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Rightmost Non-zero Digit of \(n!\) | Brilliant Math & Science Wiki
n!
Tapas Mazumdar, Rahil Sehgal, Ben A, and
Finding the number of trailing zeroes in
n!
is an elementary and standard approach which gives us the number of zeroes at the end of
n!
. But finding the rightmost non-zero digit is one such commonly asked problem that many fewer people know about.
n!
n!
\mathfrak{D} (n!) = {\left\lfloor \frac n5 \right\rfloor}! \times 2^{{}^{\big\lfloor {\large \frac{n}{5}} \big\rfloor}} \times { ( n \bmod{5} ) }!.
{\left\lfloor \frac n5 \right\rfloor}!
is sufficiently small, then the units digit of
\mathfrak{D} (n!)
gives us the value of the rightmost non-zero digit of
n!
Otherwise, we find the units digit of
2^{{}^{\big\lfloor {\large \frac{n}{5}} \big\rfloor}} \times ( n \bmod{5} )!
\mathfrak{d}_1
. Now we solve for
\mathfrak{D} \left( {\left\lfloor \frac n5 \right\rfloor}! \right)
. Again, if we arrive at a sufficiently small value of
{\left\lfloor \frac 15 \cdot {\big\lfloor \frac n5 \big\rfloor} \right\rfloor}!
, then we evaluate the units digit of
\mathfrak{D} \left( {\left\lfloor \frac n5 \right\rfloor}! \right)
(
call this
\mathfrak{d})
and the final answer is
\mathfrak{d}_1 \times \mathfrak{d}
If such a case doesn't arise, then we keep on iterating for
\mathfrak{D} \bigg( \overbrace{ {\bigg\lfloor \frac 15 \cdot {\cdots \left\lfloor \frac 15 \cdot {\left\lfloor \frac 15 \cdot {\left\lfloor \frac 15 \cdot {\big\lfloor \frac n5 \big\rfloor} \right\rfloor} \right\rfloor} \right\rfloor } \cdots \bigg\rfloor}! }^{\text{value of expression } = \, m!} \bigg)
and record the units digit of the product
2^{a_i} \times (b_i \bmod{5}) !
\big(
a_i
b_i
are to be followed from the original definition of
\mathfrak{D} (n!)\big)
and call them
\{\mathfrak{d}_1 , \mathfrak{d}_2, \cdots, \mathfrak{d}_k\}
. Once we arrive at a sufficiently small value for
m!
, we find the units digit of
\mathfrak{D} (m)
(
\mathfrak{d})
\prod_{i=1}^k \mathfrak{d_i} \times \mathfrak{d}.
\big(
Here sufficiently small value for
m!
m \in \{1,2,3,4\}.\big)
We first expand
n!
n! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times \cdots \times (n-2) \times (n-1) \times n.
Now our objective is to collect all the multiples of 5 contained in
n!
and group the rest of the terms in groups of four consecutive integers. Note that there will be
\left\lfloor \frac n5 \right\rfloor
multiples of 5 in
n!
\begin{aligned} n! &= \bigg( 5 \times 10 \times 15 \times 20 \times \cdots \times \left(5 \left\lfloor \dfrac n5 \right\rfloor \right) \bigg) \times (1 \times 2 \times 3 \times 4) \times (6 \times 7 \times 8 \times 9) \times \cdots \\ &= \bigg( 5^{{}^{\big\lfloor {\large \frac{n}{5}} \big\rfloor}} \times {\left\lfloor \dfrac n5 \right\rfloor}! \bigg) \times (1 \times 2 \times 3 \times 4) \times (6 \times 7 \times 8 \times 9) \times \cdots. \end{aligned}
We know that in the value of
n!
the distribution of lower primes are always greater than or equal to the distribution of higher primes. Thus we know that we can find a multiple
2^{{}^{\big\lfloor {\large \frac{n}{5}} \big\rfloor}}
(or some higher exponent of 2) from each of the groups of the four consecutive integers which are non-integer multiples of 5.
Also note that for consecutive non-multiples of 5, we have
(5k+1)(5k+2)(5k+3)(5k+4) = {\color{#3D99F6} 625\big(k^4+k^2\big) + 10\big(125k^3+25k^2+25k\big)} + 24. \qquad {\color{#20A900} (*)}
k^4+k^2
is an even integer for both even and odd
k
625\big(k^4+k^2\big)
will be divisible by
10
for all non-negative integers
k
and thus the expression shown above in
\color{#3D99F6} \text{blue}
10
k
. Hence, the product of four consecutive non-integer multiples of 5, when formulated, will always have 4 in the units place.
So, if we extract all multiples of 2 from each of the given groups of four
\big(
which are also
{\big\lfloor \frac n5 \big\rfloor}
in number
\big),
we know that each group will now contribute to a product whose units digit is 2.
\begin{aligned} \big(\text{Rightmost non-zero digit of } n!\big) &= 5^{{}^{\big\lfloor {\large \frac{n}{5}} \big\rfloor}} \times {\left\lfloor \dfrac n5 \right\rfloor}! \times 2^{{}^{\big\lfloor {\large \frac{n}{5}} \big\rfloor}} \times \bigg( 2 \times 2 \times 2 \times 2 \times \cdots {\left\lfloor \dfrac n5 \right\rfloor} \text{ times} \bigg) \times {\color{#D61F06} \big( n \bmod{5} \big)!} \\\\ &= {\left\lfloor \dfrac n5 \right\rfloor}! \times 2^{{}^{\big\lfloor {\large \frac{n}{5}} \big\rfloor}} \times {\color{#D61F06} \big( n \bmod{5} \big)!}. \end{aligned}
\big(
{\big\lfloor \frac n5 \big\rfloor}!
is not sufficiently small, then iterations are to be followed.
\big)
The final term
(
shown in
{\color{#D61F06} \text{red}})
comes from the fact that
n
can be of any of the forms
5k,5k+1,5k+2,5k+3,5k+4
and hence may or may not be a part of a group of four. In such a case when it is not in a group of 4, it will be in a group of
(n \bmod{5} )
n=12,
then the groups of 4 are
(1,2,3,4)
(6,7,8,9)
. The remaining numbers are
(11,12),
in which 12 is included in the group of
( 12 \equiv 2 \bmod{5} )
. The multiplication of the integers in the final group that contains
n
will yield a product whose units digit will be equal to the units digit of
( n \bmod{5} )!
. This result can be proved using a similar method as in
{\color{#20A900} (*)}
_\square
Find the rightmost non-zero digit of
501!
Using our method of continued iterations, we have
\begin{aligned} \mathfrak{D} (501!) &=& {\left\lfloor \dfrac{501}{5} \right\rfloor}! \times 2^{\lfloor {501}/{5} \rfloor} \times \big( 501 \bmod{5} \big)! &=& 100! \times 2^{100} \times 1! \\ \mathfrak{D} (100!) &=& {\left\lfloor \dfrac{100}{5} \right\rfloor}! \times 2^{\lfloor {100}/{5} \rfloor} \times \big( 100 \bmod{5} \big)! &=& 20! \times 2^{20} \\ \mathfrak{D} (20!) &=& {\left\lfloor \dfrac{20}{5} \right\rfloor}! \times 2^{\lfloor {20}/{5} \rfloor} \times \big( 20 \bmod{5} \big)! &=& 4! \times 2^{4}. \end{aligned}
Combining the above three results, we have
\begin{aligned} \mathfrak{D} (501!) &= \big(\text{Units digit of } 2^{100}\big) \times \big(\text{Units digit of } 2^{20}\big) \times \big(\text{Units digit of } 4! \cdot 2^4\big)\\ & = \big(\text{Units digit of } 6 \times 6 \times 4 \big) \\ &= 4.\ _\square \end{aligned}
Add an example here.
It is a common exercise to determine the number of trailing zeros of a factorial. For example,
20! = 2\; 432\; 902\; 008\; 176\; 64{\color{#20A900}{0}}\; {\color{#20A900}{000}}
has 4 trailing zeros (as highlighted in green above).
However, finding the rightmost non-zero digit of a factorial is much harder. For example, the rightmost non-zero digit of
20!
is 4, as shown above.
Problem: Find the rightmost non-zero digit of
10000!
Find the right most non-zero digit of
45!
400!
Cite as: Rightmost Non-zero Digit of
n!
. Brilliant.org. Retrieved from https://brilliant.org/wiki/rightmost-non-zero-digit-of-n/
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Inverse Laplace Transform without using equations I'm stuck with a question
Inverse Laplace Transform without using equations
I'm stuck with a question about Inverse Laplace Transform here, but the use of inverse laplace transform equation is forbidden.
Y\left(s\right)=\frac{{e}^{-\pi s}}{s\left[{\left(s+1\right)}^{2}+1\right]}
dagars5nx
The inverse Laplace transform of
\frac{1}{{s}^{2}+1}
\mathrm{sin}t
{\mathcal{L}}^{-1}\left\{\frac{1}{\left(s+1{\right)}^{2}+1}\right\}={e}^{-t}\mathrm{sin}\left(t\right)=\frac{1}{2i}\left({e}^{\left(-1+i\right)t}-{e}^{\left(-1-i\right)t}\right)
{\mathcal{L}}^{-1}\left\{\frac{1}{s}\frac{1}{\left(s+1{\right)}^{2}+1}\right\}={\int }_{0}^{t}{e}^{-u}\mathrm{sin}\left(u\right)du=\frac{1}{2i}\left(\frac{{e}^{\left(-1+i\right)t}-1}{-1+i}-\frac{{e}^{\left(-1-i\right)t}-1}{-1-i}\right)
=\frac{1}{2}-\frac{1}{2}{e}^{-t}\left(\mathrm{sin}\left(t\right)+\mathrm{cos}\left(t\right)\right)
The last step is to use the shift theorem
{\mathcal{L}}^{-1}\left\{{e}^{-as}F\left(s\right)\right\}=f\left(t-a\right)\mathrm{\Theta }\left(t-a\right)
{\mathcal{L}}^{-1}\left\{{\mathrm{e}}^{-\mathrm{\pi s}}\frac{1}{\mathrm{s}}\frac{1}{\left(\mathrm{s}+1{\right)}^{2}+1}\right\}=\left[\frac{1}{2}-\frac{1}{2}{\mathrm{e}}^{-\left(\mathrm{t}-\mathrm{\pi }\right)}\left(\mathrm{sin}\left(\mathrm{t}-\mathrm{\pi }\right)+\mathrm{cos}\left(\mathrm{t}-\mathrm{\pi }\right)\right)\right]\mathrm{\Theta }\left(\mathrm{t}-\mathrm{\pi }\right)
=\left(\frac{1}{2}+\frac{1}{2}{\mathrm{e}}^{\mathrm{\pi }-\mathrm{t}}\left(\mathrm{cos}\left(\mathrm{t}\right)+\mathrm{sin}\left(\mathrm{t}\right)\right)\right)\mathrm{\Theta }\left(\mathrm{t}-\mathrm{\pi }\right)
Solve the following differential equations for a general solution:
y3{y}^{\prime }+2y={e}^{x}
Solve the following differential equations using the Laplace transform and the unit step function
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}
y\frac{dy}{dx}={e}^{x}
Solve it by separating variables like this:
ydy={e}^{x}dx
\int ydy=\int {e}^{x}dx
\frac{{y}^{2}}{2}={e}^{x}+c
Find the inverse Laplace transform f(t) of the following transforms F(s). Match each of the following to the value of f(1.234) to two decimal places.
Find the solution of the integral equation f(t) using Laplace transforms
f\left(t\right)=36t+5{\int }_{0}^{t}f\left(t-u\right)\mathrm{sin}\left(5u\right)du
How to solve this differential equation?
y{}^{″}-\frac{{y}^{\prime }}{x}=4{x}^{2}y
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Quadratic Formula - Course Hero
College Algebra/Solving Quadratic Equations and Inequalities/Quadratic Formula
The quadratic formula is used to solve a quadratic equation of the form
ax^2+bx+c=0
a\neq0
, by using its coefficients:
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
The quadratic formula can be used to solve any quadratic equation. It is found by completing the square for the standard form of a quadratic equation.
How to Derive the Quadratic Formula
ax^2+bx+c=0
a\neq0
The equation is in standard form.
\begin{aligned}\frac{a}{a}x^2+\frac{b}{a}x+\frac{c}{a}&=\frac{0}{a}\\x^2+\frac{b}{a}x+\frac{c}{a}&=0\end{aligned}
a
x^2+\frac{b}{a}x=-\frac{c}{a}
\frac{c}{a}
x^2+\frac{b}{a}x+\left ( \frac{b}{2a} \right )^{2}=-\frac{c}{a}+\left ( \frac{b}{2a} \right )^{2}
\left ( \frac{b}{2a} \right )^{2}
to the left side to complete the square. Add
\left ( \frac{b}{2a} \right )^{2}
to the right side to keep the equation balanced.
\left (x+\frac{b}{2a}\right )^2=\frac{b^2}{4a^2}-\frac{c}{a}
\begin{aligned}\left (x+\frac{b}{2a}\right )^2&=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}\\ \left (x+\frac{b}{2a}\right )^2&=\frac{b^2-4ac}{4a^2}\end{aligned}
Use a common denominator on the right side, and simplify.
\begin{aligned}x+\frac{b}{2a}&=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\x+\frac{b}{2a}&=\pm\frac{\sqrt{b^2-4ac}}{2a}\end{aligned}
Take the square root of both sides, and simplify.
x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}
\frac{b}{2a}
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
The discriminant is the part of the quadratic formula that is under the radical sign:
b^2-4ac
Recall that a quadratic equation may have zero, one, or two real solutions. The value of the discriminant determines how many real solutions a quadratic equation has:
b^2-4ac
is positive, then the equation has two real solutions.
b^2-4ac
is zero, then the equation has one real solution.
b^2-4ac
is negative, then the equation has zero real solutions.
Using the Quadratic Formula with Two Real Solutions
x^2-4x-6=0
The equation is in the standard form, with
a=1
b=-4
c=-6
\begin{aligned}ax^2+bx+c&=0\\x^2-4x-6&=0\end{aligned}
The discriminant from the quadratic formula is the expression under the radical symbol:
x=\frac{-b\pm\sqrt{{\color{#c42126}{b^2-4ac}}}}{2a}
Use the discriminant from the quadratic formula to determine the number of solutions:
\begin{gathered}b^2-4ac\\(-4)^2-4(1)(-6)\\16+24\\40\end{gathered}
The discriminant is positive. So, the equation has two real solutions.
Using the quadratic formula, substitute the value of the discriminant under the radical. Then substitute the values for
and
b
\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x&=\frac{-(-4)\pm\sqrt{40}}{2(1)}\end{aligned}
\begin{aligned}x&=\frac{4\pm\sqrt{40}}{2}\\x&=\frac{4\pm2\sqrt{10}}{2}\\x&=2\pm\sqrt{10}\end{aligned}
The two solutions of the equation are:
x=2+\sqrt{10}\;\;\;\;\;\text{or}\;\;\;\;\;x=2-\sqrt{10}
Using the Quadratic Formula with One Real Solution
Solve the quadratic formula:
2x^2+8x+8=0
a=2
b=8
c=8
\begin{aligned}ax^2+bx+c&=0\\2x^2+8x+8&=0\end{aligned}
x=\frac{-b\pm\sqrt{{\color{#c42126}{b^2-4ac}}}}{2a}
\begin{gathered}b^2-4ac\\(8)^2-4(2)(8)\\64-64\\0\end{gathered}
The discriminant is zero. So, the equation has 1 real solution.
and
b
\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x&=\frac{-8\pm\sqrt{0}}{2(2)}\end{aligned}
\begin{aligned}x&=\frac{-8\pm0}{4}\\x&=\frac{-8}{4}\\x&=-2\end{aligned}
x=-2
Using the Quadratic Formula with No Real Solutions
x^2-6x+11=0
a=1
b=-6
c=11
\begin{aligned}ax^2+bx+c&=0\\x^2-6x+11&=0\end{aligned}
x=\frac{-b\pm\sqrt{{\color{#c42126}{b^2-4ac}}}}{2a}
\begin{gathered}b^2-4ac\\(-6)^2-4(1)(11)\end{gathered}
Evaluate the discriminant.
\begin{gathered}(-6)^2-4(1)(11)\\(-6)(-6)-(4)(11)\\36-44\\-8\end{gathered}
The discriminant is negative. So, the equation has zero real solutions. There is no need to use the quadratic formula.
<Completing the Square>Solving Quadratic Equations
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\mathrm{LerchPhi}\left(z,a,v\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{z}^{n}}{{\left(v+n\right)}^{a}}
|z|<1
|z|=1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{and}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}1<\mathrm{ℜ}\left(a\right)
z
and
v
v
are positive integers, LerchPhi(z, a, v) has a branch cut in the
z
z=1
z=1
a
z
1-a
z=1
1<\mathrm{ℜ}\left(a\right)
\mathrm{ℜ}\left(a\right)\le 1
0\le \mathrm{ℜ}\left(a\right)
a
\mathrm{LerchPhi}\left(3,4,1\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{polylog}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)}{\textcolor[rgb]{0,0,1}{3}}
\mathrm{LerchPhi}\left(0,7,4\right)
\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{16384}}
\mathrm{LerchPhi}\left(4,0,3\right)
\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}}
\mathrm{LerchPhi}\left(z,a,1\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{polylog}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\right)}{\textcolor[rgb]{0,0,1}{z}}
\mathrm{LerchPhi}\left(1,z,1\right)
\textcolor[rgb]{0,0,1}{\mathrm{\zeta }}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)
\mathrm{diff}\left(\mathrm{LerchPhi}\left(z,3,4\right),z\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{LerchPhi}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)}{\textcolor[rgb]{0,0,1}{z}}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{LerchPhi}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)}{\textcolor[rgb]{0,0,1}{z}}
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Frequency (statistics) - Wikipedia
(Redirected from Frequency distribution)
Number of occurrences in an experiment or study
In statistics, the frequency (or absolute frequency) of an event
{\displaystyle i}s the number
{\displaystyle n_{i}}
of times the observation occurred/recorded in an experiment or study.[1]: 12–19 These frequencies are often depicted graphically or in tabular form.
2 Depicting Frequency distributions
2.5 Joint frequency distributions
The cumulative frequency is the total of the absolute frequencies of all events at or below a certain point in an ordered list of events.[1]: 17–19
The relative frequency (or empirical probability) of an event is the absolute frequency normalized by the total number of events:
{\displaystyle f_{i}={\frac {n_{i}}{N}}={\frac {n_{i}}{\sum _{j}n_{j}}}.}
{\displaystyle f_{i}}
for all events
{\displaystyle i}
can be plotted to produce a frequency distribution.
{\displaystyle n_{i}=0}
{\displaystyle i}
, pseudocounts can be added.
Depicting Frequency distributions[edit]
A frequency distribution shows us a summarized grouping of data divided into mutually exclusive classes and the number of occurrences in a class. It is a way of showing unorganized data notably to show results of an election, income of people for a certain region, sales of a product within a certain period, student loan amounts of graduates, etc. Some of the graphs that can be used with frequency distributions are histograms, line charts, bar charts and pie charts. Frequency distributions are used for both qualitative and quantitative data.
Histogram of travel time (to work), US 2000 census.
Bar chart, with 'Country' as the categorical variable for the discrete data set.
Pie chart of world population by country
Different ways of depicting frequency distributions
Decide the number of classes. Too many classes or too few classes might not reveal the basic shape of the data set, also it will be difficult to interpret such frequency distribution. The ideal number of classes may be determined or estimated by formula:
{\displaystyle {\text{number of classes}}=C=1+3.3\log n}
(log base 10), or by the square-root choice formula
{\displaystyle C={\sqrt {n}}}
where n is the total number of observations in the data. (The latter will be much too large for large data sets such as population statistics.) However, these formulas are not a hard rule and the resulting number of classes determined by formula may not always be exactly suitable with the data being dealt with.
Calculate the range of the data (Range = Max – Min) by finding the minimum and maximum data values. Range will be used to determine the class interval or class width.
Decide the width of the classes, denoted by h and obtained by
{\displaystyle h={\frac {\text{range}}{\text{number of classes}}}}
(assuming the class intervals are the same for all classes).
Generally the class interval or class width is the same for all classes. The classes all taken together must cover at least the distance from the lowest value (minimum) in the data to the highest (maximum) value. Equal class intervals are preferred in frequency distribution, while unequal class intervals (for example logarithmic intervals) may be necessary in certain situations to produce a good spread of observations between the classes and avoid a large number of empty, or almost empty classes.[2]
Decide the individual class limits and select a suitable starting point of the first class which is arbitrary; it may be less than or equal to the minimum value. Usually it is started before the minimum value in such a way that the midpoint (the average of lower and upper class limits of the first class) is properly[clarification needed] placed.
Take an observation and mark a vertical bar (|) for a class it belongs. A running tally is kept till the last observation.
Find the frequencies, relative frequency, cumulative frequency etc. as required.
The following are some commonly used methods of depicting frequency:[3]
A histogram is a representation of tabulated frequencies, shown as adjacent rectangles or squares (in some of situations), erected over discrete intervals (bins), with an area proportional to the frequency of the observations in the interval. The height of a rectangle is also equal to the frequency density of the interval, i.e., the frequency divided by the width of the interval. The total area of the histogram is equal to the number of data. A histogram may also be normalized displaying relative frequencies. It then shows the proportion of cases that fall into each of several categories, with the total area equaling 1. The categories are usually specified as consecutive, non-overlapping intervals of a variable. The categories (intervals) must be adjacent, and often are chosen to be of the same size.[4] The rectangles of a histogram are drawn so that they touch each other to indicate that the original variable is continuous.[5]
Bar graphs[edit]
A bar chart or bar graph is a chart with rectangular bars with lengths proportional to the values that they represent. The bars can be plotted vertically or horizontally. A vertical bar chart is sometimes called a column bar chart.
Frequency distribution table[edit]
A frequency distribution table is an arrangement of the values that one or more variables take in a sample. Each entry in the table contains the frequency or count of the occurrences of values within a particular group or interval, and in this way, the table summarizes the distribution of values in the sample.
This is an example of a univariate (=single variable) frequency table. The frequency of each response to a survey question is depicted.
A different tabulation scheme aggregates values into bins such that each bin encompasses a range of values. For example, the heights of the students in a class could be organized into the following frequency table.
less than 5.0 feet 25 25
Joint frequency distributions[edit]
Bivariate joint frequency distributions are often presented as (two-way) contingency tables:
Two-way contingency table with marginal frequencies
The total row and total column report the marginal frequencies or marginal distribution, while the body of the table reports the joint frequencies.[6]
Under the frequency interpretation of probability, it is assumed that as the length of a series of trials increases without bound, the fraction of experiments in which a given event occurs will approach a fixed value, known as the limiting relative frequency.[7][8]
This interpretation is often contrasted with Bayesian probability. In fact, the term 'frequentist' was first used by M. G. Kendall in 1949, to contrast with Bayesians, whom he called "non-frequentists".[9][10] He observed
3....we may broadly distinguish two main attitudes. One takes probability as 'a degree of rational belief', or some similar idea...the second defines probability in terms of frequencies of occurrence of events, or by relative proportions in 'populations' or 'collectives'; (p. 101)
12. It might be thought that the differences between the frequentists and the non-frequentists (if I may call them such) are largely due to the differences of the domains which they purport to cover. (p. 104)
Statistical hypothesis testing is founded on the assessment of differences and similarities between frequency distributions. This assessment involves measures of central tendency or averages, such as the mean and median, and measures of variability or statistical dispersion, such as the standard deviation or variance.
A frequency distribution is said to be skewed when its mean and median are significantly different, or more generally when it is asymmetric. The kurtosis of a frequency distribution is a measure of the proportion of extreme values (outliers), which appear at either end of the histogram. If the distribution is more outlier-prone than the normal distribution it is said to be leptokurtic; if less outlier-prone it is said to be platykurtic.
Letter frequency distributions are also used in frequency analysis to crack ciphers, and are used to compare the relative frequencies of letters in different languages and other languages are often used like Greek, Latin, etc.
Multiset multiplicity as frequency analog
^ a b Kenney, J. F.; Keeping, E. S. (1962). Mathematics of Statistics, Part 1 (3rd ed.). Princeton, NJ: Van Nostrand Reinhold.
^ Manikandan, S (1 January 2011). "Frequency distribution". Journal of Pharmacology & Pharmacotherapeutics. 2 (1): 54–55. doi:10.4103/0976-500X.77120. ISSN 0976-500X. PMC 3117575. PMID 21701652.
^ Carlson, K. and Winquist, J. (2014) An Introduction to Statistics. SAGE Publications, Inc. Chapter 1: Introduction to Statistics and Frequency Distributions
^ Howitt, D. and Cramer, D. (2008) Statistics in Psychology. Prentice Hall
^ Charles Stangor (2011) "Research Methods For The Behavioral Sciences". Wadsworth, Cengage Learning. ISBN 9780840031976.
^ Earliest Known Uses of Some of the Words of Probability & Statistics
^ Kendall, Maurice George (1949). "On the Reconciliation of Theories of Probability". Biometrika. Biometrika Trust. 36 (1/2): 101–116. doi:10.1093/biomet/36.1-2.101. JSTOR 2332534.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Frequency_(statistics)&oldid=1087807143"
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Title: Quantum Matrix Algebras and their applications
Quantum Matrix Algebras are very interesting objects from algebraic viewpoint. Particular examples of these algebras are related to Drinfeld-Jimbo Quantum Groups. Some of these QMA admit defining analogs of partial derivatives. In a limit it is possible to develop a new calculus on the enveloping algebras
{\displaystyle U(gl(N))}
Slides: independ20220330.pdf
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Understanding Excess Returns
Excess Return vs. Riskless Rates
Excess Return & Risk Concepts
Excess Return & Optimal Portfolios
Excess returns are returns achieved above and beyond the return of a proxy. Excess returns will depend on a designated investment return comparison for analysis. Some of the most basic return comparisons include a riskless rate and benchmarks with similar levels of risk to the investment being analyzed.
Excess returns are returns achieved above and beyond the return of a proxy.
Excess returns will depend on a designated investment return comparison for analysis.
The riskless rate and benchmarks with similar levels of risk to the investment being analyzed are commonly used in calculating excess return.
Alpha is a type of excess return metric that focuses on performance return in excess of a closely comparable benchmark.
Excess return is an important consideration when using modern portfolio theory which seeks to invest with an optimized portfolio.
Excess returns are an important metric that helps an investor to gauge performance in comparison to other investment alternatives. In general, all investors hope for positive excess return because it provides an investor with more money than they could have achieved by investing elsewhere.
Excess return is identified by subtracting the return of one investment from the total return percentage achieved in another investment. When calculating excess return, multiple return measures can be used. Some investors may wish to see excess return as the difference in their investment over a risk-free rate.
Other times, excess return may be calculated in comparison to a closely comparable benchmark with similar risk and return characteristics. Using closely comparable benchmarks is a return calculation that results in an excess return measure known as alpha.
In general, return comparisons may be either positive or negative. Positive excess return shows that an investment outperformed its comparison, while a negative difference in returns occurs when an investment underperforms. Investors should keep in mind that purely comparing investment returns to a benchmark provides an excess return that does not necessarily take into consideration all of the potential trading costs of a comparable proxy.
For example, using the S&P 500 as a benchmark provides an excess return calculation that does not typically take into consideration the actual costs required to invest in all 500 stocks in the Index or management fees for investing in an S&P 500 managed fund.
Riskless and low risk investments are often used by investors seeking to preserve capital for various goals. U.S. Treasuries are typically considered the most basic form of riskless securities. Investors can buy U.S. Treasuries with maturities of one month, two months, three months, six months, one year, two years, three years, five years, seven years, 10-years, 20-years, and 30-years.
Each maturity will have a different expected return found along the U.S. Treasury yield curve. Other types of low risk investments include certificates of deposits, money market accounts, and municipal bonds.
Investors can determine excess return levels based on comparisons to risk free securities. For example, if the one year Treasury has returned 2.0% and the technology stock Meta (formerly Facebook) has returned 15%, then the excess return achieved for investing in Meta is 13%.
Oftentimes, an investor will want to look at a more closely comparable investment when determining excess return. That’s where alpha comes in. Alpha is the result of a more narrowly focused calculation that includes only a benchmark with comparable risk and return characteristics to an investment. Alpha is commonly calculated in investment fund management as the excess return a fund manager achieves over a fund’s stated benchmark.
Broad stock return analysis may look at alpha calculations in comparison to the S&P 500 or other broad market indexes like the Russell 3000. When analyzing specific sectors, investors will use benchmark indexes that include stocks in that sector. The Nasdaq 100 for example can be a good alpha comparison for large cap technology.
In general, active fund managers seek to generate some alpha for their clients in excess of a fund’s stated benchmark. Passive fund managers will seek to match the holdings and return of an index.
Consider a large-cap U.S. mutual fund that has the same level of risk as the S&P 500 index. If the fund generates a return of 12% in a year when the S&P 500 has only advanced 7%, the difference of 5% would be considered as the alpha generated by the fund manager.
Excess Return vs. Risk Concepts
As discussed, an investor has the opportunity to achieve excess returns beyond a comparable proxy. However the amount of excess return is usually associated with risk. Investment theory has determined that the more risk an investor is willing to take the greater their opportunity for higher returns. As such, there are several market metrics that help an investor to understand if the returns and excess returns they achieve are worthwhile.
Beta is a risk metric quantified as a coefficient in regression analysis that provides the correlation of an individual investment to the market (usually the S&P 500). A beta of one means that an investment will experience the same level of return volatility from systematic market moves as a market index.
A beta above one indicates that an investment will have higher return volatility and therefore higher potential for gains or losses. A beta below one means an investment will have less return volatility and therefore less movement from systematic market effects with less potential for gain but also less potential for loss.
Beta is an important metric used when generating an Efficient Frontier graph for the purposes of developing a Capital Allocation Line which defines an optimal portfolio. Asset returns on an Efficient Frontier are calculated using the following Capital Asset Pricing Model:
\begin{aligned} &R_a = R_{rf} + \beta \times (R_m - R_{rf}) \\ &\textbf{where:} \\ &R_a = \text{Expected return on a security} \\ &R_{rf} = \text{Risk-free rate} \\ &R_m = \text{Expected return of the market} \\ &\beta = \text{Beta of the security} \\ &R_m - R_{rf} = \text{Equity market premium} \\ \end{aligned}
Ra=Rrf+β×(Rm−Rrf)where:Ra=Expected return on a securityRrf=Risk-free rateRm=Expected return of the marketβ=Beta of the securityRm−Rrf=Equity market premium
Beta can be a helpful indicator for investors when understanding their excess return levels. Treasury securities have a beta of approximately zero. This means that market changes will have no effect on the return of a Treasury and the 2.0% earned from the one year Treasury in the example above is riskless.
Meta on the other hand has a beta of approximately 1.29 so systematic market moves that are positive will lead to a higher return for Meta than the S&P 500 Index overall and vice versa.
In active management, fund manager alpha can be used as a metric for evaluating the performance of a manager overall. Some funds provide their managers a performance fee which offers extra incentive for fund managers to exceed their benchmarks. In investments there is also a metric known as Jensen’s Alpha. Jensen’s Alpha seeks to provide transparency around how much of a manager’s excess return was related to risks beyond a fund’s benchmark.
Jensen’s Alpha is calculated by:
\begin{aligned} &\text{Jensen's Alpha} = R_i - (R_f + \beta (R_m - R_f)) \\ &\textbf{where:} \\ &R_i = \text{Realized return of the portfolio or investment} \\ &R_f = \text{Risk-free rate of return for the time period} \\ &\beta = \text{Beta of the portfolio of investment} \\ &\text{with respect to the chosen market index} \\ &R_m = \text{Realized return of the appropriate market index} \\ \end{aligned}
Jensen’s Alpha=Ri−(Rf+β(Rm−Rf))where:Ri=Realized return of the portfolio or investmentRf=Risk-free rate of return for the time periodβ=Beta of the portfolio of investmentwith respect to the chosen market indexRm=Realized return of the appropriate market index
A Jensen’s Alpha of zero means that the alpha achieved exactly compensated the investor for the additional risk taken on in the portfolio. A positive Jensen’s Alpha means the fund manager overcompensated its investors for the risk and a negative Jensen’s Alpha would be the opposite.
In fund management, the Sharpe Ratio is another metric that helps an investor understand their excess return in terms of risk.
The Sharpe Ratio is calculated by:
\begin{aligned} &\text{Sharpe Ratio} = \frac{ R_p - R_f }{ \text{Portfolio Standard Deviation} } \\ &\textbf{where:} \\ &R_p = \text{Portfolio return} \\ &R_f = \text{Riskless rate} \\ \end{aligned}
Sharpe Ratio=Portfolio Standard DeviationRp−Rfwhere:Rp=Portfolio returnRf=Riskless rate
The higher the Sharpe Ratio of an investment the more an investor is being compensated per unit of risk. Investors can compare Sharpe Ratios of investments with equal returns to understand where excess return is more prudently being achieved. For example, two funds have a one year return of 15% with a Sharpe Ratio of 2 vs. 1. The fund with a Sharpe Ratio of 2 is producing more return per one unit of risk.
Critics of mutual funds and other actively managed portfolios contend that it is next to impossible to generate alpha on a consistent basis over the long term, as a result investors are then theoretically better off investing in stock indexes or optimized portfolios that provide them with a level of expected return and a level of excess return over the risk free rate.
This helps to make the case for investing in a diversified portfolio that is risk optimized to achieve the most efficient level of excess return over the risk free rate based on risk tolerance.
This is where the Efficient Frontier and Capital Market Line can come in. The Efficient Frontier plots a frontier of returns and risk levels for a combination of asset points generated by the Capital Asset Pricing Model. An Efficient Frontier considers data points for every available investment an investor may wish to consider investing in. Once an efficient frontier is graphed, the capital market line is drawn to touch the efficient frontier at its most optimal point.
With this portfolio optimization model developed by financial academics, an investor can choose a point along the capital allocation line for which to invest based on their risk preference. An investor with zero risk preference would invest 100% in risk free securities.
The highest level of risk would invest 100% in the combination of assets suggested at the intersect point. Investing 100% in the market portfolio would provide a designated level of expected return with excess return serving as the difference from the risk-free rate.
As illustrated from the Capital Asset Pricing Model, Efficient Frontier, and Capital Allocation Line, an investor can choose the level of excess return they wish to achieve above the risk free rate based on the amount of risk they wish to take on.
Treasury Direct. "Treasury Notes." Accessed Dec. 3, 2021.
TreasuryDirect. "Treasury Securities & Programs." Accessed Dec. 3, 2021.
U.S. Department of the Treasury. "Daily Treasury Yield Curve Rates." Accessed Dec. 3, 2021.
Zacks. "Meta Platforms, Inc. (FB)." Accessed Dec. 3, 2021.
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Online exam in Data representation | Online Exam
Online exam in Data representation,Computer for Competitive exams like UPSC & State Level PSCs Exams, SSC, Railways, IBPS, SBI & Its Associates, RBI, Bank PO/MT, Clerical, RRBs, MBA, MCA, BBA, BCA, BBS
A hexadecimal number is represented by
Decimal number system is the group of _____ numbers
Hexadecimal number system has _____ base
Hexadecimal number system consists of
1 gigabyte is equivalent to 1024 kilobytes
Octal number system has 8 digits
The number system based on 0 and 1 only,is known as
Which of the following is the octal number equivalent to binary number
{\left(110101\right)}_{2}
Which of the following is hexadecimal number equivalent to binary number
{\left(1111 1001\right)}_{2}?
Which of the following is an octal number equal to decimal number
{\left(896\right)}_{10}?
Which of the following is invalid hexadecimal number?
Which of the following is a hexadecimal number equal to 3431 octal number?
There are how many types of number system ?
How many values can be represented by single byte?
Modern computers represents characters and numbers internally using one of the following number system ?
Which of the following is not a computer code?
The coding system allows non-english characters and special characters to be represented
MSD refers as
Most Significant Digits
Many Significant Digits
Multiple Significant Digits
Most Significant Decimal
Binary system is also called as
base one system
The negative number in the binary system can be represented by
single magnitude
Today’s mostly used coding system is/are
{\left(1010\right)}_{2} equivalent decimal number is
Data representation is base on the ____ number system,which uses two numbers to represent all data
Most commonly used codes for representing bits are
Previous articleDesign Tools and Programming Languages -test 1
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Polynomial satisfying p(x)=3^{x} for x\in\mathbb{N}
haguemarineo6h 2022-04-25 Answered
Polynomial satisfying
p\left(x\right)={3}^{x}
x\in \mathbb{N}
p\left(x+1\right)=3p\left(x\right)
\mathbb{N}⇒p\left(x+1\right)=3p\left(x\right)⇒p\left(-1\right)=\frac{1}{3}
p\left(x\right)\in \mathbb{Z}\left[x\right]
Or, continuing,
\text{ }p\left(-n\right)={3}^{-n}:⇒\text{ }p\left(x\right):p\left(-x\right)=1
\mathrm{ℕ}\text{ }⇒\text{ }p\left(-x\right):p\left(x\right)=1
contra degree comparison. This proof works over much more general coefficient rings, e.g.
\mathbb{Q}
bobthemightyafm
{a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\cdots +{a}_{0}={3}^{x}
x\in \mathbb{N}
{a}_{n}{\left(x+1\right)}^{n}+{a}_{n-1}{\left(x+1\right)}^{n-1}+\cdots +{a}_{0}=3{a}_{n}{x}^{n}+3{a}_{n-1}{x}^{n-1}\cdot +\cdots +3{a}_{0}
But the coefficient of
{x}^{n}
on the LHS is
{a}_{n}
and on the RHS it is 3an.
This contradicts the assumption.
Show that an intersection of normal subgroups of a group G is again a normal subgroup of G
H\le G
{x}^{-1}{y}^{-1}xy\in H\mathrm{\forall }x
y\in G⇔H⊴G
\frac{G}{H}
Z×5
Z×8
is not a cyclic group.
Finding the Units in the Ring
\mathbb{Z}\left[t\right]\left[\sqrt{{t}^{2}-1}\right]
Use the isomorphism theorem to determine the group
G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)
G{L}_{2}\left(\mathbb{R}\right)
is the group of
2×2
matrices with determinant not equal to 0, and
S{L}_{2}\left(\mathbb{R}\right)
2×2
matrices with determinant 1. In the first part of the problem, I proved that
S{L}_{2}\left(\mathbb{R}\right)
G{L}_{2}\left(\mathbb{R}\right)
. Now it wants me to use the isomorphism theorem. I tried using
|G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)|=\frac{|G{L}_{2}\left(\mathbb{R}\right)|}{|S{L}_{2}\left(\mathbb{R}\right)|},
but since both groups have infinite order, I don't think I can use this here.
\frac{\mathbb{Z}}{p}\mathbb{Z}
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Quantitative Aptitude Exam papers | Online Exam
Home Quantitative Aptitude Mensuration Mensuration-Test 1
Mensuration-Test 1
Chapter :- Mensuration-Test 1
The perimeter of a rectangle is 42 m. If the area of the square formed on the diagonal
of the rectangle as its side is 1 1/12 % more than the area of the rectangle, find the
longer side of the rectangle.
At the rate of Rs. 2 per sq m, cost of painting a rectangular floor is Rs 5760. If the length of the floor is 80% more than its breadth, then what is the length of the floor?
Let the length and the breadth of the floor
be l m and b m respectively.
l = b + 80% of b = l + 0.8 b = 1.8b
Area of the floor = 5760/2 = 2880 sq m
l*b = 2880 i.e., l * l/1.8 = 2880
A 7 m wide path is to be made around a circular garden having a diameter of 7 m.
What will be the area of the path in square metre?
Area of the path = Area of the outer circle –
Area of the inner circle = π{7/2 + 7}2 – π[7/2]2
The perimeter of a rectangle of length 62 cm and breadth 50 cm is four times
perimeter of a square. What will be the circumference of a semicircle whose
diameter is equal to the side of the given square?
Parameter of the rectangle = 2(62 + 50) =
224 cm Parameter of the square = 56 cm
Diameter, d of the semicircle = 14 cm
Circumference of the semicircle = 1/2(π)(r)+ d
= 1/2(22/7)(7) + 14 = 25 cm
A cone with diameter of its base as 30 cm is formed by melting a spherical ball of
diameter 10 cm. What is the approximate height of the cone?
Radius of cone = 30/2 = 15, radius of ball = 10/2 = 5
Volumes will be equal, so
\left(1/3\right){\mathrm{\pi r}}^{2}\mathrm{h}=\left(4/3\right){\mathrm{\pi r}}^{3}\phantom{\rule{0ex}{0ex}}{15}^{2}\mathrm{h}=4*53\phantom{\rule{0ex}{0ex}}\mathrm{so} \mathrm{h}=2.2
\left(1/3\right){\mathrm{\pi r}}^{2}\mathrm{h}=\left(4/3\right){\mathrm{\pi r}}^{3}\phantom{\rule{0ex}{0ex}}{15}^{2}\mathrm{h}=4*53\phantom{\rule{0ex}{0ex}}\mathrm{so} \mathrm{h}=2.2
A cylinder whose base of circumference is 6 m can roll at a rate of 3 rounds per second. How much distance will the cylinder cover in 9 seconds?
Distance covered in one round = 2 x π x r = 6 m
Distance covered in 1 second = 3 x 6 = 18m
So distance covered in 9 seconds = 18×9= 162 m
A container is formed by surmounting a hemisphere on a right circular cylinder of same radius as that of hemisphere. If the volume of the container is
576{\mathrm{\pi m}}^{3}
and radius of cylinder is 6 m, then find the height of the container.
Volume of the container = Volume of the cylinder + Volume of the hemisphere
Volume of the container =
\mathrm{\pi }{6}^{2}\mathrm{h}+\left(2/3\right)\mathrm{\pi }{6}^{3}=576\mathrm{\pi }
= π 36 (h + 4) = 576π
Solving we get h = 12
So the height of the container = 12 + 6 = 18m
\mathrm{\pi }{6}^{2}\mathrm{h}+\left(2/3\right)\mathrm{\pi }{6}^{3}=576\mathrm{\pi }
The radii of two cylinders are in the ratio 3: 2 and their curved surface areas are in the
ratio 3 : 5. What is the ratio of their volumes?
A right circular cone is exactly fitted inside a cube in such a way that the edges of the
base of the cone are touching the edge of one of the faces of the cube and the vertex is on the opposite face of the cube. If the volumes of cube is 216 cm3 , what is the volume of the cone (approximately)?
r= 3 cm; height of the cone= 6cm (as it is
fitted in this cube of side 6 cm, hence its
height will also be 6 cm)
Volume of cone= 1/3 π*r2 * h =56
All have equal area
In such case the area in descending order is: Circle> Square> Rectangle
A cylinder has some water at height 20 cm. If a sphere of radius 6 cm is poured into it
then find the rise in height of water if the radius of cylinder is 4 cm.
Volume of ball= volume of rising water in
4/3*{\mathrm{\pi r}}^{3}=\mathrm{\pi }*{\mathrm{r}}^{2}*\mathrm{h}
4/3*{\mathrm{\pi r}}^{3}=\mathrm{\pi }*{\mathrm{r}}^{2}*\mathrm{h}
A sphere of 5 cm radius is melted and small sphere of radius 1 cm is made from it. Find
the number of sphere that can be made from it.
If radius of cone decrease by 50% and height increase by 20%. Then find the
percentage change in the volume.A) 70% decrease
The parameter of a square is equal to the perimeter of a rectangle of length 14 cm and
breadth 20 cm. Find the circumference of a semicircle (approx.) whose diameter is equal to the side of the square.
So circumference of a semicircle = πr = 22/7 *17/2 = 27 cm
So circumference of smaller circle = 2 * 22/7 * 9= 56.5 cm
The barrel of a fountain pen is cylindrical in shape which radius of base as 0.7 cm and is 5
cm long. One such barrel in the pen can be used to write 300 words. A barrel full of ink which has a capacity of 14 cu cm can be used to write how many words approximately?
Volume of the barrel of pen = πr2h = 22/7 *0.7*0.7 * 5 = 7.7 cu cm
So which has capacity 14 cu cm can write =
300/7.7 * 14 = 545 words
A car has wheels of diameter 70 m. How many revolutions can the wheel complete in 20
minutes if the car is travelling at a speed of 110 m/s?
Distance travelled in one revolution = 2πr = 2 *22/7 * 35 = 220 cm
So total distance travelled = distance travelled in
one revolution * number of revolutions
The diameters of the internal and external surfaces of a hollow spherical shell are 10cm
and 6 cm respectively. If it is melted and recasted into a solid cylinder of length 8/3 cm,
find the diameter of the cylinder.
C.S.A=1/2*(perimeter of base)*l20+10+(20*10)/100=32%
A man wants to make small sphere of size 1 cm of radius from a large sphere of size
of 6 cm of radius. Find out how many such sphere can be made?
Volume of sphere1/volume of sphere 2=required number of sphere
from height 20 cm to upwards.
The length of the perpendicular drawn from any point in the interior of an equilateral
triangle to the respective sides are P1, P2 and P3. Find the length of each side of the
2/√3 *(P1 + P2 + P3)
1/3 * (P1 + P2 + P3)
Assume that a drop of water is spherical and its diameter is one tenth of a cm. A
conical glass has equal height to its diameter of rim. If 2048000 drops of water
fill the glass completely then find the height of the glass.
diameter of drop of water=1/10 =>
radius=1/20
volume of 204800 drop of
water=204800*4/3* π*1/20 *1/20*1/20 =
1024 π/3
Volume of cone=1024 π/3 = 1/3 * π *r2 * h
(r=h/2)
If the radius of a sphere increase by 4 cm then the surface area increase by 704 cm2 .
The radius of the sphere initially was?
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Revision as of 09:08, 29 July 2013 by NikosA (talk | contribs) (→Automatising Conversions: units="..." was too long in the script)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
{\displaystyle [0,255]}
{\displaystyle [0,2047]}
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===== in Python =====
A custom python script, performing the operations of interest, might be like [https://github.com/NikosAlexandris/i.quickbird.toar i.quickbird.toar (for GRASS 7.x)]
===== in bash =====
Am example bash shell script, might look like the one that follows. '''Note,''' however, constants, band parameters and acquisition related metadata are hard-coded! Reviewing the code and altering appropriately is required, i.e. checking for the parameters <code>ESD</code>, <code>SEA</code>, <code>PanTDI</code>, <code>K_BAND</code>.
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L_{\lambda {\text{Pixel, Band}}}={\frac {K_{\text{Band}}*q_{\text{Pixel, Band}}}{\Delta \lambda _{\text{Band}}}}}
{\displaystyle L_{\lambda {\text{Pixel,Band}}}}
{\displaystyle K_{\text{Band}}}
{\displaystyle q_{\text{Pixel,Band}}}
{\displaystyle \Delta _{\lambda _{\text{Band}}}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
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Combine terms of identical algebraic structure - MATLAB combine - MathWorks Deutschland
Powers of the Same Base
Powers of the Same Exponent
Terms with Logarithms
Terms with Sine and Cosine Function Calls
Terms with Integrals
Terms with Inverse Tangent Function Calls
Terms with Calls to Gamma Function
Multiple Input Expressions in One Call
Combine terms of identical algebraic structure
Y = combine(S) rewrites products of powers in the expression S as a single power.
Y = combine(S,T) combines multiple calls to the target function T in the expression S. Use combine to implement the inverse functionality of expand with respect to the majority of the applied rules.
Y = combine(___,'IgnoreAnalyticConstraints',true) simplifies the output by applying common mathematical identities, such as log(a) + log(b) = log(a*b). These identities might not be valid for all values of the variables, but applying them can return simpler results.
Combine powers of the same base.
Combine powers of numeric arguments. To prevent MATLAB® from evaluating the expression, use sym to convert at least one numeric argument into a symbolic value.
Here, sym converts 1 into a symbolic value, preventing MATLAB from evaluating the expression e1.
Combine powers with the same exponents in certain cases.
combine does not usually combine the powers because the internal simplifier applies the same rules in the opposite direction to expand the result.
Combine terms with logarithms by specifying the target argument as log. For real positive numbers, the logarithm of a product equals the sum of the logarithms of its factors.
Try combining log(a) + log(b). Because a and b are assumed to be complex numbers by default, the rule does not hold and combine does not combine the terms.
Apply the rule by setting assumptions such that a and b satisfy the conditions for the rule.
For future computations, clear the assumptions set on variables a and b by recreating them using syms.
Alternatively, apply the rule by ignoring analytic constraints using 'IgnoreAnalyticConstraints'.
Rewrite products of sine and cosine functions as a sum of the functions by setting the target argument to sincos.
Rewrite sums of sine and cosine functions by setting the target argument to sincos.
Rewrite a cosine squared function by setting the target argument to sincos.
combine does not rewrite powers of sine or cosine functions with negative integer exponents.
Combine terms with exponents by specifying the target argument as exp.
Combine terms with integrals by specifying the target argument as int.
Combine integrals with the same limits.
Combine two calls to the inverse tangent function by specifying the target argument as atan.
Combine two calls to the inverse tangent function. combine simplifies the expression to a symbolic value if possible.
For further computations, clear the assumptions:
Combine multiple gamma functions by specifying the target as gamma.
combine simplifies quotients of gamma functions to rational expressions.
Evaluate multiple expressions in one function call by using a symbolic matrix as the input parameter.
S — Input expression
symbolic expression | symbolic vector | symbolic matrix | symbolic function
Input expression, specified as a symbolic expression, function, or as a vector or matrix of symbolic expressions or functions.
combine works recursively on subexpressions of S.
If S is a symbolic matrix, combine is applied to all elements of the matrix.
T — Target function
Target function, specified as 'atan', 'exp', 'gamma', 'int', 'log', 'sincos', or 'sinhcosh'. The rewriting rules apply only to calls to the target function.
Y — Expression with combined functions
symbolic variable | symbolic number | symbolic expression | symbolic vector | symbolic matrix
Expression with the combined functions, returned as a symbolic variable, number, expression, or as a vector or matrix of symbolic variables, numbers, or expressions.
combine applies the following rewriting rules to the input expression S, depending on the value of the target argument T.
When T = 'exp', combine applies these rewriting rules where valid,
{e}^{a}{e}^{b}={e}^{a+b}
{\left({e}^{a}\right)}^{b}={e}^{ab}.
When T = 'log',
\mathrm{log}\left(a\right)+\mathrm{log}\left(b\right)=\mathrm{log}\left(ab\right).
If b < 1000,
b\mathrm{log}\left(a\right)=\mathrm{log}\left({a}^{b}\right).
When b >= 1000, combine does not apply this second rule.
The rules applied to rewrite logarithms do not hold for arbitrary complex values of a and b. Specify appropriate properties for a or b to enable these rewriting rules.
When T = 'int',
a\int f\left(x\right)dx=\int af\left(x\right)dx
\int f\left(x\right)dx+\int g\left(x\right)dx=\int f\left(x\right)+g\left(x\right)dx
{\int }_{a}^{b}f\left(x\right)dx+{\int }_{a}^{b}g\left(x\right)dx={\int }_{a}^{b}f\left(x\right)+g\left(x\right)dx
{\int }_{a}^{b}f\left(x\right)dx+{\int }_{a}^{b}g\left(y\right)dy={\int }_{a}^{b}f\left(y\right)+g\left(y\right)dy
{\int }_{a}^{b}yf\left(x\right)dx+{\int }_{a}^{b}xg\left(y\right)dy={\int }_{a}^{b}yf\left(c\right)+xg\left(c\right)dc.
When T = 'sincos',
\mathrm{sin}\left(x\right)\mathrm{sin}\left(y\right)=\frac{\mathrm{cos}\left(x-y\right)}{2}-\frac{\mathrm{cos}\left(x+y\right)}{2}.
combine applies similar rules for sin(x)cos(y) and cos(x)cos(y).
A\mathrm{cos}\left(x\right)+B\mathrm{sin}\left(x\right)=A\sqrt{1+\frac{{B}^{2}}{{A}^{2}}}\mathrm{cos}\left(x+{\mathrm{tan}}^{-1}\left(\frac{-B}{A}\right)\right).
When T = 'atan' and -1 < x < 1, -1 < y < 1,
\mathrm{atan}\left(x\right)+\mathrm{atan}\left(y\right)=\mathrm{atan}\left(\frac{x+y}{1-xy}\right).
When T = 'sinhcosh',
\mathrm{sinh}\left(x\right)\mathrm{sinh}\left(y\right)=\frac{\mathrm{cosh}\left(x+y\right)}{2}-\frac{\mathrm{cosh}\left(x-y\right)}{2}.
combine applies similar rules for sinh(x)cosh(y) and cosh(x)cosh(y).
combine applies the previous rules recursively to powers of sinh and cosh with positive integral exponents.
When T = 'gamma',
a\Gamma \left(a\right)=\Gamma \left(a+1\right).
\frac{\Gamma \left(a+1\right)}{\Gamma \left(a\right)}=a.
For positive integers n,
\Gamma \left(-a\right)\Gamma \left(a\right)=-\frac{\pi }{\mathrm{sin}\left(\pi a\right)}.
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Revision as of 09:06, 29 July 2013 by NikosA (talk | contribs) (→Deriving Physical Quantities: bash scripted conversions from to DN to Radiances & Reflectances)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
The conversion process ca be scripted to avoid repeating the same steps for each band separately. In bash, such a script might be as the following example. Note, however, in this example script constants, band parameters and acquisition related metadata are hard-coded!
{\displaystyle [0,255]}
{\displaystyle [0,2047]}
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Write a multivariable function (i.e z=f(x,y)) for a linear function that contains the points (−1,0),
Write a multivariable function (i.e z=f(x,y)) for a linear function that contains the points (−1,0),(0,2),(1,−1). Ive
chukizosv
The general form of a linear function of two variables is given by
f\left(x,y\right)=Ax+By+C
Plugging in the points you were given should give you three equations in the three unknowns
\left(A,B,C\right)
, which you can then solve.
P\left(x\right)=-12{x}^{2}+2136x-41000
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
\overline{)\begin{array}{cc}Venture\text{ }A& \\ Earnings& Probability\\ -20& 3\\ 50& 4\\ 50& 3\end{array}}
\overline{)\begin{array}{cc}Venture\text{ }B& \\ Earnings& Probability\\ -15& 2\\ 30& 5\\ 40& 3\end{array}}
Find the variance/covariance matriz for the following data set
\begin{array}{|cccc|}\hline Student& Algebra& Combinatorics& Analysis\\ 1& 90& 80& 90\\ 2& 30& 60& 30\\ 3& 60& 60& 50\\ 4& 90& 60& 30\\ 5& 40& 70& 40\\ \hline\end{array}
f\left(x,y\right)=x\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)-y\mathrm{sin}\left(x\right)+\frac{{y}^{2}}{2}
{f}_{x}^{\prime }=\left(x-y\right)\mathrm{cos}\left(x\right)
{f}_{y}^{\prime }=y-\mathrm{sin}\left(x\right)
x=y
x-y\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)=y
\frac{\pi }{4}\le \theta \le \frac{3\pi }{4},0\le r\le 1
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The symmetric difference of A and B, denoted
The symmetric difference of A and B, denoted by A B, is the set containing those elements in
either A or B, but not in both A and B.
a) Find the symmetric difference o f { 1, 3 , 5 } and { 1, 2, 3 } .
b) Determine whether the symmetric difference is associative; that is, if A, B, and C are sets, does it follow that
A (B C) = (A B) C?
