican0558-44
2. Lockwood, M. and Wild, M.N. (1993) On the Quasi-Periodic Nature of Magnetopause Flux Transfer Events. Journal of Geophysical Research, 98, 5935-5940. https://doi.org/10.1029/92JA02375
3. Fear, R.C., Trenchi, L., Coxon, J.C. and Milan, S.E. (2017) How Much Flux Transfer Does a Flux Transfer Event Transfer. JGR Space Physics, 122, 12,310-12,327. https://doi.org/10.1002/2017JA024730
4. Milan, S.E., Lester, M., Greenwald, R.A. and Sofko, G. (1999) The Ionospheric Signature of Transient Dayside Reconnection and the Associated Pulsed Convection Return Flow. Annales Geophysicae, 17, 1166-1171. https://doi.org/10.1007/s00585-999-1166-2
5. Milan, S.E., Lester, M., Cowley, S.W.H. and Brittnacher, M. (2000) Convection and Auroral Response to a Southward Turning of the IMF: Polar UVI, CUTLASS, and IMAGE Signatures of Transient Magnetic Flux Transfer at the Magnetopause. Journal of Geophysical Research, 105, 15,741-15,755. https://doi.org/10.1029/2000JA900022
6. Ahrens, T.J. (1995) Global Earth Physics A Handbook of Physical Constants. American Geophysics Union, Washington DC, 93.
7. Poole, G. (2018) Dynamo Speed Control and Tectonics—Modeling Earth as a Shunt Wound DC Machine. Journal of High Energy Physics, Gravitation and Cosmology, 4, 152-165. https://doi.org/10.4236/jhepgc.2018.41014
8. Brumbach, M.E. and Clade, J. (2003) Industrial Maintenance. Thomson Delmar Learning, Library of Congress, Washington DC, 400.
9. Raymond, H.M. (1910) Cyclopedia of Mechanical Engineering: A General Reference Work on Machine Shop.... Chicago American Technical Society, Chicago, Volume 3, 10.
10. Seeds, M. (2009) The Solar System. Thomson Higher Education, Belmont, 83.
11. Breithaupt, J. (2004) Physics for OCR. Nelson Thomas Ltd., Nashville.
12. Dennis, J.T. (2003) The Complete Idiots Guide to Physics. Alpha Books, Indianapolis, 103.
13. Shadowitz, A. (1975) The Electromagnetic Field. McGraw-Hill Book Company, Pennsylvania, 88.
14. Buffett, B. (2010) Tidal Dissipation and the Strength of the Earths Internal Magnetic Field. Nature, 468, 952-954. https://doi.org/10.1038/nature09643
15. Poole, G. (2018) Cosmic Wireless Power Transfer System and the Equation for Everything E = mc2 = vc2/60 = a3/T = G(M1+ M2)/4π2 = (KE+PE)/1.0E15 = Q = PA/F = λ/hc = 1/2q = VI = 1/2LI2 = 1/2CV = I2R = … Journal of High Energy Physics, Gravitation and Cosmology, 4, 588-650. https://doi.org/10.4236/jhepgc.2018.44036
16. Laithwaite, E. (1968) Video: The Circle of Magnetism. Imperial College of London in Association with the Royale Institute, London.
17. Zocholl, S. (2003) AC Motor Protection. Schweitzer Engineering Laboratories, Pullman.
18. Poole, G. (2017) Theory of Electromagnetism and Gravity. Journal of High Energy Physics, Gravitation and Cosmology, 3, 663-692. https://doi.org/10.4236/jhepgc.2017.34051
19. Russell, C.T. and Elphic, R.C. (1979) ISEE Observations of Flux Transfer Events at the Dayside Magnetopause. Geophysical Research Letters, 6, 33-36. https://doi.org/10.1029/GL006i001p00033
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LimitPoints - Maple Help
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : RegularChains : AlgebraicGeometryTools Subpackage : LimitPoints
compute the limit points of a regular chain
LimitPoints(rc, R)
LimitPoints(rc, R, L)
LimitPoints(rc, R, coefficient=real)
LimitPoints(rc, R, output=rootof)
LimitPoints(rc, R, output=chain)
The command LimitPoints(rc, R) returns the non-trivial limit points of the quasi-component given by the regular chain rc in the Zariski topology. Non-trivial refers to the limit points that are not points of that same quasi-component.
The returned limit points forum a zero-dimensional variety which, by default, is given as the union of the zero sets of regular chains.
Each limit point returned by LimitPoints(rc, R) is obtained by following a branch (given by a Puiseux series solution) of rc associated with a root of the product of the initials of rc.
If the optional argument L is present, then only the limit points obtained from a branch associated with a root of a polynomial in L are returned.
If the option coefficient=real is present, then only the points obtained from a real branch are returned.
If the option output=chain is present, then the returned limit points are given as solutions of zero-dimensional regular chains; this is the default representation for the returned limit points.
If the option output=rootof is present, then RootOf expressions are are used (instead of regular chains) to represent the coordinates of the limit points.
This command is part of the RegularChains[AlgebraicGeometryTools] package, so it can be used in the form LimitPoints(..) only after executing the command with(RegularChains[AlgebraicGeometryTools]). However, it can always be accessed through the long form of the command by using RegularChains[AlgebraicGeometryTools][LimitPoints](..).
\mathrm{with}\left(\mathrm{RegularChains}\right):
\mathrm{with}\left(\mathrm{ChainTools}\right):
\mathrm{with}\left(\mathrm{AlgebraicGeometryTools}\right):
R≔\mathrm{PolynomialRing}\left([x,y,z]\right)
\textcolor[rgb]{0,0,1}{R}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{polynomial_ring}}
\mathrm{rc}≔\mathrm{Chain}\left([{y}^{5}-{z}^{4},xz-{y}^{2}],\mathrm{Empty}\left(R\right),R\right)
\textcolor[rgb]{0,0,1}{\mathrm{rc}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}
\mathrm{Display}\left(\mathrm{rc},R\right)
{\begin{array}{cc}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ {\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{\ne }\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\end{array}
\mathrm{lm}≔\mathrm{LimitPoints}\left(\mathrm{rc},R\right)
\textcolor[rgb]{0,0,1}{\mathrm{lm}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}]
\mathrm{Display}\left(\mathrm{lm},R\right)
[{\begin{array}{cc}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\end{array}]
\mathrm{rc}≔\mathrm{Chain}\left([{y}^{5}-{z}^{4}{\left(z+1\right)}^{5},xz{\left(z+1\right)}^{2}-{y}^{2}],\mathrm{Empty}\left(R\right),R\right)
\textcolor[rgb]{0,0,1}{\mathrm{rc}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}
\mathrm{lm}≔\mathrm{LimitPoints}\left(\mathrm{rc},R\right);
\mathrm{Display}\left(\mathrm{lm},R\right)
\textcolor[rgb]{0,0,1}{\mathrm{lm}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}]
[{\begin{array}{cc}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\end{array}\textcolor[rgb]{0,0,1}{,}{\begin{array}{cc}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\end{array}\textcolor[rgb]{0,0,1}{,}{\begin{array}{cc}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\end{array}]
\mathrm{lm}≔\mathrm{LimitPoints}\left(\mathrm{rc},R,[z]\right):
\mathrm{Display}\left(\mathrm{lm},R\right)
[{\begin{array}{cc}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\end{array}]
\mathrm{rc}≔\mathrm{Chain}\left([{y}^{3}-2{y}^{3}+{y}^{2}+{z}^{5},{z}^{4}x+{y}^{3}-{y}^{2}],\mathrm{Empty}\left(R\right),R\right)
\textcolor[rgb]{0,0,1}{\mathrm{rc}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}
\mathrm{lm}≔\mathrm{LimitPoints}\left(\mathrm{rc},R,\mathrm{coefficient}=\mathrm{complex}\right)
\textcolor[rgb]{0,0,1}{\mathrm{lm}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}]
\mathrm{Display}\left(\mathrm{lm},R\right)
[{\begin{array}{cc}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\end{array}\textcolor[rgb]{0,0,1}{,}{\begin{array}{cc}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\end{array}]
\mathrm{lm}≔\mathrm{LimitPoints}\left(\mathrm{rc},R,\mathrm{coefficient}=\mathrm{real}\right)
\textcolor[rgb]{0,0,1}{\mathrm{lm}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{regular_semi_algebraic_system}}]
\mathrm{Display}\left(\mathrm{lm},R\right)
[{\begin{array}{cc}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\\ \textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{}\end{array}]
The RegularChains[AlgebraicGeometryTools][LimitPoints] command was introduced in Maple 2020.
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Use the two-way table below and answer the questions. 1)
Use the two-way table below and answer the questions. 1) P (< 3 cars | 3 to 5 kids) = ? 2) P ( > 3 kids or >= 2 cars) = ?
Use the two-way table below and answer the questions.
1) P (< 3 cars | 3 to 5 kids) = ?
2) P ( > 3 kids or
\ge
\begin{array}{}& & Pet\\ Gender& Male& 33& 8\\ & Female& 35& 11\end{array}
One hundred adults and children were randomly selected and asked whether they spoke more than one language fluently. The data were recorded in a two-way table. Maria and Brennan each used the data to make the tables of joint relative frequencies shown below, but their results are slightly different. The difference is shaded. Can you tell by looking at the tables which of them made an error?
Please, tell whether the statement is TRUE or FALSE.
1) The F distribution is symmetric, and its mean is close to 1.
2) A small P-value indicates that the two variables in a two-way table are associated.
3) If the null hypothesis of an ANOVA F-test is true, the F statistic is greater than 1.
4) The more similar the conditional distributions in a two-way table are, the smaller the P-value of the chi-square test becomes.
5) In a two-way table, if the expected counts are about the same as the observed counts, we reject the null hypothesis.
6) The larger the chi-square statistic, the stronger the evidence against the null hypothesis.
Statistics students at a state college compiled the following two-way table from a sample of randomly selected students at their college:
\begin{array}{|ccc|}\hline & \text{Play chess}& \text{Don`t play chess}\\ \text{Male students}& 25& 162\\ \text{Female students}& 19& 148\\ \hline\end{array}\phantom{\rule{0ex}{0ex}}
Answer the following questions about the table. Be sure to show any calculations.
What question about the population of students at the state college would this table attempt to answer?
{H}^{0}
{H}^{1}
for the test related to this table.
Tell whether the statement is TRUE or FALSE.
1. The F distribution is not symmetric but skewed to the right (or skewed positively).
2. If the null hypothesis of an ANOVA F-test is true, the F statistic is close to one.
3. In a two-way table, if the expected counts are about the same as the observed counts, we fail to reject the null hypothesis.
4. If the conditional distributions in a two-way table are similar to each other, the P-value of the test is close to zero.
5. A small P-value indicates that the two variables in a two-way table are not associated.
P\left(A\cup B\right)
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Cross_ventilation Knowpia
Cross-ventilation pertains to wind, fresh air or a breeze entering through an opening (namely a window) that flows directly through the occupied space and out through an opening on the opposite side of the building, where the air pressure is lower, whereby creating a flow of cool air and as well as a current of air across the room from the exposed area to the sheltered area. Windows or vents positioned on opposite sides of the room allow passive breezes a pathway through the structure, which circulate the air and provide passive cooling.[1]
Cross-breezes work when two windows are opposite of each other.
Cross-ventilation is a wind-driven effect and requires no energy, in addition to being the most effective method of wind ventilation. A commonly used technique to remove pollutants and heat in an indoor environment, cross-ventilation can also decrease or even obviate the need for an air-conditioner and can improve indoor air quality. Others terms used for the effect include, cross-breeze, natural cross-ventilation, cross-draft, through-draft, wind-driven ventilation, wind effect ventilation and cross-flow ventilation.[2]
The phenomena occurs when openings in an environment (including vehicles) or building (houses, factories, sheds, etc) are set on opposite or adjoining walls, which allow air to enter and exit, thus creating a current of air across the interior environment. There is also a pressure difference between the opposite sides of the establishment. The effect is mostly driven by the wind, whereby the air is pulled into the building on the high pressure windward part and is pushed out on the low pressure downwind side of the establishment (because of the pressure difference between the openings). A wind's effect on a structure creates regions that have positive pressure on the building's upwind area and a negative pressure on the downwind side. Thus, the building shape and local wind patterns are critical in making wind pressures that force airflow through its openings.[3][4]
If the windows on both sides of the buildings are opened, the overpressure on the side facing the wind, and/or low pressure on the adjacent protected side, will make a current of air through the room from the uncovered side towards the sheltered side. If there are windows on both sides in a building, cross ventilation is appropriate where the width of the room is up to five times the floor-to-ceiling height. If openings are only one side then wind-driven ventilation is more suited for structures where the width is around 2.5 times the floor to ceiling height.[5]
Window size is important for cross ventilation
Cross ventilation relies on many factors, such as the tightness of the establishment, wind direction and how much wind is available, its potential travel through chimneys, vents and other openings in the home. Casement windows can be installed to improve to improve cross-breezes. Air quality may also affect cross ventilation. Although cross ventilation is generally more direct at its job than stack ventilation, its cons include its effects being unproductive on hot, still days, when it is most necessary. Moreover, cross ventilation is generally only suitable for narrow buildings. The contrasting height of the openings (walls, sill, panels or furniture) ordered by the space also immediately influence the level and velocity of ventilation.[1]
EffectivityEdit
Cross ventilation works well in climates with hotter temperatures, where the system allows continual changes of the air within the building, refreshing it and reducing the temperature inside the structure and also when the window on the windward side of the building is not opened as much as the one on the leeward side. Cross ventilation will not be efficacious if the windows are more than 12m apart and if a window is behind a door that is regularly shut.[6]
An opened window that faces a prevailing wind and is conjugated with another window on the opposite side of a building will supply natural ventilation for fresh air. A decent and effective cross ventilation will remove heat from the interior and keep indoor air temperatures approximately 1.5 C° (2.7 F°) below the outdoor air temperatures, ensuring that there is a steady inflow and outflow of fresh air inside the building.[7]
Besides windows, brise soleils, doors, louvers or ventilation grills and ducts can also work as effective ventilation openings, though an awning window provides the least effectivity. The wind surrounding building structures is important when it comes to assessing the air quality and thermal comfort indoors since both air and heat exchange rely heavily on the wind pressure on the exterior of the building. For the best airflow, the windward windows of the occupied space should not be opened as much as those on the leeward side.[8] Disadvantages of wind-driven ventilation include capricious wind speeds and directions (which may create a strong unpleasant draft), and the polluted air from the outside that may tarnish the indoor air quality.[9]
In a windcatcher, wind is forced down on the windward side and exits on the leeward side, using the stack effect.
There are four different types of cross ventilation:[10]
Single-sided ventilation: This method depends on the pressure contrasts between different openings within the occupied space. For rooms that only feature a single opening, the ventilation is impelled by turbulence, thereby creating a pumping activity on that lone opening, causing small inflows and outflows. It is worth noting that single-sided ventilation has a weak effect. It is preferable when cross ventilation is not achievable, where it uses windows or vents at the other side of the space to control air pressure.
Cross-ventilation (single spaces): Being unsophisticated and efficacious, this type of ventilation is a horizontal process that is driven by pressure differences between the windward and leeward sides of the occupied indoor environment. Ventilation here is generally provided using windows and vents at either side of a building where the variation in pressure draw air in and out.
Cross-ventilation (double-banked spaces): Involving banked rooms, this method features openings in the hallway structure. The openings allow a way for noise to move between spaces. It can provide a much higher air-exchange rate in comparison with single-sided ventilation.
Stack ventilation: This ventilation is a vertical process and it's beneficiary for taller buildings with central atriums. It draws cooler air in at a lower level, whereby the air rises thereafter due to heat exposure before it is ventilated out at a higher level. Benefits from temperature compartmentalization and related pressure quality of the air, whereby warm air loses density when it rises and the cooler air supplants it.
For a simple volume with two openings, the cross wind flow rate can be calculated using the following equation:[11]
{\displaystyle Q=U_{\textrm {wind}}{\sqrt {\frac {C_{\textrm {p1}}-C_{\textrm {p2}}}{1/\left(A_{\textrm {1}}^{2}C_{\textrm {1}}^{2}\right)+1/\left(A_{\textrm {2}}^{2}C_{\textrm {2}}^{2}\right)}}}\qquad {}\left(1\right)}
{\displaystyle U_{\textrm {wind}}}
is the far-field wind speed;
{\displaystyle C_{\textrm {p1}}}
is a local pressure drag coefficient for the building, defined at the location of the upstream opening;
{\displaystyle C_{\textrm {p2}}}
is a local pressure drag coefficient for the building, defined at the location of the downstream opening;
{\displaystyle A_{\textrm {1}}}
is the cross-sectional area of the upstream opening;
{\displaystyle A_{\textrm {2}}}
is the cross-sectional area of the downstream opening;
{\displaystyle C_{\textrm {1}}}
is the discharge coefficient of the upstream opening; and
{\displaystyle C_{\textrm {2}}}
is the discharge coefficient of the downstream opening.
For rooms with single opening, the calculation of ventilation rate is more complicated than cross-ventilation due to the bi-directional flow and strong turbulent effect. The ventilation rate for single-sided ventilation can be accurately predicted by combining different models for mean flow, pulsating flow and eddy penetration.[12] The mean flow rate for single-sided ventilation is determined by:
{\displaystyle {\bar {Q}}={\frac {C_{d}\;l\;{\sqrt {Cp}}\;\int \limits _{z_{0}}^{h}{\sqrt {-{\frac {2\;\Delta \;P(z)}{\rho }}}}\,\mathrm {d} z}{z_{ref}^{1/7}}}\;{\bar {U}}}
l = width of the window;
h = elevation of the top edge of the window;
z0 = elevation of neural level (where inside and outside pressure balance);
zref = reference elevation where the wind velocity is measured (at 10 m) and
{\displaystyle {\bar {U}}}
= mean wind velocity at the reference elevation.
As observed in the equation (1), the air exchange depends linearly on the wind speed in the urban place where the architectural project will be built. CFD (Computational Fluid Dynamics) tools and zonal modelings are usually used to design naturally ventilated buildings. Windcatchers can assist wind-driven ventilation by guiding air in and out of structures.
^ a b Wind ventilation and cross ventilation Connection Magazines
^ Cross-ventilation in a generic isolated building equipped with louvers: Wind-tunnel experiments and CFD simulations Building and Environment Volume 154, May 2019, Pages 263-280. Katarina Kosutova, Twanvan Hooff, Christina Vanderwel, Bert Blocken, Jan Hensena.
^ Cross Ventilation, the Chimney Effect and Other Concepts of Natural Ventilation ArchDaily 2008-2022
^ Basics of Natural Ventilation CoolVent
^ Cross ventilation Designing Buildings Ltd. 2022
^ The Importance of Cross-ventilation HACK architecture | Newcastle Architects
^ CROSS VENTILATION 2030 PALETTE®
^ Natural Ventilation Strategies Window Master
^ Analysis of Ventilation Efficiency and Effective Ventilation Flow Rate for Wind-driven Single-sided Ventilation Buildings by Junli Zhou, Yong Hua, Yuan Xiao, Cheng Ye, Wei Yang. Control Techniques and Strategy. August 1, 2020.
^ ENERGY EFFICIENCY MEASURES - HME09 – NATURAL VENTILATION International Finance Corporation
^ Wang, Haojie; Chen, Qingyan (2012). "A New Empirical Model for Predicting Single-Sided, Wind-Driven Natural Ventilation in Buildings". Energy and Buildings. 54: 386–394. doi:10.1016/j.enbuild.2012.07.028.
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a. In how many ways can the letters in the word CARLETON be arranged so that it
a. In how many ways can the letters in the word CARLETON be arranged so that it contains either CA or AC as sub-words?
The word CARLETON have 8 letters. We want to arranged this word so that it contains either CA of AC as sub-words.
The letters C and A can be grouped and considered as a single letter. that is, (CA)RLETON.
Hence we can assume total letters as
1+6=7
and all are different. Number of ways to arrange these letters is
7!=5040
. Now, CA or AC can be arranged themselves in
2!=2ways
, Therefore, the total number of arrangements is
5040×2=10080
Thus, the total number of arranged so that the word contains either CA or AC as sub-words are 10080.
\cap \cup
\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)
\left(a,b\right)\in R
Find the number of elements of the set
P=\left\{n\mid n\in Z,1\le n\le 400,3\mathrm{¬}\left\{\mid \right\}n
7\mathrm{¬}\left\{\mid \right\}n
13\mathrm{¬}\left\{\mid \right\}n\right\}
2 points) Page Turner loves discrete mathematics. She has 8 "graph theory" books, 6 books about combinatorics, and 4 "set theory" books.
How many ways can she place her discrete mathematics books on the same shelf in a row if:
a) there are no restrictions.
b) graph theory books are next to each other but the others could be anywhere on the shelf.
c) books are organized by their topic (same kinds are next to each other).
Restate each proposition in the form
p\to q
A. Joey will pass the symbolic logic exam if he studies hard.
B. A sufficient condition for Katrina to take the algorithms course is that she pass discrete mathematics.
C. A necessary condition for Fernando to buy a computer is that he obtain $2000.
D. When better cars are built, Toyota will build them.
E. The program is readable only if it is well structured.
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Data - Mean Practice Problems Online | Brilliant
What is the mean of the following three numbers:
1, 9, 14?
\{ 5, 8, 10, 15, 27 \}
What is the mean of the above set?
Species Number of Legs
Sea pig 12
Using the table above, determine the mean number of legs of the group.
A group of students were asked how many family members they each have in their family, and the above graph is the frequency distribution from the survey. Determine the mean number of their family members.
\{ 4, 8, 30 \}
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Transition Metals and Coordination Chemistry - Course Hero
General Chemistry/Transition Metals and Coordination Chemistry
Transition metals consist of the elements in groups 3–12 of the periodic table (Sc through Zn). In their elemental states, they display metallic luster (shiny appearance) and are good conductors of heat and electricity. Transition metals generally act as Lewis acids, accepting electron pairs from ions or polyatomic compounds (molecules or ions) that often function as Lewis bases called ligands. These Lewis acid/base donor-acceptor interactions lead to the formation of covalent bonds between transition metals and ligands. These are called coordinate covalent bonds or dative bonds. The entire molecule (or ion) is called a coordination complex. The simplest transition metal complexes contain a central metal center, usually a cation, that has one or more covalent bonds to several ligands. The three-dimensional shape of a transition metal complex can often be predicted using valence shell electron repulsion theory concepts. The metal ion with attached ligands is a coordination complex, and its shape depends on how ligands bond. The same ligands can bond with the central metal ion in different geometries, forming isomers, with distinct properties. Hemoglobin and chlorophyll are examples of two coordination complexes found in biological systems.
The crystal field theory description of bonding is a simple electrostatic model that is used to predict numbers of unpaired electrons and in some cases the colors of transition metal complexes. Electrons occupy
n
d orbitals according to Aufbau filling principles. Most complexes maximize the number of unpaired electrons due to electron-electron repulsion; pairing of electrons is generally avoided unless all other
n
d orbitals contain one (or more) electrons, according to Hund’s rule (electrons first singly occupy empty orbitals before pairing in an orbital).
Transition metals have electrons in d orbitals. They have incomplete d subshells and form cations. Inner transition elements have 4f or 5f electrons in their valence shell.
Transition metal–ligand bonding leads to the formation of numerous coordination complexes, whose molecular shapes can often be predicted using VSEPR concepts.
The naming of coordination complexes follows five rules: name the cation, name the ligands, use prefixes for multiple ligands, name the central metal, and give the central metal's oxidation state.
The magnetic properties of a coordination complex are affected by the arrangement of electrons, ligand field strength, electron-electron repulsion, and pairing energies.
