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Unlimited growth is generally impossible, because birth and death rates are affected by factors such as lack of food, water and land. We now incorporate the effect of the environment on population size into the exponential model: $\frac {\mathrm{d}N}{\mathrm{d}t} =rN(t)$
Below are two derivations of this new model, called the logistic equation. It was originally devised by the mathematician Thomas Malthus. Notes on the discrete form of this equation can be found here.
The
carrying capacity, K, is the largest population that can be supported indefinitely, given the resources available in the environment. When the population size is far below K, its growth is exponential. As the population approaches K, it begins to be affected by the reduced ability of the environment to provide necessary resources.
In order to have this effect, we consider the term $\frac {K-N}{K}$ . Note that when $N< < K$ then $\frac {K-N}{K}\approx 1$ and when $N \rightarrow K$ then $\frac {K-N}{K} \rightarrow 0$.
Including the above term, we now get the equation: $$\frac {\mathrm{d}N}{\mathrm{d}t}=r \frac{K-N}{K} N(t)= rN\left(1-\frac{N}{K}\right)$$Note that each individual added to the population reduces the rate of increase of the whole population.
Consider a population of lions of size
y, and begin by assuming the simple exponential equation $\frac {\mathrm{d}y}{\mathrm{d}t}=r y$ , where r is the intrinsic growth rate.
If the probability of food being found by an individual lion is
y, then the probability of some food being found by two lions is proportional to $y^2$. Fighting will occur within the species for these limited resources, so the death rate due to fighting can be represented by $\gamma y^2$ . Our equation then becomes: $$\frac {\mathrm{d}y}{\mathrm{d}t}=ry-\gammay^2=ry\left(1-\frac{y}{Y}\right)$$where $Y=\frac{r}{\gamma}$. Again this is the logistic equation.
Question: Graph the logistic equation to find the equilibrium points. Then solve the equation using standard integrals, showing that the solution is given by $N(t)=\frac {K N_0 e^{rt}}{K+N_0 (e^{rt}-1)}$
Graphically we describe this kind of population growth by a sigmoid, or S-shaped growth curve:
By looking at the blue curve, we see the size of the population begins to level off and reaches a stable value, which must be less than the carrying capacity of the environment (marked in red).
Mathematically, we can show this by looking at the above solution for $N(t)$ and showing $\lim_{t\to\infty} N(t)=K$
Question: A small lake has a carrying capacity of 100 geese. Starting with a pair of geese, how would the population change over 70 years if $r=0.1$? Draw a sigmoid graph of this change.
Recall that
r is the intrinsic growth rate. How will this affect the time taken for the number of geese to reach the carrying capacity of the lake? Perhaps start by drawing graphs with different r-values. |
The
Kelvin–Helmholtz mechanism is an astronomical process that occurs when the surface of a star or a planet cools. The cooling causes the pressure to drop, and the star or planet shrinks as a result. This compression, in turn, heats up the core of the star/planet. This mechanism is evident on Jupiter and Saturn and on brown dwarfs whose central temperatures are not high enough to undergo nuclear fusion. It is estimated that Jupiter radiates more energy through this mechanism than it receives from the Sun, but Saturn might not. The latter process causes Jupiter to shrink at a rate of two centimetres each year. [1]
The mechanism was originally proposed by Kelvin and Helmholtz in the late 19th century to explain the source of energy of the Sun. By the mid-19th century, conservation of energy had been accepted, and one consequence of this law of physics is that the Sun must have some energy source to continue to shine. Because nuclear reactions were unknown, the main candidate for the source of solar energy was gravitational contraction.
However, it soon was recognized by Sir Arthur Eddington and others that the total amount of energy available through this mechanism only allowed the Sun to shine for millions of years rather than the billions of years that the geological and biological evidence suggested for the age of the Earth. (Kelvin himself had argued that the Earth was millions, not billions, of years old.) The true source of the Sun's energy remained uncertain until the 1930s, in which it was shown by Hans Bethe to be nuclear fusion.
It was theorised that the gravitational potential energy from the contraction of the Sun could be its source of power. To calculate the total amount of energy that would be released by the Sun in such a mechanism (assuming uniform density), it was approximated to a perfect sphere made up of concentric shells. The gravitational potential energy could then be found as the integral over all the shells from the centre to its outer radius.
Gravitational potential energy from Newtonian mechanics is defined as:
U = -\frac{Gm_1m_2}{r},
where
G is the gravitational constant, and the two masses in this case are that of the thin shells of width dr, and the contained mass within radius r as one integrates between zero and the radius of the total sphere. This gives: U = -G\int_0^R \frac{m(r) 4 \pi r^2 \rho}{r}\, dr,
where
R is the outer radius of the sphere, and m( r) is the mass contained within the radius r. Changing m( r) into a product of volume and density to satisfy the integral, U = -G\int_0^R \frac{4 \pi r^3 \rho 4 \pi r^2 \rho}{3r}\, dr = -\frac{16}{15}G \pi^2 \rho^2 R^5.
Recasting in terms of the mass of the sphere gives the total gravitational potential energy as
U = -\frac{3M^2G}{5R}.
Then, applying the virial theorem, half of this energy is radiated during the collapse, giving the total radiated energy:
U_\text{r} = \frac{3M^2G}{10R}.
While uniform density is not correct, one can get a rough order of magnitude estimate of the expected age of our star by inserting known values for the mass and radius of the Sun, and then dividing by the known luminosity of the Sun (note that this will involve another approximation, as the power output of the Sun has not always been constant):
\frac{U_\text{r}}{L_\odot} \approx \frac{1.1 \times 10^{41}~\text{J}}{3.9 \times 10^{26}~\text{W}} \approx 8\,900\,000~\text{years},
where L_\odot is the luminosity of the Sun. While giving enough power for considerably longer than many other physical methods, such as chemical energy, this value was clearly still not long enough due to geological and biological evidence that the Earth was billions of years old. It was eventually discovered that thermonuclear energy was responsible for the power output and long lifetimes of stars.
[3] References
^ Patrick G. J. Irwin (2003). Giant Planets of Our Solar System: Atmospheres, Composition, and Structure. Springer. ^ == BW Carroll & DA Ostlie (2007). An Introduction to Modern Astrophysics (2nd Ed.). Pearson Addison Wesley. pp. 296–298. ^ R. Pogge (2006-01-15). "The Kelvin-Helmholtz Mechanism". Lecture 12: As Long as the Sun Shines. Ohio State University. Retrieved 2009-11-05.
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It’s been a year since I started this math blog!!
More than 500 problems were posted during a year (July 19th 2016-July 19th 2017).
I made a list of the 10 math problems on this blog that have the most views.
Can you solve all of them?
The level of difficulty among the top 10 problems.
【★★★】 Difficult (Final Exam Level)
【★★☆】 Standard(Midterm Exam Level)
【★☆☆】 Easy (Homework Level)
Read solution Add to solve later
Problem 512 (a) Prove that the matrix $A=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}$ does not have a square root. Namely, show that there is no complex matrix $B$ such that $B^2=A$.
Add to solve later
(b) Prove that the $2\times 2$ identity matrix $I$ has infinitely many distinct square root matrices. Problem 509
Using the numbers appearing in
\[\pi=3.1415926535897932384626433832795028841971693993751058209749\dots\] we construct the matrix \[A=\begin{bmatrix} 3 & 14 &1592& 65358\\ 97932& 38462643& 38& 32\\ 7950& 2& 8841& 9716\\ 939937510& 5820& 974& 9 \end{bmatrix}.\]
Prove that the matrix $A$ is nonsingular.Add to solve later
Problem 506
Let $A$ be an $n\times n$ invertible matrix. Then prove the transpose $A^{\trans}$ is also invertible and that the inverse matrix of the transpose $A^{\trans}$ is the transpose of the inverse matrix $A^{-1}$.
Namely, show that \[(A^{\trans})^{-1}=(A^{-1})^{\trans}.\] Problem 505
Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.
Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula: \[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]
Using the formula, calculate the inverse matrix of $\begin{bmatrix}
2 & 1\\ 1& 2 \end{bmatrix}$. Problem 500
10 questions about nonsingular matrices, invertible matrices, and linearly independent vectors.
The quiz is designed to test your understanding of the basic properties of these topics.
You can take the quiz as many times as you like.
The solutions will be given after completing all the 10 problems.
Click the View question button to see the solutions. Problem 498
Let $T:\R^2 \to \R^2$ be a linear transformation of the $2$-dimensional vector space $\R^2$ (the $x$-$y$-plane) to itself which is the reflection across a line $y=mx$ for some $m\in \R$.
Then find the matrix representation of the linear transformation $T$ with respect to the standard basis $B=\{\mathbf{e}_1, \mathbf{e}_2\}$ of $\R^2$, where
\[\mathbf{e}_1=\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}.\] |
Tagged: determinant of a matrix Problem 718
Let
\[ A= \begin{bmatrix} 8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2 \end{bmatrix} . \] Notice that $A$ contains every integer from $1$ to $9$ and that the sums of each row, column, and diagonal of $A$ are equal. Such a grid is sometimes called a magic square.
Compute the determinant of $A$.Add to solve later
Problem 686
In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not.
(a) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=0$.
Add to solve later
(b) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=-1$. Problem 582
A square matrix $A$ is called
nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.
Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample. Problem 571
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold). The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.
Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let \[\mathbf{a}_1=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \mathbf{a}_2=\begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}, \mathbf{b}=\begin{bmatrix} 0 \\ a \\ 2 \end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5. Find the inverse matrix of \[A=\begin{bmatrix} 0 & 0 & 2 & 0 \\ 0 &1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}\] if it exists. If you think there is no inverse matrix of $A$, then give a reason. Problem 6. Consider the system of linear equations \begin{align*} 3x_1+2x_2&=1\\ 5x_1+3x_2&=2. \end{align*} (a) Find the coefficient matrix $A$ of the system. (b) Find the inverse matrix of the coefficient matrix $A$. (c) Using the inverse matrix of $A$, find the solution of the system.
(
Linear Algebra Midterm Exam 1, the Ohio State University) Read solution Problem 546
Let $A$ be an $n\times n$ matrix.
The $(i, j)$
cofactor $C_{ij}$ of $A$ is defined to be \[C_{ij}=(-1)^{ij}\det(M_{ij}),\] where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column.
Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$.
The matrix $\Adj(A)$ is called the adjoint matrix of $A$.
When $A$ is invertible, then its inverse can be obtained by the formula
For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula.
(a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$. (b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$. Problem 509
Using the numbers appearing in
\[\pi=3.1415926535897932384626433832795028841971693993751058209749\dots\] we construct the matrix \[A=\begin{bmatrix} 3 & 14 &1592& 65358\\ 97932& 38462643& 38& 32\\ 7950& 2& 8841& 9716\\ 939937510& 5820& 974& 9 \end{bmatrix}.\]
Prove that the matrix $A$ is nonsingular.Add to solve later
Problem 505
Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.
Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula: \[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]
Using the formula, calculate the inverse matrix of $\begin{bmatrix}
2 & 1\\ 1& 2 \end{bmatrix}$. Problem 486
Determine whether there exists a nonsingular matrix $A$ if
\[A^4=ABA^2+2A^3,\] where $B$ is the following matrix. \[B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.\]
If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$.
(
The Ohio State University, Linear Algebra Final Exam Problem) Read solution Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$.
Add to solve later
(b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. |
Most people see a trigonometric function and instantly start to experience anxiety, either moderate or extreme. Primarily, this anxiety stems from lack of understanding. Let's start at the basics and work our way to a deeper understanding of the sine, cosine, and tangent functions.
Trigonometric functions are basically comparing similar right triangles. Similar means that the triangles' angles are congruent (the same measure), and their side lengths are proportional. Look at the similar right triangles CAT and DOG. The fact that the triangles are proportional means that we can set up a proportion (equal ratios, or fractions) of corresponding parts. For example, side AT corresponds to side OG, and side CT corresponds to side DG. Here is the proportion that is important for understanding trigonometric functions.
These fractions are equal! This will always happen with similar triangles!
The main point is that we can pick any two sides of a triangle, make a ratio (fraction) out of the side lengths, and say that it is equal to the corresponding ratio from the other triangle. Here are two other proportions that we can set up. Again, when we evaluate the ratios, they are equal.
\frac{AT}{AC} = \frac{OG}{OD}
\frac{CT}{CA} = \frac{DG}{DO}
\frac{6}{8} = \frac{3}{4}
\frac{10}{8} = \frac{5}{4}
If you know that triangles are similar, you can find out missing side lengths if you know only one of the three side lengths. For example, let's look at triangle MUT and compare it to triangle DOG.
Because triangle MUT is similar to triangle DOG, side UT corresponds to side OG,side MU corresponds to side DO, and side MT corresponds to side DG. We can set up three different proportions of corresponding parts.
\frac{UT}{MT} = \frac{OG}{DG} \
\frac{UT}{MU} = \frac{OG}{DO} \
\frac{MU}{MT} = \frac{DO}{DG}
\frac{
m}{ u} = \frac{3}{5}
\frac{
m}{9} = \frac{3}{4}
\frac{9}{
u} = \frac{4}{5}
We could cross multiply to solve for the variables, thus finding the lengths of unknown side lengths
u and m.
4 \cdot
m=9\cdot 3
9 \cdot 5=4 \cdot
u
4 \cdot
m=27
45=4 \cdot
u m= \frac{27}{4}
\frac{45}{4}=
u m= 6 \frac{3}{4}
\frac{1}{4}=
u
Now we know everthing that there is to know about the triangle MUT.
Any triangle with the angles measureing 36.9 ^\circ, 53.1 ^\circ , and 90 ^\circ will always have ratios of side lengths that reduce to \frac{3}{5}, \frac{3}{4}, and \frac{4}{5} . The idea that similar figures always have equal ratios gave birth to the idea behind trigonometric functions. Since there are three ratios for every right triangle, there are three trigonometric functions. (Technically, we can take the reciprocals of the ratios, which makes it six ratios total. But, it is redundant and isn't important yet. Basically, this is where the sec, tan, and csc trigonometric functions come from.)
In order to distinguish between the three different ratios, we have to be able to look at any triangle and specify which two side lengths we should use to make our ratio. To do this, each side of the right triangle is called either opposite, adjacent, or the hypotenuse, depending on which angle we will focus on.
In triangle BEN, we are first going to focus on \angle B.
Side EN is opposite \angle B. Side BE is adjacent to \angle B. The hypotenuse of the triangle is side BN.
But, if we focus on \angle N, the relationship to the sides changes.
Side BE is opposite \angle N. Side EN is adjacent to \angle N. The hypotenuse of the triangle is still side BN.
We won't focus on the 90^\circ angle because trigonometric functions are defined by right triangles. The fact that there is a 90^\circ does not give us any additional information. Whereas we could compute the third angle if we are told that we have a right triangle that also has an angle of some specified measurement in degrees, such as \angle B and \angle N.
The angle we use and which trigonometric function we use tells us which side lengths we need to make the trigonometric ratio.