Given: A={ 1, 3 , 5 } and B={ 1, 2, 3 }
The formula of symmetric difference of two sets:
A+B=\left(A\cup B\right)-\left(A\cap B\right)
\left\{1,3,5\right\}\cup \left\{1,2,3\right\}=\left\{1,2,3,5\right\}
\left\{1,3,5\right\}\cap \left\{1,2,3\right\}=\left\{1,3\right\}
\left\{1,3,5\right\}+\left\{1,2,3\right\}=\left\{1,2,3,5\right\}-\left\{1,3\right\}=\left\{2,5\right\}
A-B=\left\{x|x\in A but x\notin B\right\}
n\mid \varphi \left({a}^{n}-1\right)
in "Topics in Algebra 2nd Edition" by I. N. Herstein. Any natural solution that uses
Aut\left(G\right)
The probability of an event to happen in the next week is p and It is constant throughout the week What is the probability of this event to happen in the next 2 days?
\sqrt{ }
A closely wound rectangular coil of 100 turns has dimension 10 cm x 30 cm. In
time t = 0.1s the plane of the coil is rotated from a position where it makes an angle of 60° with
a magnetic field of 1.5 T to a position perpendicular the field. What is the average emf induced
in the coil?
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What is P ( A ∪<!-- ∪ --> ( B ∩<!-- ∩ --> C ) ) ?
P\left(A\cup \left(B\cap C\right)\right)
The question says it all. I know
P\left(A\cap \left(B\cup C\right)\right)=P\left(A\cap B\right)+P\left(A\cap C\right).
Would this mean,
P\left(A\cup \left(B\cap C\right)\right)=P\left(A\cup B\right)+P\left(A\cup C\right)?
P\left(A\cup \left(B\cap C\right)\right)=P\left(A\cup B\right)+P\left(A\cup C\right)
will almost certainly not hold. This is because
P\left(A\cup \left(B\cap C\right)\right)=P\left(\left(A\cup B\right)\cap \left(A\cup C\right)\right)\le P\left(A\cup B\right)
P\left(A\cup \left(B\cap C\right)\right)\le P\left(A\cup C\right).
For the stated equation to hold, both
P\left(A\cup B\right)
P\left(A\cup C\right)
would have to be zero, showing that
P\left(A\right)=P\left(B\right)=P\left(C\right)=0.
However, if this condition is satisfied, then the stated equation will hold.
Okay, so, my teacher gave us this worksheet of "harder/unusual probability questions", and Q.5 is real tough. I'm studying at GCSE level, so it'd be appreciated if all you stellar mathematicians explained it in a way that a 15 year old would understand. Thanks!
So, John has an empty box. He puts some red counters and some blue counters into the box.
The ratio of the number of red counters to blue counters is 1:4
Linda takes out, at random, 2 counters from the box.
The probability that she takes out 2 red counters is 6/155
I came across the following question in a textbook (bear in mind that this is the only information given)-
There is a 50 percent chance of rain today. There is a 60 percent chance of rain tomorrow. There is a 30 percent chance that it will not rain either day. What is the chance that it will rain both today and tomorrow?
My instinct was to multiply the probabilities together for today and tomorrow to arrive at an answer of 30 percent. However, the answer given is 40 percent (based on taking the addition of the individual probabilities and subtracting their union). Can someone explain to me why the multiplication rule does not apply here? Does it have to do with independence? Bear in mind, I am trying to relearn probability theory from scratch. Thanks.
20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?
I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.I started out this problem by thinking about the two scenarios:
1) the first two are yellow
2) the first two are blue
I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?
Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?
I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:
1- [P(no yellow) or P(first two are different)]
I'm studying the Cartesian product, which is bound to the idea of a binary relation. Even with Cartesian products of several sets, n-ary Cartesian products, we have to think combinatorically as two sets at a time, recursively. Is there any sort of operation where there is, say, a triary operation happening. Obviously, when we have
3-2-1
we have to think of associative rules, forcing subtraction to be a binary operator "step-through" affair. There is no
3\oplus 2\oplus 1
operator that does something in one fell swoop to all three numbers is there? Lisp has (+ 1 2 3) and grade-school math has vertical addition with the plus-sign and a line under the lowest number, but these are not "internally" non-binary. The only thing that takes "three at once" is, yes, a product-based operator, again, an n-ary Cartesian product. Correct? Another example would be playing poker and being dealt five cards. The cards were shuffled, which is a combinatoric permutation of the stack of cards, which are then dealt into hands. Is there anything from probability (or anywhere) that doesn't start with a shuffle permutation therefore binary, rather, just looks at the five cards coming together non-permutation-, non-binary-wise?
Getting 2 or 5 in two throws should be
P\left(2\right)+P\left(5\right)
P\left(2\right)=1/6,P\left(5\right)=1/6
so the combined so it should be 1/3.
I tried to visualize but not able to do so correctly.
11,12,13,14,15,16, 21,22,23,24,25,26,31,32, ....6,6
total of 36 possibilities.
out of which 20 possibilities, so the probability should be 20/36 which is not 1/3.
I know that there is the addition rule of probability, but I want to understand the intuition behind it. Specifically, why does OR signifies addition in probability theory?
Alfonso and Colin each bought one raffle ticket at the state fair. If 50 tickets were randomly sold, what is the probability that Alfonso got ticket 14 and Colin got ticket 23?
The answer should be
\frac{1}{2450}
which presumably comes from
\frac{1}{50}×\frac{1}{49}
. But it seems that the order does not count. I did not assume that Alfonso got ticket 14 first then Colin got ticket 23 second.
Update: What is wrong with this reasoning. When I said that I did not assume order, I meant that it's possible
1. Alfonso got ticket 14 first, then Colin got ticket 23,
2. Colin got ticket 23 first, then Alfonso got ticket 14.
Both of these possibilities are possible before the tickets are given out, so we can make an 'or' statement. Label the event Alfonso got ticket 14 by
{A}_{14}
and Colin got ticket 23 by
{A}_{23}
. Then by the addition rule
Pr\left(\text{ (}{A}_{14}\text{ first and }{C}_{23}\text{ second) or (}{C}_{23}\text{ first and }{A}_{14}\text{ second}\right)\right)\phantom{\rule{0ex}{0ex}}=Pr\left({A}_{14}\right)×Pr\left({C}_{23}\mid {A}_{14}\right)+Pr\left({C}_{23}\right)×Pr\left({A}_{14}\mid {C}_{23}\right)=\frac{1}{50}×\frac{1}{49}×2.
I realize that once the tickets are sold, then only one of
\left\{{A}_{14}{C}_{23}\text{ },\text{ }{C}_{23}{A}_{14}\right\}
must occur, but before the tickets are sold both possibilities are plausible. Why would the probability change before and after the tickets are sold.
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IsSupersoluble - Maple Help
Home : Support : Online Help : Mathematics : Group Theory : IsSupersoluble
attempt to determine whether a group is supersoluble
IsSupersoluble( G )
IsSupersolvable( G )
G
is supersoluble if it has a normal series with cyclic quotients. That is, there is a normal series
G={G}_{0}▹{G}_{1}▹\dots ▹{G}_{r}=1
with each subgroup
{G}_{i}
G
, and for which each of the quotients
\frac{{G}_{i}}{{G}_{i+1}}
It follows that every supersoluble group is soluble but, as the examples below illustrate, the converse is not true.
The IsSupersoluble( G ) command attempts to determine whether the finite group G is supersoluble. It returns true if G is supersoluble and returns false otherwise.
The IsSupersolvable( G ) command is provided as an alias.
\mathrm{with}\left(\mathrm{GroupTheory}\right):
\mathrm{IsSupersoluble}\left(\mathrm{DihedralGroup}\left(4\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
4
is soluble, but is not supersoluble.
\mathrm{IsSupersoluble}\left(\mathrm{Alt}\left(4\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{IsSoluble}\left(\mathrm{Alt}\left(4\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
The GroupTheory[IsSupersoluble] command was introduced in Maple 2019.
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Marmo G. A Manifold view point of Lie Algebras (abstract) - Geometry of Differential Equations
Marmo G. A Manifold view point of Lie Algebras (abstract)
Title: A Manifold view point of Lie Algebras
Lie algebra brackets play a prominent role in the description of evolution (equations of motion) for any physical system, be it classical or quantum. On the other hand, the advent of general relativity has called for a description of physical systems in a coordinate independent manner. Special relativity has introduced the need for a composition law for velocities whose expression
{\displaystyle (v+w)/[1+(vw/c^{2})]}
is alternative to the usual Galilean one
{\displaystyle v+w}
, which appears to be a "contraction" of previous one when
{\displaystyle c}
, the speed of light, goes to infinity. The composition law of special relativity is not Archimedean and is not associative in usual space-time (1+3). However in 1+1 space-time it defines an alternative "linear structure".
What we learn from these observations is that from the point of view of Physics it would be convenient to have a "tensorial" presentation of Lie algebras where not only the binary, bilinear product is represented by a tensor but the linear structure itself be represented by a tensor field. These aspects have been tackled in some previous papers by our research group and, as we shall show, have a strong relation with some of latest published papers by Alexander Vinogradov on the classification of finite dimensional real Lie algebras [Particle-like structure of Lie algebras].
Slides: MarmoAMVconf2021slides.pdf
Retrieved from "https://gdeq.org/w/index.php?title=Marmo_G._A_Manifold_view_point_of_Lie_Algebras_(abstract)&oldid=6509"
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This question concerns the vector field F = −xy2
This question concerns the vector fieldF = −xy2 i − x2y j + z(x2 + y2) k.(a) Calculate the divergence of F. Use the divergence theorem to show thatthe flux of F over any closed surface is equal to zero. (b) A cylinder of height h and radius R has its base in the xy-plane and itsaxis of symmetry along the z-axis, as shown in the diagram below.Calculate the flux of F over the curved portion S of the surface of thecylinder by using the result of part (a) to relate the flux of F over S tothe flux of F over the flat top surface T and the flux of F over the flatbottom surface B.
Your solution is below, rate my answer please
Equation of the line tangent to
\left\{\begin{array}{l}y={x}^{2}\\ {z}^{2}=16-y\end{array}\right\
, at the point (4,16,0) ?
\left\{\begin{array}{l}4x=y\\ z=0\end{array}\right\
\left\{\begin{array}{l}x=4\\ y=16\end{array}\right\
\left\{\begin{array}{l}y-x=12\\ z=0\end{array}\right\
\left\{\begin{array}{l}x+y=20\\ z=0\end{array}\right\
use squeeze theorem find the following lim x,y-0,0 y^2(1-cos2x) /x^4 +y^2
U
is a random variable that follows Uniform distribution with interval
\left(0,1\right). {X}_{1},...,{X}_{n}
follows Bernoulli distribution with mean
U
. How do I find the function
f:\left(0,1{\right)}^{n}->R
E\left[\left(U-f\left({X}_{1},...,{X}_{n}\right){\right)}^{2}\right)\right]
? I have no idea where to start solving. Thank you in advance.
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String metric - Wikipedia
"String distance" redirects here. For the distance between strings and the fingerboard in musical instruments, see Action (music).
In mathematics and computer science, a string metric (also known as a string similarity metric or string distance function) is a metric that measures distance ("inverse similarity") between two text strings for approximate string matching or comparison and in fuzzy string searching. A requirement for a string metric (e.g. in contrast to string matching) is fulfillment of the triangle inequality. For example, the strings "Sam" and "Samuel" can be considered to be close.[1] A string metric provides a number indicating an algorithm-specific indication of distance.
The most widely known string metric is a rudimentary one called the Levenshtein distance (also known as edit distance).[2] It operates between two input strings, returning a number equivalent to the number of substitutions and deletions needed in order to transform one input string into another. Simplistic string metrics such as Levenshtein distance have expanded to include phonetic, token, grammatical and character-based methods of statistical comparisons.
String metrics are used heavily in information integration and are currently used in areas including fraud detection, fingerprint analysis, plagiarism detection, ontology merging, DNA analysis, RNA analysis, image analysis, evidence-based machine learning, database data deduplication, data mining, incremental search, data integration, Malware Detection, [3] and semantic knowledge integration.
1 List of string metrics
2 Selected string measures examples
List of string metrics[edit]
Levenshtein distance, or its generalization edit distance
Variational distance[4]
Skew divergence[4]
Confusion probability[4]
Fellegi and Sunters metric (SFS)[4]
Maximal matches[4]
Grammar-based distance[5]
TFIDF distance metric[6]
Selected string measures examples[edit]
Hamming distance "karolin" and "kathrin" is 3.
Levenshtein distance and Damerau–Levenshtein distance kitten and sitting have a distance of 3.
Jaro–Winkler distance JaroWinklerDist("MARTHA","MARHTA") =
{\displaystyle d_{j}={\frac {1}{3}}\left({\frac {m}{|s_{1}|}}+{\frac {m}{|s_{2}|}}+{\frac {m-t}{m}}\right)={\frac {1}{3}}\left({\frac {6}{6}}+{\frac {6}{6}}+{\frac {6-{\frac {2}{2}}}{6}}\right)=0.944}
{\displaystyle m}
{\displaystyle t}
is half the number of transpositions("MARTHA"[3]!=H, "MARHTA"[3]!=T).
Most frequent k characters MostFreqKeySimilarity('research', 'seeking', 2) = 2
^ Lu, Jiaheng; et al. (2013). "String similarity measures and joins with synonyms". Proceedings of the 2013 ACM SIGMOD International Conference on Management of Data: 373–384. doi:10.1145/2463676.2465313. ISBN 9781450320375.
^ Navarro, Gonzalo (2001). "A guided tour to approximate string matching". ACM Computing Surveys. 33 (1): 31–88. doi:10.1145/375360.375365. hdl:10533/172862.
^ Shlomi Dolev; Mohammad, Ghanayim; Alexander, Binun; Sergey, Frenkel; Yeali, S. Sun (2017). "Relationship of Jaccard and edit distance in malware clustering and online identification". 16th IEEE International Symposium on Network Computing and Applications: 369–373.
^ a b c d e Sam's String Metrics - Computational Linguistics and Phonetics
^ Russell, David J., et al. "A grammar-based distance metric enables fast and accurate clustering of large sets of 16S sequences." BMC bioinformatics 11.1 (2010): 1-14.
^ Cohen, William; Ravikumar, Pradeep; Fienberg, Stephen (2003-08-01). "A Comparison of String Distance Metrics for Name-Matching Tasks": 73–78. {{cite journal}}: Cite journal requires |journal= (help)
https://web.archive.org/web/20070304092115/http://www.dcs.shef.ac.uk/~sam/stringmetrics.html#qgram A fairly complete overview Archive index at the Wayback Machine
Carnegie Mellon University open source library
StringMetric project a Scala library of string metrics and phonetic algorithms
Natural project a JavaScript natural language processing library which includes implementations of popular string metrics
Retrieved from "https://en.wikipedia.org/w/index.php?title=String_metric&oldid=1084267733"
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How to solve a cyclic quintic in radicals? Galois theory tells us that \frac{z^{11}-1}{z-
\frac{{z}^{11}-1}{z-1}={z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z}^{6}+{z}^{5}+{z}^{4}+{z}^{3}+{z}^{2}+z+1
{\zeta }^{1},{\zeta }^{2},\dots ,{\zeta }^{10}
\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta }^{2}{x}_{3}+{\zeta }^{3}{x}_{4}+{\zeta }^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta }^{2}{x}_{2}+{\zeta }^{4}{x}_{3}+\zeta {x}_{4}+{\zeta }^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta }^{3}{x}_{2}+\zeta {x}_{3}+{\zeta }^{4}{x}_{4}+{\zeta }^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta }^{4}{x}_{2}+{\zeta }^{3}{x}_{3}+{\zeta }^{2}{x}_{4}+\zeta {x}_{5}\end{array}
{A}_{0},\dots ,{A}_{4}
{x}_{1},\dots ,{x}_{5}
{A}_{0}
\tau
{A}_{j}
{\zeta }^{-j}{A}_{j}
{A}_{j}^{5}
{A}_{j}^{5}
{A}_{j}^{5}
\zeta
\zeta
{A}_{1}^{5}
{A}_{1}
{A}_{j}
{A}_{1}
Landyn Whitney
The solution of a solvable quintic,
{x}^{5}+a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+e=0
x=\frac{1}{5}\left(-a+{z}_{1}^{\frac{1}{5}}+{z}_{2}^{\frac{1}{5}}+{z}_{3}^{\frac{1}{5}}+{z}_{4}^{\frac{1}{5}}\right)
{z}_{i}
are the roots of the quartic. However, there is also a form where one takes a fifth root only once.
p=11
{x}^{5}+{x}^{4}-4{x}^{3}-3{x}^{2}+3x+1=0
The five roots
{x}_{k}
k=0,1,2,3,4
are,
{x}_{k}=\frac{-1}{5}\left(\frac{1}{{\beta }_{k}^{-1}}+\frac{1}{{\beta }_{k}^{0}}+\frac{11}{{\beta }_{k}^{1}}+\frac{a}{{\beta }_{k}^{2}}+\frac{b}{{\beta }_{k}^{3}}\right)
a=t\frac{11}{4}\left(-1+5\sqrt{5}+\sqrt{-10\left(5+\sqrt{5}\right)}\right)
b=t\frac{11}{4}\left(-31+5\sqrt{5}-\sqrt{-10\left(85+31\sqrt{5}\right)}\right)
{\beta }_{k}={\zeta }_{5}^{k},{\left(\frac{ab}{11}\right)}^{\frac{1}{5}}
{\zeta }_{5}={e}^{2\pi ,\frac{i}{5}}
p=31
As discussed in this post,
{x}^{5}+{x}^{4}-12{x}^{3}-21{x}^{2}+x+5=0
{x}_{k}=\frac{1}{5}\left(\frac{1}{{\beta }_{k}^{-1}}-\frac{1}{{\beta }_{k}^{0}}+\frac{31}{{\beta }_{k}^{1}}+\frac{a}{{\beta }_{k}^{2}}+\frac{b}{{\beta }_{k}^{3}}\right)
a=t\frac{31}{4}\left(11+5\sqrt{5}+\sqrt{-10\left(25-11\sqrt{5}\right)}\right)
b=t\frac{31}{4}\left(-1-5\sqrt{5}+\sqrt{-10\left(1525-\sqrt{5}\right)}\right)
{\beta }_{k}={\zeta }_{5}^{k},{\left(\frac{ab}{31}\right)}^{\frac{1}{5}}
and so on for other '
p=10n+1
Jayla Matthews
{\omega }_{1}=\sqrt{5}\left\{\left(\frac{66}{3125}{\zeta }_{5}+\frac{451}{3125}{\zeta }_{5}^{2}+\frac{176}{3125}{\zeta }_{5}^{3}+\frac{286}{3125}{\zeta }_{5}^{4}\right)\right\}
{\omega }_{2}=\sqrt{2}\left\{-\frac{11}{20}+\left(\frac{1}{4}{\zeta }_{5}^{4}\right){\omega }_{1}+\left(-\frac{5}{44}{\zeta }_{5}+\frac{15}{44}{\zeta }_{5}^{2}+\frac{5}{44}{\zeta }_{5}^{3}+\frac{5}{44}{\zeta }_{5}^{4}\right){\omega }_{1}^{2}+\left(\frac{25}{121}{\zeta }_{5}-\frac{75}{242}{\zeta }_{5}^{2}-\frac{75}{484}{\zeta }_{5}^{3}+\frac{75}{242}{\zeta }_{5}^{4}\right){\omega }_{1}^{3}+\left(-\frac{375}{2662}{\zeta }_{5}+\frac{625}{1331}{\zeta }_{5}^{2}+\frac{625}{2662}{\zeta }_{5}^{3}+\frac{4375}{5324}{\zeta }_{5}^{4}\right){\omega }_{1}^{4}\right\}
-\frac{1}{10}+\frac{1}{2}{\omega }_{1}+\left(-\frac{5}{22}{\zeta }_{5}^{2}-\frac{5}{11}{\zeta }_{5}^{3}+\frac{5}{11}{\zeta }_{5}^{4}\right){\omega }_{1}^{2}+\left(\frac{75}{242}{\zeta }_{5}+\frac{150}{121}{\zeta }_{5}^{2}+\frac{75}{121}{\zeta }_{5}^{3}+\frac{125}{121}{\zeta }_{5}^{4}\right){\omega }_{1}^{3}+\left(\frac{1625}{1331}{\zeta }_{5}+\frac{1000}{1331}{\zeta }_{5}^{2}+\frac{5125}{2662}{\zeta }_{5}^{3}+\frac{375}{1331}{\zeta }_{5}^{4}\right){\omega }_{1}^{4}-{\omega }_{2}
{\int }_{0}^{\infty }\frac{\mathrm{cos}\left(xt\right)}{1+{t}^{2}}dt
\text{□}
or goodness of fit, homogeneity or independence?
x=a\mathrm{cos}t,\phantom{\rule{1em}{0ex}}y=b\mathrm{sin}t,\phantom{\rule{1em}{0ex}}0\le t\le 2\pi
=4a{\int }_{0}^{\pi /2}\sqrt{1-{e}^{2}{\mathrm{cos}}^{2}t}dt
where e
is the ellipse's eccentricity. The integral in this formula, called an elliptic integral, is non elementary except when e=0
or 1 a. Use the Trapezoidal Rule with n=10
to estimate the length of the ellipse when a=1
and e=1/2
f\left(t\right)=\sqrt{1-{e}^{2}{\mathrm{cos}}^{2}t}
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Revision as of 20:43, 4 August 2013 by NikosA (talk | contribs) (→Automatising Conversions: some "wrong" comment removed from script)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L_{\lambda {\text{Pixel, Band}}}={\frac {K_{\text{Band}}*q_{\text{Pixel, Band}}}{\Delta \lambda _{\text{Band}}}}}
{\displaystyle L_{\lambda {\text{Pixel,Band}}}}
{\displaystyle K_{\text{Band}}}
{\displaystyle q_{\text{Pixel,Band}}}
{\displaystyle \Delta _{\lambda _{\text{Band}}}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
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Convert transfer function filter parameters to zero-pole-gain form - MATLAB tf2zp - MathWorks Nordic
Zeros, Poles, and Gain of Continuous-Time System
[z,p,k] = tf2zp(b,a) finds the matrix of zeros z, the vector of poles p, and the associated vector of gains k from the transfer function parameters b and a. The function converts a polynomial transfer-function representation
H\left(s\right)=\frac{B\left(s\right)}{A\left(s\right)}=\frac{{b}_{1}{s}^{n-1}+\cdots +{b}_{n-1}s+{b}_{n}}{{a}_{1}{s}^{m-1}+\cdots +{a}_{m-1}s+{a}_{m}}
of a single-input/multi-output (SIMO) continuous-time system to a factored transfer function form
H\left(s\right)=\frac{Z\left(s\right)}{P\left(s\right)}=k\frac{\left(s-{z}_{1}\right)\left(s-{z}_{2}\right)\cdots \left(s-{z}_{m}\right)}{\left(s-{p}_{1}\right)\left(s-{p}_{2}\right)\cdots \left(s-{p}_{n}\right)}.
Use tf2zp when working with positive powers (s2 + s + 1), such as in continuous-time transfer functions. A similar function, tf2zpk, is more useful when working with transfer functions expressed in inverse powers (1 + z–1 + z–2).
Generate a system with the following transfer function.
H\left(s\right)=\frac{2{s}^{2}+3s}{{s}^{2}+\frac{1}{\sqrt{2}}s+\frac{1}{4}}=\frac{2\phantom{\rule{0.16666666666666666em}{0ex}}\left(s-0\right)\phantom{\rule{0.16666666666666666em}{0ex}}\left(s-\left(-\frac{3}{2}\right)\right)}{\left(s-\frac{-1}{2\sqrt{2}}\left(1-j\right)\right)\phantom{\rule{0.16666666666666666em}{0ex}}\left(s-\frac{-1}{2\sqrt{2}}\left(1+j\right)\right)}
Find the zeros, poles, and gain of the system. Use eqtflength to ensure the numerator and denominator have the same length.
a = [1 1/sqrt(2) 1/4];
Plot the poles and zeros to verify that they are in the expected locations.
text(real(z)+.1,imag(z),'Zero')
text(real(p)+.1,imag(p),'Pole')
Transfer function numerator coefficients, specified as a vector or matrix. If b is a matrix, then each row of b corresponds to an output of the system. b contains the coefficients in descending powers of s. The number of columns of b must be less than or equal to the length of a.
Transfer function denominator coefficients, specified as a vector. a contains the coefficients in descending powers of s.
sos2zp | ss2zp | tf2sos | tf2ss | tf2zpk | zp2tf
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Karleigh Moore, Ethan W, Ejun Dean, and
An Enigma machine is a famous encryption machine used by the Germans during WWII to transmit coded messages. An Enigma machine allows for billions and billions of ways to encode a message, making it incredibly difficult for other nations to crack German codes during the war — for a time the code seemed unbreakable. Alan Turing and other researchers exploited a few weaknesses in the implementation of the Enigma code and gained access to German codebooks, and this allowed them to design a machine called a Bombe machine, which helped to crack the most challenging versions of Enigma. Some historians believe that the cracking of Enigma was the single most important victory by the Allied powers during WWII. Using information that they decoded from the Germans, the Allies were able to prevent many attacks. However, to avoid Nazi suspicion that they had insight to German communications, the Allies had to allow some attacks to be carried out despite the fact that they had the knowledge to stop them.
Military Enigma machine, model "Enigma 1", used during the late 1930s and during the war [1]
Enigma machines use a form of substitution encryption.
Substitution encryption is a straightforward way of encoding messages, but these codes are fairly easy to break. A simple example of a substitution encryption scheme is a Caesar cipher. A Caesar cipher shifts each letter of the alphabet some number of places. A Caesar cipher with a shift of
1
would encode an A as a B, an M as an N, and a Z as an A, and so on.
Below is an image of a Caesar cipher with a shift of
3
Using a Caesar cipher with a shift of 5, encode the message “math is fun”. Hint: it might be useful to construct a table showing the encoding.
“Math is fun” can be encrypted by this scheme as “rfym nx kzs”.
But Enigma machines are much more powerful than a simple Caesar cipher.
Imagine that each time a letter was mapped to another, the entire encoding scheme changed. After each button press, the rotors move and repressing that same button routes current along a different path to a different revealed letter.
So for the first press of a key, one encoding (like the table in the example above) is generated, and when the second key is pressed, another encoding is generated, and so on. This greatly increases the number of possible encoding configurations. Each time a key is pressed on an Enigma machine, the rotors turn, and the code changes. This means that the message “AA” could be encoded as something like “TU” even though the same key was pressed twice.
An Enigma machine is made up of several parts including a keyboard, a lamp board, rotors, and internal electronic circuitry. Some machines, such as the ones used by the military, have additional features such as a plugboard.
Enigma Machine at the Imperial War Museum, London. [3]
Encoded messages would be a particular scramble of letters on a given day that would would translate to a comprehendible sentence when unscrambled.
When a key on the keyboard is pressed, one or more rotors move to form a new rotor configuration which will encode one letter as another. Current flows through the machine and lights up one display lamp on the lamp board, which shows the output letter. So if the "K" key is pressed, and the Enigma machine encodes that letter as a "P," the "P" would light up on the lamp board.
Each month, Enigma operators received codebooks which specified which settings the machine would use each day. Every morning the code would change.
For example, one one day, the codebook may list the settings described in the day-key below:
1.
Plugboard settings: A/L – P/R – T/D – B/W – K/F – O/Y
A plugboard is similar to an old-fashioned telephone switch board that has ten wires, each wire having two ends that can be plugged into a slot. Each plug wire can connect two letters to be a pair (by plugging one end of the wire to one letter’s slot and the other end to another letter). The two letters in a pair will swap over, so if “A” is connected to “Z,” “A” becomes “Z” and “Z” becomes “A.” This provides an extra level of scrambling for the military.
To implement this day-key first you would have to swap the letters A and L by connecting them on the plugboard, swap P and R by connecting them on the plugboard, and then the same with the other letter pairs listed above. Essentially, a one end of a cable would be plugged into the "A" slot and the other end would be plugged into the L slot. Before any further scrambling happens by the rotors, this adds a first layer of scrambling where the letters connected by the cable are encoded as each other. For example, if I were to encode the message APPLE after connecting only the "A" to the "L", this would be encoded as LPPAE.
The plugboard is positioned at the front of an Enigma machine, below the keys.[4]
2.
Rotor (or scrambler) arrangement: 2 — 3 —1
The Enigma machines came with several different rotors, each rotor providing a different encoding scheme. In order to encode a message, the Enigma machines took three rotors at a time, one in each of three slots. Each different combination of rotors would produce a different encoding scheme. Note: most military Enigma machines had three rotor slots though some had more.
To accomplish the configuration above, place rotor #2 in the 1st slot of the enigma, rotor #3 in the 2nd slot, and rotor #1 in the 3rd slot.
3.
Rotor orientations: D – K –P
On each rotor, there is an alphabet along the rim, so the operator can set in a particular orientation. For this example, the operator would turn the rotor in slot 1 so that D is displayed, rotate the second slot so that K is displayed, and rotate the third slot so that P is displayed.
Enigma wheels within alphabet rings in position in an Enigma scrambler[6]
As mentioned in above sections, Enigma uses a form of substitution ciphers.
Each of the three rotors will display a number or letter (the rotors in the image above have letters), and when the rotors turn, a new set of three numbers/letters appears. With the initial set of three numbers/letters (meaning the numbers/letters on the sender’s machine when they began to type the message), a message recipient can decode the message by setting their (identical) Enigma machine to the initial settings of the sender’s Enigma machine. Each rotor has
26
numbers/letters on it. An Enigma machine takes three rotors at a time, and the Germans could interchange rotors, choosing from a set of five, resulting in thousands of possible configurations. For example, one configuration of rotors could be: rotor #5 in slot one, rotor #2 in slot two, and rotor #1 in slot three.
How many configurations are possible with an Enigma machine with these specifications?
In the first slot, there are
5
rotors to pick from, in the second there are
4
rotors to pick from, and in the third slot there are
3
rotors to pick from. So there are
5 \times 4 \times 3 = 60
ways to configure the five rotors.
26
starting positions for each rotor, so there are
26 \times 26 \times 26 = 17,576
choices for initial configurations of the rotors’ numbers/letters.
760 10 60 940 10! 720 362
How many rotor configurations would an Enigma machine encrypter be able to select from if they needed to choose 3 rotors from a set of 10 rotors? (The order of the rotors matters).
Here is a set of three Enigma rotors
Image Credit: Emigma rotor set Wapcaplet.
The features above describe the components of commercial Enigma machines, but military-grade machines have additional features, such as a plugboard, which allow for even more configuration possibilities.
26
letters in the alphabet, there are
26!
ways to arrange the letters, but the plugboard can only make
10
pairs, so there are
20
letters involved with the pairings, and
6
leftover that must be divided out. Furthermore, there are
10
pairs of letters, and it does not matter what order the pairs are in, so divide also by
10!
, and the order of the letters in the pair does not matter, so divide also by
2^10
. The resulting number of combinations yielded by the plugboard is as follows:
\frac{26!}{6! \times 10! \times 2^{10}} = 150,738,274,937,250.
All of the components put together yields:
60 \times 17576 \times 150,738,274,937,250 = 158,962,555,217,826,360,000
total number of ways to set a military-grade Enigma machine. [7]
A major flaw with the Enigma code was that a letter could never be encoded as itself. In other words, an “M” would never be encoded as an “M.” This was a huge flaw in the Enigma code because it gave codebreakers a piece of information they could use to decrypt messages. If the codebreakers could guess a word or phrase that would probably appear in the message, they could use this information to start breaking the code. Because the Germans always sent a weather report at the beginning of the message, and usually included the phrase “Heil Hitler” at the end of the message, there were phrases decrypters knew to look for. [8]Decoders could compare a given phrase to the letters in the code, and if a letter in the phrase matched up with a letter in the code, they knew that that part of the code did not contain the phrase. The decoders could then begin cracking the code with a process of elimination approach.
Find possible contenders for the encoding of the word “RAIN” in the coded string below. (The symbol % is used to denote unknown letters).
Coded Message E R W N I K O L K M M M M
Phrase R A I N % % % % % % % % %
RAIN cannot be encoded as ERWN because the N in RAIN and the N in ERWN match up. Since N cannot be encoded as itself, this isn’t the encoding.
Let’s shift our message one slot to the right, and see if the result is a valid encoding.
Phrase % R A I N % % % % % % % %
RAIN cannot be encoded as RWNI because the R in RAIN matches with the R in RWNI. Let’s shift again.
Phrase % % R A I N % % % % % % %
RAIN cannot be encoded as WNIK because the I in RAIN matches with the I in WNIK.
Phrase % % % R A I N % % % % % %
RAIN can be encoded as NIKO because the two phrases have no letters that match up. So NIKO is a possible encoding of RAIN.
If we repeat this process, we will find that NIKO, IKOL, KOLK, OLKM, LKMM, KMMM, and MMMM are all possible encodings of RAIN since no letters match up between RAIN and the encoding. It is okay that MMMM could encode RAIN even though this means that M would encode R, A, I, and N because remember that at each key press, the letter mapping in an Enigma machine changes.
It is not guaranteed that RAIN is encoded in this string at all, though, but it gave decoders a good starting point for decrypting messages.
Alan Turing and Gordon Welchman designed a machine called the Bombe machine which used electric circuits to solve an Enigma encoded message in under 20 minutes. The Bombe machine would try to determine the settings of the rotors and the plugboard of the Enigma machine used to send a given coded message.
The standard British Bombe machine was essentially 36 Enigma machines wired together, this way, the Bombe machine would simulate several Enigma machines at once. Most Enigma machines had three rotors and to represent this in the Bombe, each of the Enigma simulators in the Bombe had three drums, one for each rotor.
Bombe machine drums[9]
The Bombe's drums were color coded to correspond with which rotor they were simulating. While an 3-rotor Enigma machine only used three rotors at a time, there are more to choose from. The drums were arranged so that the top one of the three simulated the left-hand rotor of the Enigma scrambler, the middle one simulated the middle rotor, and the bottom one simulated the right-hand rotor. The drums would turn to try out a new configuration. For each full rotation of the top drums, the middle drums were incremented by one position, and likewise for the middle and bottom drums, giving the total of 26 × 26 × 26 = 17,576 positions of the 3-rotor Enigma scrambler.[10]
Then, for a given rotor configuration (at each turn of the drums), the Bombe machine would make a guess about a plugboard setting, say “A is connected to Z.” It then ran through and determined what all of the other letters must be set to on the plugboard. If any contradictions arose, say, it deduced that “A was connected to W,” then it must be that A is not connected to Z on the plugboard, a contradiction arises. Since the other letter mappings the machine just figured out were determined based off of a false assumption (namely the assumption that A is connected to Z), all of those combinations are invalid, and the Bombe machine knows not to waste time checking any of those combinations later. So, say that the machine guessed that A is connected to Z, and then the machine deduces that if A is connected to Z, then B must be connected to E. If it later determines that A is not connected to Z, it knows that B is not connected to E. After such a contradiction arises, the Bombe machine will not guess that A is connected to Z again, and it knows not to guess that B is connected to E, and so on. The Bombe machine shifts the rotor positions, and chooses a new guess and repeats this process until a satisfying arrangement of settings appears. Because electric circuits can perform computations very quickly, the Bombe machine can go through all the rotor combinations in about 20 minutes.
At each position of the drums, the configuration would be tested to see if the configuration lead to a logical contradiction, ruling out that setting. If the test did not lead to a contradiction, the machine would stop and the decoder would note that configuration as a candidate solution. Then, the machine is restarted and more configurations are tested. These tests would narrow down the list of possible configurations and the candidate solutions would be tested further to eliminate ones that wouldn't work. There were usually many unsuccessful candidate solutions before the correct one was found.[10]
Nassiri, A. Enigma (crittografia) - Museo scienza e tecnologia Milano. Retrieved April 30, 2016, from https://en.wikipedia.org/wiki/File:Enigma_(crittografia)_-_Museo_scienza_e_tecnologia_Milano.jpg
Sperling, K. EnigmaMachineLabeled. Retrieved April 30, 2016, from https://en.wikipedia.org/wiki/File:EnigmaMachineLabeled.jpg
Lord, B. Enigma-plugboard. Retrieved April 30, 2016, from https://en.wikipedia.org/wiki/File:Enigma-plugboard.jpg
, M. File:Enigma-rotor-windows.jpg. Retrieved July 20, 2016, from https://commons.wikimedia.org/wiki/File:Enigma-rotor-windows.jpg
, T. Enigma rotors with alphabet rings. Retrieved April 30, 2016, from https://en.wikipedia.org/wiki/File:Enigma_rotors_with_alphabet_rings.jpg
, N. 158,962,555,217,826,360,000 (Enigma Machine). Retrieved April 28, 2016, from https://www.youtube.com/watch?v=G2_Q9FoD-oQ
, N. Flaw in the Enigma Code. Retrieved April 28, 2016, from https://www.youtube.com/watch?annotation_id=annotation_786414&feature=iv&src_vid=G2_Q9FoD-oQ&v=V4V2bpZlqx8
, T. File:Rotating upper bombe drum.jpg. Retrieved July 21, 2016, from https://en.wikipedia.org/wiki/File:Rotating_upper_bombe_drum.jpg
, . Bombe. Retrieved July 21, 2016, from https://en.wikipedia.org/wiki/Bombe
Cite as: Enigma Machine. Brilliant.org. Retrieved from https://brilliant.org/wiki/enigma-machine/
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The Effect of China’s Family Structure on Household Nutrition
1Department of Agribusiness and Applied Economics, North Dakota State University, Fargo, ND, USA
2Beijing Food Safety Policy & Strategy Research Base, College of Economics and Management, China Agricultural University, Beijing, China
3Department of Food and Resource Economics, University of Florida, Gainesville, FL, USA
{M}_{jk}
\begin{array}{l}{S}_{ij}=\beta \frac{{Q}_{ij}}{{M}_{jL}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{Q}_{ij}<{M}_{jL}\\ {S}_{ij}=-\beta \frac{{Q}_{ij}-{M}_{jH}}{{M}_{jL}}+\beta ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{M}_{jL}<{Q}_{ij}<\left({M}_{jL}+{M}_{jH}\right)\\ {S}_{ij}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{Q}_{ij}>\left({M}_{jL}+{M}_{jH}\right)\end{array}
{S}_{ij}
\beta
{Q}_{ij}
{M}_{jk}
{S}_{f}={\beta }_{0f}+{\beta }_{jf}{H}_{jf}+{\beta }_{m}{W}_{m}+{\beta }_{kf}{D}_{kf}\ast \left({B}_{f}+{R}_{f}+{A}_{f}\right)+{\epsilon }_{f}
{S}_{if}
{H}_{jf}
{W}_{m}
{D}_{kf}
{B}_{f}=1
{R}_{f}=1
{A}_{f}=1
{\epsilon }_{f}
Wang, Y., Wahl, T.I., Seale, J.L. and Bai, J.J. (2019) The Effect of China’s Family Structure on Household Nutrition. Food and Nutrition Sciences, 10, 198-206. https://doi.org/10.4236/fns.2019.102015
1. NCD Risk Factor Collaboration (NCD-RisC) (2016) Trends in Adult Body-Mass Index in 200 Countries from 1975 to 2014: A Pooled Analysis of 1698 Population-Based Measurement Studies with 19.2 Million Participants. The Lancet, 387, 1377-1396. https://doi.org/10.1016/S0140-6736(16)30054-X
2. Victora, C.G., Adair, L., Fall, C., Hallal, P.C., Martorell, R., Richter, L. and Sachdev, H.S. (2008) Maternal and Child Undernutrition: Consequences for Adult Health and Human Capital. Lancet, 371, 340-357. https://doi.org/10.1016/S0140-6736(07)61692-4
3. Ma, L.F., Guo, H.W. and He, A.N. (2007) Analysis of the Hazards and Correlative Factors of Overweight and Obesity in community Adults. Journal of Environmental & Occupational Medicine, 24.
4. Mohsen, N., Wang, H.D., Lozano, R., Liang, A.D., Zhou, X.F., et al. (2014) Global, Regional, and National Age-Sex Specific All-Cause and Cause-Specific Mortality for 240 Causes of Death, 1990-2013: A Systematic Analysis for the Global Burden of Disease Study 2013. The Lancet, 385, 117-171.
5. Popkin, B.M. (2008) Will China’s Nutrition Transition Overwhelm Its Health Care System and Slow Economic Growth? Health Affairs, 27, 1064-1076. https://doi.org/10.1377/hlthaff.27.4.1064
6. Yan, S., Li., J., Li, S. and Zhang, B. (2012) The Expanding Burden of Cardiometabolic Risk in China: The China Health and Nutrition Survey. Obesity, 13, 810-821. https://doi.org/10.1111/j.1467-789X.2012.01016.x
7. Wang, S.S., Lay, S., Yu, H.N. and Shen, S.R. (2016) Dietary Guidelines for Chinese Residents (2016): Comments and Comparisons. Journal of Zhejiang University. Science B, 17, 649-656. https://doi.org/10.1631/jzus.B1600341
8. Chinese Nutrition Society. http://www.cnsoc.org/
9. US Department of Agriculture (USDA) and US Department of Health and Human Services (USDHHS) (2010) Dietary Guidelines for Americans, 2010. 7th Edition, US Government Printing Office, Washington DC. http://www.cnpp.usda.gov/dietaryguidelines.htm
10. Yuan, Y.Q., Li, F., Dong, R.H., Chen, J.S., He, G.S., Li, S.G. and Chen, B. (2017) The Development of a Chinese Healthy Eating Index and Its Application in the General Population. Nutrients, 9, 977. https://doi.org/10.3390/nu9090977
11. Bai, J., Zhang, C., Qiao, F. and Wahl, T. (2012) Disaggregating Household Expenditures on Food away from Home in Beijing by Type of Food Facility and Type of Meal. China Agricultural Economic Review, 4, 18-35. https://doi.org/10.1108/17561371211196757
12. Bai, J., Wahl, T.I., Lohmar, B.T. and Huang, J. (2010) Food Away from Home in Beijing: Effects of Wealth, Time and “Free” Meals. China Economic Review, 21, 432-441. https://doi.org/10.1016/j.chieco.2010.04.003
13. Atkin, D. (2013) Trade, Tastes, and Nutrition in India. American Economic Review, 103, 1629-1663. https://doi.org/10.1257/aer.103.5.1629
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Social return on investment (SROI) is a method for measuring values that are not traditionally reflected in financial statements, including social, economic, and environmental factors. They can identify how effectively a company uses its capital and other resources to create value for the community. While a traditional cost-benefit analysis is used to compare different investments or projects, SROI is used more to evaluate the general progress of certain developments, showing both the financial and social impact the corporation can have.
Social return on investment (SROI) is a method of accounting for the social, economic, and environmental value created by a company.
Companies issue financial statements that show investors the revenue, sales, net profits, debts, and other key metrics, but SROI is not a factor.
The purpose of issuing SROI is for corporations to be able to look at their social impact in financial terms.
The factors that go into calculating the SROI are the social impact value and the initial investment amount.
SROI is useful to corporations because it can improve program management through better planning and evaluation. It can also increase the corporation’s understanding of its effect on the community and allow better communication regarding the value of the corporation’s work (both internally and to external stakeholders). Philanthropists, venture capitalists, foundations, and other non-profits may use SROI to monetize their social impact, in financial terms.
A general formula used to calculate SROI is as follows:
\begin{aligned} &SROI = \dfrac{SIV - IIA}{IIA \times 100\%}\\ &\textbf{where:}\\ &SIV = \text{social impact value}\\ &IIA = \text{initial investment amount}\\ \end{aligned}
SROI=IIA×100%SIV−IIAwhere:SIV=social impact valueIIA=initial investment amount
Assigning a dollar value to the social impact can present problems, and various methodologies have been developed to help quantify the results. The Analytical Hierarchy Process (AHP), for example, is one method that converts and organizes qualitative information into quantitative values.
Inputs, or resources investments in your activity (such as the costs of running, say, a job-readiness program)
Outputs, or the direct and tangible products from the activity (for example, the number of people trained by the program)
Outcomes, or the changes to people resulting from the activity (i.e., new jobs, better income, improved quality of life for the individuals; increased taxes for, and reduced support from, the government)
Impact, or the outcome less an estimate of what would have happened anyway (For example, if 20 people got new jobs but five of them would have been hired in any event, the impact is based on the 15 people who got jobs directly as a result of the job-readiness program.)
Dorothy A. Johnson Center for Philanthropy. "Valuing SROI: Social Return on Investment Techniques and Organizational Implementation In the Netherlands and United States," Pages 7-14. Accessed Oct. 17, 2021.
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1 How many lines can be drawn through a given point 2 In how many points - Maths - Introduction to Euclid\s Geometry - 8144955 | Meritnation.com
1. How many lines can be drawn through a given point ?
2. In how many points two distinct lines can intersect ?
3. In how many lines tow distinct planes can intersect ?
4. In how many least no. of distinct points determine a unique plane ?
5. If B lies between A and C and AC=10, BC=6, what is AB2 ?
1\right)\phantom{\rule{0ex}{0ex}}\mathrm{As} \mathrm{we} \mathrm{know} \mathrm{from} \mathrm{Euclid}\text{'}\mathrm{s} \mathrm{axioms} \mathrm{that} \mathrm{Infinte} \mathrm{number} \mathrm{of} \mathrm{lines} \mathrm{can} \mathrm{be} \mathrm{drawn} \mathrm{from} \mathrm{given} \mathrm{point} \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}5\right)\phantom{\rule{0ex}{0ex}}\mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{from} \mathrm{Euclid}\text{'}\mathrm{s} \mathrm{axioms}, \phantom{\rule{0ex}{0ex}}\mathrm{AC}+\mathrm{BC}=\mathrm{AB}\phantom{\rule{0ex}{0ex}}\mathrm{Things} \mathrm{which} \mathrm{coincide} \mathrm{with} \mathrm{one} \mathrm{another} \mathrm{aree} \mathrm{qual} \mathrm{to} \mathrm{one} \mathrm{a}\mathrm{nother}.\phantom{\rule{0ex}{0ex}}\mathrm{So} \mathrm{AB}=10+6=16\phantom{\rule{0ex}{0ex}}\mathrm{So} {\mathrm{AB}}^{2}={16}^{2}=256.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}
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How do you evaluate 10P8 using a calculator?
using the definition for
\text{using the definition for}\phantom{\rule{1ex}{0ex}}{\text{}}^{n}P{\text{}}_{r}
\text{that is}\phantom{\rule{1ex}{0ex}}{\text{}}^{n}P{\text{}}_{r}=\frac{n!}{\left(n-r\right)!}
\text{where}\phantom{\rule{1ex}{0ex}}n!=n\left(n-1\right)\left(n-2\right)\left(n-3\right)........×1
⇒{\text{}}^{10}P{\text{}}_{8}=\frac{10!}{\left(10-8\right)!}=\frac{10!}{2!}←\phantom{\rule{1ex}{0ex}}\text{cancel}\phantom{\rule{1ex}{0ex}}2!
=10×9×8×7×6×5×4×3
=1814400
What is the Kolmogorov formula for conditional probability?
A family has 6 children. I've figured out that there are 64 different ways the genders of the kids can work out. What's the probability that there's at least 1 boy in the family?
What is the probability that a five-card poker hand does not contain the queen of hearts?
Max is studying for a probability exam. He attempts 703 problems in total while practicing. He gets 560 of those problems correct. 216 of the total problems involve conditional probability. Of those 216 conditional probability problems, he gets 134 correct. A random question that Max got wrong is selected. What is the probability that it involved conditional probability?
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xy'ln(y/x)=x+yln(y/x) find general solution of the homogenous equation
\cap \cup
\left(a,b\right)\in R
\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)
3n+1
A 5-card hand is dealt from a perfectly shuffled deck of playing cards. What is the probability that the hand is a full house? A full house has three cards of the same rank and another pair of the same rank. For example,
\left\{4\mathrm{♠},\text{ }4\mathrm{♡},\text{ }4\mathrm{♢},\text{ }J\mathrm{♠},\text{ }J\mathrm{♣}\right\}
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Gaussian elimination - Wikipedia
(Redirected from Gauss–Jordan elimination)
Algorithm for solving systems of linear equations
Swapping two rows,
Multiplying a row by a nonzero number,
Adding a multiple of one row to another row. (subtraction can be achieved by multiplying one row with -1 and adding the result to another row)
{\displaystyle {\begin{bmatrix}1&3&1&9\\1&1&-1&1\\3&11&5&35\end{bmatrix}}\to {\begin{bmatrix}1&3&1&9\\0&-2&-2&-8\\0&2&2&8\end{bmatrix}}\to {\begin{bmatrix}1&3&1&9\\0&-2&-2&-8\\0&0&0&0\end{bmatrix}}\to {\begin{bmatrix}1&0&-2&-3\\0&1&1&4\\0&0&0&0\end{bmatrix}}}
1 Definitions and example of algorithm
1.2 Echelon form
1.3 Example of the algorithm
3.1 Computing determinants
3.3 Computing ranks and bases
4 Computational efficiency
Definitions and example of algorithm[edit]
See also: Elementary matrix
Swap the positions of two rows.
Multiply a row by a non-zero scalar.
Add to one row a scalar multiple of another.
Echelon form[edit]
Main article: Row echelon form
{\displaystyle {\begin{bmatrix}0&\color {red}{\mathbf {2} }&1&-1\\0&0&\color {red}{\mathbf {3} }&1\\0&0&0&0\end{bmatrix}}.}
Example of the algorithm[edit]
{\displaystyle {\begin{alignedat}{4}2x&{}+{}&y&{}-{}&z&{}={}&8&\qquad (L_{1})\\-3x&{}-{}&y&{}+{}&2z&{}={}&-11&\qquad (L_{2})\\-2x&{}+{}&y&{}+{}&2z&{}={}&-3&\qquad (L_{3})\end{alignedat}}}
{\displaystyle {\begin{alignedat}{4}2x&{}+{}&y&{}-{}&z&{}={}&8&\\-3x&{}-{}&y&{}+{}&2z&{}={}&-11&\\-2x&{}+{}&y&{}+{}&2z&{}={}&-3&\end{alignedat}}}
{\displaystyle \left[{\begin{array}{rrr|r}2&1&-1&8\\-3&-1&2&-11\\-2&1&2&-3\end{array}}\right]}
{\displaystyle {\begin{alignedat}{4}2x&{}+{}&y&{}-{}&z&{}={}&8&\\&&{\tfrac {1}{2}}y&{}+{}&{\tfrac {1}{2}}z&{}={}&1&\\&&2y&{}+{}&z&{}={}&5&\end{alignedat}}}
{\displaystyle {\begin{aligned}L_{2}+{\tfrac {3}{2}}L_{1}&\to L_{2}\\L_{3}+L_{1}&\to L_{3}\end{aligned}}}
{\displaystyle \left[{\begin{array}{rrr|r}2&1&-1&8\\0&{\frac {1}{2}}&{\frac {1}{2}}&1\\0&2&1&5\end{array}}\right]}
{\displaystyle {\begin{alignedat}{4}2x&{}+{}&y&{}-{}&z&{}={}&8&\\&&{\tfrac {1}{2}}y&{}+{}&{\tfrac {1}{2}}z&{}={}&1&\\&&&&-z&{}={}&1&\end{alignedat}}}
{\displaystyle L_{3}+-4L_{2}\to L_{3}}
{\displaystyle \left[{\begin{array}{rrr|r}2&1&-1&8\\0&{\frac {1}{2}}&{\frac {1}{2}}&1\\0&0&-1&1\end{array}}\right]}
{\displaystyle {\begin{alignedat}{4}2x&{}+{}&y&&&{}={}7&\\&&{\tfrac {1}{2}}y&&&{}={}{\tfrac {3}{2}}&\\&&&{}-{}&z&{}={}1&\end{alignedat}}}
{\displaystyle {\begin{aligned}L_{2}+{\tfrac {1}{2}}L_{3}&\to L_{2}\\L_{1}-L_{3}&\to L_{1}\end{aligned}}}
{\displaystyle \left[{\begin{array}{rrr|r}2&1&0&7\\0&{\frac {1}{2}}&0&{\frac {3}{2}}\\0&0&-1&1\end{array}}\right]}
{\displaystyle {\begin{alignedat}{4}2x&{}+{}&y&\quad &&{}={}&7&\\&&y&\quad &&{}={}&3&\\&&&\quad &z&{}={}&-1&\end{alignedat}}}
{\displaystyle {\begin{aligned}2L_{2}&\to L_{2}\\-L_{3}&\to L_{3}\end{aligned}}}
{\displaystyle \left[{\begin{array}{rrr|r}2&1&0&7\\0&1&0&3\\0&0&1&-1\end{array}}\right]}
{\displaystyle {\begin{alignedat}{4}x&\quad &&\quad &&{}={}&2&\\&\quad &y&\quad &&{}={}&3&\\&\quad &&\quad &z&{}={}&-1&\end{alignedat}}}
{\displaystyle {\begin{aligned}L_{1}-L_{2}&\to L_{1}\\{\tfrac {1}{2}}L_{1}&\to L_{1}\end{aligned}}}
{\displaystyle \left[{\begin{array}{rrr|r}1&0&0&2\\0&1&0&3\\0&0&1&-1\end{array}}\right]}
{\displaystyle {\begin{aligned}L_{2}+{\tfrac {3}{2}}L_{1}&\to L_{2},\\L_{3}+L_{1}&\to L_{3}.\end{aligned}}}
Computing determinants[edit]
Swapping two rows multiplies the determinant by −1
Multiplying a row by a nonzero scalar multiplies the determinant by the same scalar
Adding to one row a scalar multiple of another does not change the determinant.
{\displaystyle \det(A)={\frac {\prod \operatorname {diag} (B)}{d}}.}
Computationally, for an n × n matrix, this method needs only O(n3) arithmetic operations, while using Leibniz formula for determinants requires O(n!) operations (number of summands in the formula), and recursive Laplace expansion requires O(2n) operations (number of sub-determinants to compute, if none is computed twice). Even on the fastest computers, these two methods are impractical or almost impracticable for n above 20.