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Chebyshev polynomials of the first kind - MATLAB chebyshevT
First Five Chebyshev Polynomials of the First Kind
Chebyshev Polynomials for Numeric and Symbolic Arguments
Evaluate Chebyshev Polynomials with Floating-Point Numbers
Plot Chebyshev Polynomials of the First Kind
chebyshevT(n,x)
chebyshevT(n,x) represents the nth degree Chebyshev polynomial of the first kind at the point x.
Find the first five Chebyshev polynomials of the first kind for the variable x.
chebyshevT([0, 1, 2, 3, 4], x)
[ 1, x, 2*x^2 - 1, 4*x^3 - 3*x, 8*x^4 - 8*x^2 + 1]
Depending on its arguments, chebyshevT returns floating-point or exact symbolic results.
Find the value of the fifth-degree Chebyshev polynomial of the first kind at these points. Because these numbers are not symbolic objects, chebyshevT returns floating-point results.
chebyshevT(5, [1/6, 1/4, 1/3, 1/2, 2/3, 3/4])
Find the value of the fifth-degree Chebyshev polynomial of the first kind for the same numbers converted to symbolic objects. For symbolic numbers, chebyshevT returns exact symbolic results.
chebyshevT(5, sym([1/6, 1/4, 1/3, 1/2, 2/3, 3/4]))
[ 361/486, 61/64, 241/243, 1/2, -118/243, -57/64]
Floating-point evaluation of Chebyshev polynomials by direct calls of chebyshevT is numerically stable. However, first computing the polynomial using a symbolic variable, and then substituting variable-precision values into this expression can be numerically unstable.
Find the value of the 500th-degree Chebyshev polynomial of the first kind at 1/3 and vpa(1/3). Floating-point evaluation is numerically stable.
chebyshevT(500, 1/3)
chebyshevT(500, vpa(1/3))
Now, find the symbolic polynomial T500 = chebyshevT(500, x), and substitute x = vpa(1/3) into the result. This approach is numerically unstable.
T500 = chebyshevT(500, x);
subs(T500, x, vpa(1/3))
-3293905791337500897482813472768.0
Approximate the polynomial coefficients by using vpa, and then substitute x = sym(1/3) into the result. This approach is also numerically unstable.
subs(vpa(T500), x, sym(1/3))
Plot the first five Chebyshev polynomials of the first kind.
fplot(chebyshevT(0:4,x))
ylabel('T_n(x)')
legend('T_0(x)','T_1(x)','T_2(x)','T_3(x)','T_4(x)','Location','Best')
title('Chebyshev polynomials of the first kind')
n — Degree of polynomial
nonnegative integer | symbolic variable | symbolic expression | symbolic function | vector | matrix
Degree of the polynomial, specified as a nonnegative integer, symbolic variable, expression, or function, or as a vector or matrix of numbers, symbolic numbers, variables, expressions, or functions.
x — Evaluation point
number | symbolic number | symbolic variable | symbolic expression | symbolic function | vector | matrix
Evaluation point, specified as a number, symbolic number, variable, expression, or function, or as a vector or matrix of numbers, symbolic numbers, variables, expressions, or functions.
Chebyshev polynomials of the first kind are defined as Tn(x) = cos(n*arccos(x)).
These polynomials satisfy the recursion formula
T\left(0,x\right)=1,\text{ }T\left(1,x\right)=x,\text{ }T\left(n,x\right)=2\text{ }x\text{ }T\left(n-1,x\right)-T\left(n-2,x\right)
Chebyshev polynomials of the first kind are orthogonal on the interval -1 ≤ x ≤ 1 with respect to the weight function
w\left(x\right)=\frac{1}{\sqrt{1-{x}^{2}}}
\underset{-1}{\overset{1}{\int }}\frac{T\left(n,x\right)T\left(m,x\right)}{\sqrt{1-{x}^{2}}}\text{\hspace{0.17em}}dx=\left\{\begin{array}{cc}0& \text{if }n\ne m\\ \pi & \text{if }n=m=0\\ \frac{\pi }{2\text{\hspace{0.17em}}}& \text{if }n=m\ne 0.\end{array}
Chebyshev polynomials of the first kind are special cases of the Jacobi polynomials
T\left(n,x\right)=\frac{{2}^{2n}{\left(n!\right)}^{2}}{\left(2n\right)!}P\left(n,-\frac{1}{2},-\frac{1}{2},x\right)
and Gegenbauer polynomials
T\left(n,x\right)=\left\{\begin{array}{cc}\frac{1}{2}\underset{a\to 0}{\mathrm{lim}}\frac{n+a}{a}G\left(n,a,x\right)& \text{if }n\ne 0\\ \underset{a\to 0}{\mathrm{lim}}G\left(0,a,x\right)=1& \text{if }n=0\end{array}
chebyshevT returns floating-point results for numeric arguments that are not symbolic objects.
chebyshevT acts element-wise on nonscalar inputs.
At least one input argument must be a scalar or both arguments must be vectors or matrices of the same size. If one input argument is a scalar and the other one is a vector or a matrix, then chebyshevT expands the scalar into a vector or matrix of the same size as the other argument with all elements equal to that scalar.
[1] Hochstrasser, U. W. “Orthogonal Polynomials.” Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. (M. Abramowitz and I. A. Stegun, eds.). New York: Dover, 1972.
[2] Cohl, Howard S., and Connor MacKenzie. “Generalizations and Specializations of Generating Functions for Jacobi, Gegenbauer, Chebyshev and Legendre Polynomials with Definite Integrals.” Journal of Classical Analysis, no. 1 (2013): 17–33. https://doi.org/10.7153/jca-03-02.
chebyshevU | gegenbauerC | hermiteH | jacobiP | laguerreL | legendreP
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Frequency-Domain Linear Regression - MATLAB & Simulink - MathWorks Deutschland
This example shows how to use the discrete Fourier transform to construct a linear regression model for a time series. The time series used in this example is the monthly number of accidental deaths in the United States from 1973 to 1979. The data are published in Brockwell and Davis (2006). The original source is the U. S. National Safety Council.
Enter the data. Copy the exdata matrix into the MATLAB® workspace.
exdata = [
10017 8714 9387 8634 8890 9115
8927 8680 8034 8647 8796 9240];
exdata is a 12-by-6 matrix. Each column of exdata contains 12 months of data. The first row of each column contains the number of U.S. accidental deaths for January of the corresponding year. The last row of each column contains the number of U.S. accidental deaths for December of the corresponding year.
Reshape the data matrix into a 72-by-1 time series and plot the data for the years 1973 to 1978.
ts = reshape(exdata,72,1);
years = linspace(1973,1979,72);
plot(years,ts,'o-','MarkerFaceColor','auto')
ylabel('Number of Accidental Deaths')
A visual inspection of the data indicates that number of accidental deaths varies in a periodic manner. The period of the oscillation appears to be roughly 1 year (12 months). The periodic nature of the data suggests that an appropriate model may be
X\left(n\right)=\mu +\sum _{k}\left({A}_{k}\mathrm{cos}\frac{2\pi kn}{N}+{B}_{k}\mathrm{cos}\frac{2\pi kn}{N}\right)+\epsilon \left(n\right),
\mu
is the overall mean,
N
is the length of the time series, and
\epsilon \left(n\right)
is a white noise sequence of independent and identically-distributed (IID) Gaussian random variables with zero mean and some variance. The additive noise term accounts for the randomness inherent in the data. The parameters of the model are the overall mean and the amplitudes of the cosines and sines. The model is linear in the parameters.
To construct a linear regression model in the time domain, you have to specify which frequencies to use for the cosines and sines, form the design matrix, and solve the normal equations in order to obtain the least-squares estimates of the model parameters. In this case, it is easier to use the discrete Fourier transform to detect the periodicities, retain only a subset of the Fourier coefficients, and invert the transform to obtain the fitted time series.
Perform a spectral analysis of the data to reveal which frequencies contribute significantly to the variability in the data. Because the overall mean of the signal is approximately 9,000 and is proportional to the Fourier transform at 0 frequency, subtract the mean prior to the spectral analysis. This reduces the large magnitude Fourier coefficient at 0 frequency and makes any significant oscillations easier to detect. The frequencies in the Fourier transform are spaced at an interval that is the reciprocal of the time series length, 1/72. Sampling the data monthly, the highest frequency in the spectral analysis is 1 cycle/2 months. In this case, it is convenient to look at the spectral analysis in terms of cycles/year so scale the frequencies accordingly for visualization.
tsdft = fft(ts-mean(ts));
freq = 0:1/72:1/2;
plot(freq.*12,abs(tsdft(1:length(ts)/2+1)),'o-', ...
'MarkerFaceColor','auto')
ax.XTick = [1/6 1 2 3 4 5 6];
Based on the magnitudes, the frequency of 1 cycle/12 months is the most significant oscillation in the data. The magnitude at 1 cycle/12 months is more than twice as large as any other magnitude. However, the spectral analysis reveals that there are also other periodic components in the data. For example, there appears to be periodic components at harmonics (integer multiples) of 1 cycle/12 months. There also appears to be a periodic component with a period of 1 cycle/72 months.
Based on the spectral analysis of the data, fit a simple linear regression model using a cosine and sine term with a frequency of the most significant component: 1 cycle/year (1 cycle/12 months).
Determine the frequency bin in the discrete Fourier transform that corresponds to 1 cycle/12 months. Because the frequencies are spaced at 1/72 and the first bin corresponds to 0 frequency, the correct bin is 72/12+1. This is the frequency bin of the positive frequency. You must also include the frequency bin corresponding to the negative frequency: -1 cycle/12 months. With MATLAB indexing, the frequency bin of the negative frequency is 72-72/12+1.
Create a 72-by-1 vector of zeros. Fill the appropriate elements of the vector with the Fourier coefficients corresponding to a positive and negative frequency of 1 cycle/12 months. Invert the Fourier transform and add the overall mean to obtain a fit to the accidental death data.
freqbin = 72/12;
freqbins = [freqbin 72-freqbin]+1;
tsfit = zeros(72,1);
tsfit(freqbins) = tsdft(freqbins);
tsfit = ifft(tsfit);
mu = mean(ts);
tsfit = mu+tsfit;
Plot the original data along with the fitted series using two Fourier coefficients.
plot(years,tsfit,'linewidth',2)
legend('Data','Fitted Model')
The fitted model appears to capture the general periodic nature of the data and supports the initial conclusion that data oscillate with a cycle of 1 year.
To assess how adequately the single frequency of 1 cycle/12 months accounts for the observed time series, form the residuals. If the residuals resemble a white noise sequence, the simple linear model with one frequency has adequately modeled the time series.
To assess the residuals, use the autocorrelation sequence with 95%-confidence intervals for a white noise.
resid = ts-tsfit;
[xc,lags] = xcorr(resid,50,'coeff');
lconf = -1.96*ones(51,1)/sqrt(72);
uconf = 1.96*ones(51,1)/sqrt(72);
plot(lags(51:end),lconf,'r')
plot(lags(51:end),uconf,'r')
ylabel('Correlation Coefficient')
title('Autocorrelation of Residuals')
The autocorrelation values fall outside the 95% confidence bounds at a number of lags. It does not appear that the residuals are white noise. The conclusion is that the simple linear model with one sinusoidal component does not account for all the oscillations in the number of accidental deaths. This is expected because the spectral analysis revealed additional periodic components in addition to the dominant oscillation. Creating a model that incorporates additional periodic terms indicated by the spectral analysis will improve the fit and whiten the residuals.
Fit a model which consists of the three largest Fourier coefficient magnitudes. Because you have to retain the Fourier coefficients corresponding to both negative and positive frequencies, retain the largest 6 indices.
tsfit2dft = zeros(72,1);
[Y,I] = sort(abs(tsdft),'descend');
indices = I(1:6);
tsfit2dft(indices) = tsdft(indices);
Demonstrate that preserving only 6 of the 72 Fourier coefficients (3 frequencies) retains most of the signal's energy. First, demonstrate that retaining all the Fourier coefficients yields energy equivalence between the original signal and the Fourier transform.
norm(1/sqrt(72)*tsdft,2)/norm(ts-mean(ts),2)
The ratio is 1. Now, examine the energy ratio where only 3 frequencies are retained.
norm(1/sqrt(72)*tsfit2dft,2)/norm(ts-mean(ts),2)
Almost 90% of the energy is retained. Equivalently, 90% of the variance of the time series is accounted for by 3 frequency components.
Form an estimate of the data based on 3 frequency components. Compare the original data, the model with one frequency, and the model with 3 frequencies.
tsfit2 = mu+ifft(tsfit2dft,'symmetric');
plot(years,tsfit2,'linewidth',2)
legend('Data','1 Frequency','3 Frequencies')
Using 3 frequencies has improved the fit to the original signal. You can see this by examining the autocorrelation of the residuals from the 3-frequency model.
resid = ts-tsfit2;
Using 3 frequencies has resulted in residuals that more closely approximate a white noise process.
Demonstrate that the parameter values obtained from the Fourier transform are equivalent to a time-domain linear regression model. Find the least-squares estimates for the overall mean, the cosine amplitudes, and the sine amplitudes for the three frequencies by forming the design matrix and solving the normal equations. Compare the fitted time series with that obtained from the Fourier transform.
X(:,2) = cos(2*pi/72*(0:71))';
X(:,3) = sin(2*pi/72*(0:71))';
X(:,4) = cos(2*pi*6/72*(0:71))';
X(:,5) = sin(2*pi*6/72*(0:71))';
X(:,6) = cos(2*pi*12/72*(0:71))';
X(:,7) = sin(2*pi*12/72*(0:71))';
beta = X\ts;
tsfit_lm = X*beta;
max(abs(tsfit_lm-tsfit2))
The two methods yield identical results. The maximum absolute value of the difference between the two waveforms is on the order of 10-12. In this case, the frequency-domain approach was easier than the equivalent time-domain approach. You naturally use a spectral analysis to visually inspect which oscillations are present in the data. From that step, it is simple to use the Fourier coefficients to construct a model for the signal consisting of a sum cosines and sines.
For more details on spectral analysis in time series and the equivalence with time-domain regression see (Shumway and Stoffer, 2006).
While spectral analysis can answer which periodic components contribute significantly to the variability of the data, it does not explain why those components are present. If you examine these data closely, you see that the minimum values in the 12-month cycle tend to occur in February, while the maximum values occur in July. A plausible explanation for these data is that people are naturally more active in summer than in the winter. Unfortunately, as a result of this increased activity, there is an increased probability of the occurrence of fatal accidents.
Brockwell, Peter J., and Richard A. Davis. Time Series: Theory and Methods. New York: Springer, 2006.
Shumway, Robert H., and David S. Stoffer. Time Series Analysis and Its Applications with R Examples. New York: Springer, 2006.
fft | ifft | xcorr
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FIFTH SERIES, VOLUME XXIV.
FROM THE BEGINNING, VOL. CXXXIX.
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No. 1790. — October 5, 1878.
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Cyprus. Part II., Macmillan's Magazine, 3
II. An Indiscretion in the Life of an Heiress, New Quarterly Review, 11
III. Child's Play, Cornhill Magazine, 18
IV. Mr. Froude's "Life and Times of Thomas Becket". By Edward A. Freeman. Part IV., Contemporary Review, 32
V. The Chinese as Colonists, Nineteenth Century, 50
VI. The Relation of Memory to Will, Spectator, 56
VII. Garden-Parties, Spectator, 59
VIII. Invalids, Spectator, 61
The Creature's Song of Expectation, 2 "Ein Jungling liebt ein Madchen", 2
Lament of Marjory Cockburne, 2 "Ein Fichtenbaum steht einsam", 2
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Henri Greville's Sketches of Russian Life. By W. R. S. Ralston, Nineteenth Century, 67
II. An Indiscretion in the Life of an Heiress. By Thomas Hardy. Conclusion, New Quarterly Review, 76
III. The Public Career and Personal Character of Francis Bacon. By James Rowley, Fraser's Magazine, 91
IV. Selling the Soul, Contemporary Review, 104
V. A Fetish City, Blackwood's Magazine, 111
VI. Sark, and its Caves, Gentleman's Magazine, 116
VII. The Habit of Reading, Saturday Review, 124
VIII. An American Zollverein, Pall Mall Gazette, 127
The Spray of Seaweed, 66 "So Wandl' ich wieder den Alten Weg", 66
The Sea-Horse, 66
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. The Ceremonial Use of Flowers, Nineteenth Century, 131
II. Sir Gibbie. By George MacDonald, author of "Malcolm," "The Marquis of Lossie," etc., Advance Sheets, 143
III. French Home Life. The Idea of Home, Blackwood's Magazine, 152
IV. Within the Precincts. By Mrs. Oliphant. Part XV., Advance Sheets, 161
V. Holidays in Eastern France, Fraser's Magazine, 172
VI. Apologies, Spectator, 186
VII. For and Against Norway, Spectator, 189
VIII. Decorative Coloring In Freshwater Fleas, Nature, 192
La Pellegrina Rondinella, 130 Emblem of True Philosophy, 130
"Dein Angesicht, so lieb und schon", 130 "My Wife and Child come close to me", 130
The Woods in Autumn, 130
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Martin Joseph Routh, Quarterly Review, 195
II. Macleod of Dare. By William Black. Part XVII., Advance Sheets, 216
III. Among the Burmese. Part III., Fraser's Magazine, 232
IV. Sir Gibbie. By George MacDonald, author of "Malcolm," "The Marquis of Lossie," etc. Part II., Advance Sheets, 240
V. A New Author's Grievance, Spectator, 250
VI. The Sorrows of the Slow, Spectator, 252
VII. Calculating Boys, Spectator, 255
My Friend: a Portrait, 194 To the Finest of Fruits, 194
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
No. 1794. — November 2, 1878.
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Hallucinations of the Senses, Fortnightly Review, 259
II. Within the Precincts. By Mrs. Oliphant. Part XVI., Advance Sheets, 269
III. Holidays in Eastern France. Part II., Fraser's Magazine, 274
IV. "Fred": a Tale from Japan, Blackwood's Magazine, 283
V. Color in Painting, Cornhill Magazine, 287
VI. Sir Walter Scott and the Romantic Reaction, Contemporary Review, 298
VII. "Through the Dark Continent" in 1720, Macmillan's Magazine, 313
VIII. The Buzzing of Insects, Public Opinion, 319
IX. Public Opinion, Spectator, 320
At Nightfall, 258 True, 258
The Wrong Time, 258 With a Present, 258
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. The Sixteenth Century Arraigned before the Nineteenth. A Study on the Reformation. By Rt. Hon. W. E. Gladstone, Contemporary Review 323
II. A Doubting Heart. By Miss Keary, author of "Castle Daly," "Oldbury," etc. Part IX., Advance Sheets, 343
III. The Troubles of a Scots Traveller, Blackwood's Magazine, 359
IV. Sir Gibbie. By George MacDonald, author of "Malcolm," "The Marquis of Lossie," etc. Part III., Advance Sheets, 371
V. Lighthouses, Argosy, 377
VI. Bookworms, Saturday Review, 381
Commissioned, 322 We Well to Mourn? 322
Holyday, 322
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. The Brothers Chambers, London Quarterly Review, 387
II. Sir Gibbie. By George MacDonald, author of "Malcolm," "The Marquis of Lossie," etc. Part IV., Advance Sheets, 400
III. Faith and Verification. By W. H. Mallock, Nineteenth Century, 410
IV. Within the Precincts. By Mrs. Oliphant. Part XVII., Advance Sheets, 423
V. The Things We Have Not, Saturday Review, 432
VI. The North American Indians, London Times, 434
VII. The Art of Going Away, Saturday Review, 438
VIII. Social Hypocrites, Saturday Review, 440
IX. Character and Position, Spectator, 442
IX. Weather Prophecies, Spectator, 445
X. Endurance and Fatalism, Spectator, 447
In Harvest Day, 386 The Unburied Church, 386
"Aus Alten Marchen winkt es", 386
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{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. What is Going on at the Vatican. A Voice from Rome, Contemporary Review, 643
II. Sir Gibbie. By George MacDonald, author of "Malcolm," "The Marquis of Lossie," etc. Part VI., Advance Sheets, 653
III. Holidays in Eastern France. Part III., Fraser's Magazine, 666
IV. Macleod of Dare. By William Black. Part XIX., Advance Sheets, 677
V. Miss Ferrier's Novels, Temple Bar, 693
Love and Loss, 642
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Epping Forest. By Alfred Russell Wallace, Fortnightly Review, 707
II. Within the Precincts. By Mrs. Oliphant. Part XIX., Advance Sheets, 717
III. Among the Burmese. Part V., Fraser's Magazine, 734
IV. Racine and his Works. By the author of "Mirabeau," etc., Temple Bar, 740
V. A New Method of Social Evolution, Blackwood's Magazine, 750
VI. The Undefinable in Art, Cornhill Magazine, 760
Aberglaube, 706 Stanzas, 706
Olden Times and Present, 706
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Petrarch, Quarterly Review, 771
II. Bianca. By W. E. Norris, Belgravia, 787
III. A Red-Cross Ride through Snow and Death, Temple Bar, 798
IV. The Cottage by the River, Blackwood's Magazine, 803
V. The Sorrows of Lord Penzance, Spectator, 819
VI. Clerical Self-Conceit, Spectator, 821
VII. A Rajput Chief of the Old School. By A. C. Lyall, Fortnightly Review, 823
⁂ Title and Index to Volume CXXXIX.
"Baby's Dead", 770 Cardinal-Flowers, 770
After the Concert, 770 A Rajput Chief of the Old School, 823
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Dilution gauging - FIThydrowiki
Figure 1: Dilution gauging in a fishway: making the dilution of the tracer (Rhodamine Wt) (source: Itagra.ct).
Figure 2: Dilution gauging in a fishway: Mariotte device for continuous injection into the fishway (source: Itagra.ct).
Developed by:Francisco Javier Bravo, Itagra.ct
This method requires the release of a known tracer concentration in a section of the river or fishway and the subsequent determination of the tracer concentration in a downstream section. It is based on the dilution relationships between the injection of the tracer and the discharge we want to know. There are two main different methods for the injection of the tracers into the flow: (a) instantaneous or integration method and (b) continuous method.
The discharge for (a) is calculated as:
{\displaystyle Q={\frac {c_{0}}{c}}*{\frac {V}{T}}}
{\displaystyle Q={c_{0}}}
the tracer concentration which is introduced in the watercourse with discharge Q. c is the concentration of the sample in volume V and T is equal to the time needed for the tracer to be transported downstream.
The discharge formula for (b):
{\displaystyle Q=q*{\frac {c_{1}}{c_{2}}}}
{\displaystyle {c_{1}}}
the tracer concentration in the injection, q the constant injection flow for the tracer (e.g. using a Mariotte device) and
{\displaystyle {c_{2}}}
the tracer concentration at the downstream point.
The tracer choice depends on several factors: chemical characteristics of the water, suspended sediment, distance between the injection and measuring section, type of flow to be measured, sensitivity of the tracer measurement devices and possible environmental impact of the tracer. Common tracers used: chemical (NaCl, NaI, NH4Cl) and fluorescent (fluorescein and Rhodamine WT).
It is important to comply with some conditions: the tracer cannot be absorbed and the solution must be well mixed. Emphasizing the last one, it is necessary to assure a good mixing length, which is the distance between the injection and measure sections, ensuring a stable concentration of the diluted tracer.
This method has been used for Itagra.ct to measure discharge into the fishways (Bravo-Córdoba y Sanz-Ronda, 2011; Fuentes-Pérez et al., 2016 and 2017; Sanz-Ronda et al., 2016; Bravo-Córdoba et al., 2018). One of the main aims was to improve the accuracy of the discharge coefficients related to technical fishways into the field. A continuous method, using Rodamine WT as a tracer and a portable fluorometer to measure the fluorescence of the samples,was applied (Figure 1 and 2).