Here are the definitions of the trigonometric functions.
sine of an angle = \frac{opposite}{hypotenuse} cosine of an angle = \frac{adjacent}{hypotenuse} tangent of an angle = \frac{opposite}{adjacent}
Let's find all of the trigonometric ratios for the triangle BEN.
sin(B) = \frac{opposite \angle B}{hypotenuse} = \frac{7.5}{12.5} = 0.6 = \frac{3}{5} cos(B) = \frac{adjacent \angle B}{hypotenuse} = \frac{10}{12.5} = 0.8 = \frac{4}{5} tan(B) = \frac{opposite \angle B}{adjacent \angle B} = \frac{7.5}{10} = 0.75 = \frac{3}{4}
Notice that the ratios are \frac{3}{5}, \frac{4}{5}, and \frac{3}{4} as we predicted would happen. Also notice that if you evaluate sin(36.9^\circ) with a calulator you get 0.6004. Similarly, cos(36.9^\circ)= 0.7997 and tan(36.9^\circ) = 0.7508.
Similarly, we can evaluate the trigonometric functions with respect to \angle N.
sin(N) = \frac{10}{12.5} = 0.8 = \frac{4}{5} cos(N) = \frac{7.5}{12.5} = 0.6 = \frac{3}{5} tan(N) = \frac{10}{7.5} = 1.3 = \frac{4}{3}
All of the examples we have looked at are similar triangles with 36.9^\circ, 53.1^\circ , and 90^\circ angles. The same idea works with any right triangle you might have. Think of trigonometric ratios as a huge list of ratios of every possible right triangle you can come up with. Some textbooks even have a list of the trigonometric ratios for all of the angles ranging from 0^\circ to 90^\circ. |
The Fourier transform of a constant exists. Can anyone please tell me what the $\mathcal{Z}$-transform of a constant is? Thanks in advance.
if you define the constant like this: $$ x[n] = C ~~~,~~~\text{ for all } n $$
Then its $\mathcal{Z}$ transform $$ X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} = \sum_{n=-\infty}^{\infty} C z^{-n} $$
does not converge for any value of $z$. Hence its $\mathcal{Z}$ transform does not exist.
Note that $\mathcal{Z}$ transform must be an analytic function with continuous derivatives of all orders. Hence unlike the DTFT which is not necesarily an analytic function and therefore admit the use of impulses in its expressions which would for example define the DTFT of the constant $x[n] =C$ to be $C 2\pi \delta(\omega)$, we canot allow impulses in $\mathcal{Z}$ transform.
Had you defined the constant as the
practical signal like$$ y[n] = C u[n]$$ where $u[n] = \begin{cases} 1 &,& \text{ for } n \geq 0 \\ 0 &,& \text{ for } n < 0\\ \end{cases} $ is the unit step function, then its $\mathcal{Z}$-Transform would exist:$$ Y(z) = \sum_{n=-\infty}^{\infty} y[n]z^{-n} = \sum_{n=0}^{\infty} C z^{-n} = \frac{C}{ 1- z^{-1} } ~~~, ~~~ ROC~~ |z|>1$$
The distinction is clear I suppose.
Instead of considering a constant signal, let us consider a rectangular window of length $2 \ell + 1$
$$w (n) = \begin{cases} 1 & \text{if } |n| \leq \ell\\ 0 & \text{otherwise} \end{cases}$$
whose $\mathcal Z$-transform is
$$W (z) = z^{-\ell} \left( 1 + z + z^2 + \cdots + z^{2 \ell} \right) = \dfrac{1}{z^\ell} \left(\dfrac{z^{2\ell+1} - 1}{z-1}\right)$$
which has a simple pole at $z = 1$, an $\ell$-th order pole at the origin, and $2 \ell + 1$ simple zeros on the unit circle at angles that are integer multiples of $ \frac{2 \pi}{2 \ell + 1}$. The pole at $z = 1$ cancels the zero at $z = 1$ and we are left with $2 \ell$ zeros on the unit circle. Note that
$$W (1) = 2 \ell + 1$$
As $\ell \to \infty$, the window $w (n)$ approaches a constant signal of unit amplitude, the unit circle of the complex plane gets "perforated" with more and more zeros, and $W (1) \to \infty$. |
Category: Group Theory
Group Theory Problems and Solutions.
Popular posts in Group Theory are:
Problem 625
Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$.
(a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.
Add to solve later
(b) Prove that a group cannot be written as the union of two proper subgroups. Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$.
Add to solve later
(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. Problem 613
Let $m$ and $n$ be positive integers such that $m \mid n$.
(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined. (b) Prove that $\phi$ is a group homomorphism. (c) Prove that $\phi$ is surjective.
Add to solve later
(d) Determine the group structure of the kernel of $\phi$. If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 497
Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively. Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.
Also determine whether the statement is true if $G$ is a non-abelian group.Add to solve later |
2019-05-20 15:18 Detaljerad journal - Similar records 2019-01-23 09:13
nuSTORM at CERN: Feasibility Study / Long, Kenneth Richard (Imperial College (GB)) The Neutrinos from Stored Muons, nuSTORM, facility has been designed to deliver a definitive neutrino-nucleus scattering programme using beams of $\bar{\nu}_e$ and $\bar{\nu}_\mu$ from the decay of muons confined within a storage ring. The facility is unique, it will be capable of storing $\mu^\pm$ beams with a central momentum of between 1 GeV/c and 6 GeV/c and a momentum spread of 16%. [...] CERN-PBC-REPORT-2019-003.- Geneva : CERN, 2019 - 150. Detaljerad journal - Similar records 2019-01-23 08:54 Detaljerad journal - Similar records 2019-01-15 15:35
Report from the LHC Fixed Target working group of the CERN Physics Beyond Colliders forum / Barschel, Colin (CERN) ; Bernhard, Johannes (CERN) ; Bersani, Andrea (INFN e Universita Genova (IT)) ; Boscolo Meneguolo, Caterina (Universita e INFN, Padova (IT)) ; Bruce, Roderik (CERN) ; Calviani, Marco (CERN) ; Carassiti, Vittore (Universita e INFN, Ferrara (IT)) ; Cerutti, Francesco (CERN) ; Chiggiato, Paolo (CERN) ; Ciullo, Giuseppe (Universita e INFN, Ferrara (IT)) et al. Several fixed-target experiments at the LHC are being proposed and actively studied. Splitting of beam halo from the core by means of a bent crystal combined with a second bent crystal after the target has been suggested in order to study magnetic and electric dipole moments of short-lived particles. [...] CERN-PBC-REPORT-2019-001.- Geneva : CERN, 2019 Fulltext: PDF; Detaljerad journal - Similar records 2018-12-20 13:45 Detaljerad journal - Similar records 2018-12-18 14:08
Physics Beyond Colliders at CERN: Beyond the Standard Model Working Group Report / Beacham, J. (Ohio State U., Columbus (main)) ; Burrage, C. (U. Nottingham) ; Curtin, D. (Toronto U.) ; De Roeck, A. (CERN) ; Evans, J. (Cincinnati U.) ; Feng, J.L. (UC, Irvine) ; Gatto, C. (INFN, Naples ; NIU, DeKalb) ; Gninenko, S. (Moscow, INR) ; Hartin, A. (U. Coll. London) ; Irastorza, I. (U. Zaragoza, LFNAE) et al. The Physics Beyond Colliders initiative is an exploratory study aimed at exploiting the full scientific potential of the CERN’s accelerator complex and scientific infrastructures through projects complementary to the LHC and other possible future colliders. These projects will target fundamental physics questions in modern particle physics. [...] arXiv:1901.09966; CERN-PBC-REPORT-2018-007.- Geneva : CERN, 2018 - 150 p. Full Text: PDF; Fulltext: PDF; Detaljerad journal - Similar records 2018-12-17 18:05
PBC technology subgroup report / Siemko, Andrzej (CERN) ; Dobrich, Babette (CERN) ; Cantatore, Giovanni (Universita e INFN Trieste (IT)) ; Delikaris, Dimitri (CERN) ; Mapelli, Livio (Universita e INFN, Cagliari (IT)) ; Cavoto, Gianluca (Sapienza Universita e INFN, Roma I (IT)) ; Pugnat, Pierre (Lab. des Champs Magnet. Intenses (FR)) ; Schaffran, Joern (Deutsches Elektronen-Synchrotron (DE)) ; Spagnolo, Paolo (INFN Sezione di Pisa, Universita' e Scuola Normale Superiore, Pisa (IT)) ; Ten Kate, Herman (CERN) et al. Goal of the technology WG set by PBC: Exploration and evaluation of possible technological contributions of CERN to non-accelerator projects possibly hosted elsewhere: survey of suitable experimental initiatives and their connection to and potential benefit to and from CERN; description of identified initiatives and how their relation to the unique CERN expertise is facilitated.. CERN-PBC-REPORT-2018-006.- Geneva : CERN, 2018 - 31. Fulltext: PDF; Detaljerad journal - Similar records 2018-12-14 16:17
AWAKE++: The AWAKE Acceleration Scheme for New Particle Physics Experiments at CERN / Gschwendtner, Edda (CERN) ; Bartmann, Wolfgang (CERN) ; Caldwell, Allen Christopher (Max-Planck-Institut fur Physik (DE)) ; Calviani, Marco (CERN) ; Chappell, James Anthony (University of London (GB)) ; Crivelli, Paolo (ETH Zurich (CH)) ; Damerau, Heiko (CERN) ; Depero, Emilio (ETH Zurich (CH)) ; Doebert, Steffen (CERN) ; Gall, Jonathan (CERN) et al. The AWAKE experiment reached all planned milestones during Run 1 (2016-18), notably the demonstration of strong plasma wakes generated by proton beams and the acceleration of externally injected electrons to multi-GeV energy levels in the proton driven plasma wakefields. During Run~2 (2021 - 2024) AWAKE aims to demonstrate the scalability and the acceleration of electrons to high energies while maintaining the beam quality. [...] CERN-PBC-REPORT-2018-005.- Geneva : CERN, 2018 - 11. Detaljerad journal - Similar records 2018-12-14 15:50
Particle physics applications of the AWAKE acceleration scheme / Wing, Matthew (University of London (GB)) ; Caldwell, Allen Christopher (Max-Planck-Institut fur Physik (DE)) ; Chappell, James Anthony (University of London (GB)) ; Crivelli, Paolo (ETH Zurich (CH)) ; Depero, Emilio (ETH Zurich (CH)) ; Gall, Jonathan (CERN) ; Gninenko, Sergei (Russian Academy of Sciences (RU)) ; Gschwendtner, Edda (CERN) ; Hartin, Anthony (University of London (GB)) ; Keeble, Fearghus Robert (University of London (GB)) et al. The AWAKE experiment had a very successful Run 1 (2016-8), demonstrating proton-driven plasma wakefield acceleration for the first time, through the observation of the modulation of a long proton bunch into micro-bunches and the acceleration of electrons up to 2 GeV in 10 m of plasma. The aims of AWAKE Run 2 (2021-4) are to have high-charge bunches of electrons accelerated to high energy, about 10 GeV, maintaining beam quality through the plasma and showing that the process is scalable. [...] CERN-PBC-REPORT-2018-004.- Geneva : CERN, 2018 - 11. Fulltext: PDF; Detaljerad journal - Similar records 2018-12-13 13:21
Summary Report of Physics Beyond Colliers at CERN / Jaeckel, Joerg (CERN) ; Lamont, Mike (CERN) ; Vallee, Claude (Centre National de la Recherche Scientifique (FR)) Physics Beyond Colliders is an exploratory study aimed at exploiting the full scientific potential of CERN's accelerator complex and its scientific infrastructure in the next two decades through projects complementary to the LHC, HL-LHC and other possible future colliders. These projects should target fundamental physics questions that are similar in spirit to those addressed by high-energy colliders, but that require different types of beams and experiments. [...] arXiv:1902.00260; CERN-PBC-REPORT-2018-003.- Geneva : CERN, 2018 - 66 p. Fulltext: PDF; PBC summary as submitted to the ESPP update in December 2018: PDF; Detaljerad journal - Similar records |
TL;DR: "Scientifically correct" (according to current established science) and "faster-than-light travel" cannot be used in the same context without some form of negation. What you are asking for is not possible within the boundaries of science as we know it.
Here's why:
Our best model for this type of effects, insofar as I know, is special and general relativity. Special relativity postulates that colinear velocities are added according to the formula $$ s = \cfrac{v+u}{1 + \cfrac{vu}{c^2}} $$ for an initial velocity $v$ and a total acceleration $u$ (over some period of time) yielding a final velocity $s$.
For small values of $v$ and $u$, this behaves like we are used to, because for such values, the fraction $\frac{vu}{c^2}$ is very small, so the term $1 + \frac{vu}{c^2}$ is very close to 1 giving $s \approx v+u$. Of course, in some situations, even with everyday velocities this approximation might not be good enough.
However, look what happens if we set $v = 0.90c$ and $u = 0.10c$ (meaning that in an intertial reference frame, our initial velocity is 0.90 times the speed of light, and we increase our velocity by 0.10 times the speed of light). Intuitively, the velocity would come out as $(0.90 + 0.10)c = c$, but it turns out that
this is not the case at all. Rather, using units of $c$ for simplicity:
$$ s = \cfrac{0.90 + 0.10}{1 + \cfrac{0.90 \times 0.10}{1^2}} \approx 0.9174 $$
See what happens? In an
inertial reference frame, our velocity only rose from $0.900c$ to about $0.917c$, an increase of 1.9%, even though we tried to raise the velocity by 11% ($0.10c$ out of $0.90c$).
This effect becomes even more pronounced as your initial velocity approaches $c$ ($v \to c$). For example, look what happens if we are moving at $0.99c$ and increase our velocity by $0.10c$ (yes, I really mean that): $$ s = \cfrac{0.99 + 0.10}{1 + \cfrac{0.99 \times 0.10}{1^2}} \approx 0.9918 $$ for a 0.18% increase for the same effort that got us 1.9% starting at 90% of $c$.
And of course, in the real world, these are both absurdly high values for $u$, reminiscent of instantaneous acceleration. Instead, we should be working with $u \to 0$ (because in the real world, the time over which we measure acceleration goes to 0), but since that's difficult to show in a single equation, I'll settle for $u = 10^{-12}c \approx 0.3~\text{mm/s}$ which isn't a totally unrealistic change of velocity over a short period of time given something resembling a real-world device trying to propel itself. Now look what we get if we start out at $0.90c$:
$$ s = \cfrac{0.90 + 10^{-12}}{1 + \cfrac{0.90 \times 10^{-12}}{1^2}} = 0.900~000~000~000~189~999... $$
Our velocity increase, which we tried to make $\frac{10^{-12}}{0.90} \approx 1 \times 10^{-12}$, became $\frac{0.90000000000019 - 0.90}{0.90} \approx 2 \times 10^{-13}$. We only got 1/5 of the increase that we spent the effort for, and at 90% of $c$, we are
still a good long ways away from $c$. It only gets worse from there.
Eventually, this means that the energy cost of increasing your velocity grows in an exponential fashion. If you work the math all the way, effectively solving $$\lim \limits_{v \to c, u \to 0} \cfrac{v+u}{1 + \cfrac{vu}{c^2}}$$ you end up with an
energy requirement that grows toward infinity as you get closer and closer to the speed of light. Because instantaneous velocity changes are not possible (because of inertia, for one thing), you can't simply "jump past" the difficult part of the acceleration curve.
Because your spacecraft will, at every instant, have an instantaneous velocity (along some vector) and an instantaneous acceleration ($\vec{a} = \vec{\Delta v} / \Delta t$ for some $\Delta t \to 0$), you will thus only ever be able to (with humungous energy expenditure)
approach the speed of light, but you will never be able to reach the speed of light. As your velocity increases, the marginal utility of any given acceleration (within the local frame of reference) decreases; you get less and less (inertial reference frame) acceleration out of any given amount of effort. Because inertial reference frames are what we are generally concerned with when going places, this means you work exponentially harder but get exponentially less utility for your efforts.