{\displaystyle }
Finding the inverse of a matrix[edit]
See also: Invertible matrix
{\displaystyle A={\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}}.}
{\displaystyle [A|I]=\left[{\begin{array}{ccc|ccc}2&-1&0&1&0&0\\-1&2&-1&0&1&0\\0&-1&2&0&0&1\end{array}}\right].}
{\displaystyle [I|B]=\left[{\begin{array}{rrr|rrr}1&0&0&{\frac {3}{4}}&{\frac {1}{2}}&{\frac {1}{4}}\\0&1&0&{\frac {1}{2}}&1&{\frac {1}{2}}\\0&0&1&{\frac {1}{4}}&{\frac {1}{2}}&{\frac {3}{4}}\end{array}}\right].}
Computing ranks and bases[edit]
{\displaystyle T={\begin{bmatrix}a&*&*&*&*&*&*&*&*\\0&0&b&*&*&*&*&*&*\\0&0&0&c&*&*&*&*&*\\0&0&0&0&0&0&d&*&*\\0&0&0&0&0&0&0&0&e\\0&0&0&0&0&0&0&0&0\end{bmatrix}},}
Computational efficiency[edit]
h := 1 /* Initialization of the pivot row */
k := 1 /* Initialization of the pivot column */
while h ≤ m and k ≤ n
/* Find the k-th pivot: */
i_max := argmax (i = h ... m, abs(A[i, k]))
/* No pivot in this column, pass to next column */
swap rows(h, i_max)
/* Do for all rows below pivot: */
for i = h + 1 ... m:
f := A[i, k] / A[h, k]
/* Fill with zeros the lower part of pivot column: */
/* Do for all remaining elements in current row: */
A[i, j] := A[i, j] - A[h, j] * f
/* Increase pivot row and column */
h := h + 1
Fangcheng (mathematics)
^ Grcar, Joseph F. (2011-05-01). "How ordinary elimination became Gaussian elimination". Historia Mathematica. 38 (2): 163–218. doi:10.1016/j.hm.2010.06.003. ISSN 0315-0860. S2CID 14259511.
^ Calinger (1999), pp. 234–236
^ Timothy Gowers; June Barrow-Green; Imre Leader (8 September 2008). The Princeton Companion to Mathematics. Princeton University Press. p. 607. ISBN 978-0-691-11880-2.
^ Grcar (2011a), pp. 169-172
^ Grcar (2011b), pp. 783–785
^ Lauritzen, p. 3
^ Grcar (2011b), p. 789
^ Althoen, Steven C.; McLaughlin, Renate (1987), "Gauss–Jordan reduction: a brief history", The American Mathematical Monthly, Mathematical Association of America, 94 (2): 130–142, doi:10.2307/2322413, ISSN 0002-9890, JSTOR 2322413
^ Farebrother (1988), p. 12.
^ Fang, Xin Gui; Havas, George (1997). "On the worst-case complexity of integer Gaussian elimination" (PDF). Proceedings of the 1997 international symposium on Symbolic and algebraic computation. ISSAC '97. Kihei, Maui, Hawaii, United States: ACM. pp. 28–31. doi:10.1145/258726.258740. ISBN 0-89791-875-4. [permanent dead link]
^ J. B. Fraleigh and R. A. Beauregard, Linear Algebra. Addison-Wesley Publishing Company, 1995, Chapter 10.
^ Golub & Van Loan (1996), §3.4.6.
^ Hillar, Christopher; Lim, Lek-Heng (2009-11-07). "Most tensor problems are NP-hard". arXiv:0911.1393 [cs.CC].
Atkinson, Kendall A. (1989), An Introduction to Numerical Analysis (2nd ed.), New York: John Wiley & Sons, ISBN 978-0471624899 .
Bolch, Gunter; Greiner, Stefan; de Meer, Hermann; Trivedi, Kishor S. (2006), Queueing Networks and Markov Chains: Modeling and Performance Evaluation with Computer Science Applications (2nd ed.), Wiley-Interscience, ISBN 978-0-471-79156-0 .
Lauritzen, Niels, Undergraduate Convexity: From Fourier and Motzkin to Kuhn and Tucker .
Grcar, Joseph F. (2011a), "How ordinary elimination became Gaussian elimination", Historia Mathematica, 38 (2): 163–218, arXiv:0907.2397, doi:10.1016/j.hm.2010.06.003, S2CID 14259511
Grcar, Joseph F. (2011b), "Mathematicians of Gaussian elimination" (PDF), Notices of the American Mathematical Society, 58 (6): 782–792
Higham, Nicholas (2002), Accuracy and Stability of Numerical Algorithms (2nd ed.), SIAM, ISBN 978-0-89871-521-7 .
Kaw, Autar; Kalu, Egwu (2010). "Numerical Methods with Applications: Chapter 04.06 Gaussian Elimination" (PDF) (1st ed.). University of South Florida.
The Wikibook Linear Algebra has a page on the topic of: Gaussian elimination
Interactive didactic tool
Retrieved from "https://en.wikipedia.org/w/index.php?title=Gaussian_elimination&oldid=1083031630"
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1.4 Flow Geometry
The Application Challenge TA 2 - AC 7, "Confined Double Annular Jet" proposed by the VUB, Belgium, is an axisymmetric double annular flow field generated by a burner, and discharging into a confined combustion chamber (see Fig. 1 below). The flow is studied in cold conditions, and in a nozzle region going from the nozzle to 1.5 diameters downstream. The resulting mean flow is axisymmetric ; it can be qualified as a complex turbulent flow. The flow possesses a central vortex and two small toroidal contra rotating vortices. There are stagnation points and lines, mixing regions and recirculation regions (see Fig. 2 below). This is thus a good AC, since there is a relatively simple geometry (axisymmetric), and the ability of CFD codes to reproduce several properties of complex flows can be tested.
Figure 1: An overall display of the facility. Inlet arrangement, a cut on the burner, and the combustion chamber are shown. The nozzle region corresponding to the measurements is indicated.
Experimental data #1: Air flow with a 156 mm largest diameter jet, with a maximum velocity of 6.3 m/s, corresponding to a Reynolds number of 6 104. For this flow, 2D LDV data have been recorded, providing axial and radial mean velocities, axial and radial turbulence intensities and Reynolds stress. The data are provided on a dense experimental grid composed of 36 sections, each of which possessing 100 to 200 measurement points.
Experimental data #2: The flow field, generated in the combustion chamber model by water flow through the model, is measured with a 2D2C PIV system. A total of 271 instantaneous vector fields, each containing 5040 vectors in the near nozzle region, are presented. The flow rate through the burner was 4.09x10-5 m3/s giving a Reynolds number 180, based on nozzle opening dr (2 mm) and mean exit velocity Ue (0.09 m/s). This produces a central toroidal vortex in the transitional state.
Since international norms in matter of pollution are getting more and more binding, it is important to be able to modify the burner in order to minimize the emission of pollutant species. Since burners are usually designed and modified empirically by the manufacturer, the modifications cannot be optimal. It is therefore important to be able to better understand the flow field and the combustion process associated to burners. The turbulent diffusion of species is faster than the combustion process, so that the first step to understand a burner is to study the flow field in cold conditions. It is essential to understand the behavior of the various jets and their interaction with the neighboring jets and surrounding flows to successfully predict the performance of the device.
Figure 2: Streamlines of the flow, and designations of specific position points (DOAP).
Since the L/D of the channels of the burner are large, the fully-developed flows dominate over any influence from the inlet geometries. Therefore in this AC the geometry is determined only by dimensions of the exit of the burner and of the combustion chamber: see Table 1 and Figure 3 below.
Orientation of axis and sign conventions: since the geometry is axisymmetric, in the following a cylindrical coordinate system is considered (x,z,q), where x is the axial coordinate (the distance from the exit of the burner, along its axis), z is the radial coordinate (the distance from the axis) and q the tangential coordinate, not considered here due to axisymmetry.
Table 1 Definition of Some Geometric Parameters
r1 radius of the primary jet
r2 radius of the secondary jet
dr width of annular jets
rc radius of the combustion chamber
{\displaystyle \beta ={\frac {r_{2}}{r_{1}}}}
secondary to primary radius ratio
{\displaystyle {Y}={\frac {A_{e}}{A_{1}}}={\frac {r_{c}^{2}}{2r_{1}dr}}}
exit area to primary area ratio
Figure 3: The diameters of the boundaries of the streams produced by the burner.
The mean flow is axisymmetric. Only the flow in the nozzle region has been recorded. For this region, the jet presents a complex flow: it possesses recirculation regions, several toroidal vortices, a stagnation point and stagnation lines. First the primary and secondary streams merge, creating a vortex bubble between them. This bubble corresponds to a pair of contrarotating toroidal vortices. Then the simple annular stream becomes a central jet, creating a big central vortex bubble. This central vortex is less stable than the toroidal one (see the streamlines in Figure 2).
The databases correspond to a turbulent, incompressible, isothermal flow with recirculation.
The following tables provide different informations characterizing the flow: The Problem Definition Parameters (PDP) are dimensional quantities characterizing the experiment, and Governing Non-Dimensional Parameters (GNDP) are non-dimensional numbers governing and characterizing the physics of the flow.
First, some useful fluid dynamics parameters are presented in Table 2. The PDP introduced here characterize the geometry of the burner, and determine the inflow (through the velocity ratio of primary to secondary bulk velocity). They are also presented in Table 2.
The GNDP are the Reynolds number and the Craya-Curtet numbers, presented in Table 2. The Reynolds number is defined using the bulk-averaged velocity of the secondary jet and its diameter. The Craya-Curtet number characterizes the recirculation of confined jet flows. It was shown experimentally for some simple annular jet flows that the flow and mixing properties of the classical axisymmetric confined jet depend only on the Craya-Curtet number. It is assumed here that this number is also governing the physics of the double annular jet flow. The Craya-Curtet number can be defined as a normalization of the stagnation pressure loss in the system:
{\displaystyle {\frac {g\Delta {P}^{\star }}{U_{e}^{2}\rho }}={\frac {1}{C_{t}^{2}}}\qquad \qquad \qquad \qquad (1)}
where the different parameters in this equation are defined in Table 2. A small value of the Craya-Curtet number corresponds to large static pressure loss, indicating recirculation.
Table 2 Different Parameters of the AC
Fluid Dynamic Parameters U1 Bulk average velocity of the primary jet
U2 Bulk average velocity of the secondary jet
Ue Bulk average velocity of the average stream
{\displaystyle \Delta P^{\star }=P_{e}^{\star }-P_{i}^{\star }}
Stagnation pressure loss in the system (difference between exit and entrance stagnation pressure)
Problem Definition Parameters (PDPs)
{\displaystyle \alpha ={\frac {U_{2}}{U_{1}}}}
Secondary to primary bulk-average velocity ratio
{\displaystyle \beta ={\frac {r_{2}}{r_{1}}}}
{\displaystyle \gamma ={\frac {A_{e}}{A_{1}}}={\frac {r_{c}^{2}}{2r_{1}dr}}}
Governing Non-Dimensional Parameters (GNDPs)
{\displaystyle {\text{Re}}={\frac {2U_{2}r_{2}}{\nu }}}
{\displaystyle C_{t}={\frac {1+\beta }{\sqrt {\gamma (1+a^{2}\beta )}}}}
Craya-Curtet number
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Solve this: Draw the sign scheme1) x - 22 x - 3x - 1 02) x - - Maths - Linear Inequalities - 12660113 | Meritnation.com
Draw the sign scheme\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}1\right) \frac{{\left(x - 2\right)}^{2} \left(x - 3\right)}{x - 1} > 0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2\right) \frac{{\left(x - 2\right)}^{2} \left(x - 3\right)}{x - 1} \ge 0
\frac{{\left(x-2\right)}^{2} \left(x-3\right)}{x-1}>0\phantom{\rule{0ex}{0ex}}Note: Critical points are given by\phantom{\rule{0ex}{0ex}}x-2=0\phantom{\rule{0ex}{0ex}}\mathbit{x}\mathbf{=}\mathbf{2}\phantom{\rule{0ex}{0ex}}x-3=0\phantom{\rule{0ex}{0ex}}\mathbit{x}\mathbf{=}\mathbf{3}\phantom{\rule{0ex}{0ex}}x-1=0\phantom{\rule{0ex}{0ex}}\mathbit{x}\mathbf{=}\mathbf{1}\phantom{\rule{0ex}{0ex}}Sign will change around the point whose expression has odd power and will remain same around the point whose expression has even power.
x\in \left(-\infty ,1\right)\cup \left[3,\infty \right)\cup \left\{2\right\}\phantom{\rule{0ex}{0ex}}Note: We have included 2 because at x=2, expression is 0
Shubh Ashish answered this
Just use wavy curve method. Its too easy. If you are not aware of the method, i can help.
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Mathematics in medieval Islam - Wikipedia
(Redirected from Islamic mathematics)
Overview of the role of mathematics in medieval Islam
1.7 Double false position
2 Other major figures
Omar Khayyám's "Cubic equations and intersections of conic sections" the first page of the two-chaptered manuscript kept in Tehran University
The study of algebra, the name of which is derived from the Arabic word meaning completion or "reunion of broken parts",[3] flourished during the Islamic golden age. Muhammad ibn Musa al-Khwarizmi, a Persian scholar in the House of Wisdom in Baghdad was the founder of algebra, is along with the Greek mathematician Diophantus, known as the father of algebra. In his book The Compendious Book on Calculation by Completion and Balancing, Al-Khwarizmi deals with ways to solve for the positive roots of first and second degree (linear and quadratic) polynomial equations. He introduces the method of reduction, and unlike Diophantus, also gives general solutions for the equations he deals with.[4][5][6]
Al-Khwarizmi's algebra was rhetorical, which means that the equations were written out in full sentences. This was unlike the algebraic work of Diophantus, which was syncopated, meaning that some symbolism is used. The transition to symbolic algebra, where only symbols are used, can be seen in the work of Ibn al-Banna' al-Marrakushi and Abū al-Ḥasan ibn ʿAlī al-Qalaṣādī.[7][6]
On the work done by Al-Khwarizmi, J. J. O'Connor and Edmund F. Robertson said:[8]
"Perhaps one of the most significant advances made by Arabic mathematics began at this time with the work of al-Khwarizmi, namely the beginnings of algebra. It is important to understand just how significant this new idea was. It was a revolutionary move away from the Greek concept of mathematics which was essentially geometry. Algebra was a unifying theory which allowed rational numbers, irrational numbers, geometrical magnitudes, etc., to all be treated as "algebraic objects". It gave mathematics a whole new development path so much broader in concept to that which had existed before, and provided a vehicle for the future development of the subject. Another important aspect of the introduction of algebraic ideas was that it allowed mathematics to be applied to itself in a way which had not happened before."
Several other mathematicians during this time period expanded on the algebra of Al-Khwarizmi. Abu Kamil Shuja' wrote a book of algebra accompanied with geometrical illustrations and proofs. He also enumerated all the possible solutions to some of his problems. Abu al-Jud, Omar Khayyam, along with Sharaf al-Dīn al-Tūsī, found several solutions of the cubic equation. Omar Khayyam found the general geometric solution of a cubic equation.
Sharaf al-Dīn al-Ṭūsī (? in Tus, Iran – 1213/4) developed a novel approach to the investigation of cubic equations—an approach which entailed finding the point at which a cubic polynomial obtains its maximum value. For example, to solve the equation
{\displaystyle \ x^{3}+a=bx}
, with a and b positive, he would note that the maximum point of the curve
{\displaystyle \ y=bx-x^{3}}
{\displaystyle x=\textstyle {\sqrt {\frac {b}{3}}}}
, and that the equation would have no solutions, one solution or two solutions, depending on whether the height of the curve at that point was less than, equal to, or greater than a. His surviving works give no indication of how he discovered his formulae for the maxima of these curves. Various conjectures have been proposed to account for his discovery of them.[12]
The earliest implicit traces of mathematical induction can be found in Euclid's proof that the number of primes is infinite (c. 300 BCE). The first explicit formulation of the principle of induction was given by Pascal in his Traité du triangle arithmétique (1665).
In between, implicit proof by induction for arithmetic sequences was introduced by al-Karaji (c. 1000) and continued by al-Samaw'al, who used it for special cases of the binomial theorem and properties of Pascal's triangle.
In the twelfth century, Latin translations of Al-Khwarizmi's Arithmetic on the Indian numerals introduced the decimal positional number system to the Western world.[16] His Compendious Book on Calculation by Completion and Balancing presented the first systematic solution of linear and quadratic equations. In Renaissance Europe, he was considered the original inventor of algebra, although it is now known that his work is based on older Indian or Greek sources.[17][18] He revised Ptolemy's Geography and wrote on astronomy and astrology. However, C.A. Nallino suggests that al-Khwarizmi's original work was not based on Ptolemy but on a derivative world map,[19] presumably in Syriac or Arabic.
The spherical law of sines was discovered in the 10th century: it has been attributed variously to Abu-Mahmud Khojandi, Nasir al-Din al-Tusi and Abu Nasr Mansur, with Abu al-Wafa' Buzjani as a contributor.[13] Ibn Muʿādh al-Jayyānī's The book of unknown arcs of a sphere in the 11th century introduced the general law of sines.[20] The plane law of sines was described in the 13th century by Nasīr al-Dīn al-Tūsī. In his On the Sector Figure, he stated the law of sines for plane and spherical triangles and provided proofs for this law.[21]
In the 9th century, Islamic mathematicians were familiar with negative numbers from the works of Indian mathematicians, but the recognition and use of negative numbers during this period remained timid.[22] Al-Khwarizmi did not use negative numbers or negative coefficients.[22] But within fifty years, Abu Kamil illustrated the rules of signs for expanding the multiplication
{\displaystyle (a\pm b)(c\pm d)}
.[23] Al-Karaji wrote in his book al-Fakhrī that "negative quantities must be counted as terms".[22] In the 10th century, Abū al-Wafā' al-Būzjānī considered debts as negative numbers in A Book on What Is Necessary from the Science of Arithmetic for Scribes and Businessmen.[23]
By the 12th century, al-Karaji's successors were to state the general rules of signs and use them to solve polynomial divisions.[22] As al-Samaw'al writes:
Between the 9th and 10th centuries, the Egyptian mathematician Abu Kamil wrote a now-lost treatise on the use of double false position, known as the Book of the Two Errors (Kitāb al-khaṭāʾayn). The oldest surviving writing on double false position from the Middle East is that of Qusta ibn Luqa (10th century), an Arab mathematician from Baalbek, Lebanon. He justified the technique by a formal, Euclidean-style geometric proof. Within the tradition of medieval Muslim mathematics, double false position was known as hisāb al-khaṭāʾayn ("reckoning by two errors"). It was used for centuries to solve practical problems such as commercial and juridical questions (estate partitions according to rules of Quranic inheritance), as well as purely recreational problems. The algorithm was often memorized with the aid of mnemonics, such as a verse attributed to Ibn al-Yasamin and balance-scale diagrams explained by al-Hassar and Ibn al-Banna, who were each mathematicians of Moroccan origin.[24]
Sally P. Ragep, a historian of science in Islam, estimated in 2019 that "tens of thousands" of Arabic manuscripts in mathematical sciences and philosophy remain unread, which give studies which "reflect individual biases and a limited focus on a relatively few texts and scholars".[25][full citation needed]
'Abd al-Hamīd ibn Turk (fl. 830) (quadratics)
Abu'l-Hasan al-Uqlidisi (952–953) (arithmetic)
'Abd al-'Aziz al-Qabisi (d. 967)
Engraving of Abū Sahl al-Qūhī's perfect compass to draw conic sections.
^ Katz (1993): "A complete history of mathematics of medieval Islam cannot yet be written, since so many of these Arabic manuscripts lie unstudied... Still, the general outline... is known. In particular, Islamic mathematicians fully developed the decimal place-value number system to include decimal fractions, systematised the study of algebra and began to consider the relationship between algebra and geometry, studied and made advances on the major Greek geometrical treatises of Euclid, Archimedes, and Apollonius, and made significant improvements in plane and spherical geometry."
^ Smith (1958), Vol. 1, Chapter VII.4: "In a general way it may be said that the Golden Age of Arabian mathematics was confined largely to the 9th and 10th centuries; that the world owes a great debt to Arab scholars for preserving and transmitting to posterity the classics of Greek mathematics; and that their work was chiefly that of transmission, although they developed considerable originality in algebra and showed some genius in their work in trigonometry."
^ Lumpkin, Beatrice; Zitler, Siham (1992). "Cairo: Science Academy of the Middle Ages". In Van Sertima, Ivan (ed.). Golden age of the Moor, Volume 11. Transaction Publishers. p. 394. ISBN 1-56000-581-5. "The Islamic mathematicians exercised a prolific influence on the development of science in Europe, enriched as much by their own discoveries as those they had inherited by the Greeks, the Indians, the Syrians, the Babylonians, etc."
^ "algebra". Online Etymology Dictionary.
^ Boyer 1991, p. 228.
^ a b Gullberg, Jan (1997). Mathematics: From the Birth of Numbers. W. W. Norton. p. 298. ISBN 0-393-04002-X.
^ O'Connor, John J.; Robertson, Edmund F., "al-Marrakushi ibn Al-Banna", MacTutor History of Mathematics archive, University of St Andrews
^ O'Connor, John J.; Robertson, Edmund F., "Arabic mathematics: forgotten brilliance?", MacTutor History of Mathematics archive, University of St Andrews
^ a b Boyer 1991, pp. 241–242.
^ Berggren, J. Lennart; Al-Tūsī, Sharaf Al-Dīn; Rashed, Roshdi (1990). "Innovation and Tradition in Sharaf al-Dīn al-Ṭūsī's al-Muʿādalāt". Journal of the American Oriental Society. 110 (2): 304–309. doi:10.2307/604533. JSTOR 604533.
^ a b Sesiano, Jacques (2000). Helaine, Selin; Ubiratan, D'Ambrosio (eds.). Islamic mathematics. Mathematics Across Cultures: The History of Non-western Mathematics. Springer. pp. 137–157. ISBN 1-4020-0260-2.
^ O'Connor, John J.; Robertson, Edmund F., "Abu Mansur ibn Tahir Al-Baghdadi", MacTutor History of Mathematics archive, University of St Andrews
^ Allen, G. Donald (n.d.). "The History of Infinity" (PDF). Texas A&M University. Retrieved 7 September 2016.
^ Rosen 1831, p. v–vi.
^ Toomer, Gerald (1990). "Al-Khwārizmī, Abu Ja'far Muḥammad ibn Mūsā". In Gillispie, Charles Coulston (ed.). Dictionary of Scientific Biography. Vol. 7. New York: Charles Scribner's Sons. ISBN 0-684-16962-2 – via Encyclopedia.com.
^ Nallino 1939.
^ O'Connor, John J.; Robertson, Edmund F., "Abu Abd Allah Muhammad ibn Muadh Al-Jayyani", MacTutor History of Mathematics archive, University of St Andrews
^ Berggren 2007, p. 518.
^ a b c d e Rashed, R. (1994-06-30). The Development of Arabic Mathematics: Between Arithmetic and Algebra. Springer. pp. 36–37. ISBN 9780792325659.
^ a b Mat Rofa Bin Ismail (2008), "Algebra in Islamic Mathematics", in Helaine Selin (ed.), Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures, vol. 1 (2nd ed.), Springer, p. 115, ISBN 9781402045592
^ Schwartz, R. K. (2004). Issues in the Origin and Development of Hisab al-Khata'ayn (Calculation by Double False Position) (PDF). Eighth North African Meeting on the History of Arab Mathematics. Radès, Tunisia. Archived from the original (PDF) on 2014-05-16. Retrieved 2012-06-08. "Issues in the Origin and Development of Hisab al-Khata'ayn (Calculation by Double False Position)". Archived from the original (.doc) on 2011-09-15.
^ "Science Teaching in Pre-Modern Societies", in Film Screening and Panel Discussion, McGill University, 15 January 2019.
Berggren, J. Lennart (2007). "Mathematics in Medieval Islam". In Victor J. Katz (ed.). The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook (2nd ed.). Princeton, New Jersey: Princeton University Press. ISBN 978-0-691-11485-9.
Nallino, C.A. (1939), "Al-Ḥuwārismī e il suo rifacimento della Geografia di Tolomeo", Raccolta di scritti editi e inediti (in Italian), vol. V, Rome: Istituto per l'Oriente, pp. 458–532
Review: Toomer, Gerald J.; Berggren, J. L. (1988). "Episodes in the Mathematics of Medieval Islam". American Mathematical Monthly. Mathematical Association of America. 95 (6): 567. doi:10.2307/2322777. JSTOR 2322777.
Review: Hogendijk, Jan P.; Berggren, J. L. (1989). "Episodes in the Mathematics of Medieval Islam by J. Lennart Berggren". Journal of the American Oriental Society. American Oriental Society. 109 (4): 697–698. doi:10.2307/604119. JSTOR 604119.
Ronan, Colin A. (1983). The Cambridge Illustrated History of the World's Science. Cambridge University Press. ISBN 0-521-25844-8.
Cooke, Roger (1997). "Islamic Mathematics". The History of Mathematics: A Brief Course. Wiley-Interscience. ISBN 0-471-18082-3.
Joseph, George Gheverghese (2000). The Crest of the Peacock: Non-European Roots of Mathematics (2nd ed.). Princeton University Press. ISBN 0-691-00659-8. (Reviewed: Katz, Victor J.; Joseph, George Gheverghese (1992). "The Crest of the Peacock: Non-European Roots of Mathematics by George Gheverghese Joseph". The College Mathematics Journal. Mathematical Association of America. 23 (1): 82–84. doi:10.2307/2686206. JSTOR 2686206. )
Marcus du Sautoy (presenter) (2008). "The Genius of the East". The Story of Maths. BBC.
Hogendijk, Jan P. (January 1999). "Bibliography of Mathematics in Medieval Islamic Civilization".
O'Connor, John J.; Robertson, Edmund F. (1999), "Arabic mathematics: forgotten brilliance?", MacTutor History of Mathematics archive, University of St Andrews
Retrieved from "https://en.wikipedia.org/w/index.php?title=Mathematics_in_medieval_Islam&oldid=1084983047"
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Use the Divergence Theorem to calculate the surface
Use the Divergence Theorem to calculate the surface integral F · dS, that is, calculate the flux of F across S.
F\left(x,y,z\right)={x}^{3}+cosyi+{y}^{3}sinxzj+{z}^{3}+2ek
S is the surface of the solid bounded by the cylinder
{y}^{2}+{z}^{2}=4
x=0
x=4
a. The work required to move an object around a closed curve C in the presence of a vector force field is the circulation of the force field on the curve.
b. If a vector field has zero divergence throughout a region (on which the conditions of Green’s Theorem are met), then the circulation on the boundary of that region is zero.
c. If the two-dimensional curl of a vector field is positive throughout a region (on which the conditions of Green’s Theorem are met), then the circulation on the boundary of that region is positive (assuming counterclockwise orientation).
A particle moves along line segments from the origin to the points(3,0,0),(3,3,1),(0,3,1), and back to the origin under the influence of the force field
F\left(x,y,z\right)={z}^{2}i+3xyj+4{y}^{2}k
Use Stokes' Theorem to find the work done.
Use Green's Theorem to find
{\int }_{C}\stackrel{\to }{F}\cdot d\stackrel{\to }{r}
\stackrel{\to }{F}=⟨{y}^{3},-{x}^{3}⟩
and C is the circle
{x}^{2}+{y}^{2}=3
Evaluate the line integral by the two following methods. y) dx + (x+y)dy C os counerclockwise around the circle with center the origin and radius 3(a) directly (b) using Green's Theorem.
Use Greens
How do Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem relate to the Fundamental Theorem of Calculus for ordi-nary single integral?
Use Green's Theorem to evaluate
{\int }_{C}\left({e}^{x}+{y}^{2}\right)dx+\left({e}^{y}+{x}^{2}\right)dy
where C is the boundary of the region(traversed counterclockwise) in the first quadrant bounded by
y={x}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=4
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The function g is related to one of
jamesvivian911 2022-02-08 Answered
g\left(x\right)={x}^{2}+6
The parent function,
f\left(x\right)={x}^{2}
⇒
g\left(x\right)=f\left(x\right)+6
f\left(x\right)=\text{ }-\frac{1}{{x}^{2}}
f\left(x\right)={\mathrm{log}}_{2}x
y=\mathrm{cos}x
, sketch the function of
y=-\mathrm{cos}\left(3x-\frac{\pi }{6}\right)+4
y=-\mathrm{log}3
x a. Use transformations of the graphs of
y=\mathrm{log}2x
y=\mathrm{log}3x
x to graph the given functions. b. Write the domain and range in interval notation. c. Write an equation of the asymptote.
\begin{array}{|cccccc|}\hline x& -2& -1& 0& 1& 2\\ f\left(x\right)& -1& -3& 4& 2& 1\\ \hline\end{array}
\begin{array}{|cccccc|}\hline x& -3& -2& -1& 0& 1\\ g\left(x\right)& -1& -3& 4& 2& 1\\ \hline\end{array}
\begin{array}{|cccccc|}\hline x& -2& -1& 0& 1& 2\\ h\left(x\right)& -2& -4& 3& 1& 0\\ \hline\end{array}
y=\mathrm{sin}x
3\cdot \mathrm{sin}\left(\frac{x}{2}\right)
You have $30000 to invest. write the expression to find the total value of the investment if: a) you invest at 6% interest for 7 years compounded quarterly b) you invest for 7 years compounded continuously.
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Find the discriminant b2−4acb2-4ac, and use it to determine
Find the discriminant b2−4acb2-4ac, and use it to determine the number of real solutions to the equation. a2+8=4aa2+8=4a Discriminant: Real Solution:
{a}^{2}+8=4a
{a}^{2}-4a+8=0
D={b}^{2}-4ac
=\left(-4{\right)}^{2}-4\cdot 1\cdot 8=16-32=-16
x
{x}^{2}-4x+8=0
Using the quadratic formula, solve for x.
x=\frac{4±\sqrt{{\left(-4\right)}^{2}-4×8}}{2}=\frac{4±\sqrt{16-32}}{2}=\frac{4±\sqrt{-16}}{2}:
x=\frac{4+\sqrt{-16}}{2}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=\frac{4-\sqrt{-16}}{2}
\sqrt{-16}
in terms of i.
\sqrt{-16}=\sqrt{-1}\sqrt{16}=i\sqrt{16}:
x=\frac{4+i\sqrt{16}}{2}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=\frac{4-i\sqrt{16}}{2}
\sqrt{16}=\sqrt{{2}^{4}}=4=4:
x=\frac{4+i×4}{2}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=\frac{4-i×4}{2}
Factor the greatest common divisor (gcd) of
4,4i
and 2 from
4+4i
Factor 2 from
4+4i
4+4i:
x=\frac{1}{2}4+4
i or
x=\frac{4-4i}{2}
Cancel common terms in the numerator and denominator.
\frac{4+4i}{2}=2+2i:
x=2+2i
x=\frac{4-4i}{2}
4,-4i
4-4i.
4-4i
4-4i:
x=2+2i
x=\frac{1}{2}4-4i
\frac{4-4i}{2}=2-2i:
x=2+2i
x=2-2i
A batter hits a pitched ball when the center of the ball is 1.22 m above the ground. The ball leaves the bat at an angle of
{45}^{\circ }
with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107 m.
a) Does the ball clear a 7.32-m-high fence that is 97.5 m horizontally from the launch point?
b) At the fence, what is the distance between the fence top and the ball center?
Find the x- and y-intercepts of the graph of the equation.
y=14x-6
To calculate:The average cost when 1000 books are produced and 32000 books are produced.
Refer to the system of linear equations -2x+3y=5. 6x+7y=4 Is the augmented matrix row-equivalent to its reduced row-echelon form?
Find the x-and y-intercepts of the rational function
f\left(x\right)=\frac{6}{{x}^{2}+4x-7}
Starting from the point
\left(3,-2,-1\right)
, reparametrize the curve
x\left(t\right)=\left(3-3t,-2-t,-1-t\right)
in terms of arclength.
To calculate: The probability that the selected student received an "A" in the course.
A table depicting the grade distribution for a college algebra class based on age and grade.
|
Transforming Linear Graphs - Course Hero
College Algebra/Linear Functions and Modeling/Transforming Linear Graphs
Translations of Linear Functions
A translation of a linear function shifts its graph vertically or horizontally.
The parent function, or most basic function in the linear family, is the linear function
f(x) = x
. Its graph is a line that passes through the origin and has a slope of
m=1
. Other linear functions can be graphed as a transformation of the parent function
f(x) = x
A translation, or shift, is a transformation in which a graph is moved vertically or horizontally. For any function
f(x)
k
units or horizontally
h
h\gt0
k\gt0
f(x)+k
f(x)
translated up
k
f(x)-k
f(x)
translated down
k
f(x-h)
f(x)
translated right
h
f(x+h)
f(x)
translated left
h
A translation of the parent linear function
f(x) = x
can be viewed as either a horizontal or vertical translation.
For example, the parent function of a line is:
f(x) = x
y
-intercept is zero. Its
x
-intercept is also zero. One translation of the parent function is:
f(x)=x+4
It is a vertical translation of 4 units up or a horizontal translation 4 units to the left. The
y
-intercept is 4. The
x
-intercept is –4. Another translation is:
f(x)=x-5
It is a vertical translation 5 units down or a horizontal translation 5 units to the right. The
y
x
-intercept is 5. All functions have a slope of 1.
Translations of a parent function, such as
f(x)=x
, consists of a shift up, down, left, or right. For instance, a shift of the parent function 4 units up or to the left is represented by
f(x)=x+4
. A shift 5 units down or to the right is represented by
f(x)=x-5
Stretches, Compressions, and Reflections of Linear Functions
f(x)
, the function can be stretched, compressed, or reflected across an axis. A reflection is a transformation in which a figure is flipped across a line.
a>0
a
is a scaling factor:
af(x)
is a vertical stretch or compression of the graph of
f(x)
a
a \gt 1
f(x)
is stretched. For
0\lt a\lt 1
f(x)
is compressed.
-f(x)
f(x)
x
f(x) = x
Its slope is 1. The function that shows the vertical stretch of the parent function by a factor of 4 is:
f(x)=4x
The slope of the vertical stretch function is 4. The function that shows a vertical compression of the parent function by a factor of
\frac{1}{4}
f(x)=\frac{1}{4}x
The slope of the vertical compression function is
\frac{1}{4}
The function that shows a reflection across the
x
-axis is:
f(x)=-x
The slope of the reflection function is –1. All three functions have a
y
Stretches, compressions, and reflections of parent functions largely depend on the slope. For a parent function, such as
f(x) = x
, a vertical stretch of 4 is represented by
f(x)=4x
. A vertical compression of
\frac{1}{4}
f(x)=\frac{1}{4}x
. A reflection of the parent function over the
x
-axis is represented by
f(x)=-x
Combined Transformations of Linear Functions
Sometimes, a parent function may be transformed in more than one way. In general, the order of transformations is like the order of operations: Perform stretches, compressions, and reflections first (multiplication and division), followed by translations (addition and subtraction).
Graphing a Linear Function with Combined Transformations
Graph the given linear function by using transformations of its parent function
f(x)=x
f(x)=2x+3
Identify the types of transformations of the parent function
f(x)=x
that are used in the given function:
f(x)=2x+3
The first term in the linear function is
2x
. It shows that the parent function is multiplied by 2, so the graph will be stretched vertically by a factor of 2.
The next term in the given linear function is 3. It shows that 2 is added to the vertically stretched function f(x)=2x. So, the vertically stretched graph will be translated 3 units up.
Graph the first transformation:
f(x)=2x
The graph of the parent function is vertically stretched by a factor of 2.
Then graph the second transformation. The function
f(x)=2x
is translated up 3 units. This is represented by the graph of the function:
f(x)=2x+3
<Graphing Linear Functions>Linear Modeling
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Revision as of 10:36, 24 October 2014 by NikosA (talk | contribs) (Added link to working script i.fusion.hpf (for GRASS-GIS ver. 6.4))
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
{\displaystyle [0,255]}
{\displaystyle [0,2047]}
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Disc Method | Brilliant Math & Science Wiki
Mei Li, Beakal Tiliksew, Aditya Virani, and
Given a region under a curve
y=f(x)
a \leq x \leq b
, revolving this region around the
x
-axis gives a three-dimensional figure called a solid of revolution.
To find the volume of this solid of revolution, consider a thin vertical strip with thickness
\Delta x
y=f(x)
and consider revolving this thin strip around the
x
-axis. This strip generates a thin circular disk with radius
y=f(x)
\Delta x
, which has volume
\Delta V = \pi y^2 \Delta x = \pi (f(x))^2 \Delta x.
The disk method calculates the volume of the full solid of revolution by summing the volumes of these thin circular disks from the left endpoint
a
to the right endpoint
b
as the thickness
\Delta x
0
in the limit. This gives the volume of the solid of revolution:
V = \int_a^b dV = \int_a^b \pi \big(f(x)\big)^2 dx.
A cylinder with height
h
r
can be thought of as the solid of revolution obtained by revolving the line
y=r
x
-axis. Using the disk method, find the volume of the cylinder of height
h
r
Using the disk method, we revolve the line
y=r
x
x=0
x=h
\Delta V = \pi y^2 = \pi r^2 \Delta x
and the volume of the right circular cylinder is
\begin{aligned} V &= \int_{0}^{h} dV \\&= \int_{0}^{h} \pi r^2 dx \\ & = \left[ \pi r^2 x \right]^{h}_{0} \\ & = \pi r^2 h. \ _\square \end{aligned}
A sphere can be thought of as the solid of revolution obtained by revolving a semicircle around the
x
-axis. Using the disk method, find the volume of the sphere of radius
r
We can consider the semicircle to be centered at the origin with radius
r
, which has equation
y = \sqrt{r^2 - x^2}
\Delta V = \pi y^2 = \pi \big(r^2 - x^2\big) \Delta x
and the volume of the sphere is
\begin{aligned} V &= \int_{-r}^{r} dV \\&= \int_{-r}^{r} \pi \big(r^2-x^2\big) dx \\ & = \left[ \pi r^2 x - \frac{\pi x^3}{3} \right]^{r}_{-r}\\ &= \pi r^2 \big(r-(-r)\big) - \left( \frac{\pi r^3}{3} - \frac{\pi (-r)^3}{3} \right)\\ & = 2 \pi r^3 - \left(\frac{2 \pi r^3}{3} \right)\\ & = \frac{4\pi r^3}{3}. \ _\square \end{aligned}
Volume of a right circular cone:
A right circular cone can be thought of as the solid of revolution obtained by revolving a right triangle around the
x
-axis. Using the disk method, find the volume of the right circular cone of height
h
r
The hypotenuse of the right triangle we revolve around the
x
y = \frac{r}{h}x
\Delta V = \pi y^2 = \pi \left( \frac{r}{h}x \right)^2 \Delta x
and the volume of the right circular cone is
\begin{aligned} V &= \int_{0}^{h} dV \\&= \int_{0}^{h} \pi \frac{r^2}{h^2 }x^2 dx \\ & = \left[ \pi \frac{r^2}{h^2} \cdot \frac{1}{3} x^3\right]^{h}_{0}\\ & = \frac{ \pi r^2 h}{3}. \ _\square \end{aligned}
As shown in the examples above, we have the following relationships between the volumes of the cone and sphere as fractional parts of the volumes of their respective circumscribed cylinders:
\begin{aligned} V_\text{cone} & = \frac{1}{3} V_\text{cylinder}\\\\ V_\text{sphere} &= \frac{2}{3} V_\text{cylinder}. \end{aligned}
When the curve
\displaystyle{y=\frac{1}{x}}\ \ (1\le x \le \infty)
x
-axis, a funnel-shaped surface is formed. What is the volume of that revolution?
Using the disk method gives
\Delta V = \pi y^2 = \pi\cdot \frac{1}{x^2} \Delta x,
and the volume of the revolution is
\begin{aligned} V &= \int_{1}^{\infty} dV \\&= \int_{1}^{\infty} \pi\cdot\frac{1}{x^2}dx \\ & =\left.-\pi\cdot\frac{1}{x}\right|_{1}^{\infty}\\ &= -\pi(0-1)\\ & = \pi. \end{aligned}
It is interesting to note that the volume is finite.
_\square
Let there be a region
R:\{(x,y) \ | \ x^{1/4}+y^{4} \leq 1\}
. What is the volume of the solid generated when
R
is rotated around the line
x=0?
Note: You may use a calculator for the final step of your calculation.
This question was posed by my calculus teacher as the last and higher value exercise on the third Calculus I test.
A bullet is formed by revolving the area bounded by the the curve
y = \ln(x)
x = 1
x = e
x
It is then shot straight into a very thick wall (i.e. it does not pierce through the other side at all) making a closed cylindrical hole until it stops moving. Then the bullet is carefully extracted without affecting the hole at all, leaving an empty hole with a pointy end where the bullet once was.
The length of the entire hole is
e+1
. If the volume of the hole can be expressed as
\pi i e,
is a constant, find the value of
i
2 \pi r L
\pi r L
r L
2 r L
The above figure consists of a rectangle with a semicircle cut out of one end and added to the other end, where
L
is the width of the rectangle, and the curved length of a semicircle is
\pi r
To calculate the area of the shaded figure, Svatejas applies the disc method as follows:
Consider the axis of integration to be the semicircular arc, which has length
\pi r
. For each horizontal strip, we have an area element (technically length element) of
L
. Hence, the area is
\int_{R} L \, dR = \pi r \times L.
Inspiration, see solution comments.
Cite as: Disc Method. Brilliant.org. Retrieved from https://brilliant.org/wiki/disc-method/
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Discrete Random Variables - Definition | Brilliant Math & Science Wiki
Alexander Katz, Ognjen Vukadin, and Jimin Khim contributed
A random variable is a variable that takes on one of multiple different values, each occurring with some probability. When there are a finite (or countable) number of such values, the random variable is discrete. Random variables contrast with "regular" variables, which have a fixed (though often unknown) value.
For instance, a single roll of a standard die can be modeled by the random variable
X = \begin{cases} 1: \text{ die shows 1} \\ 2: \text{ die shows 2} \\ 3: \text{ die shows 3} \\ 4: \text{ die shows 4} \\ 5: \text{ die shows 5} \\ 6: \text{ die shows 6}, \end{cases}
where each case occurs with probability
\frac{1}{6}
Random variables are important in several areas of statistics; in fact, the standard probability distributions can be viewed as random variables. They also model situations involving conditional probability as well as problems involving linearity of expectation.
A probability space consists of
a sample space
\Omega
, the set of all possible outcomes,
\mathcal{F}
, where each event is a subset of
\Omega,
P: \mathcal{F} \rightarrow \mathbb{R}
, which assigns a probability to each event.
For example, consider flipping a fair coin. The sample space is the set
\{\text{heads}, \text{tails}\}
, and the possible events are
\{\}, \{\text{heads}\}, \{\text{tails}\}, \{\text{heads}, \text{tails}\}
. Here the empty set refers to neither heads nor tails occurring, and the set
\{\text{heads}, \text{tails}\}
refers to either one of heads or tails occurring. The function
P
is thus defined by
P\big(\{\}\big)=0,\quad P\big(\{\text{heads}\}\big)=\frac{1}{2},\quad P\big(\{\text{tails}\}\big)=\frac{1}{2},\quad P\big(\{\text{heads},\text{tails}\}\big)=1.
\mathcal{F}
consists of selected subsets of
\Omega
which represent relevant events of the experiment. For example, if the experiment consists of rolling a 6-sided die
\big(\Omega = \{1,2,3,4,5,6\}\big),
one can take
\mathcal{F}
to be the set of all subsets of
\Omega
\{1,3,5\}\in \mathcal{F}
is the event which can be described as "an odd number is rolled." If the experiment produced the outcome
\omega\in \Omega
, then each event
F\in \mathcal{F}
\omega\in F
is said to have happened. For example, if a 6-sided die is rolled and number
2
is the outcome, some of the events that have happened are
\{2\}
2
is rolled"),
\{1,2,3\}
("a number less than
4
is rolled"), and
\{2,4,6\}
("an even number is rolled").
\mathcal{F}
need not contain all possible subsets of
\Omega
. The restrictions on
\mathcal{F}
\Omega\in \mathcal{F};
A\in \mathcal{F}
\Omega\setminus A \in \mathcal{F};
\{A_i\}_{i\in I}
is a countable family of subsets of
\Omega
A_i\in \mathcal{F}
i\in I
\displaystyle\bigcup_{i\in I} A_i \in \mathcal{F}
P
P(\Omega) = 1
and is required to be countably additive, that is, for any countable collection of pairwise disjoint sets
\{F_i\}_{i\in I},
P\bigg(\displaystyle\bigcup_{i\in I} F_i\bigg) = \displaystyle\sum_{i\in I} P(F_i).
In this way, the probability space
(\Omega, \mathcal{F}, P)
becomes a measure space with measure
P
and the collection of all measurable sets
\mathcal{F}
The requirements on the probability space reflect basic rules of probability. For example, if
A\in \mathcal{F},
A^{c}:= \Omega\setminus A \in \mathcal{F}
A, A^{c}
are disjoint sets with
A\cup A^{c} = \Omega
P
is countably additive,
1 = P(\Omega) = P(A\cup A^{c}) = P(A) + P(A^{c}),
which is the complement rule of probability.
X
is formally defined as a measurable function from the sample space
\Omega
to another measurable space
S
. The requirement that
X
is measurable means that the inverse image of each measurable set
B
S
X
is a function taking values in
\mathbb{R}
which describes some property of the outcomes of the probability space. For example, a random variable
X
that denotes the number of heads in a single coin flip would have
X\big(\{\text{heads}\}\big)=1,\quad X\big(\{\text{tails}\}\big)=0.
Note that the random variable
X
does not return a probability by itself; rather, it is the probability space itself that assigns probabilities. This is perhaps more obvious when considering multiple coin flips; for instance, the probability space describing 3 consecutive coin flips has 8 possible outcomes and the probability that exactly two heads are flipped is the probability that the event
\{\text{heads}, \text{heads}, \text{tails}\}\cup \{\text{heads}, \text{tails}, \text{heads}\}\cup \{\text{tails}, \text{heads}, \text{heads}\}\in \mathcal{F}
happened:
P\big(\{\text{heads}, \text{heads}, \text{tails}\}\cup \{\text{heads}, \text{tails}, \text{heads}\}\cup \{\text{tails}, \text{heads}, \text{heads}\}\big) = \frac{3}{8}.
X
is a variable that denotes the number of heads flipped,
\text{Pr}(X=2) = \frac{3}{8}.
A discrete random variable is a random variable which takes only finitely many or countably infinitely many different values.
However, this does not imply that the sample space must have at most countably infinitely many outcomes. For example, if a point
a
is chosen uniformly at random in the interval
[-1,1]
, consider the random variable
X
which takes the value
-1
-1\leq a<0
1
otherwise. Although the sample space is the interval
[-1,1]
which is infinite and uncountable, the variable
X
is discrete since it takes only finitely many (i.e. 2) values.
Cite as: Discrete Random Variables - Definition. Brilliant.org. Retrieved from https://brilliant.org/wiki/discrete-random-variables-definition/
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Egyptian Fractions | Brilliant Math & Science Wiki
Ameya Daigavane, Worranat Pakornrat, Geoff Pilling, and
An Egyptian fraction is the sum of finitely many rational numbers, each of which can be expressed in the form
\frac{1}{q},
q
For example, the Egyptian fraction
\frac{61}{66}
\frac{61}{66} = \frac12 + \frac13 + \frac{1}{11}.
Greedy Algorithm for Finding Egyptian Fractions
Using Egyptian Fractions to Evenly Divide an Object into Equal Shares
Methods for Decomposing an Egyptian Fraction
Relevance to Modern Number Theory
\frac{m}{n}
m, n
are integers, there exist unit fractions (with numerator 1, and integer denominators) such that the sum of these unit fractions is equal to
\frac{m}{n}
\dfrac{2}{3}= \dfrac{1}{2} + \dfrac{1}{6}.
However, the representation of
\frac{m}{n}
in this form is not unique because of the identity
1 = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}.
\dfrac{2}{3}= \dfrac{1}{2} + \dfrac{1}{6} = \dfrac{1}{2} + \dfrac{1}{6}\left(\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}\right) = \dfrac{1}{2} + \dfrac{1}{12} + \dfrac{1}{18} + \dfrac{1}{36}
as well, and we can keep on decomposing any fraction as such.
Dating back to over 2,000 B.C., the Egyptians had developed their fraction system for mathematical division and calculation for any rational numbers. The Egyptian mathematicians exclusively used only unit fractions in their perception and did not seem to accept the idea of vulgar fraction, where the numerator is divided by denominator, as we do today. For example, if a cylindrical bucket has
\frac{3}{5}
of water full, we can understand that by dividing the bucket into 5 equal heights, the water level will reach the brim of such 3 heights. However, for the Egyptians, other numerators apart from 1 were simply inconsistent with their algorithm. Instead, they would see such volume as half a bucket full of water plus one-tenth of that, for
\frac{1}{2}+\frac{1}{10}=\frac{3}{5}
Moreover, they specifically designated the hieroglyphics for various denominators or rather symbols of equal dividing for any natural number. As such, their numerator would be assumed to be 1 though there were rare exceptions for
\frac{2}{3}
\frac{3}{4}
, for which they invent special symbols. The classic examples of the Egyptian fractions could be found in the ancient image of the Eye of Horus as shown below:
The reasons behind the Egyptian usage of such system remained unclear, but they might have perceived these unit fractions as "units" to be added as we similarly build up integers by summing the decimal units system. Their further study also contributed vastly to numerous writings, problems, and solutions. One of the most well-known work was called the Rhind Papyrus, written by Ahmes, which contained a table of converting
\frac{2}{n}
to the sets of Egyptian fractions of various terms. This proved that the Egyptian fractions were indeed foundation of many Egyptian mathematical development and civilization during that period of time.
One of the simplest algorithms to understand for finding Egyptian fractions is the greedy algorithm. With this algorithm, one takes a fraction
\frac{a}{b}
and continues to subtract off the largest fraction
\frac{1}{n}
until he/she is left only with a set of Egyptian fractions.
Find the Egyptian fraction representation of
\frac{8}{9}
The greatest unit fraction less than
\frac{8}{9}
\frac{1}{2}
The remainder is
\frac{7}{18}
\frac{7}{18}
\frac{1}{3}
\frac{1}{18}
This is a unit fraction, so the answer is given by
\dfrac{8}{9} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{18}.\ _\square
A recursive implementation of this algorithm in a perl script called "egyptian" is as follows:
($a,$b) = @ARGV;
$i = 2; while (1/$i > $a/$b) {$i++}
print "Greatest unit fraction: 1/$i\n";
$lcm = lcm($b, $i);
$a2 = $a * $lcm / $b;
$itop = $lcm / $i;
$remainder = $a2 - $itop;
$gcd = gcd($remainder,$lcm);
$r2 = $remainder/$gcd;
$lcm2 = $lcm/$gcd;
print "Remainder: $r2/$lcm2\n";
if ($r2 == 1) {
print "Greatest unit fraction: $r2/$lcm2\n";
print `egyptian $r2 $lcm2`;
while ($a) { ($a, $b) = ($b % $a, $a) }
($a && $b) and $a / gcd($a, $b) * $b or 0
The Egyptian fractions were particularly useful when dividing a number of objects equally for more number of people. This was practically important because many of the Egyptian structures required massive labor work. Any uneven distribution of food ration among the labors could easily kindle dispute and disrupt their work process.
For example, if the Egyptian boss were to give 5 Kubz (Arabic pancake-like bread) rations for 5 workmen, and then he just realized there was an extra man in the site, how would he divide 5 Kubz among 6 men? The answer is, as we know,
\frac{5}{6}
Using common sense, we would simply divide each Kubz into 6 equal portions, resulting in 30 pieces, and each workman would happily pick 5 of them. Surely, the problem was solved, but the Egyptian mathematician would apply a more elegant way to divide the Kubz by following the Egyptian fractions:
\dfrac{5}{6} = \dfrac{1}{2}+\dfrac{1}{3}.
This way, each workman would get half a Kubz plus one-third of a Kubz. To do so, first we take 3 Kubz and divide each in halves, resulting in 6 equal half portions, and each man would obtain each half. Then we take the remaining 2 Kubz and divide each into 3 equal portions, resulting in 6 equal one-third portions, and again each man would get one. By using this method, only a few cuts are needed for each Kubz without any need to go through all the same 6-denominator!
Naturally, the easiest way to break down vulgar fraction into Egyptian summation is to follow through the greedy algorithm, where the unit fraction nearest to the vulgar fraction is put in the equation first, followed by another nearest one to the remained, and so on. As in the previous example,
\frac{1}{2}
is the nearest fraction to
\frac{5}{6}
as shown. Nonetheless, this algorithm may not give the shortest way of the summation and thus may result in more unit fraction terms to be used.
You were having a party and had made 11 congruent pancakes for 11 people before an unexpected guest showed up, and there was no flour left to make another pancake.
As a result, for an even distribution to the 12 people, you decided to cut the pancakes under the condition that each cut would be along a pancake's diameter only. For instance, to divide a pancake into 12 equal slices, 6 cuts would be made, as shown above right.
What would be the least number of cuts you need to make to achieve equal shares for everyone?
No stacking of pancakes is allowed.
Each cut must be done on one pancake at a time, where the pancake must be divided into equal slices.
Some useful algebraic identities for the Egyptian fraction decomposition are
\begin{aligned} \dfrac{a}{ab - 1} &= \dfrac{1}{b} + \dfrac{1}{b(ab - 1)} \\\\ \dfrac{1}{a} &= \dfrac{1}{a + 1} + \dfrac{1}{a(a + 1)}. \end{aligned}
These are easily proved by the laws of addition of fractions.
\frac{8}{11}
\dfrac{8}{11} = \dfrac{6}{11} + \dfrac{2}{11}.
11 = 2 \cdot 6 - 1.
Thus, using the above identity
\dfrac{a}{ab - 1} = \dfrac{1}{b} + \dfrac{1}{b(ab - 1)},
\begin{aligned} \dfrac{6}{11} &= \dfrac{1}{2} + \dfrac{1}{22} \\\\ \dfrac{2}{11} &= \dfrac{1}{6} + \dfrac{1}{66} \\\\ \Rightarrow \dfrac{8}{11} &= \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{22} + \dfrac{1}{66}.\ _\square \end{aligned}
and
b
are distinct positive integers satisfying
\frac13 = \frac1a+\frac1b
a+b
Bonus: Generalize it for any unit fraction
\frac1c
c
\large abc = 10 (a + b + c )
a,b,c
be pairwise coprime positive integers greater than 1 that satisfy the equation above. What is the minimum value of
ab+bc+ca
Egyptian fractions show up in all sorts of places! Below are some problems that involve Egyptian fraction representations. First, we prove a useful lemma:
r
s
are two integers, such that
\gcd(r,s) = 1
a_1, a_2, \ldots, a_n
\frac{r}{s} = \sum\limits_{i = 1} ^n \frac{1}{a_i},
s
a_1\times a_2 \times \cdots \times a_n
By summing up the fractions on the right-hand side, we have
\frac{r}{s} = \frac{\sum\limits_\text{cyc} {(a_1\times a_2 \times \cdots \times a_{n-1})}}{a_1\times a_2 \times \cdots \times a_n}.