ISO 9555:1994. Measurement of liquid flow in open channels — Tracer dilution methods for the measurement of steady flow
Tazioli, A. (2011). Experimental methods for river discharge measurements: Comparison among tracers and current meter. Hydrological Sciences Journal – Journal des Sciences Hydrologiques. 56. 1314-1324.
Bravo‐Córdoba FJ, Sanz‐Ronda FJ. 2011. Evaluación de la eficiencia biológica de una escala de hendiduras verticales para la trucha autóctona (Salmo trutta L.) en la Cuenca del Duero. Master’s thesis. University of Valladolid, Spain.
Bravo‐Córdoba FJ, Sanz‐Ronda FJ, Ruiz‐Legazpi J, Celestino LF, Makrakis S. 2018. Fishway with two entrance branches: Understanding its performance for potamodromous Mediterranean barbels. Fisheries Management and Ecology 25(1): 12–21.
Fuentes-Pérez JF, Sanz-Ronda FJ, Martínez de Azagra-Paredes A, García-Vega A, Martínez de Azagra A, García-Vega A. 2016. Nonuniform hydraulic behavior of pool-weir fishways: a tool to optimize its design and performance. Ecol Eng 86: 5–12.
Fuentes-Pérez JF, García-Vega A, Sanz-Ronda FJ, Martínez de Azagra Paredes A. 2017. Villemonte’s approach: a general method for modelling uniform and non-uniform performance in stepped fishways. Knowl. Manag. Aquat. Ecosyst., 418, 23.
Sanz-Ronda FJ, Bravo-Córdoba FJ, Fuentes-Pérez JF, Castro-Santos T. 2016. Ascent ability of brown trout, Salmo trutta, and two Iberian cyprinids− Iberian barbel, Luciobarbus bocagei, and northern straight-mouth nase, Pseudochondrostoma duriense− in a vertical slot fishway. Knowledge and Management of Aquatic Ecosystems (417): 10.
Retrieved from "https://www.fithydro.wiki/index.php?title=Dilution_gauging&oldid=6081"
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0 5 g of nonvolatile solute is dissolved in 100 g of ethyl acetate in 20C the vapour pressure of - Chemistry - Solutions - 9880542 | Meritnation.com
0.5 g of nonvolatile solute is dissolved in 100 g of ethyl acetate in 20C.the vapour pressure of the solution and pure ethyl acetate are 72 and 72.8 torr rezpectively at 20C. Calculate the molecular weight of the solute
Please find below the solution to the query posted by you.
Mass of non-volatile solute, m1 = 0.5 g
Mass of solvent (ethyl acetate, CH3COOCH2CH3), m2 = 100 g
Molar mass of solvent, M2 = 88 g
Vapour pressure of pure ethyl acetate, p2o = 72.8 torr
Vapour pressure of solution, ps = 72 torr
Mole fraction of the solute can be determined by using the formula for relative lowering of vapour pressure
\frac{{\mathrm{p}}_{2}^{\mathrm{o}} - {\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}_{2}^{\mathrm{o}}}=\frac{{\mathrm{m}}_{1} × {\mathrm{M}}_{2}}{{\mathrm{M}}_{1} × {\mathrm{m}}_{2}}\phantom{\rule{0ex}{0ex}}\frac{72.8 - 72}{72.8} = \frac{0.5 × 88}{{\mathrm{M}}_{1} × 100}\phantom{\rule{0ex}{0ex}}{\mathrm{M}}_{1} = \frac{0.5 × 88 × 72.8}{0.8 × 100}= 40.04 \mathrm{g} {\mathrm{mol}}^{-1}
Hence, the molar mass of the non-volatile solute is 40.04 g mol−1.
Hope this information will clear your doubt regarding relative lowering of vapour pressure.
In case of any doubt just post your query on the forum and our experts will try to help you as soon as possible.
Ayush Agrawal answered this
72.8-72/72.8 = mole fraction
thus 0 .8/72.8 = 0.5 * 88 / m* 100 ( as mm of ethyl acetate = 88 ad .5 <<100)
thus m = 12.54
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FIFTH SERIES, VOLUME XV.
FROM THE BEGINNING, VOL. CXXX.
Volume XV.
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
No. 1673. — July 1, 1876.
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
Vol. CXXX.
I. The Intellectual Revival of the Middle Ages, Westminister Review, 3
II. What She Came Through. By Sarah Tytler, author of "Lady Bell," etc. Part IX., Good Words, 16
III. Natural Religion. Part VI., Macmillan's Magazine, 25
IV. Carita. By Mrs. Oliphant, author of "Chronicles of Carlingford," "Zaidee," etc., Cornhill Magazine, 34
V. The Kings of the Renaissance. Charles VIII. and Louis XII. By the author of "Mirabeau," etc., Temple Bar, 43
VI. Early Autumn on the Lower Yang-Tze, Fortnightly Review, 53
VII. Zachary Macaulay. By W. G. Blaikie, D.D., Sunday Magazine, 59
VIII. Tricks of Memory, Queen, 63
The Lament of the Covenant, 1876, 2 A Ballad of Past Meridian, 2
The Death of the Violet, 2
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. The Courses of Religious Thought. By Rt. Hon. W. E. Gladstone, Contemporary Review, 67
II. For Pity's Sake. By the author of "Robert Holt's Illusion," etc. Conclusion, Sunday Magazine, 82
III. James Northcote, R.A., Fortnightly Review, 89
IV. Janet Mason's Troubles. A Story of Town and Country. Conclusion, Sunday Magazine, 99
V. Remarks on Modern Warfare. By a Military Officer, Fraser's Magazine, 113
VI. Visit to a Spanish Prison, Temple Bar, 117
VII. The "Venus" of Quinipily, Macmillan's Magazine, 122
VIII. Pocket-Money, Saturday Review, 126
IX. Norwegian Deep-Sea Explorations, Academy, 128
The Conscience and Future
Judgment, 66 Epicede, 66
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
No. 1675. — July 15, 1876.
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. The Cruise of the "Challenger", Nature, 131
II. What She Came Through. By Sarah Tytler, author of "Lady Bell," etc. Part X., Good Words, 143
III. Russian Village Communities, Macmillan's Magazine, 153
IV. Carita. By Mrs. Oliphant, author of "Chronicles of Carlingford," "Zaidee," etc. Part II., Cornhill Magazine, 162
V. Quakers and Quakerism, Macmillan's Magazine, 166
VI. The Originals of Werther, Temple Bar, 172
VII. Sketch of a Journey across Africa. By Verney Lovett Cameron, Lieutenant Royal Navy, Good Words, 176
VIII. Unbreakable or Toughened Glass. By Perry F. Nursey, C.E., Popular Science Review, 185
IX. Johnsonese Poetry, Spectator, 190
Harvest, 130 Exiled, 130
Nobody! 130 Blue Eyes, 130
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Physics and Physiology of Harmony, Westminster Review, 195
II. The Lady Candidate. Conclusion, Blackwood's Magazine, 207
III. Ordeals and Oaths, Macmillan's Magazine, 220
IV. Walther von der Vogelweide, Cornhill Magazine, 229
V. Leigh Hunt and Lord Brougham. With Original Letters. By S. R. Townshend Meyer, Temple Bar, 239
VI. Partial Deafness, Spectator, 247
VII. The Extradition Quarrel, Economist, 249
VIII. Lunar Studies, Spectator, 252
IX. The Remington Type-Writing Machine, Nature, 254
X. Physical Influences upon Character, Victoria Magazine, 255
XI. A Free Spanish University, Nature, 256
To a Young Lady on the Approach
of the Season, 194 Leisure and Love, 194
"Rejected", 194
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Jonathan Swift, British Quarterly Review, 259
II. Infallible, Chamber's Journal, 276
III. Thoughts of an Outsider: International Prejudices, Cornhill Magazine, 284
IV. Carita. By Mrs. Oliphant, author of "Chronicles of Carlingford," "Zaidee," etc., Part III., Cornhill Magazine, 294
V. Quakers and Quakerism. Part II., Macmillan's Magazine, 301
VI. Sketch of a Journey across Africa. By Verney Lovett Cameron, Lieutenant Royal Navy. Part II., Good Words, 308
VII. Harriet Martineau's Autobiography, Spectator, 316
VIII. Periodicity of the Fresh-water Lakes of Australia, Nature, 318
Two French Hymns, 258 Church Music, 258
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Turkey, Contemporary Review, 323
II. What She Came Through. By Sarah Tytler, author of "Lady Bell," etc., Part XI., Good Words, 337
III. In a Studio. By W. W. Story. Part VIII., Blackwood's Magazine, 346
IV. Carita. By Mrs. Oliphant, author of "Chronicles of Carlingford," "Zaidee," etc., Part IV., Cornhill Magazine, 355
V. The Last of the Grand School of Connoisseurs, Temple Bar, 361
VI. Dull Sermons, Macmillan's Magazine, 373
VII. Sick-Nursing, an Employment for Educated Women, Chambers' Journal, 377
VIII. Leeches, Chambers' Journal, 379
IX. Work for Gentlewomen, Victoria Magazine, 381
X. A Perfect Lawn, Gardener's Magazine, 382
Utinam, 322 Milkweed, 322
A Lullaby, 322 The House Beautiful, 382
Rosenlied, 322
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Adam Smith as a Person, Fortnightly Review, 387
II. What She Came Through. By Sarah Tytler, author of "Lady Bell," etc. Part XII., Good Words, 402
III. In a Studio. By W. W. Story. Part IX., Blackwood's Magazine, 411
IV. Lady Adelaide. A Study, Blackwood's Magazine, 418
V. Autobiography of a Vegetarian. A True Narrative of a Successful Career. Reported by C. O. Groom Napier, of Merchiston, F.G.S., Fraser's Magazine, 427
VI. Modern Warfare. By John Ruskin, Fraser's Magazine, 434
VII. A Lady's Visit to the Herzegovinian Insurgents, Cornhill Magazine, 436
VIII. Lightning-Prints, Chambers' Journal, 445
Saints, 386 The Song-Bird, 386
Dreamland, 386
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Ticknor's Memoirs, Quarterly Review, 451
II. John's Hero, Blackwood's Magazine, 473
III. Italian Art and Literature before Giotto and Dante, Macmillan's Magazine, 480
IV. How I Went to the Levee, Macmillan's Magazine, 490
V. Memoir of Norman Macleod, D.D., Church Quarterly Review, 498
VI. London under an Indian Climate, Spectator, 505
VII. The Tasmanians, Nature, 507
VIII. The American Centenary, Spectator, 510
To a Caged Bob-o-link in the City, 450 Lilies, 450
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Lord Macaulay, Quarterly Review, 515
II. What She Came Through. By Sarah Tytler, author of "Lady Bell," etc., Part XIII., Good Words, 542
III. Moresby's "New Guinea and Polynesia", Edinburgh Review, 553
IV. Carita. By Mrs. Oliphant, author of "Chronicles of Carlingford," "Zaidee," etc., Part V., Cornhill Magazine, 563
V. The Poetry of Leading Law Cases, Spectator, 572
VI. The First Appearance of Gipsies, Temple Bar, 575
A Summer's Ghost, 514 My Treasures, 514
A Water-Lily at Evening, 514 To Kate, 514
The Lily of the Valley, 514
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
No. 1682. — September 2, 1876.
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. The Comte de Paris' Campaign on the Potomac, Edinburgh Review, 579
II. Carita. By Mrs. Oliphant, author of "Chronicles of Carlingford," "Zaidee," etc., Part VI., Cornhill Magazine, 593
III. When the Sea was Young, Cornhill Magazine, 597
IV. What She Came Through. By Sarah Tytler, author of "Lady Bell," etc., Part XIV., Good Words, 608
V. Wordsworth's Ethics, Cornhill Magazine, 615
VI. Sketch of a Journey across Africa. By Verney Lovett Cameron, Lieutenant Royal Navy. Part III., Good Words, 629
VII. Secret-Service Money under George I., Academy, 635
VIII. Eccentricity, Good Words, 636
IX. Resources of Servia and Bosnia, Nature, 639
The Seven-Nights' Watch, 578 When we are Parted, 578
One of the Sevens, 578
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Clarendon. Part II., Contemporary Review, 643
II. The Rev. Adam Cameron's Visit to London, Cornhill Magazine, 656
III. The Territorial Expansion of Russia, Fortnightly Review, 677
IV. Narrative of the Wreck of the "Strathmore". By one of the Survivors, Chambers' Journal, 690
V. Whewell's Writings and Correspondence, Nature, 696
VI. The Turkish Atrocities, Economist, 698
VII. The Whole Duty of Woman from a Chinese Point of View, Pall Mall Budget, 700
VIII. On Talkers, Examiner, 702
IX. Servia, Leisure Hour, 703
Autumn in the Woods, 642 Two Seasons, 642
A Suicide, 642
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
No. 1684. — September 16, 1876.
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Social Relations of England and America, Quarterly Review, 707
II. The Rev. Adam Cameron's Visit to London. Conclusion, Cornhill Magazine, 728
III. From Belgrade to Constantinople Overland. By Humphry Sandwith, C.B., Fraser's Magazine, 741
IV. The Great Lord Fairfax, Athenæum, 755
V. Retribution as a Faith and a Superstition, Spectator, 757
VI. Collapse, Saturday Review, 758
VII. Temporary Duty, Saturday Review, 761
VIII. A Letter of Miss Martineau's, Athenæum, 763
IX. The Native Races of the Pacific States of North America, Academy, 764
Sometime, 706 Ivy, 706
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}}
No. 1685. — September 23 & 30, 1876.
{\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}}
I. Modern Philosophers on the Probable Age of the World, Quarterly Review, 771
II. Carita. By Mrs. Oliphant, author of "Chronicles of Carlingford," "Zaidee," etc., Part VII., Cornhill Magazine, 787
III. Charlotte Bronte. A Monograph, Macmillan's Magazine, 801
IV. An Excursion in Formosa, Contemporary Review, 816
V. The Swarming of Mankind, Spectator, 821
VI. The Limits of Fiction, Saturday Review, 823
⁂ Title and Index to Volume CXXX.
To Summer, 770 Just a Few Words, 770
Retrieved from "https://en.wikisource.org/w/index.php?title=Littell%27s_Living_Age/Volume_130&oldid=6739506"
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Temperature Stimulation of Effective Value of Density of the Current of Semiconductor Solar Cells
Temperature Stimulation of Effective Value of Density of the Current of Semiconductor Solar Cells
Aliev Rayimjon1, Ikromov Rustamjon Gulomjonovich2, Alinazarova Mahfuza Alisherovna3
1Department of Physics, Andizhan State University, Andizhan, Uzbekistan
2Department of Physics, Namangan Engineering and Technology Institute, Namangan, Uzbekistan
3Department of Physics, Namangan State University, Namangan, Uzbekistan
Influence of temperature on effective value of a current density of a semiconductor solar cell is investigated. New equations for explanation of dependence of photovoltaic characteristics of solar cell from temperature are received.
Semiconductor Solar Сell, Effective Value of a Current, Temperature Dependence of Photovoltaic Parameters, Photovoltaic Characteristics
The semiconductor photo-electric power takes a perspective place among set of renewable energy. Thus on a way of wide application of solar cells (SC) in power supply programs, there are some problems. Most considerable of them are the high cost price of solar energy and low efficiency of SC [1] .
Semiconductor SC with high cost can be used without an exception in various special conditions. For example, at their use as an energy source of space flying devices, they are exposed to essential cooling [2] . At use of SC for conversation of the concentrated radiation is inevitable their strong heating [3] . Therefore, research of temperature properties of SC is one of important applied problems of a modern science.
Now is spent theoretical and experimental researches influence of ambient temperature on the cores Photovoltaic parameters of SC [2] . For interpretation of the temperature properties of SC defined experimentally scientists often address to the received by theoretical way laws. Thus quite often there are inexplicable known theoretical expressions, temperature dependences, the cores of Photovoltaic parameters of SC. Such situation demands necessity of working out of more suitable semi-empirical expressions for the description of experimental dependences. The given work is devoted this problem―to definition of semi-empirical expressions for more exact definition of temperature properties of semiconductor SC with p-n-structure.
Influence of ambient temperature and, hence, the most SC on its photovoltaic characteristics and efficiency of photoelectric conversation by semiconductor structures is the most important object of experimental and theoretical research. It is known that one of the basic photovoltaic characteristics of SC, defining efficiency of conversation of solar energy is effective power which is traditionally expressed by formula [1] :
{P}_{m}={j}_{m}{U}_{m}
where Jm and Um are maximum (effective) values of according to photocurrent density and voltage.
Experimental results show that Jm and Um are depends on temperature. But in the scientific literature the expressions allowing preliminary to define these dependence almost are not presented.
Therefore in this work the results, the research conducted by us by a semi-empiric method definition of dependence of effective value of Jm and Um from temperature are stated.
On Figure 1 the qualitative method of definition of a point of effective power, from experimental I - V (3-curve) of SC is shown. For this purpose in the beginning define a straight line passing through two points of I - V, voltage of idling and density of a current of short circuit, i.e. (Uoc whence are defined; 0) and (0; Jsc) (1-straight line). By parallel shifting of this line to a effective point of SC I - V it is possible to receive a 2-straight line. From this I - V point defines effective power of SC. Being based on this method it is possible to deduce the formula for defining effective value of density of a photo current of SC developed.
As it is known the equation for a straight line which is passing through two points has the following appearance
\frac{j-{j}_{1}}{{j}_{2}-{j}_{1}}=\frac{U-{U}_{1}}{{U}_{2}-{U}_{1}}
From this Equation (2) it is possible to receive for a 1-straight line (Figure 1):
j=-\frac{{j}_{sc}}{{U}_{oc}}U+{j}_{sc}
It is visible that the angular factor of this straight line matters
a=-\frac{{j}_{sc}}{{U}_{oc}}
. For a 2-straight line also looks like
j=aU+b
. As, the 2-straight line is a tangent to I - V, therefore angular factor of this line it is possible to define from I - V:
Figure 1. A method of definition of a point of effective power from experimental I - V of SC.
{j}_{ph}={j}_{0}\left(\mathrm{exp}\left(\frac{eU}{nkT}\right)-1\right)-{j}_{sc}
where jph―photocurrent density, j0―density of a saturation current, e―a electron charge, k―constant of Boltzmann, T―temperature.
It is known that value of factor not idealities (n) of SC I - V is define by type of an electric current [4] . Therefore, it is possible to assume that leaving voltage of SC does not depend on this parameter. We define a derivative on pressure from SC I - V and we will find its value at U = Um then for and we will receive:
a={{j}^{\prime }}_{ph}\left({U}_{m}\right)=-{j}_{0}\frac{e}{nkT}\mathrm{exp}\left(\frac{e{U}_{m}}{nkT}\right)
From a condition of parallelism of two straight lines, the angular factor of these lines should be identical:
\frac{{j}_{sc}}{{U}_{oc}}={j}_{0}\frac{e}{nkT}\mathrm{exp}\left(\frac{e{U}_{m}}{nkT}\right)
From equality (6) for effective value of SC voltage (Um) we will receive:
{U}_{m}=\frac{nkT}{e}\mathrm{ln}\left(\frac{{j}_{sc}}{{j}_{0}}\frac{nkT}{e{U}_{oc}}\right)
Substituting the expression, defining effective value of SC voltage (Um), we receive for effective value of density of a current:
{j}_{m}={j}_{0}\left(\mathrm{exp}\left(\frac{e{U}_{m}}{nkT}\right)-1\right)-{j}_{sc}
{j}_{m}={j}_{sc}\left(\frac{nkT}{e{U}_{oc}}-1-\frac{{j}_{0}}{{j}_{sc}}\right)
As it is known, both j0 saturation, and jsc short circuits, and Uoc―voltage of idling depend on temperature [5] [6] :
{j}_{0}={j}_{00}\mathrm{exp}\left(-\frac{e\phi }{k}\left(\frac{1}{{T}_{0}}-\frac{1}{T}\right)\right)
{j}_{sc}={j}_{00}\mathrm{exp}\left[\frac{e\phi }{k}\left(\frac{1}{{T}_{0}}-\frac{1}{T}\right)\right]\left[\mathrm{exp}\left[\frac{e\phi }{nk{T}_{0}}\left(\frac{{U}_{0oc}}{\phi }-1+\frac{{T}_{0}}{T}\right)\right]-1\right]
{U}_{oc}=\left({U}_{0oc}-\phi \right)\frac{T}{{T}_{0}}+\phi
where U0oc―open circuit voltage at room temperature. As shown in work [7] of I - V ideality factor does not depend on temperature in the range of 100 < Т < 500 K.
On the other hand, dependence of height of a potential barrier φ from temperature of SC at not so low temperatures can be described in a kind:
\phi ={\phi }_{0}-\gamma T
{\phi }_{0}
―height of a potential barrier at temperature Т = 0, γ―temperature factor of height of the potential barrier, having identical sense with temperature factor of width of the band gap [5] . Value
{\phi }_{0}
can be defined, from the formula (11), by extrapolating dependence of voltage of idling on temperature Uoc (T) to T = 0 K. From a temperature dependence of width of the band gap knows that value of temperature factor of height of a potential barrier lies in an interval 10−3 ÷ 10−5 V/К [8] .
On Figure 2 results of experiment (curve-1) and calculation (curve-2) the currents of short circuit for SC which made on the base of AlGaAs-GaAs [1] received by (11) temperature dependence are presented. Conformity of results of experiment and calculation (section A) is received for dates: n = 1.0028; φo = 1.42 V and γ = 5 × 10−5 V/К. From drawing, it is visible that expression (11) not only explains results of experiences but also supplements them (Section A and C). Calculations have shown that short circuit currents of these SC at T = 500 K have a weak maximum (Jsc ≈ 122 mА), at temperatures more low T = 120 K and above T = 620 K start to decrease strongly. It is defined also that short circuit currents at temperatures T < 110 K and T > 770 K will be equal to zero (Jsc = 0). These results once again show correctness of the expressions received in works [5] [6] .
Now we pass to calculation of temperature dependence of effective value of SC current density. Delivering expressions for currents density of jo―saturation and Jshc―short circuit, and Uoc―voltage of idling in (9) is possible to receive the formula defining temperature dependence of effective value of density of SC current.
On Figure 3 results of experiment (curve-1) and calculation (curve-2), received by this formula for SC on the base of AlGaAs-GaAs [2] are resulted. Conformity of these results (Section B) are received at values n = 2.5; φo = 1.42 V
Figure 2. Dependence of density of a current of short circuit (mA) on temperature SC. 1-experiment [2] and 2-results of calculation from expression (11) when n = 1.0028; φo = 1.42 V and γ = 5 × 10−5 V/К.
Figure 3. Dependence of effective value of SC current density on temperature. 1-experiment [2] and 2-results of calculation received by expression (9) at n = 2.5; φo = 1.42 V and γ = 2 × 10−4 V/K.
and γ = 5 × 10−5 V/К. It is visible that in the expressions received for temperature dependence of short circuit current density and of current effective value from temperature has distinction only in values of I - V not ideality factor (n) of SC. This says that electric current type in these points of I - V differs from each other. From drawing it is visible that, as well as the density, and a short circuit current, and effective value of current density of SC have a weak maximum at T = 200 K (Jm ≈ 119 mА). Value of this photovoltaic characteristic in temperatures more low T = 120 K (Section A) and above T = 470 K (Section C) begin will strongly decrease. Calculations also have shown that at T < 110 K and T > 540 K value of effective density of a current equals to zero (Jm = 0).