If you want a single formula that explains why faster-than-light travel is impossible in the real world as we currently understand it, the mass-energy equivalence $E=mc^2$ (as suggested by AndreiROM) isn't what you are looking for (in fact, it might even to a limited extent be your friend, if you can figure out how to do the mass-energy conversion); rather, the one you want is the relativistic colinear velocity addition formula and an understanding of how it behaves as $v \to c$. |
The Gaussian Distribution is pretty common in the case of continuous probability distribution. The distribution is frequently used in statistics and it is generally required in natural or social sciences to showcase the real-valued random variables. The probability density formula for Gaussian Distribution in mathematics is given as below –
\[\large f(x,\mu , \sigma )=\frac{1}{\sigma \sqrt{2\pi}}\; e^{\frac{-(x- mu)^{2}}{2\sigma ^{2}}}\]
Where,
x is the variable μ is the mean sigma is the standard deviation
You must be wondering what is the usage of Gaussian functions in statistics. They are used to describe the normal distributions and signal processing for images. They are also required in the heat transfer and diffusion of equations to define the wire transformation.
If the number of events is very large then Gaussian distribution is particularly suitable for various physical events too. This is a continuous function that give you an idea of exact distribution of binomial events.
This concept is valid for normalized equations too when sun over all values of x define the probability of one. It is somehow related to the standard deviation of mean too. The other name for Gaussian distribution is the normal distribution that is usually defined as the bell-shaped curve.
The formula is designed to evaluate different mathematical concepts like mean value, standard deviation, and the value of distribution function too where the value of x is supplied. The concept is valid for a large number of distributions too for evaluating the precise results.
Further, the concept is necessary in discrete applications too particularly to process the digital signatures. The simple answer to each of your problem is Gaussian distribution yielding the best outcomes as needed. |
The main objective in statistics is testing a hypothesis. For example, you could conduct a test to find either a drug is effective or not for treating headaches. If you repeat the experiment then no one would take the results seriously. A hypothesis is an educated guess that analyze the things around you and it could be tested either through some experiment or observation. For example –
If you wanted to propose a hypothesis then this is necessary to write a statement first and it should be well explained to understand the circumstances. A good statement always includes If and then statements. It involves both independent or dependent variables. It should be based on research or data collected earlier.
The statement must have specific design criteria that could suit the engineering or programming projects perfectly. The hypothesis testing formula in mathematics is given as below –
\[\large z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\]
Where,
$\overline{x}$ is the sample mean $\mu$ is the population mean $\sigma$ is the standard deviation and n is the sample size.
Hypothesis testing in statistics helps you to identify either result of a survey or experiment is meaningful or not. You can also check either your results find out the odds that happened by chance only. The experiments that has to repeat, they are of little use only. The concept could be confusing in beginning but a depth understanding always gives you the accurate results. |
This question already has an answer here:
Defining multivariable differentiation is just linear algebra. However, defining integration is complicated measure theory. Why are these efforts so different?
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This question already has an answer here:
Defining multivariable differentiation is just linear algebra. However, defining integration is complicated measure theory. Why are these efforts so different?
I don't see these two concepts as so different.
Multivariate differentiation is not "just linear algebra". It involves limits. A strict definition of a differential would be
If $f:\Omega\subseteq\mathbb R^m \to \mathbb R^n$ and $x\in\Omega$, then the differential of $f$ at $x$ is the matrix $D_f(x)\in \mathbb R^{n\times m}$ such that $$f(x+d)=f(x) + D_f(x)\cdot d + o(d)$$ and $$\lim_{d\to 0}\frac{o(d)}{\|d\|} = 0$$
Note that the limits themselves include some "for all $\epsilon>0$, there exists some $\delta>0$" things hidden inside.
On the other hand, integration does not need to involve measure theory. The (arguably) simplest way to define integrals is the Riemann integral which defines the integral as something very similar to a limit. A strict definition of the Riemann integral is
If $f:\Omega\subseteq\mathbb R^m \to \mathbb R$, then the integral of $f$ over $\Omega$ is $L$ if and only if, for every $\epsilon > 0$, there exists some $\delta >0$ such that for every partition $P$ of $\Omega$ into hyperboxes of dimension at most $\epsilon$, the Riemann sum $R_P$ of the function $f$ is at most $\epsilon$-away from $L$, i.e. $$|R_P-L|<\epsilon.$$
I don't see this definition as being much more complicated than the first. It's saying that something very similar to a limit, and actually has
no algebra, so it's even simpler. |
A few hints on how to do this calculation:
1) write down the equation for the total energy emitted per unit time at 0.1 K (Stefan-Boltzmann law).
2) write down the equation for the shape of the spectrum (Planck's Law)
Note that Planck's Law gives the
energy in an interval: you would have to divide that by the energy of a photon at each wavelength in order to get the probability of an emission in unit time. Integrate (Planck)/$\hbar\lambda$ between the limits of wavelength of interest: this is the probability of an emission, $p$.
To get to 95% confidence that you have an emission, you need to consider that you have $(1-p)$ probability of no emission per unit time. This is a number that must get below 5%, so solve for $s$:
$$(1-p)^s = 0.05$$
Confirm by integrating (2) , over all wavelengths, and show you get (1).
You will want to do this using numerical integration. Pick a small enough step size, and be careful of rounding errors (numerical instability).
If you do the above and work it into your question (or write it as an answer yourself) we can help you figure out any missing parts without violating the "homework and exercises" policy.
UPDATE
I took a quick look at the magnitude of the numbers, and there is a problem...
Most of your math programs will not want to evaluate Planck's Law in the range of temperatures and wavelengths: the exponential term will overflow. You therefore have to make a simple approximation so you can evaluate the logarithm:
$$\begin{align}\\B(\lambda, T) &= \frac{hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kT}}-1}\\&\approx\frac{hc^2}{\lambda^5e^{\frac{hc}{\lambda kT}}}\end{align}$$
when $\frac{hc}{\lambda kT} \gg 1$. Taking the natural logarithm, you find
$$\log(B) = \log(hc^2) - 5\log(\lambda) - \frac{hc}{\lambda kT}$$
I wrote a few lines of Python to evaluate this over the wavelength range of interest:
zooming in a lot more, you get this:
The probability of photon emission is very, very small indeed. |
Can I be a pedant and say that if the question states that $\langle \alpha \vert A \vert \alpha \rangle = 0$ for every vector $\lvert \alpha \rangle$, that means that $A$ is everywhere defined, so there are no domain issues?
Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of:We can always detect uniform motion with respect to a medium by a positive result to a Michelson...
Hmm, it seems we cannot just superimpose gravitational waves to create standing waves
The above search is inspired by last night dream, which took place in an alternate version of my 3rd year undergrad GR course. The lecturer talks about a weird equation in general relativity that has a huge summation symbol, and then talked about gravitational waves emitting from a body. After that lecture, I then asked the lecturer whether gravitational standing waves are possible, as a imagine the hypothetical scenario of placing a node at the end of the vertical white line
[The Cube] Regarding The Cube, I am thinking about an energy level diagram like this
where the infinitely degenerate level is the lowest energy level when the environment is also taken account of
The idea is that if the possible relaxations between energy levels is restricted so that to relax from an excited state, the bottleneck must be passed, then we have a very high entropy high energy system confined in a compact volume
Therefore, as energy is pumped into the system, the lack of direct relaxation pathways to the ground state plus the huge degeneracy at higher energy levels should result in a lot of possible configurations to give the same high energy, thus effectively create an entropy trap to minimise heat loss to surroundings
@Kaumudi.H there is also an addon that allows Office 2003 to read (but not save) files from later versions of Office, and you probably want this too. The installer for this should also be in \Stuff (but probably isn't if I forgot to include the SP3 installer).
Hi @EmilioPisanty, it's great that you want to help me clear out confusions. I think we have a misunderstanding here. When you say "if you really want to "understand"", I've thought you were mentioning at my questions directly to the close voter, not the question in meta. When you mention about my original post, you think that it's a hopeless mess of confusion? Why? Except being off-topic, it seems clear to understand, doesn't it?
Physics.stackexchange currently uses 2.7.1 with the config TeX-AMS_HTML-full which is affected by a visual glitch on both desktop and mobile version of Safari under latest OS, \vec{x} results in the arrow displayed too far to the right (issue #1737). This has been fixed in 2.7.2. Thanks.
I have never used the app for this site, but if you ask a question on a mobile phone, there is no homework guidance box, as there is on the full site, due to screen size limitations.I think it's a safe asssumption that many students are using their phone to place their homework questions, in wh...
@0ßelö7 I don't really care for the functional analytic technicalities in this case - of course this statement needs some additional assumption to hold rigorously in the infinite-dimensional case, but I'm 99% that that's not what the OP wants to know (and, judging from the comments and other failed attempts, the "simple" version of the statement seems to confuse enough people already :P)
Why were the SI unit prefixes, i.e.\begin{align}\mathrm{giga} && 10^9 \\\mathrm{mega} && 10^6 \\\mathrm{kilo} && 10^3 \\\mathrm{milli} && 10^{-3} \\\mathrm{micro} && 10^{-6} \\\mathrm{nano} && 10^{-9}\end{align}chosen to be a multiple power of 3?Edit: Although this questio...
the major challenge is how to restrict the possible relaxation pathways so that in order to relax back to the ground state, at least one lower rotational level has to be passed, thus creating the bottleneck shown above
If two vectors $\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}$ and$\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}$, have angle $\theta$ between them then the dot product (scalar product) of $\vec{A}$ and $\vec{B}$ is$$\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos \theta$$$$\vec{A}\cdot\...
@ACuriousMind I want to give a talk on my GR work first. That can be hand-wavey. But I also want to present my program for Sobolev spaces and elliptic regularity, which is reasonably original. But the devil is in the details there.
@CooperCape I'm afraid not, you're still just asking us to check whether or not what you wrote there is correct - such questions are not a good fit for the site, since the potentially correct answer "Yes, that's right" is too short to even submit as an answer |
Answer
4.5 percent
Work Step by Step
We use the margin of error equation to find: $$=100\times \frac{1}{\sqrt{n}}\\ = 100 \times \frac{1}{\sqrt{500}} \\ =4.5\text{%}$$
You can help us out by revising, improving and updating this answer.Update this answer
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Homework Helper
1,417 1
My book does not make sense to me. Here is what it says:
I know that: [tex]e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n}, 0 < \theta < 1[/tex] If e is rational then [tex]e = \frac{m}{n}; m, n \in Z[/tex] And the greatest common factor of m, n is 1. [tex]\Leftrightarrow 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n} = \frac{m}{n}[/tex] Multiple both sides by n!, the right side is an integer, while the left side is an integer plus [itex]\frac{\theta}{n}[/itex] This makes the contradiction, therefore e is irrational. My question is: Why they say [tex]e = \frac{m}{n}; m, n \in Z[/tex], so that they can multiple this by n! and get an integer. Can [tex]e = \frac{p}{q}; p, q \in Z[/tex]? Is there a better way of proving this? Thanks a lot, Viet Dao,
I know that:
[tex]e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n}, 0 < \theta < 1[/tex]
If e is rational then [tex]e = \frac{m}{n}; m, n \in Z[/tex]
And the greatest common factor of m, n is 1.
[tex]\Leftrightarrow 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n} = \frac{m}{n}[/tex]
Multiple both sides by n!, the right side is an integer, while the left side is an integer plus [itex]\frac{\theta}{n}[/itex]
This makes the contradiction, therefore e is irrational.
My question is: Why they say [tex]e = \frac{m}{n}; m, n \in Z[/tex], so that they can multiple this by n! and get an integer. Can [tex]e = \frac{p}{q}; p, q \in Z[/tex]?
Is there a better way of proving this?
Thanks a lot,
Viet Dao,
Last edited: |
Suppose in our double slit experimental setup with the usual notations $d,D$, we have a beam of light of known frequency $(\nu)$ and wavelength $(\lambda)$ - so we can describe it as $$ξ_0 = A\sin(kx-\omega t). \tag{1}$$ It passes through the two holes and moves ahead doing the usual interference stuff, so the final form of the wave will be $$ξ = ξ_1 + ξ_2 = 2A\cos(u/2)\sin(kx-\omega t+0.5*u) \tag{2}$$ where $u$ is the phase difference.
We can convert phase difference $u$ to path difference $q$. Now we choose the point of interest on the screen $(s)$ ,(which depends on path difference q and hence phase difference u). The amplitude at $s$ will be $$ξ = 2A\cos(as)\sin(kx-wt+as), \tag{3}$$ where $a$ is constant.
Now this amplitude is a set of waves which interfere with different phases, and is function of the variables $s,x,t$. Since I placed the screen at some fix distance $x=D$ from the wall with slits, $ξ$ reduces to a function of two variables $s,t$. Rewriting $$ξ_D=2A\cos(as)\sin(as -wt +kD), \tag{4}$$ this is also a wave description (but with different meaning).
The screen is along our $x$-axis (or to be precise $s$-axis). The intensity obtained on the screen is proportional to to absolute square of the wave amplitude written above, which in turn depends on $s$ (and t as well).
But the intensity is also proportional to number of photons. So we postulate that the probability that a photon hit a certain $s$ is proportional to the $$\text {intensity} = |\text {amplitude}|^2. \tag{5}$$
Now, the function $ξ$ I have written above is the wave function ($\psi$) from the quantum mechanics with $s$ acting as $x$ (in $\psi$)? If not, then what is the relation between them? (I will have some additional things to ask depending upon your response.) Thank you! |
Here is a general result on promoting identities for Lie groups from linear groups to all connected groups.
Let $\mathcal{LG}$ be the category of connected Lie groups and local isomorphisms ($=$ coverings) as morphisms. Assigning to a group $G$ the topological space $G^m \times \mathfrak{g}^n$ defines a functor from $\mathcal{LG}$ to topological spaces, with the property that it maps all coverings to coverings. It makes sense to ask for a natural subset $U(G) \subset G^m \times \mathfrak{g}^n$. In the example from the original question, $U= \mathfrak{g}^2$, in Tom's example, it is an algebraic set, but it could be the set of all $X \in \mathfrak{g}$ with $ad$-spectral radius bounded by $1$, for example. Now let $(F_i)_G:U(G) \to G$ (one could look at maps $\to \mathfrak{g}$ as well, which is slightly easier),$i=0,1$, be continuous natural transformations.
If you unwind this definition, $F_i$ is a function that is composed out of the data that are available for all Lie groups: the group operations,the identity element, vector space operation and Lie bracket on $\mathfrak{g}$, the adjoint representation, all operations in the matrix algebra $End (\mathfrak{g})$ (as taking power series, characteristic polynomials etc), the exponential map and taking derivatives of curves through the identity element in $G$ (no attempt to make a complete list).
Theorem: ''With the above notations, assume that
$F_0=F_1$ holds for all linear groups.
$U(G)$ is path-connected for all $G$ (recall that I assumed all groups to be connected);
For each covering $H \to G$, the map $U(H) \to U(G)$ is surjective;
There exists a natural transformation $u:\ast \to U$ of functors such that $F_0 (u)=F_1(u)$ holds for all connected Lie groups.
Then $F_0=F_1$ is true for all connected Lie groups.''