\sum\limits_\text{cyc} {(a_1\times a_2 \times \cdots \times a_{n-1})} = t
is an integer. Thus, we have
st = r (a_1\times a_2 \times \cdots \times a_n),
r (a_1\times a_2 \times \cdots \times a_n)
s.
s
cannot divide
r
\gcd(r,s) = 1
s
a_1\times a_2 \times \cdots \times a_n
_\square
[IMO 1979 Problem 1]
p
q
\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \cdots -\frac{1}{1318}+\frac{1}{1319},
p
1979
\begin{aligned} \dfrac{p}{q} &= 1 + \dfrac{1}{2}+ \dfrac{1}{3} + \dfrac{1}{4}+ \cdots + \dfrac{1}{1318}+\dfrac{1}{1319} - 2\left(\dfrac{1}{2} + \dfrac{1}{4} + \cdots + \dfrac{1}{1318}\right) \\ &= 1 + \dfrac{1}{2}+ \dfrac{1}{3} + \dfrac{1}{4}+ \cdots + \dfrac{1}{1318}+\dfrac{1}{1319} - \left(1 + \dfrac{1}{2}+ \dfrac{1}{3} + \dfrac{1}{4}+ \cdots + \dfrac{1}{659}\right) \\ &= \dfrac{1}{660}+ \dfrac{1}{661} + \dfrac{1}{662}+ \cdots + \dfrac{1}{1318}+\dfrac{1}{1319}. \end{aligned}
\dfrac{1}{660} + \dfrac{1}{1319} = \dfrac{660 + 1319}{660 \cdot 1319} = \dfrac{1979}{660 \cdot 1319}.
Similarily,
\dfrac{1}{661} + \dfrac{1}{1318} = \dfrac{661 + 1318}{661 \cdot 1318} = \dfrac{1979}{661 \cdot 1318}.
We can pair up all the fractions as such (as there are an even number of them).
\frac{p}{q} = 1979 \cdot \frac{r}{s},
\frac{r}{s} = \dfrac{1}{660 \cdot 1319} + \dfrac{1}{661 \cdot 1318} + \cdots + \dfrac{1}{989 \cdot 990}.
p = 1979 \cdot \frac{qr}{s}.
1979
is prime! Therefore, as
p
s
1979
(
as
660 \times 661 \times \cdots \times 1319
, using our lemma above
)
p
1979
_\square
The next problem is from the USA Mathematical Olympiad.
[USAMO 2010 Problem 5]
p
q = \frac{3p-5}{2}
S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7} + \cdots + \frac{1}{q(q+1)(q+2)}
\frac{1}{p}-2S_q = \frac{m}{n}
m
n
m - n
p
\dfrac{2017}{n} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}
n
be an integer greater than 2017 and
a, b, c
be distinct pairwise coprime positive integers satisfying the equation above.
What is the least possible value of
n?
Cite as: Egyptian Fractions. Brilliant.org. Retrieved from https://brilliant.org/wiki/egyptian-fractions/
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Using Exponential Decay to Explain Amplitude Decreases | Brilliant Math & Science Wiki
Consider the equation of motion of the underdamped harmonic oscillator:
x(t) = Ae^{-\frac{b}{2m} t} e^{ i\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }t} + Be^{-\frac{b}{2m} t} e^{ -i\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }t}.
This solution describes rapid oscillation within an envelope of exponentially decaying envelope. The amplitudes of the critically damped and overdamped harmonic oscillators similarly decay exponentially.
Several parameters are used throughout physics and engineering literature to describe how the amplitude of a damped harmonic oscillator decays over time. The most mathematically straightforward parameter is the
1/e
decay time, often denoted as
\tau
. Suppose a damped harmonic oscillator starts at amplitude
x_0
; then the amplitude of the damping envelope is
x_0 e^{-\frac{b}{2m} t}
1/e
decay time is defined as the time
\tau
for which the amplitude has decreased to
x_0 / e \approx .368 x_0
. This is equivalent to the exponent in the decay envelope taking value
-1
-\frac{b}{2m} \tau = -1 \implies \tau = \frac{2m}{b} .
10 \text{ kg}
mass is attached to a spring of spring constant
10 \text{ N}/\text{m}
. The entire system is submerged in water, which exerts a viscous damping force on the mass
F_d = -(2 \text{ N}\cdot \text{s}/\text{m}) \:v
. The mass is pulled so that the spring is displaced from equilibrium by
.1 \text{ m}
and is released. Find the
1/e
decay time of oscillation in seconds.
A more sophisticated parameter is the quality factor
Q
Q = \frac{\text{energy stored}}{\text{energy dissipated per radian}}.
As a mnemonic for understanding and remembering the name, a high quality crystal will ring for a very long time when struck. Damped harmonic oscillators with large quality factors are underdamped and have a slowly decaying amplitude and vice versa. Critical damping occurs at
Q = \frac12
, marking the boundary of the two damping regimes.
What is the quality factor of a damped harmonic oscillator in terms of
k
m
b
The stored energy in the damped harmonic oscillator is the "spring potential energy":
E(t) = \frac12 kA(t)^2
A(t)
is the amplitude of the harmonic oscillator. Recalling that the damped harmonic oscillator has a
e^{-\frac{b}{2m} t}
decay envelope, this is equal to:
E(t) = \frac12 k A^2 e^{-\frac{b}{m} t} = E_0 e^{-\frac{b}{m} t}.
The energy dissipated per radian is:
\Delta E = \left|\frac{dE}{dt}\right| \Delta t,
\Delta t
giving the time it takes to oscillate through one radian, equal to
\frac{1}{\omega}
\frac{dE}{dt} = -\frac{b}{m} E(t)
, so the energy dissipated is:
\Delta E = \frac{b}{m \omega} E,
and finally the quality factor is:
Q = \frac{E}{b/(m\omega) E} = \frac{m\omega}{b},
\omega = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }
is the frequency of the damped harmonic oscillator. For highly underdamped systems,
\omega \approx \sqrt{\frac{k}{m}}
and the quality factor is
Q \approx \frac{\sqrt{km}}{b}
. From this, it is apparent that critical damping occurs at
Q = \frac12
by squaring both sides and comparing to the criterion for critical damping.
A last metric that is more common in engineering literature for describing amplitude attenuation in damped oscillators is the logarithmic decrement
\delta
, defined as the
Challenge problem: show that
\delta = \frac{\pi}{Q}
It is important to note that the viscous damping model is a good model only for intermolecular forces in certain fluids. It is not a good model for dry friction, the usual friction force from rubbing against solid objects governed by the equation
F_f = \mu N
\mu
the coefficient of friction and
N
the normal force. Interestingly, simple models of dry friction are solvable and demonstrate a linear damping envelope rather than an exponential.
Cite as: Using Exponential Decay to Explain Amplitude Decreases. Brilliant.org. Retrieved from https://brilliant.org/wiki/using-exponential-decay-to-explain-amplitude/
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Heat Transfer | Brilliant Math & Science Wiki
Ashish Menon, Andrew Ellinor, Sravanth C., and
This wiki is the subject of a wiki collaboration party set to be held on Saturday, April 9th at 8:30pm IST (8:00am PST). Please add examples to the appropriate headings or under the examples heading.
Contribute wherever you would like! Here are the author breakdowns:
Types of heat transfer: Sravanth/Julien
Phases of matter: Ashish
Steady-state heat transfer: Ameya
Specific heat: Ashish
Applications (green house effect, thermos): Sravanth
Examples for Wiki Collaboration
How much heat energy is gained when
5\text{ kg}
10\ ^\circ\text{C}
is brought to its boiling point?
(Heat capacity of water) =
4200\text{ J kg}^{-1}\ ^\circ\text{C}^{-1}
\begin{aligned} \text{Rise in temperature} ({\theta}_R) & = (100 - 10)\ ^\circ\text{C}\\ & = 90\ ^\circ\text{C}\\ \\ \text{Rise in temperature} & = m × c × {\theta}_R\\ & = 5 × 4200 × 90\\ & = 1890000 \text{ J}. \ _\square \end{aligned}
A waterfall is
840\text{ m}
high. If the initial temperature of water at the top of the waterfall is
15\ ^\circ\text{C}
, what is its temperature when it reaches the bottom?
Assume that all the energy of falling water changes into heat energy. Take
g = 10\text{ m/s}^2
and specific heat capacity of water as
4200\text{ J kg}^{-1}\ ^\circ\text{C}^{-1}
\begin{aligned} (\text{K.E. of falling water at the bottom}) & = (\text{P.E. of water at the top})\\ & = m × g × h\\ & = (\text{Heat energy produced by water})\\ \\ m × g × h & = m × c × {\theta}_R\\ \\ \Rightarrow {\theta}_R & = \dfrac{g × h}{c}\\ & = \dfrac{10 × 840}{4200}\\ & = 2\ ^\circ\text{C}\\ \\ (\text{Temperature of water at the bottom}) & = (15 + 2)\ ^\circ\text{C}\\ & = 17\ ^\circ\text{C}. \ _\square \end{aligned}
Transfer of heat occurs in three different ways. As discussed earlier, a physical object can be transferred via three modes, but even energy can be transferred in similar modes.
Conduction is the transfer of heat which happens through the particles in the medium as a result of a series of collisions, without the actual movement of particles, i.e. the particles do not act as the carriers of energy but they pass the energy with the help of vibrations.
Heat conduction may be described quantitatively as the time rate of heat flow in a material for a given temperature difference.
Main article: Phase changes
How much heat energy is required to melt
5 × {10}^6\text{ kg}
of iron?
\big(
Specific latent heat (SLH) of iron
\big)
2.7 × {10}^5\text{ J/kg}.
\begin{aligned} (\text{Heat energy required to melt iron}) & = m × L\\ & = 5 × {10}^6 × 2.7 × {10}^5\\ & = 13.5 × {10}^{11}\\ & = 1.35 × {10}^{12}. \ _\square \end{aligned}
70\text{ g}
-30\ ^\circ\text{C}
is heated by a burner such that it forms an equal amount of water at
40\ ^\circ\text{C}
. Find the amount of heat energy that should be supplied for this to happen.
(SHC of ice) =
2.1\text{ J}/g\ ^\circ\text{C},
(SLH of ice) =
336\text{ J}/g,
and (SHC of water) =
4.2\text{ J}/g\ ^\circ\text{C}
\begin{aligned} (\text{Heat energy required to raise the temperature of ice to 0°C}) & = m × c × {\theta}_R\\ & = 70 × 2.1 × 30\\ & = 4410\text{ J}\\ \\ (\text{Heat energy required to melt the ice}) & = 70 × 336\\ & = 23520\text{ J}\\ \\ (\text{Heat energy required to raise the temp of ice at 0°C to water at 40°C}) & = m × c × {\theta}_R\\ & = 11760\text{ J}\\ \\ \Rightarrow (\text{Total heat energy required}) & = (4410 + 23520 + 11760)\text{ J}\\ & = 39690 \text{ J}.\ _\square \end{aligned}
300\text{ g}
-25\ ^\circ\text{C}
is contained in a vessel of mass
10\text{ g}
and specific heat capacity
0.4\text{ J}/g\ ^\circ\text{C}
. How much water at
50\ ^\circ\text{C}
should be added to the vessel such that the final temperature of the combination is
15\ ^\circ\text{C}?
Assume that the vessel is at the same initial temperature as that of the ice contained in it.
336\text{ J}/g,
2.1\text{ J}/g\ ^\circ\text{C},
4.2\text{ J}/g\ ^\circ\text{C}.
Let amount of water required be
x.
\begin{aligned} (\text{Amount of heat lost by water}) & = m × c × {\theta}_F\\ & = x × 4.2 × 35\\ & = 147x\\ \\ (\text{Heat energy required to raise the temperature of ice to 0°C}) & = m × c × {\theta}_R\\ & = 300 × 2.1 × 25\\ & = 15750\text{ J}\\ \\ (\text{Heat energy required to melt the ice}) & = 300 × 336\\ & = 10080\text{ J}\\\\ (\text{Heat energy required to raise temp. of ice at 0°C to water at 15°C}) & = m × c × {\theta}_R\\ & = 18900\text{ J}\\ \\ (\text{Heat energy required to raise the temp. of vessel}) & = m × c × {\theta}_R\\ & = 10 × 0.4 × (25 + 15)\\ & = 160\text{ J}. \end{aligned}
According to the law of conservation of energy, heat gained is equal to heat lost. Thus,
\begin{aligned} 147x & = (15750 + 10080 + 18900 + 160)\text{ J}\\ & = 44890\text{ J}\\ x & = 305.37415\text{ J}.\ _\square \end{aligned}
A solid initially at
20\ ^\circ\text{C}
is heated. The graph shows variation in temperature with the amount of heat energy supplied to it. Now, if the specific heat capacity of the solid is
2\text{ J}{g}^{-1}\ ^\circ\text{C}^{-1},
(i) the mass of the solid and
(ii) the specific latent heat of fusion of the solid.
Submit your answer as the sum of (i) (in grams) and (ii)
(
\text{ J/g}).
500\text{ g}
of a metal of specific heat capacity
0.25 \text{ Jg}^{-1}\ ^\circ\text{C}^{-1}
125\ ^\circ\text{C}
is added to an equal amount of ice at
0\ ^\circ\text{C}
. Did all the ice melt?
If so, enter 555.
If not, enter the quantity of unmelted ice in grams (up to 2 decimal places).
Note: Specific latent heat of ice =
336\text{ Jg}^{-1}.
\text{Q = Cm}\Delta\text T
5\text{ kg}
10\text{ }^\circ\text{C}
Specific heat capacity of water is equal to
4200 \text{ J kg}^{-1}{\text{ }^\circ\text{C}}^{-1}.
\begin{aligned} \text{Rise in temperature}({\theta}_R) & = (100 - 10)\text{ }^\circ\text{C}\\ & = 90\text{ }^\circ\text{C}\\ \\ \Rightarrow (\text{Rise in temperature}) & = m × c × {\theta}_R\\ & = 5 × 4200 × 90\\ & = 1890000\text{ J}. \ _\square \end{aligned}
The heat capacity of a solid of mass
200\text{ g}
500\text{ J}/ \, ^\circ\text{C}
. Calculate the specific heat capacity of the solid.
\begin{aligned} (\text{Heat capacity}) & = \text{m × c}\\ 500 & = 200 × c\\ \Rightarrow c & = \dfrac{500}{200}\\ & = 2.5 \text{ Jg}^{-1}{\text{ }^\circ\text{C}}^{-1}. \ _\square \end{aligned}
Water at
80\text{ }^\circ\text{C}
is poured in a bucket which contains
5\text{ kg}
of crushed ice, such that all the ice melts and the final temperature recorded is
0\text{ }^\circ\text{C}
. Calculate the amount of hot water added to ice.
Take specific heat capacity of water is
4200\text{ J kg}^{-1}{\text{ }^\circ\text{C}}^{-1}.
Also, take specific latent heat of fusion of ice is
336000\text{ J kg}^{-1}
Let the mass of hot water be
x\text{ kg}.
\begin{aligned} (\text{Heat given out by hot water}) & = m × c × {\theta}_F\\ & = x × 4200 × 80\\ & = 336000 x\text{ J} \\ \\ (\text{Heat absorbed by ice to form water at 0°C}) & = m × L\\ & = 5 × 336000\\ & = 1680000\text{ J}. \end{aligned}
\begin{aligned} (\text{Heat given out by hot water}) & = (\text{Heat absorbed by cold water})\\ 336000x &= 1680000\\ \\ \Rightarrow x & = \dfrac{1680000}{336000} =5 \text{ (kg)}.\ _\square \end{aligned}
A 50-watt immersion heater just keeps
375\text{ g}
of a molten material at its melting point. If I switch off the heater, then the temperature starts falling 5 minutes later. Calculate the specific latent heat of fusion of the metal.
Submit your answer in
\text{ Jg}^{-1}
100\text{ g}
of solid
A
of specific heat capacity
50\text{ J/g}\, ^\circ\text{C}
90\ ^\circ\text{C}
is placed in a superconducting vessel of mass
10\text { g}
1\text { J/g}\, ^\circ\text{C}
. Then some amount of water at
40\ ^\circ\text{C}
is added to the vessel to cool the mixture to
55\ ^\circ\text{C}
Now, the above mixture after cooling down obtains a specific heat capacity of
25\text{ J/g}\, ^\circ\text{C}
. Then, to the above mixture is added a solid
B
20 \text{ J/g}\, ^\circ\text{C}
10\ ^\circ\text{C}
such that the final temperature of the overall combination becomes
30\ ^\circ\text{C}
How much amount of solid
B
is added (up to 2 decimal places)?
Main article: Calorimetry
100\text{ g}
50\ ^\circ\text{C}
is poured into a vessel containing
120\text{ g}
10\ ^\circ\text{C}
. The final temperature recorded is
20\ ^\circ\text{C}
. Calculate the thermal capacity of the vessel. Consider that the vessel is at the same initial temperature as the cold water.
\text{Substance}
\hspace{10mm} \text{Mass} \hspace{10mm}
\hspace{10mm} \text {S.H.C} \hspace{10mm}
\hspace{5mm} \text{Initial temperature} \hspace{5mm}
\hspace{10mm} \text{Change in temperature}\hspace{10mm}
\text{Hot water}
100\text{ g}
4.2\text{ J/g}\, ^\circ\text{C}
50\ ^\circ\text{C}
(50 - 20)\ ^\circ\text{C} = 30\ ^\circ\text{C}
\text{Cold water}
120\text{ g}
4.2\text{ J/g}\, ^\circ\text{C}
10\ ^\circ\text{C}
(20 - 10)\ ^\circ\text{C} = 10\ ^\circ\text{C}
\text{Vessel}
?
?
10\ ^\circ\text{C}
(20 - 10)\ ^\circ\text{C} = 10\ ^\circ\text{C}
\begin{aligned} (\text{The thermal capacity of the vessel}) & = m × c = x (\text{say})\\ \\ (\text{Heat lost by hot water}) & = m × c × {\theta}_F\\ & = 100 × 4.2 × 30\\ & = 12600\text{ J}\\ \\ (\text{Heat gained by cold water}) & = m × c × {\theta}_R\\ & = 120 × 4.2 × 10\\ & = 5040\text{ J}\\ \\ (\text{Heat gained by the vessel}) & = m × c × {\theta}_R\\ & = x × 10 (\text{substituting the value of }mc = x)\\ & = 10x \text{ J}. \end{aligned}
\begin{aligned} (\text{Heat gained}) & = (\text{Heat lost})\\ 5040 + 10x & = 12600\\ 10x & = 7560\\ \Rightarrow x & = 756. \end{aligned}
Therefore, the thermal capacity of the vessel is
756\text{ J/}^\circ\text{C}. \ _\square
There are quite a few applications of heat transfer between particles. We will be discussing two of them.
Cite as: Heat Transfer. Brilliant.org. Retrieved from https://brilliant.org/wiki/heat-transfer/
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Seminar talk, 11 May 2022 - Geometry of Differential Equations
Seminar talk, 11 May 2022
Speaker: Ilia Gaiur
Title: Hamiltonian reduction and rational Calogero system
In my talk I am going to give an introduction to the theory of the moment map for the Hamiltonian group action on the symplectic manifolds with the focus on Hamiltonian reduction and integrable systems. In particular, I will show how to translate symmetries of the Hamiltonian system to the first integrals using the moment map and what kind of systems we may obtain by performing such reduction. As the main example, I will demonstrate how to obtain a rational Calogero system from the free particle system on the cotangent bundle to the Lie algebra
{\displaystyle su(n)}
Retrieved from "https://gdeq.org/w/index.php?title=Seminar_talk,_11_May_2022&oldid=6649"
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VisualEditor:Test - MediaWiki
3World languages
6Shape of this world
This photo of this world is called The Blue Marble.
This world is a planet where a society of people has formed.
Authors sometimes invent new worlds. The authors use these worlds as the setting for their stories. Some authors invent worlds that have magic.
Nobody knows whether there are intelligent beings on other worlds.
6 Shape of this world
World literature[edit]
World literature is literature that is read by many people all over this world. World literature is different from national literature.
Worlds in literature[edit]
Creating a different world is a literary device used by authors to illustrate ideas. By placing the story in the setting of a different world, the author can change the way that things happen in the world. For example, the author might imagine a world that has very little water or a world that has very little dry land. Deciding what the world looks like and how the world works is called world-building. Thinking about the world helps the author make good choices about what happens to the characters in the story. Some authors think about many details, such as what languages the characters speak and what the architecture is on the world.
Worlds in science fiction[edit]
Science fiction stories often use different worlds. Frank Herbert's famous Dune series focused on a world called Arrakis that produces a rare chemical substance.
Often a science-fiction story will be involve multiple worlds. The Foundation series by Isaac Asimov was set in a galaxy with thousands of populated worlds. The Star Wars movies had a several important worlds, and characters traveled between them. Some authors of science fiction worlds try to make them scrupulously obey the laws of physics.
Fantasy worlds[edit]
J.R.R. Tolkien Middle Earth The Lord of the Rings triology Middle-earth has some qualities similar to Mediæval Europe. The author added magical creatures like elves and wizards. At the end of the story, some magical creatures leave the world.
Some languages spoken in many parts of the world. These are called world languages. As of 2016, English is the most common world language. Previously, French was the most popular language in the West. Chinese was used by traders in all of East Asia for centuries. Arabic is common in the Middle East, Northern Africa, and other parts of the world.
World health[edit]
The World Health Organisation (French: Organisation mondiale de la santé) is an international organisation for public health.[1] It is part of the United Nations. The World Health Organisation began in 1948.[1] It wants people to be healthy and safe. It studies public health and tells governments and other organisations how to help people become healthy.
The organisation counts the number of people with health problems. These health problems include influenza, HIV infection, and depression. It also counts the number of people who experience other problems. These problems include dirty water, violence, and hunger.
The World Bank is also interested in health.[3] Health affects economic prospects.
Shape of this world[edit]
This world is not a perfect sphere. It is slightly flattened. This is the mathematical formula for measuring the flatness of a sphere:
{\displaystyle {\begin{aligned}f&={\frac {a-b}{a}}.\end{aligned}}}
For this world,
{\textstyle f\,\!}
{\textstyle f\,\!}
{\textstyle f\,\!}
Local planets[edit]
A world is on a planet. There are different types of planets. There are several types of planets in this solar system:
These are the planets in this solar system:
Four large satellites
63 other satellites
62 satellites. Some are very small. The largest, called Titan, is larger than the planet Mercury.
Seven are large.
Five satellites
One satellite, called Triton
Comparison of local planets (not to scale)
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Lando S. Polynomial graph invariants and the KP hierarchy (abstract) - Geometry of Differential Equations
Lando S. Polynomial graph invariants and the KP hierarchy (abstract)
Title: Polynomial graph invariants and the KP hierarchy
We describe a large family of polynomial graph invariants whose average value is a
{\displaystyle \tau }
-function for the Kadomtsev-Petviashvili hierarchy of partial differential equations. In particular, this is valid for Stanley's symmetrized chromatic polynomial, as well as for the Abel polynomial for graphs we introduce. The key point here is a Hopf algebra structure on the space spanned by graphs and the behavior of the invariants on its primitive space.
The talk is based on a joint paper with S. Chmutov and M. Kazarian.
Retrieved from "https://gdeq.org/w/index.php?title=Lando_S._Polynomial_graph_invariants_and_the_KP_hierarchy_(abstract)&oldid=6496"
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Fees - Balancer
There are a few different types of fees on Balancer, each collected to support a healthy ecosystem. For example, Liquidity Providers collect swap fees as users trade with pools; this acts as an incentive for them to continue providing liquidity, which is useful to facilitate trades.
Traders pay swap fees when they trade with a pool. The fees ultimately go to Liquidity Providers in exchange for them putting their tokens in the pool to facilitate trades. Trade fees are collected at the time of a swap, and it goes directly into the pool, growing the pool's balance. For a trade with a given
inputToken
outputToken
, the amount collected by the pool as a fee is
Amount_{fee} = Amount_{inputToken} * swapFee
. As the pool collects fees, Balancer Pool Tokens automatically collect fees because they represent a proportional share of the pool.
Let's say Alice, Bob, Chuck, and Diana all provide liquidity in the same pool starting out with a total value of $100. After some time, the pool has collected many trade fees and is now worth $200. The pool itself grows while the Liquidity Providers' proportional shares stay the same.
Value After Trading
At the time of pool creation, the pool creator defines the fee and the owner. If the owner is set to an uncontrolled address (typically the zero address) then the swap fee is static. If the owner sets it to a controlled address, that address can set the fee at will, therefore making the swap fee dynamic.
There are two different ways that a pool can have Dynamic Fees, either the pool owner manually sets fees, or the pool creator delegates fees to a governance-approved controller by setting the pool owner to the delegate address.
Governance approved Gauntlet to have Delegated Fee Control on many pool types. They have been updating fees regularly based on models designed to optimize fee volume to Liquidity Providers. To learn more about how they determine fees, check out their Medium post.
Protocol Swap Fees
Protocol Swap Fees are a percentage of swap fees collected by pools. These go to the DAO Treasury to be used or held as the DAO sees fit. At deployment, Protocol Fees for swaps were set to 0%. These fees were activated by a governance vote and were later raised to 50% by a subsequent proposal. For the most up-to-date protocol fee data, please call getSwapFeePercentage() on the ProtocolFeesCollector contract.
Protocol Swap Fees are a percent of the already collected swap fees; the traders would see no change in the amount collected. The Liquidity Providers, however, would see a small change. For example, if a pool has a 1% swap fee, and there was a 10% protocol swap fee, 0.9% of each trade would be collected for the LPs, and 0.1% would be collected to the protocol fee collector contract.
Flash Loan fees are a type of Protocol Fee on Balancer. The fees collected as interest on flash loans go to the DAO Treasury. At deployment, flash loan fees were set to zero, and as of this writing they have not been activated by governance.
Stable Math
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Vulnerability Assessment of Climate Change on Sea Level Rise Impacts on Some Economic Sectors in Binh Dinh Province, Vietnam
Department of Climate Change, Ministry of Natural Resources and Environment (MoNRE), Hanoi, Vietnam
V=f\left(E,S,AC\right)
↑
{x}_{ij}=\frac{{X}_{ij}-\underset{i}{\mathrm{min}}\left\{{X}_{ij}\right\}}{\underset{i}{\mathrm{max}}\left\{{X}_{ij}\right\}-\underset{i}{\mathrm{min}}\left\{{X}_{ij}\right\}}
↓
{y}_{ij}=\frac{\underset{i}{\mathrm{max}}\left\{{X}_{ij}\right\}-{X}_{ij}}{\underset{i}{\mathrm{max}}\left\{{X}_{ij}\right\}-\underset{i}{\mathrm{min}}\left\{{X}_{ij}\right\}}
\stackrel{¯}{{y}_{i}}={\displaystyle \underset{j=1}{\overset{K}{\sum }}{w}_{j}\times {x}_{ij}}
0<w<\text{1}
\underset{j=1}{\overset{K}{\sum }}{w}_{j}=1
{w}_{j}=\frac{c}{\sqrt{\underset{i}{\mathrm{var}}\left({x}_{ij}\right)}}
c={\left[\underset{j=1}{\overset{K}{\sum }}\frac{1}{\sqrt{\underset{i}{\mathrm{var}}\left({x}_{ij}\right)}}\right]}^{-1}
{w}_{E}+{w}_{S}+{w}_{A}=1
{V}_{i}={E}_{i}\times {w}_{E}+{S}_{i}\times {w}_{S}+{A}_{i}\times {w}_{A}
f\left(z\right)=\frac{{z}^{a-1}{\left(1-z\right)}^{b-1}}{\beta \left(a,b\right)}
0<z<1
a,b>0
\beta \left(a,b\right)
\beta \left(a,b\right)={\int }_{0}^{1}{x}^{a-1}{\left(1-x\right)}^{b-1}\text{d}x
Lien, M.K. (2019) Vulnerability Assessment of Climate Change on Sea Level Rise Impacts on Some Economic Sectors in Binh Dinh Province, Vietnam. American Journal of Climate Change, 8, 302-324. https://doi.org/10.4236/ajcc.2019.82017
1. Hinkel, J. (2011) Indicators of Vulnerability and Adaptive Capacity: Towards a Clarification of the Science Policy Interface. Global Environmental Change, 21, 198-208. https://doi.org/10.1016/j.gloenvcha.2010.08.002
2. Cutter, S.L., Boruff, B. and Shirley, W. (2003) Social Vulnerability to Environmental Hazards. Social Science Quarterly, 84, 242-261. https://doi.org/10.1111/1540-6237.8402002
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5. Smit, B. and Wandel, J. (2006) Adaptation, Adaptive Capacity and Vulnerability. Global Environmental Change, 16, 282-292. https://doi.org/10.1016/j.gloenvcha.2006.03.008
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12. Harmeling, S. (2011) Global Climate Risk Index 2011. Who Suffers Most from Extreme Weather Events? Weather Related Loss Events in 2009 and 1990 to 2009. Germanwatch and Climate Action Europe, Bonn.
13. Ellison, J.C. and Zouh, I. (2012) Vulnerability to Climate Change of Mangroves: Assessment from Cameroon, Central Africa. Biology, 1, 617-638. https://doi.org/10.3390/biology1030617
14. Young, B., Byers, E., Gravuer, K., Hall, K., Hammerson, G. and Redder, A. (2011) The NatureServe Climate Change Vulnerability Index. NatureServe, Arlington.
15. Klein, R.J.T. and Nicholls, R.J. (1999) Assessment of Coastal Vulnerability to Climate Change. Ambio, 28, 182-187. https://www.jstor.org/stable/4314873
16. Messner, F. and Meyer, V. (2005) Flood Damage, Vulnerability and Risk Perception Challenges for Flood Damage Research. Discussion Paper 13/2005, Department of Economics, UFZ-Umweltforschungszentrum Leipzig-Halle, Leipzig.
17. Antle, J., Capalbo, S., Elliot, E. and Paustian, K. (2004) Adaptation, Spatial Heterogeneity and the Vulnerability of Agricultural Systems to Climate Change and CO2 Fertilization: An Integrated Assessment Approach. Climate Change, 64, 289-315. https://doi.org/10.1023/B:CLIM.0000025748.49738.93
18. Eakin, H. (2005) Institutional Change, Climate Risk and Rural Vulnerability: Cases from Central México. World Development, 33, 1923-1938. https://doi.org/10.1016/j.worlddev.2005.06.005
19. Hareau, A., Hofstadter, R. and Saizar, A. (1999) Vulnerability to Climate Change in Uruguay: Potential Impacts on the Agricultural and Coastal Resource Sectors and Response Capabilities. Climate Research, 12, 185-193. https://doi.org/10.3354/cr012185
20. Luers, A.L., Lobell, D.B., Skar, L.S., Addams, C.L. and Matson, P.A. (2003) A Method for Quantifying Vulnerability, Applied to the Agricultural System of the Yaqui Valley, México. Global Environmental Change, 13, 255-267. https://doi.org/10.1016/S0959-3780(03)00054-2
21. Wheeler, D. (2011) Quantifying Vulnerability to Climate Change: Implications for Adaptation Assistance. Center for Global Development, Washington DC. https://www.cgdev.org/sites/default/files/1424759_file_Wheeler_Quantifying_Vulnerability_FINAL.pdf https://doi.org/10.2139/ssrn.1824611
22. Zhao, H.X., Wu, S.H. and Jiang, L.G. (2007) Research Advances in Vulnerability Assessment of Natural Ecosystem Response to Climate Change. Chinese Journal of Applied Ecology, 18, 445-450.
23. Johnson, J.E. and Marshall, P.A. (2007) Climate Change and the Great Barrier Reef: A Vulnerability Assessment. Great Barrier Reef Marine Park Authority and Australian Greenhouse Office, Townsville.
24. Nitschke, C.R. and Innes, J.L. (2008) Integrating Climate Change into Forest Management in South-Central British Columbia: An Assessment of Landscape Vulnerability and Development of a Climate-Smart Framework. Forest Ecology and Management, 256, 313-327. https://doi.org/10.1016/j.foreco.2008.04.026
25. Glick, P. and Stein, B.A. (2010) Scanning the Conservation Horizon: A Guide to Climate Change Vulnerability Assessment. National Wildlife Federation, Washington DC.
26. Bell, J.D., Johnson, J.E. and Hobday, A.J. (2011) Vulnerability of Fisheries and Aquaculture in the Pacific to Climate Change. Secretariat of the Pacific Community, Noumea.
27. Remling, E. and Persson, A. (2014) Who Is Adaptation for? Vulnerability and Adaptation Benefits in Proposals Approved by the UNFCCC Adaptation Fund. Climate and Development, 7, 16-34. https://doi.org/10.1080/17565529.2014.886992
28. Brooks, N., Adger, N. and Kelly, M. (2005) The Determinants of Vulnerability and Adaptive Capacity at the National Level and the Implications for Adaptation. Global Environmental Change, 15, 151-163. https://doi.org/10.1016/j.gloenvcha.2004.12.006
29. Moss, R.H., Brenkert, A.L. and Malone, E.L. (2001) Vulnerability to Climate Change: A Quantitative Approach. U.S. Department of Energy, Washington DC.
30. O’Brien, K.L., Leichenko, R.M., Kelkar, U., Venema, H.M., Aandahl, G., Tompkins, H., Javed, A., Bhadwal, S., Barg, S., Nygaard, L. and West, J. (2004) Mapping Vulnerability to Multiple Stressors: Climate Change and Globalization un India. Global Environmental Change, 14, 303-313. https://doi.org/10.1016/j.gloenvcha.2004.01.001
31. Iyengar, N.S. and Sudarshan, P. (1982) A Method of Classifying Regions from Multivariate Data. Economic and Political Weekly, 17, 2047-2052. https://www.jstor.org/stable/4371674
32. Palanisami, K., Ranganathan, C.R., Nagothu, U.S. and Kakumanu, K.R. (2013) Climate Change and Agriculture in India. Routledge, New Delhi, 344. https://doi.org/10.4324/9781315734088
33. Kumar, P.N.S., Paul, K.S.R. and Rao, D.V.S. (2014) Assess the Vulnerability of Climate Change in Krishna River Basin of Andhra Pradesh. International Journal of Development Research, 4, 1059-1061.
34. Behnassi, M., Syomiti Muteng’e, M., Ramachandran, G. and Shelat, K.N. (2014) Vulnerability of Agriculture, Water and Fisheries to Climate Change. Springer, Berlin. https://doi.org/10.1007/978-94-017-8962-2
35. Kumar, N.S.P., Radha, Y., Rao, D.V.S., Rao, V.S. and Gopikrishna, T. (2018) Vulnerability to Climate Change in Andhra Pradesh State, India. International Journal of Current Microbiology and Applied Sciences, 7, 495-502. https://doi.org/10.20546/ijcmas.2018.711.059
36. IPCC (2001) Impacts, Adaptation and Vulnerability. Third Assessment Report. Climate Change 2001. Intergovernmental Panel on Climate Change/Cambridge University Press, Geneva/Cambridge.
37. IPCC (2007) Impacts, Adaptation and Vulnerability. Fourth Assessment Report. Climate Change 2007. Intergovernmental Panel on Climate Change/Cambridge University Press, Geneva/Cambridge. http://www.ipcc.ch
38. Szlafsztein, C. and Sterr, H. (2007) A GIS-Based Vulnerability Assessment of Coastal Natural Hazards, State of Pará, Brazil. Journal of Coastal Conservation, 11, 53-66. https://doi.org/10.1007/s11852-007-0003-6
39. Kantamaneni, K., Sudha Rani, N.N.V., Rice, L., Sur, K., Thayaparan, M., Kulatunga, U., Rege, R., Yenneti, K. and Campos, L.C. (2019) A Systematic Review of Coastal Vulnerability Assessment Studies along Andhra Pradesh, India: A Critical Evaluation of Data Gathering, Risk Levels and Mitigation Strategies. Water, 11, 393. https://doi.org/10.3390/w11020393
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Shemyakova E. On super Plücker embedding and cluster algebras (abstract) - Geometry of Differential Equations
Shemyakova E. On super Plücker embedding and cluster algebras (abstract)
Speaker: Ekaterina Shemyakova
There has been active work towards definition of super cluster algebras (Ovsienko, Ovsienko-Shapiro, and Li-Mixco-Ransingh-Srivastava), but the notion is still a mystery. As it is known, the classical Plücker map of a Grassmann manifold into projective space provides one of the model examples for cluster algebras. In the talk, we present our construction of "super Plücker embedding" for Grassmannian of
{\displaystyle r|s}
{\displaystyle n|m}
There are two cases. The first one is of completely even planes in a super space, i.e., the Grassmannian
{\displaystyle G_{r|0}(n|m)}
. It admits a straightforward algebraic construction similar to the classical case. In the second, general case of
{\displaystyle r|s}
-planes, a more complicated construction is needed.
Our super Plücker map takes the Grassmann supermanifold
{\displaystyle G_{r|s}(V)}
to a "weighted projective space"
{\displaystyle P_{1,-1}(\Lambda ^{r|s}(V)\oplus \Lambda ^{s|rs}(\Pi V))}
, with weights
{\displaystyle +1,-1}
{\displaystyle \Lambda ^{r|s}(V)}
{\displaystyle (r|s)}
th exterior power of a superspace
{\displaystyle V}
{\displaystyle \Pi }
is the parity reversion functor. We identify the super analog of Plücker coordinates and show that our map is an embedding. We obtain the super analog of the Plücker relations and consider applications to conjectural super cluster algebras.
Based on a joint work with Th. Voronov.
Retrieved from "https://gdeq.org/w/index.php?title=Shemyakova_E._On_super_Plücker_embedding_and_cluster_algebras_(abstract)&oldid=6502"
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If u, v, w ∈ R n ,
Let u and v be vectors in
{\mathrm{ℝ}}^{n}
. span(u,v) is defined as the set of a linear combinations of vectors u and v
w\in span\left(u,v\right)
This implies that the vector w can be written as a Linear Combination of vectors u and v
w={c}_{1}u+{c}_{2}v
span\left(u,v\right)=span\left(u,v,w\right)
Consider the RHS of the above equation
span\left(u,v,w\right)={k}_{1}u+{k}_{2}v+{k}_{3}w
Now, substitute w in terms of u and v implies
span\left(u,v,w\right)={k}_{1}u+{k}_{2}v+{k}_{3}w\left[{c}_{1}u+{c}_{2}v\right]=\left({k}_{1}+{k}_{3}{c}_{1}\right)u+\left({k}_{2}+{k}_{3}{c}_{2}\right)v
span\left(u,v,w\right)={K}_{1}u+{K}_{2}v=span\left(u,v\right)
n\mid \varphi \left({a}^{n}-1\right)
Aut\left(G\right)
\left\{\begin{array}{l}y={x}^{2}\\ {z}^{2}=16-y\end{array}\right\
\left\{\begin{array}{l}4x=y\\ z=0\end{array}\right\
\left\{\begin{array}{l}x=4\\ y=16\end{array}\right\
\left\{\begin{array}{l}y-x=12\\ z=0\end{array}\right\
\left\{\begin{array}{l}x+y=20\\ z=0\end{array}\right\
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Makna lagu Bungong Jeumpa
Video Seni Budaya SMP Teknik dan Gaya Bernyanyi dalam Musik Tradisi
Gaya Bernyanyi Lagu Daerah Teknik dan Gaya Bernyanyi dalam Musik Tradisi Seni Budaya Kurikulum 2013 K13
This is Louville, one of the downtown areas in London. There are many buildings in this district. The buildings are mostly shops and stores, so there are lots of businessmen around. There is a popular restaurant called George’s Family Restaurant. People like to have scrambled eggs for breakfast and fish and chips for lunch. Louville also has a movie theatre. The citizens usually go there on Sunday for entertainment. The town is noisy but British people enjoy living there.
The sentence “There are many buildings in this district†is written to ….
Perhatikan kedudukan keempat garis pada koordinat cartesius di bawah ini :
Pada gambar tersebut, bagaimana kedudukan garis t terhadap garis n?
x
y
merupakan penyelesaian dari sistem persamaan 5
x
+
y
=
x
+
y
=
8. Nilai dari 7
x
-
y
Choose the best answer to complete the dialogue.
Anna : Ollie and Joan are absent today because they are sick. Can we play for our basketball team just for today?
Elsa : ________________.
Anna : That's great.
IPA Level 8 Biologi Sistem Gerak pada Makhluk Hidup Sistem Gerak pada Hewan Kelas VIII Kurikulum 2013 K13 IPA
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Triple Angle Identities | Brilliant Math & Science Wiki
Pranshu Gaba, Mei Li, Pi Han Goh, and
\sin 3 \theta = 3 \sin \theta - 4 \sin ^3 \theta
\cos 3\theta = 4 \cos ^ 3 \theta - 3 \cos \theta
To prove the triple-angle identities, we can write
\sin 3 \theta
\sin(2 \theta + \theta)
\begin{aligned} \sin 3 \theta & = \sin (2 \theta + \theta)\\ & = \sin 2 \theta \cos \theta + \cos 2 \theta \sin \theta \\ & = (2 \sin \theta \cos \theta) \cos \theta + \big(1 - 2 \sin^2 \theta\big) \sin \theta \\ & = 2 \sin \theta \cos^2 \theta + \sin \theta - 2 \sin^3 \theta \\ & = 2 \sin \theta \big(1 - \sin^2 \theta\big) + \sin \theta - 2 \sin^3 \theta \\ & = 2 \sin \theta - 2 \sin^3 \theta + \sin \theta - 2 \sin^3 \theta \\ & = 3 \sin \theta - 4 \sin^3 \theta.\ _\square \end{aligned}
The triple angle identity of
\cos 3 \theta
\sin^3(\theta)
\cos^3 (\theta)
\sin^3(\theta) = \frac{3 \sin (\theta) - \sin \left( 3 \theta \right) }{4},\quad \cos^3 (\theta) = \frac{\cos(3\theta) + 3 \cos (\theta)}{4}.
To remember the cosine formula, the trick that I like to use is to read cosine as "dollar." Then, we say
"Dollar thirty is equal to four dollar thirty minus three dollar."
\begin{array} {l l l l l } \$1. 30 & = \$ 4.30 & - \$ 3 \\ 1 \cos 3 \theta & = 4 \cos ^3 \theta & - 3 \cos \theta \\ \end{array}
Just leaving a mark here,
\begin{aligned} \tan x \tan(60^\circ - x) \tan (60^\circ + x) &=& \tan(3x) \\ 4\sin x \sin(60^\circ - x) \sin (60^\circ + x) &=&\sin(3x) \\ 4\cos x \cos(60^\circ - x) \cos(60^\circ + x) &=& \cos(3x) \\ \tan(3x) &=& \frac {3\tan x - \tan^3 x}{1 - 3\tan^2 x}. \end{aligned}
\tan (6^\circ) = \tan (12^\circ) \tan(24^\circ) \tan(48^\circ ).
Cite as: Triple Angle Identities. Brilliant.org. Retrieved from https://brilliant.org/wiki/triple-angle-identities/
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Prove instability using Lyapunov function x' = x^3 + xy y' =
hadaasyj 2022-05-01 Answered
{x}^{\prime }={x}^{3}+xy
{y}^{\prime }=-y+{y}^{2}+xy-{x}^{3}
{x}^{6}+{x}^{3}y+{y}^{2}
. This term is positive (complete the square). All other terms are small when x,y, are small, so they can be controlled. Here are the details.
{V}^{\prime }\left(x,y\right)={x}^{6}+{x}^{4}y+{y}^{2}-{y}^{3}-x{y}^{2}+{x}^{3}y
={x}^{6}+\left(1+x\right){x}^{3}y+\left(1-x-y\right){y}^{2}
\ge {x}^{6}+\left(1+x\right){x}^{3}y+\frac{1}{2}{y}^{2}\phantom{\rule{1em}{0ex}}\text{if}|x|+|y|\le \frac{1}{4}
\ge {x}^{6}-\frac{5}{4}{|x|}^{3}|y|+\frac{1}{2}{y}^{2}\phantom{\rule{1em}{0ex}}\text{since}|x|\le \frac{1}{4}
={\left({|x|}^{3}-\frac{5}{8}|y|\right)}^{2}+\frac{7}{64}{y}^{2}
\ge 0
with equality iff
x=y=0
\frac{dw}{dt}
\frac{dw}{dt}
w=x\mathrm{sin}y,\text{ }x={e}^{t},\text{ }y=\pi -t
t=0
Finding the extrema of a functional (calculus of variations)
I\left(y\right)={\int }_{0}^{1}{y}^{\prime 2}-{y}^{2}+2xy dx
\begin{array}{rl}\frac{\mathrm{\partial }F}{\mathrm{\partial }{y}^{\text{'}}}& =2{y}^{\text{'}}\phantom{\rule{0ex}{0ex}}⟹\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }{y}^{\text{'}}}\right)=2{y}^{\text{'}\text{'}}\\ \frac{\mathrm{\partial }F}{\mathrm{\partial }y}& =-2y+2x\end{array}
Which gives the Euler equation
2y{ }^{″}+2y-2x=0
(I think, unless I messed up my math).
Ok now I need to try to find solutions to this differential equation, however, I only know how to solve linear second order DE's. The x term is throwing me off and I am not sure how to solve this.
y\left(x\right)=x
is a solution by inspection, but inspection is a poor man's approach to solving DE's.
y{}^{″}+16y=-\frac{2}{\mathrm{sin}\left(4x\right)}
{r}^{2}+16=0
{u}_{1}\left(x\right)=\mathrm{sin}\left(4x\right),{u}_{2}\left(x\right)=\mathrm{cos}\left(4x\right)
y\left(x\right)={c}_{1}\left(x\right)\mathrm{sin}\left(4x\right)+{c}_{2}\left(x\right)\mathrm{cos}\left(4x\right)
{c}_{1}^{\prime }\left(x\right)\mathrm{sin}\left(4x\right)+{c}_{2}^{\prime }\mathrm{cos}\left(4x\right)=0
{c}_{1}^{\prime }\left(x\right)\mathrm{cos}\left(4x\right)-{c}_{2}^{\prime }\mathrm{sin}\left(4x\right)=-\frac{2}{\mathrm{sin}\left(4x\right)}
\mathrm{cos}\left(4x\right)
{c}_{1}^{\prime }\left(\mathrm{sin}\left(4x\right)\mathrm{cos}\left(4x\right)-\mathrm{cos}\left(4x\right)\mathrm{sin}\left(4x\right)\right)+{c}_{2}^{\prime }\left({\mathrm{cos}}^{2}\left(4x\right)+{\mathrm{sin}}^{2}\left(4x\right)\right)=2
{c}_{2}^{\prime }=2⇒{c}_{2}=2x
{c}_{1}^{\prime }=-\frac{2\mathrm{cos}\left(4x\right)}{\mathrm{sin}\left(4x\right)}=-2\mathrm{cot}\left(4x\right)⇒{c}_{1}=-\frac{1}{2}\mathrm{ln}|\mathrm{sin}\left(4x\right)|.
My textbook states we need locally Lipschitz continuous with respect to the second argument and uniformly with respect to the first argument of
f\left(t,\text{ }x\right)
. This is said to be equivalent to the result that for all compact subsets
V\subset U
U\subseteq {\mathbb{R}}^{n+1}
is open we have
L\phantom{\rule{0.222em}{0ex}}={\supset }_{\left(t,x\right)\ne \left(t,y\right)\in V\subset U}\frac{|f\left(t,x\right)-f\left(t,y\right)|}{|x-y|}<\mathrm{\infty }
I understand that locally Lipschitz means there is a neighborhood around each point where f is Lipschitz, but what does the uniform part mean with respect to t?
Nonlinear ODE initial value problem
A student came to me with a problem I couldn't solve. It's the beginning of the semester in his Intro DiffEq class, and so the solution shouldn't be too difficult. But it completely stumped me, and now I can't let it go! Here it is:
Problem. Find all solutions to the IVP:
\sqrt{{\left(\frac{dy}{dx}\right)}^{2}-4{x}^{2}}=\sqrt{{x}^{4}-{y}^{2}},;;;;y\left(0\right)=0.
I thought about maybe just doing a sort of guess-and-check method, but the existence/uniqueness theorem doesn't apply, so I'm not sure that would help even if I could find a single solution, which I cannot in any case.
{x}^{2}+xy+{y}^{3}=1
find the value of y''' at the point where x = 1
{y}^{\prime }=v
to write each second-order equation as a system of two first-order differential equations (planar system).
{y}^{{}^{″}}+\mu \left({t}^{2}-1\right){y}^{{}^{″}}+y=0
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Hidden Markov Models | Brilliant Math & Science Wiki
John McGonagle, Ayush G Rai, and Eli Ross contributed
A simple hidden Markov model with 3 states and 4 observation tokens. The states are represented by the labels
x_1
x_2
x_3
, while the transition probabilities between states are represented by the edge label
a_{ij}
for the probability of moving from state
x_i
x_j
. Observations are represented by the labels
y_1
y_2
y_3
y_4
. The probability of state
x_i
generating observation
y_j
is represented by the edge label
b_{ij}
A hidden Markov model is a type of graphical model often used to model temporal data. Unlike traditional Markov models, hidden Markov models (HMMs) assume that the data observed is not the actual state of the model but is instead generated by the underlying hidden (the H in HMM) states. While this would normally make inference difficult, the Markov property (the first M in HMM) of HMMs makes inference efficient.
Because of their flexibility and computational efficiency, hidden Markov models have found a wide application in many different fields. They are known for their use in temporal pattern recognition and generation such as speech recognition, handwriting recognition, and speech synthesis.
Often, when a person makes an observation of a system, what is being observed is not the state of the system but instead some tokens or data generated by the system's underlying hidden states. For example, when a person hears someone else speak, the sounds entering their ears are not the state of the system generating the sound. The true state of that system would be the collection of parameters directly determining which sounds are generated, such as the shape of the speaker's mouth, the frequency of their vocal chords, and the semantic meaning behind the sounds.
However, it is impossible to know the exact state of the underlying system from the observations alone, since many underlying states may correspond to the same observation. Thus, it is possible for the function mapping states to observations
f(s)=o
s
o
is an observation generated by the function
f(x)
s
, to have two different states
s_1 \ne s_2
that generate the same observation
f(s_1) = f(s_2)
. This is known as a many-to-one function, which in general is impossible to perfectly invert, since
f^{-1}(o)
s_1
s_2
Consider the many-to-one function
y=x^2
. Given only
, one cannot recover
x
exactly since both
-x
x
y
when squared. An HMM is similar in that, given an observation sequence, one can only infer a distribution over the states at any point in time, since many different sequences of underlying states can give rise to the same observation sequence.
In the case of speech recognition, knowing the underlying semantic meaning which generated the sounds would make the problem easy. This is because the whole problem of speech recognition is discovering the model states (semantic meaning) behind the observations (sounds). Thus, being able to model and infer the underlying states of an HMM given the observations provides a powerful technique for many kinds of problems.
Hidden Markov models evolve according to two rules:
The first rule is that the model moves from the current state to the next state, which may be the same state, according to some probability distribution that depends only on the current state, i.e.
p(s_t|s_{t-1})=p(s_t|s_{t-1}, \dots, s_0)
. This is known as the Markov property. Intuitively, this rule states that the system evolves without regard to past states of the system and only depends on the current state.
The second rule is that after each transition, the model emits an observation whose distribution depends only on the current state, i.e.
p(o_t|s_t)=p(o_t|s_t, o_{t-1}, s_{t-1}, \dots, o_0, s_0)
. Since the model only emits the observations and the states generating the observations are unknown to the observer, the states generating those observations are called hidden states.
A hidden Markov model is fully specified by the following parameters:
1) State Transition Probabilities
The probability of transition from state
s_i
s_j
a_{ij}
2) Observation Emission Probabilities
The probability of emitting observation
o_t
s_i
P(o_t|s_i)
. If the set of observations is discrete, then the probability of emitting token
o_j
from state
s_i
b_{ij}
3) State Initialization Probability
The probability of the HMM starting in state
s_i
\pi_i
Knowing the above parameters lets us generate observation sequences very easily. We start by picking an initial hidden state
s_0
according to the distribution
\pi
. Then, we pick an observation
o_0
s_0
. Next, the model transitions to a new hidden state
s_1
according to the state transition probability for state
s_0
, after which the model emits a new observation
o_1
. This continues until the desired number of observations is generated.
Since HMMs calculate a probabilistic sequence of hidden states and outputs, it is natural to ask what the likelihood of a particular observation sequence
O
is. Calculating the probability of an observation sequence involves a sum over the hidden states of the HMM, which for large HMMs, makes the naive calculation very slow. Luckily, applying techniques from dynamic programming can make this calculation tractable.
Imagine two speakers, Bob and Alice. If each has an HMM trained to model their voice, then we can calculate the likelihood of any sound sequence given each of their trained HMMs. If we calculate the likelihood of Bob's voice given his HMM, we would expect it to be higher than the likelihood of his voice given Alice's HMM. Intuitively, this is because observation sequences coming from Bob's HMM should sound like Bob, or at least more like Bob than Alice since it was trained on Bob's voice. Likewise, Alice's voice should have a higher likelihood on her HMM than on Bob's HMM. Knowing this, we can perform some basic speaker recognition, just by calculating the likelihood of an observation sequence given some known HMMs. The most likely speaker is the one whose HMM best generates their actual voice data.
, M. Hidden-markov-model-abc. Retrieved September 11, 2013, from https://commons.wikimedia.org/wiki/File:Hidden-markov-model-abc.svg)
, T. HiddenMarkovModel. Retrieved August 29, 2007, from https://commons.wikimedia.org/wiki/File:HiddenMarkovModel.png
Cite as: Hidden Markov Models. Brilliant.org. Retrieved from https://brilliant.org/wiki/hidden-markov-models/
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Peak of Electron Density in F2-Layer Parameters Variability at Quiet Days on Solar Minimum
Peak of Electron Density in F2-Layer Parameters Variability at Quiet Days on Solar Minimum
Emmanuel Nanéma1,2*, Moustapha Konaté2, Frédéric Ouattara2
1Centre National de la Recherche Scientifique et Technologique (CNRST), Institut de Recherche en Sciences Appliquées et Technologies (IRSAT), Ouagadougou, Burkina Faso
2Université Norbert ZONGO, Laboratoire de Recherche en Météorologie de l’Espace (LAREME), Koudougou, Burkina Faso
This study deals with Peak of electron density in F2-layer sensibility scale during quiet time on solar minimum. Peaks of electron density in F2-layer (NmF2) values at the quietest days are compared to those carried out from the two nearest days (previous and following of quietest day). The study uses International Reference Ionosphere (IRI) for ionosphere modeling. The located station is Ouagadougou, in West Africa. Solar minimum of phase 22 is considered in this study. Using three core principles of ionosphere modeling under IRI running conditions, the study enables to carry out Peak of electron density in F2-layer values during the quietest days of the characteristic months for the four different seasons. These parameters are compared to those of the previous and the following of the quietest days (the day before and following each quietest selected day) at the same hour. The knowledge of NmF2 values at the quietest days and at the two nearest days enables to calculate the relative error that can be made on this parameter. This calculation highlights insignificant relative errors. This means that NmF2 values at the two nearest days of each quietest day on solar minimum can be used for simulating the quietest days’ behavior. NmF2 values obtained by running IRI model have good correlation with those carried out by Thermosphere-Ionosphere-Electrodynamics-General Circulation Model (TIEGCM).