In the scientific point of view, influence of temperature on effective value of current density of a semiconductor SC is investigated. The new expression allowing more approximately explain of experimental dependence of SC photovoltaic characteristics from temperature is received. We will notice that the new formulas received in this work for definition of dependence of the basic photo-galvanic characteristics of solar cells from temperature can use for SC on any semiconductor material basis.
From the practical point of view, it is established that SC, made on semiconductor alloy AlGaAs-GaAs, is not expedient to use at temperatures lower T = 110 K and above T = 540 K (that follows because of equality to zero of effective value of current density of such SC).
Rayimjon, A., Gulomjonovich, I.R. and Alisherovna, A.M. (2019) Temperature Stimulation of Effective Value of Density of the Current of Semiconductor Solar Cells. Energy and Power Engineering, 11, 92-97. https://doi.org/10.4236/epe.2019.112006
1. Natarova, J.V., Gnatenko, A.S. and Galat, A.B. (2018) Research of Photo-Electric Converters on the Basis of Various Semiconductor Materials. Journal of Nano- and Electronic Physics, 10, 04023. (In Russian)
2. Bjub, F.A. (1987) Solar Cells (the Theory and Experiment). М, Energoatomizdat, 278 p. (In Russian)
3. Bavin, M.R. (2014) Development and Research of Refracting Photoelectric Installations. The Dissertation on Competition of a Scientific Degree of Cand. Tech. Sci. a Speciality 05.14.08, Power Installations on the Basis of Renewable Energy. Moscow, 139 p. (In Russian)
4. Zi, S. (1984) Physics of Semiconductor Devices. The Book 1, М, Mir, 456 p. (In Russian)
5. Zaynobiddinov, S., Ikramov, R.G., Aliev, R., Ismanova, O.T., Niyazov, О. and Nuritdinova, M.A. (2003) Influence of Temperature on Photoelectric Characteristics of Solar Cells from Amorphous Silicon. Geliotecknica, №3, 19-22. (In Russian)
6. Aliev, R., Zaynobiddinov, С., Ikramov, R.G., Ismanova, O.T. and Nuritdinova, M.A. (2003) An Estimation of Photoelectric Parameters of Solar Cells on the Basis of Amorphous Silicon. Geliotecknica, № 1, 18-22. (In Russian)
7. Aliev, R., Alinazarova, M.А., Ikramov, R.G. and Ismanova, O.T. (2009) Features of Influence of Temperature on Photocurrent of Solar Cell, Created on the Basis of the Amorphous Semiconductor. Geliotecknica, № 3, 15-20. (In Russian)
8. Aut, I., Gentsov, D. and Herman, T. (1980) The Photoelectric Phenomena. М, Mir, 208 p. (In Russian)
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I am having difficulty finding if this series converges or diverges: \sum_{n=0}^\infty\frac{(-2)^{n+1}}{n^{n-1}}Z
I am having difficulty finding if this series converges or d
I am having difficulty finding if this series converges or diverges:
\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-2\right)}^{n+1}}{{n}^{n-1}}
I am unsure of which test to use. At first, I thought I should use alternating series test, but I am unable to manipulate the numerator to
{\left(-1\right)}^{n+1}
. I thought I may be able to use root test, but I have no clue how to manipulate the series to do that.
What series test would I use, and how?
\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-2\right)}^{n+1}}{{n}^{n-1}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n+1}\frac{{2}^{n+1}}{{n}^{n-1}}
By the alternating series test this converges if
{a}_{n}>0
: we can easily this is satisfied
{a}_{n}
is monotic:
{a}_{n}
is decreasing monotonically
\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=0
I will leave this one to you.
|\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-2\right)}^{n+1}}{{n}^{n-1}}|\le \sum _{n=0}^{\mathrm{\infty }}\frac{{2}^{n+1}}{{n}^{n-1}}\le \sum _{n=0}^{3}{\left(\frac{2}{n}\right)}^{n+1}+\sum _{n=4}^{\mathrm{\infty }}{\left(\frac{2}{n}\right)}^{n+1}
The last series is a geometric series and therefore converges.
Hint: what can you say about
\frac{{2}^{n-1}}{{n}^{n-1}}
Can you show that
\frac{{2}^{n-1}}{{n}^{n-1}}\le \frac{1}{{n}^{2}}
n\ge N
Making the problem more general, let
S=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}\text{ }with\text{ }{a}_{n}=\frac{{x}^{n+1}}{{n}^{n-1}}
\mathrm{log}\left({a}_{n}\right)=\left(n+1\right)\mathrm{log}\left(x\right)-\left(n-1\right)\mathrm{log}\left(n\right)
\mathrm{log}\left({a}_{n+1}\right)-\mathrm{log}\left({a}_{n}\right)=\left(n-1\right)\mathrm{log}\left(n\right)-n\mathrm{log}\left(n+1\right)+\mathrm{log}\left(x\right)
Using Taylor for large values of n
\mathrm{log}\left({a}_{n+1}\right)-\mathrm{log}\left({a}_{n}\right)=\mathrm{log}\left(\frac{x}{ne}\right)+\frac{1}{2n}+O\left(\frac{1}{{n}^{2}}\right)
\frac{{a}_{n+1}}{{a}_{n}}={e}^{\mathrm{log}\left({a}_{n+1}\right)-\mathrm{log}\left({a}_{n}\right)}\sim \frac{x}{ne}\to 0
\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n+1}{\left(1+n\right)}^{\frac{1}{n}}}{n}
Prove the following inequality for every
n\ge 1
\sum _{k=1}^{n}\frac{1}{{k}^{2}+3k+1}\le \frac{13}{20}
Show that the following infinite series
\sum _{i=1}^{\mathrm{\infty }}\frac{f\left(n\right)}{{n}^{2}}
\pi =\sqrt{12}\sum _{n\ge 0}\frac{\left(-1{\right)}^{n}}{{3}^{n}\left(2n+1\right)}
Calculate the sun of the infinite series:
\sum _{n=0}^{\mathrm{\infty }}\frac{n}{{4}^{n}}
Does the following series converge or diverge?
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{n}+\sqrt{n+1}}
\sum _{n=1}^{\mathrm{\infty }}\left(\frac{n}{500}{\right)}^{n}
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Parabolic_partial_differential_equation Knowpia
A parabolic partial differential equation is a type of partial differential equation (PDE). Parabolic PDEs are used to describe a wide variety of time-dependent phenomena, including heat conduction, particle diffusion, and pricing of derivative investment instruments.
To define the simplest kind of parabolic PDE, consider a real-valued function
{\displaystyle u(x,y)}
of two independent real variables,
{\displaystyle x}
{\displaystyle y}
. A second-order, linear, constant-coefficient PDE for
{\displaystyle u}
{\displaystyle Au_{xx}+2Bu_{xy}+Cu_{yy}+Du_{x}+Eu_{y}+F=0,}
and this PDE is classified as being parabolic if the coefficients satisfy the condition
{\displaystyle B^{2}-AC=0.}
{\displaystyle x}
represents one-dimensional position and
{\displaystyle y}
represents time, and the PDE is solved subject to prescribed initial and boundary conditions.
The name "parabolic" is used because the assumption on the coefficients is the same as the condition for the analytic geometry equation
{\displaystyle Ax^{2}+2Bxy+Cy^{2}+Dx+Ey+F=0}
to define a planar parabola.
The basic example of a parabolic PDE is the one-dimensional heat equation,
{\displaystyle u_{t}=\alpha \,u_{xx},}
{\displaystyle u(x,t)}
{\displaystyle t}
and at position
{\displaystyle x}
along a thin rod, and
{\displaystyle \alpha }
is a positive constant (the thermal diffusivity). The symbol
{\displaystyle u_{t}}
signifies the partial derivative of
{\displaystyle u}
with respect to the time variable
{\displaystyle t}
{\displaystyle u_{xx}}
is the second partial derivative with respect to
{\displaystyle x}
. For this example,
{\displaystyle t}
{\displaystyle y}
in the general second-order linear PDE:
{\displaystyle A=\alpha }
{\displaystyle E=-1}
The heat equation says, roughly, that temperature at a given time and point rises or falls at a rate proportional to the difference between the temperature at that point and the average temperature near that point. The quantity
{\displaystyle u_{xx}}
measures how far off the temperature is from satisfying the mean value property of harmonic functions.
The concept of a parabolic PDE can be generalized in several ways. For instance, the flow of heat through a material body is governed by the three-dimensional heat equation,
{\displaystyle u_{t}=\alpha \,\Delta u,}
{\displaystyle \Delta u:={\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}+{\frac {\partial ^{2}u}{\partial z^{2}}}}
denotes the Laplace operator acting on
{\displaystyle u}
. This equation is the prototype of a multi-dimensional parabolic PDE.
{\displaystyle -\Delta }
is an elliptic operator suggests a broader definition of a parabolic PDE:
{\displaystyle u_{t}=-Lu,}
{\displaystyle L}
is a second-order elliptic operator (implying that
{\displaystyle L}
must be positive; a case where
{\displaystyle u_{t}=+Lu}
is considered below).
A system of partial differential equations for a vector
{\displaystyle u}
can also be parabolic. For example, such a system is hidden in an equation of the form
{\displaystyle \nabla \cdot (a(x)\nabla u(x))+b(x)^{\text{T}}\nabla u(x)+cu(x)=f(x)}
if the matrix-valued function
{\displaystyle a(x)}
has a kernel of dimension 1.
Parabolic PDEs can also be nonlinear. For example, Fisher's equation is a nonlinear PDE that includes the same diffusion term as the heat equation but incorporates a linear growth term and a nonlinear decay term.
Under broad assumptions, an initial/boundary-value problem for a linear parabolic PDE has a solution for all time. The solution
{\displaystyle u(x,t)}
{\displaystyle x}
for a fixed time
{\displaystyle t>0}
, is generally smoother than the initial data
{\displaystyle u(x,0)=u_{0}(x)}
For a nonlinear parabolic PDE, a solution of an initial/boundary-value problem might explode in a singularity within a finite amount of time. It can be difficult to determine whether a solution exists for all time, or to understand the singularities that do arise. Such interesting questions arise in the solution of the Poincaré conjecture via Ricci flow.[citation needed]
Backward parabolic equationEdit
One occasionally encounters a so-called backward parabolic PDE, which takes the form
{\displaystyle u_{t}=Lu}
(note the absence of a minus sign).
An initial-value problem for the backward heat equation,
{\displaystyle {\begin{cases}u_{t}=-\Delta u&{\textrm {on}}\ \ \Omega \times (0,T),\\u=0&{\textrm {on}}\ \ \partial \Omega \times (0,T),\\u=f&{\textrm {on}}\ \ \Omega \times \left\{0\right\}.\end{cases}},}
is equivalent to a final-value problem for the ordinary heat equation,
{\displaystyle {\begin{cases}u_{t}=\Delta u&{\textrm {on}}\ \ \Omega \times (0,T),\\u=0&{\textrm {on}}\ \ \partial \Omega \times (0,T),\\u=f&{\textrm {on}}\ \ \Omega \times \left\{T\right\}.\end{cases}}}
Similarly to a final-value problem for a parabolic PDE, an initial-value problem for a backward parabolic PDE is usually not well-posed (solutions often grow unbounded in finite time, or even fail to exist). Nonetheless, these problems are important for the study of the reflection of singularities of solutions to various other PDEs.[1]
^ Taylor, M. E. (1975), "Reflection of singularities of solutions to systems of differential equations", Comm. Pure Appl. Math., 28 (4): 457–478, CiteSeerX 10.1.1.697.9255, doi:10.1002/cpa.3160280403
Perthame, Benoît (2015), Parabolic Equations in Biology : Growth, Reaction, Movement and Diffusion, Springer, ISBN 978-3-319-19499-8
Evans, Lawrence C. (2010) [1998], Partial differential equations, Graduate Studies in Mathematics, vol. 19 (2nd ed.), Providence, R.I.: American Mathematical Society, doi:10.1090/gsm/019, ISBN 978-0-8218-4974-3, MR 2597943
"Parabolic partial differential equation", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
"Parabolic partial differential equation, numerical methods", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
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Supported Equity Derivative Functions - MATLAB & Simulink - MathWorks í•œêµ
\mathrm{max}\left(0,{S}_{av}âX\right)
\mathrm{max}\left(0,Xâ{S}_{av}\right)
\mathrm{max}\left(0,Sâ{S}_{av}\right)
\mathrm{max}\left(0,{S}_{av}âS\right)
{S}_{av}
S
X
{S}_{av}
A double barrier option is similar to the standard single barrier option except that they have two barrier levels: a lower barrier (LB) and an upper barrier (UB). The payoff for a double barrier option depends on whether the underlying asset remains between the barrier levels during the life of the option. Double barrier options are less expensive than single barrier options as the probability of being knocked out is higher. Because of this, double barrier options allow investors to achieve reduction in the option premiums as and match an investor’s belief about the future movement of the underlying price process.
\mathrm{max}\left(âWiâSiâK;0\right)
\mathrm{max}\left(âKâWiâSi;0\right)
A chooser option has a specified decision time t1 where the holder has to make the decision whether the option is a call or put. At the expiration time t2 the option expires. If the holder chooses a call option, the payout is max ( S − K , 0). For the choice of a put option, the payout is max ( K − S , 0) where K is the strike price of the option and S is the equity price at expiry.
The payoff formulas for compound options are too complex for this discussion. If you are interested in the details, consult the paper by Mark Rubinstein entitled “Double Trouble,†published in Risk 5 (1991).
Coupon — The coupon in convertible bonds are typically lower than coupons in vanilla bonds since investors are willing to take the lower coupon for the opportunity to participate in the company’s stock via the conversion.
\mathrm{max}\left(0,{S}_{\mathrm{max}}âX\right)
\mathrm{max}\left(0,Xâ{S}_{\mathrm{min}}\right)
\mathrm{max}\left(0,Sâ{S}_{\mathrm{min}}\right)
\mathrm{max}\left(0,{S}_{\mathrm{max}}âS\right)
{S}_{\mathrm{max}}
{S}_{\mathrm{min}}
S
X
\mathrm{max}\left(StâK,0\right)
\mathrm{max}\left(KâSt,0\right)
\mathrm{max}\left(X1âX2âK,0\right)
{f}_{T}={S}_{T}âK
{f}_{T}=Kâ{S}_{T}
F\left(t,T\right)
F\left(s,T\right)âF\left(t,T\right)
F\left(T,T\right)
F\left(T,T\right)
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The centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a
Margin of error (E) = 0.02
Population proportion (p) = 0.26
The sample size could be calculated using the following formula:
n=p\left(1-p\right){\left(\frac{{Z}_{\frac{\alpha }{2}}}{E}\right)}^{2}
The sample size at 95% confidence level can be calculated as:
\frac{{Z}_{\alpha }}{2}
corresponding to 95% confidence level is 1.96
n=p\left(1-p\right){\left(\frac{{Z}_{\frac{\alpha }{2}}}{E}\right)}^{2}
=0.26\left(1-0.26\right){\left(\frac{1.96}{0.02}\right)}^{2}
=0.1924×9604
\approx 1848
Here, X = 520 and n = 1848.
The point estimate can be calculated as:
\stackrel{^}{p}=\frac{X}{n}
=\frac{520}{1848}
The 95% confidence interval for the proportion of smokers in the population can be calculated as:
CI=\stackrel{^}{p}±{Z}_{\frac{\alpha }{2}}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}
=0.2814±1.96×\sqrt{\frac{0.2814\left(1-0.2814\right)}{1848}}
=0.2814±0.0205
=(0.2609,0.3019)
Know these terms. Explain in one sentence each what “randomized,” “double-blind,” and “placebo-controlled” mean in the description of the design of the study.
Experiment basics. Identify the subjects, the explanatory variable, and several response variables for this study.
With the aid of examples briefly explain the difference between fixed effect and random effects models in experimental design.
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1School of Economics and Management, Shanghai Maritime University, Shanghai, China.
2Glorious Sun School of Business and Management, Donghua University, Shanghai, China.
3School of Health and Management, Abdou Moumouni Dioffo University, Niamey, Niger.
Abdoulkarim, H.T., Fatouma, S.H. and Hassane, H.T. (2019) Assessment of Dry Port Efficiency in Africa Using Data Envelopment Analysis. Journal of Transportation Technologies, 9, 193-203. doi: 10.4236/jtts.2019.92012.
\mathrm{max}\frac{{\sum }_{k=1}^{s}{u}_{k}{y}_{kp}}{{\sum }_{j=1}^{m}{v}_{i}{x}_{ip}}
s.t=\frac{{\sum }_{k=1}^{s}{u}_{k}{y}_{ki}}{{\sum }_{j=1}^{s}{v}_{j}{x}_{i,j}}\le 1\text{and}{u}_{k}{v}_{j}\ge 0\text{}\forall k,j
{y}_{ki}
{\text{DMU}}_{i}
{x}_{ji}
{\text{DMU}}_{i}
{u}_{k}
{v}_{j}
\mathrm{max}{\sum }_{k=1}^{s}{u}_{k}{y}_{kp}={\theta }_{p}
{\sum }_{j=1}^{m}{v}_{j}{y}_{jp}=1
{\sum }_{k=1}^{s}{u}_{k}{y}_{ki}-{\sum }_{j=1}^{m}{v}_{j}{x}_{ij}\le 0\text{}\forall i,{u}_{k},{v}_{j}\ge 0\text{}{\forall }_{k,j}
{\varphi }_{k}
s.t{\sum }_{i=1}^{n}{\lambda }_{j}{x}_{ij}\le {\varphi }_{r}{x}_{ir}\text{\hspace{0.17em}}i=1,2,\cdots ,m
{\sum }_{j=1}^{n}{\lambda }_{j}{y}_{rk}\le {\varphi }_{r}{x}_{ir}\text{\hspace{0.17em}}r=1,2,\cdots ,s
{\lambda }_{j}\ge 0,\forall j
{\sum }_{j=1}^{n}{\lambda }_{i}=1
{\varphi }_{k}
[1] PIDG (2015) PIDG Project Fact Sheet Africa: Niger Dry Port.
[2] Council, N.S. (2016) Dry Ports. https://www.shipperscouncil.gov.ng/inland-dry-port
[3] Elshaday, W.G. (2016) Assessment of the Performance of Dry Ports in Ethiopia Using SCOR Method. Addis Ababa Institute of Technology, Addis Ababa.
[4] World Port Source.
http://www.worldportsource.com/ports/commerce/KEN_Port_of_Mombasa_1365.php
[5] Bentaleb, F., Mouhsene, F., Charif, M. and Alami, S. (2016) Dry Port-Seaport System Development: Application of the Product Life Cycle Theory. Journal of Transportation and Logistics, 1, 115-128.
https://doi.org/10.22532/jtl.267840
[6] Farrell, M. (1957) Measurement of Productive Efficiency. Journal of Royal Statistical Efficiency, 120, 253-281.
[7] Szczepura, A., Davis, C., Fletcher, J. and Bousoffiane, A. (1992) Applied Data Envelopment Analysis in Healthcare: The Relative Efficiency of NHS General Practices. Warwick Business School Research Bureau, Coventry.
[9] Wang, T., Song, D.-W. and Cullinane, K. (2003) Container Port Production Efficiency: A Comparative Study of DEA and FDH Approaches. Journal of the Eastern Asia Society for Transportation Studies, 5, 698-713.
[10] Martinez, B.E., Diaz, A.R., Navarro, I.M. and Ravelo, M.T. (1999) A Study of the Efficiency of Spanish Port Authorities Using Data Envelopment Analysis. International Journal of Transport Economics, 26, 237-257.
[11] Tongzon, J. (2001) Efficiency Measurement of Selected Australian and Other International Ports Using DEA. Transportation Research Part A: Policy and Practice, 35, 107-122.
[12] Itoh, H. (2002) Efficiency Changes at Major Container Ports in Japan: A Window Application of Data Envelopment Analysis. Review of Urban and Regional Development Studies, 14, 133-152.
[13] Baros, C. (2003) The Measurement of Efficiency of Portuguese Seaport Authorities with DEA. International Journal of Transport Economics, 30, 335-354.
[14] Barros, C. and Athanassiou, M. (2004) Efficiency in European Seaports with DEA: Evidence from Greece and Portugal. Maritime Economics and Logistics, 6, 122-140.
[15] Min, H. and Park, B. (2005) Evaluating the Inter-Temporal Efficiency Trends of International Container Terminals Using Data Envelopment Analysis. International Journal of Integrated Supply Management, 1, 258-277.
[16] Cullinane, K. and Wang, T. (2006) The Efficiency of European Container Ports: A Cross-Sectional Data Envelopment Analysis. International Journal of Logistics Research and Applications, 9, 19-31.
[17] Judit, O., Steffen, N., Thomas, N. and József, P. (2018) Ranking of Dry Ports in Europe-Benchmarking. Periodica Polytechnica Transportation Engineering, 46, 95-100.
[18] Hercules, H., Girish, G. and Mukul, J. (2011) Dry Port Efficiency and Container Security. IAME Latin America, Santiago, 26-28 October 2011, 1-16.
[19] Banker, R., Charnes, A. and Cooper, W. (1984) Some Models for Estimating Technical and Scale Inefficiencies in Data Envelopment Analysis. Management Science, 30, 1078-1092.
[20] Tulkens, H. and Van den Eeckhaut, P. (1995) Non-Frontier Measures of Efficiency, Progress and Regress for Time Series Data. International Journal of Production Economics, 33, 83-97.
[21] Pjevčević, D., Radonjić, A., Hrle, Z. and Čolić, V. (2012) DEA Window Analysis for Measuring Port Efficiencies in Serbia. PROMET-Traffic & Transportation, 24, 63-72.
https://doi.org/10.7307/ptt.v24i1.269
[22] Chung, S., Lee, A., Kang, H. and Lai, C. (2008) A DEA Window Analysis on the Product Family Mix Selector for Semi-Conductor Fabricator. Expert Systems with Applications, 35, 379-388.
[23] George, K.V. (2015) Assessment of Port Efficiency in West Africa Using Data Envelopment Analysis. American Journal of Industrial and Business Management, 5, 208-218.
[24] Citizen (2013) Isaka Dry Port on Track. The Citizen Newspaper.
https://www.thecitizen.co.tz/News/Isaka-dry-port-project-remains-on-track/1840340-1990596-irr9jm/index.html
[25] Gerald, W.W. and Jin, Z. (2015) A Comparative Study of Dry Ports in East Africa and China. IISTE, 7-17.
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On a moving train, a passenger who is seated facing the front, tosses a coin up in the air. Given that the coin falls into the lap of the person sitting behind this person, what can we say about the motion of the train?
All the people are facing the front of the train.
All the windows/doors are shut.
It is accelerating It is moving along a circular track It is slowing down It is moving at uniform velocity
You are hovering in a helicopter well above the Earth's surface such that the helicopter is suspended in mid air, stationary with respect to the Earth. The Earth is rotating at the same time. Why should you fly to my destination, if it will come to you due to Earth's rotation?
Helicopter will crash because of the rotatory forces that bind it and the clouds. I can definitely reach my destination without wasting fuel. I will reach my destination but it will take me many years. My helicopter is rotating along with the Earth, though I do not notice it.
by Harshit Dahiya
As shown in the picture, water flows from the bottle into the cup. What explains this phenomena?
Note, the picture is upside down. The Earth is in the top of the photo and the bottom of the plane is facing the atmosphere.
Gravity pulls the water down toward the stationary cup. A normal force from the bottle pushes the water out. A bumpy ride due to turbulence forces water from the bottle. The plane and water accelerate toward the Earth, but the plane accelerates faster.
A particle is moving on an elliptical path as shown, at a constant speed. At which point is the particle's acceleration at maximum?