Proof: ''Denote by $P(G)$ the statement that the Theorem holds for $G$. For each connected Lie group $G$, there exist, by Ado's theorem,a Lie group $H$, a linear Lie group $L$ and coverings $H \to G$ and $p:H \to L$. Naturality and assumption $3$ show the implication $P(H) \Rightarrow P(G)$ and the tricky part is $P(L) \Rightarrow P(H)$. Observe that $p \circ (F_0)_H = (F_0)_L \circ p_U = (F_1)_L \circ p_U = p \circ (F_1)_H$; the second equality is assumption $1$. Therefore, $(F_0)_H$ and $(F_1)_H$ are both lifts of the map $(F_0)_L \circ p_U$ through the covering $H \to L$. By assumption $2$, they have to coincide once they coincide at one point, which is the content of assumption $4$. qed.''
The proof makes clear that one can restrict to the subcategory of connected Lie groups whose Lie algebra is isomorphic to one of a fixed set $S$ of Lie algebras,say $S=\{\mathfrak{sl}_2 (\mathbb{R})\}$.
The set $U$ in Toms example has at least two components: the boring one is $\{(0,0\}$ and other one contains those $(X,Y)$ that generate$\mathfrak{sl}_2 (\mathbb{R})$ (this is probably connected). It is assumption $3$ that fails for this component.
The proof also shows that the target of the transformations $F_i$ can be more generally any functor from $\mathcal{LG}$ that maps local isomorphisms to coverings. |
This is a very general theoretical question about classical and weak solutions.
If I know that there exists a unique classical solution to a PDE. What can I say about the solutions to the corresponding variational (or "weak") problem?
Consider for example the common Poisson equation:
$$ \cases{- \Delta u = f \ \ \text{in } \ \ \Omega \\ u = 0 \ \ \text{on } \ \ \partial \Omega} $$
And its corresponding variational formulation: Find $u \in H^1_0(\Omega)$ such that
$$ (\nabla u, \nabla v) = (f,v) \ \ \forall v \in H^1_0(\Omega). $$
Say I know that there exists a unique classical solution. Then this is also a solution to the variational formulation. But can I deduce something about the uniqueness of a solution to the variational problem?
I know that using Lax-Milgram it is possible to deduce the both existence and uniqueness of a weak solution. But given that I already know the existence of such a solution and also the uniqueness of a classical solution, can I somehow deduce that the weak solution is also unique? Does this only hold under some conditions?
If anything is unclear, don't hesitate to ask. Thanks in advance! |
dyld: Library not loaded: /sw/lib/libpng12.0.dylib Referenced from: /sw/bin/latex Reason: Incompatible library version: latex requires version 30.0.0 or later, but libpng12.0.dylib provides version 26.0.0 sh: line 1: 81964 Trace/BPT trap latex -output-directory=/var/folders/71/71wgK3o7FtGClPSEzjU3v++++TM/-Tmp-/inkscape-9yWR6s -halt-on-error /var/folders/71/71wgK3o7FtGClPSEzjU3v++++TM/-Tmp-/inkscape-9yWR6s/eq.tex > /var/folders/71/71wgK3o7FtGClPSEzjU3v++++TM/-Tmp-/inkscape-9yWR6s/eq.out invalid LaTeX input: \(\displaystyle\frac{\pi^2}{6}=\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{k^2}\) temporary files were left in: /var/folders/71/71wgK3o7FtGClPSEzjU3v++++TM/-Tmp-/inkscape-9yWR6s
Curiously, when I print out the version number of "/sw/lib/libpng12.0.dylib" via otool, it tells me that I already use the most recent version, which is 32.0.0, so there should be no problems.
I just found out what goes wrong: The Inkscape .dmg application package which I got from the Inkscape homepage comes with an own version of libpng12.0.dylib, which lies in "Inkscape.app/Contents/Resources/lib/", that has version number 26.0.0. It seems that Inkscape uses this file although the error message above explicitely shows the Fink path "/sw/lib/" !!
So I just renamed the libpng12.0.dylib which comes with the application package so that Inkscape doesn't find / use this file anymore and now the LaTeX effect works fine.
Just wanted to tell you in case someone else has a similar problem...
bluefloyd
P.S.: My version of Inkscape is 0.46.
P.P.S.: The identical problem occurs when using the textext extension. The solution is the same. An additional comment concerning textext: On a Mac there are actually two places where you can put the files of the textext extension:
1) ~/inkscape/extensions/
2)/Applications/Inkscape.app/Contents/Resources/extensions/
I propose putting the files in 2), since the extension is looking for inkex.py which has to be in the same directory as the extensions itself; inkex.py lies in 2), not in 1). Another possibility would be to use 1) and copy inkex.py (and probably some other needed files?) also from 2) to 1). |
If you have an understanding of Fourier transforms then you probably already have a conceptual model of transforming signals into the frequency domain. The Laplace transform provides an alternative frequency domain representation of the signal - usually referred to as the "S domain" to differentiate it from other frequency domain transforms (such as the Z transform - which is essentially a descretised equivalent of the Laplace transform).
What is the moment of a signal?
As you are no doubt aware the Laplace transform gives us a description of a signal from it's moments, similar to how the Fourier transform gives us a description from phase and amplitudes.
Broadly speaking a moment can be considered how a sample diverges from the mean value of a signal - the first moment is actually the mean, the second is the variance etc... (these are known collectively as "moments of a distribution")
Given our function F(t) we can calculate the n'th derivative at t=0 to give our n'th moment. Just as a signal can be described completely using phase and amplitude, it can be described completely by all of its derivatives.
Why is the fourier transform a special case of the laplace transform?
If we look at the bilateral laplace transform:
$${\int_{-\infty}^\infty}e^{-st}f(t)dt$$
It should be quite apparent that a substitution $s=i\omega$ will yield the familiar Fourier transform equation:
$${\int_{-\infty}^\infty}e^{-i\omega t}f(t)dt$$
There are some notes about this relationship (http://en.wikipedia.org/wiki/Laplace_transform#Fourier_transform) but the mathematics should be quite transparent. |
I would not use a
formula for the understanding of the so called bandpass sampling (or undersampling) operation. Instead try to analyse the situation by yourself considering the signal spectrum, sampling operation and the definition of aliasing which defines the permitted range of sampling frequencies.
First, we state the
fundamental principle of sampling: in order to represent a signal $x(t)$ perfectly with a set of samples $x[n]$ taken uniformly at a sampling rate $f_s$ samples per second, there should be no aliasing (spectral overlap) in the sampled signal $x_s(t)$'s spectrum $X_s(\Omega)$.
Then we define the ideally sampled signal as $x_s(t) = x(t) \sum_{k=-\infty}^{\infty} \delta(t - k T_s)$ and its associated CTFT spectrum as:$$X_s(\Omega) = \frac{2\pi}{T_s} \sum_{k=-\infty}^{\infty} X(\Omega - k \frac{2\pi}{T_s}) $$
Finally we ask, given the consequences of sampling on the spectrum $X_s(\Omega)$ of the sampled signal $x_s(t)$, which set of frequencies $f_s$ can satisfy the fundamental principle of no-aliasing. Then we try to determine the
minimum of this set of valid sampling frequencies.
Lets apply this to your problem:Given your real bandpass signal with a spectrum as in the figure-1.
First assume that for some $k=m$ the spectrum is shifted to right by $m f_s$ such that the shifted left-piece stands closest from left to the original right piece in its original place. Then consider the next shift at $k=m+1$ so that the shifted left piece jumps over the original right piece and stands closest (from right) to the original right piece. These two conditions are mathematicaly yielding the following constraints on the sampling frequency $f_s$:
For the case one with $k=m$ we have:$$ -f_L + m f_s \leq f_L $$
For the case $k=m+1$ we have:$$ -f_H + (m+1) f_s \geq f_H $$
Combining these two yields:$$ f_s \geq 2(f_H - f_L) $$
In your case $f_H-f_L$ is $10$ kHz, so we conclude that $$f_{smin} = 20 \text{kHz}$$
Note that this is the minimum sampling rate for
real valued samples. If you are allowd to perform complex valued sampling, then you can use only one side of the spectrum (via complex analytic signal generation) and sample it at half the real sampling rate, but then you will be getting twice the number samples per second nevertheless)
Note also that this particular analysis just defines the minimum valid sampling frequency without aliasing. However, unlike the baseband sampling theorem which states that any frequency greater than the minimum Nyquist rate will also be a valid sampling frequency, in the case of bandpass sampling this is not so. For example you can easily verify that for the frequency $f_s = 2 f_L + B/2$ which is can be larger than the minimum we computed above, the resulting spectrum shifts will
overlapp; there will be aliasing. So bandpass sampling frequencies are selected with care on the resulting spectral shifts. |
It is well know that quantum Yang-Mills theory has a periodic vacuum structure. Consider electroweak theory. For a single generation of fermions, the theory is CP invariant. I would like to know if the periodic vacua of the theory are also CP invariant.
One expects the trivial vacuum with topological charge $n=0$ to be CP invariant, where the topological charge, defined as
\begin{equation}n= \int d^4x \mathcal{P}(x),\end{equation}is the integral of the Pontryagin density $\mathcal{P}(x)$ over the spacetime manifold. However, since $\mathcal{P}(x) \sim tr\left( F \tilde{F} \right)$ is odd under CP, one generally expects gauge configurations that have non-zero topological charge to be odd under CP (?), and this includes the non-trivial vacua.
On the other hand, electroweak theory is different from QCD in that there is no explicit theta angle term in the action, so it seems to me that, as opposed to QCD, there should be no way for us to physically distinguish between the vacua and therefore they should all have the same properties under C, P and T (note however that $\textit{changing}$ vacua via sphaleron and instanton processes does have physical significance, but that is not the main concern of the question).
What is the resolution of this apparent paradox?
Will the pure gauge configurations that can be connected to the trivial vacuum via large gauge transformations (and hence have non-zero topological charge) be even under CP or odd?This post imported from StackExchange Physics at 2019-07-25 09:06 (UTC), posted by SE-user Optimus Prime |
Years ago I found this identity through experimentation playing with data and transformations. After explaining it to my statistics professor he came in the next class with a one-page proof using vector and matrix notation. Unfortunately I lost the paper he gave me. (This was back in 2007)
Is anyone able to reconstruct a proof?
Let $(x_i,y_i)$ be your original data points. Define a new set of data points by rotating the original set by angle $\theta$; call these points $(x'_i,y'_i)$.
The R squared value of the original set of points is equal to the negative product of the derivative with respect to $\theta$ of the natural log of the standard deviation for each coordinate of the new set of points, each evaluated at $\theta=0$
$r^2= - \left(\left.\frac{d}{d\theta}\ln(\sigma_{x'})\right|_{\theta=0} \right) \left(\left.\frac{d}{d\theta}\ln(\sigma_{y'})\right|_{\theta=0} \right)$ |
Take a functional $S[\varphi]$ on a $d$-dimensional space-time of the form
$S[\varphi]=\int d^d x\, L(x,\varphi,\partial_\mu \varphi,\partial_\mu \partial_\nu \varphi, \dots)$
where $\varphi(x)$ is a scalar function of $x^\mu$, and $L$ is a polynomial function of $\varphi$ and its derivatives, with $x^\mu$-dependent tensor-valued coefficients, and contains only even powers in $\varphi$ e.g.
$L = c_1(x) \varphi^2+ c^\mu_2(x)\varphi^3 \partial_\mu \varphi+c_3(x)^{\mu\nu}\partial_\mu \varphi \partial_\nu \varphi+c_4(x)^{\mu\nu}\varphi^3 \partial_\mu\partial_\nu \varphi+\cdots\ .$
Can one show that
$S[\varphi] \geq 0 \quad\text{for all }\varphi\qquad \Rightarrow\qquad L+\partial_\mu K^\mu\geq 0\quad\text{for all }\varphi ,$
where we consider configurations of $\varphi$ such that $\varphi$ and its derivatives vanish sufficiently fast at space and time infinity, and where $K^\mu$ is some vector that depends on $x^\mu$, $\varphi$ and the derivatives of $\varphi$?
In case of negative answer, is there any hope for the statement to be true if one restricts to more specific forms of $S[\varphi]$ (e.g. if one assumes the coefficients $c_1,\ c_2^\mu,\ \dots$ to be independent of $x^\mu$)? |
The spherical cap is no different but a part of the sphere itself that falls over the plane of a sphere. If you know the values of the base area, radius, and the height of a sphere then this is easy to calculate the area of the shape. The other popular name for the spherical cap is spherical dome. Sometimes, you will be given the base radius and sometimes you will be given the value of sphere radius.
Make sure that you know the difference between these two terms and apply the respective formula accordingly. Here is given a diagram to give you an accurate idea of the concept. Also, we have given the spherical cap volume formulas to make the thing easier for you.
\[\large V=\frac{\pi}{3}\,H^{2}(3R^{2}=H^{2})\]
Where,
H = height S = sphere radius A = base radius R = sphere radius
Based on the formula given earlier, H will be the height, R is the sphere radius and a is the base radius. The value of pi is fixed in mathematics.
The shape of a spherical sector is similar to the ice cream cone. In mathematics, a portion of the sphere with vertices in the middle and the conical boundary will be named as the spherical sector. The base of a spherical sector is termed as the zone as well. The other names for the bases are the union of a cone or spherical cap. The formula of a spherical sector is given as below –
\[\large surface\;Area\;of\;a\;sector\;of\;a\;sphere=\pi r(2h+a)\]
\[\large volume\;of\;a\;sector\;of\;a\;sphere=\frac{2\pi r^{2}h}{3}\]
If you know the height and radius of a spherical sector then calculating the volume of a spherical sector would be easier.
A spherical segment is a solid shape defined by cutting the shape into parallel planes. It can be taken as the spherical cap when the top is truncated, so it looks like a spherical frustum too. The base of the spherical segment is named as the zone. At some places, zone, spherical cap, and spherical segment terms are used interchangeably. So, you have to make sure of each of the concept and respective formulas too. The volume of a spherical segment is given as below –
\[\large surface\;Area\;of\;a\;segment\;of\;a\;sphere=2\pi Rh\]
\[\large volume\;of\;a\;segment\;of\;a\;sphere=\frac{\pi h}{6}(3r_{1}^{2}+r_{2}^{2}+h^{2})\]
Where R (r1 + r2) is the radius and ‘h’ is the height of a spherical segment.
When semi-circle is revolved around the diameter with less than 360-degree angle then the solid shape will be named as the spherical wedge. At the same time, Spherical Lune is the total space occupied within a sphere and surrounded by two half circles that will meet further at antipodal points. You must be shocked what we meant by antipodal here.
\[\large Volume=\frac{2}{3}\:R^{3}\theta\]
\[\large Surface\;Area=2R^{2}\theta\]
\[\large Arc\;Length\;at\;the\;equator=R\theta\]
This is the point marked on the surface of a sphere which is diametrically opposite to each other. When a line is drawn from one point to another it will make a straight line. Here, we have given the formulas for spherical wedge and Lune to calculate the Volume, surface area, and the arc length. |
Definition and Examples
Let a family of curves be given by the equation
\[g\left( {x,y} \right) = C,\]
where \(C\) is a constant. For the given family of curves, we can draw the orthogonal trajectories, that is another family of curves \(f\left( {x,y} \right) = C\) that cross the given curves at right angles.
For example, the orthogonal trajectory of the family of straight lines defined by the equation \(y = kx,\) where \(k\) is a parameter (the slope of the straight line), is any circle having center at the origin (Figure \(1\)):
\[{x^2} + {y^2} = {R^2},\]
where \(R\) is the radius of the circle.