Ionosphere, Peak of Electron Density in F2-Layer, Solar Cycle, Quiet Day, International Reference Ionosphere Model
Ionosphere layer is an important site for radio waves reflection because of its composition in particles. This layer moves like a plasma and so, is electrically neutral. Solar radiations hit the particles in this layer and causes ionization. Ionization of particles in ionosphere layer due to solar radiations creates electrons and ions in the layer. Ionosphere composition in particles enables to determine the critical frequency of radio waves frequency of transmitters. Critical frequency of radio waves is closely linked to the density of electron in the F2-layer. Many works about ionosphere parameters determining have been done during these last years [1] - [10] . The main purpose of these different studies is to carry out ionosphere parameters for telecommunication, navigation, electrical disturbance predictions. Several models have been developed for ionosphere investigation. In previous studies, we used International Reference Ionosphere model, Thermosphere-Ionosphere-Electrodynamics General Circulation Model, and data [11] - [17] to carry out ionosphere parameters. The present study is focused on the calculation of relative error on Peak of electron density in F2-layer values at the limits of the quietest days for different seasons during solar minimum, compared to those of the quietest days. In this work, we calculate the relative error value on NmF2. The study is based on quiet time variation of solar cycle 22 at Ouagadougou station and uses International Reference Ionosphere (IRI) model for ionosphere investigation. 2012-version of IRI is used to run the model.
2. Methodology―Fundamentals
In this study, the minimum year (1985) of solar cycle 22 is considered for Peak of electron density in F2-layer behavior. Ionosphere modeling using IRI study is focused on the following three core principles: 1) The characteristic months are March, June, September and December for spring, summer, autumn and winter respectively. 2) The five quietest days of each characteristic month are used. 3) Solar minimum is characterized by sunspot number Rz < 20 and Aa ≤ 20 nT. During quiet time conditions, the five quietest days characterize the whole month in each season. We consider the two nearest days (previous and following) of each quietest day in the characteristic month as the boundaries. Ouagadougou is located in West Africa. The following input parameters are used for running IRI model during solar minimum at the station: Year = 1985, Longitude = 358.5˚E, Latitude = 12.5˚N, Height = 500, Stepsize = 1. With above input parameters, NmF2 time values are obtained on “List Model data”. These values are exported in an Excel file for plotting.
Table 1 highlights the five quietest days selected in each season on solar minimum.
Figures 1-4 present NmF2 time variation carried out by running IRI model under
Table 1. Retain days during solar minimum (1985) of cycle 22.
Figure 1. NmF2 time variation on March 1985.
Figure 2. NmF2 time variation on June 1985.
Figure 3. NmF2 time variation on September 1985.
Figure 4. NmF2 time variation on December 1985.
its 2012-version for the retain days previously found in Table 1. On each figure, we present NmF2 time profiles on five different panels. Each panel highlights NmF2 time variations for three days that are the quietest day, the previous and the following day.
Relative error on Peak of electron density parameter in F2-layer is calculated by the following Equation (1):
\frac{\Delta \text{NmF}{2}_{\text{i}}}{\text{NmF}{2}_{\text{i}}}=\frac{\left|\text{NmF}{2}_{\text{i}}\left(\text{D}\right)-\text{NmF}{2}_{\text{i}}\left(\text{D}\pm 1\right)\right|}{\text{NmF}{2}_{\text{i}}\left(\text{D}\right)}
In Equation (1), NmF2i(D) is the Peak of electron density value at i hour for D-day. D-day is the quietest day.
\left(\text{D}\pm 1\right)
are respectively the previous (D − 1) and the following (D + 1) day of the quietest day.
\text{i}\in \left[0,24\right]
. The step for i values variation is 1.
Calculating the relative error on Peak of electron density in F2-layer for Figures 1-4 corresponding to March (spring), June (summer), September (autumn) and December (winter) gives the following ranges:
On March 1985:
\frac{\Delta \text{NmF}{2}_{\text{i}}}{\text{NmF}{2}_{\text{i}}}\in \left[0%,4%\right]
On June 1985:
\frac{\Delta \text{NmF}{2}_{\text{i}}}{\text{NmF}{2}_{\text{i}}}\in \left[0%,2%\right]
On September 1985:
\frac{\Delta \text{NmF}{2}_{\text{i}}}{\text{NmF}{2}_{\text{i}}}\in \left[0%,4%\right]
On December 1985:
\frac{\Delta \text{NmF}{2}_{\text{i}}}{\text{NmF}{2}_{\text{i}}}\in \left[0%,1%\right]
The range of
\frac{\Delta \text{NmF}{2}_{\text{i}}}{\text{NmF}{2}_{\text{i}}}
variation given by (2), (3), (4) and (5) is inferior to 4%. Peak of electron density in F2-layer values vary slightly from the quietest day to its two nearest selected days. This means that NmF2D-1 ~ NmF2D ~ NmF2D+1 at the same time Calculated relative error at each hour between the quietest day and its two nearest days is inferior to 4% for NmF2. However, NmF2 value is a multiple of 105 cm−3. In each panel, a variation of 4% is insensible on NmF2 value. This enables to conclude that a 4% variation on NmF2 does not influence measurably Peak of electron density in F2-layer. So, using NmF2 value at the nearest days of quietest day doesn’t introduce a sensitive error on this parameter.
In this study, Peak of electron density in F2-layer parameter is carried out by the use of IRI model. The study highlights insignificant relative errors on NmF2 carried out from the quietest days on solar minimum of phase 22 at Ouagadougou station. This means that using NmF2 values at the nearest days of each quietest day instead of those of the quietest day doesn’t introduce significant error. The contribution of this study is the possibility to neglect the relative error on Peak of electron density in F2-layer parameter due by considering the value of this parameter at the limits of each quietest day on solar minimum at low latitudes. This means that we can use either NmF2 carried from the quietest day or those from the nearest days to characterize the quietest day behavior. For a next study, this method will be applied to NmF2 variability on solar maximum.
Nanéma, E., Konaté, M. and Ouattara, F. (2019) Peak of Electron Density in F2-Layer Parameters Variability at Quiet Days on Solar Minimum. Journal of Modern Physics, 10, 302-309. https://doi.org/10.4236/jmp.2019.103021
1. Pedatella, N.M., Forbes, J.M., Maute, A., Richmond, A.D., Fang, T.-W., Larson, K.M. and Millward, G. (2011) Journal of Geophysical Research, 116, A12309. https://doi.org/10.1029/2011JA016600
2. Roble, R.G., Ridley, E.C., Richmond, A.D. and Dickinson, R.E. (1988) Geophysics Research Letter, 15, 1325-1328. https://doi.org/10.1029/GL015i012p01325
3. Wang, W., Wiltberger, M., Burns, A.G., Solomon, S.C., Killeen, T.L., Maruyama, N. and Lyon, J.G. (2004) Journal of Atmospheric and Solar-Terrestrial Physics, 66, 1425-1441. https://doi.org/10.1016/j.jastp.2004.04.008
4. Richmond, A.D., Ridley, E.C. and Roble, R.G. (1992) Geophysics Research Letter, 19, 601-604. https://doi.org/10.1029/92GL00401
5. Burns, A.G., Wang, W., Killen, T.L. and Solomon, S.C. (2004) Journal of Atmospheric and Solar-Terrestrial Physics, 66, 1457-1468. https://doi.org/10.1016/j.jastp.2004.04.009
6. Weimer, D.R. (2005) Journal of Geophysical Research, 110, A05306. https://doi.org/10.1029/2004JA010884
7. Jin, S. and Park, J.U. (2007) Earth Planet Space, 59, 287-292. https://doi.org/10.1186/BF03353106
8. Bittencourt, J.A. and Chryssafidis, M. (1994) Journal of Atmospheric and Solar-Terrestrial Physics, 56, 995-1009. https://doi.org/10.1016/0021-9169(94)90159-7
9. Ouattara, F. and Rolland, F. (2011) Scientific Research and Essays, 6, 3609-3622. https://doi.org/10.5897/SRE10.1050
10. Qian, L., Burns, A.G., Chamberlin, P.C. and Solomon, S.C. (2010) Journal of Geophysical Research, 115, A09311. https://doi.org/10.1029/2009JA015225
11. Ouattara, F. and Nanéma, E. (2014) Physical Science International Journal, 4, 892-902. https://doi.org/10.9734/PSIJ/2014/9748
12. Ouattara, F. (2013) Archives of Physics Research, 4, 12-18.
13. Ouattara, F. and Nanéma E. (2013) Archives of Applied Science Research, 5, 55-61.
14. Nanéma, E., Ouédraogo, I., Zoundi, C. and Ouattara, F. (2018) International Journal of Geosciences, 9, 572-578. https://doi.org/10.4236/ijg.2018.99033
15. Nanéma, E., Gnabahou, D.A., Zoundi, C. and Ouattara, F. (2018) International Journal of Astronomy and Astrophysics, 8, 163-170. https://doi.org/10.4236/ijaa.2018.820311
16. Nour, A.M., Frédéric, O., Louis, Z.J., Frédéric, G.A.M., Emmanuel, N. and François, Z. (2015) International Journal of Geosciences, 6, 201-208.https://doi.org/10.4236/ijg.2015.63014
17. Nanéma, E., Zerbo, J.L., Konaté, M. and Ouattara, F. (2018) Journal of Scientific and Engineering Research, 5, 62-68.
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How to prove that the inverse Laplace Transform
Emilia Hoffman 2022-04-06 Answered
{L}^{-1}\left\{0\right\}=0
Buizzae77t
L\left(f\right)=F
{L}^{-1}\left(F\right)=f.\phantom{\rule{1em}{0ex}}L\left(0\right)=0
because L is a linear operator. Or you can actually compute L(0) using the definition.
{y}^{″}+2{y}^{\prime }+y=4
y\left(0\right)=3
{y}^{\prime }\left(0\right)=0
y3{y}^{\prime }+2y=\delta \left(t-1\right),y\left(0\right)=1,{y}^{\prime }\left(0\right)=0
Solve the following differential equations. Use the method of Bernoulli’s Equation.
x\left(2{x}^{3}+y\right)dy-6{y}^{2}dx=0
Determine the Laplace transform of the following functions.
\frac{2}{t}\mathrm{sin}3at
where a is any positive constant and s>a
Solve the following second-order linear differential equation by the method of reduction of order.
\left({D}^{2}+1\right)y=\mathrm{csc}\text{ }x
The following differential equations appear similar but have very different solutions:
dydx=x
dydx=y
Solve both subject to the initial condition
y\left(1\right)=2
A simple question about the solution of homogeneous equation to this differential equation
t,1+t,{t}^{2},-t
are the solutions to
y{}^{‴}+a\left(t\right)y{}^{″}+b\left(t\right)y{}^{″}+c\left(t\right)y=d\left(t\right)
, what is the solution of homogeneous equation to this differential equation? What i have done is tried the properties of linear differential equation that
L\left(t\right)=L\left(1+t\right)=L\left({t}^{2}\right)=L\left(-t\right)=d\left(t\right)
so the homogeneous solution should be independent and i claim that
1,t,{t}^{2}
should be the solution. However, i am not sure hot can i actually conclude that these are the solutions? It seems that it can be quite a number of sets of solution by the linearity.
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An Analysis of the Effect of Lower Extremity Strength on Impact Severity During a Backward Fall | J. Biomech Eng. | ASME Digital Collection
Reuben Sandler,
Biomechanics Laboratory, Department of Orthopaedic Surgery, San Francisco General Hospital and University of California, San Francisco, San Francisco, CA 94110
Contributed by the Bioengineering Division for publication in the JOURNAL OF BIOMECHANICAL ENGINEERING. Manuscript received by the Bioengineering Division July 2, 1999; revised manuscript received May 16, 2001. Associate Editor: M. G. Pandy.
Sandler , R., and Robinovitch, S. (May 16, 2001). "An Analysis of the Effect of Lower Extremity Strength on Impact Severity During a Backward Fall ." ASME. J Biomech Eng. December 2001; 123(6): 590–598. https://doi.org/10.1115/1.1408940
At least 280,000 hip fractures occur annually in the U.S., at an estimated cost of $9 billion. While over 90 percent of these are caused by falls, only about 2 percent of all falls result in hip fracture. Evidence suggests that the most important determinants of hip fracture risk during a fall are the body’s impact velocity and configuration. Accordingly, protective responses for reducing impact velocity, and the likelihood for direct impact to the hip, strongly influence fracture risk. One method for reducing the body’s impact velocity and kinetic energy during a fall is to absorb energy in the lower extremity muscles during descent, as occurs during sitting and squatting. In the present study, we employed a series of inverted pendulum models to determine: (a) the theoretical effect of this mechanism on impact severity during a backward fall, and (b) the effect on impact severity of age-related declines (or exercise-induced enhancements) in lower extremity strength. Compared to the case of a fall with zero energy absorption in the lower extremity joints, best-case falls (which involved 81 percent activation of ankle and hip muscles, but only 23 percent activation of knees muscles) involved 79 percent attenuation (from 352 J to 74 J) in the body’s vertical kinetic energy at impact
KEv,
and 48 percent attenuation (from 3.22 to 1.68 m/s) in the downward velocity of the pelvis at impact
vv.
Among the mechanisms responsible for this were: (1) eccentric contraction of lower extremity muscles during descent, which resulted in up to 150 J of energy absorption; (2) impact with the trunk in an upright configuration, which reduced the change in potential energy associated with the fall by 100 J; and (3) knee extension during the final stage of descent, which “transferred” up to 90 J of impact energy into horizontal (as opposed to vertical) kinetic energy. Declines in joint strength reduced the effectiveness of mechanisms (1) and (3), and thereby increased impact severity. However, even with reductions of 80 percent in available torques,
KEv
was attenuated by 50 percent. This indicates the importance of both technique and strength in reducing impact severity. These results provide motivation for attempts to reduce elderly individuals’ risk for fall-related injury through the combination of instruction in safe falling techniques and exercises that enhance lower extremity strength.
geriatrics, biomechanics, bone, muscle, fracture, physiological models, impact strength, Falls, Hip Fracture, Strength, Mathematical Modeling, Biomechanics
Biomechanics, Fracture (Materials), Fracture (Process), Hip fractures, Kinetic energy, Knee, Muscle, Potential energy, Risk, Simulation, Torque, Wounds, Absorption, Rotation, Pendulums, Bone, Equations of motion
United States Department of Health and Human Services, Public Health Service, 1993. Healthy People 2000: National Health Promotion and Disease Prevention Objectives. Washington, DC.
Risk Factors for Falls as a Cause of Hip Fracture in Women. The Northeast Hip Fracture Study Group
Meislin
Secular Trends in the Incidence of Postmenopausal Vertebral Fractures
Praemar, A., Furner, S., and Rice, D. P., 1992, “Costs of Musculoskeletal Conditions,” in: Musculoskeletal Conditions on the United States, American Academy of Orthopaedic Surgeons, Park Ridge, IL, pp. 143–170.
Prediction of Upper Extremity Impact Forces During Falls on the Outstretched Hand
Sebeny
Correlations Between Photon Absorption Properties and Failure Load of the Distal Radius in Vitro
Age-Related Reductions in the Strength of the Femur Tested in a Fall-Loading Configuration
Non-skeletal Determinants of Fractures: The Potential Importance of the Mechanics of Falls. Study of Osteoporotic Fractures Research Group
, Suppl. 1, pp.
Type of Fall and Risk of Hip and Wrist Fractures: The Study of Osteoporotic Fractures
Characteristics of Falls and Risk of Hip Fracture in Elderly Men
Fall Direction, Bone Mineral Density, and Function: Risk Factors for Hip Fracture in Frail Nursing Home Elderly
Distribution of Contact Force During Falls on the Hip
Body Height and Hip Fracture: A Cohort Study of 90,000 Women
Physiological Factors Associated With Falls in Older Community Dwelling Women
Risk Factors for Recurrent Nonsyncopal Falls: A Prospective Study
A Hypothesis: The Cause of Hip Fractures
Impact Severity in Self-Initiated Sits and Falls Associates With Center-of-Gravity Excursion During Descent
Common Protective Movements Govern Unexpected Falls From Standing Height
Hip Impact Velocities and Body Configurations for Voluntary Falls From Standing Height
Winter, D. A., 1990, Biomechanics and Motor Control of Human Movement, Wiley, New York.
Force Attenuation and Energy Absorption by Soft Tissues During Falls on the Hip
Effects of Loading Rate on Strength of the Proximal Femur
Angeles, J., 1997, Fundamentals of Robotic Mechanical Systems: Theory, Methods, and Algorithms, Springer, New York.
Effect of Upper and Lower Extremity Control Strategies on Predicted Injury Risk During Simulated Forward Falls: A Study in Healthy Young Adults
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Convert geocentric latitude to geodetic latitude - Simulink - MathWorks Switzerland
Output altitude
The Geocentric to Geodetic Latitude block converts a geocentric latitude (λ) into geodetic latitude (μ). The function uses an iteration-method of Bowring's formula is used to calculate the geodetic latitude. For more information, see Algorithms.
This implementation generates a geodetic latitude that lies between ±90 degrees.
Geocentric latitude, specified as a scalar, in degrees. Latitude values can be any value. However, values of +90 and -90 may return unexpected values because of singularity at the poles.
Radius from center of the planet to the center of gravity, specified as a scalar.
Geodetic latitude, specified as a scalar, in degrees.
Mean sea-level altitude (MSL), returned as a scalar.
To enable this port, select Output altitude.
Output altitude — Enable mean sea-level altitude
Select this check box to output the mean sea-level altitude (MSL).
Select this check box to enable the h port.
Block Parameter: outputAltitude
The Geocentric to Geodetic Latitude block converts a geocentric latitude (λ) into geodetic latitude (μ), where:
r — Radius from the center of the planet
Given geocentric latitude (λ) and the radius (r) from the center of the planet, this block first converts the desired points into the distance from the polar axis (ρ) and the distance from the equatorial axis (z).
\begin{array}{l}\rho =r\left(\mathrm{cos}\left(\lambda \right)\right)\\ z=r\left(\mathrm{sin}\left(\lambda \right)\right).\end{array}
It then calculates the geometric properties of the planet:
\begin{array}{l}b=a\left(1-f\right)\\ {e}^{2}=f\left(2-f\right)\\ e{\text{'}}^{2}=\frac{{e}^{2}}{\left(1-{e}^{2}\right)}.\end{array}
And then uses the fixed-point iteration of Bowring's formula to calculate μ. This formula typically converges in three iterations.
\begin{array}{l}\beta ={\mathrm{tan}}^{-1}\left(\frac{\left(1-f\right)\mathrm{sin}\left(\mu \right)}{\mathrm{cos}\left(\mu \right)}\right)\\ \mu ={\mathrm{tan}}^{-1}\left(\frac{z+b{e}^{{\text{'}}^{2}}\mathrm{sin}{\left(\beta \right)}^{3}}{\rho -a{e}^{2}\mathrm{cos}{\left(\beta \right)}^{3}}\right).\end{array}
[1] Jackson, E. B., Manual for a Workstation-based Generic Flight Simulation Program (LaRCsim) Version 1.4, NASA TM 110164, April, 1995.
[2] Hedgley, D. R., Jr. "An Exact Transformation from Geocentric to Geodetic Coordinates for Nonzero Altitudes." NASA TR R-458, March, 1976.
[3] Clynch, J. R. "Radius of the Earth - Radii Used in Geodesy." Naval Postgraduate School, Monterey, California, 2002.
[5] Edwards, C. H., and D. E. Penny. Calculus and Analytical Geometry 2nd Edition, Prentice-Hall, Englewood Cliffs, New Jersey, 1986.
ECEF Position to LLA | Flat Earth to LLA | Geodetic to Geocentric Latitude | LLA to ECEF Position
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Change in Entropy of Spinning Black Holes Due to Corresponding Change in Mass in XRBs
Dipo Mahto1, Anuradha Kumari2
1 Department of Physics, Marwari College, T. M. B. University, Bhagalpur, India.
2 Department of Physics, T. M. B. University, Bhagalpur, India.
Abstract: The present paper gives a theoretical model for the change in entropy of spinning black holes due to change in mass to use the first law of black hole mechanics for unit spinning parameter and angular momentum in XRBs. This shows that the entropy change with respect to the mass of uncharged spinning black holes is essentially the function of mass and their surface gravity are lesser than to that of the non-spinning black holes.
Keywords: Entropy Change, Spinning Parameters and Angular Momentum
Classically, the black holes are perfect absorbers and do not emit anything; their temperature is absolute zero. However, in quantum theory black holes emit Hawking radiation with a perfect thermal spectrum [1] . The laws of black hole Physics describe the behaviour of a black hole in close analogy to the laws thermodynamics by relating mass to energy, area to entropy, and surface gravity to temperature [2] . The general formula for the entropy due to Bekenstein and Hawking provides a deep connection between quantum mechanics, general relativity and thermodynamics [3] . In 2006, Zhao Ren et al. provided a general method for the discussion on quantum corrections to Bekenstein and Hawking Entropy [4] . In 2011, Mahto et al. proposed a model for the change in energy and entropy of Non-spinning black holes to use the first law of black hole mechanics and mass-energy equivalence relation [5] . In 2013, Mahto et al. proposed also a model for the change in entropy of Non-spinning black holes with respect to the radius of event horizon in XRBs by using the first law of the black hole mechanics with the relation for the entropy change
\text{d}S=8\text{πd}M
[6] . In 2015, Mahto et al. gave a model for classical statistical entropy of black hole using M-B Statistics and showed that the classical statistical entropy of black hole is directly proportional to the area of event horizon [7] .
In the present paper, we have proposed a model for the change in entropy of spinning black holes due to change in mass to use the first law of black hole mechanics for unit spinning parameter and angular momentum and also calculated their values for different test spinning black holes in XRBs.
The first law of black hole mechanics is simply an identity relating the change in mass M, angular momentum J, horizon area A and charge Q, of a black hole. The first order variations of these quantities in the vacuum satisfy [8] .
\delta M=\frac{k}{\text{8π}}\delta A+\Omega \delta J-\upsilon \delta Q
\Omega
= Angular velocity of the horizon.
\upsilon
= difference in the electrostatic potential between infinity and horizon.
In this more general context, the first law of black hole mechanics is seen to be a direct consequence of an identity holding for the variation of the Noether current. The general form of the first law takes the form [1]
\delta M=\frac{k}{\text{2π}}\delta S+\Omega \delta J
\delta S=\frac{\text{2π}}{\kappa }\left[\delta M-\Omega \delta J\right]
The above equation shows a model for change in entropy due to corresponding change in mass and angular momentum of black holes and depends on the surface gravity.
3. Surface Gravity
The surface gravity as an entity playing a role in black hole physics is analogous to temperature in thermodynamics and can be defined for stationary asymmetric black holes in asymptotically flat space time. The surface gravity for the Kerr-Newman solution is [8]
\kappa =\frac{\sqrt{{M}^{2}-{Q}^{2}-{J}^{2}/{M}^{2}}}{2{M}^{2}-{Q}^{2}+2M\sqrt{{M}^{2}-{Q}^{2}-{J}^{2}/{M}^{2}}}
where Q is the electric charge, J is the angular momentum.
For uncharged black holes
Q=0
Using the above equation into Equation (4), we have
\kappa =\frac{\sqrt{{M}^{2}-{J}^{2}/{M}^{2}}}{2{M}^{2}+2M\sqrt{{M}^{2}-{J}^{2}/{M}^{2}}}
\kappa =\frac{\sqrt{{M}^{4}-{J}^{2}}}{2{M}^{3}+2M\sqrt{{M}^{4}-{J}^{2}}}
\sqrt{{M}^{4}-{J}^{2}}={\left({M}^{4}-{J}^{2}\right)}^{1/2}={M}^{2}{\left(1-\frac{{J}^{2}}{{M}^{4}}\right)}^{1/2}
The square root expression then expanded by use of the binomial theorem:
={M}^{2}\left(1-\frac{1}{2}\frac{{J}^{2}}{{M}^{4}}+\frac{1}{8}{\left(\frac{{J}^{2}}{{M}^{4}}\right)}^{2}-\frac{1}{16}{\left(\frac{{J}^{2}}{{M}^{4}}\right)}^{3}+\cdots \right)
Neglecting the higher power terms, we have
\sqrt{{M}^{4}-{J}^{2}}={M}^{2}\left(1-\frac{{J}^{2}}{2{M}^{4}}\right)
2{M}^{3}+2M\sqrt{{M}^{4}-{J}^{2}}=2{M}^{3}+2{M}^{3}\left(1-\frac{{J}^{2}}{{M}^{4}}\right)
2{M}^{3}+2M\sqrt{{M}^{4}-{J}^{2}}=4{M}^{3}\left(1-\frac{{J}^{2}}{4{M}^{4}}\right)
Putting (10) and (11) in Equation (7), we have
\kappa =\frac{{M}^{2}\left(1-\frac{{J}^{2}}{2{M}^{4}}\right)}{4{M}^{3}\left(1-\frac{{J}^{2}}{4{M}^{4}}\right)}=\frac{\left(1-\frac{{J}^{2}}{2{M}^{4}}\right)}{4M\left(1-\frac{{J}^{2}}{4{M}^{4}}\right)}
\kappa =\frac{\left(1-\frac{{J}^{2}}{2{M}^{4}}\right){\left(1-\frac{{J}^{2}}{4{M}^{4}}\right)}^{-1}}{4M}=\frac{\left(1-\frac{{J}^{2}}{2{M}^{4}}\right)\left(1+\frac{{J}^{2}}{4{M}^{4}}\right)}{4M}
The solution of above equation gives the following result.
\kappa =\frac{1}{4M}\left[1-\frac{{J}^{2}}{4{M}^{4}}+\frac{{J}^{4}}{8{M}^{8}}\right]
For massive/super massive black holes, the term J4/8M4 can be neglected.
\kappa =\frac{1}{4M}\left[1-\frac{{J}^{2}}{4{M}^{4}}\right]
For a black holes of zero angular momentum, J = 0
\kappa =\frac{1}{4M}
The above equation shows the surface gravity of black hole in Schwarzschild case [9] .
\frac{1}{\kappa }=4M\left[1+\frac{{J}^{2}}{2{M}^{4}}\right]
The angular momentum of black hole is defined by the following equation [10]
J={a}_{\ast }G{M}^{2}/c
Throughout this research paper, assuming G = c = 1 [9] [10] , we have
J={a}_{\ast }{M}^{2}
The radius is smaller in the case of spinning black holes, tending to GM/c2 as a* = 1 [9] and hence, we have
J={M}^{2}
Using the above equation into Equation (17), we have
\frac{1}{\kappa }=6M
\kappa =\frac{1}{6M}
The Equation (16) shows the surface gravity of non-spinning black holes/Schwarzschild black holes, while Equation (22) shows the surface gravity of spinning black holes with spin parameter a* = 1. This means that the surface gravity of spinning BH with a* = 1 is lesser than to that of non-spinning black holes of the same mass.
Using the above Equation (22) into Equation (3), we have
\delta S=12\text{π}M\left[\delta M-\Omega \delta J\right]
For unit angular velocity, the above equation takes place
\delta S=12\text{π}M\left[\delta M-\delta J\right]
Differentiating equation (20), we have
\delta J=2M\delta M
\delta S=12\text{π}M\left[\delta M-2M\delta M\right]
\delta S=12\text{π}M\left[1-2M\right]\delta M
\frac{\delta S}{\delta M}=12\text{π}M\left[1-2M\right]
The term 1 in compared with the mass of black holes is negligible and can be neglected.
\frac{\delta S}{\delta M}=-24\text{π}{M}^{2}
|\frac{\delta S}{\delta M}|=24\text{π}{M}^{2}
|\frac{\delta S}{\delta M}|\alpha {M}^{2}
4. Data in Support for the Mass of Black Holes
There are two categories of black holes classified on the basis of their masses clearly very distinct from each other, with very different masses M ~ 5 20 Mʘ for stellar-mass black holes in X-ray binaries [10] .
On the basis of the data mentioned above regarding the mass of sun and black holes in XRBs, we have calculated change in entropy of spinning black holes due to corresponding change in mass for different test spinning black holes.
In the present work, firstly we have started our work with the general form of first law of black hole mechanics to give a model for change in entropy of
uncharged spinning black holes by the relation
\delta S=\frac{\text{2π}}{\kappa }\left[\delta M-\Omega \delta J\right]
Secondly, we have calculated the surface gravity of uncharged spinning black holes by putting dQ = 0 in the Kerr solution and obtained its value by this
\kappa =\frac{1}{6M}
Thirdly, we have used a relation between the angular momentum (J) and mass of spinning black holes to convert the change in the angular momentum in terms of change in mass of the spinning black holes for unit spin parameter and angular velocity.
The work is further extended with proper mathematical operation to obtain the change in entropy of uncharged spinning black hole with respect to the
change in mass (dS/dM) in terms of mass by the equation
|\frac{\delta S}{\delta M}|=24\text{π}{M}^{2}
The final Equation (30) shows that the magnitude of the change in entropy due to change in mass of spinning black holes in XRBs is directly proportional to the square of the mass of the spinning black holes.
Fourthly, we have plotted the graph with the help of Table 1, between the mass of spinning black holes and corresponding the magnitude of the change in entropy with change in mass of spinning black holes in XRBs as shown in the Figure 1 and shows that the magnitude of change in entropy with change in mass increases with the increasing the mass of different test spinning black holes.
We have concluded the following facts during the research work:
1) The surface gravity of uncharged spinning black holes is lesser than to that of the non-spinning black holes.
2) The magnitude of the change in entropy with change in mass of spinning black holes is directly proportional to the square of their masses.
3) The magnitude of the change in entropy with change in mass of uncharged spinning black holes is essentially the function of their masses.
Table 1. Change in mass of spinning black holes due to corresponding change in angular momentum in XRBs for a* = 1 and (Ω = 1).
Figure 1. The graph plotted between the mass of black holes in terms of solar mass and their corresponding change in entropy relative to the change in mass for XRBs.
4) On increasing the mass of spinning black holes, their corresponding change in magnitude of the entropy with change in mass increases in the case of unit spin parameter and angular velocity.
The authors are very grateful to reviewers to find out the errors in the original manuscript and providing the constructive suggestions. This paper is also devoted in memory of great scientist Stephen Hawking (March 14, 2018).
Cite this paper: Mahto, D. and Kumari, A. (2018) Change in Entropy of Spinning Black Holes Due to Corresponding Change in Mass in XRBs. International Journal of Astronomy and Astrophysics, 8, 171-177. doi: 10.4236/ijaa.2018.82012.
[1] Wald, R.M. (2001) The Thermodynamics of Black Holes. Living Rev. in Relativity.
https://www.springer.com/cn/livingreviews
[2] Hawking, S.W. (1975) Particle Creation by Black Hole. Communications in Mathematical Physics, 43, 199-220.
[3] Dabholkar, A. (2006) Black Hole Entropy in String Theory: A Window into the Quantum Structure of Gravity. In: The Legacy of Albert Einstein, Current Science, 47-68.
[4] Ren, Z., Hai-Xia, Z. and Shuang-Qi, H. (2006) General Logarithmic Corrections to Bekenstein and Hawking Entropy. arxiv-gr-qc0609080.
[5] Mahto, D., Kumari, K., Sah, R.K. and Singh, K.M. (2011) Study of Non-Spinning Black Holes with Reference to the Change in Energy and Entropy. Astrophysics and Space Science, 337, 685-691.
[6] Mahto, D., Prakash, V., Prasad, U. And Singh, B.K. (2013) Change in Entropy of Non-Spinning Black Holes w.r.t. the Radius of Event Horizon in XRBs. Astrophysics and Space Science, 343, 153-159.
[7] Mahto, D., Prakash, V., Singh, K.M. and Kumar, B. (2015) Classical Statistical Entropy of Black Hole. American Journal of Theoretical and Applied Statistics, 4, 1-18.
https://doi.org/10.11648/j.ajtas.s.2015040101.13
[9] Transchen, J. (2000) An Introduction to Black Hole Evaporation. arXiv: gr-qc/0010055V.
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Dyadic analysis filter bank - MATLAB - MathWorks Switzerland
dsp.DyadicAnalysisFilterBank
Subband Ordering For Asymmetric Tree Structure Using Dyadic Analysis Filter Bank
Subband Ordering For Symmetric Tree Structure Using Dyadic Analysis Filter Bank
The dsp.DyadicAnalysisFilterBank System object™ decomposes a broadband signal into a collection of subbands with smaller bandwidths and slower sample rates. The System object uses a series of highpass and lowpass FIR filters to provide approximate octave band frequency decompositions of the input. Each filter output is downsampled by a factor of two. With the appropriate analysis filters and tree structure, the dyadic analysis filter bank is a discrete wavelet transform (DWT) or discrete wavelet packet transform (DWPT).
To obtain approximate octave band frequency decompositions of the input:
Create the dsp.DyadicAnalysisFilterBank object and set its properties.
dydan = dsp.DyadicAnalysisFilterBank
dydan = dsp.DyadicAnalysisFilterBank(Name,Value)
dydan = dsp.DyadicAnalysisFilterBank constructs a dyadic analysis filter bank object, dydan, that computes the level-two discrete wavelet transform (DWT) of a column vector input. For a 2-D matrix input, the object transforms the columns using the Daubechies third-order extremal phase wavelet. The length of the input along the first dimension must be a multiple of 4.
dydan = dsp.DyadicAnalysisFilterBank(Name,Value) returns a dyadic analysis filter bank object, with each property set to the specified value.
Filter — Type of filter used in subband decomposition
Specify the type of filter used to determine the high and lowpass FIR filters in the dyadic analysis filter bank as Custom , Haar, Daubechies, Symlets, Coiflets, Biorthogonal, Reverse Biorthogonal, or Discrete Meyer. All property values except Custom require Wavelet Toolbox™ software. If the value of this property is Custom, the filter coefficients are specified by the values of the CustomLowpassFilter and CustomHighpassFilter properties. Otherwise, the dyadic analysis filter bank object uses the Wavelet Toolbox function wfilters to construct the filters. The following table lists supported wavelet filters and example syntax to construct the filters:
Syntax for Analysis Filters
Haar N/A [Lo_D,Hi_D]=wfilters('haar');
Daubechies extremal phase WaveletOrder=3; [Lo_D,Hi_D]=wfilters('db3');
Symlets (Daubechies least-asymmetric) WaveletOrder=4; [Lo_D,Hi_D]=wfilters('sym4');
Coiflets WaveletOrder=1; [Lo_D,Hi_D]=wfilters('coif1');
Biorthogonal FilterOrder='[3/1]'; [Lo_D,Hi_D,Lo_R,Hi_R]=... wfilters('bior3.1');
Reverse biorthogonal FilterOrder='[3/1]'; [Lo_D,Hi_D,Lo_R,Hi_R]=... wfilters('rbior3.1');
Discrete Meyer N/A [Lo_D,Hi_D]=wfilters('dmey');
Specify a vector of lowpass FIR filter coefficients, in powers of z-1. Use a half-band filter that passes the frequency band stopped by the filter specified in the CustomHighpassFilter property. The default specifies a Daubechies third-order extremal phase scaling (lowpass) filter.
Specify a vector of highpass FIR filter coefficients, in powers of z-1. Use a half-band filter that passes the frequency band stopped by the filter specified in the CustomLowpassFilter property. The default specifies a Daubechies 3rd-order extremal phase wavelet (highpass) filter.
WaveletOrder — Order for orthogonal wavelets
FilterOrder — Analysis and synthesis filter orders for biorthogonal filters
Specify the order of the analysis and synthesis filter orders for biorthogonal filter banks as 1 / 1, 1 / 3, 1 / 5, 2 / 2, 2 / 4, 2 / 6, 2 / 8, 3 / 1, 3 / 3, 3 / 5, 3 / 7, 3 / 9, 4 / 4, 5 / 5, or 6 / 8. Unlike orthogonal wavelets, biorthogonal wavelets require different filters for the analysis (decomposition) and synthesis (reconstruction) of an input. The first number indicates the order of the synthesis (reconstruction) filter. The second number indicates the order of the analysis (decomposition) filter.
NumLevels — Number of filter bank levels used in analysis (decomposition)
Specify the number of filter bank analysis levels a positive integer greater than or equal to 1. A level-N asymmetric structure produces N+1 output subbands. A level-N symmetric structure produces 2N output subbands. The size of the input along the first dimension must be a multiple of 2N, where N is the number of levels.
Specify the structure of the filter bank as Asymmetric or Symmetric. The asymmetric structure decomposes only the lowpass filter output from each level. The symmetric structure decomposes the highpass and lowpass filter outputs from each level. If the analysis filters are scaling (lowpass) and wavelet (highpass) filters, the asymmetric structure is the discrete wavelet transform, while the symmetric structure is the discrete wavelet packet transform.
When this property is Symmetric, the output has 2N subbands each of size M/2N. In this case, M is the length of the input along the first dimension and N is the value of the NumLevels property. When this property is Asymmetric, the output has N+1 subbands. The following equation gives the length of the output in the kth subband in the asymmetric case:
{M}_{k}=\left\{\begin{array}{ll}\frac{M}{{2}^{k}}\hfill & 1\le k\le N\hfill \\ \frac{M}{{2}^{N}}\hfill & k=N+1\hfill \end{array}
y = dydan(x)
y = dydan(x) computes the subband decomposition of the input x and outputs the dyadic subband decomposition in y as a single concatenated column vector or matrix of coefficients.
Data input, specified as a column vector or a matrix. Each column of x is treated as an independent input, and the number of rows of x must be a multiple of
{2}^{N},
where N is the number of levels specified by the NumLevels property.
y — Dyadic subband decomposition output
Dyadic subband decomposition output, returned as a column vector or a matrix. The elements of y are ordered with the highest-frequency subband first followed by subbands in decreasing frequency.
When TreeStructure is set to Symmetric, the output has 2N subbands each of size M/2N. In this case, M is the length of the input along the first dimension, and N is the value of the NumLevels property. When TreeStructure is set to Asymmetric, the output has N+1 subbands. The following equation gives the length of the output in the kth subband in the asymmetric case:
{M}_{k}=\left\{\begin{array}{ll}\frac{M}{{2}^{k}}\hfill & 1\le k\le N\hfill \\ \frac{M}{{2}^{N}}\hfill & k=N+1\hfill \end{array}
Sampling frequency 1 kHz input length 1024
t = 0:.001:1.023;
x = square(2*pi*30*t);
Default asymmetric structure with order 3 extremal phase wavelet
dydan = dsp.DyadicAnalysisFilterBank;
Y = dydan(xn);
Level 2 yields 3 subbands (two detail-one approximation) Nyquist frequency is 500 Hz
D1 = Y(1:512); % subband approx. [250, 500] Hz
D2 = Y(513:768); % subband approx. [125, 250] Hz
Approx = Y(769:1024); % subband approx. [0,125] Hz
Sampling frequency 1 kHz input length 1024.
dydan = dsp.DyadicAnalysisFilterBank('TreeStructure',...
'Symmetric');
D1 = Y(1:256); % subband approx. [375,500] Hz
D2 = Y(257:512); % subband approx. [250,375] Hz
Approx = Y(769:1024); % subband approx. [0, 125] Hz
This object implements the algorithm, inputs, and outputs described on the Dyadic Analysis Filter Bank block reference page. The object properties correspond to the block parameters, except:
The dyadic analysis filter bank object always concatenates the subbands into a single column vector for a column vector input, or into the columns of a matrix for a matrix input. This behavior corresponds to the block's behavior when you set the Output parameter to Single port.
dsp.DyadicSynthesisFilterBank | dsp.SubbandAnalysisFilter
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Proving single solution of initial value problem is increasing Given the
Libby Boone 2022-05-01 Answered
Proving single solution of initial value problem is increasing
{y}^{\prime }\left(x\right)=y\left(x\right)-\mathrm{sin}\left\{y\left(x\right)\right\},y\left(0\right)=1
wellnesshaus4n4
We use the simple fact that
v\left(y\right)=y-\mathrm{sin}y
is increasing and positive on
y>0
{y}_{0}\left(x\right)=0
is a solution implies that
y\left(x\right)>0
for all x (by uniqueness). Then
{y}^{\prime }\left(x\right)=v\left(y\left(x\right)\right)>0.
\underset{x\to -\mathrm{\infty }}{lim}y\left(x\right)
exists and is nonnegative. If it's not zero, then
y\left(x\right)\ge {y}_{1}>0
{y}_{1}
{y}^{\prime }\left(x\right)=v\left(y\left(x\right)\right)\ge v\left({y}_{1}\right)>0.
That would imply for any
x<0
(by the mean value theorem and that
y\left(0\right)=1
y\left(x\right)=y\left(x\right)-y\left(0\right)+y\left(0\right)={y}^{\prime }\left({x}_{0}\right)x+1\le v\left({y}_{1}\right)x+1
this is impossible as
y\left(x\right)>0
Giovanny Howe
Let us introduce the new variable z defined by
z=-x
{y}^{\prime }\left(x\right)=\frac{dy}{dx}
=\frac{dy}{dz}\frac{dz}{dx}
=-\frac{dy}{dz}
Therefore, we have the differential equation with the reversed variable direction
\frac{dy}{dz}=-y+\mathrm{sin}y
This is in fact the dynamics of a simple damped pendulum. Let
V\left(y\right)=\frac{{y}^{2}}{2}
\frac{dV}{dz}=y\frac{dy}{dz}
=-{y}^{2}+y\mathrm{sin}y
\frac{dV}{dz}<0
\frac{y\in \mathbb{R}}{\left\{0\right\}}
\frac{dV}{dz}=0
y=0
\mathbb{R}
is positively invariant with respect to the ODE, meaning that every solution starting at
z={z}_{0}
zgiven{z}_{0}
. Let v be the solution of the ODE with
y\left(0\right)=1
. Then, we have that
\underset{z⇒\mathrm{\infty }}{lim}v\left(z\right)=0
. Finally, we conclude that
\underset{x⇒-\mathrm{\infty }}{lim}u\left(x\right)=0
We can also show that
v\left(z\right)
cannot change its sign and monotonically decreases whenever
v\left(z\right)>0
\frac{dw}{dt}
using the appropriate Chain Rule. Evaluate
\frac{dw}{dt}
at the given value of t. Function:
w=x\mathrm{sin}y,\text{ }x={e}^{t},\text{ }y=\pi -t
t=0
Limit of a separable equation:
\frac{dy}{dt}=k\left(a-y\right)\left(b-y\right)
where a,b and k are constants. Assuming
y\left(0\right)=0
a) Solve for y(t) when
a=b
b) Solve for the case
0<a<b
c) By considering the limit
b\to a
in (b) show that the two results are consistent
How to solve the ODE
y\text{'}=\frac{x-{e}^{x}}{x+{e}^{y}}
Write new boundary conditions to a system of ODEs
Suppose we have the following system of equations:
\left\{\begin{array}{l}{\stackrel{¨}{x}}_{1}=f\left({x}_{1},{x}_{2}\right)\\ {\stackrel{¨}{x}}_{2}=g\left({x}_{1},{x}_{2}\right)\\ {x}_{1}\left(0\right)=\alpha \\ {x}_{2}\left(0\right)=\beta \\ {\stackrel{˙}{x}}_{1}\left(0\right)=\gamma \\ {\stackrel{˙}{x}}_{2}\left(0\right)=\delta \end{array}
where all the derivatives are made in a time variable t.
Since I would like to work with a system of first order equations, I would go like this:
\left\{\begin{array}{l}{\stackrel{˙}{x}}_{1}={x}_{3}\\ {\stackrel{˙}{x}}_{3}=f\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right)\\ {\stackrel{˙}{x}}_{2}={x}_{4}\\ {\stackrel{˙}{x}}_{4}=g\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right)\end{array}
How can I apply the starting conditions at time
t=0
at the new system?
Solution to Linear Time-invariant Matrix ODE
\frac{d\mathbf{P}}{dt}=\mathbf{A}\mathbf{P}+\mathbf{B}
x\left(2y+1\right)dx=y\left({x}^{2}-3x+2\right)dy
\frac{dy}{dx}=y+1
Solving the above given differential equation, yields the following general solution.
y+1={e}^{x+C}
y=C{e}^{x}-1
⇒
y=-1
C=0
Can I say that
y=-1
is a singular solution?
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Interpreting the Return on Investment (ROI)
An Alternative Return on Investment (ROI) Calculation
Investments and Annualized ROI
Combining Leverage With Return on Investment (ROI)
The Problem of Unequal Cash Flows
Return on investment (ROI) is a financial metric that is widely used to measure the probability of gaining a return from an investment. It is a ratio that compares the gain or loss from an investment relative to its cost. It is as useful in evaluating the potential return from a stand-alone investment as it is in comparing returns from several investments.
In business analysis, ROI and other cash flow measures—such as internal rate of return (IRR) and net present value (NPV)—are key metrics that are used to evaluate and rank the attractiveness of a number of different investment alternatives. Although ROI is a ratio, it is typically expressed as a percentage rather than as a ratio.
Return on investment (ROI) is an approximate measure of an investment's profitability.
ROI has a wide range of applications; it can be used to measure the profitability of a stock investment, when deciding whether or not to invest in the purchase of a business, or evaluate the results of a real estate transaction.
ROI is relatively easy to calculate and understand, and its simplicity means that it is a standardized, universal measure of profitability.
One disadvantage of ROI is that it doesn't account for how long an investment is held; so, a profitability measure that incorporates the holding period may be more useful for an investor that wants to compare potential investments.
ROI can be calculated using two different methods.
ROI = \frac{\text{Net\ Return \ on \ Investment}}{\text{Cost \ of \ Investment}}\times 100\%
ROI=Cost of InvestmentNet Return on Investment×100%
ROI = \frac{\text{Final Value of Investment}\ -\ \text{Initial Value of Investment}}{\text{Cost of Investment}}\times100\%
ROI=Cost of InvestmentFinal Value of Investment − Initial Value of Investment×100%
When interpreting ROI calculations, it's important to keep a few things in mind. First, ROI is typically expressed as a percentage because it is intuitively easier to understand (as opposed to when expressed as a ratio). Second, the ROI calculation includes the net return in the numerator because returns from an investment can be either positive or negative.
When ROI calculations yield a positive figure, it means that net returns are in the black (because total returns exceed total costs). Alternatively, when ROI calculations yield a negative figure, it means that net returns are in the red because total costs exceed total returns. (In other words, this investment produces a loss.) Finally, to calculate ROI with the highest degree of accuracy, total returns and total costs should be considered. For an apples-to-apples comparison between competing investments, annualized ROI should be considered.
Assume an investor bought 1,000 shares of the hypothetical company Worldwide Wickets Co. at $10 per share. One year later, the investor sold the shares for $12.50. The investor earned dividends of $500 over the one-year holding period. The investor also spent a total of $125 on trading commissions in order to buy and sell the shares.
The ROI for this investor can be calculated as follows:
ROI = ([($12.50 - $10.00) * 1000 + $500 - $125] ÷ ($10.00 * 1000)) * 100 = 28.75%
Here is a step-by-step analysis of the calculation:
To calculate net returns, total returns and total costs must be considered. Total returns for a stock result from capital gains and dividends. Total costs would include the initial purchase price as well as any commissions paid.
In the above calculation, the gross capital gain (before commissions) from this trade is ($12.50 - $10.00) x 1,000. The $500 amount refers to the dividends received by holding the stock, while $125 is the total commissions paid.
If you further dissect the ROI into its component parts, it is revealed that 23.75% came from capital gains and 5% came from dividends. This distinction is important because capital gains and dividends are taxed at different rates in most jurisdictions.
ROI = Gross Capital Gains % - Commission % + Dividend Yield
Gross Capital Gains = $2500 ÷ $10,000 * 100 = 25.00%Commissions = $125 ÷ $10,000 * 100 = 1.25%Dividend Yield = $500 ÷ $10,000 * 100 = 5.00%ROI = 25.00% - 1.25% + 5.00% = 28.75%
A positive ROI means that net returns are positive because total returns are greater than any associated costs; a negative ROI indicates that net returns are negative—total costs are greater than returns.
If, for example, commissions were split, there is an alternative method of calculating this hypothetical investor's ROI for their Worldwide Wickets Co. investment. Assume the following split in the total commissions: $50 when buying the shares and $75 when selling the shares.
FVI = $12,500 + $500 - $75 = $12,925
ROI = [($12,925 - $10,050) ÷ $10,000] * 100 = 28.75%
In this formula, IVI refers to the initial value of the investment (or the cost of the investment). FVI refers to the final value of the investment.
Annualized ROI helps account for a key omission in standard ROI—namely, how long an investment is held.
The annualized ROI calculation provides a solution for one of the key limitations of the basic ROI calculation; the basic ROI calculation does not take into account the length of time that an investment is held, also referred to as the holding period. The formula for calculating annualized ROI is as follows:
\begin{aligned} &\text{Annualized } ROI = [(1 + ROI) ^{1/n} - 1]\times100\%\\ &\textbf{where:}\\ &\begin{aligned} n=\ &\text{Number of years for which the investment}\\ &\text{is held} \end{aligned} \end{aligned}
Annualized ROI=[(1+ROI)1/n−1]×100%where:
Assume a hypothetical investment that generated an ROI of 50% over five years. The simple annual average ROI of 10%–which was obtained by dividing ROI by the holding period of five years–is only a rough approximation of annualized ROI. This is because it ignores the effects of compounding, which can make a significant difference over time. The longer the time period, the bigger the difference between the approximate annual average ROI, which is calculated by dividing the ROI by the holding period in this scenario, and annualized ROI.
\begin{aligned} &\text{From the formula above,}\\ &\text{Annualized ROI}=[(1+0.50)^{1/5}-1]\times100\%=8.45\% \end{aligned}
From the formula above,
This calculation can also be used for holding periods of less than a year by converting the holding period to a fraction of a year.
Assume an investment that generated an ROI of 10% over six months.
\text{Annualized ROI}=[(1+0.10)^{1/0.5}-1]\times100\%=21.00\%
Annualized ROI=[(1+0.10)1/0.5−1]×100%=21.00%
In the equation above, the numeral 0.5 years is equivalent to six months.
Annualized ROI is especially useful when comparing returns between various investments or evaluating different investments.
Assume that an investment in stock X generated an ROI of 50% over five years, while an investment in stock Y returned 30% over three years. You can determine what the better investment was in terms of ROI by using this equation:
\begin{aligned} &AROIX=[(1+0.50)^{1/5}-1]\times100\%=8.45\%\\ &AROIY=[(1+0.30)^{1/3}-1]\times100\%=9.14\%\\ &\textbf{where:}\\ &AROIX = \text{Annualized ROI for stock }X\\ &AROIY = \text{Annualized ROI for stock }Y \end{aligned}
AROIX=[(1+0.50)1/5−1]×100%=8.45%AROIY=[(1+0.30)1/3−1]×100%=9.14%where:AROIX=Annualized ROI for stock X
According to this calculation, stock Y had a superior ROI compared to stock X.
Leverage can magnify ROI if the investment generates gains. However, by the same token, leverage can also amplify losses if the investment proves to be a losing investment.
Assume that an investor bought 1,000 shares of the hypothetical company Worldwide Wickets Co. at $10 per share. Assume also that the investor bought these shares on a 50% margin (meaning they invested $5,000 of their own capital and borrowed $5,000 from their brokerage firm as a margin loan). Exactly one year later, this investor sold their shares for $12.50. They earned dividends of $500 over the one-year holding period. They also spent a total of $125 on trading commissions when they bought and sold the shares. In addition, their margin loan carried an interest rate of 9%.
When calculating the ROI on this specific, hypothetical investment, there are a few important things to keep in mind. First, in this example, the interest on the margin loan ($450) should be considered in total costs. Second, the initial investment is now $5,000 because of the leverage employed by taking the margin loan of $5,000.
ROI = [($12.50 - $10)*1000 + $500 - $125 - $450] ÷ [($10*1000) - ($10*500)] * 100 = 48.5%
Thus, even though the net dollar return was reduced by $450 on account of the margin interest, ROI is still substantially higher at 48.50% (compared with 28.75% if no leverage was employed).
As an additional example, consider if the share price fell to $8.00 instead of rising to $12.50. In this situation, the investor decides to cut their losses and sell the full position. Here is the calculation for ROI in this scenario:
\begin{aligned} \text{ROI}=&\frac{[(\$8.00-\$10.00)\times1,000]+\$500-\$125-\$450}{(\$10.00\times1,000)-(\$10.00\times500)}\\ &\times100\%=-\frac{\$2,075}{\$5,000} =-41.50\% \end{aligned}
ROI=($10.00×1,000)−($10.00×500)[($8.00−$10.00)×1,000]+$500−$125−$450
In this case, the ROI of -41.50% is much worse than an ROI of -16.25%, which would have occurred if no leverage was employed.
When evaluating a business proposal, it's possible that you will be contending with unequal cash flows. In this scenario, ROI may fluctuate from one year to the next.
This type of ROI calculation is more complicated because it involves using the internal rate of return (IRR) function in a spreadsheet or calculator.
Assume you are evaluating a business proposal that involves an initial investment of $100,000. (This figure is shown under the "Year 0" column in the "Cash Outflow" row in the following table.) This investment will generate cash flows over the next five years; this is shown in the "Cash Inflow" row. The row called "Net Cash Flow" sums up the cash outflow and cash inflow for each year.
Using the IRR function, the calculated ROI is 8.64%.
The final column shows the total cash flows over the five-year period. Net cash flow over this five-year period is $25,000 on an initial investment of $100,000. If this $25,000 was spread out equally over five years, the cash flow table would then look like this:
In this case, the IRR is now only 5.00%.
The substantial difference in the IRR between these two scenarios—despite the initial investment and total net cash flows being the same in both cases—has to do with the timing of the cash inflows. In the first case, substantially larger cash inflows are received in the first four years. Because of the time value of money, these larger inflows in the earlier years have a positive impact on IRR.
The biggest benefit of ROI is that it is a relatively uncomplicated metric; it is easy to calculate and intuitively easy to understand. ROI's simplicity means that it is often used as a standard, universal measure of profitability. As a measurement, it is not likely to be misunderstood or misinterpreted because it has the same connotations in every context.
There are also some disadvantages of the ROI measurement. First, it does not take into account the holding period of an investment, which can be an issue when comparing investment alternatives. For example, assume investment X generates an ROI of 25%, while investment Y produces an ROI of 15%.
One cannot assume that X is the superior investment unless the time frame of each investment is also known. It's possible that the 25% ROI from investment X was generated over a period of five years, but the 15% ROI from investment Y was generated in only one year. Calculating annualized ROI can overcome this hurdle when comparing investment choices.
Second, ROI does not adjust for risk. It is common knowledge that investment returns have a direct correlation with risk: the higher the potential returns, the greater the possible risk. This can be observed firsthand in the investment world, where small-cap stocks typically have higher returns than large-cap stocks (but are accompanied by significantly greater risk).