A B C Same Everywhere
A point-mass (particle) is shot up an inclined plane that makes an angle
\theta
with the ground. It slides up the incline to a certain distance, and then slides all the way back down. However, due to friction, the time it takes for the particle to slide from the maximum distance it reaches up the incline back to the bottom is twice the time it takes for it to go to the maximum distance up the incline from the bottom.
\mu
between the particle and the incline can be written as
\frac{a}{b}\tan\theta
and
b
are co-prime positive integers.
a + b
by Milly Choochoo
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On Fresnel's Theory of the Aberration of Light - Wikisource, the free online library
On Fresnel's Theory of the Aberration of Light
Philosophical Magazine, Volume 28: 76-81, Online
1496993On Fresnel's Theory of the Aberration of Light1846George Gabriel Stokes
XVII. On Fresnel's Theory of the Aberration of Light. By G. G. Stokes, M.A., Fellow of Pembroke College, Cambridge[1].
THE theory of the aberration of light, and of the absence of any influence of the motion of the earth on the laws of refraction, &c., given by Fresnel in the ninth volume of the Annales de Chimie, p. 57, is really very remarkable. If we suppose the diminished velocity of propagation of light within refracting media to arise solely from the greater density of the æther within them, the elastic force being the same as without, the density which it is necessary to suppose the æther within a medium of refractive index
{\displaystyle \mu }
to have is
{\displaystyle \mu ^{2}}
, the density in vacuum being taken for unity. Fresnel supposes that the earth passes through the æther without disturbing it, the æther penetrating the earth quite freely. He supposes that a refracting medium moving with the earth carries with it a quantity of æther, of density
{\displaystyle \mu ^{2}-1}
, which constitutes the excess of density of the æther within it over the density of the æther in vacuum. He supposes that light is propagated through this æther, of which part is moving with the earth, and part is at rest in space, as it would be if the whole were moving with the velocity of the centre of gravity of any portion of it, that is, with a velocity
{\displaystyle \left(1-{\tfrac {1}{\mu ^{2}}}\right)v}
{\displaystyle v}
being the velocity of the earth. It may be observed however that the result would be the same if we supposed the whole of the æther within the earth to move together, the æther entering the earth in front, and being immediately condensed, and issuing from it behind, where it is immediately rarefied, undergoing likewise sudden condensation or rarefaction in passing from one refracting medium to another. On this supposition, the evident condition that a mass
{\displaystyle v}
of the æther must pass in a unit of time across a plane of area unity, drawn anywhere within the earth in a direction perpendicular to that of the earth's motion, gives
{\displaystyle \left(1-{\tfrac {1}{\mu ^{2}}}\right)v}
for the velocity of the æther within a refracting medium. As this idea is rather simpler than Fresnel's, I shall adopt it in considering his theory. Also, instead of considering the earth as in motion and the æther outside it as at rest, it will be simpler to conceive a velocity equal and opposite to that of the earth impressed both on the earth and on the æther. On this supposition the earth will be at rest; the æther outside it will be moving with a velocity
{\displaystyle v}
, and the æther in a refracting medium with a velocity
{\displaystyle {\tfrac {v}{\mu ^{2}}}}
, in a direction contrary to that of the earth's real motion. On account of the smallness of the coefficient of aberration, we may also neglect the square of the ratio of the earth's velocity to that of light; and if we resolve the earth's velocity in different directions, we may consider the effect of each resolved part separately.
In the ninth volume of the Comptes Rendus of the Academy of Sciences, p. 774, there is a short notice of a memoir by M. Babinet, giving an account of an experiment which seemed to present a difficulty in its explanation. M. Babinet found that when two pieces of glass of equal thickness were placed across two streams of light which interfered and exhibited fringes, in such a manner that one piece was traversed by the light in the direction of the earth's motion, and the other in the contrary direction, the fringes were not in the least displaced. This result, as M. Babinet asserts, is contrary to the theory of aberration contained in a memoir read by him before the Academy in 1829, as well as to the other received theories on the subject. I have not been able to meet with this memoir, but it is easy to show that the result of M. Babinet's experiment is in perfect accordance with Fresnel's theory.
{\displaystyle T}
be the thickness of one of the glass plates,
{\displaystyle V}
the velocity of propagation of light in vacuum, supposing the æther at rest. Then
{\displaystyle {\tfrac {V}{\mu }}}
would be the velocity with which light would traverse the glass if the æther were at rest; but the æther moving with a velocity
{\displaystyle {\tfrac {v}{\mu ^{2}}}}
, the light traverses the glass with a velocity
{\displaystyle {\tfrac {V}{\mu }}\pm {\tfrac {v}{\mu ^{2}}}}
, and therefore in a time
{\displaystyle T\div \left({\frac {V}{\mu }}\pm {\frac {v}{\mu ^{2}}}\right)={\frac {\mu T}{V}}\left(1\pm {\frac {v}{\mu V}}\right)}
But if the glass were away, the light, travelling with a velocity
{\displaystyle V\pm v}
, would pass over the space
{\displaystyle T}
{\displaystyle T\div (V\pm v)={\frac {T}{V}}\left(1\pm {\frac {v}{V}}\right)}
Hence the retardation, expressed in time,
{\displaystyle =(\mu -1){\tfrac {T}{V}}}
, the same as if the earth were at rest. But in this case no effect would be produced on the fringes, and therefore none will be produced in the actual case.
I shall now show that, according to Fresnel's theory, the laws of reflexion and refraction in singly refracting media are uninfluenced by the motion of the earth. The method which I employ will, I hope, be found simpler than Fresnel's; besides it applies easily to the most general case. Fresnel has not given the calculation for reflexion, but has merely stated the result; and with respect to refraction, he has only considered the case in which the course of the light within the refracting medium is in the direction of the earth's motion. This might still leave some doubt on the mind, as to whether the result would be the same in the most general case.
If the æther were at rest, the direction of light would be that of a normal to the surfaces of the waves. When the motion of the æther is considered, it is most convenient to define the direction of light to be that of the line along which the same portion of a wave moves relatively to the earth. For this is in all cases the direction which is ultimately observed with a telescope furnished with cross wires. Hence, if
{\displaystyle A}
is any point in a wave of light, and if we draw
{\displaystyle AB}
normal to the wave, and proportional to
{\displaystyle V}
{\displaystyle {\tfrac {V}{\mu }}}
, according as the light is passing through vacuum or through a refracting medium, and if we draw
{\displaystyle BC}
in the direction of the motion of the æther, and proportional to
{\displaystyle v}
{\displaystyle {\tfrac {v}{\mu ^{2}}}}
, and join
{\displaystyle AC}
, this line will give the direction of the ray. Of course, we might equally have drawn
{\displaystyle AD}
{\displaystyle BC}
and in the opposite direction, when
{\displaystyle DB}
would have given the direction of the ray.
Let a plane
{\displaystyle P}
be drawn perpendicular to the reflecting or refracting surface and to the waves of incident light, which in this investigation may be supposed plane. Let the velocity
{\displaystyle v}
of the æther in vacuum be resolved into
{\displaystyle p}erpendicular to the plane
{\displaystyle P}
{\displaystyle q}
in that plane; then the resolved parts of the velocity
{\displaystyle {\tfrac {v}{\mu ^{2}}}}
of the æther within a refracting medium will be
{\displaystyle {\frac {p}{\mu ^{2}}},{\frac {q}{\mu ^{2}}}}
. Let us first consider the effect of the velocity
{\displaystyle p}
It is easy to see that, as far as regards this resolved part of the velocity of the æther, the directions of the refracted and reflected waves will be the same as if the æther were at rest. Let
{\displaystyle BAC}
(fig. I) be the intersection of the refracting surface and the plane
{\displaystyle P}
{\displaystyle DAE}
a normal to the refracting surface;
{\displaystyle AF,AG,AH}
normals to the incident, reflected and refracted waves. Hence
{\displaystyle AF,AG,AH}
will be in the plane
{\displaystyle P}
{\displaystyle \angle GAD=FAD,\ \mu \sin HAE=\sin FAD}
{\displaystyle AG=AF,\ AH={\frac {1}{\mu }}AF}
Draw
{\displaystyle Gg,Hh}
perpendicular to the plane
{\displaystyle P}
, and in the direction of the resolved part
{\displaystyle p}
of the velocity of the æther, and
{\displaystyle Ff}
in the opposite direction; and take
{\displaystyle Ff:Hh:FA::P:{\frac {p}{\mu ^{2}}}:V}
{\displaystyle Gg=Ff}
and join
{\displaystyle A}
{\displaystyle f,g}
{\displaystyle h}
{\displaystyle fA,Ag,Ah}
will be the directions of the incident, reflected and refracted rays. Draw
{\displaystyle FD,HE}
{\displaystyle DE}
{\displaystyle fD,hE}
{\displaystyle fDF,hEH}
will be the inclinations of the planes
{\displaystyle fAD,hAE}
{\displaystyle P}
{\displaystyle \tan FDf={\frac {p}{V\sin FAD}},\ \tan HEh={\frac {\mu ^{-2}p}{\mu ^{-1}V\sin HAE}}}
{\displaystyle FAD=\mu \sin HAE}
{\displaystyle FDf=\tan HEh}
, and therefore the refracted ray
{\displaystyle Ah}
lies in the plane of incidence
{\displaystyle fAD}
. It is easy to see that the same is true of the reflected ray
{\displaystyle Ag}
{\displaystyle \angle gAD=fAD}
; and the angles
{\displaystyle fAD,hAE}
are sensibly equal to
{\displaystyle FAD,HAE}
respectively, and we therefore have without sensible error,
{\displaystyle \sin fAD=\mu \sin hAE}
. Hence the laws of reflexion and refraction are not sensibly affected by the velocity
{\displaystyle p}
Let us now consider the effect of the velocity
{\displaystyle q}
. As far as depends on this velocity, the incident, reflected and refracted rays will all be in the plane
{\displaystyle P}
{\displaystyle AH,AK,AL}
be the intersections of the plane
{\displaystyle P}
with the incident, reflected and refracted waves. Let
{\displaystyle \psi ,\psi _{'},\psi ^{'}}
be the inclinations of these waves to the refracting surface; let
{\displaystyle NA}
be the direction of the resolved part
{\displaystyle q}
of the velocity of the æther, and let the angle
{\displaystyle NAC=\alpha }
The resolved part of
{\displaystyle q}
in a direction perpendicular to
{\displaystyle AH}
{\displaystyle q\sin(\psi +\alpha )}
. Hence the wave
{\displaystyle AH}
travels with the velocity
{\displaystyle V+q\sin(\psi +\alpha )}
; and consequently the line of its intersection with the refracting surface travels along
{\displaystyle AB}
{\displaystyle \mathrm {cosec} \ \psi \left\{V+q\sin(\psi +\alpha )\right\}}
. Observing that
{\displaystyle {\tfrac {q}{\mu ^{2}}}}
is the velocity of the æther within the refracting medium, and
{\displaystyle {\tfrac {V}{\mu }}}
the velocity of propagation of light, we shall find in a similar manner that the lines of intersection of the refracting surface with the reflected and refracted waves travel along
{\displaystyle AB}
with velocities
{\displaystyle \mathrm {cosec} \ \psi _{'}\left\{V+q\sin(\psi _{'}+\alpha )\right\},\ \mathrm {cosec} \ \psi '\left\{{\frac {V}{\mu }}+{\frac {q}{\mu ^{2}}}\sin(\psi '+\alpha )\right\}}
But since the incident, reflected and refracted waves intersect the refracting surface in the same line, we must have
{\displaystyle \left.{\begin{aligned}\sin \psi _{'}\left\{V+q\sin(\psi +\alpha )\right\}=&\sin \psi \left\{V+q\sin(\psi _{'}+\alpha )\right\},\\\mu \sin \psi '\left\{V+q\sin(\psi +\alpha )\right\}=&\sin \psi \left\{V+{\frac {q}{\mu }}\sin(\psi '+\alpha )\right\}.\end{aligned}}\right\}}
{\displaystyle HS}
{\displaystyle AH}
{\displaystyle ST}
{\displaystyle NA}
{\displaystyle ST:HS::q:V}
{\displaystyle HT}
{\displaystyle HT}
is the direction of the incident ray; and denoting the angles of incidence, reflexion and refraction by
{\displaystyle \phi ,\phi _{'},\phi ^{'}}
{\displaystyle \phi -\psi =SHT={\frac {ST\sin S}{SH}}={\frac {1}{V}}\times }
resolved part of
{\displaystyle q}
{\displaystyle AH}
{\displaystyle ={\frac {q}{V}}\cos(\psi +\alpha )}
{\displaystyle \phi _{'}-\psi _{'}={\frac {q}{V}}\cos \left(\psi _{'}-\alpha \right),\ \phi '-\psi '={\frac {q}{\mu V}}\cos \left(\psi '+\alpha \right):}
{\displaystyle {\begin{aligned}\sin \psi =&\sin \phi -{\frac {q}{V}}\cos \phi \cos(\phi +\alpha ),\\\sin \psi _{'}=&\sin \phi _{'}-{\frac {q}{V}}\cos \phi _{'}\cos(\phi _{'}-\alpha ),\\\sin \psi '=&\sin \phi '-{\frac {q}{V}}\cos \phi '\cos(\phi '+\alpha ).\end{aligned}}}
On substituting these values in equations (A), and observing that in the terms multiplied by
{\displaystyle q}
we may put
{\displaystyle \phi _{'}=\phi ,\ \mu \sin \phi '=\sin \phi }
, the small terms destroy each other, and we have
{\displaystyle \sin \phi _{'}=\sin \phi ,\ \mu \sin \phi '=\sin \phi }
. Hence the laws of reflexion and refraction at the surface of a refracting medium will not be affected by the motion of the æther.
In the preceding investigation it has been supposed that the refraction is out of vacuum into a refracting medium. But the result is the same in the general case of refraction out of one medium into another, and reflexion at the common surface. For all the preceding reasoning applies to this case if we merely substitute
{\displaystyle {\tfrac {p}{\mu '^{2}}},{\tfrac {q}{\mu '^{2}}}}
{\displaystyle p,q,{\tfrac {V}{\mu '}}}
{\displaystyle V}
{\displaystyle {\tfrac {\mu }{\mu '}}}
{\displaystyle \mu }
{\displaystyle \mu '}
being the refractive index of the first medium. Of course refraction out of a medium into vacuum is included as a particular case.
It follows from the theory just explained, that the light coming from any star will behave in all cases of reflexion and ordinary refraction precisely as it would if the star were situated in the place which it appears to occupy in consequence of aberration, and the earth were at rest. It is, of course, immaterial whether the star is observed with an ordinary telescope, or with a telescope having its tube filled with fluid. It follows also that terrestrial objects are referred to their true places. All these results would follow immediately from the theory of aberration which I proposed in the July number of this Magazine; nor have I been able to obtain any result, admitting of being compared with experiment, which would be different according to which theory we adopted. This affords a curious instance of two totally different theories running parallel to each other in the explanation of phænomena. I do not suppose that many would be disposed to maintain Fresnel's theory, when it is shown that it may be dispensed with, inasmuch as we would not be disposed to believe, without good evidence, that the æther moved quite freely through the solid mass of the earth. Still it would have been satisfactory, if it had been possible, to have put the two theories to the test of some decisive experiment.
Retrieved from "https://en.wikisource.org/w/index.php?title=On_Fresnel%27s_Theory_of_the_Aberration_of_Light&oldid=4443751"
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Wheel and axle - 3D CAD Models & 2D Drawings
Wheel and axle (13005 views - Mechanical Engineering)
The wheel and axle is one of six simple machines identified by Renaissance scientists drawing from Greek texts on technology.[1] The wheel and axle consists of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other. A hinge or bearing supports the axle, allowing rotation. It can amplify force; a small force applied to the periphery of the large wheel can move a larger load attached to the axle.
The wheel and axle can be viewed as a version of the lever, with a drive force applied tangentially to the perimeter of the wheel and a load force applied to the axle, respectively, that are balanced around the hinge which is the fulcrum. The mechanical advantage of the wheel and axle is the ratio of the distances from the fulcrum to the applied loads, or what is the same thing the ratio of the diameter of the wheel and axle.[2] A major application is in wheeled vehicles, in which the wheel and axle are used to reduce friction of the moving vehicle with the ground. Other examples of devices which use the wheel and axle are capstans, belt drives and gears.
Further information: Wheel
The Halaf culture of 6500–5100 BCE has been credited with the earliest depiction of a wheeled vehicle.[3]
Precursors of wheels, known as "tournettes" or "slow wheels", were known in the Middle East by the 5th millennium BCE (one of the earliest examples was discovered at Tepe Pardis, Iran, and dated to 5200–4700 BCE). These were made of stone or clay and secured to the ground with a peg in the center, but required effort to turn. True (freely-spinning) wheels were apparently in use in Mesopotamia by 3500 BCE and possibly as early as 4000 BCE,[4] and the oldest surviving example, which was found in Ur (modern day Iraq), dates to approximately 3100 BCE.
Evidence of wheeled vehicles appears in the second half of the 4th millennium BCE, near-simultaneously in Mesopotamia (Sumerian civilization), the Northern Caucasus (Maykop culture) and Central Europe (Cucuteni-Trypillian culture).
An early well-dated depiction of a wheeled vehicle (a wagon—four wheels, two axles) is on the Bronocice pot, a ca. 3635–3370 BCE ceramic vase, excavated in a Funnelbeaker culture settlement in southern Poland.[5]
The oldest known example of a wooden wheel and its axle was found in 2002 at the Ljubljana Marshes some 20 km south of Ljubljana, the capital of Slovenia. According to radiocarbon dating, it is between 5,100 and 5,350 years old. The wheel was made of ash and oak and had a radius of 70 cm and the axle is 120 cm long and made of oak.[6]
In Roman Egypt, Hero of Alexandria identified the wheel and axle as one of the simple machines used to lift weights.[7] This is thought to have been in the form of the windlass which consists of a crank or pulley connected to a cylindrical barrel that provides mechanical advantage to wind up a rope and lift a load such as a bucket from the well.[8]
The simple machine called a wheel and axle refers to the assembly formed by two disks, or cylinders, of different diameters mounted so they rotate together around the same axis.The thin rod which needs to be turned is called the axle and the wider object fixed to the axle, on which we apply force is called the wheel. A tangential force applied to the periphery of the large disk can exert a larger force on a load attached to the axle, achieving mechanical advantage. When used as the wheel of a wheeled vehicle the smaller cylinder is the axle of the wheel, but when used in a windlass, winch, and other similar applications (see medieval mining lift to right) the smaller cylinder may be separate from the axle mounted in the bearings. It cannot be used separately.[9][10]
Assuming the wheel and axle does not dissipate or store energy, that is it has no friction or elasticity, the power input by the force applied to the wheel must equal the power output at the axle. As the wheel and axle system rotates around its bearings, points on the circumference, or edge, of the wheel move faster than points on the circumference, or edge, of the axle. Therefore, a force applied to the edge of the wheel must be less than the force applied to the edge of the axle, because power is the product of force and velocity.[11]
Let a and b be the distances from the center of the bearing to the edges of the wheel A and the axle B. If the input force FA is applied to the edge of the wheel A and the force FB at the edge of the axle B is the output, then the ratio of the velocities of points A and B is given by a/b, so the ratio of the output force to the input force, or mechanical advantage, is given by
{\displaystyle MA={\frac {F_{B}}{F_{A}}}={\frac {a}{b}}.}
The mechanical advantage of a simple machine like the wheel and axle is computed as the ratio of the resistance to the effort. The larger the ratio the greater the multiplication of force (torque) created or distance achieved. By varying the radii of the axle and/or wheel, any amount of mechanical advantage may be gained.[2] In this manner, the size of the wheel may be increased to an inconvenient extent. In this case a system or combination of wheels (often toothed, that is, gears) are used. As a wheel and axle is a type of lever, a system of wheels and axles is like a compound lever.[12]
The mechanical advantage of a wheel and axle with no friction is called the ideal mechanical advantage (IMA). It is calculated with the following formula:
{\displaystyle \mathrm {IMA} ={F_{\text{out}} \over F_{\text{in}}}={\mathrm {Radius} _{\text{wheel}} \over \mathrm {Radius} _{\text{axle}}}}
All actual wheels have friction, which dissipates some of the power as heat. The actual mechanical advantage (AMA) of a wheel and axle is calculated with the following formula:
{\displaystyle \mathrm {AMA} ={F_{\text{out}} \over F_{\text{in}}}=\nu \cdot {\mathrm {Radius} _{\text{wheel}} \over \mathrm {Radius} _{\text{axle}}}}
{\displaystyle \nu ={P_{\text{out}} \over P_{\text{in}}}}
is the efficiency of the wheel, the ratio of power output to power input
^ Wheel and Axle, The World Book Encyclopedia, World Book Inc., 1998, pp. 280-281
^ a b Bowser, Edward Albert, 1890, An elementary treatise on analytic mechanics: with numerous examples. (Originally from the University of Michigan) D. Van Nostrand Company, pp. 190
^ D. T. Potts (2012). A Companion to the Archaeology of the Ancient Near East. p. 285.
^ Anthony, David A. (2007). The horse, the wheel, and language: how Bronze-Age riders from the Eurasian steppes shaped the modern world. Princeton, N.J: Princeton University Press. p. 67. ISBN 0-691-05887-3.
^ Aleksander Gasser (March 2003). "World's Oldest Wheel Found in Slovenia". Government Communication Office of the Republic of Slovenia. Retrieved 19 August 2010.
^ Usher, Abbott Payson (1988). A History of Mechanical Inventions. USA: Courier Dover Publications. p. 98. ISBN 048625593X.
^ Elroy McKendree Avery, Elementary Physics, New York : Sheldon & Company, 1878.
^ Prater, Edward L. (1994), Basic Machines, Naval Education and Training Professional Development and Technology Center, NAVEDTRA 14037
^ Bureau of Naval Personnel, 1971, Basic Machines and How They Work, Dover Publications.