Similarly, the orthogonal trajectories of the family of ellipses
\[{\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{c^2} – {a^2}}} = 1,\;\;}\kern0pt{\text{where}\;\;}\kern-0.3pt{0 \lt a \lt c,}\]
are confocal hyperbolas satisfying the equation:
\[{\frac{{{x^2}}}{{{b^2}}} – \frac{{{y^2}}}{{{b^2} – {c^2}}} = 1,\;\;}\kern0pt{\text{where}\;\;}\kern-0.3pt{0 \lt c \lt b.}\]
Both families of curves are sketched in Figure \(2.\) Here \(a\) and \(b\) play the role of parameters describing the family of ellipses and hyperbolas, respectively.
General Method of Finding Orthogonal Trajectories
The common approach for determining orthogonal trajectories is based on solving the partial differential equation:
\[\nabla f\left( {x,y} \right) \cdot \nabla g\left( {x,y} \right) = 0,\]
where the symbol \(\nabla\) means the gradient of the function \(f\left( {x,y} \right)\) or \(g\left( {x,y} \right)\) and the dot means the dot product of the two gradient vectors.
Using the definition of gradient, one can write:
\[{\nabla f\left( {x,y} \right) = \mathbf{grad}\,f\left( {x,y} \right) }={ \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right),}\]
\[{\nabla g\left( {x,y} \right) = \mathbf{grad}\,g\left( {x,y} \right) }={ \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}}} \right).}\]
Hence, the partial differential equation is written in the form:
\[
{\nabla f\left( {x,y} \right) \cdot \nabla g\left( {x,y} \right) = 0,\;\;}\Rightarrow {\left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right) \cdot \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}}} \right) = 0,\;\;}\Rightarrow {\frac{{\partial f}}{{\partial x}}\frac{{\partial g}}{{\partial x}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial g}}{{\partial y}} = 0.} \]
Solving the last PDE, we can determine the equation of the orthogonal trajectories \(f\left( {x,y} \right) = C.\)
A Practical Algorithm for Constructing Orthogonal Trajectories
Below we describe an easier algorithm for finding orthogonal trajectories \(f\left( {x,y} \right) = C\) of the given family of curves \(g\left( {x,y} \right) = C\) using only ordinary differential equations. The algorithm includes the following steps:
Construct the differential equation \(G\left( {x,y,y’} \right) = 0\) for the given family of curves \(g\left( {x,y} \right) = C.\) See the web page Differential Equations of Plane Curves about how to do this. Replace \(y’\) with \(\left( { – \large\frac{1}{{y’}}\normalsize} \right)\) in this differential equation. As a result, we obtain the differential equation of the orthogonal trajectories. Solve the new differential equation to determine the algebraic equation of the family of orthogonal trajectories \(f\left( {x,y} \right) = C.\) Solved Problems
Click a problem to see the solution. |
The answer is $90 ^{\circ}$Let's denote the points of tangency of the inscribed circle with $BC, AC, AB$ by $T_a, T_b, T_c$.
Lemma. ("
A-bisector, B-midline, C-touchchord") The point P lies on $T_cT_a$
This is a well-known fact and can be proven by mass point geometry. It will come in handy later.
Now, let's denote by $Q$ the point of intersection of $T_bI$ with perpendicular to $MN$ at $P$ and it remains to prove that $QN$ is indeed parallel to $BI$.
Let the angles $A, B, C$ be $\alpha, \beta, \gamma$.
The angle $BIQ$ is $90^{\circ} - \gamma - \frac{\beta}{2}$
So, in order to prove that $QN || BI$ we need to prove that $\angle T_bQN = 90^{\circ} - \gamma - \frac{\beta}{2}$.
Now, $Q, P, T_b, N$ lie on the same circle with diameter QN, so $\angle T_bQN=\angle T_bPN$ = angle between $BC$ and $PT_b$
Let's do a symmetry about the line $CI$. Under that symmetry the line $BC$ maps to the line $AC$ and the line $PT_b$ maps to $PT_a$.
So, our angle is the angle between $AC$ and $PT_a$. Now, by the "bisector, midline, touchchord " lemma, the line $PT_a$ is just the line $T_cT_a$.
The angle between $T_cT_a$ and $AC$ is easy to compute. It equals $90^{\circ} - \beta/2 - \gamma$, which finishes the proof. |
Advanced Monitoring Strategy
The algorithm incorporated in OptiLayer takes various criteria into account:
The strategy is available at:
Results --> Monitor --> Strategy button --> Strategy 4
\[ A=V_\max-V_\min \]
\[ S_{in}= \frac{V_\max-V_{in}}{A}\cdot 100\%, \]if the first extremum is maximum;\[ S_{in}= \frac{V_{in}-V_\min}{A}\cdot 100\%, \]otherwise.
\[ S_{fin}= \frac{V_\max-V_{fin}}{A}\cdot 100\%, \]if the last extremum is maximum;
\[ S_{fin}= \frac{V_{fin}-V_\min}{A}\cdot 100\%, \]otherwise, where \(V_{in}\) is the signal level at the start of the layer deposition, \(V_{fin}\) is the termination level.
In the case of a thin layer, having no extrema of the monitoring signal inside, OptiLayer considers continuously increasing thickness of the monitored layer above its nominal thickness until the next two extrema appears on the monitoring curve. These two virtual extrema are considered as the first and the last one in the conditions above.
Evidently, swing values close to 0% and 100% mean that signal is close to extremum values. Initial swing for the first layer is always equal to zero.
Another important parameter is the difference between the termination level \(V_{fin}\) and the next signal extremum \(E_{next}\):
\[\Delta=|V_{fin}-E_{next}|\]
OptiLayer chooses monitoring wavelengths in order to satisfy the following five monitoring conditions:
The values \(a_1, a_2, b_1, b_2, \epsilon, \delta\) are the parameters of the algorithm. The conditions are connected with the requirement that termination levels should be located at signal slopes with enough steepness. For example, reasonable values of these parameters are \(a_1=a_2=15\%,\) \(b_1=b_2=85\%\), \(\epsilon=4\%\), \(\delta=4\%\).
Generally, it may be not possible to find a sequence of wavelengths so that all conditions are satisfied simultaneously for all coating layers. Our algorithm chooses monitoring wavelength(s) so that the conditions are satisfied as close as possible.
It may happen that some conditions are more important than the others. For such cases, condition weights are introduced. The weights allow you to adjust relative importance of different conditions.
Bad Monitoring control allows you to visualize layers having monitoring signal with bad quality for reliable monitoring purposes.
Illustrating example: a model coating consisting of two layers on a Suprasil substrate; layer refractive indices equal to 2.35 and 1.45, layer physical thicknesses equal to 150 nm and 100 nm.
The corresponding signal for the case when monitoring wavelength is 500 nm is shown in Fig. 1 (upper panel). The corresponding parameters for Layer 1 are:
\[ A=93.21\%-64.63\%=28.58\%>\epsilon=4\%,\]
\[ S_{in}=0, \; \mbox{(excluded)}, \]
\[ S_{fin}=\frac{93.21\%-66.21\%}{28.58\%}\cdot 100\%=94.47\%>a_2=85\%,\]
\[\Delta=66.21\%-64.63\%=1.58\%<\delta=4\%.\]
The parameters for Layer 2 are:
\[ A=89.75\%-65.95\%=23.8\%>\epsilon=4\%,\]
\[ S_{in}=\frac{66.23\%-65.95\%}{23.8\%}\cdot 100\%=1.2\%<a_1=15\%, \]
\[ S_{fin}=\frac{89.75\%-89.25\%}{23.8\%}\cdot 100\%=2.1\%<b_1=15\%,\]
\[\Delta=89.25\%-65.95\%=23.3\%>\delta=4\%.\]
It is seen that some conditions are are satisfied, and others are violated; termination levels are very close to signal extrema.
Application of our algorithm gives the monitoring wavelength of 588 nm. The corresponding monitoring signal is shown in Fig. 1 (lower pane). The parameters are now for Layer 1 as follows:
\[ A=93.28\%-64.57\%=28.71\%>\epsilon=4\%,\]
\[ S_{in}=0, \; \mbox{(excluded)}, \]
\[ b_1\le S_{fin}=\frac{93.28\%-80.95\%}{28.71\%}\cdot 100\%=42.9\%\le b_2=85\%,\]
\[\Delta=80.95\%-64.57\%=16.38\%>\delta=4\%;\]
and for Layer 2:
\[ A=95.56\%-78.27\%=17.29\%>\epsilon=4\%,\]
\[ a_1=15\%\le S_{in}=\frac{80.95\%-78.27\%}{17.29\%}\cdot 100\%=15.5\%\le a_2=85\%, \]
\[ b_1=15\%\le S_{fin}=\frac{91.52\%-78.27\%}{17.29\%}\cdot 100\%=76.6\%\le b_2=85\%,\]
\[\Delta=95.56\%-91.52\%=4.04\%>\delta=4\%.\]
All conditions for both layers are satisfied.
A monitoring wavelength the wavelength of 700 nm is located in the middle of the antireflection spectral range.The corresponding signal is depicted in Fig. 2 (upper panel).
The advanced algorithm proposes the wavelength of 621 nm (corresponding monitoring signal is plotted in the lower panel of Fig. 2.
In order to compare the monitoring wavelengths of 700 nm and 621 nm from the practical point of view we perform a series of computational manufacturing experiments in order to estimate production yields.
Production yields for the wavelengths of 700 nm and 621 nm are estimated as 0.2% and 98.6%, respectively. See the details in Monochromatic Monitoring Simulation. |
I'm asked to show that $f(x)=\frac{1}{1+x^2}$ is Lebesgue integrable over the real line and that its integral is $\pi$.
I can bound this function by the measurable function $\phi=\sum_{n=0}^{\infty}\frac{1}{1+n^2}\chi_{I_n}$ where $I_n = (n-1,n]\cup [n,n+1)$ and $\chi$ means characteristic function. This function $\phi$ can be written as the limit of its increasing sequence of partial sums, so I know that its integral is $2\sum_{n=0}^{\infty}\frac{1}{1+n^2}$ which converges. This ensures that the integral of $f$ is bounded, and hence $f$ is Lebesgue integrable.
I'm a bit stuck on how to show that $\int f$ is $\pi$ however. I suspect that I'm intended to write $\int f$ as a limit of series like that above. By cutting up those characteristic functions ever more finely, I can express the integral of $f$ as $$\lim_{k\to \infty} \frac{2}{k} \sum_{n=0}^{\infty} \frac{1}{1+(n/k)^2}$$ but I don't at all know how to evaluate that limit (though apparently it does evaluate to $\pi$).
So, I suppose my question is whether this is a good way to go about this (the overall point is to use the theory of Lebesgue measure and Lebesgue integration), and whether this last limit I've written is somehow easily seen to converge to $\pi$? Or, should I try to express the function as a limit of some other functions? |
I know that the integral $\int_0^1 \frac{(x+1)^n-1}{x} dx,$ for $n \in \mathbb{Z}^+$, can be evaluated by expanding the numerator with the binomial theorem and integrating term by term. You get the nice expression $$\int_0^1 \frac{(x+1)^n-1}{x} dx = \sum_{k=1}^n \binom{n}{k} \frac{1}{k}.$$ My question is this:
Is there some other "nice" expression for $$\int_0^1 \frac{(x+1)^n-1}{x} dx,$$ when $n$ is a positive integer?
Mathematica and Wolfram Alpha both tell me that $$\int_0^1 \frac{(x+1)^n-1}{x} dx = n \text{HypergeometricPFQ}[\{1, 1, 1 - n\}, \{2, 2\}, -1],$$ but the hypergeometric function is just another way of writing $\sum \limits_{k=1}^n \binom{n}{k} \frac{1}{k}.$
I'm led to think that there might be such a nice expression because we do get one for the same integrand but slightly different bounds: $$\int_{-1}^0 \frac{(x+1)^n-1}{x} dx = \sum_{k=1}^{n} \binom{n}{k} \frac{(-1)^{k+1}}{k} = H_n,$$ where $H_n$ is the $n$th harmonic number. On the other hand, alternating binomial sums often evaluate to simpler expressions than the corresponding sums that don't alternate, so perhaps the existence of the simpler expression here doesn't mean much for my question.
I would accept a known special function as an answer (other than the hypergeometric one I already mention), or a simple expression in terms of well-known numbers, like the harmonic numbers. |
Al-Zamil, Qusay and Montaldi, James (2012)
Witten-Hodge theory for manifolds with boundary and equivariant cohomology. Differential Geometry and its Applications, 30 (2). pp. 179-194. ISSN 0926-2245 This is the latest version of this item.
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Abstract
We consider a compact, oriented, smooth Riemannian manifold $M$ (with or without boundary) and we suppose $G$ is a torus acting by isometries on $M$. Given $X$ in the Lie algebra and corresponding vector field $X_M$ on $M$, one defines Witten's inhomogeneous coboundary operator $\d_{X_M} = \d+\iota_{X_M}: \Omega_G^\pm \to\Omega_G^\mp$ (even/odd invariant forms on $M$) and its adjoint $\delta_{X_M}$. Witten \cite{Witten} showed that the resulting cohomology classes have $X_M$-harmonic representatives (forms in the null space of $\Delta_{X_M} = (\d_{X_M}+\delta_{X_M})^2$), and the cohomology groups are isomorphic to the ordinary de Rham cohomology groups of the set $N(X_M)$ of zeros of $X_M$. Our principal purpose is to extend these results to manifolds with boundary. In particular, we define relative (to the boundary) and absolute versions of the $X_M$-cohomology and show the classes have representative $X_M$-harmonic fields with appropriate boundary conditions. To do this we present the relevant version of the Hodge-Morrey-Friedrichs decomposition theorem for invariant forms in terms of the operators $\d_{X_M}$ and $\delta_{X_M}$. We also elucidate the connection between the $X_M$-cohomology groups and the relative and absolute equivariant cohomology, following work of Atiyah and Bott. This connection is then exploited to show that every harmonic field with appropriate boundary conditions on $N(X_M)$ has a unique $X_M$-harmonic field on $M$, with corresponding boundary conditions. Finally, we define the $X_M$-Poincar\'{e} duality angles between the interior subspaces of $X_M$-harmonic fields on $M$ with appropriate boundary conditions, following recent work of DeTurck and Gluck.
Item Type: Article Uncontrolled Keywords: Hodge theory, manifolds with boundary, equivariant cohomology, Killing vector fields Subjects: MSC 2010, the AMS's Mathematics Subject Classification > 35 Partial differential equations
MSC 2010, the AMS's Mathematics Subject Classification > 53 Differential geometry
MSC 2010, the AMS's Mathematics Subject Classification > 55 Algebraic topology
MSC 2010, the AMS's Mathematics Subject Classification > 57 Manifolds and cell complexes
Depositing User: Dr James Montaldi Date Deposited: 29 Apr 2011 Last Modified: 20 Oct 2017 14:12 URI: http://eprints.maths.manchester.ac.uk/id/eprint/1613 Available Versions of this Item Witten-Hodge theory for manifolds with boundary. (deposited 11 Apr 2010) Witten-Hodge theory for manifolds with boundary and equivariant cohomology. (deposited 29 Apr 2011) [Currently Displayed] Witten-Hodge theory for manifolds with boundary and equivariant cohomology. (deposited 29 Apr 2011) Actions (login required)
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So I am considering a projectile fired with velocity components $U_x$ and $U_y$ from position $(0,0)$ at time $t=0$ with drag forces proportional to the square of the speed in each direction (x and y directions) so I let the drag force in the x direction be $-cV_x^2$ and in the y direction be $-bV_y^2$. My first question is:
Are these drag forces typical of projectile motion through air only? Or, for the case in air, is dray force simply proportional to the speed (not the square)? If so, what are examples of where the resistive force would be proportional to the square of the speed?