An investor who is targeting a portfolio return of 12%, for example, would have to assume a substantially higher degree of risk than an investor whose goal is a return of only 4%. If an investor hones in on only the ROI number without also evaluating the associated risk, the eventual outcome of the investment decision may be very different from the expected result.
Third, ROI figures can be exaggerated if all the expected costs are not included in the calculation. This can happen either deliberately or inadvertently. For example, in evaluating the ROI on a piece of real estate, all associated expenses should be considered. These include mortgage interest, property taxes, insurance, and all costs of maintenance. These expenses can subtract a large amount from the expected ROI; without including all of them in the calculation, a ROI figure can be grossly overstated.
Finally, like many profitability metrics, ROI only emphasizes financial gains when considering the returns on an investment. It does not consider ancillary benefits, such as social or environmental goods. A relatively new ROI metric, known as social return on investment (SROI), helps to quantify some of these benefits for investors.
Return on investment (ROI) is a simple and intuitive metric of the profitability of an investment. There are some limitations to this metric, including that it does not consider the holding period of an investment and is not adjusted for risk. However, despite these limitations, ROI is still a key metric used by business analysts to evaluate and rank investment alternatives.
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Electrostatic Mechanism for Depolymerization-Based Poleward Force Generation at Kinetochores
Electrostatic Mechanism for Depolymerization-Based Poleward Force Generation at Kinetochores ()
L. John Gagliardi1, Daniel H. Shain2
1Department of Physics, Rutgers The State University of New Jersey, Camden, NJ, USA.
2Department of Biology, Rutgers The State University of New Jersey, Camden, NJ, USA.
DOI: 10.4236/ojbiphy.2019.93014 PDF HTML XML 309 Downloads 565 Views Citations
Experiments implicating bound volume positive charge at kinetochores interacting with negative charge at microtubule free ends have prompted our calculation of the force at kinetochores for chromosome poleward motility during mitosis. We present here a corroborating force calculation between positively charged Hec1 tails in kinetochores and negatively charged C-termini at microtubule free ends. Based on experimentally-known charge magnitudes on Hec1 tails and C-termini at microtubule free ends, an ab initio calculation of poleward (tension) force per microtubule that falls within the experimental range is demonstrated. Due to the locations of C-termini charges on concave sides of splaying microtubules, this attractive force between subsets of low curvature splaying microtubule protofilaments C-termini eventually fails for subsets of protofilaments with more pronounced curvature, thus generating poleward force as microtubules depolymerize in a dynamic coupling, as observed experimentally. The mechanism by which kinetochores establish and maintain a dynamic coupling to microtubules for force production during the complex motions of mitosis remains elusive, and force generation at kinetochores has emerged as a signature problem in chromosome motility. In agreement with experiment, two separate calculations show that attractive electrostatic interactions over nanometer distances account for poleward chromosome forces at kinetochores.
Mitosis, Chromosome, Motility, Force, Electrostatics
Gagliardi, L. and Shain, D. (2019) Electrostatic Mechanism for Depolymerization-Based Poleward Force Generation at Kinetochores. Open Journal of Biophysics, 9, 198-203. doi: 10.4236/ojbiphy.2019.93014.
Force generation at kinetochores has emerged as one of the signature problems in mitotic movements. Consistent with theoretical predictions made over a decade ago [1] [2], electrostatic interactions at kinetochores between negatively charged microtubule plus ends and positive charge at kinetochores have more recently been proposed for chromosome motility during mitosis [3] . A number of currently advanced models involve interactions that are fundamentally electrostatic, including mechanisms for chromosome movements based on protofilament-end splaying. A brief review of current models for force production at kinetochores is given elsewhere [4], where we support the experimental work of Miller et al. [3] with an ab initio calculation of the force between bound volume positive charge distributions at kinetochores interacting electrostatically with bound negative charge at free ends of microtubules.
Our purpose here is to mathematically corroborate that calculation with one that is based on direct interactions between positively charged unstructured Ndc80Hec1 tails in kinetochores and negatively charged C-termini at the free ends of microtubules, supporting an electrostatic-based model that explains poleward force generation.
Miller et al. [3], advances Ndc80/Hec1 as responsible for electrostatics-based force production at kinetochores. They propose that the force-producing interaction is electrostatic since an unstructured positively charged Hec1 tail cannot bind microtubules lacking negatively charged C-termini, concluding that “… the highest affinity interactions between kinetochores and microtubules are ionic attractions between two unstructured domains”. Our approach supports the role of Hec1 as bound volume charge distributions—“positively charged Hec1 tails” [3] —at kinetochores, interacting electrostatically with bound negative charge at and near the free ends of microtubules—“ionic attractions between two unstructured domains” [3] .
Chromosomes can move toward a proximal pole only when their kinetochores are connected to microtubules coming from that pole [5] . Microtubule polymerization and depolymerization follow a pattern characterized as “dynamic instability.” This means that at any given time, some of the microtubules are growing, while others are undergoing rapid breakdown. The rate at which microtubules undergo net assembly, or disassembly (depolymerization), varies with mitotic stage [6] . In the present context, depolymerization-based electrostatic attractions are responsible for poleward force generation at kinetochores; electrostatic interactions for poleward force production at centrosomes are treated elsewhere [7] .
The electrostatic properties of tubulin have been well-studied [8] [9] [10] [11] . Large-scale computer calculations have determined the dipole moment to be as large as 1800 Debye [9] [12] . Tubulin has a large overall charge of −20 (electron charges) at pH 7, and up to 40% of the charge resides on C-termini [13] . This large net charge on C-termini is integral to electrostatics-based force production at kinetochores (see below).
In the context of force generation for chromosome motility at kinetochores, Miller et al. [3] state that “… our data argue strongly that the Hec1 tail is the critical attachment for deploymerization-coupled movements of chromosomes”; and conclude “… the highest affinity interactions between kinetochores and microtubules are ionic attractions between two unstructured domains.” This essentially proposes that bound, oppositely charged distributions are the underlying cause for poleward chromosome motions. As mentioned above, we recently published a force calculation between Hec1 charges, modeled as an experimentally known bound volume positive charge—“unstructured” positive charge—at kinetochores, and experimentally known negative charge at kinetochore microtubule free plus ends that agrees with experimental measurements of the poleward force for chromosome motility [4] . Here we provide a force calculation between positively charged Hec1 tails in kinetochores and negatively charged C-termini at and near microtubule free ends that confirms our previous proposal.
Since the lengths of Hec1 tails are much longer than the location volumes of C-termini charge distributions, Hec1 tails will be modeled as very long linear charges, with a linear charge density λ C/m (Coulombs/meter). A simple application of Gauss’s law [14] for an infinitely long line charge distribution gives the electric field magnitude at a distance r from the line charge as
E=\lambda /2\pi \epsilon r
where ε (=kε0) is the kinetochore permittivity, ε0 = 8.85 pF/m (picoFarads/meter), and k is the kinetochore dielectric constant. Note that the relatively small contributions from edge effects near the ends of the Hec1 tails are neglected in this calculation.
The N-terminal tail of Hec1 contains an equivalent positive charge Q of 10 (electron charges, e) [3], distributed over a distance l of 55 nm [15], giving a linear charge density λ = Q/l of 10 e/55 nm = 29 pC/m (picoCoulombs per meter). The force generating interaction is between positively charged Hec1 tails and negatively charged C-termini on concave sides of splaying protofilaments.
For the force per protofilament, we have:
Fpf=qE=ne\lambda /2\pi \epsilon r
where q = ne is the charge of n electrons on C-termini of a protofilament interacting with a Hec1 tail. Consistent with their open structures, a cytosol-saturated kinetochore is expected to have a dielectric constant midway between the kinetochore dry value and cytoplasmic water [16] . Since most condensed-matter (dry) dielectric constants are between 1 and 5, the value for cytoplasmic water dominates, and a conservative midpoint value k = 45 ((80 + 10)/2) will be assumed [4] . Substituting this value in (2), with λ = 29 pC/m, and the distance of the effective charge centers of C-termini charges, r = 3 nm, we have Fpf = 0.6n pN/pf (picoNewtons per protofilament).
Kinetochores generally number at least 8 Hec1 proteins per microtubule [17], and there are 13 protofilaments per microtubule. It will therefore be conservatively assumed that four protofilaments in a microtubule are interacting with a Hec1 tail at any given moment. These subsets would be constantly changing among the microtubules penetrating a kinetochore. Thus, the total force per microtubule F = 4(0.62) n = 2.5 n pN/MT (picoNewtons per microtubule). Equating this to the experimental range 1 - 5 pN/MT [18], we find that n = 0.4 - 2.5 electron charges. This result, like that of the previous calculation [4], falls well within the observed experimental range [9] [13] [19], and the agreement represents a successful ab initio theoretical derivation of this force magnitude.
Since microtubule C-termini are on the concave sides of progressively splaying microtubules, increasing protofilament curvature will lead to a separation of the charges on Hec1 tails and C-termini. Subsets of low curvature splaying protofilaments produce poleward force, while other subsets of protofilaments with more pronounced curvature in later stages of depolymerization fail to bind. Accordingly, poleward forces are generated as microtubules depolymerize, in agreement with observation.
Electrostatic fields within the cytosol are subject to strong attenuation due to screening by oppositely charged ions (counterion screening), decreasing exponentially to much smaller values over a distance of several Debye lengths. The Debye length within cells is typically given to be of order 1 nm [20], and since cells have much larger dimensions, one is tempted to conclude that electrostatic force could not be a major factor in providing the cause for chromosome motility in biological cells. However the presence of microtubules challenges that notion. Microtubules can be thought of as intermediaries that extend the reach of the electrostatic interaction over cellular distances, making the second most powerful force in nature available to cells in spite of their ionic nature.
Cellular electrostatics is also strongly influenced by reduced counterion screening due to layered water adhering to charged molecules. Such water layering – with consequent reduction or elimination of Debye screening – at charged proteins has long been theorized [21] [22], and has been confirmed by experiment [23] . Additionally, water between sufficiently close (up to 3 nm) charged proteins has a dielectric permittivity that is considerably reduced from the bulk value far from charged surfaces [24] [25] [26] . The combination of these effects (or conditions)—water layering and reduced dielectric constant—can significantly influence cellular electrostatics in a number of important ways. This is especially true in relation to mitosis [26] .
Given positive charge at kinetochores and negative charge on plus ends of microtubules, it is difficult to conceptualize there not being an attractive electrostatic poleward-directed force between these structures. A direct calculation of the electrostatic force between positively charged Hec1 tails and negatively charged C-termini at and near the free ends of microtubules supports an electrostatic force generating mechanism for poleward chromosome motions during mitosis. A singular strength of the present calculation is that the disassembly rate of microtubules at kinetochores is explicitly shown to be correlated with force production at kinetochores. In a broader context, understanding the underlying forces and mechanisms that dictate chromosome movements through mitosis will be critical to the development of approaches to circumvent anomalous cell divisions (e.g., cancer).
[1] Gagliardi, L.J. (2002) Electrostatic Force in Prometaphase, Metaphase, and Anaphase-A Chromosome Motions. Physical Review E, 66, Article ID: 011901.
[2] Gagliardi, L.J. (2005) Electrostatic Force Generation in Chromosome Motions during Mitosis. Journal of Electrostatics, 63, 309-327.
[3] Miller, S.A., Johnson, M.L. and Stukenberg, P.T. (2008) Kinetochore Attachments Require an Interaction between Unstructured Tails on Microtubules and Ndc80/Hec1. Current Biology, 18, 1785-1791.
[4] Gagliardi, L.J. and Shain, D.H. (2016) Electrostatic Forces Drive Poleward Chromosome Motions at Kinetochores. Cell Division, 11, 14.
[5] Nicklas, R.B. and Kubai, D.F. (1985) Microtubules, Chromosome Movement, and Reorientation after Chromosomes Are Detached from the Spindle by Micromanipulation. Chromosoma, 92, 313-324.
[6] Alberts, B., et al. (1994) Molecular Biology of the Cell. 3rd Edition, Garland, New York, 920.
[7] Gagliardi, L.J. and Shain, D.H. (2014) Polar Electrostatic Forces Drive Poleward Chromosome Motions. Cell Division, 9, 5.
[8] Sataric, M.V., Tuszynski, J.A. and Zakula, R.B. (1993) Kinklike Excitations as an Energy Transfer Mechanism. Physical Review E, 48, 589-597.
[9] Brown, J.A. and Tuszynski, J.A. (1997) Dipole Interactions in Axonal Microtubules as a Mechanism of Signal Propagation. Physical Review E, 56, 5834-5840.
[10] Baker, N.A., et al. (2001) Electrostatics of Nanosystems: Applications to Microtubules and the Ribosome. Proceedings of the National Academy of Sciences, 98, 10037-10041.
[11] Tuszynski, J.A., Brown, J.A. and Hawrylak, P. (1998) Dielectric Polarization, Electrical Conduction, Information Processing and Quantum Computation in Microtubules: Are They Plausible Philosophical Transactions of the Royal Society of London. Series A, 356, 1897-1926.
[12] Tuszynski, J.A., et al. (1995) Ferroelectric Behavior in Microtubule Dipole Lattices: Implications for Information Processing, Signaling and Assembly/Disassembly. Journal of Theoretical Biology, 174, 371-380.
[13] Tuszynski, J.A., et al. (2002) Electrostatic Properties of Tubulin and Microtubules. In: Crowley, J.M., Zaretsky, M. and Kazkaz, G., Eds., Proceedings of the Electrostatics Society of America and Institute of Electrostatics—Japan, Laplacian Press, Morgan Hill, 41-50.
[14] Halliday, D. and Resnick, R. (1962) Physics. Vol. 2, John Wiley & Sons, New York, 698.
[15] Wilson-Kubalek, E.M., et al. (2008) Orientation and Structure of the Ndc80 Complex on the Microtubule Lattice. The Journal of Cell Biology, 182, 1055-1061.
[16] Schelkunoff, S.A. (1963) Electromagnetic Fields. Blaisdell, New York, 29.
[17] Joglekar, A.P., et al. (2006) Molecular Architecture of a Kinetochore-Microtubule Attachment Site. Nature Cell Biology, 8, 581-598.
[18] Grishchuk, E.L., et al. (2005) Force Production by Disassembling Microtubules. Nature, 438, 384-388.
[19] Stracke, R., et al. (2002) Analysis of the Migration Behavior of Single Microtubules in Electric Fields. Biochemical and Biophysical Research Communications, 293, 602-609.
[20] Benedek, G.B. and Villars, F.M.H. (2000) Physics: With Illustrative Examples from Medicine and Biology: Electricity and Magnetism. Springer-Verlag, New York, 403.
[21] Jordan-Lloyd, D. and Shore, A. (1938) The Chemistry of Proteins. JA Churchill Publishing Company, London.
[22] Pauling, L. (1945) The Adsorption of Water by Proteins. Journal of the American Chemical Society, 67, 555-557.
[23] Toney, M.F., et al. (1994) Voltage-Dependent Ordering of Water Molecules at an Electrode-Electrolyte Interface. Nature, 368, 444-446.
[24] Bockris, J.O. and Reddy, A.K.N. (1977) Modern Electrochemistry. Plenum, New York.
[25] Teschke, O., Ceotto, G. and De Souza, E.F. (2001) Interfacial Water Dielectric Permittivity-Profile Measurements Using Atomic Force Spectroscopy. Physical Review E, 64, Article ID: 011605.
[26] Gagliardi, L.J. (2009) Electrostatic Considerations in Mitosis. iUniverse Publishers, Bloomington, IN.
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Home : Support : Online Help : Mathematics : Calculus : Transforms : inttrans : Examples : addtable
addtable (inttrans Package)
The integral transforms package contains functionality that allows great flexibility in the patterns that can be entered into the tables, as well as the ability to match these patterns quickly and efficiently.
\mathrm{restart}
\mathrm{with}\left(\mathrm{inttrans}\right):
\mathrm{addtable}\left(\mathrm{fourier},f\left(t\right),\mathrm{Ff}\left(s\right),t,s\right)
\mathrm{fourier}\left(f\left(x\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Ff}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{addtable}\left(\mathrm{fouriercos},f\left(t\right),\mathrm{Fc}\left(s\right),t,s\right)
\mathrm{fouriercos}\left(f\left(x\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Fc}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{addtable}\left(\mathrm{fouriersin},f\left(t\right),\mathrm{Fs}\left(s\right),t,s\right)
\mathrm{fouriersin}\left(f\left(x\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Fs}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{addtable}\left(\mathrm{hankel},f\left(t\right),\mathrm{Fha}\left(s,\mathrm{ν}\right),t,s,\mathrm{hankel}=\mathrm{ν}::\left(\mathrm{Range}\left(-\mathrm{∞},\mathrm{∞}\right)\right)\right)
\mathrm{hankel}\left(f\left(x\right),x,y,\mathrm{μ}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Fha}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{μ}}\right)
\mathrm{addtable}\left(\mathrm{hilbert},f\left(t\right),\mathrm{Fhi}\left(s\right),t,s\right)
\mathrm{hilbert}\left(f\left(x\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Fhi}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{addtable}\left(\mathrm{invlaplace},f\left(t\right),\mathrm{Fil}\left(s\right),t,s\right)
\mathrm{invlaplace}\left(f\left(x\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Fil}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{addtable}\left(\mathrm{invmellin},f\left(t\right),\mathrm{Fim}\left(s\right),t,s,\mathrm{invmellin}=-\mathrm{∞}..\mathrm{∞}\right)
\mathrm{invmellin}\left(f\left(x\right),x,y,-\mathrm{∞}..\mathrm{∞}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Fim}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{addtable}\left(\mathrm{laplace},f\left(t\right),\mathrm{Fl}\left(s\right),t,s\right)
\mathrm{laplace}\left(f\left(x\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Fl}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{addtable}\left(\mathrm{mellin},f\left(t\right),\mathrm{Fm}\left(s\right),t,s\right)
\mathrm{mellin}\left(f\left(x\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Fm}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
Functions with Several Parameters
This functionality distinguishes between user-defined symbolic constants and variable parameters.
In the following case,
a
is a variable parameter:
\mathrm{addtable}\left(\mathrm{fourier},g\left(ta\right),\mathrm{Gf}\left(s-a\right),t,s,\left\{a\right\}\right)
\mathrm{fourier}\left(g\left(x\right),x,y\right);
\textcolor[rgb]{0,0,1}{\mathrm{Gf}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
Whereas, in the following situation, alpha is a user-defined symbolic constant, and it must be matched exactly by "alpha" for the pattern to be successfully returned. The first call to fouriercos will not return an answer for this reason--because 4 is not the same as alpha--while the second will.
\mathrm{addtable}\left(\mathrm{fouriercos},g\left(at,\mathrm{α}\right),\mathrm{Gc}\left(s-\mathrm{α}\right)+a,t,s,\left\{a\right\}\right)
\mathrm{fouriercos}\left(g\left(3x,4\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{fouriercos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{fouriercos}\left(g\left(3x,\mathrm{α}\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Gc}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{α}}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}
The rest of these examples use variable parameters as opposed to symbolic constants:
\mathrm{addtable}\left(\mathrm{fouriersin},g\left(at+b\right),\mathrm{Gs}\left(as-b\right),t,s,\left\{a,b\right\}\right)
\mathrm{fouriersin}\left(g\left(\mathrm{α}x+\mathrm{β}\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Gs}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{α}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{β}}\right)
\mathrm{addtable}\left(\mathrm{hankel},g\left(ta\right),\mathrm{Gha}\left(s,\mathrm{ν}-a\right),t,s,\left\{a\right\},\mathrm{hankel}=\mathrm{ν}::\left(\mathrm{Range}\left(-\mathrm{∞},\mathrm{∞}\right)\right)\right)
\mathrm{hankel}\left(g\left(x\right),x,y,\mathrm{μ}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Gha}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{μ}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{addtable}\left(\mathrm{hilbert},g\left(t+b\right),\mathrm{Ghi}\left(s+\mathrm{sin}\left(b\right)\right),t,s,\left\{b\right\}\right)
\mathrm{hilbert}\left(g\left(x+b\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Ghi}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{b}\right)\right)
\mathrm{addtable}\left(\mathrm{invlaplace},g\left(t+a\right),\frac{\mathrm{Gil}\left(s\right)}{1-a},t,s,\left\{a\right\}\right)
\mathrm{invlaplace}\left(g\left(x+3\right),x,y\right)
\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Gil}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{addtable}\left(\mathrm{invmellin},g\left(ta\right),\mathrm{Gim}\left(s-a\right),t,s,\left\{a\right\},\mathrm{invmellin}=-\mathrm{∞}..\mathrm{∞}\right)
\mathrm{invmellin}\left(g\left(x\right),x,y,-\mathrm{∞}..\mathrm{∞}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Gim}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{addtable}\left(\mathrm{laplace},g\left(at\right),\mathrm{Gl}\left(\frac{a}{s+a}\right),t,s,\left\{a\right\}\right)
\mathrm{laplace}\left(g\left(x\right),x,y\right)
\textcolor[rgb]{0,0,1}{\mathrm{Gl}}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}}\right)
\mathrm{addtable}\left(\mathrm{mellin},g\left(tb\right),{\mathrm{Gm}\left(s\right)}^{b},t,s,\left\{b\right\}\right)
\mathrm{mellin}\left(g\left(6x\right),x,y\right)
{\textcolor[rgb]{0,0,1}{\mathrm{Gm}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)}^{\textcolor[rgb]{0,0,1}{6}}
You can specify conditions on the parameters to be matched:
\mathrm{addtable}\left(\mathrm{hankel},h\left(ta\right),\mathrm{Hha}\left(s,\mathrm{ν}-a\right),t,s,\left\{a\right\},a::\left(\mathrm{Range}\left(-3,5\right)\right),\mathrm{hankel}=\mathrm{ν}::\left(\mathrm{Range}\left(-\mathrm{∞},\mathrm{∞}\right)\right)\right)
\mathrm{hankel}\left(h\left(x\right),x,y,\mathrm{μ}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Hha}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{μ}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{addtable}\left(\mathrm{invlaplace},h\left(t+a\right),\frac{\mathrm{Hil2}\left(s\right)}{1-a},t,s,\left\{a\right\},a::\mathrm{integer}\right)
\mathrm{addtable}\left(\mathrm{invlaplace},h\left(t+a\right),\frac{\mathrm{Hil1}\left(s\right)}{1-a},t,s,\left\{a\right\},a::\left(\mathrm{Not}\left(\mathrm{integer}\right)\right)\right)
\mathrm{invlaplace}\left(h\left(x+\frac{3}{2}\right),x,y\right)
\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Hil1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{invlaplace}\left(h\left(x+3\right),x,y\right)
\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Hil2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{addtable}\left(\mathrm{mellin},h\left(t,b\right),{\mathrm{Hm1}\left(s\right)}^{b},t,s,\left\{b\right\},\mathrm{_testeq}\left(\mathrm{_signum}\left(b\right)+1\right)\right)
\mathrm{addtable}\left(\mathrm{mellin},h\left(t,b\right),{\mathrm{Hm2}\left(s\right)}^{b},t,s,\left\{b\right\},\mathrm{_testeq}\left(\mathrm{_signum}\left(b\right)-1\right)\right)
\mathrm{mellin}\left(h\left(x,6\right),x,y\right)
{\textcolor[rgb]{0,0,1}{\mathrm{Hm2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)}^{\textcolor[rgb]{0,0,1}{6}}
\mathrm{mellin}\left(h\left(x,-6\right),x,y\right)
\frac{\textcolor[rgb]{0,0,1}{1}}{{\textcolor[rgb]{0,0,1}{\mathrm{Hm1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\right)}^{\textcolor[rgb]{0,0,1}{6}}}
invmellin and hankel
With invmellin and hankel, arguments are added to the addtable function, as arguments are added to the transforms themselves. These arguments can be varied, introducing greater flexibility in the information that can be added to user-defined tables.
\mathrm{addtable}\left(\mathrm{hankel},i\left(ta\right),\mathrm{Iha2}\left(s,\mathrm{ν}-a\right),t,s,\left\{a\right\},a::\left(\mathrm{Range}\left(-3,5\right)\right),\mathrm{hankel}=\mathrm{ν}::\left(\mathrm{Range}\left(0,\mathrm{∞}\right)\right)\right)
\mathrm{addtable}\left(\mathrm{hankel},i\left(ta\right),\mathrm{Iha1}\left(s,\mathrm{ν}-a\right),t,s,\left\{a\right\},a::\left(\mathrm{Range}\left(-3,5\right)\right),\mathrm{hankel}=\mathrm{ν}::\left(\mathrm{Range}\left(-\mathrm{∞},0\right)\right)\right)
\mathrm{assume}\left(0<\mathrm{μ}\right)
\mathrm{hankel}\left(i\left(x\right),x,y,\mathrm{μ}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Iha2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{μ~}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{assume}\left(\mathrm{μ}<0\right)
\mathrm{hankel}\left(i\left(x\right),x,y,\mathrm{μ}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Iha1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{μ~}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{addtable}\left(\mathrm{invmellin},i\left(t,a\right),\mathrm{Iim1}\left(s-a\right),t,s,\left\{a\right\},\mathrm{_testeq}\left(\mathrm{_signum}\left(a\right)+1\right),\mathrm{invmellin}=0..\mathrm{∞}\right)
\mathrm{addtable}\left(\mathrm{invmellin},i\left(t,a\right),\mathrm{Iim2}\left(s-a\right),t,s,\left\{a\right\},\mathrm{_testeq}\left(\mathrm{_signum}\left(a\right)+1\right),\mathrm{invmellin}=-\mathrm{∞}..0\right)
\mathrm{addtable}\left(\mathrm{invmellin},i\left(t,a\right),\mathrm{Iim3}\left(s-a\right),t,s,\left\{a\right\},\mathrm{_testeq}\left(\mathrm{_signum}\left(a\right)-1\right),\mathrm{invmellin}=0..\mathrm{∞}\right)
\mathrm{addtable}\left(\mathrm{invmellin},i\left(t,a\right),\mathrm{Iim4}\left(s-a\right),t,s,\left\{a\right\},\mathrm{_testeq}\left(\mathrm{_signum}\left(a\right)-1\right),\mathrm{invmellin}=-\mathrm{∞}..0\right)
\mathrm{invmellin}\left(i\left(x,7\right),x,y,3..7\right)
\textcolor[rgb]{0,0,1}{\mathrm{Iim3}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{7}\right)
\mathrm{invmellin}\left(i\left(x,-7\right),x,y,3..7\right)
\textcolor[rgb]{0,0,1}{\mathrm{Iim1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{7}\right)
\mathrm{invmellin}\left(i\left(x,7\right),x,y,-7..-3\right)
\textcolor[rgb]{0,0,1}{\mathrm{Iim4}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{7}\right)
\mathrm{invmellin}\left(i\left(x,-7\right),x,y,-\mathrm{∞}..0\right)
\textcolor[rgb]{0,0,1}{\mathrm{Iim2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{7}\right)
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Unit Circle | Brilliant Math & Science Wiki
Akshay Yadav, Andy Hayes, Aareyan Manzoor, and
Contribute to this wiki by adding relevant examples! For more information, see here.
The unit circle is a circle of radius 1 unit that is centered on the origin of the coordinate plane.
The unit circle is fundamentally related to concepts in trigonometry. The trigonometric functions can be defined in terms of the unit circle, and in doing so, the domain of these functions is extended to all real numbers.
The unit circle is also related to complex numbers. A unit circle can be graphed in the complex plane, and all roots of unity will lie on this circle.
Relation to Right Triangles
Coordinates in the unit circle
Every point on the unit circle corresponds to a right triangle with vertices at the origin and the point on the unit circle. The right triangle has leg lengths that are equal to the absolute values of the
x
y
coordinates, respectively.
This right triangle is used to apply trigonometric relations.
\begin{array}{rll} \sin (\theta) & = \frac{\text{opposite}}{\text{hypotenuse}} & = \frac{b}{c} \\ \\ \cos (\theta) & = \frac{\text{adjacent}}{\text{hypotenuse}} & = \frac{a}{c} \\ \\ \tan (\theta) & = \frac{\text{opposite}}{\text{adjacent}} & = \frac{b}{a} \\ \\ \end{array}
Since the hypotenuse of the right triangle is always 1 unit long, the values of the
x
y
coordinates of a point on the circle are always equal to the cosine and sine (respectively) of the angle
\theta.
This angle is measured in a unit called radians, which corresponds to the distance around the unit circle from the point
(1,0).
The circumference of the unit circle is
2\pi,
2\pi
radians is the same as 360
^\circ.
Any other angle less than 360
^\circ
can be represented as some fraction of
2\pi
radians. For example, A 90
^\circ
angle is the same as
\frac{1}{4}
of the way around the circle, which would be
\frac{2\pi}{4}=\frac{\pi}{2}.
Some possible values of
\theta
are listed below, along with their corresponding values of sine and cosine.
\begin{array} { | c | c | c | } \hline \text{angle measure, } \theta & \sin \theta & \cos \theta\\ \hline 0 & 0 & 1 \\ \hline \dfrac{\pi}{6} & \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\ \hline \dfrac{\pi}{4} & \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2} \\ \hline \dfrac{\pi}{3} & \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ \hline \dfrac{\pi}{2} & 1 & 0 \\ \hline \end{array}
The trigonometry we are familiar with so far is based on only right triangles and acute angles. However, with help of Unit Circle we can extend our understanding of trigonometric functions plus also become familiar with the use of non-acute angles.
More information about the circular system of angle measurement can be found on its wiki page.
An angle on Unit Circle is always measured from the positive
x
-axis, with its vertex at the origin. It is measured to a point on the unit circle. The ray that begins at the origin and contains the point on the unit circle is called the terminal side.
An angle is said to be positive if it is measured by going in anticlockwise direction from the positive
x
-axis and negative if it is measured by going in clockwise direction from the
x
2\pi \text{rad}=360^\circ,
any degree measurement can be converted to radians, and vice versa.
d
be an angle's measurement in degrees, and let
r
be that same angle's measurement in radians.
r = \frac{\pi d}{180}
d = \frac{180r}{\pi}
AOB
A
lies on the Cartesian plane such that
\overline{OA}
x
-axis , point
O
lies on the origin and point
B
lies anywhere on the Unit Circle. Note that
OB=1
The sine and cosine trigonometric functions are given below. When defining these functions in terms of the unit circle, it is possible to have negative lengths. If
\overline{OA}
is along the negative
x
OA
is considered to be negative. Likewise, if
\overline{AB}
extends below the
x
AB
is considered to be negative.
\begin{aligned} \sin (\theta) &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{OB} = \frac{AB}{1} = AB\\ \cos (\theta) &= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{OA}{OB} = \frac{OA}{1} = OA. \end{aligned}
By this convention, the sine of an angle is considered to be the
y
-coordinate of a point on the unit circle given by that angle. Likeways, the cosine of an angle is considered to be the
x
-coordinate of a point on the unit circle given by that angle. In general, to compute the sine or cosine of any angle
\theta,
look at the coordinates of the point on unit circle made by that angle.
Main Article: Special Angles on Unit Circle
The special angles are angles on the unit circle for which the coordinates are well-known. These coordinates can be solved for with right-triangle relationships.
\text{The sixteen special angles (measured in radians) on the unit circle, each labeled at the terminal point.}
Given that a line passes through the unit circle with the angle
\theta=\dfrac{\pi}{4}
find the x,y. Refer to the given diagram.
\begin{cases} \cos(\theta)= x \\ \sin(\theta)=y\end{cases}
. put the value of theta to find
\begin{cases} x=\dfrac{\sqrt{2}}{2}\\y=\dfrac{\sqrt{2}}{2}\end{cases}
We are done
_\square
A line passes through the unit circle at the point at
x=\dfrac{1}{2}
\tan^2(\theta)
first, we know that
x^2=\cos^2(\theta)=\dfrac{1}{4}
. we also know
y^2=\sin^2(\theta)=1-cos^2(\theta)=\dfrac{3}{4}
. using another trig identity we have
\tan^2(\theta)=\dfrac{\sin^2(\theta)}{\cos^2(\theta)}=\boxed{3}
_\square
R=0
R≤0
R<0
R>0
R = \sin 130^\circ+\cos 130^\circ
, then which of the following is true?
Image Credit: Wikimedia Geek3.
Cite as: Unit Circle. Brilliant.org. Retrieved from https://brilliant.org/wiki/unit-circle-basic-concept-for-higher-trigonometry/
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How would you use permutations to find the arrangements possible
Line of 4 boys:
4×3×2×1=4!=24
Line of 4 girls: also
4!=24
Combination of the two:
4!×4!={24}^{2}=576
posibilities.
If it doesn't matter whether the first is a boy or a girl, we'll have to double that to
2×576=1152
A card is drawn randomly from a deck of cards.
a. n(S) =____________
b. What is the probability that:
1. It is from the heart suit?
2. It is a queen?
3. The number on the card is greater than 2 but less than 9?
Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood.
f two dice are rolled once and one die rolled a second time if the first two rolls have same number, what is the probability of getting a sum of 14?
A normal distribution has a mean of 32 and a standard deviation of 4. Find the probability that a randomly selected xx -value from the distribution is at most 35. Round to four decimal places.
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Let f be the function from R to
DAWIT BERIHUN 2022-04-02
We want to prove that
\left({Z}_{7},\oplus \right)
is group. I have difficulty proving associativity axiom. The solution reads
a\in {\mathbb{Z}}_{7},b\in {\mathbb{Z}}_{7}
c\in {\mathbb{Z}}_{7}
. By Theorem 3.4.10 we only need to show
\left(a+\left(b+c\right)\right)b\text{mod}7=\left(\left(a+b\right)+c\right)b\text{mod}7.
This holds since
a+\left(b+c\right)=\left(a+b\right)+c
for all integers a, b, and c by the associative property of the integers. Hence
\oplus
is associative.
Therorem 3.4.10. Let a and b be integers, and let m be a natural number. Then
\left(a+b\right)modm=\left(\left(amodm\right)+\left(bmodm\right)\right)modm
y\le 4
individual plays a game of tossing a coin where he wins Rs 2 if head turns up and nothing if tail turns up.On the basis of the given information, find (i) The expected value of the game. (4) (ii) The risk premium this person will be willing to pay to avoid the risk associated with the game.
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Quotient rule - Wikipedia
Formula for the derivative of a ratio of functions
In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions.[1][2][3] Let
{\displaystyle f(x)=g(x)/h(x),}
where both g and h are differentiable and
{\displaystyle h(x)\neq 0.}
The quotient rule states that the derivative of f(x) is
{\displaystyle f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.}
2.1 Proof from derivative definition and limit properties
2.2 Proof using implicit differentiation
2.3 Proof using the chain rule
3 Higher order formulas
{\displaystyle {\begin{aligned}{\frac {d}{dx}}{\frac {e^{x}}{x^{2}}}&={\frac {\left({\frac {d}{dx}}e^{x}\right)(x^{2})-(e^{x})\left({\frac {d}{dx}}x^{2}\right)}{(x^{2})^{2}}}\\&={\frac {(e^{x})(x^{2})-(e^{x})(2x)}{x^{4}}}\\&={\frac {e^{x}(x-2)}{x^{3}}}.\end{aligned}}}
The quotient rule can be used to find the derivative of
{\displaystyle f(x)=\tan x={\tfrac {\sin x}{\cos x}}}
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\tan x&={\frac {d}{dx}}{\frac {\sin x}{\cos x}}\\&={\frac {\left({\frac {d}{dx}}\sin x\right)(\cos x)-(\sin x)\left({\frac {d}{dx}}\cos x\right)}{\cos ^{2}x}}\\&={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}=\sec ^{2}x.\end{aligned}}}
Proof from derivative definition and limit properties[edit]
{\displaystyle f(x)={\frac {g(x)}{h(x)}}.}
Applying the definition of the derivative and properties of limits gives the following proof.
{\displaystyle {\begin{aligned}f'(x)&=\lim _{k\to 0}{\frac {f(x+k)-f(x)}{k}}\\&=\lim _{k\to 0}{\frac {{\frac {g(x+k)}{h(x+k)}}-{\frac {g(x)}{h(x)}}}{k}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k\cdot h(x)h(x+k)}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k}}\cdot \lim _{k\to 0}{\frac {1}{h(x)h(x+k)}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)+g(x)h(x)-g(x)h(x+k)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)}{k}}-\lim _{k\to 0}{\frac {g(x)h(x+k)-g(x)h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(h(x)\lim _{k\to 0}{\frac {g(x+k)-g(x)}{k}}-g(x)\lim _{k\to 0}{\frac {h(x+k)-h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}
Proof using implicit differentiation[edit]
{\displaystyle f(x)={\frac {g(x)}{h(x)}},}
{\displaystyle g(x)=f(x)h(x).}
The product rule then gives
{\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x).}
{\displaystyle f'(x)}
and substituting back for
{\displaystyle f(x)}
{\displaystyle {\begin{aligned}f'(x)&={\frac {g'(x)-f(x)h'(x)}{h(x)}}\\&={\frac {g'(x)-{\frac {g(x)}{h(x)}}\cdot h'(x)}{h(x)}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}
Proof using the chain rule[edit]
{\displaystyle f(x)={\frac {g(x)}{h(x)}}=g(x)h(x)^{-1}.}
Then the product rule gives
{\displaystyle f'(x)=g'(x)h(x)^{-1}+g(x)\cdot {\frac {d}{dx}}(h(x)^{-1}).}
To evaluate the derivative in the second term, apply the power rule along with the chain rule:
{\displaystyle f'(x)=g'(x)h(x)^{-1}+g(x)\cdot (-1)h(x)^{-2}h'(x).}
Finally, rewrite as fractions and combine terms to get
{\displaystyle {\begin{aligned}f'(x)&={\frac {g'(x)}{h(x)}}-{\frac {g(x)h'(x)}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}
Higher order formulas[edit]
Implicit differentiation can be used to compute the nth derivative of a quotient (partially in terms of its first n − 1 derivatives). For example, differentiating
{\displaystyle fh=g}
twice (resulting in
{\displaystyle f''h+2f'h'+fh''=g''}
) and then solving for
{\displaystyle f''}
{\displaystyle f''=\left({\frac {g}{h}}\right)''={\frac {g''-2f'h'-fh''}{h}}.}
Retrieved from "https://en.wikipedia.org/w/index.php?title=Quotient_rule&oldid=1077983992"
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Revision as of 16:19, 29 July 2013 by NikosA (talk | contribs) (→Importing data: creating another Mapset, copying over material and optionally renaming)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
{\displaystyle [0,255]}
{\displaystyle [0,2047]}
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Proved that question: S(n) = \sum_{k \geq 1} \frac{n!}{k (n-k)! n^k}
Proved that question:
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)
Cloplerotly1gu
Here is a rough argument. Filling in the details should be elementary, but challenging. I find myself reusing several ideas from this earlier question.
\sum _{k=1}^{n}\frac{1}{k}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k}{n}\right)
As discussed in the earlier answer,
\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k}{n}\right)\approx {e}^{-\frac{{k}^{2}}{2n}}
So the sum is roughly
\sum _{k\ge 1}\frac{1}{k}{e}^{-\frac{{k}^{2}}{2n}}=\sum _{k\ge 1}\frac{1}{\sqrt{n}}\frac{\sqrt{n}}{k}{e}^{-\frac{{\left(\frac{k}{\sqrt{n}}\right)}^{2}}{2}}
This is a Riemann sum approximation (with interval size
\frac{1}{\sqrt{n}}
) to the integral
\left[x<1\right]
be 1 for
x<1
x\ge 1
. Then the integral is
{\int }_{\frac{1}{\sqrt{n}}}^{1}\frac{dx}{x}+{\int }_{\frac{1}{\sqrt{n}}}^{\mathrm{\infty }}\frac{{e}^{-\frac{{x}^{2}}{2}}-\left[x<1\right]}{x}dx
The first integral is
\left(\frac{1}{2}\right)\mathrm{log}n
. The second approaches
{\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-\frac{{x}^{2}}{2}}-\left[x<1\right]}{x}dx
, which is convergent.
So our final integral is
\left(\frac{1}{2}\right)\mathrm{log}n+O\left(1\right)
\frac{{\pi }^{3}}{32}=1-\sum _{k=1}^{\mathrm{\infty }}\frac{2k\left(2k+1\right)\zeta \left(2k+2\right)}{{4}^{2k+2}}
How to find the Maclaurin series for
{e}^{x}
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\sqrt[n]{n}}
Does the following series absolutely converge, conditionally converge or diverge?
\sum _{n=1}^{+\mathrm{\infty }}\mathrm{sin}\left({n}^{2}\right)\mathrm{sin}\left(\frac{1}{{n}^{2}}\right)
Find the region of convergence and absolute convergence of the series.
\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{ln}}^{n}x}{n}
\prod _{n=1}^{\mathrm{\infty }}\left(1+\frac{1}{{\pi }^{2}{n}^{2}}\right)
Fix a positive number
\alpha
and consider the series
\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\left(k+1\right){\left[\mathrm{ln}\left(k+1\right)\right]}^{\alpha }}
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tobeint39r 2022-04-30 Answered
M\in End\left(V\right)
M\otimes M
End\left(V\right)\otimes End\left(V\right)
This looks like it can get complicated. Generically, at least over an algebraically closed field, a matrix M will have distinct eigenvalues
{m}_{1},\dots ,{m}_{n}
, and generically
M\otimes M
will have distinct eigenvalues
{m}_{1}^{2},\dots ,{m}_{n}^{2}
with multiplicity one, and
{m}_{1}{m}_{2},{m}_{1}{m}_{3},\dots ,{m}_{n-1}{m}_{n}
with multiplicity two. Thus the centralizer will have dimension
n+4\left(\genfrac{}{}{0}{}{n}{2}\right)=2{n}^{2}-n
But there are many degenerate cases: for instance if M has eigenvalues 1
1,a,\dots ,{a}^{n-1}
M\otimes M
will have eigenvalues
1,\text{ }a,\dots ,{a}^{2n-2}
1,2,\dots n-1,n,n-1,\dots ,1
. Things can get more complicated still.
Then M might have non-trivial Jordan blocks, and then the real fun starts!
card\left(G\right)={p}^{2}q
p<q
{s}_{q}
{s}_{q}\in \left\{1, {p}^{2}\right\}
\underset{S\in Sy{l}_{q}\left(G\right)}{\cup }S\setminus \left\{1\right\}={\stackrel{˙}{\cup }}_{S\in Sy{l}_{q}\left(G\right)}S\setminus \left\{1\right\}
S,\text{ }T\in Sy{l}_{q}\left(G\right)
S\ne T
\mathrm{S}\setminus \left\{1\right\}\cap \mathrm{T}\setminus \left\{1\right\}=\varnothing
\frac{K}{F}
is Galois, and the coefficients of the monic
f\in K\left[x\right]
generates K, then
\prod _{\sigma \in Gal\left(\frac{K}{F}\right)}\sigma f\in F\left[x\right]
How does one see that
g\in F\left[x\right]
Write formulas for the isometries in terms of a complex variable z = x + iy.
n\ge 8
big-\Theta
\Theta \left(n\right)\Theta \left(n8\right)\Theta \left(8\right)\Theta \left(8n\right)\Theta \left(8n\right)
Is there a particular characterization of
Aut\left({\mathbb{Q}}^{3}\right)
what is abelin group?
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In many cases, MATSim uses a terminology that is different from the mainstream terminology. In most cases, the reason is that the concepts are only similar, but not identical, and we wanted to avoid the confusion of using the same term for aspects that are similar but not identical. The following attempts some commented approximate ”translations” from more standard teminology to MATSim terminology.
Choice set → “plan set” of an agent
During MATSim iterations, agent accumulate plans. This can be interpreted as building a choice set over time. A problem is that the process that generates the choice set at this point is not systematic.
Possible future developments: Once it has been made explicit that ”plans generation” means ”choice set generation”, the terminology may be made standard.
Choice set generation → Time mutation/re-route/... ; ”innovation”
As said above, the set of MATSim plans can be seen as this agent’s choice set. MATSim generates new plans ”on-the-fly”, i.e. while the simulation is running. We sometimes call this ”innovation”, since agents create new plans (= add entries to the choice set), rather than choosing between existing plans.
Choice set generation, choice → replanning
In MATSim, there is no strict separation between ”choice set generation” and ”choice”: at the replanning step, for each agent, a replanning strategy is randomly choosen. This strategy may consist in selecting a random plan to use to generate a new plan by mutation (”choice set generation” part), or just to select a past plan based on the experienced score (”choice” part).
Convergence → learning rate
Scores in matsim are computed as
scorenew = (1 − α) · scoreold + α · scoresim ,
where scoresim is the score that is obtained from the execution of the plans (= network loading).
μ (logit model scaling factor) → beta brain
MATSim scoring function:
\mathrm{BrainExpBeta}\cdot \sum _{i}{\beta }_{i}\phantom{\rule{0.167em}{0ex}}{x}_{i}
Typical logit model formulation:
\mu \cdot \sum _{i}{\beta }_{i}\phantom{\rule{0.167em}{0ex}}{x}_{i}
As is well known, μ or βi are not independently identifiable from estimation. For simulation, they are hence somewhat arbitrary. The default value for ”BrainExpBeta” is 2 for historical reasons, but it should be set to 1 if the parameters of the scoring function are estimated rather than hand-calibrated.
Possible future development: Default value of BrainExpBeta should be set to 1 instead of 2.
Multinomial logit → ExpBetaPlanSelector
The main problem is that one needs to keep in mind how the choice set is constructed (see above).
In most simulations, we use ExpBetaPlanChanger instead, which is a Metropolis Monte Carlo variant of making multinomial logit draws
Possible future developments: None of this is ideal, since, after the introduction of a policy, it is not clear which behavioral switches are due to the policy, and which are due to sampling. In theory, one should have unbiased samples before and after the introduction of the policy, but at this point this is not implemented and it is also computationally considerably more expensive than what is done now.
Network loading → mobsim, mobility simulation, physical simulation
The standard terminology has the ”network loading” on the ”supply side”. In my (KN’s) view, the ”simulation of the physical system” is not the supply side, but what in economics is called ”technology”. This can for example be seen in the fact that ”lane changing” is part of the mobsim, but this is, in my view, not a ”supply side” aspect.
Possible future developments: May switch to ”network loading” if there is agreement that this is a better name.
Stationary → relaxed
“stationary” means that the probability distribution does not shift any more. However, as long as ”innovation” is still switched in on MATSim (new routes, new times, ...), the result is not truly stationary. Thus we avoid the word. If innovation is switched off, the result is indeed a statinary process, but limited to the set of plans that every agent has at that point in time.
Possible future developments: not clear. Minimally, publications should be precise.
<module name="strategy" >
<!-- iteration after which module will be disabled. ... -->
<param name="ModuleDisableAfterIteration_1" value="null" />
<param name="ModuleDisableAfterIteration_2" value="950" />
<!-- probability that a strategy is applied to a given person. ... -->
<param name="ModuleProbability_1" value="0.9" />
<!-- name of strategy ... -->
<param name="Module_1" value="ChangeExpBeta" />
<param name="Module_2" value="ReRoute" />
<!-- maximum number of plans per agent ... -->
<param name="maxAgentPlanMemorySize" value="4" />
StrategyModule ”ReRoute” (= innovative Module, produces plans with new routes) is switched off after iteration 950.
StrategyModule ”ChangeExpBeta” (= non-innovative Module, switches between existing plans) is never switched off.
If an agent ever ends up with more than 4 plans, plans are deleted until she is back to 4 plans. (Deletion goes via a ”PlanSelectorForRemoval”, which affects the choice set, and thus more thought needs to go into this. Currently, the plan with the worst score is removed.)
Utility ↔ score
Configuration: At least when using random utility models (such as multinomial logit aka ExpBeta...), the score has the same function as the deterministic utility.
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Fenwick Tree | Brilliant Math & Science Wiki
Agnishom Chattopadhyay, Pranjal Jain, Debarghya Adhikari, and
A Fenwick tree or binary indexed tree is a data structure that helps compute prefix sums efficiently. Computing prefix sums are often important in various other algorithms, not to mention several competitive programming problems. For example, they are used to implement the arithmetic coding algorithm. Fenwick trees were invented by Peter M. Fenwick in 1994.
A Bit Manipulation Trick
n^\text{th}
prefix sum of an array is the sum of the first
n
elements of the array.
This idea is also referred to as partial sums or cumulative sums.
What is the prefix sum of positive integers?
They are triangular numbers as shown in the diagram below.
The simplest way to compute the
i^\text{th}
prefix sum is to just scan over the first
i
elements of the array whilst adding them.
def prefix(arr,i):
This computes the prefix table in
O(n)
time. However, this is a little too much when the table has to be frequently updated, since the prefix tables have to be computed over and over again.
We can do better using a Fenwick tree.
Let's say that we have a 16 element array which forms the leaves of the following binary tree:
The leaves are labelled from 0000 to 1111.
Let us imagine that the parent nodes contain the sum of the children nodes everywhere throughout the tree. That'd mean that the root node has the sum of all the elements in the array.
Key Idea: The interesting observation is that to produce the sum of the first
i
elements we only need to know (and check) just
O(\log i)
We only need to preserve the red nodes. The yellow nodes are only around for the demonstration. In fact, the red nodes are what the Fenwick tree consists of.
How would you find the
1010^\text{th}
prefix sum?
We use the red nodes labelled 1010, 100 and 0.
Now, let us look into how this can help when the values in the leaves change. When the value in a leaf is incremented, we just need to notify all the red nodes from the root to the leaf. Since the height of the tree is in
O(\log n)
, we just need to perform
O(\log n)
This section can be hard unless you have the intuition.
For the sake of simplicity, we will assume that we are working with 1-based arrays.
Here is what the Fenwick tree on the array [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16] would look like when embedded in an 1-based array:
We have already seen the motivation behind the data structure. Now, we need to see how it can be built, updated and queried.
To increment the
i^\text{th}
position by
x
1.
i_1 = i = 2^{j_1} + 2^{j_2} + \cdots + 2^{j_n}, \text{ where } j_1 < j_2 < \cdots < j_n.
\hspace{1cm} 1.1
Increment the
{i_1}^\text{th}
position of the tree by
x.
2.
i_2 = 2^{j_1 + 1} + 2^{j_2} + \cdots + 2^{j_n},
j
's are set as before.
3.
i_1
i_2
and loop from step
1
i_1
exceeds the length of the array/tree.
As discussed before, this runs in
O(\log N,)
N
is the length of the Fenwick tree.
The above can be a little confusing, but the essential idea here is that we're notifying the parent "responsible" nodes of the changes. If you are really baffled, sketching through the idea with pen and paper can really help.
To create a Fenwick tree for an array, simply create and fill an empty tree with zeroes and then update the tree with all the array positions one by one. Thus, for an array with
N
entries, it takes
O(N \log N)
operations to setup its Fenwick tree.
To query the
i^\text{th}
prefix sum:
1.
Set the accumulator to 0.
2.
i = 2^{j_1} + 2^{j_2} + \cdots + 2^{j_n}, \text{ where } j_1 < j_2 < \cdots < j_n.
\hspace{1cm} 2. 1.
Increment the accumulator by the value in the
\left(2^{j_1} + 2^{j_2} + \cdots + 2^{j_n}\right)^\text{th}
position of the tree.
\hspace{1cm} 2. 2.
\left(2^{j_2} + \cdots + 2^{j_n}\right)^\text{th}
position of the tree\hspace{1.3cm}.
\hspace{1.3cm}.
\hspace{1.3cm}.
\hspace{1cm}2. n.
Increment accumulator by the value in the
\left(2^{j_n}\right)^\text{th}
3.
Return the current value of the accumulator.
This is no surprise, just simply adding up whatever we already stored during the update. This runs in
O(\lg N)
The happy coincidence that computers internally use the binary system to represent integers brings us to a very interesting bit manipulation trick that makes it easy for us to implement the above algorithms.
Notice that we need to express integers as powers of two very frequently, and usually work with the lowest power in the above algorithms. Now, this lowest power is really the last set bit of the integer.
12 = 2^3 + 2^2 = (1100)_2
Here, the place value of the last set bit is
100.
Obviously, we could write a loop to find this but it turns out that there is a better (simpler) way to do this. It turns out that a & (-a) does a very good job of the same.
Let us work out the same thing with 12.
a =
0b1100, the complement of which is 0b0011.
-a =
0b0011 + 0b0001 or 0b0100.
>>> 12&-12
This should work for any integer. Here is a generalisation
The highest power of two which is not more than a is given by a & (-a).
_\square
a
s 1 0^n.
a
s' 0 1^n.
-a
a
s' 1 0^n
and
-a
1 0^n
s
s'
_\square
The implementations are straightforward, except that using the 1-based array can be easily mishandled.
class Fenwick():
def update(self, i, x): #add x to the ith position
while i <= self.x:
self.BIT[i-1] += x #because we're working with an 1-based array
i += i & (-i) #magic! don't touch!
def query(self, i): #find the ith prefix sum
s += self.BIT[i-1]
def __init__(self, l=[]): #initialize the fenwick tree
self.N = len(l)
self.BIT = [0 for i in xrange(self.N)]
for i in xrange(1,self.N+1):
self.update(i, l[i-1])
public class Fenwick{
int* MyTree;
public static int Fenwick(int a[],int N){
this.MyTree=new int[(this.N)+1];
MyTree[i]=0;
c=i+a[i-1];
int getCum(int i){
int ax=i;
while(ax>0){
sum+=MyTree[ax];
ix=ix+(ix&-ix);
void addValue(int i, int v){
while(ax<=N){
MyTree[ax]+=v;
ax+=(ax&-ax);
int get(int i){
return getCum(i)-getCum(i-1);
class Fenwick{ /*1-indexed*/
Fenwick(int a[],int N){ /*constructor to initialize Fenwick Tree*/
this->MyTree=new int[(this->N)+1];
addValue(i,a[i-1]);
int getCum(int i){ /*get i-th Prefix Sum*/
int ix=i;
while(ix>0){
sum+=MyTree[ix];
ix=ix-(ix&-ix);
void addValue(int i,int v){ /*add v to i-th value*/
while(ix<=N){
MyTree[ix]+=v;
return getCum(i)-getCum(i-1); /*get i-th value of array*/
Cite as: Fenwick Tree. Brilliant.org. Retrieved from https://brilliant.org/wiki/fenwick-tree/
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-7x=-a-4
Divide each term in -7x=-a-4 by -7 and simplify.
x=\frac{a}{7}+\frac{4}{7}
P\left(x\right)=-12{x}^{2}+2136x-41000
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
{y}^{\prime }-xy+y=2y
Write down the definition of the complex conjugate,
\stackrel{―}{z}\text{ }\text{ ,if }\text{ }z=x+iy
where z,y are real numbers. Hence prove that, for any complex numbers w and z,
\stackrel{―}{w\stackrel{―}{z}}=\stackrel{―}{w}z
Decision analysis. After careful testing and analysis, an oil company is considering drilling in two different sites. It is estimated that site A will net $20 million if successful (probability.2) and lose $3 million if not (probability .8); site B will net $80 million if successful (probability .1) and lose $7 million if not (probability.9). Which site should the company choose according to the expected return from each site?