^ Baker, C.E. The Principles and Practice of Statics and Dynamics … for the Use of Schools and Private Students. London: John Weale, 59, High Holborn. 1851 pp. 26-29 read online or download full text
Basic Machines and How They Work, United States. Bureau of Naval Personnel, Courier Dover Publications 1965, pp. 3–1 and following preview online
Mechanical engineeringAlloy wheelLjubljana Marshes WheelWire wheelRotation around a fixed axisLocking differentialDriving wheelPropulsionAxle
This article uses material from the Wikipedia article "Wheel and axle", 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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Use the Table of Integrals to evaluate the integral. \int \frac{\tan^{3}(1/z)}{z^{2}}dz
\int \frac{{\mathrm{tan}}^{3}\left(\frac{1}{z}\right)}{{z}^{2}}dz
The value of the given integral
\int \frac{{\mathrm{tan}}^{3}\left(\frac{1}{z}\right)}{{z}^{2}}dz
The formula for the cube of the tangent function is given by
\int {\mathrm{tan}}^{3}\left(x\right)dx=\frac{1}{2}{\mathrm{tan}}^{2}x+\mathrm{ln}|\mathrm{cos}\left(x\right)|+C
\int \frac{{\mathrm{tan}}^{3}\left(\frac{1}{z}\right)}{{z}^{2}}dz
By using the substitution,
u=\frac{1}{z}
Differentiate both sides of the above equation with respect to z,
\frac{du}{dz}=\frac{-1}{{z}^{2}}
Multiply both sides by dz,
du=\frac{-1}{{z}^{2}}dz
⇒dz=-{z}^{2}du
Substitute the value of dz in the given integral
\int \frac{{\mathrm{tan}}^{3}\left(u\right)}{{z}^{2}}\left(-{z}^{2}\right)du=\int -{\mathrm{tan}}^{3}\left(u\right)du
Taking a negative sign outside the integral,
\int -{\mathrm{tan}}^{3}\left(u\right)du=-\int {\mathrm{tan}}^{3}\left(u\right)du
By using the formula for the cube of the tangent function,
-\int {\mathrm{tan}}^{3}\left(u\right)du=-\left[\frac{1}{2}{\mathrm{tan}}^{2}u+\mathrm{ln}|\mathrm{cos}\left(u\right)|\right]
Distributing the negative sign,
-\int {\mathrm{tan}}^{3}\left(u\right)du=-\frac{1}{2}{\mathrm{tan}}^{2}u-\mathrm{ln}|\mathrm{cos}\left(u\right)|
Resubstitute u = 1/z,
\int \frac{{\mathrm{tan}}^{3}\left(\frac{1}{z}\right)}{{z}^{2}}dz=-\frac{1}{2}{\mathrm{tan}}^{2}\left(\frac{1}{z}\right)-\mathrm{ln}|\mathrm{cos}\left(\frac{1}{z}\right)|+C
\int \left(\frac{{\left(\mathrm{tan}\left(\frac{1}{z}\right)\right)}^{3}}{{z}^{2}}dz
Let us put the expression
-\frac{1}{{z}^{2}}
\left(-\frac{1}{{z}^{2}}\right)dz=d\left(\frac{1}{z}\right),t=\frac{1}{z}
Then the initial integral can be written as follows:
\int \left(-{\mathrm{tan}\left(t\right)}^{3}\right)\cdot dt
\int -{\mathrm{tan}\left(z\right)}^{3}dz
\mathrm{tan}\left(z\right)=
dt=\frac{1}{1+{t}^{2}}
\int -\frac{{t}^{3}}{{t}^{2}+1}\cdot dt
\int \frac{-{z}^{3}}{{z}^{2}+1}dz
Degree the numerator P (x) is greater than or equal to the degree of the denominator Q (x), so we divide the polynomials.
\frac{-{z}^{3}}{{z}^{2}+1}=-z+\frac{z}{{z}^{2}+1}
Integrating the integer part, we get:
\int \left(-z\right)dz=-\frac{{z}^{2}}{2}
\int \frac{z}{{z}^{2}+1}dz=\frac{\mathrm{ln}\left({z}^{2}+1\right)}{2}
-\frac{{z}^{2}}{2}+\frac{\mathrm{ln}\left({z}^{2}+1\right)}{2}+C
-\frac{{z}^{2}}{2}+\mathrm{ln}\left(\sqrt{{z}^{2}+1}\right)+C
\left(t=\mathrm{tan}\left(z\right)\right)
I=\frac{\mathrm{ln}\left({\mathrm{tan}\left(x\right)}^{2}+1\right)}{2}-\frac{{\mathrm{tan}\left(z\right)}^{2}}{2}+C
To write down the final answer, it remains to substitute 1 / z instead of t.
\frac{\mathrm{ln}\left({\mathrm{tan}\left(\frac{1}{z}\right)}^{2}+1\right)}{2}-\frac{{\mathrm{tan}\left(\frac{1}{z}\right)}^{2}}{2}+C
\begin{array}{}\int \frac{\mathrm{tan}\left(\frac{1}{z}{\right)}^{3}}{{z}^{2}}dz\\ \int -\mathrm{tan}\left(t{\right)}^{3}dt\\ -\int \mathrm{tan}\left(t{\right)}^{3}dt\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}-\int \mathrm{tan}\left(t\right)dt\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}-\int \frac{\mathrm{sin}t}{\mathrm{cos}t}dt\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}-\int -\frac{1}{u}du\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}+\int \frac{1}{u}du\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}+\mathrm{ln}\left(|u|\right)\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}+\mathrm{ln}\left(|\mathrm{cos}\left(t\right)|\right)\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(\frac{1}{z}{\right)}^{2}+\mathrm{ln}\left(|\mathrm{cos}\left(\frac{1}{z}\right)|\right)\right)\\ -\frac{\mathrm{tan}\left(\frac{1}{z}{\right)}^{2}}{2}-\mathrm{ln}\left(|\mathrm{cos}\left(\frac{1}{z}\right)|\right)\\ -\frac{\mathrm{tan}\left(\frac{1}{z}{\right)}^{2}}{2}-\mathrm{ln}\left(|\mathrm{cos}\left(\frac{1}{z}\right)|\right)+C\end{array}
\int \left(x+7\right)dx
\int \sqrt{5{x}^{2}-1}dx
\int \frac{1}{\mathrm{cos}\left(x\right)}dx
u=\mathrm{tan}\left(\frac{x}{2}\right)
\frac{dx}{dt}=t\left(x-2\right)
How do you solve x''(t)+x3=0?
\int {\left({x}^{6}-7x\right)}^{4}dx
Trigonometric integrals Evaluate the following integrals.
\int {\mathrm{tan}}^{2}xdx
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Determine the following indefinite integral. \int \frac{x^{5}-3}{x}dx
\int \frac{{x}^{5}-3}{x}dx
\int \frac{{x}^{5}-3}{x}dx
\frac{{x}^{5}-3}{x}\text{ }as\text{ }{x}^{4}-\frac{3}{x}
\int \frac{{x}^{5}-3}{x}dx=\int \left({x}^{4}-\frac{3}{x}\right)dx
=\int {x}^{4}dx-\int \frac{3}{x}dx
=\frac{{x}^{5}}{5}-3\mathrm{ln}\left(|x|\right)+C
\int \frac{{x}^{5}-3}{x}dx=\frac{{x}^{5}}{5}-3\mathrm{ln}\left(|x|\right)+C
\frac{{x}^{5}-3}{x}\to {x}^{4}-\frac{3}{x}
\int {x}^{4}-\frac{3}{x}dx
\int f\left(x\right)+g\left(x\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx
\int {x}^{4}dx-\int \frac{3}{x}dx
\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C
\frac{{x}^{5}}{5}-\int \frac{3}{x}dx
\int cf\left(x\right)dx=f\int f\left(x\right)dx
\frac{{x}^{5}}{5}-3\int \frac{1}{x}dx
Step 5: The derivative of
\mathrm{ln}x\text{ }is\text{ }\frac{1}{x}
\frac{{x}^{5}}{5}-3\mathrm{ln}x
\frac{{x}^{5}}{5}-3\mathrm{ln}x+C
How to compute the integral of
\int \mathrm{sin}\left(1-{x}^{2}\right)dx
I tried substitution,
u=1-{x}^{2},\text{ }du=-2xdx,\text{ }dx=-\frac{du}{2x}
, yielding:
\int \mathrm{sin}u\left(-\frac{du}{2}\right)
=-\frac{1}{2}\int \mathrm{sin}\left(u\right)du
=-\frac{1}{2}\mathrm{cos}u
=-\frac{1}{2}\mathrm{cos}\left(1-{x}^{2}\right)
WolframAlpha is showing something completely different, however. What's wrong with my solution?
Write the integral in terms of u and du. Then evaluate:
\int {\left(x+25\right)}^{-2}dx,u=x+25
\int {e}^{x}+\frac{{e}^{3x}}{{e}^{2x}}dx
How Can you factor 1/2 out of integral of
\mathrm{cos}2x
\int \mathrm{cos}2xdx
\frac{1}{2}\int \mathrm{cos}udu
where u=2x
My Question is where does the
\frac{1}{2}
that is factored out come from?
It would make more sense for a factor of 1/2 to be extracted if the integral was:
\int \frac{\mathrm{cos}x}{2}dx
\int \sqrt{x}\sqrt{1-x}dx
\frac{1}{\tau }{\int }_{0}^{\tau }A\mathrm{sin}\left(\mathrm{\Omega }t\right)\cdot A\mathrm{sin}\left(\mathrm{\Omega }\left(t-\lambda \right)\right)dt
\frac{{A}^{2}}{\tau }{\int }_{0}^{\tau }\frac{{e}^{j\mathrm{\Omega }t}-{e}^{-j\mathrm{\Omega }t}}{2j}\cdot \frac{{e}^{j\mathrm{\Omega }\left(t-\lambda \right)}-{e}^{-j\mathrm{\Omega }\left(t-\lambda \right)}}{2j}dt
\frac{1}{2j}
\frac{-{A}^{2}}{4\tau }{\int }_{0}^{\tau }\left({e}^{j\mathrm{\Omega }t}-{e}^{-j\mathrm{\Omega }t}\right)\cdot \left({e}^{j\mathrm{\Omega }\left(t-\lambda \right)}-{e}^{-j\mathrm{\Omega }\left(t-\lambda \right)}\right)dt
\frac{1}{2j}\to \frac{-1}{4}
{\int }_{3}^{0}\left(x+2\right)dx
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Gammatone filter bank - MATLAB - MathWorks í•œêµ
Specific to gammatoneFilterBank
Apply Gammatone Filter Bank
Create Gammatone Spectrogram
gammatoneFilterBank decomposes a signal by passing it through a bank of gammatone filters equally spaced on the ERB scale. Gammatone filter banks were designed to model the human auditory system.
To model the human auditory system:
Create the gammatoneFilterBank object and set its properties.
gammaFiltBank = gammatoneFilterBank
gammaFiltBank = gammatoneFilterBank(range)
gammaFiltBank = gammatoneFilterBank(range,numFilts)
gammaFiltBank = gammatoneFilterBank(range,numFilts,fs)
gammaFiltBank = gammatoneFilterBank(___,Name,Value)
gammaFiltBank = gammatoneFilterBank returns a gammatone filter bank. The object filters data independently across each input channel over time.
gammaFiltBank = gammatoneFilterBank(range) sets the Range property to range.
gammaFiltBank = gammatoneFilterBank(range,numFilts) sets the NumFilters property to numFilts.
gammaFiltBank = gammatoneFilterBank(range,numFilts,fs) sets the SampleRate property to fs.
gammaFiltBank = gammatoneFilterBank(___,Name,Value) sets each property Name to the specified Value. Unspecified properties have default values.
Example: gammaFiltBank = gammatoneFilterBank([62.5,12e3],'SampleRate',24e3) creates a gammatone filter bank, gammaFiltBank, with bandpass filters placed between 62.5 Hz and 12 kHz. gammaFiltBank operates at a sample rate of 24 kHz.
Frequency range of the filter bank in Hz, specified as a two-element row vector of monotonically increasing values.
Number of filters in the filter bank, specified as a positive integer scalar.
audioOut = gammaFiltBank(audioIn)
audioOut = gammaFiltBank(audioIn) applies the gammatone filter bank on the input and returns the filtered output.
audioIn — Audio input to filter bank
Audio input to the filter bank, specified as a scalar, vector, or matrix. If specified as a matrix, the columns are treated as independent audio channels.
audioOut — Audio output from filter bank
Audio output from the filter bank, returned as a scalar, vector, matrix, or 3-D array. The shape of audioOut depends on the shape of audioIn and NumFilters. If audioIn is an M-by-N matrix, then audioOut is returned as an M-by-NumFilters-by-N array. If N is 1, then audioOut is returned as a matrix.
getBandwidths Get filter bandwidths
Create a default gammatone filter bank for a 16 kHz sample rate.
gammaFiltBank = gammatoneFilterBank('SampleRate',fs)
gammaFiltBank =
gammatoneFilterBank with properties:
FrequencyRange: [50 8000]
Use fvtool to visualize the response of the filter bank.
fvtool(gammaFiltBank)
y = gammaFiltBank(x);
This example illustrates a nonoptimal but simple approach to analysis and synthesis using gammatoneFilterBank.
Create a default gammatoneFilterBank. The default frequency range of the filter bank is 50 to 8000 Hz. Frequencies outside of this range are attenuated in the reconstructed signal.
Pass the audio signal through the gammatone filter bank. The output is 32 channels, where the number of channels is set by the NumFilters property of the gammatoneFilterBank.
[N,numChannels] = size(audioOut)
To reconstruct the original signal, sum the channels. Listen to the result.
reconstructedSignal = sum(audioOut,2);
The gammatone filter bank introduced various group delays for the output channels, which results in poor reconstruction. To compensate for the group delay, remove the beginning delay from the individual channels and zero-pad the ends of the channels. Use info to get the group delays. Listen to the group delay-compensated reconstruction.
infoStruct = info(gammaFiltBank);
groupDelay = round(infoStruct.GroupDelays); % round for simplicity
audioPadded = [audioOut;zeros(max(groupDelay),gammaFiltBank.NumFilters)];
for i = 1:gammaFiltBank.NumFilters
audioOut(:,i) = audioPadded(groupDelay(i)+1:N+groupDelay(i),i);
Read in an audio signal and convert it to mono for easy visualization.
audio = mean(audio,2);
Create a gammatoneFilterBank with 64 filters that span the range 62.5 to 20,000 Hz. Pass the audio signal through the filter bank.
gammaFiltBank = gammatoneFilterBank('SampleRate',fs, ...
'NumFilters',64, ...
'FrequencyRange',[62.5,20e3]);
audioOut = gammaFiltBank(audio);
Calculate the energy-per-band using 50 ms windows with 25 ms overlap. Use dsp.AsyncBuffer to divide the signals into overlapped windows and then to log the RMS value of each window for each channel.
samplesPerFrame = round(0.05*fs);
samplesOverlap = round(0.025*fs);
buff = dsp.AsyncBuffer(numel(audio));
write(buff,audioOut.^2);
sink = dsp.AsyncBuffer(numel(audio));
while buff.NumUnreadSamples > 0
currentFrame = read(buff,samplesPerFrame,samplesOverlap);
write(sink,mean(currentFrame,1));
Convert the energy values to dB. Plot the energy-per-band over time.
gammatoneSpec = read(sink);
D = 20*log10(gammatoneSpec');
timeVector = ((samplesPerFrame-samplesOverlap)/fs)*(0:size(D,2)-1);
cf = getCenterFrequencies(gammaFiltBank)./1e3;
surf(timeVector,cf,D,'EdgeColor','none')
axis([timeVector(1) timeVector(end) cf(1) cf(end)])
caxis([-150,-60])
A gammatone filter bank is often used as the front end of a cochlea simulation, which transforms complex sounds into a multichannel activity pattern like that observed in the auditory nerve.[2] The gammatoneFilterBank follows the algorithm described in [1]. The algorithm is an implementation of an idea proposed in [2]. The design of the gammatone filter bank can be described in two parts: the filter shape (gammatone) and the frequency scale. The equivalent rectangular bandwidth (ERB) scale defines the relative spacing and bandwidth of the gammatone filters. The derivation of the ERB scale also provides an estimate of the auditory filter response which closely resembles the gammatone filter.
The ERB scale was determined using the notched-noise masking method. This method involves a listening test wherein notched noise is centered on a tone. The power of the tone is tuned, and the audible threshold (the power required for the tone to be heard) is recorded. The experiment is repeated for different notch widths and center frequencies.
The notched-noise method assumes the audible threshold corresponds to a constant signal-to-masker ratio at the output of the theoretical auditory filter. That is, the ratio of the power of the fc tone and the shaded area is constant. Therefore, the relationship between the audible threshold and 2Δf (the notch bandwidth) is linearly related to the relationship between the noise passed through the filter and 2Δf.
The derivative of the function relating Δf to the noise passed through the filter estimates the auditory filter shape. Because Δf has an inverse relationship with the noise power passed through the filter, the derivative of the function must be multiplied by –1. The resulting auditory filter shape is usually approximated as a roex filter.
\text{ERB}=24.7\left(0.00437{f}_{c}+1\right)
The ERB scale (ERBs) is an extension of the relationship between ERB and center frequency, derived by integrating the reciprocal of the ERB function:
\text{ERBs}=21.4{\mathrm{log}}_{10}\left(0.00437f+1\right)
To design a gammatone filter bank, [2] suggests distributing the center frequencies of the filters in proportion to their bandwidth. To accomplish this, gammatoneFilterBank defines the center frequencies as linearly spaced on the ERB scale, covering the specified frequency range with the desired number of filters. You can specify the frequency range and desired number of filters using the FrequencyRange and NumFilters properties.
g\left(t\right)=a{t}^{nâ1}{e}^{â2\mathrm{Ï}bt}\mathrm{cos}\left(2\mathrm{Ï}{f}_{\text{c}}t+\mathrm{Ï}\right)
The gammatone filter is similar to the roex filter derived from the notched-noise experiment. gammatoneFilterBank implements the digital filter as a cascade of four second-order sections, as described in [1].
[2] Patterson, R. D., K. Robinson, J. Holdsworth, D. McKeown, C. Zhang, and M. Allerhand. "Complex Sounds and Auditory Images." Auditory Physiology and Perception. 1992, pp. 429–446.
[4] Glasberg, Brian R., and Brian C. J. Moore. "Derivation of Auditory Filter Shapes from Notched-Noise Data." Hearing Research. Vol. 47. Issue 1-2, 1990, pp. 103 –138.
octaveFilterBank | crossoverFilter
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Linear Equations: Level 1 Challenges Practice Problems Online | Brilliant
If $80 is divided among three people so that the second person will have twice as much as the first, and the third person will have $5 less than the second, then how many dollars will the third person have?
by Samarth Badyal
I went to the candy store last weekend and I bought a total of 10 packs of candy. I bought M&Ms for $1 per pack and Twizzlers for $0.50 per pack. I spent a total of $6. How many packs of M&Ms did I buy?
Consider the collection of points such that the sum of each point's coordinates is 1. The graph of these points in the coordinate plane forms what shape?
A line A hyperbola A parabola A circle An ellipse Another different function
\$110
\$100
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Based on a poll, 40% of adults believe in reincarnation.
Based on a poll, 40% of adults believe in reincarnation. Assume that 6 adults ar
Based on a poll, 40% of adults believe in reincarnation. Assume that 6 adults are randomly selected, and find the indicated probability. What is the probability that at least 5 of the selected adults believe in reincarnation? The probability that at least 5 of the selected adults believe in reincarnation is |_| (Round to three decimal places as needed.)
Jenny sells ‘honey tea’ to raise fund for homeless people. The probability of making a sale is, independently, 0.3 for each person she approaches
Based on Jenny’s experience selling ‘honey tea’ for the last 2 weeks, the probability that the person she approaches will buy 0, 1, 2 or more than 2 cups are 0.7, 0.15, 0.10 and 0.05 respectively. Jenny plans to approached 20 persons. What is the probability that 10 will not buy, 3 will buy 1 cup, 4 will buy 2 cups and the rest will buy than 2 cups?
n\left(A\cap B\right)
n\left(A\right)=7,n\left(B\right)=9,
n\left(A\cup B\right)=13
The probability that a person selected at random from a populetion will exhibit the classic symptom of a certain disease is 0.2, and the probability that aperson selected at random has the disease is 0.23. The probability that a person who has the symptom also has the disease is 0.18. A person selected at random from the population does not have the symptom. What is the probability that the person has the disease?
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diffalg(deprecated)/reduced - Maple Help
Home : Support : Online Help : diffalg(deprecated)/reduced
test if a differential polynomial is reduced with respect to a set of differential polynomials
reduced (p, F, R, code)
reduced (p, P, code)
differential polynomial or list/set of differential polynomials in R
(optional) name; 'fully' or 'partially'
The function reduced returns true if p is reduced with respect to F or with respect to the equations of P. It returns false otherwise.
A differential polynomial p is said to be partially reduced with respect to a polynomial q if no proper derivative of the leader of q appears in p.
A differential polynomial p is said to be fully reduced with respect to a polynomial q if it is partially reduced with respect to q and if its degree in the leader of q is less than the degree of q in this leader.
A differential polynomial p is said to be reduced with respect to a set of differential polynomials F if it is reduced with respect to each element of F.
If code is omitted, it is assumed to be 'fully'.
If the second form of the function is used and P is a radical differential ideal defined by a list of characterizable differential ideals then the function is mapped over all the components of the ideal.
The command with(diffalg,reduced) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{diffalg}\right):
R≔\mathrm{differential_ring}\left(\mathrm{derivations}=[x,y],\mathrm{ranking}=[u]\right)
\textcolor[rgb]{0,0,1}{R}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{PDE_ring}}
q≔{u[x]}^{2}-4u[]
\textcolor[rgb]{0,0,1}{q}\textcolor[rgb]{0,0,1}{≔}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}[]
\mathrm{reduced}\left({u[x]}^{3},q,R\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{reduced}\left({u[x]}^{3},q,R,'\mathrm{partially}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
P≔\mathrm{Rosenfeld_Groebner}\left([q],R\right)
\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{characterizable}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{characterizable}}]
\mathrm{equations}\left(P\right)
[[{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}[]]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{u}[]]]
\mathrm{reduced}\left(u[x],P\right)
[\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{false}}]
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Shadow zone - Wikipedia
A seismic shadow zone is an area of the Earth's surface where seismographs cannot detect direct P waves and/or S waves from an earthquake. This is due to liquid layers or structures within the Earth's surface. The most recognized shadow zone is due to the core-mantle boundary where P waves are refracted and S waves are stopped at the liquid outer core; however, any liquid boundary or body can create a shadow zone. For example, magma reservoirs with a high enough percent melt can create seismic shadow zones.