Now my main problem is with signs. For motion in the x direction, the situation is simple as the net acceleration is
$a_{net}=\frac{-bV_x^2}{m}$
which can be integrated simply using separation of variables.
However when I am considering the y direction, I noticed that saying
$F_{drag}=-cV_y^2$
is wrong because $V_y^2$ is a positive quuantity, so then my frictional force does not always act in the opposite direction to motion. I would instead have to make
$F_{drag}=-c\mid V_y\mid V_y$
But then integrating the acceleration to find velocity would require integration of a term proportional to
$\int \frac{1}{\mid V_y\mid V_y}dV_y$
which I don't know how to approach. Is there a better way of doing this to get the correct signs? |
What are Polymers?
A polymer is a large molecule or a macromolecule which essentially is a combination of many subunits. The term polymer in Greek means ‘many parts’. Polymers can be found all around us. From the strand of our DNA which is a naturally occurring biopolymer to polypropylene which is used throughout the world as plastic.
Polymers may be naturally found in plants and animals (
natural polymers) or may be man-made ( synthetic polymers). Different polymers have a number of unique physical and chemical properties due to which they find usage in everyday life. Table of Content Classification of Polymers Structure Types Properties Polymers and their Monomers Polymerization Reactions Molecular Mass of Polymers Uses of Polymers FAQs
Polymers are all created by the process of polymerization wherein their constituent elements called monomers, are reacted together to form polymer chains i.e 3-dimensional networks forming the polymer bonds.
The type of polymerization mechanism used depends on the type of functional groups attached to the reactants. In biological contexts, almost all macromolecules are either completely polymeric or are made up of large polymeric chains.
Classification of Polymers
Polymers cannot be classified under one category because of their complex structures, different behaviours, and vast applications. We can, therefore, classify polymers based on the following considerations.
Classification of Polymers based on the Source of Availability
There are
three types of classification under this category, namely, Natural, Synthetic, and Semi-synthetic Polymers. Natural Polymers:
They occur naturally and are found in plants and animals. For example proteins, starch, cellulose, and rubber. To add up, we also have biodegradable polymers which are called biopolymers.
Semi-synthetic Polymers:
They are derived from naturally occurring polymers and undergo further chemical modification. For example, cellulose nitrate, cellulose acetate.
Synthetic Polymers:
These are man-made polymers. Plastic is the most common and widely used synthetic polymer. It is used in industries and various dairy products. For example, nylon-6, 6, polyether’s etc.
Also Read: Natural Polymers vs Synthetic Polymers
Classification of Polymers based on the Structure of the Monomer Chain
This category has the following classifications:
Linear Polymers
The structure of polymers containing long and straight chains fall into this category. PVC, i.e. poly-vinyl chloride is largely used for making pipes and electric cables is an example of a linear polymer.
Branched-chain Polymers
When linear chains of a polymer form branches, then, such polymers are categorized as branched chain polymers. For example, Low-density polythene.
Cross-linked Polymers
They are composed of bifunctional and trifunctional monomers. They have a stronger covalent bond in comparison to other linear polymers. Bakelite and melamine are examples in this category.
Other Ways to Classify Polymers Classification Based on Polymerization Addition Polymerization:Example, poly ethane, Teflon, Polyvinyl chloride (PVC) Condensation Polymerization:Example, Nylon -6, 6, perylene, polyesters. Classification Based on Monomers Homomer:In this type, a single type of monomer unit is present. For example, Polyethene Heteropolymer or co-polymer:It consists of different type of monomer units. For example, nylon -6, 6 Classification Based on Molecular Forces Elastomers:These are rubber-like solids weak interaction forces are present. For example, Rubber. Fibres:Strong, tough, high tensile strength and strong forces of interaction are present. For example, nylon -6, 6. Thermoplastics:These have intermediate forces of attraction. For example, polyvinyl chloride. Thermosetting polymers:These polymers greatly improve the material’s mechanical properties. It provides enhanced chemical and heat resistance. For example, phenolics, epoxies, and silicones. Structure of Polymers
Most of the polymers around us are made up of a
hydrocarbon backbone. A Hydrocarbon backbone being a long chain of linked carbon and hydrogen atoms, possible due to the tetravalent nature of carbon.
A few examples of a hydrocarbon backbone polymer are polypropylene, polybutylene, polystyrene. Also, there are polymers which instead of carbon have other elements in its backbone. For example, Nylon, which contains nitrogen atoms in the repeated unit backbone.
Types of Polymers
On the basis of the type of the backbone chain, polymers can be divided into:
Organic Polymers: Carbon backbone. Inorganic Polymers: Backbone constituted by elements other than carbon.
On the basis of their synthesis:
Biodegradable Polymers
The polymers which are degraded and decayed by microorganisms like bacteria are known as
biodegradable polymers. These types of polymers are used in surgical bandages, capsule coatings and in surgery. For example, Poly hydroxybutyrate co vel [PHBV] High-Temperature Polymers
These polymers are stable at high temperatures. Due to their high molecular weight, these are not destroyed even at very high temperatures. They are extensively used in the healthcare industries, for making sterilization equipment and in the manufacturing of heat and shock-resistant objects.
Few of the important polymers are: Polypropylene: It is a type of polymer that softens beyond a specific temperature allowing it to be moulded and on cooling it solidifies. Due to its ability to be easily moulded into various shapes, it has a lot of applications.
A few of which are in stationary equipment’s, automotive components, reusable containers speakers and much more. Due to its relatively low energy surface, the polymer is fused with the welding process and not using glue.
Polyethene: It is the most common type of plastic found around us. Mostly used in packaging from plastic bags to plastic bottles. There are different types of polyethene but their common formula being (C 2H 4) n. Properties of Polymers Physical Properties As chain length and cross-linking increases the tensile strength of the polymer increases. Polymers do not melt, they change state from crystalline to semi-crystalline. Chemical Properties Compared to conventional molecules with different side molecules, the polymer is enabled with hydrogen bonding and ionic bonding resulting in better cross-linking strength. Dipole-dipole bonding side chains enable the polymer for high flexibility. Polymers with Van der Waals forces linking chains are known to be weak, but give the polymer a low melting point. Optical Properties Due to their ability to change their refractive index with temperature as in the case of PMMA and HEMA: MMA, they are used in lasers for applications in spectroscopy and analytical applications. Some Polymers and their Monomers Polypropene, also known as polypropylene, is made up of monomer propene. Polystyrene is an aromatic polymer, naturally transparent, made up of monomer styrene. Polyvinyl chloride (PVC) is a plastic polymer made of monomer vinyl chloride. The urea-formaldehyde resin is a non-transparent plastic obtained by heating formaldehyde and urea. Glyptal is made up of monomers ethylene glycol and phthalic acid. Bakelite or polyoxybenzylmethylenglycolanhydride is a plastic which is made up of monomers phenol and aldehyde. Types of Polymerization Reactions Addition Polymerization
This is also called as chain growth polymerization. In this, small monomer units joined to form a giant polymer. In each step length of chain increases. For example, Polymerization of ethane in the presence of Peroxides
Condensation Polymerization
In this type small molecules like H
2O, CO, NH 3 are eliminated during polymerization (step growth polymerization). Generally, organic compounds containing bifunctional groups such as idols, -dials, diamines, dicarboxylic acids undergo this type of polymerization reaction. For example, Preparation of nylon -6, 6. What is Copolymerization? How to Calculate Molecular Mass of Polymers?
There are two types of average molecular masses of Polymers:
Number Average Molecular Masses Weight Average Molecular Mass Number Average Molecular Masses:
If N1, N2, N3…. are the number of macromolecular with molecular masses. M1, M2, M3….. Respectively then the number average molecular masses of the polymer is given by\(\bar{M_n} = \frac{N_1M_1+N_2M_2+N_3M_3+….\sum N_iM_i}{N_1+N_2+N_3+…..\sum N_i}\)
The number average molecular mass \(\bar{M_n}\) is determined by colligative properties such as Osmotic Pressure.
Weight Average Molecular Mass:
If m1, m2, m3…. Are the masses of a macromolecule with molecular masses M1, M2, M3… respectively, Then weight average molecular mass of the polymer is given by\({{\overline{M}}_{\omega }}=\frac{m{}_{1}{{M}_{1}}+{{m}_{2}}{{M}_{2}}+{{m}_{3}}{{M}_{3}}+…….}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\)
⇒ \(=\frac{\sum{miMi}}{\sum{mi}}\)
⇒ \({{\overline{M}}_{\omega }}=\frac{\sum{NiMi\times Mi}}{\sum{NiMi}}\)
⇒ \({{\overline{M}}_{\omega }}=\frac{\sum{NiM{{i}^{2}}}}{\sum{NiMi}}\)
\(PDI=\frac{\overline{M}w}{\overline{M}n}\). For natural polymers PDI = 1.
Polydispersive index: It is the ratio of weight average molecular mass and number average molecular mass of Polymers. Uses of Polymers
Here we will list some of the important uses of polymers in our everyday life.
Polypropene finds usage in a broad range of industries such as textiles, packaging, stationery, plastics, aircraft, construction, rope, toys, etc. Polystyrene is one of the most common plastic, actively used in the packaging industry. Bottles, toys, containers, trays, disposable glasses and plates, tv cabinets and lids are some of the daily-used products made up of polystyrene. It is also used as an insulator. The most important use of polyvinyl chloride is the manufacture of sewage pipes. It is also used as an insulator in the electric cables. Polyvinyl chloride is used in clothing and furniture and has recently become popular for the construction of doors and windows as well. It is also used in vinyl flooring. Urea-formaldehyde resins are used for making adhesives, moulds, laminated sheets, unbreakable containers, etc. Glyptal is used for making paints, coatings, and lacquers. Bakelite is used for making electrical switches, kitchen products, toys, jewellery, firearms, insulators, computer discs, etc. Commercial Uses of Polymers
Polymer Monomer Uses of Polymer Rubber Isoprene (1, 2-methyl 1 – 1, 3-butadiene) Making tyres, elastic materials BUNA – S (a) 1, 3-butadiene (b) Styrene Synthetic rubber BUNA – N (a) 1, 3-butadiene (b) Vinyl Cyanide Synthetic rubber Teflon Tetra Flouro Ethane Non-stick cookware – plastics Terylene (a) Ethylene glycol (b) Terephthalic acid Fabric Glyptal (a) Ethylene glycol (b) Phthalic acid Fabric Bakelite (a) Phenol (b) Formaldehyde Plastic switches, Mugs, buckets PVC Vinyl Cyanide Tubes, Pipes Melamine Formaldehyde Resin (a) Melamine (b) Formaldehyde Ceramic plastic material Nylon-6 Caprolactum Fabric FAQs on Polymers How do polymers have different physical properties? Give examples.
In polymers, monomers are bonded, by different molecular interactions. Nature of these interactions, yield polymers of varying elasticity, tensile strength, toughness, thermal stability, etc.
Monomers forming a linear chain with weak bonding. These polymers exhibit elasticity and are called elastomers. Example: Neoprene, Buna-S, Buna-R. Polymers with strong forces of interaction between the monomer in both linear and between the chains have higher tensile strength and are used as fibres. Example: Polyamides (nylon6,6), polyesters(terylene). Polymers having their intermolecular force in between the elastomers and fibres are thermoplastics. They can be repeatedly reprocessed without much change in their polymeric properties. Example: Polythene, polyvinyl. Monomers that undergo heavy branching, gets fused on heating and cannot be reused or reprocessed. Such materials are thermosetting plastics. Bakelite, Urea-formaldehyde are examples. What is the vulcanization of rubber?
Natural rubber is highly elastic to be of poor physical stability. Addition of 5% of sulphur, enhances the crosslinking of the linear chains and thus improves the stiffening of the rubber for an application like vehicle tires.
Match the column A with B.
Column A Column B 1 Buna -S a Ziegler Natta catalyst 2 Nylon 6-6 b Addition polymerization 3 High-density polyethene c Terephthalic acid ethylene glycol 4 Declon d Biodegradable polymer 5 Polymer of glycine and aminocaproic acid e Fibre Answer:
Column A Column B 1 Buna -S a Addition polymerization 2 Nylon 6-6 b Fibre 3 High-density polyethene c Ziegler Natta catalyst 4 Declon d Terephthalic acid 5 Polymer of glycine and aminocaproic acid e Biodegradable polymer What are biodegradable polymers? Give examples.
These polymers have functional groups found in the natural polymers.
Example: Poly β-hydroxybutyrate -co-β-hydroxy valerate (PHBV). This can be degraded by bacterial presence. What do you mean by Engineering plastics and synthetic metal?
These polymers possess high strength, resistance to chemical, thermal, abrasion similar to ceramics and metals for use in engineering applications. Examples: Silicone, Polycarbonate, ABS, Polysulfone.
The polymer that is similar to the metal in terms of their electrical, electronic, magnetic, and optical properties is termed as ‘intrinsically conducting polymer (ICP)’ or ‘synthetic metal’.
Stay tuned with BYJU’S to know more about polymers, biopolymers, and other interesting chemistry topics. |
Possible Duplicate: Countable set having uncountably many infinite subsets
Question:
Is it possible to find uncountably many infinite sets of natural numbers that any two of these sets have only finitely many common elements
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Possible Duplicate: Countable set having uncountably many infinite subsets
Question:
Is it possible to find uncountably many infinite sets of natural numbers that any two of these sets have only finitely many common elements
You can even get $2^\omega=\mathfrak c$ of them. Such families of sets are called
almost disjoint families.
One way to get an uncountable almost disjoint family of subsets of $\Bbb N$ is as follows. For each irrational number $x$ let $\langle q_x(k):k\in\Bbb N\rangle$ be a sequence of rational numbers converging monotonically to $x$. It’s easy to see that if $x$ and $y$ are distinct irrationals, the sequences $\langle q_x(k):k\in\Bbb N\rangle$ and $\langle q_y(k):k\in\Bbb N\rangle$ can have only finitely many terms in common. Now let $\varphi:\Bbb Q\to\Bbb N$ be any bijection, for each irrational $x$ let $D(x)=\{\varphi(q_x(k)):k\in\Bbb N\}$, and let $\mathscr{D}=\{D(x):x\in\Bbb R\setminus\Bbb Q\}$; $\mathscr{D}$ is an uncountable family of almost disjoint subsets of $\Bbb N$.
Added: Here’s another cute way. Imagine that you have an infinite plastic strip one unit wide. You pin some point on its midline to the origin, so that the strip can rotate about the origin. In each orientation it covers an infinite number of points of the integer lattice $\Bbb Z\times\Bbb Z$. For $\alpha\in[0,\pi)$ let $S_\alpha$ be the set of points in $\Bbb Z\times\Bbb Z$ covered by the strip when its midline coincides with the line $y=\alpha x$; it’s easy to see that $S\alpha\cap S_\beta$ is finite when $\alpha\ne\beta$. Now let $\varphi:\Bbb Z\times\Bbb Z\to\Bbb N$ be any injection, and for $\alpha\in[0,\pi)$ let $T_\alpha=\varphi[S_\alpha]$; the family $\{T_\alpha:\alpha\in[0,\pi)\}$ is then almost disjoint. |
I am trying to prove problem 1.59 in Sipser's book:
Introduction to the theory of computation , 2nd Edition.