1. What is the expected return for site A?
Solve the differential equation ..(b and a are variables)
\left({D}^{2}+4D+3\right)b=1+2a+3{a}^{2}
A supplement was given to patients to lower their blood pressure. The blood pressure before and after the supplement was recorded.
What kind of variables are Begin and End and what kind of scale are they following? Multiple choice. Choose one answer for scale and one answer for variables.
Note: the scale is identified as either nominal, ordinal, interval, or ratio (Choose the correct answer). Please explain why you chose the answer.
Variables are defined as either numerical (count/discrete or decimals) OR categorical (ordinal or nominal) (Choose the correct answer). Please explain why you chose the answer.
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1.1 Solution 1a
1.1.3 Step functions approximate L^1 functions well
1.2 Problem 1b
1.3 Solution 1b
Prove the following version of the Riemann-Lebesque Lemma: Let
{\displaystyle f\in L^{2}[-\pi ,\pi ]\!\,}
. Prove in detail that
{\displaystyle {\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)e^{-inx}dx\rightarrow 0\!\,}
{\displaystyle n\rightarrow \infty \!\,}
{\displaystyle n\!\,}
denotes a positive integer. You may use any of a variety of techniques, but you cannot simply cite another version of the Riemann-Lebesque Lemma.
{\displaystyle e^{-inx}=\cos(nx)-i\sin(nx)\!\,}
Hence we can equivalently show
{\displaystyle \int _{-\pi }^{\pi }f(x)\cos(nx)\rightarrow 0\!\,}
{\displaystyle n\rightarrow \infty \!\,}
{\displaystyle \psi (x)\!\,}
be a step function.
{\displaystyle \int _{-\pi }^{\pi }\psi (x)\cos(nx)\rightarrow 0\!\,}
{\displaystyle n\rightarrow \infty \!\,}
{\displaystyle {\begin{aligned}\int _{-\pi }^{\pi }\psi (x)\cos(nx)&=\sum _{i=1}^{m}c_{i}\int _{\xi _{i-1}}^{\xi _{i}}\cos(nx)dx\\&=\sum _{i=1}^{m}c_{i}{\frac {1}{n}}\underbrace {(\left.\sin(nx)\right|_{\xi _{i-1}}^{\xi _{i}})} _{<2}\\&\leq 2\max _{i}c_{i}{\frac {1}{n}}\\&\rightarrow 0{\mbox{ as }}n\rightarrow \infty \end{aligned}}\!\,}
Step functions approximate L^1 functions wellEdit
{\displaystyle f\in L^{2}[-\pi ,\pi ]\!\,}
{\displaystyle f\in L^{1}[\pi ,\pi ]\!\,}
Hence, given
{\displaystyle \epsilon >0\!\,}
{\displaystyle \psi (x)\!\,}
{\displaystyle \int _{-\pi }^{\pi }|f(x)-\psi (x)|dx<\epsilon \!\,}
{\displaystyle {\begin{aligned}\int _{-\pi }^{\pi }f(x)\cos(nx)&=\int _{-\pi }^{\pi }([f(x)-\psi (x)+\psi (x)]\cos(nx))dx\\&\leq \int _{-\pi }^{\pi }|f(x)-\psi (x)|\cos(nx)+\int _{-\pi }^{\pi }|\psi (x)|\cos(nx)\\&\leq \epsilon \cdot 2\pi +\epsilon \end{aligned}}}
{\displaystyle n_{k}\!\,}
be an increasing sequence of positive integers. Show that
{\displaystyle \{x|\lim \inf _{k\rightarrow \infty }\sin(n_{k}x)>0\}\!\,}
has measure 0.
Notes: You may take it as granted that the above set is measurable.
For the sake of brevity, let S be the set of such x. If S has positive measure, then it contains a subset of positive measure on which liminf[sin(nkx)] is bounded below by some positive constant; ie, the integral of liminf[sin(nkx)] over S will be positive. If S has zero measure, then the same integral will be zero. Thus, we must only compute an appropriate integral to show that m(S)=0.
Since we do not know that S has finite measure, take a sequence of functions fk(x)=2-|x|*sin(nkx), each of which is clearly integrable. By Fatou's lemma, the integral of liminf[fk(x)] over S <= the liminf of the integrals of fk(x) over S. However, each fk is the L1 function 2-|x|*sin(nkx). By the Riemann-Lebesgue Lemma, this goes to 0 as nk goes to infinity.
Hence our original integral of a strictly positive function over S is bounded above by 0, so m(S)=0.
{\displaystyle (x^{p}+{\frac {1}{x^{p}}})f\in L^{2}(0,\infty )}
{\displaystyle p>1/2}
{\displaystyle f\in L^{1}(0,\infty )}
{\displaystyle A=||(x^{p}+{\frac {1}{x^{p}}})f||_{L^{2}(0,\infty )}}
{\displaystyle {\begin{aligned}||f||_{L^{1}(0,\infty )}=\int _{0}^{\infty }|f|&=\int _{0}^{1}|f{\frac {1}{x^{p}}}||x^{p}|+\int _{1}^{\infty }|{\frac {1}{x^{p}}}||fx^{p}|\\&=||f{\frac {1}{x^{p}}}||_{L^{2}(0,1)}||x^{p}||_{L^{2}(0,1)}+||{\frac {1}{x^{p}}}||_{L^{2}(1,\infty )}||fx^{p}||_{L^{2}(1,\infty )}\\&\leq A||x^{p}||_{L^{2}(0,1)}+A||fx^{p}||_{L^{2}(1,\infty )}<\infty \end{aligned}}}
{\displaystyle f\in L^{1}(0,\infty )}
{\displaystyle f\in L^{1}(\mathbb {R} )}
{\displaystyle F(x)=\int _{\mathbb {R} }f(t){\frac {\sin xt}{t}}\,dt.}
{\displaystyle F}
is differentiable a.e. and find
{\displaystyle F'(x)}
(b) Is
{\displaystyle F}
absolutely continuous on closed bounded intervals
{\displaystyle [a,b]}
Look at the difference quotient:
{\displaystyle F'(x)=\lim _{h\to 0}{\frac {1}{h}}(F(x+h)-F(x))=\lim _{h\to 0}\int f(t){\frac {[\sin((x+h)t)-\sin(xt)]}{ht}}\,dt}
We can justify bringing the limit inside the integral. This is because for every
{\displaystyle x}
{\displaystyle |\sin(xt)/t|<1}
. Hence, our integrand is bounded by
{\displaystyle 2f(t)}
and hence is
{\displaystyle L^{1}}
{\displaystyle n}
. Then by Lebesgue Dominated Convergence, we can take the pointwise limit of the integrand. to get
{\displaystyle F'(x)=\int f(t)\cos(xt)\,dt.}
{\displaystyle F'(x)}
is bounded (specifically by
{\displaystyle ||f||_{L}^{1}}
) which implies that
{\displaystyle F}
is Lipschitz continuous which implies that it is absolutely continuous.
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Linearization around the equilibrium point for a system of differential equations \frac{dN}{d
pacificak8m 2022-05-01 Answered
Linearization around the equilibrium point for a system of differential equations
\frac{dN}{dt}=\mu N\left(1-\frac{N}{K}\right)+N{\int }_{0}^{\mathrm{\infty }}p\left(a,t\right)da
\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp\left(a,t\right).
Linearizing it around
\left(N=K,p=0\right)
, I am getting the system
\frac{dN}{dt}=h-\mu N+N{\int }_{0}^{\mathrm{\infty }}p\left(a,t\right)da
\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp\left(a,t\right),
h=\mu K
The reason I am getting this system is because I am considering
N{\int }_{0}^{\mathrm{\infty }}p\left(a,t\right)da
as linear term. Is this linearisation right or the correct system is
\frac{dN}{dt}=h-\mu N
\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp\left(a,t\right),
h=\mu K
eslasadanv3
I agree with the linearisation of the second equation. For the first, you are close but not quite correct. To find the linearisation around
N=K,p=0
N=K+\epsilon \stackrel{~}{N}
p=\epsilon \stackrel{~}{p}
into your equations and throw out everything that isn't order
\epsilon
. I got:
\begin{array}{rl}\epsilon \frac{d\stackrel{~}{N}}{dt}=\frac{dN}{dt}& =\mu N\left(1-\frac{N}{K}\right)+N{\int }_{0}^{\mathrm{\infty }}p\left(a,t\right)\phantom{\rule{0ex}{0ex}}da\\ & =-\mu \left(K+\epsilon \stackrel{~}{N}\right)\epsilon \left(\frac{\stackrel{~}{N}}{K}\right)+\epsilon \left(K+\epsilon \stackrel{~}{N}\right){\int }_{0}^{\mathrm{\infty }}\stackrel{~}{p}\left(a,t\right)\phantom{\rule{0ex}{0ex}}da\\ & =\left(-\mu \stackrel{~}{N}+K{\int }_{0}^{\mathrm{\infty }}\stackrel{~}{p}\left(a,t\right)\phantom{\rule{0ex}{0ex}}da\right)\epsilon +o\left(\epsilon \right).\end{array}
Thus, the linearised equation is (dropping the tildes):
\frac{dN}{ dt }=-\mu N+K{\int }_{0}^{\mathrm{\infty }}p\left(a,t\right),da
Edit: Answer to question in the comments. Let
N\cdot ,p\cdot
be steady state solutions to your system. Replacing p with
{p}^{\ast }+\epsilon p
the equation for p becomes:
\begin{array}{rl}k{p}^{\ast }+\epsilon kp=k\left({p}^{\ast }+\epsilon p\right)& =\frac{\mathrm{\partial }{p}^{\ast }}{\mathrm{\partial }t}+\epsilon \frac{\mathrm{\partial }p}{\mathrm{\partial }t}+\frac{\mathrm{\partial }{p}^{\ast }}{\mathrm{\partial }a}+\epsilon \frac{\mathrm{\partial }p}{\mathrm{\partial }a}\\ & =\frac{\mathrm{\partial }{p}^{\ast }}{\mathrm{\partial }a}+\epsilon \left(\frac{\mathrm{\partial }p}{\mathrm{\partial }t}+\frac{\mathrm{\partial }p}{\mathrm{\partial }a}\right)\end{array}
which implies the linearised equation is
kp=\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}
(not surprising since the original equation was linear). For the equation for
N
N
and p with
{N}^{\ast }+\epsilon N
{p}^{\ast }+\epsilon p
\begin{array}{rl}\epsilon \frac{dN}{dt}& =\frac{d\left({N}^{\ast }+\epsilon N\right)}{dt}\\ & =\mu \left({N}^{\ast }+\epsilon N\right)\left(1-\frac{{N}^{\ast }+\epsilon N}{K}\right)+\left({N}^{\ast }+\epsilon N\right){\int }_{0}^{\mathrm{\infty }}\left({p}^{\ast }\left(a\right)+\epsilon p\left(a,t\right)\right)\phantom{\rule{0ex}{0ex}}da\\ & =\mu {N}^{\ast }\left(1-\frac{{N}^{\ast }}{K}\right)+{N}^{\ast }{\int }_{0}^{\mathrm{\infty }}{p}^{\ast }\phantom{\rule{0ex}{0ex}}da\\ & \phantom{\rule{2em}{0ex}}+\epsilon \left[{N}^{\ast }{\int }_{0}^{\mathrm{\infty }}p\phantom{\rule{0ex}{0ex}}da+N{\int }_{0}^{\mathrm{\infty }}{p}^{\ast }\phantom{\rule{0ex}{0ex}}da+\mu N\left(1-\frac{2{N}^{\ast }}{K}\right)\right]\\ & \phantom{\rule{2em}{0ex}}+{\epsilon }^{2}\left[N{\int }_{0}^{\mathrm{\infty }}p\phantom{\rule{0ex}{0ex}}da-\frac{\mu {N}^{2}}{K}\right]\\ & =\epsilon \left[{N}^{\ast }{\int }_{0}^{\mathrm{\infty }}p\phantom{\rule{0ex}{0ex}}da+N{\int }_{0}^{\mathrm{\infty }}{p}^{\ast }\phantom{\rule{0ex}{0ex}}da+\mu N\left(1-\frac{2{N}^{\ast }}{K}\right)\right]\\ & \phantom{\rule{2em}{0ex}}+{\epsilon }^{2}\left[N{\int }_{0}^{\mathrm{\infty }}p\phantom{\rule{0ex}{0ex}}da-\frac{\mu {N}^{2}}{K}\right]\end{array}
\left({N}^{\ast },{p}^{\ast }\right)
is the steady state solution. Thus, the linearised equation is:
\frac{dN}{ dt }={N}^{\ast }{\int }_{0}^{\mathrm{\infty }}p,da+N{\int }_{0}^{\mathrm{\infty }}{p}^{\ast },da+\mu N\left(1-\frac{2{N}^{\ast }}{K}\right).
\frac{dw}{dt}
\frac{dw}{dt}
w=x\mathrm{sin}y,\text{ }x={e}^{t},\text{ }y=\pi -t
t=0
Please solve this differential equation
x+y{y}^{\prime }=3{y}^{2}{y}^{\prime }\sqrt{{x}^{2}+{y}^{2}}
Consider the following system of differential equations:
\left\{\begin{array}{l}{u}^{\prime }=-u+uv\\ {v}^{\prime }=-2v-{u}^{2}\end{array}
I'm able to prove that solutions must tend to 0 if
t\to 0
by the use of Lyapunov function
L\left(x,y\right)={x}^{2}+{y}^{2}
but I'm unable to prove that the function:
I\left(t\right)=\frac{{x}^{2}\left(t\right)+2{y}^{2}\left(t\right)}{{x}^{2}\left(t\right)+{y}^{2}\left(t\right)}
must admit limit for
t\to +\mathrm{\infty }
Instability of a parameter varying system whose parameters belong to a compact set:
Suppose, there is a system
\stackrel{˙}{x}=f\left(t,{\gamma }_{p}\left(t\right),x\right)
x\in {\mathbb{R}}^{2}
. For my specific case, parameter vector
{\gamma }_{p}
is a scalar and known monotonic function with a compact image set (for example
{\gamma }_{p}\left(t\right)={e}^{-t}
{\gamma }_{p}\in \left[1,0\right)
). I would like to show instability of this system
I\left(y\right)={\int }_{0}^{1}{y}^{\prime 2}-{y}^{2}+2xy dx
\begin{array}{rl}\frac{\mathrm{\partial }F}{\mathrm{\partial }{y}^{\text{'}}}& =2{y}^{\text{'}}\phantom{\rule{0ex}{0ex}}⟹\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }{y}^{\text{'}}}\right)=2{y}^{\text{'}\text{'}}\\ \frac{\mathrm{\partial }F}{\mathrm{\partial }y}& =-2y+2x\end{array}
2y{ }^{″}+2y-2x=0
y\left(x\right)=x
x\left(2y+1\right)dx=y\left({x}^{2}-3x+2\right)dy
f\left(t,\text{ }x\right)
V\subset U
U\subseteq {\mathbb{R}}^{n+1}
L\phantom{\rule{0.222em}{0ex}}={\supset }_{\left(t,x\right)\ne \left(t,y\right)\in V\subset U}\frac{|f\left(t,x\right)-f\left(t,y\right)|}{|x-y|}<\mathrm{\infty }
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Find the Laplace transform of \(\displaystyle{f{{\left({t}\right)}}}={10}{t}{e}^{{-{5}{t}}}\)
Jakobe Norton 2022-04-15 Answered
f\left(t\right)=10t{e}^{-5t}
stecchiniror7
\mathcal{L}\left(t\right)=\frac{1}{{s}^{2}}
\mathcal{L}\left(10t{e}^{-5t}\right)=10F\left(s+5\right)=\frac{10}{{\left(s+5\right)}^{2}}
Note: we made use of the Shifting Theorem, which was mentioned to you in another post.
granfury90210birm
L\left\{{e}^{at}f\left(t\right)\right\}=F\left(s-a\right)\text{ }\text{ if }\text{ }L\left\{f\left(t\right)\right\}=F\left(s\right)
L\left({t}^{n}\right)=\frac{\mathrm{\Gamma }\left(n+1\right)}{{s}^{n+1}}
\mathrm{\Gamma }\left(n+1\right)=n!
if n is non-negative integer
solve the given initial value problem
y{}^{‴}+3y{}^{″}-13{y}^{\prime }-15y=0,y\left(0\right)={y}^{\prime }\left(0\right)=0,y{}^{″}\left(0\right)=1
find the given inverse Laplace transform
{L}^{-1}\left\{\frac{{s}^{2}+1}{s\left(s-1\right)\left(s+1\right)\left(s-2\right)}\right\}
Solve the Exact Differential Equations of GIven Problem #2 only, ignore the other given problem.
Please answer all of the problems in the attached photo. I will rate you with “like/upvote” after; if you answer all of the problems. If not, I will rate you “unlike/downvote.” Kindly show the complete step-by-step solution.
\left(x{e}^{x}y+{e}^{x}y\right)dx+\left(8+x{e}^{x}\right)dy=0;y\left(0\right)=-4
\left(\frac{x+y}{{y}^{2}+1}\right)dy+\left(\mathrm{arctan}y+x\right)dx=0
Find the inverse laplace transforms for the following functions
F\left(s\right)=\frac{{e}^{-3s}}{{s}^{2}}
F\left(s\right)=\frac{2{e}^{-5s}}{{s}^{2}+4}
4y\prime \prime -4y\prime -3y=0
Help with a generating function and differential equation
I have a generating function that Im
\text{Misplaced &}
use properties of Laplace transform
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Discrete Random Variables - Cumulative Distribution Function | Brilliant Math & Science Wiki
The cumulative distribution function of a random variable
X
F_X
that, when evaluated at a point
x
, gives the probability that the random variable will take on a value less than or equal to
x:
\text{Pr}[X \leq x]
. For example, a random variable representing a single dice roll has cumulative distribution function
\text{Pr}[X \leq x] = F_X(x) = \begin{cases} \frac{1}{6} & x = 1 \\ \frac{2}{6} & x = 2 \\ \frac{3}{6} & x = 3 \\ \frac{4}{6} & x = 4 \\ \frac{5}{6} & x = 5 \\ \frac{6}{6} & x = 6. \end{cases}
For a discrete random variable, the cumulative distribution is defined by
F(x) = P(X \leq x) = \sum_{x_i \leq x}P(X = x_i) = \sum_{x_i \leq x}p(x_i),
p
is the p.d.f. of
X
. This distribution is not continuous, and is constant between the
x_i
Furthermore, the p.d.f. is related to the cumulative distribution by
\text{Pr}[X=x] = F_X(x) - F_X(x'),
x'
is the next smallest possible value of
x
. In the case of a random variable defined on integers (as is typical),
x'=x-1
. This forms the intuition for the relationship between the continuous p.d.f. and continuous cumulative distribution, where the p.d.f. is the derivative of the c.d.f.
Cite as: Discrete Random Variables - Cumulative Distribution Function. Brilliant.org. Retrieved from https://brilliant.org/wiki/discrete-random-variables-cumulative-distribution/
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Optimization of Bleaching Process of Crude Palm Oil by Activated Plantain (Musa paradisiaca) Peel Ash Using Response Surface Methodology
Optimization of Bleaching Process of Crude Palm Oil by Activated Plantain (Musa paradisiaca) Peel Ash Using Response Surface Methodology ()
Wuraola Abake Raji*, Rowland Ugochukwu Azike, Fredericks Wirsiy Ngubi
Department of Chemical Engineering, Igbinedion University, Okada, Nigeria.
The bleaching of crude palm oil using activated plantain peel ash (APPA) was studied in this work. Historical data design (HDD) in response surface methodology (RSM) experimental design was employed to optimize and correlate the process operating parameters (temperature, time and adsorbent dosage) to the percentage bleaching efficiency. The analysis of the results showed that the quadratic effects of the operating parameters were significant. The optimum condition for the maximum adsorption efficiency of 70.04% was obtained at 160°C temperature, 60 minutes reaction time, and 4 g adsorbent dosage. The predicted bleaching efficiency of 74% was in good agreement with the optimum experimental yield. This study has revealed that APPA is a good source of adsorbent for palm oil bleaching.
Bleaching, Activated Plantain Peel Ash, Adsorption, Bleaching Efficiency, Response Surface Methodology
Raji, W. , Azike, R. and Ngubi, F. (2019) Optimization of Bleaching Process of Crude Palm Oil by Activated Plantain (Musa paradisiaca) Peel Ash Using Response Surface Methodology. Open Journal of Optimization, 8, 38-46. doi: 10.4236/ojop.2019.81004.
Palm oil is an edible vegetable oil, produced by extracting its oil from the fresh fleshy fruits of oil palms which serves as an important ingredient for the food industry because of its superior characteristics and attributes [1] . Palm oil in its raw form contains impurities such as organic pigments, oxidation metals, trace metals and trace soaps [2] . For palm oil to be edible these impurities which negatively influence the taste and smell of the oil as well as its appearance and shelf life stability must be removed through the bleaching process. The bleaching of the oil is accomplished via adsorption of the impurities onto a surface-active micro porous adsorbent material or bleaching agent for a period of time by Vander Waal forces [3] [4] .
The refining of palm oil through adsorptive bleaching remains inevitable in the palm oil refining industry. The major adsorbent used in the industry is bleaching earth, but it is expensive in terms of production cost. To avert this, utilization of agro-wastes as adsorbents is currently receiving wide attention because of their abundant availability and low cost owing to relatively high fixed carbons and presence of porous structures [5] . Unripe plantain peel is a carbonaceous waste material that is currently disposed of either by burning in the open or left in the field to decay. Both disposal methods contribute to environmental degradation [6] . Bleaching of palm oils with activated unripe plantain peel ash (APPA) therefore merits more investigations in order to combat environmental pollution and to reduce cost of production in the refining industry.
Optimization of the crude palm oil process variables is viable through the application of statistical experimental design techniques in adsorption process development that results in improved product yields and reduced development time and overall costs [7] . Regular methods of studying a process by maintaining other parameters involved at an undesignated constant level do not describe the combined effect of all the parameters involved [8] . This is referred as one factor at a time. This method is time consuming and requires a large number of experiments to determine optimum levels, which are unreliable [3] . These limitations can be avoided by optimizing all the parameters collectively by statistical experimental design such as response surface methodology [9] .
Response surface methodology (RSM) is based on polynomial surface analysis and it is a collection of mathematical and statistical techniques that are useful for the modelling and analysis of problems in which a response of interest is influenced by several variables [10] . It also employs multiple regression and correlation analyses as tools to assess the effects of two or more independent factors on the dependent variables [11] . RSM has been applied in researches into complex variable processes by many authors to optimize operational conditions of processes [3] [8] [12] [13] [14] [15] . In this study, RSM was used to optimize the bleaching process variables such as temperature, contact time and adsorbent dosage in order to obtain optimum conditions for the bleaching of palm oil using Historical Data Design (HDD).
The unripe plantain peels and crude palm olein (CPO) used in this study were obtained from the open market, Benin City, Edo State, Nigeria. The analytical grade reagents used was procured from LUCO-Consult Limited, Benin City. A constant gram of CPO was bleached with varying dosage of APPA at different temperatures and time. The heated mixture of CPO and APPA formed a slurry. The slurry was cooled at room temperature in a water bath and filtered with Whatman’s filter paper for effective separation of the bleached oil and spent adsorbent. The Labscience 721A UV-Spectrophotometer was used to analyse the bleached oil. The percentage bleached oil was determined using Equation (1) [3] .
\%\text{Bleachedoil}=\frac{{A}_{0}-A}{{A}_{0}}\times 100
where A0 is the absorbance of unbleached palm oil and A is the absorbance of bleached palm oil.
The experimental design and statistical analysis were performed according to the response surface analysis method using Design Expert 7.0.0 (Stat-Ease Inc., Minneapolis, MN, USA) trial version software. Historical data design (HDD) was used for the optimization of bleached palm oil in order to examine the combined effect of the three different factors (independent variables): bleaching temperature, bleaching time and adsorbent dosage on bleaching performance (dependent variable). The statistical analysis was performed using: Analysis of Variance (ANOVA), regression analysis and response surface plots of the interaction effects of the factors to evaluate optimum conditions for the bleaching process. The effects of the process variables on the bleaching efficiency were calculated and their respective significant evaluated by ANOVA test. The p-value was used as the yardstick for measuring the significance of the regression coefficients, values of p greater than 0.05 signified that the coefficient is significant. The experimental data were fitted to the second-order polynomial regression model and the adequacy of the model tested by the coefficient of determination R-Squared (R2) value as compared to the adjusted R-Squared (R2) value.
3.1. Regression Model Equation
The developed model correlated the response to the process variables using second degree polynomial. The model selected was based on the highest order model where the additional terms were significant and the model was not aliased and there was reasonable agreement between Adjusted R-Squared and predicted R-squared (within 0.2 of each other). The predictive model was given in Equation (2) in a coded term. The coded equation is useful for identifying the relative significance of the factors by comparing factor coefficients. Base on the experimental design and the experimental results obtained, the second order response functions representing Y is the response for bleaching performance, A is the coded value of variable temperature (˚C), B is the coded value of variable adsorbent dosage (gram) and C is the coded value of variable contact time (mins). From the model equation, time had highest effect on the bleaching performance with coefficient of 37.95 followed by temperature and lastly, adsorbent dosage.
\begin{array}{l}Y=+49.29+16.34\ast A+8.44\ast B+37.95\ast C+2.29\ast AB+13.25\ast AC\\ \text{}-3.15\ast BC-88.39\ast A2-1.71\ast B2+3.72\ast C2\end{array}
Table 1 shows the result of analysis of variance in which the Model F-value of 42.87 implies the model is significant. There is only a 0.01% chance that a “Model F-Value” this large could occur due to noise. Values of “Prob > F” less than 0.0500 indicate model terms are significant. In this case A, B, C, AC, BC, A2, A2C are significant model terms. Values greater than 0.1000 indicate the model terms are not significant. In addition, the model did not show lack of fit and presented high determination coefficients, R2 = 0.94871 indicating that 94.8% of the variability was explained by the model shown in Table 2. The “Pred R-Squared” of 0.8852 is in reasonable agreement with the “Adj R-Squared” of 0.9266. “Adeq Precision” measures the signal to noise ratio. A ratio greater than 4 is desirable. Your ratio of 22.636 indicates an adequate signal. This model can be used to navigate the design space.
3.3. Model Diagnostic Plots
3.3.1. Normal Plot of Residuals
Figure 1 depicts normal probability plot which indicates normal distribution of residuals, in which the points follows a straight line. Some moderate scatter points maybe expected even with normal data. Transformation of the response maybe needed to provide a better analysis if patterns look like an “S-shaped” curve.
Table 1. The analysis of variance (ANOVA) result.
Figure 1. Normal plot of residuals.
3.3.2. Predicted vs. Actual
A graph of the actual response values versus the predicted response values is depicted in Figure 2. It helps to detect a value, or group of values, that are not easily predicted by the model. The data points should be spitted evenly by the 45 degree line as it is seen in the graph.
3.4. Model Graphs
Figure 3 shows the response surface plot for bleaching performance as a function of bleaching temperature and adsorbent dosage. The interacting effect of temperature and adsorbent dosage has a positive influence on the crude palm oil bleaching. It was observed that increasing the APPA dosage from 1.0 gram to 4.0 grams increased the bleaching efficiency. The results clearly indicate that the bleaching efficiency increases to an optimum value at adsorbent dosage of 4.0 g, this result is in agreement with the reports of [6] [13] . It is also observed that the bleaching efficiency increases as temperature increased. An optimum value of 70% bleaching performance was attained as the temperature reached 160˚C, it was observed that further increase in temperature darkens the colour of oil.
The effect of contact time and temperature on the bleaching efficiency is depicted in Figure 4. From the plot, bleaching does not proceed to any appreciable
Figure 2. A plot of predicted vs. actual.
Figure 3. Effect of temperature and adsorbent dosage on the bleaching efficiency of crude palm oil.
Figure 4. Effect of contact time and temperature on the bleaching performance of crude palm oil.
Figure 5. Effect of adsorbent dosage and contact time on the bleaching performance of crude palm oil.
degree at low contact time but as the time increased from 15 mins to 60 mins, the bleaching performance increased showing that reaction time favours and promotes access to further adsorption sites. It can be deduced that for essentially all bleaching temperatures investigated, the bleaching efficiency increases with contact time. A further increase of the temperature and time beyond 160˚C and 60 minutes respectively may not have significant effect on the bleaching process. It is worthy to note that time and temperature factors in bleaching of palm oil improves the bleaching efficiency.
The surface response curve of the calculated response (bleaching performance) against time and adsorbent dosage at constant temperature is shown in Figure 5. The adsorbent dosage was varied from 1.00 g to 4.00 g. The increase in bleaching performance of APPA on the oil was as a result of an increase in the number of active sites available for adsorption. Noticeable change in the bleaching of the crude oil was observed as the time increases from 15 minutes to 60 minutes. This is as a result of the palm oil having more interfacial contact with the adsorbent which improves the bleaching process.
The use of low-cost adsorbent developed from activated plantain peel ash (APPA) in adsorptive bleaching of palm oil has been investigated in this work. HDD in response surface methodology was successfully applied in the experimental design in order to optimize the process operating parameters such as time, temperature and adsorbent dosage on the bleaching efficiency. The optimum bleaching efficiency of 70% was obtained at 60 mins reaction time, 4 g adsorbent dosage and 160˚C reaction temperature. Hence, the activated plantain peel ash has proved to be a good source of local adsorbent by performing optimally in the bleaching of crude palm oil.
[1] Chigozie, U.F. and Stone, O.R. (2014) Experimental Process Design for Sorption Capacity of Kogi And Ibusa Clay Activated with HNO3 and H2SO4 in Palm Oil Bleaching. Journal of Emerging Trends in Engineering and Applied Sciences (JETEAS), 5, 174-182.
[2] Usman, M.A., Ekweme, V.I., Alaye, T.O. and Mohammed, A.O. (2012) Characterization, Acid Activation and Bleaching Performance of Ibeshe Clay. ISRN Ceramics, 2012, Article ID: 658508.
[3] Nwabanne, J.T. and Ekwu, F.C. (2013) Decolourization of Palm Oil by Nigerian Local Clay, a Study of Adsorption Isotherms and Bleaching Kinetics. International Journal of Multidisciplinary Sciences and Engineering, 4, 20-27.
[4] Usman, M.A., Oribayo, O. and Adebayo, A.A. (2013) Bleaching of Palm Oil by Activated Local Bentonite and Kaolin Clay from Afashion, Edo—Nigeria. Chemical and Process Engineering Research, 10, 1-12.
[5] Mohammed, S.G., Ahmed, S.M., Badawi, A.F.M. and El-Desonki, D.S. (2015) Activated Carbon Derived from Egyptian Banana Peels for Removal of Cadmium from Water. Journal of Applied Life Science International, 3, 77-88.
[6] Agatemor, C. (2008) Some Aspects of Palm Oil Bleaching with Activated Plantain Peel Ash. Food Science and Technology Research, 14, 301-305.
[7] Annadurai, G., Juang, R.S. and Lee, D.J. (2002) Factorial Design Analysis of Adsorption of Activated Carbon on Activated Carbon Incorporated with Calcium Aginate. Advances in Environmental Research, 6, 191-198.
[8] Kumar, A., Prasad, B. and Mishra, I.M. (2008) Adsorptive Removal of Acrylonitrile Using Powered Activated Carbon. Journal of Environmental Protection Science, 2, 54-62.
[9] Ko, D.C.K., Porter, J.F. and McKay, G. (2000) Optimized Correlations for the Fixed Bed Adsorption of Metal Ions on Bone Char. Chemical Engineering Science, 55, 5819-5829.
[10] Park, K. and Ahn, J.H. (2004) Design of Experiment Considering Two-Way Interactions and Its Application to Injection Molding Processes with Numerical Analysis. Journal of Materials Processing Technology, 146, 221-227.
[11] Jeong, G.T., Yang, H.S. and Park, D.H. (2009) Optimization of Transesterification of Animal Fat Ester Using Response Surface Methodology. Bioresource Technology, 100, 25-30.
[12] EjikemeEbere, M., Egbuna, S.O. and Ejikeme, P.C.N. (2013) Optimal Bleaching Performance of Acid Activated Ngwulangwu Clay. International Journal of Engineering and Innovative Technology (IJEIT), 3, 13-19.
[13] Egbuna, S.O., Mbah, C.N. and Chime, T.O. (2015) Determination of the Optimal Process Conditions for the Acid Activation of Ngwo Clay in the Bleaching of Palm Oil. International Journal of Computational Engineering Research (IJCER), 5, 17-29.
[14] Larouci, M., Safa, M., Meddah, B., Aoues, A. and Sonnet P., (2015) Response Surface Modeling of Acid Activation of Raw Dialomite using Sunflower Oil Bleaching by: Box-Behnken Experimental Design. Journal of Food Science and Technology, 52, 1677-1683.
[15] Chong, W.-T., Tan, C.-P., Cheah, Y.-K., Lajis, A.F.B., Habi Mat Dian, N.L., Kanagaratnam, S., et al. (2018) Optimization of Process Parameters in Preparation of Tocotrienol-Rich Red Palm Oil-Based Nanoemulsion Stabilized by Tween80-Span 80 Using Response Surface Methodology. PLoS ONE, 13, 1-22.
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Khudaverdian H. Non-linear homomorphisms of algebras of functions are induced by thick morphisms (abstract) - Geometry of Differential Equations
Khudaverdian H. Non-linear homomorphisms of algebras of functions are induced by thick morphisms (abstract)
Speaker: Hovhannes Khudaverdian
Title: Non-linear homomorphisms of algebras of functions are induced by thick morphisms
In 2014 Th. Voronov introduced the notion of thick morphisms of (super)manifolds as a tool for constructing
{\displaystyle L_{\infty }}
-morphisms of homotopy Poisson algebras. Thick morphisms generalise ordinary smooth maps, but are not maps themselves. Nevertheless, they induce pull-backs on
{\displaystyle C^{\infty }}
functions. These pull-backs are in general non-linear maps between the algebras of functions which are so-called "non-linear homomorphisms". By definition, this means that their differentials are algebra homomorphisms in the usual sense. The following conjecture was formulated: an arbitrary non-linear homomorphism of algebras of smooth functions is generated by some thick morphism. We prove here this conjecture in the class of formal functionals.
Slides: KhudaverdianAMVconf2021slides
Retrieved from "https://gdeq.org/w/index.php?title=Khudaverdian_H._Non-linear_homomorphisms_of_algebras_of_functions_are_induced_by_thick_morphisms_(abstract)&oldid=6448"
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Algebra Shortcuts Warmup Practice Problems Online | Brilliant
\large \frac{ \color{#20A900}{2015} \times \color{#3D99F6}{1999} -\color{#20A900}{2015} \times \color{#D61F06}{1379}} { \color{#3D99F6}{1999} - \color{#D61F06}{1379} } .
Hint: Use the distributive property.
a = 2 , b = 7, c = 9
1 + a + b + c + ab + bc + ca + abc ?
\Large \color{#D61F06}{87} ^ 2 - \color{#EC7300}{86} ^ 2.
Hint: Use the difference of two squares identity.
\large \frac {\color{#D61F06}6! \times \color{#3D99F6}7!}{\color{#69047E}{10}!} = \ \color{#20A900}?
2
1
\frac{1}{3}
\frac{1}{2}
\large \frac{ \color{#3D99F6}{42} ^ 2 - \color{#20A900}{38} ^ 2 } { \color{#3D99F6}{42} - \color{#20A900}{38} }.
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Want to know more about Green's, Stokes', and the divergence theorem?
Determine the lowest common multiple (LCM) of the following 4𝑥2𝑦2 , 6𝑥𝑦 , 10𝑥𝑦2
A nonhomogeneous second-order linear equation and a complementary function
{y}_{c}
are given. Find a particular solution of the equation.
4{x}^{2}y{ }^{″}-4x{y}^{\prime }+3y=8{x}^{\frac{3}{4}}
A simple random sample of pulse rates of 25 women from a normally distributed population results in a standard deviation of 12.7 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pullse rates of women have a standard deviation equal to 10 beats per minute. Complete parts (a) through (d) below.
a. Identify the null and alternative hypotheses. Choose the correct answer below.
{H}_{0}:\sigma \ne 10
beats per minute,
{H}_{1}:\sigma =10
{H}_{0}:\sigma =10
{H}_{1}:\sigma \ne 10
beats per minuteSNKC.
{H}_{0}:\sigma =10
{H}_{1}:\sigma <10
{H}_{0}:\sigma \ge 10
{H}_{1}:\sigma <10
b. Compute the test statistic.
{\chi }^{2}=\square
c. Find the P-value of the test statistic.
The P-value of the test statistic is
\square
d. State the conclusion.
'Reject'/'Fail to reject'
{H}_{0}
'is'/'is not' sufficiert evidence to warrant rejection of the claim that pulse rates of women have a standard deviation equal to 10 beats per minute.
\sqrt[3]{\pi }
is transcendental
In a family of 6, what is the probability of getting 1 boy and 5 girls
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Abstract algebra questions, solved and explained
Abstract algebra questions and answers
Recent questions in Abstract algebra
Paula Good 2022-03-25 Answered
How can I show these polynomials are not co'?
x,\text{ }x-1
x+1
{\mathbb{Z}}_{6}\left[x\right]
siliciooy0j 2022-03-24 Answered
Let k be a field, V a finite-dimensional k-vectorspace and
M\in End\left(V\right)
. How can I determine Z, the centralizer of
M\otimes M
End\left(V\right)\otimes End\left(V\right)
Find all zero divisors of z_81
Pelizzolaf40 2022-03-19 Answered
Calculate e and f for
{\mathbb{Q}}_{2}\left(\sqrt{3},\text{ }\sqrt{2}\right)
Caroline Carey 2022-03-18 Answered
If we have an algebraic number α with (complex) absolute value 1, it does not follow that α is a root of unity (i.e., that
\alpha n=1
for some n). For example,
\left(\frac{3}{5}+\frac{4}{5i}\right)
is not a root of unity.But if we assume that
\alpha
is an algebraic integer with absolute value 1, does it follow that
\alpha
is a root of unity?
fun9vk 2022-03-18 Answered
Applications of the concept of homomorphism
What are some interesting applications of the concept of homomorphism?
Example: If there is a homorphism from a ring R to a ring r then a solution to a polynomial equation in R gives rise to a solution in r. e.g. if
f:R\to r
{X}^{2}+{Y}^{2}=0
f\left({X}^{2}+{Y}^{2}\right)=f\left(0\right),f\left({X}^{2}\right)+f\left({Y}^{2}\right)=0,{f\left(X\right)}^{2}+{f\left(Y\right)}^{2}=0,{x}^{2}+{y}^{2}=0
Juliet Jackson 2022-03-17 Answered
\mathrm{\forall }n\in \mathbb{N}:\left[\mathbb{Q}\left(\mathrm{exp}\left(2\pi \frac{i}{n}\right):\mathbb{Q}\right]=n-1
f\in \mathbb{Q}\left[X\right]
be irreducible,
deg\left(f\right)=n
|Gal\left(f\right)|=n
\frac{K\left(a\right\}}{K}
be an algebraic field extension. Then
Gal\left(\frac{K\left(a\right)}{K}\right)
is commutative
\frac{L}{K}
be a finite field extension and
|Gal\left(\frac{L}{K}\right)|=1
\frac{L}{K}
Tianna Costa 2022-03-17 Answered
Can we find a binary operation s.t. the map
\varphi \left(n\right)=n+1
becomes an isomorphism from
⟨B\mathbf{Z},\cdot ⟩
⟨B\mathbf{Z},.⟩
pinka1hf 2022-03-14 Answered
When does the product of two polynomials
={x}^{k}
f,g\in \mathbb{C}\left[x\right]
fg=\alpha {x}^{n+m}
\alpha \in \mathbb{C}
Thaddeus Nolan 2022-03-14 Answered
Jerimiah Boone 2022-03-14 Answered
Why are the only (associative) division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?
Here a division algebra is an associative algebra where every nonzero number is invertible (like a field, but without assuming commutativity of multiplication).
Kienastsrx 2022-03-12 Answered
\mathbb{Q}\left(t,\sqrt{{t}^{3}-t}\right)
not a purely transcendental extension of
\mathbb{Q}
Vandinimo3 2022-03-09 Answered
Why require R to be a field to obtain that
\frac{R\left[\alpha \right]}{\left(2{\alpha }^{2}\right)}
is either an exterior algebra or a polynomial algebra?
Without actually computing the orders, explain why the two elements in each of the following pairs of elements from Z30 must have the same order: {2, 28}, {8, 22}. Do the same for the following pairs of elements from U(15): {2, 8}, {7, 13}.
ordanlarroque1xm 2022-03-08 Answered
Why the ring
{S}^{-1}R
is Noetherian if S is multiplicative?
Umaiza Hutton 2022-03-03 Answered
In the group
GL\left(2,{\mathbb{Z}}_{7}\right)
, inverse of
A-\left(\begin{array}{cc}4& 5\\ 6& 3\end{array}\right)
How does one prove that ring of Hamilton Quaternions with coefficients coming from the field
\frac{\mathbb{Z}}{p}\mathbb{Z}
is not a divison ring.
Junaid Ayala 2022-03-01 Answered
If G is a finite group with
|G|<180
and G has subgroups of orders 10, 18 and 30 then the order of G is:
Erik Sears 2022-03-01 Answered
\gamma =\left(124\right)\left(425\right)\left(64\right)
as a product of disjoint cycles.
Havlishkq 2022-02-28 Answered
Compute the kernel for the given homomorphism
\varphi
\varphi :\mathbb{Z}\to {\mathbb{Z}}_{8}
phI\left(1\right)=6
Coming up with good abstract algebra examples is essential for those who are trying to come up with the answers to theoretical questions both in Engineering and Data Science disciplines. The college students will be able to discover abstract algebra questions and answers provided by our friendly experts that will help you to understand abstract algebra questions with various examples based on high-energy physics, cryptography, and the number theory. Remember that the trick is to use number sequences to generalize the set of various integers and transformations of functions equation problems at play. Don’t forget about the application of the Algebraic number theory studies as well.
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Monte-Carlo Simulation | Brilliant Math & Science Wiki
Beakal Tiliksew, Alex Chumbley, Ethan W, and
Mayle Dump
Monte Carlo simulations define a method of computation that uses a large number of random samples to obtain results. They are often used in physical and mathematical problems and are most useful when it is difficult or impossible to use other mathematical methods. Monte Carlo methods are mainly used in three distinct problem classes: optimization, numerical integration, and generating draws from probability distributions.
Monte Carlo simulations are often used when the problem at hand has a probabilistic component. An expected value of that probabilistic component can be studied using Monte Carlo due to the law of large numbers. With enough data, even though it's sampled randomly, Monte Carlo can hone in on the truth of the problem. For example, we can estimate the value of pi by simply throwing random needles into a circle inscribed within a square that is drawn on the ground. After many needles are dropped, one quadrant of the circle is then examined. The ratio of the number of needles that are inside the square to the number of needles inside the circle is a very good approximation of pi. The more needles that are dropped, the closer the approximation gets.
Rayleigh Taylor Instability uses Monte Carlo to predict how two liquids will interact
The most comforting thing about Newtonian mechanics is that everything happens for a reason. For example, if you push one end of the lever, the other end goes up. You throw an object, and it travels in a parabolic path. The physical world is a completely deterministic place: all the future states are derived form the previous ones. For centuries we had this prevailing scientific wisdom, and then came the Copenhagen doctrine, led by Bohr and Heisenberg. The proponents of this doctrine argued that at the most fundamental level, the behavior of a physical system cannot be determined. This led to a serious debate regarding the validity of causal non-determinism, i.e. every event is not caused by previous events. People like Einstein and Schrodinger found this philosophy unacceptable as exemplified by Einstein's often repeated comment: "God does not play dice." The question of causal non-determinism is still unsettled but there is ample evidence to prove that certain systems can only be modeled accurately by stochastic processes. A process is called stochastic if its next step depends on both previous states and some random event.
Most of us are able to calculate the basic probability of occurrence of certain events, but how do we interpret the significance of the results obtained? The answer is that, in this age of fast computers, we can run multiple simulations of an event and record the outcomes and compare the results with the mathematically calculated value! Then in front of our own eyes we can see how powerful the theory of probability is!
The Monte Carlo method is a method of solving problems using statistics. Given the probability that a certain event will occur in certain conditions, a computer can be used to generate those conditions repeatedly.
Monte Carlo method process:
Pick a space of possible samples.
Randomly sample in that space using a probability distribution.
Perform defined operations on the random samples.
To tie this process to something more concrete, take the example of estimating pi that is explained in more detail below. The first step in this process is to pick the space of possible samples. That would be the physical space of the circle inscribed within a square. That space is then randomly sampled by throwing needles into it at random. A defined operation of counting the needles is then done. And finally, a ratio is calculated for those needles.
The Monte Carlo process can be a little difficult to accept. If we're just randomly gathering data, how can that help us find a correct answer?
Let's think about a coin. We want to know if the coin is fair or not. A fair coin has these properties
P(\text{heads}) = \frac{1}{2} , \quad P(\text{tails}) = \frac{1}{2} .
If you flip a coin just one time, what will happen? It will either land on heads or tails. If you flip a coin and it lands on tails, and then you walk away, have you proven that
P(\text{tails}) = 1
and that the coin is not fair? Of course not! A fair coin can also land on tails after one try, after all.
Maybe you flip the coin again and it lands on heads, thus proving that
P(\text{tails}) = P(\text{heads}) = \frac{1}{2}
, and that the coin is fair. But have you actually proven that the coin is fair? Maybe the coin isn't completely fair but it does land on both sides sometimes.
Maybe you flip tails again, and you're even more convinced that
P(\text{tails}) = 1
. Sadly, we still don't know because both a fair coin and a false coin have the ability to flip tails twice in a row.
The point is that we need to flip this coin a lot before we can be convinced that it's fair. Let's see this with some code. In the following Python snippet, there's a function that takes in the probability that a coin flips a heads (it's o.5 if the coin is fair) and a number of times to flip the coin. It then returns a guess as to the fairness of the coin (again, it's 0.5 if it's fair).
def getCoinProb(headsProb, numFlips):
currentFlip = random.random()
if currentFlip <= headsProb: #Probability of Heads
return prob / numFlips
Let's look at a fair coin first. We can flip the coin once, get tails, so the coin looks like it's actually not fair. We then do 5 flips, and it looks better, but we're still not totally convinced that the coin is completely fair. Even after 100 trials, it's close, but not 100%. It takes 100000 flips of the coin for us to start to believe that this coin is fair (and we should do even more to be completely sure).
>>> getCoinProb(.5, 1)
>>> getCoinProb(.5, 100)
>>> getCoinProb(.5, 100000)
Now let's flip a false coin. We'll say that this coin flips heads 75% of the time (so it's a really false coin). We flipped it once and actually got tails, way off our expectation! Flipping it 5 times gets us closer to our 75% mark, but it's just as far away from being fair after 5 flips as the actual fair coin. Flipping it 100 and 100000 times gets us closer to our mark.
>>> getCoinProb(.75, 1)
>>> getCoinProb(.75, 100)
>>> getCoinProb(.75, 100000)
This is the main point of the Monte Carlo method. When there's something probabilistic (like flipping a coin) that we can't predict, we need to do a lot of trials to make sure we're understanding the system correctly.
One of the most influential figures in the history of probability theory was Blaise Pascal. His interest in the field began when a friend asked him the following question:
"Would it be profitable given 24 rolls of a pair of fair dice to bet against there being at least one double six?"
Write a function that uses Monte Carlo to simulate the probability of getting a pair of 6's within twenty-four rolls of a pair of dice.
The probability of rolling a pair of 6's with just one roll is
\frac{1}{36}
. Therefore the probability of rolling a double six with
24
rolls is
1-\big(\frac{35}{36}\big)^{24}=0.49
"""returns a random int between 1 and 6"""
def Monteprob(numTrials):
numwins = 0.0
d1 = rollDie()
numwins += 1
print numwins/numTrials
Now we can test the code to see how the number of trials we make affects the probability outcome.
>>> Monteprob(100)
#10000 Trials
>>> Monteprob(10000)
#100000 Trials
>>> Monteprob(100000)
The more trial we do, the closer the answers are to each other. As the number of trials stretches to infinity, we converge on an answer. Here, we can guess from our last set of Monte Carlo runs that the answer is around 0.491 or 49.1%.
1\times 1
square. If two points are randomly picked within the square, what is the expected value (average) of the distance between them, rounded to 4 decimal places?
Monte Carlo simulation is useful for tackling problems in which nondeterminism plays a role. Interestingly, however, Monte Carlo simulation (and randomized algorithms in general) can be used to solve problems that are not inherently stochastic, i.e., for which there is no uncertainty about outcomes.
Let us imagine a rectangle of height
h
b-a
A=h(b-a)
f(x)
is within the boundaries of the rectangle. Then compute
n
pairs of random numbers
x_{i}
y_{i}
a\leq x_{i} \leq b
0\leq y_{i} \leq h
. The fraction of points
x_{i}
y_{i}
that satisfy the condition
y_{i}\leq f(x_{i})
is an estimate of
f(x)
to the area of the rectangle. Hence the estimate
G_{n}
G_{n}=A\frac{n_{h}}{n},
n_{h}
is the number of hits landed below the curve and
n
is the total number of hits on the rectangle.
In numerical integration, methods such as the trapezoidal rule use a deterministic approach. Monte Carlo integration, on the other hand, employs a non-deterministic approach: each realization provides a different outcome. In Monte Carlo, the final outcome is an approximation of the correct value. There is always some error when it comes to approximations, and the approximation of Monte Carlo is only as good as its error bounds. The bounds can be shrunk if more and more samples are collected, though, and that is the power of Monte Carlo.
Long before computers were invented, the French mathematicians Buffon (1707-1788) and Laplace (1749-1827) proposed using a stochastic simulation to estimate the value of
π
. Think about inscribing a circle in a square with sides of length
2
r
1
We have the area of the circle
\pi r^{2}=\pi
. But what’s
\pi?
Buffon suggested that he could estimate the area of a circle by a dropping a large number of needles (which he argued would follow a random path as they fell) in the vicinity of the square. The ratio of the number of needles with tips lying within the square to the number of needles with tips lying within the circle could then be used to estimate the area of the circle:
\frac { { A }_{ c } }{ { A }_{ s } } =\frac { \pi { r }^{ 2 } }{ 4{ r }^{ 2 } } \implies \pi =4\frac { { A }_{ c } }{ { A }_{ s } }.
Dropping more and more needles in the circle[1]
Notice how the more needles we drop in the circle, the closer our approximation gets to the actual value of
\pi.
trial=1000000
throws = 0
for i in range (1, trial):
throws += 1
dist = sqrt(pow(x, 2) + pow(y, 2))
hits = hits + 1.0
# hits / throws = 1/4 Pi
pi = 4 * (hits / throws)
print("pi is= ",pi)
The above algorithm simulates dropping a needle by first using random to get a pair of positive Cartesian coordinates (
x
y
values). It then uses the Pythagorean theorem to compute the hypotenuse of the right triangle with base
x
y.
This is the distance of the tip of the needle from the origin (the center of the square). Since the radius of the circle is
1
, we know that the needle lies within the circle if and only if the distance from the origin is no greater than
1
. We use this fact to count the number of needles in the circle.
Jo, C. Trie. Retrieved June 22, 2016, from https://en.wikipedia.org/wiki/Monte_Carlo_method
Cite as: Monte-Carlo Simulation. Brilliant.org. Retrieved from https://brilliant.org/wiki/monte-carlo/
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Compare accuracies of two classification models by repeated cross-validation - MATLAB testckfold - MathWorks Italia
{e}_{crk}
{\stackrel{^}{\delta }}_{rk}={e}_{1rk}-{e}_{2rk}.
{\overline{\delta }}_{r}=\frac{1}{K}\sum _{k=1}^{K}{\stackrel{^}{\delta }}_{kr}.
\overline{\delta }=\frac{1}{KR}\sum _{r=1}^{R}\sum _{k=1}^{K}{\stackrel{^}{\delta }}_{rk}.
{s}_{r}^{2}=\frac{1}{K}\sum _{k=1}^{K}{\left({\stackrel{^}{\delta }}_{rk}-{\overline{\delta }}_{r}\right)}^{2}.
{\overline{s}}^{2}=\frac{1}{R}\sum _{r=1}^{R}{s}_{r}^{2}.
{S}^{2}=\frac{1}{KR-1}\sum _{r=1}^{R}\sum _{k=1}^{K}{\left({\stackrel{^}{\delta }}_{rk}-\overline{\delta }\right)}^{2}.
{t}_{paired}^{\ast }=\frac{{\stackrel{^}{\delta }}_{11}}{\sqrt{{\overline{s}}^{2}}}.
{t}_{paired}^{\ast }
{\stackrel{^}{\delta }}_{11}
\overline{\delta }
{F}_{paired}^{\ast }=\frac{\frac{1}{RK}\sum _{r=1}^{R}\sum _{k=1}^{K}{\left({\stackrel{^}{\delta }}_{rk}\right)}^{2}}{{\overline{s}}^{2}}.
{F}_{paired}^{\ast }
{t}_{CV}^{\ast }=\frac{\overline{\delta }}{S/\sqrt{\nu +1}}.
{t}_{CV}^{\ast }
{\stackrel{^}{p}}_{1j}
{e}_{1}=\frac{\sum _{j=1}^{{n}_{test}}{w}_{j}\mathrm{log}\left(1+\mathrm{exp}\left(-2{y}_{j}^{\prime }f\left({X}_{j}\right)\right)\right)}{\sum _{j=1}^{{n}_{test}}{w}_{j}}
f\left({X}_{j}\right)
{e}_{1}=\frac{\sum _{j=1}^{{n}_{test}}{w}_{j}\mathrm{exp}\left(-{y}_{j}f\left({X}_{j}\right)\right)}{\sum _{j=1}^{{n}_{test}}{w}_{j}}.
f\left({X}_{j}\right)
{e}_{1}=\frac{\sum _{j=1}^{n}{w}_{j}\mathrm{max}\left\{0,1-{y}_{j}\prime f\left({X}_{j}\right)\right\}}{\sum _{j=1}^{n}{w}_{j}},
f\left({X}_{j}\right)
{e}_{1}=\frac{\sum _{j=1}^{{n}_{test}}{w}_{j}I\left({\stackrel{^}{p}}_{1j}\ne {y}_{j}\right)}{\sum _{j=1}^{{n}_{test}}{w}_{j}}.