Seismic shadow zone (from USGS)
2 Seismic wave properties
3 Other observations and implications
The earth is made up of different structures: the crust, the mantle, the inner core and the outer core. The crust, mantle, and inner core are typically solid; however, the outer core is entirely liquid.[1] A liquid outer core was first shown in 1906 by Geologist Richard Oldham.[2] Oldham observed seismograms from various earthquakes and saw that some seismic stations did not record direct S waves, particularly ones that were 120° away from the hypocenter of the earthquake.[3]
In 1913, Beno Gutenberg noticed the abrupt change in seismic velocities of the P waves and disappearance of S waves at the core-mantle boundary. Gutenberg attributed this due to a solid mantle and liquid outer core, calling it the Gutenberg discontinuity.[4]
Seismic wave propertiesEdit
The main observational constraint on identifying liquid layers and/or structures within the earth come from seismology. When an earthquake occurs, seismic waves radiate out spherically from the earthquake's hypocenter.[5] Two types of body waves travel through the Earth: primary seismic waves (P waves) and secondary seismic waves (S waves). P waves travel with motion in the same direction as the wave propagates and S-waves travel with motion perpendicular to the wave propagation (transverse).[6]
The P waves are refracted by the liquid outer core of the Earth and are not detected between 104° and 140° (between approximately 11,570 and 15,570 km or 7,190 and 9,670 mi) from the hypocenter.[7][8] This is due to Snell's law, where a seismic wave encounters a boundary and either refracts or reflects. In this case, the P waves refract due to density differences and greatly reduce in velocity.[7][9] This is considered the P wave shadow zone.[10]
The S waves cannot pass through the liquid outer core and are not detected more than 104° (approximately 11,570 km or 7,190 mi) from the epicenter.[7][11][12] This is considered the S wave shadow zone.[10] However, P waves that travel refract through the outer core and refract to another P wave (PKP wave) on leaving the outer core can be detected within the shadow zone. Additionally, S waves that refract to P waves on entering the outer core and then refract to an S wave on leaving the outer core can also be detected in the shadow zone (SKS waves).[7][13]
The reason for this is P wave and S wave velocities are governed by different properties in the material which they travel through and the different mathematical relationships they share in each case. The three properties are: The three properties are: incompressibility (
{\displaystyle k}
), density (
{\displaystyle p}
) and rigidity (
{\displaystyle u}
P wave velocity is equal to:
{\displaystyle {\sqrt {(k+{\tfrac {4}{3}}u)/p}}}
S wave velocity is equal to:
{\displaystyle {\sqrt {u/p}}}
S wave velocity is entirely dependent on the rigidity of the material it travels through. Liquids have zero rigidity, making the S-wave velocity when traveling through a liquid. Overall, S waves are shear waves, and shear stress is a type of deformation that cannot occur in a liquid.[11][12][14] Conversely, P waves are compressional waves and are only partially dependent on rigidity. P waves still maintain some velocity (can be greatly reduced) when traveling through a liquid.[7][8][14][15]
Other observations and implicationsEdit
Although the core-mantle boundary casts the largest shadow zone, smaller structures, such as magma bodies, can also cast a shadow zone. For example, in 1981, Páll Einarsson conducted a seismic investigation on the Krafla Caldera in Northeast Iceland.[16] In this study, Einarsson placed a dense array of seismometers over the caldera and recorded earthquakes that occurred. The resulting seismograms showed both an absence of S waves and/or small S wave amplitudes. Einarsson attributed these results to be the cause of the magma reservoir. In this case, the magma reservoir has enough percent melt to cause S waves to be directly affected.[16] In areas where there are no S waves being recorded, the S waves are encountering enough liquid, that no solid grains are touching.[17] In areas where there are highly attenuated (small aptitude) S waves, there is still a precent of melt, but enough solid grains are touching where S waves can travel through the part of the magma reservoir.[12][15][18]
Between 2014 and 2018, a geophysicist in Taiwan, Cheng-Horng Lin investigated the magma reservoir beneath the Tatun Volcano Group in Taiwan.[19][20] Lin and their research group used deep earthquakes and seismometers on or near the Tatun Volcano Group to identify changes P and S waveforms. Their results showed P wave delays and the absence of S waves in various locations. Lin attributed this finding to be due to a magma reservoir with at least 40% melt that casts an S wave shadow zone.[19][20] However, a recent study done by National Chung Cheng University used a dense array of seismometers and only saw S wave attenuation asscociated with the magma reservoir.[21] This research study investigated the cause of the S wave shadow zone Lin observed and attributed it to either a magma diapir above the subducting Philippine Sea Plate. Though it was not a magma reservoir, there was still a structure with enough melt/liquid to cause an S wave shadow zone.[21]
The existence of shadow zones, more specifically S wave shadow zones, could have implications on the eruptibility of volcanoes throughout the world. When volcanoes have enough percent melt to go below the rheological lockup (percent crystal fraction when a volcano is eruptive or not eruptive), this makes the volcanoes eruptible.[22][23] Determining the percent melt of a volcano could help with predictive modeling and assess current and future hazards. In an actively erupting volcano, Mt. Etna in Italy, a study was done in 2021 that showed both an absence of S-waves in some regions and highly attenuated S-waves in others, depending on where the receivers are located above the magma chamber.[24] Previously, in 2014, a study was done to model the mechanism leading to the December 28th, 2014 eruption. This study showed that an eruption could be triggered between 30-70% melt.[25]
^ Encyclopedia of solid earth geophysics. Harsh K. Gupta. Dordrecht: Springer. 2011. ISBN 978-90-481-8702-7. OCLC 745002805. {{cite book}}: CS1 maint: others (link)
^ Bragg, William (1936-12-18). "Tribute to Deceased Fellows of the Royal Society". Science. 84 (2190): 539–546. doi:10.1126/science.84.2190.539. ISSN 0036-8075.
^ Brush, Stephen G. (September 1980). "Discovery of the Earth's core". American Journal of Physics. 48 (9): 705–724. doi:10.1119/1.12026. ISSN 0002-9505.
^ Michael Allaby (2008). A dictionary of earth sciences (3rd ed.). Oxford. ISBN 978-0-19-921194-4. OCLC 177509121.
^ "Earthquake Glossary". earthquake.usgs.gov. Retrieved 2021-12-10.
^ Fowler, C. M. R. (2005). The solid earth: an introduction to global geophysics (2nd ed.). Cambridge, UK: Cambridge University Press. ISBN 0-521-89307-0. OCLC 53325178.
^ a b c d e "CHAPTER 19 NOTES Earth's (Interior)". uh.edu. Retrieved 2021-12-10.
^ a b "Earthquake Glossary". earthquake.usgs.gov. Retrieved 2021-12-10.
^ "Snell's Law -- The Law of Refraction". personal.math.ubc.ca. Retrieved 2021-12-10.
^ a b "Seismic Shadow Zone: Basic Introduction- Incorporated Research Institutions for Seismology". www.iris.edu. Retrieved 2021-12-10.
^ a b c "Why can't S-waves travel through liquids?". Earth Observatory of Singapore. Retrieved 2021-12-10.
^ a b c Greenwood, Margaret Stautberg; Bamberger, Judith Ann (August 2002). "Measurement of viscosity and shear wave velocity of a liquid or slurry for on-line process control". Ultrasonics. 39 (9): 623–630. doi:10.1016/S0041-624X(02)00372-4.
^ Kennett, Brian (2007), Gubbins, David; Herrero-Bervera, Emilio (eds.), "Seismic Phases", Encyclopedia of Geomagnetism and Paleomagnetism, Dordrecht: Springer Netherlands, pp. 903–908, doi:10.1007/978-1-4020-4423-6_290, ISBN 978-1-4020-4423-6, retrieved 2021-12-10
^ a b c Dziewonski, Adam M.; Anderson, Don L. (June 1981). "Preliminary reference Earth model". Physics of the Earth and Planetary Interiors. 25 (4): 297–356. doi:10.1016/0031-9201(81)90046-7.
^ a b Båth, Markus (1957). "Shadow zones, travel times, and energies of longitudinal seismic waves in the presence of an asthenosphere low-velocity layer". Eos, Transactions American Geophysical Union. 38 (4): 529–538. doi:10.1029/TR038i004p00529. ISSN 2324-9250.
^ a b Einarsson, P. (September 1978). "S-wave shadows in the Krafla Caldera in NE-Iceland, evidence for a magma chamber in the crust". Bulletin Volcanologique. 41 (3): 187–195. doi:10.1007/bf02597222. ISSN 0258-8900.
^ Asimow, Paul D. (2016), White, William M. (ed.), "Partial Melting", Encyclopedia of Geochemistry: A Comprehensive Reference Source on the Chemistry of the Earth, Cham: Springer International Publishing, pp. 1–6, doi:10.1007/978-3-319-39193-9_218-1, ISBN 978-3-319-39193-9, retrieved 2021-12-10
^ Sheriff, R. E. (1975). "Factors Affecting Seismic Amplitudes*". Geophysical Prospecting. 23 (1): 125–138. doi:10.1111/j.1365-2478.1975.tb00685.x. ISSN 1365-2478.
^ a b Lin, Cheng-Horng (2016-12-23). "Evidence for a magma reservoir beneath the Taipei metropolis of Taiwan from both S-wave shadows and P-wave delays". Scientific Reports. 6 (1): 39500. doi:10.1038/srep39500. ISSN 2045-2322.
^ a b Lin, Cheng-Horng; Lai, Ya-Chuan; Shih, Min-Hung; Pu, Hsin-Chieh; Lee, Shiann-Jong (2018-11-06). "Seismic Detection of a Magma Reservoir beneath Turtle Island of Taiwan by S-Wave Shadows and Reflections". Scientific Reports. 8 (1): 16401. doi:10.1038/s41598-018-34596-0. ISSN 2045-2322.
^ a b Yeh, Yu-Lien; Wang, Wei-Hau; Wen, Strong (2021-01-13). "Dense seismic arrays deny a massive magma chamber beneath the Taipei metropolis, Taiwan". Scientific Reports. 11 (1). doi:10.1038/s41598-020-80051-4. ISSN 2045-2322. PMC 7806728.
^ Cooper, Kari M.; Kent, Adam J. R. (2014-02-16). "Rapid remobilization of magmatic crystals kept in cold storage". Nature. 506 (7489): 480–483. doi:10.1038/nature12991. ISSN 0028-0836.
^ Marsh, B. D. (October 1981). "On the crystallinity, probability of occurrence, and rheology of lava and magma". Contributions to Mineralogy and Petrology. 78 (1): 85–98. doi:10.1007/bf00371146. ISSN 0010-7999.
^ De Gori, Pasquale; Giampiccolo, Elisabetta; Cocina, Ornella; Branca, Stefano; Doglioni, Carlo; Chiarabba, Claudio (2021-10-12). "Re-pressurized magma at Mt. Etna, Italy, may feed eruptions for years". Communications Earth & Environment. 2 (1): 1–9. doi:10.1038/s43247-021-00282-9. ISSN 2662-4435.
^ Ferlito, C.; Bruno, V.; Salerno, G.; Caltabiano, T.; Scandura, D.; Mattia, M.; Coltorti, M. (2017-07-13). "Dome-like behaviour at Mt. Etna: The case of the 28 December 2014 South East Crater paroxysm". Scientific Reports. 7 (1): 5361. doi:10.1038/s41598-017-05318-9. ISSN 2045-2322.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Shadow_zone&oldid=1085796609"
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A medical calculator or clinical calculator is a software program for calculating various clinical scores and indices such as body mass index (BMI), body surface area (BSA), coronary heart disease risk, individual drug dosing, etc. Usually calculation of clinical scores or indices involves complex formulas using several input parameters. Medical calculators typically provide a user interface to enter parameters and calculate scores using a standard formula. Users do not need to use or even know the actual formula for calculating a clinical score or index. For example, body mass index or BMI (also known as the Quetelet Index for its creator, Belgium statistician Adolphe Quetelet, 1796–1874), is the most commonly used measure of obesity internationally [54]. Our physician author has found that the
\frac{weight\left(kg\right)}{{\left(height\left(m\right)\right)}^{2}}
\frac{weight\left(lb\right)×703}{{\left(height\left(in\right)\right)}^{2}}
formula for calculating BMI, while simple to remember, is surprisingly time consuming and error-prone in the time-pressured environment of a busy clinic. The user only needs to enter a patient’s weight and height in a typical medical calculator, however, to calculate the BMI quickly and confidently.
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3Blue1Brown - The most unexpected answer to a counting puzzle
Let’s play a strange sort of mathematical croquet. We’ll have two sliding blocks and a wall. The first block starts by coming in at some velocity from the right, while the second starts out stationary. Being overly-idealistic physicists, let’s assume that there’s no friction and that all collisions are perfectly elastic, meaning no energy is lost.
The simplest case is when both blocks have the same mass.
The first block hits the second, transferring all of its momentum. Then the second one bounces off the wall, then it transfers all of its momentum back to the first, which then sails off towards infinity. Three total clacks.
What about if that first block has 100 times the mass of the second one? It’s not entirely obvious how to predict the dynamics here; I promise I’ll explain all the relevant physics in due course. But in the spirit of getting to the punchline, let’s just watch what happens.
That second one will keep bouncing back and forth between the wall and the first block with 100 times its mass, like a satisfying game of breakout, slowly and discretely redirecting the first block’s momentum to point in the opposite direction. In total, there will be 31 collisions before both blocks are sliding off to infinity, never to touch again.
What if the first block has 10,000 times the mass of the second one?
In that case, there would be quite a few more clacks, all happening very rapidly at one point, adding up to 314 collisions in all.
If the first block was 1,000,000 times the mass of the second, then again, with all our idealistic conditions, almost all clacks happen in one big burst, this time resulting in 3,141 total collisions.
Perhaps you see the pattern here, though it’s forgivable if you don’t, since it defies all expectations. When the mass of that first block is some power of 100 times the mass of the second, the number of collisions will have the same digits as the beginning of
\pi \approx 3.14159
This absolutely blew my mind when it was first shared with me. Credit to the viewer Henry Kavle for introducing me to this fact, which was originally discovered by the mathematician Gregory Galperin in 1995, and published in 2003.
Part of what I love about this is that if ever there were olympic games for algorithms computing
\pi
, this one would have to win medals both for being the most elegant, and for being the most comically inefficient. I mean, think about the algorithm:
Step 1: Implement a physics engine.
Step 2: Choose the number of digits,
d
\pi
that you’d like to compute.
Step 3: Set the mass of one block to be
100^{d - 1}
, and send it traveling on a frictionless surface towards a block of mass
1
Step 4: Count the number of collisions.
So for example, to calculate only 20 digits of
\pi
, one block would have to be 100 billion billion billion billion times the mass of the other. If the small block was 1 kilogram, that means the big one would have a mass 10 times that of the supermassive black hole at the center of the milky way.
That means you’d need to count about 31 billion billion clacks, and at one point in the virtual process, the frequency of clacks would be around 100 billion billion billion billion clacks per second. So let’s just say that you’d need very good numerical precision to get this working accurately, and it would take a very long time to run!
I’ll emphasize again that this process is way over-idealized, quickly departing from anything that could possibly happen in real physics. But of course, you and I both know that this is not interesting because of its potential as a
\pi
-computing algorithm, or as a pragmatic physics demonstration.
It’s mind-boggling because why on earth do the digits of
\pi
show up here? And it’s such a weird way for
\pi
to show up, too: Its decimal digits are counting something, whereas usually its precise value describes something continuous.
I will show you why this is true. Where there is
\pi
, there is a hidden circle. And in this case, that hidden circle comes from the conservation of energy. In fact, in the next two lessons you’ll see two separate methods for understanding this hidden
\pi
, each as stunning as the surprising fact itself.
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Mihajlo Cekić and Gabriel P. Paternain
We consider Anosov flows on closed 3-manifolds preserving a volume form
\Omega
. Following Dyatlov and Zworski (Invent. Math. 210:1 (2017), 211–229) we study spaces of invariant distributions with values in the bundle of exterior forms whose wavefront set is contained in the dual of the unstable bundle. Our first result computes the dimension of these spaces in terms of the first Betti number of the manifold, the cohomology class
\left[{\iota }_{X}\Omega \right]
X
is the infinitesimal generator of the flow) and the helicity. These dimensions coincide with the Pollicott–Ruelle resonance multiplicities under the assumption of semisimplicity. We prove various results regarding semisimplicity on 1-forms, including an example showing that it may fail for time changes of hyperbolic geodesic flows. We also study non-null-homologous deformations of contact Anosov flows, and we show that there is always a splitting Pollicott–Ruelle resonance on 1-forms and that semisimplicity persists in this instance. These results have consequences for the order of vanishing at zero of the Ruelle zeta function. Finally our analysis also incorporates a flat unitary twist in the resonant spaces and in the Ruelle zeta function.
Anosov flow, resonances, dynamical zeta functions
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Elias_M._Stein Knowpia
Elias Menachem Stein (January 13, 1931 – December 23, 2018) was an American mathematician who was a leading figure in the field of harmonic analysis. He was the Albert Baldwin Dod Professor of Mathematics, Emeritus, at Princeton University, where he was a faculty member from 1963 until his death in 2018.
Princeton Lectures on Analysis textbook series
Elly Intrator
Linear Operators on Lp Spaces (1955)
Stein was born in Antwerp Belgium, to Elkan Stein and Chana Goldman, Ashkenazi Jews from Belgium.[1] After the German invasion in 1940, the Stein family fled to the United States, first arriving in New York City.[1] He graduated from Stuyvesant High School in 1949,[1] where he was classmates with future Fields Medalist Paul Cohen,[2] before moving on to the University of Chicago for college. In 1955, Stein earned a Ph.D. from the University of Chicago under the direction of Antoni Zygmund. He began teaching at MIT in 1955, moved to the University of Chicago in 1958 as an assistant professor, and in 1963 became a full professor at Princeton.
Stein worked primarily in the field of harmonic analysis, and made contributions in both extending and clarifying Calderón–Zygmund theory. These include Stein interpolation (a variable-parameter version of complex interpolation), the Stein maximal principle (showing that under many circumstances, almost everywhere convergence is equivalent to the boundedness of a maximal function), Stein complementary series representations, Nikishin–Pisier–Stein factorization in operator theory, the Tomas–Stein restriction theorem in Fourier analysis, the Kunze–Stein phenomenon in convolution on semisimple groups, the Cotlar–Stein lemma concerning the sum of almost orthogonal operators, and the Fefferman–Stein theory of the Hardy space
{\displaystyle H^{1}}
and the space
{\displaystyle BMO}
of functions of bounded mean oscillation.
He wrote numerous books on harmonic analysis (see e.g. [1,3,5]), which are often cited as the standard references on the subject. His Princeton Lectures in Analysis series [6,7,8,9] were penned for his sequence of undergraduate courses on analysis at Princeton. Stein was also noted as having trained a high number of graduate students (he has had at least 52 students, according to the Mathematics Genealogy Project), so shaping modern Fourier analysis. They include two Fields medalists, Charles Fefferman and Terence Tao.
His honors included the Steele Prize (1984 and 2002), the Schock Prize in Mathematics (1993), the Wolf Prize in Mathematics (1999), and the National Medal of Science (2001). In addition, he had fellowships to National Science Foundation, Sloan Foundation, Guggenheim Foundation, and National Academy of Sciences. Stein was elected as a member of the American Academy of Arts and Sciences in 1982.[3] In 2005, Stein was awarded the Stefan Bergman prize in recognition of his contributions in real, complex, and harmonic analysis. In 2012 he became a fellow of the American Mathematical Society.[4]
In 1959, he married Elly Intrator,[1] a former Jewish refugee during World War II.[5] They had two children, Karen Stein and Jeremy C. Stein,[1] and grandchildren named Alison, Jason, and Carolyn. His son Jeremy is a professor of financial economics at Harvard, former adviser to Tim Geithner and Lawrence Summers, and served on the Federal Reserve Board of Governors from 2012 to 2014. Elias Stein died of complications of lymphoma in 2018, aged 87.[6]
Stein, Elias (1970). Singular Integrals and Differentiability Properties of Functions. Princeton University Press. ISBN 0-691-08079-8.
Stein, Elias (1970). Topics in Harmonic Analysis Related to the Littlewood-Paley Theory. Princeton University Press. ISBN 0-691-08067-4.
Stein, Elias; Weiss, Guido (1971). Introduction to Fourier Analysis on Euclidean Spaces. Princeton University Press. ISBN 0-691-08078-X.
Stein, Elias (1971). Analytic Continuation of Group Representations. Princeton University Press. ISBN 0-300-01428-7.
Nagel, Alexander (1979). Lectures on Pseudo-differential Operators: Regularity Theorems and Applications to Non-elliptic Problems. Princeton University Press. ISBN 978-0-691-08247-9. [7]
Stein, Elias (1993). Harmonic Analysis: Real-variable Methods, Orthogonality and Oscillatory Integrals. Princeton University Press. ISBN 0-691-03216-5. [8]
Stein, Elias; Shakarchi, R. (2003). Fourier Analysis: An Introduction. Princeton University Press. ISBN 0-691-11384-X. 2011 reprint
Stein, Elias; Shakarchi, R. (2003). Complex Analysis. Princeton University Press. ISBN 0-691-11385-8. 2010 reprint
Stein, Elias; Shakarchi, R. (2005). Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Princeton University Press. ISBN 0-691-11386-6. 2009 reprint
Stein, Elias; Shakarchi, R. (2011). Functional Analysis: An Introduction to Further Topics in Analysis. Princeton University Press. ISBN 978-0-691-11387-6.
^ a b c d e University of St Andrews, Scotland - School of Mathematics and Statistics: "Elias Menachem Stein" by J.J. O'Connor and E F Robertson February 2010
^ "Stuyvesant High School Endowment Fund". Archived from the original on 2014-01-11. Retrieved 2013-07-07.
^ "Elias Menachem Stein". American Academy of Arts & Sciences. Retrieved 2020-05-26.
^ Center for Jewish History: "AHC interview with Elly Stein" 2012
^ Chang, Kenneth (2019-01-14). "Elias M. Stein, Mathematician of Fluctuations, is Dead at 87". The New York Times.
^ Beals, Richard (1980). "Review: Lectures on pseudo-differential operators: Regularity Theorems and applications to non-elliptic problems, by Alexander Nagel and E. M. Stein" (PDF). Bull. Amer. Math. Soc. (N.S.). 3 (3): 1069–1074. doi:10.1090/s0273-0979-1980-14859-4.
^ Ricci, Fulvio (1999). "Review: Harmonic Analysis: Real-variable Methods, Orthogonality and Oscillatory Integrals, by Elias Stein". Bull. Amer. Math. Soc. (N.S.). 36 (4): 505–521. doi:10.1090/s0273-0979-99-00792-2.
This article incorporates material from Elias Stein on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.
Elias M. Stein at the Mathematics Genealogy Project
Citation for Elias Stein for the 2002 Steele prize for lifetime achievement
Elias Stein Curriculum Vitae
Fefferman, Charles; Ionescu, Alex; Tao, Terence; Wainger, Stephen (October 2020). "Analysis and applications: The mathematical work of Elias Stein". Bulletin of the American Mathematical Society. 57 (4): 523–594. doi:10.1090/bull/1691.
"Simons Foundation: Elias Stein". Simons Foundation. 2017-03-13. – Extended video interview.
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Use the following data set: 7,17,30,18,24,16,3,18,7 to answer questions The data
Use the following data set: 7,17,30,18,24,16,3,18,7 to answer questions
The data is said to be
b. has no mode
d. Unimodal
Ceitheart
use the data set 7,17,30,18,24,16,3,18,7 to answer questions the data is said to be
mode is the most repeated value of data
here values 7 repeat 2 times
18 repeat 2 times
⇒\text{mode of data}=7,18
⇒bimodal
The birthrates in the United States for the year 2003-2012 are given in the following table. (The birthrate is the number of live births/1000 population.)
\begin{array}{|cccccc|}\hline Year& 2003& 2004& 2005& 2006& 2007\\ Birthrate& 14.7& 14.0& 14.0& 14.2& 14.2\\ \hline\end{array}
\begin{array}{|cccccc|}\hline Year& 2008& 2009& 2010& 2011& 2012\\ Birthrate& 14.0& 13.8& 13.8& 13.8& 13.7\\ \hline\end{array}
a. Describe a random variable X that is associated with these data.
b. Find the probability distribution for the random variable X.
c. Compute the mean, variance, and standard deviation of X.
Explain about of the measure of central tendency that is most appropriate for these data.
For healthy adults, the mean pH level is
\mu =7.4
, a new drug for arthritis has been developed. However, it is throught that this drug may change blood pH. A random sample of 33 patients with arthritis took the drug for three months. Blood tests showed that the sample mean from these 33 patients was a pH of 8.0 and it is known that
\sigma =2.3
. (Note, there is no unit for this type of data). Use
\alpha =0.05
level of significance to test the claim that the drug has increased the mean pH level of the blood.
If we use the classical approach, what would the criterion (step 3) be?
1) -1.645 or less
2) -2.33 or less
3) 1.645 or more
4) 2.33 or more
You fit a CAPM that regresses the excess return of Coca-cola on the excess market return using 20 years of monthly data. You estimate
\alpha =0,72,\beta =1,37,{S}^{2}=20,38,{\sigma }_{X}^{2}=19,82\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }{\mu }_{x}=0,71
What are the standar errors of
\alpha \text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }\beta
An advertising campaign for a new product is tergeted to make 20% of the adult population in a metropolitan area aware of the product. After the campaign, a random sample of 400 adults in the metropolitan area is obtained.
i. Find the approximate probability that 57 or fewer adults in the sample are aware of the product.
ii. Should the approximation in part i. be accurate?