Let $M=(Q,\Sigma,\delta,q_0,A)$ be a DFA and let $q'$ be a state of $M$ called its "home". A
Synchronizing sequencefor $M$ and $q'$ is a string $s\in \Sigma^*$ where $\delta (q,s)=q'$ for every $q\in Q$. (We actually have extended $\delta$ to strings so that $\delta(q,s)$ equals the state where $M$ ends up when $M$ starts at state $q$ and reads input $s$).
Say that $M$ is
Synchronizableif it has a synchronizing sequence for some state $q'$.
Prove that, if $M$ is a $k$-state synchronizable DFA, then it has a synchronizing sequence of length at most $k^3$.
This problem was already attempted here. I don't understand the answer though:
The proof of this a standard shrinking argument: if such a word is longer than $k^{2}$, then during the runs from $q1$,$q2$ a pair of states repeats, and we can shrink $w$.
How come a pair of states repeats?
I don't see a problem with runs from $q1$ and $q2$ reaching separate states. I don't so how I could be sure that I could shrink to word without disturbing both routes. |
Talk II on Bourguignon-Lawson's 1978 paper
The stable parametrized h-cobordism theorem provides a critical link in the chain of homotopy theoretic constructions that show up in the classification of manifolds and their diffeomorphisms. For a compact smooth manifold M it gives a decomposition of Waldhausen's A(M) into QM_+ and a delooping of the stable h-cobordism space of M. I will talk about joint work with Malkiewich on this story when M is a smooth compact G-manifold.
We show $C^\infty$ local rigidity for a broad class of new examples of solvable algebraic partially hyperbolic actions on ${\mathbb G}=\mathbb{G}_1\times\cdots\times \mathbb{G}_k/\Gamma$, where $\mathbb{G}_1$ is of the following type: $SL(n, {\mathbb R})$, $SO_o(m,m)$, $E_{6(6)}$, $E_{7(7)}$ and $E_{8(8)}$, $n\geq3$, $m\geq 4$. These examples include rank-one partially hyperbolic actions. The method of proof is a combination of KAM type iteration scheme and representation theory. The principal difference with previous work
that used KAM scheme is very general nature of the proof: no specific information about unitary representations of ${\mathbb G}$ or ${\mathbb G}_1$ is required.
This is a continuation of the last talk.
A classical problem in knot theory is determining whether or not a given 2-dimensional diagram represents the unknot. The UNKNOTTING PROBLEM was proven to be in NP by Hass, Lagarias, and Pippenger. A generalization of this decision problem is the GENUS PROBLEM. We will discuss the basics of computational complexity, knot genus, and normal surface theory in order to present an algorithm (from HLP) to explicitly compute the genus of a knot. We will then show that this algorithm is in PSPACE and discuss more recent results and implications in the field.
We show that the three-dimensional homology cobordism group admits an infinite-rank summand. It was previously known that the homology cobordism group contains an infinite-rank subgroup and a Z-summand. The proof relies on the involutive Heegaard Floer homology package of Hendricks-Manolescu and Hendricks-Manolescu-Zemke. This is joint work with I. Dai, M. Stoffregen, and L. Truong.
There is a close analogy between function fields over finite fields and number fields. In this analogy $\text{Spec } \mathbb{Z}$ corresponds to an algebraic curve over a finite field. However, this analogy often fails. For example, $\text{Spec } \mathbb{Z} \times \text{Spec } \mathbb{Z} $ (which should correspond to a surface) is $\text{Spec } \mathbb{Z}$ (which corresponds to a curve). In many cases, the Fargues-Fontaine curve is the natural analogue for algebraic curves. In this first talk, we will give the construction of the Fargues-Fontaine curve.
Consider a collection of particles in a fluid that is subject to
a standing acoustic wave. In some situations, the particles tend to
cluster about the nodes of the wave. We study the problem of finding a
standing acoustic wave that can position particles in desired locations,
i.e. whose nodal set is as close as possible to desired curves or
surfaces. We show that in certain situations we can expect to reproduce
patterns up to the diffraction limit. For periodic particle patterns, we
show that there are limitations on the unit cell and that the possible
patterns in dimension d can be determined from an eigendecomposition of a
2d x 2d matrix.
Department of Mathematics
Michigan State University
619 Red Cedar Road
C212 Wells Hall
East Lansing, MI 48824
Phone: (517) 353-0844
Fax: (517) 432-1562
College of Natural Science |
Consider the following language: $$ L_1=\{uu^rv \mid u,v\in\{0,1\}^+\}.$$ that means that neither $u$ nor $v$ can be $\varepsilon$. As usual $u^r$ refers to $u$ reflected.
I think that this language is not regular, but i am not sure. Any ideas?
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Consider the following language: $$ L_1=\{uu^rv \mid u,v\in\{0,1\}^+\}.$$ that means that neither $u$ nor $v$ can be $\varepsilon$. As usual $u^r$ refers to $u$ reflected.
I think that this language is not regular, but i am not sure. Any ideas?
Nice question. It is not regular.
Notice that this language consists of words where some nonempty prefix is an even palindrome.
Intersect $L$ with $(01)^{+} (10)^{+}$. If a word of this form has a palindromic prefix: $(01)^n (10)^m = u u^R v$ and $u \neq \varepsilon$, then the center of palindrome must be between the group of $01$ and the group of $10$. For example, take the word $0101011010101010$. The only possibility of breaking a prefix into a palindrome is $(010101\cdot 101010)\cdot1010$. (I leave the proof of this statement to you.)
Therefore, $L \cap (01)^{+} (10)^{+}=\{(01)^n (10)^m : m > n \geq 1\}$. However, this language is not regular - for example, each prefix $(01)^n$ is in a different Myhill-Nerode class. |
The Akra-Bazzi method
The Akra-Bazzi method gives asymptotics for recurrences of the form: $$ T(x) = \sum_{1 \le i \le k} a_i T(b_i x + h_i(x)) + g(x) \quad \text{for $x \ge x_0$} $$ This covers the usual divide-and-conquer recurrences, but also cases in which the division is unequal. The "fudge terms" $h_i(x)$ can cater for divisions that don't come out exact, for example. The conditions for applicability are:
There are enough base cases to get the recurrence going The $a_i$ and $b_i$ are all constants For all $i$, $a_i > 0$ For all $i$, $0 < b_i < 1$ $\lvert g(x) \rvert = O(x^c)$ for some constant $c$ as $x \rightarrow \infty$ For all $i$, $\lvert h_i(x) \rvert = O(x / (\log x)^2)$ $x_0$ is a constant
Note that $\lfloor b_i x \rfloor = b_i x - \{b_i x\}$, and as the sawtooth function $\{ u \} = u - \lfloor u \rfloor$ is always between 0 and 1, replacing $\lfloor b_i x \rfloor$ (or $\lceil b_i x \rceil$ as appropiate) satisfies the conditions on the $h_i$.
Find $p$ such that: $$ \sum_{1 \le i \le k} a_i b_i^p = 1 $$ Then the asymptotic behaviour of $T(x)$ as $x \rightarrow \infty$ is given by: $$ T(x) = \Theta \left( x^p \left( 1 + \int _1^x \frac{g(u)}{u^{p + 1}} du \right) \right) $$
Example
As an example, take the recursion for $n \ge 5$, where $T(0) = T(1) = T(2) = T(3) = T(4) = 17$: $$ T(n) = 9 T(\lfloor n / 5 \rfloor) + T(\lceil 4 n / 5 \rceil) + 3 n \log n $$ The conditions are satisfied, we need $p$: $$ 9 \left( \frac{1}{5} \right)^p + \left( \frac{4}{5} \right)^p = 1 $$ As luck would have it, $p = 2$. Thus we have: $$ T(n) = \Theta \left( n^2 \left(1 + \int_1^n \frac{3 u \log u}{u^3} du \right) \right) = \Theta(n^2) $$
(The help of maxima with the algebra is gratefully aknowledged) |
Chapters
Chapter 2: Relations
Chapter 3: Functions
Chapter 4: Measurement of Angles
Chapter 5: Trigonometric Functions
Chapter 6: Graphs of Trigonometric Functions
Chapter 7: Values of Trigonometric function at sum or difference of angles
Chapter 8: Transformation formulae
Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle
Chapter 10: Sine and cosine formulae and their applications
Chapter 11: Trigonometric equations
Chapter 12: Mathematical Induction
Chapter 13: Complex Numbers
Chapter 14: Quadratic Equations
Chapter 15: Linear Inequations
Chapter 16: Permutations
Chapter 17: Combinations
Chapter 18: Binomial Theorem
Chapter 19: Arithmetic Progression
Chapter 20: Geometric Progression
Chapter 21: Some special series
Chapter 22: Brief review of cartesian system of rectangular co-ordinates
Chapter 23: The straight lines
Chapter 24: The circle
Chapter 25: Parabola
Chapter 26: Ellipse
Chapter 27: Hyperbola
Chapter 28: Introduction to three dimensional coordinate geometry
Chapter 29: Limits
Chapter 30: Derivatives
Chapter 31: Mathematical reasoning
Chapter 32: Statistics
Chapter 33: Probability
RD Sharma Mathematics Class 11 Chapter 8: Transformation formulae Chapter 8: Transformation formulae Exercise 8.10 solutions [Pages 6 - 7]
Express the following as the sum or difference of sines and cosines:
2 sin 3x cos x
Express the following as the sum or difference of sines and cosines:
2 cos 3x sin 2xa
Express the following as the sum or difference of sines and cosines:
2 sin 4 x sin 3 x
Express the following as the sum or difference of sines and cosines:
2 cos 7x cos 3x
Prove that:
Prove that:
Prove that:
Show that :
Show that :
Prove that:
cos 10° cos 30° cos 50° cos 70° = \[\frac{3}{16}\]
Prove that:
cos 40° cos 80° cos 160° = \[- \frac{1}{8}\]
Prove that:
sin 20° sin 40° sin 80° = \[\frac{\sqrt{3}}{8}\]
Prove that:
cos 20° cos 40° cos 80° = \[\frac{1}{8}\]
Prove that:
tan 20° tan 40° tan 60° tan 80° = 3
Prove that:
tan 20° tan 30° tan 40° tan 80° = 1
Prove that:
sin 10° sin 50° sin 60° sin 70° = \[\frac{\sqrt{3}}{16}\]
Prove that:
sin 20° sin 40° sin 60° sin 80° = \[\frac{3}{16}\]
Show that:
sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0
Show that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
Prove that
\[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right) = \tan 3x\]
If α + β = \[\frac{\pi}{2}\], show that the maximum value of cos α cos β is \[\frac{1}{2}\].
Chapter 8: Transformation formulae Exercise 8.20 solutions [Pages 17 - 19]
Express each of the following as the product of sines and cosines:
sin 12x + sin 4x
Express each of the following as the product of sines and cosines:
sin 5x − sin x
Express each of the following as the product of sines and cosines:
cos 12x + cos 8x
Express each of the following as the product of sines and cosines:
cos 12x - cos 4x
Express each of the following as the product of sines and cosines:
sin 2x + cos 4x
Prove that:
sin 38° + sin 22° = sin 82°
Prove that:
cos 100° + cos 20° = cos 40°
Prove that:
sin 50° + sin 10° = cos 20°
Prove that:
sin 23° + sin 37° = cos 7°
Prove that:
sin 105° + cos 105° = cos 45°
Prove that:
sin 40° + sin 20° = cos 10°
Prove that:
cos 55° + cos 65° + cos 175° = 0
Prove that:
sin 50° − sin 70° + sin 10° = 0
Prove that:
cos 80° + cos 40° − cos 20° = 0
Prove that:
cos 20° + cos 100° + cos 140° = 0
Prove that:
\[\sin\frac{5\pi}{18} - \cos\frac{4\pi}{9} = \sqrt{3} \sin\frac{\pi}{9}\]
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
sin 47° + cos 77° = cos 17°
Prove that:
cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
Prove that:
cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
Prove that:
sin A + sin 2A + sin 4A + sin 5A = 4 cos \[\frac{A}{2}\]\[\frac{3A}{2}\]
Prove that:
sin 3A + sin 2A − sin A = 4 sin A cos \[\frac{A}{2}\] \[\frac{3A}{2}\]
Prove that:
cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −\[\frac{3}{4}\]
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
cos (A + B + C) + cos (A − B + C) + cos (A + B − C) + cos (− A + B + C) = 4 cos A cos Bcos C
If cosec A + sec A = cosec B + sec B, prove that tan A tan B = \[\cot\frac{A + B}{2}\].
Prove that:
Prove that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
If cos (α + β) sin (γ + δ) = cos (α − β) sin (γ − δ), prove that cot α cot β cot γ = cot δ
If y sin ϕ = x sin (2θ + ϕ), prove that (x + y) cot (θ + ϕ) = (y − x) cot θ.
If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), prove that tan A tan B tan C + tan D = 0.
If \[x \cos\theta = y \cos\left( \theta + \frac{2\pi}{3} \right) = z \cos\left( \theta + \frac{4\pi}{3} \right)\], prove that \[xy + yz + zx = 0\]
If \[m \sin\theta = n \sin\left( \theta + 2\alpha \right)\], prove that \[\tan\left( \theta + \alpha \right) \cot\alpha = \frac{m + n}{m - n}\]
Chapter 8: Transformation formulae solutions [Pages 20 - 21]
If (cos α + cos β)
2 + (sin α + sin β) 2 = \[\lambda \cos^2 \left( \frac{\alpha - \beta}{2} \right)\], write the value of λ.
Write the value of sin \[\frac{\pi}{12}\] sin \[\frac{5\pi}{12}\].
If sin A + sin B = α and cos A + cos B = β, then write the value of tan \[\left( \frac{A + B}{2} \right)\].
If cos A = m cos B, then write the value of \[\cot\frac{A + B}{2} \cot\frac{A - B}{2}\].
Write the value of the expression \[\frac{1 - 4 \sin 10^\circ \sin 70^\circ}{2 \sin 10^\circ}\]
If A + B = \[\frac{\pi}{3}\] and cos A + cos B = 1, then find the value of cos \[\frac{A - B}{2}\].
Write the value of \[\sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{3\pi}{10}\]
If sin 2A = λ sin 2B, then write the value of \[\frac{\lambda + 1}{\lambda - 1}\]
Write the value of \[\frac{\sin A + \sin 3A}{\cos A + \cos 3A}\]
If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), then write the value of tan A tan B tan C.