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Free solutions to counting principles problems
Counting principles problems and answers
Recent questions in Counting Principles
N\left(A\cup B\cup C\right)=N\left(A\right)+N\left(B\right)+N\left(C\right)-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+N\left(A\cap B\cap C\right)
21=9+10+7-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+5
N\left(A\cap B\right)+N\left(A\cap C\right)+N\left(B\cap C\right)=10
N\left(A\cap B\cap C\right)
=N\left(A\cap B+B\cap C+A\cap C\right)-2\ast N\left(A\cap B\cap C\right)=10-2\ast 5=0
A
B
S
S=A\cup B
A\cap B=\mathrm{\varnothing }
\mathcal{P}\left(X\right)
X
|Y|
Y
|\mathcal{P}\left(A\right)|+|\mathcal{P}\left(B\right)|=|\mathcal{P}\left(A\right)\cup \mathcal{P}\left(B\right)|
\mathcal{P}\left(A\right)
\mathcal{P}\left(B\right)
\mathcal{P}\left(X\right)
|Y|
Brody Collins 2022-05-09 Answered
\left(1,2,...,n\right)
k
\left(n-k\right)!
k
\left(\genfrac{}{}{0}{}{n}{k}\right)\left(n-k\right)!=\frac{n!}{k!}
n!
n!-\left(\frac{n!}{1!}+\frac{n!}{2!}-\frac{n!}{3!}+...+\left(-1{\right)}^{n+1}\frac{n!}{n!}\right)
k
n-k
{2}^{k}\ast q
\left(x+y{\right)}^{n}=\sum _{k=0}^{n}C\left(n,k\right)\ast {x}^{n-k}\ast {y}^{k}=\frac{{n}^{2}+n}{2}
{n}_{1},{n}_{2},...,{n}_{t}
{n}_{1}+{n}_{2}+...+{n}_{t}-t+1
i=1,2,3,...,t
{n}_{i}
{26}^{2}×{10}^{4}
{}^{6}{P}_{2}
\left(5+4\right)+\left(4+4\right)+\left(3+3\right)+\left(7+4\right)=9+8+6+11=34
X=\left\{\left(123\right),\left(132\right),\left(124\right),\left(142\right),\left(134\right),\left(143\right),\left(234\right),\left(243\right)\right\}
{A}_{4}
X
x=\left(123\right)
4=|\mathcal{O}\left(x\right)|=|G|/|{G}_{x}|=12/|{G}_{x}|
{G}_{x}=\left\{1\right\}
\frac{24\cdot 1}{24\cdot 23}=\frac{1}{23}
f
A\to B
|A|=4,|B|=3
{3}^{4}-\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}+\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}
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Rubtsov V. Lie algebroid cohomology: from Der to Lie algebroid non-abelian extensions (abstract) - Geometry of Differential Equations
Rubtsov V. Lie algebroid cohomology: from Der to Lie algebroid non-abelian extensions (abstract)
Title: Lie algebroid cohomology: from Der to Lie algebroid non-abelian extensions
After a short historical reminder I consider the extension problem for Lie algebroids over schemes over a field. Given a locally free Lie algebroid
{\displaystyle Q}
over a scheme
{\displaystyle (X,{\mathcal {O}})}
, and a sheaf of finitely generated Lie
{\displaystyle {\mathcal {O}}}
{\displaystyle L}
, we determine the obstruction to the existence of extensions
{\displaystyle 0\to L\to E\to Q\to 0}
, and classify the extensions in terms of a suitable Lie algebroid hypercohomology group.
Slides: RubtsovAMVconf2021slides.pdf
Retrieved from "https://gdeq.org/w/index.php?title=Rubtsov_V._Lie_algebroid_cohomology:_from_Der_to_Lie_algebroid_non-abelian_extensions_(abstract)&oldid=6443"
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Pitchfork bifurcation - Wikipedia
Bifurcation from a system having one fixed point to three fixed points
In bifurcation theory, a field within mathematics, a pitchfork bifurcation is a particular type of local bifurcation where the system transitions from one fixed point to three fixed points. Pitchfork bifurcations, like Hopf bifurcations have two types – supercritical and subcritical.
In continuous dynamical systems described by ODEs—i.e. flows—pitchfork bifurcations occur generically in systems with symmetry.
1 Supercritical case
2 Subcritical case
Supercritical case[edit]
Supercritical case: solid lines represent stable points, while dotted line represents unstable one.
The normal form of the supercritical pitchfork bifurcation is
{\displaystyle {\frac {dx}{dt}}=rx-x^{3}.}
{\displaystyle r<0}
, there is one stable equilibrium at
{\displaystyle x=0}
{\displaystyle r>0}
there is an unstable equilibrium at
{\displaystyle x=0}
, and two stable equilibria at
{\displaystyle x=\pm {\sqrt {r}}}
Subcritical case[edit]
Subcritical case: solid line represents stable point, while dotted lines represent unstable ones.
The normal form for the subcritical case is
{\displaystyle {\frac {dx}{dt}}=rx+x^{3}.}
In this case, for
{\displaystyle r<0}
the equilibrium at
{\displaystyle x=0}
is stable, and there are two unstable equilibria at
{\displaystyle x=\pm {\sqrt {-r}}}
{\displaystyle r>0}
{\displaystyle x=0}
is unstable.
{\displaystyle {\dot {x}}=f(x,r)\,}
described by a one parameter function
{\displaystyle f(x,r)}
{\displaystyle r\in \mathbb {R} }
{\displaystyle -f(x,r)=f(-x,r)\,\,}
(f is an odd function),
{\displaystyle {\begin{aligned}{\frac {\partial f}{\partial x}}(0,r_{0})&=0,&{\frac {\partial ^{2}f}{\partial x^{2}}}(0,r_{0})&=0,&{\frac {\partial ^{3}f}{\partial x^{3}}}(0,r_{0})&\neq 0,\\[5pt]{\frac {\partial f}{\partial r}}(0,r_{0})&=0,&{\frac {\partial ^{2}f}{\partial r\partial x}}(0,r_{0})&\neq 0.\end{aligned}}}
has a pitchfork bifurcation at
{\displaystyle (x,r)=(0,r_{0})}
. The form of the pitchfork is given by the sign of the third derivative:
{\displaystyle {\frac {\partial ^{3}f}{\partial x^{3}}}(0,r_{0}){\begin{cases}<0,&{\text{supercritical}}\\>0,&{\text{subcritical}}\end{cases}}\,\,}
Note that subcritical and supercritical describe the stability of the outer lines of the pitchfork (dashed or solid, respectively) and are not dependent on which direction the pitchfork faces. For example, the negative of the first ODE above,
{\displaystyle {\dot {x}}=x^{3}-rx}
, faces the same direction as the first picture but reverses the stability.
Steven Strogatz, Non-linear Dynamics and Chaos: With applications to Physics, Biology, Chemistry and Engineering, Perseus Books, 2000.
S. Wiggins, Introduction to Applied Nonlinear Dynamical Systems and Chaos, Springer-Verlag, 1990.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Pitchfork_bifurcation&oldid=1052460642"
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1.2.7 CFD12
Numerical simulation of the flow around the wing RAE M2155 have been performed at QINETIQ, for case 1-2-3-4, and at CIRA, only for case 2, with the main aim of validation and assessment of the turbulence models employed.
QINETIQ [3] has used the following turbulence models :
1) A k-ω based variant of the Menter SST scheme
2) A k-ω based EARSM utilising a novel calibration and incorporating explicitly the variable ratio of turbulence production to dissipation rate.
3) A tensorially linear version of the k-ω based EARSM model (only case 3).
while CIRA [4,5,6] has employed the
The aerodynamic coefficients, pressure and friction coefficients, and velocity profiles at several locations are available.
Incidence (deg)
4.1x106 0.744 2.5 CP CN
4.1x106 0.806 2.5 CP, Cf, U CL,CD
Table CFD-B Summary Description of All Available Data Files and Simulated Parameters
DOAPs, or other
✔ac102c12cp02.dat ✔ac102c12cf02.dat ✔ac102c12bla.dat ✔ac102c12doap.dat
✔ac102c12cp03.dat ✔ac102c12cf03.dat ✔ac102c12blb.dat
✔ac102c12cp04.dat ✔ac102c12cf04.dat ✔ac102c12blc.dat
✔ac102c12cp05.dat ✔ac102c12cf05.dat ✔ac102c12bld.dat
✔ac102c12cp06.dat ✔ac102c12cf06.dat
✔ac102c12cp095.dat ✔ac102c12cf095.dat
✔ac102c22cp02.dat ✔ac102c22cf02.dat ✔ac102c22bla1.dat ✔ac102c22doap.dat
✔ac102c22cp03.dat ✔ac102c22cf03.dat ✔ac102c22bla2.dat
✔ac102c22cp08.dat ✔ac102c22cf08.dat ✔ac102c22blb1.dat
✔ac102c22cp095.dat ✔ac102c22blb2.dat
✔ac102c22blb3.dat
✔ac102c22bld1.dat
The code used by QINETIQ for the computations is the BAE SYSTEMS RANSMB. This is a cell centered finite volume code with a central, Jameson type, flux approximation. A multigrid with a four stage Runge-Kutta scheme is used for time stepping.
The turbulence models used in the numerical simulations are :
The computational domain includes the tunnel walls, and the computations have been performed as internal flow calculations.
The meshes used in the computations were generated at Aircraft Research Association and have approximately 1 million cells. The meshes are of constant cross section. The outflow plane is located at x=12m and the inflow plane is located at x=-12m for case 1 and 2, and at x=-2.5m for case 3 and 4. The wing leading edge at the root section is located at x=0. Therefore the entrance section of the mesh in much longer for case 1 and 2. This is due to problems encountered at ARA during the early computations.
The excessive length of the case 1 and 2 has been partly responsible for some case 2 poor results. The mesh has been recently modified at QINETIQ, so that the length of the entrance section of case 1 and 2 is the same as case 3 and 4.
The following boundary conditions have been used :
• Inflow : Velocity and density prescribed, and pressure extrapolated from interior
• Outflow : Pressure prescribed, and velocity and density extrapolated from interior
All the turbulence models used in the computations employ a low Reynolds number formulation. The SST Menter requires the use of a blending function in order to switch from a κ-g to a κ-ε formulation.
The value of y+ at the first grid point is assumed to be O(1) and the mesh is thought to be fine enough to resolve the boundary layer.
• Grid refinement
q Studies were originally conducted at ARA by using a κ-ω model.
q Studies have not performed at QINETIQ due to the high cpu time requirements.
q The solutions show a three to four orders of magnitude reduction of the average density residuals. The typical magnitude of the average density residual at convergence is 10-5
The data consist of :
• Pressure Coefficients at y/b=0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.95
• Friction Coefficients at y/b= 0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.95
• Boundary layer results at points A,B,C and D
• Lift and drag (pressure and friction) coefficients
ac102c12cp02.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns: x/c, CP)
ac102c12cp095.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns: x/c, CP)
ac102c12cf02.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns: x/c, Cf)
ac102c12cf095.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns: x/c, Cf)
ac102c12bla.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns: Z/C, U/UP, P/H)
ac102c12blb.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns: Z/C, U/UP, P/H)
ac102c12blc.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns: Z/C, U/UP, P/H)
ac102c12bld.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns: Z/C, U/UP, P/H)
ac102c12doap.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns:CL, CD( pressure), CD(friction))
[3] D.A. Peshkin. Technical Report: DERA/MSS/MSFC1/TR000228. Strictly limited circulation.
CIRA has used the RANS flow solver ZEN version 6.0
The RANS equations are discretized by means of a cell centered finite volume scheme with blended self adaptive second and fourth order artificial dissipation. The solution procedure is based on a time marching concept. The multigrid scheme is used to accelerate the convergence of the solution, and performs relaxations, by using the Runge Kutta algorithm with local time stepping and residual averaging, on different grid levels. The turbulence equations are uncoupled by the RANS equations and are solved, inside a multigrid cycle, only on the finest grid level.
The turbulence models employed in the numerical simulations are the following :
The computational domain includes the wind tunnel walls, and the simulations have been performed as internal flow calculations. Therefore wall tunnel corrections are not necessary.
The inflow plane is located at x=-4m and the outflow plane at x=4.5m. The wing leading edge at the root section is located at x=0.
The mesh used has 36 blocks with 1.2 million point and can be run on 3 different levels.
1) Inflow : General free-stream with specified enthalpy and momentum, and the pressure extrapolated
2) Outflow : General free-stream with specified pressure and extrapolated velocity and enthalpy
3) Tunnel walls : Slip
4) Wing surfaces : Adiabatic no slip with fixed transition
The tunnel walls boundary layer has shown to have an effect up to 30% wing span.
{\displaystyle {{\sqrt {\kappa \over u_{\infty }}}=0.1}}
%,
{\displaystyle {\mu \over \mu _{r}}=0.1}
At the inflow the turbulent variables are assumed to be constant, and the free stream values are derived by :
The Wilcox and TNT κ-ω turbulence models do not have a low Reynolds formulation, while the Spalart Allmaras and the κ-ε models employ the use of wall damping functions. The SST Menter κ-ω model, in order to switch from a κ-ω to a κ-ε formulation close to a solid boundary and to take into account the effect of the transport of the principal shear stress (SST formulation), requires the computation of blending functions.
The values of y+ at the first cell center is O(1)
The computations have been performed on 3 different grid levels.
For the κ-ω SST Menter turbulence model, for example, the following accuracy has been obtained :
• Grid sensitivity
q The lift coefficient changes less than 1% going from the 1st to the 2nd grid level, and less than 0.5% from the 2nd to the finest level.
q The drag coefficient changes of about 15% going from the 1st to the 2nd grid level, and less than 5% from the 2nd to the finest level.
• Convergence on the finest level
q In the last 500 iterations, the lift coefficient changes less than 0.02% and drag coefficient less than 0.03%.
• Friction coefficients at y/b= 0.2,0.3,0.4,0.5,0.6,0.7,0.8
• Velocity profiles at the stations A,B, and D
ac102c22bla1.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence, turbulence model; columns: Z/C, U/UP)
ac102c22blb1.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence, turbulence model; columns: Z/C, U/UP)
ac102c22bld1.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence, turbulence model; columns: Z/C, U/UP)
ac102c22doap.dat (ASCII file; headers: Mach number, Reynolds number, angle of incidence; columns:Turbulence model, CL, CD( pressure), CD(friction))
[4] M. Amato, P. Catalano, “An Evaluation of Stress-Strain Relationships for Aeronautical Applications”, XV AIDAA National Congress
[5] M. Amato, P.Catalano, “Non Linear κ-ε Turbulence Modeling for Industrial Applications”, ICAS 2000 Conference
[6] P. Catalano, M. Amato, “Assessment of κ-ω Turbulence modeling in the CIRA Flow Solver ZEN”, ECCOMAS CFD 2001 Conference
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Proving the generator of A = \{154a + 210b : a,b
A=\left\{154a+210b:a,b\in \mathbb{Z}\right\}
\left(154,\text{ }210\right)
For the further proof , you just need to show that
154a+210b
just generates all the common multiples of 154 and 210 for different values of a and b where
a,\text{ }b\in \mathbb{Z}
and even HCF can also be written in in the form
154a+210b
for a particular choice of a and b . And HCF obviously generate all the common multiples then.
\frac{R\left[\alpha \right]}{\left(2{\alpha }^{2}\right)}
Group theory exercise from Judson text
S=\frac{R}{-1}
and define a binary operation on S by
a×b=a+b+ab
\left(S,×\right)
f:R\to r
{X}^{2}+{Y}^{2}=0
f\left({X}^{2}+{Y}^{2}\right)=f\left(0\right),f\left({X}^{2}\right)+f\left({Y}^{2}\right)=0,{f\left(X\right)}^{2}+{f\left(Y\right)}^{2}=0,{x}^{2}+{y}^{2}=0
{x}^{2}+x+41
0\le x\le 39
\mathbb{Z}+\left(3x\right)
\mathbb{Z}\left[x\right]
\mathbb{Z}\left[x\right]⇒\mathbb{Z}+\left(3x\right)
U\left(14\right)
S3
U\left(14\right)
U\left(7\right)
U\left(8\right)
U\left(12\right)
|
ca foundation mock test in indices in Business Mathematics | Online Exam
Home Quantitative Aptitude ca foundation mock test-indices
ca foundation mock test in indices in Business Mathematics
If A = B/2 = C/5, then A : B : C is
If a/3 = b/4 = c/7, then a+b+c/c is
If p/q = r/s = 2.5/1.5, the value of ps:qr is
If x : y = z : w = 2.5 : 1.5, the value of (x+z)/(y+w) is
If (5x–3y)/(5y–3x) = 3/4, the value of x : y is
If A : B = 3 : 2 and B : C = 3 : 5, then A:B:C is
If x/2 = y/3 = z/7, then the value of (2x–5y+4z)/2y is
If x : y = 2 : 3, y : z = 4 : 3 then x : y : z is
Division of Rs. 750 into 3 parts in the ratio 4 : 5 : 6 is
The sum of the ages of 3 persons is 150 years. 10 years ago their ages were in the ratio
7 : 8 : 9. Their present ages are
If x/y = z/w, implies y/x = w/z, then the process is called
If p/q = r/s = p–r/q–s, the process is called
If a/b = c/d, implies (a+b)/(a–b) = (c+d)/(c–d), the process is called
If u/v = w/p, then (u–v)/(u+v) = (w–p)/(w+p). The process is called
12, 16, *, 20 are in proportion. Then * is
4, *, 9, 131⁄2 are in proportion. Then * is
The mean proportional between 1.4 gms and 5.6 gms is
\mathrm{If} \frac{\mathrm{a}}{4}=\frac{\mathrm{b} }{5}=\frac{\mathrm{c}}{9} \mathrm{Then} \mathrm{a}+\mathrm{b}+\mathrm{c}/\mathrm{c} \mathrm{is}
Two numbers are in the ratio 3 : 4; if 6 be added to each terms of the ratio, then the new ratio will be 4 : 5, then the numbers are
log 6 + log 5 is expressed as
If 2 log x = 4 log 3, the x is equal to
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Directions : In each of these questions a number series is given. In each series only one number is wrong. Find out the wrong number.
The series is 50 + 1^2 = 51, 51 – 2^2 = 47, 47 + 3^2 = 56, 56 – 4^2 = 40, 40 + 5^2 = 65, 65 –
6^2 = 29.
Hence, there should be 40 in place of 42.
The series is 3 × 2 + 3 = 9, 9 × 3 – 4 = 23, 23 × 4 + 5 = 97, 97 × 5 – 6 = 479, 479 × 6 + 7 =
2881, 2881 × 7 – 8 = 20159
The series is x0.5 + 0.5, x1 + 1, x 1.5 + 1.5, x 2 + 2, x 2.5 + 2.5, x 3 + 3…
Hence, there should be 5 in place of 6.
The series is 1 × 2 + 1 = 3, 3 × 2 + 0 = 6, 6 × 2 – 1 = 11, 11 × 2 – 2 = 20, 20 × 2 – 3 = 37, 37× 2 – 4 = 70.
The series is 2 × 2 + 7 = 11, 11 × 3 – 6 = 27, 27 × 4 + 5, = 113, 113 × 5 – 4 = 561, 561 × 6 +3 = 3369, 3369 × 7 – 2 = 23581.
The series is 5 × 1 + 2 = 7, 6 × 2 + 4 = 16, 7 × 3 + 6 = 27, 8 × 4 + 8 = 40, 9 × 5 + 10 = 55.
Alternate Method: +9, +11, +13, +15
The series is 9^3, 11^3, 13^3, 15^3, 17^3,
Hence, there should be 2197 in place of 2497.
The series is 9^2 – 1, 11^2 – 2, 13^2 – 3, 15^2 – 4, 17^2 – 5,
Hence, there should be 284 in place of 223.
The series is 8 + 1.5 = 9.5, 9.5 + 2 = 11.5, 11.5 + 2.5 = 14, 14 + 3 = 17
Hence, there should be 9.5 in place of 8.5.
The series is +339, +678, +1356, +2712,
17, 36, 132, 635, 3500, 21750, 153762
The series is (17 + 1^3) × 2, (36 + 2^3) × 3, (132 + 3^3) × 4, (636 + 4^3) × 5
The number series should be 636 in the place of 635.
The series is ×1 + 3, ×2 + 6, ×3 + 9, ×4 + 12, ×5 + 15
The series is (90-45) × 3, (135-40) × 3, (285-35) × 3, (750-30) × 3, (2160-25) × 3,…
The series is ×1 + 5, ×2 + 10, ×3 + 15, ×4 + 20, ×5 + 25
The number series should be 38 in the place of 40.
The series is (8+1) × 2, (18+3) × 3, (63+5) × 4, (272+7) × 5
The series is,
38 = 3+8 = 11 = 38 +11 = 49
62 = 6+2 = 8 = 62 + 8 = 70 ≠ 72
70 = 7+0 = 7 = 70+ 7 = 77
77 = 7+7 = 14 = 77+14 = 91
91 = 9+1 = 10 = 91+10 = 101
Hence, 72 is the wrong number.
First series 47, 45, 33, 3
47-(1×2) = 45
33- (5×6) = 3
Second series 44, 46, 57, 88
46 + (3×4) = 58 ≠ 57
Hence, 57 is the wrong answer.
1 8 66 460 2758 13785 55146
Here 1 × 9 – 1 = 8; 8 × 8 + 2 = 66; 66 × 7 – 3 = 459;
459 × 6 + 4 = 2758; 2758 × 5 – 5 = 13785; 13785
× 4 + 6 = 55146
56, 57, 48, 73, 24, 105, -10
Here 56 +1^2 = 57;
57 – 3^2 = 48; 48+ 5^2 = 73; 73- 7^2 = 24;
24 + 9^2 = 105; 105 -11^2 = -16
Here 2 × 3 – 4 = 2; 2 × 4 + 5 = 13;
13 × 5 – 6 = 59; 59 × 6 + 7 = 361;
361 × 7 – 8 = 2519; 2519 × 8 + 9 = 20161
3 × 1/3 = 1;
3 × 1/4 = 0.75;
0.75 × 4 = 3;
3 × 1/5 = 0.6;
0.6 × 5 = 3;
0.5 × 6 = 3.
Here 2 × 2 + 2 = 6; 6 × 2 + 1 = 13;
13 × 2 + 0 = 26; 26 × 2 – 1 = 51;
51 × 2 – 2 = 100; 100 × 2 – 3 = 197
{0}^{2} {1}^{2} {3}^{2} {6}^{2} {10}^{2 } {15}^{2} {21}^{2}
{0}^{2} {1}^{2} {3}^{2} {6}^{2} {10}^{2 } {15}^{2} {21}^{2}
The series is an alternate series, having
S 1 = 2 5 14 41; ×3 – 1 in each term
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Work | Brilliant Math & Science Wiki
July Thomas, Sravanth C., Anish Harsha, and
Work is done when a force is applied, at least partially, in the direction of the displacement of the object. If that force is constant then the work done by the force is the dot product of the force with the displacement:
W=\vec{F} \bullet \vec{d}.
The net work done on an object (the work of the net force) is equal to the energy added to the object. This is the reason for both the unit of work being the Joule,
\text{J}
, and the work-kinetic energy theorem:
W=\Delta K.
Anne spends most of her day napping on a futon. In the reference frame of the Earth, what is the work done to Anne by the futon in one hour? A futon is a couch that can also be folded flat to serve as a bed.
In the reference frame of the Earth, Anne is sitting still, so her displacement is 0
\text{m}
in any amount of time. Hence the work done in one day is
W=\vec{F} \bullet \vec{d} = \vec{F} \bullet \vec{0} = 0 \text{ J}.
Newton's laws provide a method for turning force into acceleration,
F_{net}=ma,
but this only provides the acceleration at an instant. In order to affect the motion of an object, the force must be applied over some time (impulse) or over a distance (work). For an object with displacement,
\vec{d}
, only in the x-direction being acted on only by a force in the x direction, the work is simply defined in terms of the magnitudes as
W_x=F_x d_x.
If there are also some y and z components to the force, the work of each of those forces is
W_y=F_y d_y
W_z=F_z d_z.
The net work on an object moving through space is just the sum of these individual components.
W_{net} = W_x + W_y + W_z
W_{net} = F_x d_x + F_y d_y + F_z d_z
Since the right side is the definition of the dot product, the left side must be as well.
\vec{F}\bullet\vec{d} = F_x d_x + F_y d_y + F_z d_z
This dot-product realization also produces another valuable form of the work definition.
\vec{F}\bullet\vec{d} = Fd \cos(\theta)
\vec{F} = (3 \text{ N})\hat{x} + (4 \text{ N})\hat{y}
causes a block to undergo a displacement
\vec{d} = (2 \text{ m})\hat{y}.
Find the work done.
Since the force and displacement are given in terms of their components, use the component-wise definition of work.
\vec{F}\bullet\vec{d} = F_x d_x + F_y d_y + F_z d_z = (3 \text{ N})(0 \text{ m}) + (4 \text{ N})(2 \text{ m}) +(0 \text{ N})(0 \text{ m}) = 8 \text{ J}
The Earth takes 1 year to complete its orbit around the Sun. Treating the orbit as perfectly circular, find the net work done on the Earth by the Sun in
\frac14
of a year. Give your answer to three significant figures.
Mass of the Earth =
5.97\times 10^{24} \text{kg}
Distance from Sun to Earth =
1.50\times 10^{11} \text{m}
1 Earth year = 365 Earth days
1 Earth day = 24 hours
Combining Newton's second law,
F_{net} = ma,
with the definition of work in one-dimension gives
W_x=F_x d_x = ma_x d_x.
a_x d_x
shows up in one dimensional kinematics.
2 a_x d_x = v_f^2 - v_i^2
Divide both sides by 2 to solve for the desired product on the left.
a_x d_x = \frac12 v_f^2 - \frac12 v_i^2
Finally, multiply both sides by mass to get work on the left.
ma_x d_x = \frac12 mv_f^2 - \frac12 mv_i^2
Notice, while the left is the definition of work, the right also happens to be the change in kinetic energy.
W_{net} = \Delta K
Combined with the conservation of energy,
\Delta K = -\Delta U,
provides the following relation.
Work of a conservative force
W_{net} = \Delta K = -\Delta U
The kinetic energy of a car decreases from
2.0 \times 10^6 \text{ J}
1.0 \times 10^6 \text{ J}
over a distance of 10
\text{m}
by a braking force. Find the magnitude of that braking force.
W_{net} = \Delta K
Fd = K_f - K_i
F(10\text{ m}) = 1.0 \times 10^6\text{ J} - 2.0 \times 10^6\text{ J}
10F = -10^6
F = -10^5 \text{ N}
The negative sign indicates the force is antiparallel to the displacement.
Find the mass, in
\text{kg},
of a cat whose gravitational potential energy decreases from 300
\text{J}
\text{J}
over 2
\text{m}.
g = 10 \frac{\text{m}}{\text{s}^2}.
Cite as: Work. Brilliant.org. Retrieved from https://brilliant.org/wiki/calculating-work-done-by-a-constant-force/
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If y=\arctan(\frac{2x−1}{1+x−x^2}), then \frac{dy}{dx} at x=1 is equal to? I have
stropa0u 2022-01-24 Answered
y=\mathrm{arctan}\left(\frac{2x-1}{1+x-{x}^{2}}\right)
\frac{dy}{dx}
bekiffen32
In the derivative of
y=\mathrm{arctan}\left(x\right)-\mathrm{arctan}\left(1-x\right)+k\pi
\frac{dy}{dx}=\frac{1}{1+{x}^{2}}-\left(-\frac{1}{1+{\left(1-x\right)}^{2}}\right)
which is the same as in the second approach.
Also note that the inverse tangent is an odd function, so that
\mathrm{arctan}\left(1-x\right)=-\mathrm{arctan}\left(x-1\right)
, so that there is no structural difference between the two approaches
Fallbasisz8
A computationally simpler method of directly calculating the derivative at x=1 is to write
\mathrm{tan}y=\frac{2x-1}{1+x-{x}^{2}}
\mathrm{log}\mathrm{tan}y=\mathrm{log}\left(2x-1\right)-\mathrm{log}\left(1+x-{x}^{2}\right)
Then implicit differentiation yields
\frac{{\mathrm{sec}}^{2}y}{\mathrm{tan}y}\frac{dy}{dx}=\frac{2}{2x-1}-\frac{1-2x}{1+x-{x}^{2}}
Since, when x=1, we have
\mathrm{tan}y=\frac{2-1}{1+1-1}=1
, it follows from the trigonometric identity
{\mathrm{sec}}^{2}y=1+{\mathrm{tan}}^{2}y\text{ }\text{ that }\text{ }{\mathrm{sec}}^{2}y=1+{1}^{2}=2
2{\left[\frac{dy}{dx}\right]}_{x=1}=\frac{2}{2-1}-\frac{1-2}{1+1-1}=3
hence the answer is 3/2. Note this solution uses two tactics; logarithmic differentiation and implicit differentiation, to make the calculation extremely simple.
x=\frac{{y}^{4}}{8}+\frac{1}{4{y}^{2}}\phantom{\rule{1em}{0ex}}1\le y\le 2
4\mathrm{sin}x+2=0.\left[0,2\pi \right)
\theta
\left(\frac{5}{13},\frac{12}{13}\right)
\mathrm{sin}\left(\theta \right)
{0}^{\circ }\le x\le {360}^{\circ }
{\mathrm{cos}}^{2}\left(2x\right)+\sqrt{3}\mathrm{sin}\left(2x\right)-\frac{74}{=}0
\mathrm{cos}\left(\mathrm{arctan}\left(x\right)\right)
\mathrm{sin}\left(2{\mathrm{cos}}^{-1}×\left(\frac{\sqrt{2}}{2}\right)\right)
\frac{3\pi }{4}
|
Kunshan First People’s Hospital, Kunshan, China.
Abstract: Background: To explore the therapeutic effect of Tongxinluo capsule (Tongxinluo) on patients with Syndrome X and Affective Disorder. Methods: Fifty-six patients with Syndrome X and Affective Disorder were randomly divided into a Tongxinluo capsule group and a placebo control group. The duration of treatment was 12 weeks. A 6-minute walking test, exercise load electrocardiogram and clinical symptom assessment were performed before and after treatment. After 12 weeks of treatment, the scores of the Self-Rating Anxiety Scale (SAS) and Self-Rating Depression Scale (SDS) were repeated. The levels of serum endothelin-1 (ET-1) and nitric oxide (NO) were measured before and after treatment. Results: Compared with the placebo control group, the Tongxinluo group SAS and SDS scores were lower than those before treatment (all P < 0.01), and the 6-minute walking distance increased significantly (P < 0.01). Clinical symptoms were significantly improved. The exercise test results suggested that, while improved, there was no significant difference (P > 0.05) when compared to before treatment. In the Tongxinluo treatment group, the levels of plasma endothelin-1 decreased significantly (P < 0.01) and nitric oxide levels were significantly increased (P < 0.01), with a significant difference when compared to the control group (P < 0.01). Conclusions: The Tongxinluo capsule can improve the Affective Disorder of Syndrome X, reduce the degree of anxiety and depression, increase exercise tolerance, reduce clinical symptoms, and improve vascular endothelial function.
Keywords: Syndrome X, Affective Disorder, 6-Minute Walking Test
\stackrel{¯}{x}
Cite this paper: Lv, X. , Zhao, Z. , Feng, J. and Zhu, J. (2018) Effect of Tongxinluo Capsule on Patients with Syndrome X and Affective Disorder. Chinese Medicine, 9, 55-62. doi: 10.4236/cm.2018.92004.
[1] Hu, G.L. and Bai, X.P. (2013) Progress in the Study of Coronary Microcirculation Disorder. Chinese Journal of Microcirculation, 23, 60-65.
[2] Hu, G.L., Bai, X.P. and Hou, X.L. (2014) Clinical Efficacy of Danhong Injection on Cardiac Syndrome X. Chinese Journal of Evidence-Based Cardiovascular Medicine, 6, 558-561.
[3] Asbury, E.A., Creed, F. and Collins, P. (2004) Distinct Psychosocial Differences between Women with Coronary Heart Disease and Cardiac Syndrome X. European Heart Journal, 25, 1695-1701.
[4] Vermeltfoort, I.A., Raijmakers, P.G., Odekerken, D.A., Kuijper, A., Zwijnenburg, A. and Teule, G. (2009) Association between Anxiety Disorder and the Extent of Ischemia Observed in Cardiac Syndrome X. Journal of Nuclear Cardiology, 16, 405-410.
[5] Wang, C. and Wang, J.A. (2015) Coronary Atherosclerotic Heart Disease. Internal Medicine Journal, 339.
[6] Lv, X.L., Zhao, Z.Q., Feng, Z.Q., et al. (2014) Effect of Tongxinluo Capsule on Vascular Endothelial and Inflammatory Factors in Patients with Cardiac X Syndrome. Journal of Clinical Medicine in Practice, 18, 135.
[7] Cemin, R., Erlicher, A., Fattor, B., Pitscheider, W. and Cevese, A. (2008) Reduced Coronary Flow Reserve and Parasympathetic Dysfunction in Patients with Cardiovascular Syndrome X. Coronary Artery Disease, 19, 1-7.
https://doi.org/10.1097/MCA.0b013e3282f18e8d
[8] Khawaja, I.S., Westermeyer, J.J., Gajwani, P. and Feinstein, R.E. (2009) Depression and Coronary Artery Disease: The Association, Mechanisms, and Therapeutic Implications. Psychiatry (Edgmont), 6, 38-51.
[9] Brunner, E.J., Shipley, M.J., Britton, A.R., Stansfeld, S.A., Heuschmann, P.U., Rudd, A.G., et al. (2014) Depressive Disorder, Coronary Heart Disease, and Stroke: Dose-Response and Reverse Causation Effects in the Whitehall II Cohort Study. European Journal of Preventive Cardiology, 21, 340-346.
[10] Liu, M.Y., Jiang, R.H., Hu, D.Y., et al. (2009) Emotional Disorder in Patients with Acute or Stable Coronary Heart Disease. Journal of Clinical and Experimental Medicine, 37, 904-907.
[11] Buffon, A., Rigattieri, S., Santini, S.A., Ramazzotti, V., Crea, F., Giardina, B., et al. (2000) Myocardial Ischemia-Reperfusion Damage after Pacing-Induced Tachycardia in Patients with Cardiac Syndrome X. American Journal of Physiology-Heart and Circulatory Physiology, 279, H2627-H2633.
[12] Liu, X.C., Li, S.J., Zhang, X.Y., et al. (2013) Effect of Tongxinluo Capsule on Vascular Endothelial Function, Brachial-Ankle and Pulse Wave Velocity in Stable Angina. Chinese Journal of Integrative Medicine on Cardio/Cerebrovascular Disease, 11, 408.
[13] Dong, S. and Dong, Y. (2012) Effect of Tongxinluo Capsule on Inflammatory Factors and Vascular Endothelial Function after Percutaneous Coronary Intervention in Patients with Acute Myocardial Infarction. Chinese Journal of Clinician, 6, 137-139.
[14] Liu, X.C., Chen, X.L., Zhang, Z.X., et al. (2012) Effect of Tongxinluo Capsule on the Concentration of Vascular Endothelial Growth Factor and Endothelial Progenitor cells in Coronary Heart Disease. Chinese Journal of Integrative Medicine on Cardio/Cerebrovascular Disease, 10, 551-552
[15] Guo, J. and Yin, X.G. (2016) Effects of Suanzaoren Decoction on Learning and Memory Ability and Brain Neurotransmitters Content of Senile Insomnia Model Rats. China Pharmacy, 27, 3085-3087.
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A geometric representation, showing the relationship between a whole and its part, is from Class 12 TET Previous Year Board Papers | Mathematics 2018 Solved Board Papers
\sqrt{784}
\sqrt{\overline{2\times 2}\times \overline{2\times 2}\times \overline{7\times 7}}
2\times 2\times 7
28
{\left(\frac{5}{7}\right)}^{4}\times {\left(\frac{5}{7}\right)}^{-3}={\left(\frac{5}{7}\right)}^{5x-2}
\frac{2}{5}
\frac{3}{5}
\frac{4}{5}
\frac{1}{5}
\frac{3}{5}
{\left(\frac{5}{7}\right)}^{4}
\times
{\left(\frac{5}{7}\right)}^{-3}
{\left(\frac{5}{7}\right)}^{5x-2}
\Rightarrow
{\left(\frac{5}{7}\right)}^{4-3}
{\left(\frac{5}{7}\right)}^{5x-2}
\quad \quad \left[\because {a}^{m}\times {a}^{n}={a}^{m+n}\right]\quad \quad
1=5x-2
\Rightarrow
5x=3
\Rightarrow
x=\frac{3}{5}
\frac{3}{7}+\frac{\left(-7\right)}{8}=\frac{25}{56}
\frac{4}{5}
\frac{2}{5}
\frac{1}{5}
\frac{3}{5}
\frac{3}{5}
\frac{n\left(E\right)}{n\left(S\right)}=\frac{6}{10}=\frac{3}{5}
\frac{3}{5}
\frac{2}{3}
\frac{-5}{6}
a\div \left(b+c\right)=b\div \left(a+c\right)
a+\left(b+c\right)\quad =c\quad +(a+b)
a-\left(b-c\right)=c-\left(a-b\right)
a\times \left(b+c\right)=b\times \left(a+c\right)
a+\left(b+c\right)\quad =c\quad +(a+b)
\Rightarrow
\frac{3}{5}+\left(\frac{2}{3}-\frac{5}{6}\right)=\frac{-5}{6}+\left(\frac{3}{5}+\frac{2}{3}\right)
\Rightarrow
\frac{3}{5}+\left(\frac{4-5}{6}\right)=\frac{-5}{6}+\left(\frac{9+10}{15}\right)
\Rightarrow
\frac{3}{6}-\frac{1}{6}=\frac{-5}{6}+\frac{19}{15}
\Rightarrow
\frac{18-5}{30}=\frac{-25+38}{30}
\Rightarrow
13=13
\sqrt{q}
\sqrt{91+\sqrt{70+\sqrt{121}}}
\sqrt{91+\sqrt{70+\sqrt{121}}}
\sqrt{91+\sqrt{70+11}}
\left[\because \quad \sqrt{121}\quad =11\right]
\sqrt{91+\sqrt{81}}
\sqrt{91+9}
\left[\because \sqrt{81\quad }\quad =\quad 9\quad \right]\phantom{\rule{0ex}{0ex}}
\sqrt{91+9}
\sqrt{100}
10
\left[\because \sqrt{100}\quad =10\quad \right]\phantom{\rule{0ex}{0ex}}
https://www.zigya.com/share/VE1BRU5UVDEyMTkwNTky
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xvYCC - Wikipedia
Sony's x.v.Color logo
xvYCC or extended-gamut YCbCr is a color space that can be used in the video electronics of television sets to support a gamut 1.8 times as large as that of the sRGB color space.[1][2][3] xvYCC was proposed by Sony,[4] specified by the IEC in October 2005 and published in January 2006 as IEC 61966-2-4. xvYCC extends the ITU-R BT.709 tone curve by defining over-ranged values. xvYCC-encoded video retains the same color primaries and white point as BT.709, and uses either a BT.601 or BT.709 RGB-to-YCC conversion matrix and encoding.[4] This allows it to travel through existing digital limited range YCC data paths, and any colors within the normal gamut will be compatible.[4] It works by allowing negative RGB inputs and expanding the output chroma. These are used to encode more saturated colors by using a greater part of the RGB values that can be encoded in the YCbCr signal compared with those used in Broadcast Safe Level.[4] The extra-gamut colors can then be displayed by a device whose underlying technology is not limited by the standard primaries.[4]
In a paper published by Society for Information Display in 2006, the authors mapped the 769 colors in the Munsell Color Cascade (so called Michael Pointer's gamut) to the BT.709 space and to the xvYCC space. About 55% of the Munsell colors could be mapped to the sRGB gamut, but 100% of those colors map to within the xvYCC gamut.[5] Deeper hues can be created – for example a deeper cyan by giving the opposing primary (red) a negative coefficient. The quantization range of the xvYCC601 and xvYCC709 colorimetries is always Limited Range.[6]
Camera and display technology is evolving with more distinct primaries, spaced farther apart per the CIE chromaticity diagram. Displays with more separated primaries permit a larger gamut of displayable colors, however, color data needs to be available to make use of the larger gamut color space. xvYCC is an extended gamut color space that is backwards compatible with the existing BT.709 YCbCr broadcast signal by making use of otherwise unused data portions of the signal.
The BT.709 YCbCr signal has unused code space, a limitation imposed for broadcasting purposes. In particular only 16-240 is used for the color Cb/Cr channels out of the 0-255 digital values available for 8 bit data encoding. xvYCC makes use of this portion of the signal to store extended gamut color data by using code values 1-15 and 241-254 in the Cb/Cr channels for gamut-extension. [7]
xvYCC expands the chroma values to 1-254 while keeping the luma (Y) value range at 16-235 (though Superwhite may be supported), the same as Rec. 709. First the OETF (TransferCharacteristics 11 per H.273[8] as originally specified by the first amendment to H.264) is expanded to allow negative R'G'B' inputs such that:[5]
{\displaystyle V={\begin{cases}-1.099(-L)^{0.45}+0.099&L\leq -0.018\\4.500L&-0.018<L<0.018\\1.099L^{0.45}-0.099&L\geq 0.018\end{cases}}}
Here 1.099 number has the value 1 + 5.5 * β = 1.099296826809442... and β has the value 0.018053968510807..., while 0.099 is 1.099 - 1.[8]
The YCC encoding matrix is unchanged, and can follow either Rec. 709 or Rec. 601 (MatrixCoefficients 1 and 5).[5]
The possible range for non-linear R’G’B’601 is between -1.0732 and 2.0835 and for R’G’B’709 is between -1.1206 and 2.1305.[9] That is achieved when YCC values are "1, 1, any" and "254, 254, any" in B' component.
xvYCC709 covers 37.19% of CIE 1976 u'v', while BT.709 only 33.24%.[10]
The last step encodes the values to a binary number (quantization). It is basically unchanged, except that a bit-depth n of more than 8 bits can be selected:[5]
{\displaystyle {\begin{aligned}Y_{{\rm {xv}}\ n}&=\left\lfloor 2^{n-8}(219\times Y+16)\right\rceil \\Cb_{{\rm {xv}}\ n}&=\left\lfloor 2^{n-8}(224\times Cb+128)\right\rceil \\Cr_{{\rm {xv}}\ n}&=\left\lfloor 2^{n-8}(224\times Cr+128)\right\rceil \\\end{aligned}}}
With negative primary amounts allowed, a cyan that lies outside the basic gamut of the primaries can be encoded as "green plus blue minus red".[4] Since the 16-255 Y range is used (255 value is reserved in HDMI standard for synchronization but may be in files) and since the values of Cb and Cr are only little restricted, a lot of high saturated colors outside the 0–255 RGB space can be encoded. For example, if YCbCr is 255, 128, 128, in the case of a full level YCbCr encoding (0–255), then the corresponding R'G'B' is 255, 255, 255 which is the maximum encodable luminance value in this color space. But if Y=255 and Cr and/or Cb are not 128, this codes for the maximum luminance but with an added color: one primary must necessarily be above 255 and cannot be converted to R'G'B'. Adapted software and hardware must be used during production to not clip the video data levels that are above the sRGB space. This is almost never the case for software working with an RGB core.
The more complex example is YCbCr BT.709 values 139, 151, 24 (that is RGB -21, 182, 181). That is out-of-gamut for BT.709, but is not for sYCC and xvYCC709, and to convert those values to display gamut you would convert to XYZ (0.27018, 0.40327, 0.54109) and then to display gamut.[11]
The XYZ matrix is as specified in Nvidia docs.[12]
A mechanism for signaling xvYCC support and transmitting the gamut boundary definition for xvYCC has been defined in the HDMI 1.3 Specification. No new mechanism is required for transmitting the xvYCC data itself, as it is compatible with HDMI's existing YCbCr formats, but the display needs to signal its readiness to accept the extra-gamut xvYCC values (in Colorimetry block of EDID, flags xvYCC709 and xvYCC601), and the source needs to signal the actual gamut in use in AVI InfoFrame and use gamut metadata packets to help the display to intelligently adapt extreme colors to its own gamut limitations.
This should not be confused with HDMI 1.3's other new color feature, deep color. This is a separate feature that increases the precision of brightness and color information, and is independent of xvYCC.
xvYCC is not supported by DVD-Video but is supported by the high-definition recording format AVCHD and PlayStation 3 and Blu-ray. It is also supported by some cameras, like Sony HDR-CX405, that does actually tag the video as xvYCC with BT.709 inside Sony's XAVC.[13]
On January 7, 2013, Sony announced that it would release "Mastered in 4K" Blu-ray Disc titles which are sourced at 4K and encoded at 1080p.[14] "Mastered in 4K" 1080p Blu-ray Disc titles can be played on existing Blu-ray Disc players and will support a larger color space using xvYCC.[14][15][16]
On May 30, 2013, Eye IO announced that their encoding technology was licensed by Sony Pictures Entertainment to deliver 4K Ultra HD video with their "Sony 4K Video Unlimited Service".[17][18] Eye IO encodes their video assets at 3840 x 2160 and includes support for the xvYCC color space.[17][18]
The following graphics hardware support xvYCC color space when connected to a display device supporting xvYCC:
AMD Mobility Radeon HD 4000 series and newer models
AMD Radeon HD 5000 series and newer models
AMD 785G, 880G and 890GX chipsets with integrated graphics
Intel HD Graphics integrated on some CPUs (except Pentium G6950 and Celeron G1101)
nVidia GeForce 200 series and newer models
^ "HDMI 1.3 Update" (PDF). HDMI Licensing. 2006. Retrieved 2006-08-30.
^ "新動画用拡張色空間 xvYCC(IEC61966-2-4)" (PDF).
^ "Color gamuts of Rec709 and xvYCC". res18h39.netlify.app. Retrieved 2021-01-29.
^ a b c d e f "xvYCC". Sony Global. Archived from the original on August 29, 2009. Retrieved 2009-08-13.
^ a b c d Tatsuhiko Matsumoto; Yoshihide Shimpuku; Takehiro Nakatsue; Shuichi Haga; Hiroaki Eto; Yoshiyuki Akiyama & Naoya Katoh (2006). 19.2: xvYCC: A New Standard for Video Systems using Extended-Gamut YCC Color Space. SID INTERNATIONAL SYMPOSIUM. Society for Information Display. pp. 1130–1133. doi:10.1889/1.2433175.
^ "A DTV Profile for Uncompressed High Speed Digital Interfaces (ANSI/CTA-861-H)". Consumer Technology Association®. p. 44. Retrieved 2021-03-11. {{cite web}}: CS1 maint: url-status (link)
^ Naoya Katoh (2007). "New" Extended-gamut Color Space for Video Applications; xvYCC (IEC61966-2-4) (PDF). Retrieved 2009-08-13.
^ a b "H.273: Coding-independent code points for video signal type identification". www.itu.int. Retrieved 2020-12-25.
^ "IEC 61966-2-4:2006 | IEC Webstore". webstore.iec.ch. Retrieved 2021-01-29.
^ Xu Yan; Li Yan; Li Guiling (May 2009). "A kind of nonlinear quantization method to extend the color gamut of DTV system". 2009 IEEE 13th International Symposium on Consumer Electronics: 141–143. doi:10.1109/ISCE.2009.5156953.
^ "Color Calculator". res18h39.netlify.app. Retrieved 25 January 2021.
^ "NVIDIA 2D Image And Signal Performance Primitives (NPP): RGBToXYZ". docs.nvidia.com. Retrieved 2021-01-29.
^ "XAVC-S: How to export time/date metadata to subtitles (srt)?". www.videohelp.com.
^ a b Richard Lawler (2013-01-07). "Sony to launch 4K digital distribution network this summer, 'mastered in 4K' Blu-ray discs". Engadget. Retrieved 2013-05-30.
^ Seamus Byrne (2013-05-01). "Sony 'mastered in 4K' Blu-rays a mixed blessing". CNET. Retrieved 2013-05-30.
^ "What is Mastered in 4K and does it make a difference?". Trusted Reviews. 2014-03-10. Retrieved 2020-12-24.
^ a b "eyeIO Delivers Unprecedented Viewing Experience for Sony Pictures Content on Sony 4K UltraHD TVs". Eye IO, LLC. 2013-05-30. Retrieved 2013-06-05.
^ a b Todd Sprangler (2013-05-30). "Sony Gears Up for 4K Ultra HD Internet Movie Service". Variety. Retrieved 2013-06-05.
IEC Web Store for IEC 61966-2-4
Retrieved from "https://en.wikipedia.org/w/index.php?title=XvYCC&oldid=1080732870"
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Nonlinear regression model class - MATLAB - MathWorks Italia
NonLinearModel class
Fit a Nonlinear Regression Model
Nonlinear regression model class
An object comprising training data, model description, diagnostic information, and fitted coefficients for a nonlinear regression. Predict model responses with the predict or feval methods.
Create a NonLinearModel object using fitnlm.
Diagnostics — Diagnostic information
Diagnostic information for the model, specified as a table. Diagnostics can help identify outliers and influential observations. Diagnostics contains the following fields.
Leverage Diagonal elements of HatMatrix Leverage indicates to what extent the predicted value for an observation is determined by the observed value for that observation. A value close to 1 indicates that the prediction is largely determined by that observation, with little contribution from the other observations. A value close to 0 indicates the fit is largely determined by the other observations. For a model with P coefficients and N observations, the average value of Leverage is P/N. An observation with Leverage larger than 2*P/N can be regarded as having high leverage.
CooksDistance Cook's measure of scaled change in fitted values CooksDistance is a measure of scaled change in fitted values. An observation with CooksDistance larger than three times the mean Cook's distance can be an outlier.
Fitted (predicted) values based on the input data, specified as a numeric vector. fitnlm attempts to make Fitted as close as possible to the response data.
NonLinearFormula object
Model information, specified as a NonLinearFormula object.
Display the formula of the fitted model mdl by using dot notation.
Iterative — Information about fitting process
Information about the fitting process, specified as a structure with the following fields:
InitialCoefs — Initial coefficient values (the beta0 vector)
IterOpts — Options included in the Options name-value pair argument for fitnlm.
Mean squared error, specified as a numeric value. The mean squared error is an estimate of the variance of the error term in the model.
Number of coefficients in the fitted model, specified as a positive integer. NumCoefficients is the same as NumEstimatedCoefficients for NonLinearModel objects. NumEstimatedCoefficients is equal to the degrees of freedom for regression.
Number of estimated coefficients in the fitted model, specified as a positive integer. NumEstimatedCoefficients is the same as NumCoefficients for NonLinearModel objects. NumEstimatedCoefficients is equal to the degrees of freedom for regression.
Root mean squared error, specified as a numeric value. The root mean squared error is an estimate of the standard deviation of the error term in the model.
Robust fit information, specified as a structure with the following fields:
WgtFun Robust weighting function, such as 'bisquare' (see robustfit)
Tune Value specified for tuning parameter (can be [])
Weights Vector of weights used in final iteration of robust fit
This structure is empty unless fitnlm constructed the model using robust regression.
Fit a nonlinear regression model for auto mileage based on the carbig data. Predict the mileage of an average car.
Load the sample data. Create a matrix X containing the measurements for the horsepower (Horsepower) and weight (Weight) of each car. Create a vector y containing the response values in miles per gallon (MPG).
Fit a nonlinear regression model.
Find the predicted mileage of an average car. Because the sample data contains some missing (NaN) observations, compute the mean using mean with the 'omitnan' option.
Xnew = mean(X,'omitnan')
MPGnew = 21.8073
The hat matrix H is defined in terms of the data matrix X and the Jacobian matrix J:
{J}_{i,j}={\frac{\partial f}{\partial {\beta }_{j}}|}_{{x}_{i},\beta }
Here f is the nonlinear model function, and β is the vector of model coefficients.
The Hat Matrix H is
H = J(JTJ)–1JT.
\begin{array}{l}0\le {h}_{ii}\le 1\\ \sum _{i=1}^{n}{h}_{ii}=p,\end{array}
{D}_{i}=\frac{\sum _{j=1}^{n}{\left({\stackrel{^}{y}}_{j}-{\stackrel{^}{y}}_{j\left(i\right)}\right)}^{2}}{p\text{\hspace{0.17em}}MSE},
{\stackrel{^}{y}}_{j}
{\stackrel{^}{y}}_{j\left(i\right)}
{D}_{i}=\frac{{r}_{i}^{2}}{p\text{\hspace{0.17em}}MSE}\left(\frac{{h}_{ii}}{{\left(1-{h}_{ii}\right)}^{2}}\right),
where ei is the ith residual.
GeneralizedLinearModel | LinearModel | nlinfit | fitnlm | predict
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Autoregressive Moving Average Model - MATLAB & Simulink - MathWorks Switzerland
Stationarity and Invertibility of the ARMA Model
For some observed time series, a very high-order AR or MA model is needed to model the underlying process well. In this case, a combined autoregressive moving average (ARMA) model can sometimes be a more parsimonious choice.
An ARMA model expresses the conditional mean of yt as a function of both past observations,
{y}_{t-1},\dots ,{y}_{t-p}
, and past innovations,
{\epsilon }_{t-1},\dots ,{\epsilon }_{t-q}.
The number of past observations that yt depends on, p, is the AR degree. The number of past innovations that yt depends on, q, is the MA degree. In general, these models are denoted by ARMA(p,q).
The form of the ARMA(p,q) model in Econometrics Toolbox™ is
{y}_{t}=c+{\varphi }_{1}{y}_{t-1}+\dots +{\varphi }_{p}{y}_{t-p}+{\epsilon }_{t}+{\theta }_{1}{\epsilon }_{t-1}+\dots +{\theta }_{q}{\epsilon }_{t-q},
{\epsilon }_{t}
{L}^{i}{y}_{t}={y}_{t-i}
\varphi \left(L\right)=\left(1-{\varphi }_{1}L-\dots -{\varphi }_{p}{L}^{p}\right)
\theta \left(L\right)=\left(1+{\theta }_{1}L+\dots +{\theta }_{q}{L}^{q}\right)
. You can write the ARMA(p,q) model as
\varphi \left(L\right){y}_{t}=c+\theta \left(L\right){\epsilon }_{t}.
\varphi \left(L\right)
Consider the ARMA(p,q) model in lag operator notation,
\varphi \left(L\right){y}_{t}=c+\theta \left(L\right){\epsilon }_{t}.
{y}_{t}=\mu +\frac{\theta \left(L\right)}{\varphi \left(L\right)}{\epsilon }_{t}=\mu +\psi \left(L\right){\epsilon }_{t},
\mu =\frac{c}{\left(1-{\varphi }_{1}-\dots -{\varphi }_{p}\right)}
\psi \left(L\right)
is a rational, infinite-degree lag operator polynomial,
\left(1+{\psi }_{1}L+{\psi }_{2}{L}^{2}+\dots \right)
{\psi }_{i}
\varphi \left(L\right)
, is stable, meaning all its roots lie outside the unit circle. Additionally, the process is causal provided the MA polynomial is invertible, meaning all its roots lie outside the unit circle.
Econometrics Toolbox enforces stability and invertibility of ARMA processes. When you specify an ARMA model using arima, you get an error if you enter coefficients that do not correspond to a stable AR polynomial or invertible MA polynomial. Similarly, estimate imposes stationarity and invertibility constraints during estimation.
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