In a survey of the number of trips abroad during one year was conduced among 1090 families. The survey reveals the following data:
\overline{)\begin{array}{cccccccc}\text{Number of trips}& 0& 1& 2& 3& 4& 5& 6\\ Frequency& 180& 110& 140& 150& 200& 160& 150\end{array}}
Which of the probabilities bellow (rounded to 2 decimal places) best represent the probability that a family makes 3 trips or less during one
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Find a power series representation for the function and determine
Find a power series representation for the function and determine the radius of
Find a power series representation for the function and determine the radius of convergence.
f\left(x\right)=\frac{x}{{\left(1+4x\right)}^{2}}
We will start with the power series of
\frac{1}{1-x}
\frac{1}{1-x}=\sum _{n=0}^{\mathrm{\infty }}{x}^{n}
Step 1 : Replace x with
-4x
\frac{1}{1-\left(-4x\right)}=\sum _{n=0}^{\mathrm{\infty }}{\left(-4x\right)}^{n}
\frac{1}{1+4x}=\sum _{n=0}^{\mathrm{\infty }}{\left(-4\right)}^{n}{x}^{n}
\frac{d}{dx}\left[\frac{1}{1+4x}\right]=\frac{d}{dx}\left[\sum _{n=0}^{\mathrm{\infty }}{\left(-4\right)}^{n}{x}^{n}\right]
-\frac{4}{{\left(1+4x\right)}^{2}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-4\right)}^{n}\frac{d\left[{x}^{n}\right]}{dx}
-\frac{4}{{\left(1+4x\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}{\left(-4\right)}^{n}\left(n\right){x}^{n-1}
Not that the sum in RHS is starting from
n=1
, because the first term died during differentiation
-4
\frac{1}{{\left(1+4x\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}{\left(-4\right)}^{n-1}\left(n\right){x}^{n-1}
\frac{1}{{\left(1+4x\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n-1}\cdot {4}^{n-1}\cdot \left(n\right){x}^{n-1}
Step 3 : Multiply both sides by x
\frac{x}{{\left(1+4x\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n-1}{\left(4\right)}^{n-1}\left(n\right){x}^{n}
If you want you can start the sum from
n=0
, but if you do that, you should replace n with
n+1
in the sum
\left(x,\text{ }y\right)=
Eliminate the parameter
\theta
in the following parametric equations
x=a\mathrm{tan}\theta \text{ }y=b\mathrm{sec}\theta
Let P be the vector-valued function
P\left(t\right)=\left(\mathrm{sin}t+,\mathrm{cos}t\right)
(a) What curve is traced by P? Draw a picture of this curve.
(b) Draw the vector
{P}^{\prime }\left(\frac{\pi }{3}\right)
at the point corresponding to
t=\frac{\pi }{3}
on the picture you produced in part (a).
(c) Find a vector equation for the line tangent to the curve at
t=\frac{\pi }{3}
\left\{\begin{array}{l}x–2y=-3\\ 4x-3y=10\end{array}
Polar coordinates of point P are given. Find all of its polar coordinates.
P=\left(1,-\pi /4\right)
Which of the following vectorvalued functions represent the same graph?
r\left(t\right)=\left(-3\mathrm{cos}t+1\right)i+\left(5\mathrm{sin}t+2\right)j+4k
r\left(t\right)=4i+\left(-3\mathrm{cos}t+1\right)j+\left(5\mathrm{sin}t+2\right)k
r\left(t\right)=\left(3\mathrm{cos}t-1\right)i+\left(-5\mathrm{sin}t-2\right)j+4k
r\left(t\right)=\left(-3\mathrm{cos}2t+1\right)i+\left(5\mathrm{sin}2t+2\right)j+4k
x=73-{y}^{2}
y=x-1
{y}^{2}
x=73-{y}^{2}
={y}^{2}
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Conic Sections/Parabola - Wikibooks, open books for an open world
Conic Sections/Parabola
The parabola is another commonly known conic section. The geometric definition of a parabola is the locus of all points such that they are equidistant from a point, known as the focus, and a straight line, called the directrix. In other words the eccentricity of a parabola is equal to 1. The solid formed by rotating a parabola about the axis is called a paraboloid.
1 Graphing Parabola
Graphing Parabola[edit | edit source]
The general form of a vertical parabola is
{\displaystyle (x-h)^{2}=4a(y-k)}
The axis of the parabola is the line perpendicular to the directrix which passes through the focus, and is the line
{\displaystyle x=h}
. It is also known as the line of symmetry.
The vertex is the 'base' of the parabola and is located at
{\displaystyle (h,k)}
. Therefore, a positive
{\displaystyle k}
will move the parabola upwards along its axis
{\displaystyle k}
units, while a negative one will move it downwards. This is the same regardless of whether
{\displaystyle a}
is a positive or negative number.
{\displaystyle a}
is the focal length of the parabola. The focus is located at
{\displaystyle (h,k+a)}
{\displaystyle a}
is positive the parabola opens upwards, whereas a negative
{\displaystyle a}
causes it to open downwards.
The directrix is located at
{\displaystyle y=k-a}
If the conic is horizontal, it is the same as a vertical parabola only along the x-axis rather than the y-axis. The general form is:
{\displaystyle (y-k)^{2}=4a(x-h)}
The axis of the parabola is the line
{\displaystyle y=k}
The vertex is located at
{\displaystyle (h,k)}
. A positive
{\displaystyle k}
will move the parabola towards the right along its axis
{\displaystyle k}
units, while a negative one will move it towards the left. This is the same regardless of whether
{\displaystyle a}
The focus is located at
{\displaystyle (h+a,k)}
{\displaystyle a}
is positive the parabola opens towards the right, whereas a negative
{\displaystyle a}
causes it to open leftwards.
{\displaystyle x=h-a}
Parametric Form[edit | edit source]
The parametric form of a vertical parabola is:
{\displaystyle x=h+2at}
{\displaystyle y=k+at^{2}}
For a horizontal one:
{\displaystyle x=h+at^{2}}
{\displaystyle y=k+2at}
For both of these forms:
The vertex is at (h,k)
{\displaystyle a}
units from the vertex along the major axis.
The gradient at any point on the parabola is t, which can be proved by differentiating the parametric form using the chain rule.
For information on how to graph the paramatric form, see Parametric Forms of Conic Sections.
Polar Form[edit | edit source]
The polar form of a parabola is given by
{\displaystyle r=-{\frac {2a}{1+cos\theta }}}
Parabola have a number of uses in everyday life:
Galileo showed in the 16th century that objects fired into the air at an angle to the ground followed the path of a parabola, a discovery which led to more accurate of targeting of projectiles.
In optics, parabola are particularly useful because parallel beams incident upon a parabolic mirror are all directed towards the focus, and conversely a light source placed at the focus will reflect off the mirror as a series of parallel rays. This makes parabolic mirrors useful in telescopes, headlights, satellite dishes and much more.
Another use of parabola is in training astronauts. A plane which follows a parabolic path into the air will cause passengers to experience freefall and therefore weightlessness.
http://www.tutorvista.com/math/parametric-equation-parabola
http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf
Retrieved from "https://en.wikibooks.org/w/index.php?title=Conic_Sections/Parabola&oldid=3694337"
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Electron Speed Calculator
How do I calculate the classical/non-relativistic velocity of electrons in an electric field?
How do I derive the Newtonian velocity equation for an accelerated electron?
How do I find the relativistic speed of an electron from its energy?
How to use the electron speed calculator?
Are you looking for an electron speed calculator to estimate both the classical and relativistic speed of an electron? Be our guest!
In this calculator, you can not only estimate the relativistic and non-relativistic velocities of an electron under a given accelerating potential, but you will also learn:
The events which led to the discovery of electrons;
The Newtonian velocity equation for an electron in an electric field and its derivation; and
How to find the relativistic speed of an electron from its energy.
Last but not least, you'll learn answers to some interesting questions, like "Do electrons move near the speed of light?" Let's go!
Atoms consist of three basic components: electrons, protons, and neutrons. Based on the results of 𝛼-scattering experiments, the physicist Rutherford suggested that negatively charged electrons move in circular orbits around the positively charged region called the nucleus.
An electron has a mass of 9.109 × 10-31 kg and a charge of 1.602 × 10-19 C.
Electromagnetic fields accelerate electrons as they carry a charge. If you know the value of the electric field's potential, you can calculate the speed of an electron moving under its influence using the equation for kinetic energy. Read on to learn how to calculate the speed of an electron accelerated by an electric field.
💡 Do you know that the motion of electrons in magnetic and electric fields helped determine the sign of their charge? In 1879, the English physicist and chemist William Crookes discovered that magnetic fields bend cathode rays and the direction of deflection indicated that they were negatively charged particles. Then in 1897, J. J. Thomson observed cathode rays bending towards the positive plate and deviating away from the negative plate when allowed to pass between them. Thomson’s discovery established that electricity involved the flow of negatively charged particles. Thomson called these particles electrons.
We calculate the classical or non-relativistic velocity of an electron under the influence of an electric field as:
vn = √(2eVa / m),
vn — Classical or non-relativistic velocity;
e — Elementary charge, or the charge of an electron (e = 1.602 × 10-19 C);
Va — Accelerating potential, or the potential difference that is applied to accelerate the electron; and
m — The mass of an electron (m = 9.109 × 10-31 kg).
In an electric field of potential Va, an electron experiences a force of e × E, where E is the intensity of the electric field. Therefore, the work done on the electron is:
W = force × distance = (e × E) × r.
Here, the product E × r is the electric potential Va.
Therefore, the work done on the electron is:
W = e × Va.
By the work-energy theorem, the net work done by the forces on an object equals the change in its kinetic energy.
Therefore, the kinetic energy of an electron accelerated through a potential difference Va is:
m vn2 / 2 = e Va,
vn = √(2 e Va / m).
Under the influence of electric fields, the electrons accelerate to speeds at which relativistic effects become evident such as the velocity, momentum, and energy having values different from those calculated using classical physics.
At speeds greater than the 1/10th of the speed of light, we must use the relativistic formula to find an electron's velocity in an electric field:
v_\text{rel} = c\sqrt{1-\frac{1}{(1+\frac{eV_\text{a}}{m_0c^2})^2}} .
v_\text{rel}
is the electron's relativistic speed;
c
is the velocity of light;
e
is the electron's charge;
V_\text{a}
is the accelerating potential; and
m_0
is the electron's rest mass (9.109 × 10-31 kg).
The product eVa indicates the electron's kinetic energy in the electric field.
Curious to learn more about relativistic effects? Check out our velocity addition calculator.
Good news — it's simple! To find the Newtonian (classical) and Einsteinian (relativistic) speeds of an electron under a given accelerating potential, just enter the value of accelerating potential in the respective input box.
For example, to find the speed of an electron at a potential difference of 11 kV:
Change the unit of Accelerating potential to kV - just click on the unit beside the input box for accelerating potential and select kV.
Enter 11 in the input box for accelerating potential.
Omni's electron speed calculator gives you:
Classical velocity: 62,205 km/s;
Relativistic velocity: 61,221 km/s; and
Difference between two velocities: -984 km/s.
At what speed do relativistic effects occur?
Relativistic effects occur at any speed. However, they are extremely small for speeds below 1/10th of the speed of light (30,000 km/s) and are negligible. When an object approaches speeds of 1/10th of the speed of light, relativistic effects become relevant.
Do electrons move near the speed of light?
Electrons naturally move with a speed less than 1% of the speed of light, but we can accelerate them to speeds of more than 99% of the speed of light by providing billions of electron volts of energy! Interestingly, we couldn't accelerate electrons to 100% of the speed of light as the faster the electrons move, the heavier they become. At those speeds, they require a tremendous amount of energy to increase their speed, even by 0.000000001%.
Can a magnetic field accelerate electrons?
Magnetic fields cannot accelerate an electron by changing its speed. They can only deflect an electron.
How do I calculate the kinetic energy of an electron accelerated through a potential difference of 100 V?
To obtain the kinetic energy of an electron accelerated through a potential difference of 100 V, multiply the elementary charge e (9.109 × 10-31 kg) with the given potential difference Va:
K E = e Va = 100 eV
Classical velocity
E = mc²Gravitational time dilationLength contraction… 4 more
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Caesar Shift Practice Problems Online | Brilliant
On the first day of class, a teacher writes his name on the board in a secret code:
NS. TNJUI
What is the actual last letter of his last name?
E G H R S
Julius Caesar encodes a secret message for his troops using the
+3
Caesar Shift code shown above. The encoded message is the following:
DWWDFN DW GDZQ.
Which letter from the original message does the Q in the encoded message represent?
To make their encryption harder to crack, a spy decides to use a sequence of 3 Caesar Shifts: first a
+11
Caesar Shift, then a
+24
Caesar Shift, then finally a
+17
Caesar Shift. How will the spy encode the letter X?
U X B I G
DRKYVDRKZTJ ZJ KYV HLVVE FW KYV JTZVETVJ, REU ELDSVI KYVFIP ZJ KYV HLVVE FW DRKYVDRKZTJ.
The message above was encoded using a Caesar Shift. The letter pair "KY" appears frequently in the encoded message. Which of these common letter pairs could the encoded letter pair "KY" represent?
AN HA TH IN
Which of these Caesar Shifts encodes the letter A differently than the others?
-7
+7
-19
+33
They are all the same
|
How can I evaluate \lim_{v \to \frac{\pi}{3}} \frac{1-2\cos v}{\sin(v-\frac{\pi}{3})} without
How can I evaluate
\underset{v\to \frac{\pi }{3}}{lim}\frac{1-2\mathrm{cos}v}{\mathrm{sin}\left(v-\frac{\pi }{3}\right)}
Medicim6
You can use the fact that
2\mathrm{cos}v=2\mathrm{cos}\left(\left(v-\frac{\pi }{3}\right)+\frac{\pi }{3}\right)
=\mathrm{cos}\left(v-\frac{\pi }{3}\right)-\sqrt{3}\mathrm{sin}\left(v-\frac{\pi }{3}\right)
\frac{1-2\mathrm{cos}v}{\mathrm{sin}\left(v-\frac{\pi }{3}\right)}=\frac{1-\mathrm{cos}\left(v-\frac{\pi }{3}\right)}{\mathrm{sin}\left(v-\frac{\pi }{3}\right)}-\sqrt{3}
and therefore all that remains to be done is to compute
\underset{t\to 0}{lim}\frac{1-\mathrm{cos}t}{\mathrm{sin}t}
and for this you can use the fact that
1-\mathrm{cos}t=2{\mathrm{sin}}^{2}\left(\frac{t}{2}\right)\text{ }\text{ }\text{ and that }\text{ }\mathrm{sin}t=2\mathrm{sin}\left(\frac{t}{2}\right)\mathrm{cos}\left(\frac{t}{2}\right)
just use Taylor expansions:
\mathrm{cos}x=\frac{12}{-}\frac{\sqrt{3}\cdot \left(x-\frac{\pi }{3}\right)}{2}+o\left(x-\frac{\pi }{3}\right),\text{ }x\to \frac{\pi }{3}
\mathrm{sin}x=x-\frac{\pi }{3},\text{ }o\left(x-\frac{\pi }{3}\right),\text{ }x\to \frac{\pi }{3}
\underset{x\to \frac{\pi }{3}}{lim}\frac{1-2\mathrm{cos}x}{\mathrm{sin}\left(x-\frac{\pi }{3}\right)}=\underset{x\to \frac{\pi }{3}}{lim}\frac{1-1+\sqrt{3}\left(x-\frac{\pi }{3}\right)+o\left(x-\frac{\pi }{3}\right)}{x-\frac{\pi }{3}+o\left(x-\frac{\pi }{3}\right)}=\underset{x\to \frac{\pi }{3}}{lim}\sqrt{3}+o\left(1\right)=\sqrt{3}
Use Prosthaphaeresis fromula and
\mathrm{sin}2A=2\mathrm{sin}A\mathrm{cos}A
Formulas,
2\cdot \underset{v\to \frac{\pi }{3}}{lim}\frac{\mathrm{cos}\frac{\pi }{3}-\mathrm{cos}v}{\mathrm{sin}\left(v-\frac{\pi }{3}\right)}=2\cdot \underset{v\to \frac{\pi }{3}}{lim}\frac{2\mathrm{sin}\left(\frac{v}{2}-\frac{\pi }{6}\right)\mathrm{sin}\left(\frac{v}{2}+\frac{\pi }{6}\right)}{2\mathrm{sin}\left(\frac{v}{2}-\frac{\pi }{6}\right)\mathrm{cos}\left(\frac{v}{2}-\frac{\pi }{6}\right)}
Now cancel out
\mathrm{sin}\left(\frac{v}{2}-\frac{\pi }{6}\right)\text{ }as\text{ }\mathrm{sin}\left(\frac{v}{2}-\frac{\pi }{6}\right)\ne 0\text{ }as\text{ }v\to \frac{\pi }{3},\text{ }v\ne \frac{\pi }{3}
\mathrm{sin}x+\mathrm{sin}y=a\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}x+\mathrm{cos}y=b
\mathrm{tan}\left(x-\frac{y}{2}\right)
P(t) models the distance of a swinging pendulum (In CM) from the place it has travelled t seconds after it starts to swing. Here, t is entered in radians.
P\left(t\right)=-5\mathrm{cos}\left(2\pi t\right)+5
What is the first time the pendulum reaches 3.5 CM from the place it was released?
Round your final answer to the hundredth of a second.
Ok, so I did this to get the solution:
3.5=-5\mathrm{cos}\left(2\pi t\right)+5
-1.5=-5\mathrm{cos}\left(2\pi t\right)
0.3=\mathrm{cos}\left(2\pi t\right)
\mathrm{cos}-1\left(0.3\right)=2\pi t
\frac{{\mathrm{cos}}^{-1}\left(0.3\right)}{2\pi }=t
The fact is this gives me 11.55 seconds to get to 3.5 CM, which does not sound right. Where did I go wrong and how can I fix it?
5{\mathrm{sin}}^{2}\left(x\right)+8\mathrm{sin}x\mathrm{cos}x-3=0
I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into
\mathrm{sin}2x
but then I have
{\mathrm{sin}}^{2}x
\mathrm{sin}2x
\frac{1}{sec\theta -tan\theta }+\frac{1}{sec\theta +tan\theta }
\mathrm{sin}x+\mathrm{cos}x=\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)
\sqrt{2}\left(x+\frac{\pi }{4}\right)=\sqrt{2}\left(\mathrm{sin}x\mathrm{cos}\frac{\pi }{4}+\mathrm{cos}x\mathrm{sin}\frac{\pi }{4}\right)=\mathrm{sin}x+\mathrm{cos}x
Could you solve it from opposite?
\mathrm{sin}x+\mathrm{cos}x=\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)
\underset{x\to 0}{lim}{\mathrm{cos}x}^{\frac{1}{\mathrm{sin}x}}
|
Pole placement design - MATLAB place
K = place(A,B,p)
[K,prec,message] = place(A,B,p)
Given the single- or multi-input system
\stackrel{˙}{x}=Ax+Bu
and a vector p of desired self-conjugate closed-loop pole locations, place computes a gain matrix K such that the state feedback u = –Kx places the closed-loop poles at the locations p. In other words, the eigenvalues of A – BK match the entries of p (up to the ordering).
K = place(A,B,p) places the desired closed-loop poles p by computing a state-feedback gain matrix K. All the inputs of the plant are assumed to be control inputs. The length of p must match the row size of A. place works for multi-input systems and is based on the algorithm from [1]. This algorithm uses the extra degrees of freedom to find a solution that minimizes the sensitivity of the closed-loop poles to perturbations in A or B.
[K,prec,message] = place(A,B,p) returns prec, an estimate of how closely the eigenvalues of A – BK match the specified locations p (prec measures the number of accurate decimal digits in the actual closed-loop poles). If some nonzero closed-loop pole is more than 10% off from the desired location, message contains a warning message.
You can also use place for estimator gain selection by transposing the A matrix and substituting C' for B.
l = place(A',C',p).'
p = [-1 -1.23 -5.0];
place uses the algorithm of [1] which, for multi-input systems, optimizes the choice of eigenvectors for a robust solution.
In high-order problems, some choices of pole locations result in very large gains. The sensitivity problems attached with large gains suggest caution in the use of pole placement techniques. See [2] for results from numerical testing.
[1] Kautsky, J., N.K. Nichols, and P. Van Dooren, "Robust Pole Assignment in Linear State Feedback," International Journal of Control, 41 (1985), pp. 1129-1155.
[2] Laub, A.J. and M. Wette, Algorithms and Software for Pole Assignment and Observers, UCRL-15646 Rev. 1, EE Dept., Univ. of Calif., Santa Barbara, CA, Sept. 1984.
lqr | rlocus
|
Calculate the double integral. \int\int_R\frac{1}{1+x+y}dA,\ R=[1,3]\time
\int {\int }_{R}\frac{1}{1+x+y}dA,\text{ }R=\left[1,3\right]×\left[1,2\right]
Once again use the fact that:
\int \frac{{f}^{\prime }\left(x\right)}{f\left(x\right)}dx=\mathrm{ln}|f\left(x\right)|+C
It doesn't matter do you opt to solve the first integral with respect to x or with respect to y, Here, we opt to solve the integral with respect to y first:
{\int }_{1}^{3}{\int }_{1}^{2}\frac{1}{1+x+y}dydx={\int }_{1}^{3}\mathrm{ln}\left(x+3\right)-\mathrm{ln}\left(x+2\right)
Now using the integration by parts we obtain that the integral is equal to:
\left(\left(x+3\right)\mathrm{ln}\left(x+3\right)-\left(x+3\right)\right)-\left(\left(x+2\right)\mathrm{ln}\left(x+2\right)-\left(x+2\right)\right){\mid }_{1}^{3}=6\mathrm{ln}6-5\mathrm{ln}5-4\mathrm{ln}4+3\mathrm{ln}3
6\mathrm{ln}6-5\mathrm{ln}5-4\mathrm{ln}4+3\mathrm{ln}3
x={t}^{2},y=2t,0\le t\le 5
\left({x}^{2}+2xy-4{y}^{2}\right)dx-\left({x}^{2}-8xy-4{y}^{2}\right)dy=0
Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
\int \frac{dx}{{x}^{3}\sqrt{{x}^{2}-1}},x>1
\int \left(\frac{3}{2{x}^{4}}-\frac{2}{x}+\frac{3}{2}\mathrm{sin}\left(2x\right)+\frac{5}{2}\mathrm{cos}\left(3x\right)\right)dx
r=7\mathrm{cos}\left(\theta \right)
Evaluate the integrals. The integrals are listed in random order so you need to decide which integration technique to use.
\int \left({e}^{x}-{e}^{-x}\right){\left({e}^{x}+{e}^{-x}\right)}^{3}dx
\int \frac{{e}^{x}}{{e}^{2x}+2{e}^{x}+17}dx
|
Consider two series sum_(n=1)^oo 1/2pinZ and sum_(n=0)^oo 1/pi^2n. Which o
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2}\pi nZ\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{\pi }^{2}}n
Series A:
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2}\pi n=\frac{1}{2}\pi \left\{\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots \right\}
Harmonic series – Divergent
Series B:
\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{\pi }^{2}}n=\left\{\left(\frac{1}{{\pi }^{2}}\right)+{\left(\frac{1}{{\pi }^{2}}\right)}^{2}+\dots \right\}
Geometric series, so
\frac{1}{{\pi }^{2}}<1
– Convergent
Thus, only series B converges.
Name the series:
\sum _{n=0}^{\mathrm{\infty }}\frac{{f}^{n}×a}{n!}×{\left(x–a\right)}^{n}
\sum _{n=0}^{\mathrm{\infty }}\frac{{f}^{n}×0}{n!}×{x}^{n}
f\left(0\right)=2,\text{ }{f}^{\prime }\left(0\right)=3,\text{ }f{}^{″}\left(0\right)=4,\text{ }f{}^{‴}\left(0\right)=12
\sum _{\mathrm{\infty }}^{n=1}bn
bn\to 0
Determine whether the infinite geometric series converges or diverges. If it converges, find its sum.
\sum _{k=1}^{\mathrm{\infty }}\frac{1}{3}{4}^{k=1}
Binomial expansion of
{\left(1-x\right)}^{n}
1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+\stackrel{˙}{s}
\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}×k}{6{k}^{4}+1}
|
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