Chapter 8: Transformation formulae solutions [Pages 21 - 22]
cos 40° + cos 80° + cos 160° + cos 240° =
0
1
\[\frac{1}{2}\] \[- \frac{1}{2}\]
sin 163° cos 347° + sin 73° sin 167° =
0
\[\frac{1}{2}\]
1
None of these
If sin 2 θ + sin 2 ϕ = \[\frac{1}{2}\] and cos 2 θ + cos 2 ϕ = \[\frac{3}{2}\], then cos
2 (θ − ϕ) =
\[\frac{3}{8}\] \[\frac{5}{8}\] \[\frac{3}{4}\] \[\frac{5}{4}\]
The value of cos 52° + cos 68° + cos 172° is
0
1
2
`3/2`
The value of sin 78° − sin 66° − sin 42° + sin 60° is
\[\frac{1}{2}\] \[- \frac{1}{2}\]
−1
None of these
If sin α + sin β =
a and cos α − cos β = b, then tan \[\frac{\alpha - \beta}{2}\]= \[- \frac{a}{b}\] \[- \frac{b}{a}\]
\[\sqrt{a^2 + b^2}\]
None of these
cos 35° + cos 85° + cos 155° =
0
\[\frac{1}{\sqrt{3}}\] \[\frac{1}{\sqrt{2}}\]
cos 275°
The value of sin 50° − sin 70° + sin 10° is equal to
1
0
`1/2`
2
sin 47° + sin 61° − sin 11° − sin 25° is equal to
sin 36°
cos 36°
sin 7°
cos 7°
If cos A = m cos B, then \[\cot\frac{A + B}{2} \cot\frac{B - A}{2}\]=
\[\frac{m - 1}{m + 1}\] \[\frac{m + 2}{m - 2}\] \[\frac{m + 1}{m - 1}\]
None of these
If A, B, C are in A.P., then \[\frac{\sin A - \sin C}{\cos C - \cos A}\]=
tan B
cot B
tan 2 B
None of these
If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot Care in
GP
HP
AP
None of these
If sin x + sin y = \[\sqrt{3}\] (cos y − cos x), then sin 3x + sin 3y =
2 sin 3
x
0
1
none of these
If \[\tan\alpha = \frac{x}{x + 1}\] and
\[\frac{\pi}{2}\] \[\frac{\pi}{2}\] \[\frac{\pi}{2}\] \[\frac{\pi}{2}\] Chapter 8: Transformation formulae RD Sharma Mathematics Class 11 Textbook solutions for Class 11 RD Sharma solutions for Class 11 Mathematics chapter 8 - Transformation formulae
RD Sharma solutions for Class 11 Maths chapter 8 (Transformation formulae) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Class 11 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com are providing such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 11 Mathematics chapter 8 Transformation formulae are Transformation Formulae, Values of Trigonometric Functions at Multiples and Submultiples of an Angle, Sine and Cosine Formulae and Their Applications, 180 Degree Plusminus X Function, 2X Function, 3X Function, Expressing Sin (X±Y) and Cos (X±Y) in Terms of Sinx, Siny, Cosx and Cosy and Their Simple Applications, Concept of Angle, Introduction of Trigonometric Functions, Signs of Trigonometric Functions, Domain and Range of Trigonometric Functions, Trigonometric Functions of Sum and Difference of Two Angles, Trigonometric Equations, Truth of the Identity, Negative Function Or Trigonometric Functions of Negative Angles, 90 Degree Plusminus X Function, Conversion from One Measure to Another, Graphs of Trigonometric Functions.
Using RD Sharma Class 11 solutions Transformation formulae exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 11 prefer RD Sharma Textbook Solutions to score more in exam.
Get the free view of chapter 8 Transformation formulae Class 11 extra questions for Maths and can use Shaalaa.com to keep it handy for your exam preparation |
I remember slaving over the notation in this book when I was a bad undergraduate. It brings up some interesting memories, some which may help you.
$F(x)$ is the cumulative distribution of a single bidder's valuation. $G(x)$ is the cumulative distribution of the highest bidder's valuation, given $N$ bidders.
For the example you are referring to, values being uniformly distributed along $[0,1]$ implies $F(x) = x.$ That's pretty straightforward. The chance of you having a valuation of $\frac{1}{2}$ or less for the object is, well, $\frac{1}{2}$. The chance of you having at least a value of $1$ should be a probability of $1$.
But why is $G(x) = x^{N-1}$ ?
For $n =1$, the chance of your valuation being the largest valuation or less is...well, $1 \ (= x^0)$.
For $n = 2$, the chance of your valuation being the largest valuation or less is just the chance that the other bidder has a valuation lower than you, or exactly $x$.
For $n = 3$, the chance of your valuation being the biggest or less is just the chance that both other bidders value the object less. Since you're working with independent private values, you can think of their valuations as independent (duh) events, so you can just multiply the events together. (More on conditional probability here). So for example the chance of someone having a smaller value than you is $x$, but there is another person who also has to have a smaller value than you, with the same chance $x$. The chance of both of them having smaller values than you is $x \cdot x = x^2$.
So it goes, the more bidders (independent valuations) you have.
So we have your general formula for an optimal bid under a first-price auction:
$$\beta^1(x) = x - \int^x_0 \frac{G(y)}{G(x)} dy$$
So substitute in $G(x)$:
$$\beta^1(x) = x - \int^x_0 \frac{y^{N-1}}{x^{N-1}} dy$$
Evaluate:
$$= x \ - \ \biggr\lvert \frac{y^N}{Nx^{N-1}} + c\biggr\rvert^x_0$$$$= x - \frac{x}{N} = \frac{Nx}{N} - \frac{x}{N}$$$$\boxed{\beta^1(x) = x\frac{N - 1}{N}}$$
The next example in Krishna's book has an exponential distribution, but only with two bidders. If you try using the same line of reasoning as I used above to find $G(x)$, you'll notice things appear a bit difficult, but the author doesn't explicitly state $G(x)$ in this case in fact. Try seeing for yourself if you understand why Krishna gives the helpful statement:
$$\frac{G(y)}{G(x)} = \left[\frac{F(y)}{F(x)}\right]^{N-1}$$
for the generic case where there is no functional form for the distributions. |
Problem 15
Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent?
(a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ for $i=1,2,3,4$. (b) At $0$ each of the polynomials has the value $1$. Namely $p_i(0)=1$ for $i=1,2,3,4$.
(
University of California, Berkeley) Problem 12
Let $A$ be an $n \times n$ real matrix. Prove the followings.
(a) The matrix $AA^{\trans}$ is a symmetric matrix. (b) The set of eigenvalues of $A$ and the set of eigenvalues of $A^{\trans}$ are equal. (c) The matrix $AA^{\trans}$ is non-negative definite.
(An $n\times n$ matrix $B$ is called
non-negative definite if for any $n$ dimensional vector $\mathbf{x}$, we have $\mathbf{x}^{\trans}B \mathbf{x} \geq 0$.)
Add to solve later
(d) All the eigenvalues of $AA^{\trans}$ is non-negative. Problem 11
An $n\times n$ matrix $A$ is called
nilpotent if $A^k=O$, where $O$ is the $n\times n$ zero matrix. Prove the followings. (a) The matrix $A$ is nilpotent if and only if all the eigenvalues of $A$ is zero.
Add to solve later
(b) The matrix $A$ is nilpotent if and only if $A^n=O$. Read solution Problem 9
Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues.
Show that (1) $$\det(A)=\prod_{i=1}^n \lambda_i$$ (2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$
Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix $A$.
Namely, prove that (1) the determinant of $A$ is the product of its eigenvalues, and (2) the trace of $A$ is the sum of the eigenvalues.
Read solution Problem 5
Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation.
Let $\mathbf{0}_n$ and $\mathbf{0}_m$ be zero vectors of $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively. Show that $T(\mathbf{0}_n)=\mathbf{0}_m$.
(
The Ohio State University Linear Algebra Exam)
Add to solve later
Problem 3
Let $H$ be a normal subgroup of a group $G$.
Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$.
Here $[H, G]$ is a subgroup of $G$ generated by commutators $[h,k]:=hkh^{-1}k^{-1}$.
In particular, the commutator subgroup $[G, G]$ is a normal subgroup of $G$Add to solve later |
I found day 2 much harder than day 1, and I still don't know how to solve all the problems (I am seriously impressed by those getting perfect scores). Here's what I've managed to figure out so far.Now we can consider how to construct a replacement sequence (and also to count them), which also shows that these conditions are sufficient. If the phase is not locked, pick it arbitrarily. Now the "new gondola" column is simply the numbers from n+1 up to the largest gondola, so picking a replacement sequence is equivalent to deciding which gondola replaces each broken gondola. We can assign each gondola greater than n that we can't see to a position (one where the final gondola number is larger), and this will uniquely determine the replacement sequence. We'll call such gondolas
Update: I've now solved everything (in theory), and the solutions are below. The official solutions are now also available on the IOI website. I'll try coding the solutions at some point if I get time. Gondola
This was the easiest of the three. Firstly, what makes a valid gondola sequence? In all the subtasks of this problem, there will be two cases. If you see any of the numbers 1 to n, that immediately locks in the phase, and tells you the original gondola for every position. Otherwise, the phase is unknown. So, the constraints are that
if the phase is known, every gondola up to n must appear in the correct spot if it appears; no two gondolas can have the same number. hidden.
For the middle set of subtasks, the simplest thing is to assign all hidden gondolas to one position, the one with the highest-numbered gondola in the final state. For counting the number of possible replacement sequences, each hidden gondola can be assigned independently, so we just multiply together the number of options, and also remember to multiply by n if the phase is unknown. In the last subtask there are too many hidden gondolas to deal with one at a time, but they can be handled in batches (those between two visible gondolas), using fast exponentiation.
Friend
This is a weighted maximum independent set problem. On a general graph this is NP-hard, so we will need to exploit the curious way in which the graph is constructed. I haven't figured out how to solve the whole problem, but let's work through the subtasks:
This is small enough to use brute force (consider all subsets and check whether they are independent). The graph will be empty, so the sample can consist of everyone. The graph will be complete, so only one person can be picked in a sample. Pick the best one. The graph will be a tree. There is a fairly standard tree DP to handle this case: for every subtree, compute the best answer, either with the root excluded or included. If the root is included, add up the root-excluded answers for every subtree; otherwise add up the best of the two for every subtree. This takes linear time. In this case the graph is bipartite and the vertices are unweighted. This is a standard problem which can be solved by finding the maximum bipartite matching. The relatively simple flow-based algorithm for this is theoretically \(O(n^3)\), but it is one of those algorithms that tends to run much faster in most cases, so it may well be sufficient here.
We will process the operations in reverse order. For each operation, we will transform the graph into one that omits the new person, but for which the optimal solution has the same score. Let's say that the last operation had A as the host and B as the invitee, and consider the different cases:
YourFriendsAreMyFriends: this is the simplest: any solution using B can also use A, and vice versa. So we can collapse the two vertices into one whose weight is the sum of the original weights, and use it to replace A. WeAreYourFriends: this is almost the same, except now we can use at most one of A and B, and which one we take (if either) has no effect on the rest of the graph. So we can replace A with a single vertex having the larger of the two weights, and delete B. IAmYourFriend: this is a bit trickier. Let's start with the assumption that B will form part of the sample, and add that to the output value before deleting it. However, if we later decide to use A, there will be a cost to remove B again; so A's weight decreasesby the weight of B. If it ends up with negative weight, we can just clamp it to 0. Holiday
Consider the left-most and right-most cities that Jian-Jia visits. Regardless of where he stops, he will need to travel from the start city to one of the ends, and from there to the other end. There is no point in doing any other back-tracking, so we can tell how many days he spends travelling just from the end-points. This then tells us how many cities he has time to see attractions in, and obviously we will pick the best cities within the range.
That's immediately sufficient to solve the first test case. To solve more, we can consider an incremental approach. Fix one end-point, and gradually extend the other end-point, keeping track of the best cities (and their sum) in a priority queue (with the worst of the best cities at the front). As the range is extended, the number of cities that can be visited shrinks, so items will need to be popped. Of course, the next city in the range needs to be added each time as well. Using a binary heap, this gives an \(O(n^2\log n)\) algorithm: a factor of n for each endpoint, and the \(\log n\) for the priority queue operations. That's sufficient for subtask 3. It's also good enough for subtask 2, because the left endpoint will be city 0, saving a factor of n.
For subtask 4, it is clearly not possible to consider every pair of end-points. Let's try to break things up. Assume (without loss of generality) that we move first left, then back to the start, then right. Let's compute the optimal solution for the left part and the right part separately, then combine them. The catch is that we need to know how we are splitting up our time between the two sides. So we'll need to compute the answer for each side for all possible number of days spent within each side. This seems to leave us no better off, since we're still searching within a two-dimensional space (number of days and endpoint), but it allows us to do some things differently.
We'll just consider the right-hand side. The left-hand side is similar, with minor changes because we need two days for travel (there and back) instead of one. Let f(d) be the optimal end-point if we have d days available. Then with a bit of work one can show that f is non-decreasing (provided one is allowed to pick amongst ties). If we find f(d) for d=1, 2, 3, ... in that order, it doesn't really help: we're only, on average, halving the search space. But we can do better by using a divide-and-conquer approach: if we need to find f for all \(d \in [0, D)\) then we start with \(d = \frac{D}{2}\) to subdivide the space, and then recursively process each half of the interval on disjoint subintervals of the cities. This reduces the search space to \(O(n\log n)\). This still leaves the problem of efficiently finding the total number of attractions that can be visited for particular intervals and available days. The official solution uses one approach, based on a segment tree over the cities, sorted by number of attractions rather than position. The approach I found is, I think, simpler. Visualise the recursion described above as a tree; instead of working depth-first (i.e., recursively), we work breadth-first. We make \(O(\log n)\) passes, and in each pass we compute f(d) where d is an odd multiple of \(2^i\) (with \(i\) decreasing with each pass). Each pass can be done in a single incremental process, similar to the way we tackled subpass 2. The difference is that each time we cross into the next subinterval, we need to increase \(d\), and hence bring more cities into consideration. To do this, we need either a second priority queue of excluded cities, or we can replace the priority queue with a balanced binary tree. Within each pass, d can only be incremented \(O(n)\) times, so the total running time will be \(O(n\log n)\) per pass, or \(O(n\log n \log n)\) overall.
We'll just consider the right-hand side. The left-hand side is similar, with minor changes because we need two days for travel (there and back) instead of one. Let f(d) be the optimal end-point if we have d days available. Then with a bit of work one can show that f is non-decreasing (provided one is allowed to pick amongst ties). If we find f(d) for d=1, 2, 3, ... in that order, it doesn't really help: we're only, on average, halving the search space. But we can do better by using a divide-and-conquer approach: if we need to find f for all \(d \in [0, D)\) then we start with \(d = \frac{D}{2}\) to subdivide the space, and then recursively process each half of the interval on disjoint subintervals of the cities. This reduces the search space to \(O(n\log n)\).
This still leaves the problem of efficiently finding the total number of attractions that can be visited for particular intervals and available days. The official solution uses one approach, based on a segment tree over the cities, sorted by number of attractions rather than position. The approach I found is, I think, simpler. Visualise the recursion described above as a tree; instead of working depth-first (i.e., recursively), we work breadth-first. We make \(O(\log n)\) passes, and in each pass we compute f(d) where d is an odd multiple of \(2^i\) (with \(i\) decreasing with each pass). Each pass can be done in a single incremental process, similar to the way we tackled subpass 2. The difference is that each time we cross into the next subinterval, we need to increase \(d\), and hence bring more cities into consideration. To do this, we need either a second priority queue of excluded cities, or we can replace the priority queue with a balanced binary tree. Within each pass, d can only be incremented \(O(n)\) times, so the total running time will be \(O(n\log n)\) per pass, or \(O(n\log n \log n)\) overall. |
I am familiar with the definition of the $p$-adic Tate module of an elliptic curve defined over a $p$-adic field $k$ (a finite extension of $\mathbb{Q}_p$). But I have also seen some instances where the $p$-adic Tate module of a formal group is talked about. I wanted to know what the precise definition of $T_p(F)$ is, where $F$ is any formal group, like $T_p(\mathbb{G}_m)$. I could not find a satisfactory definition anywhere. Thank you.
The answer of Alex Youcis is perfect, and applies to formal groups of any dimension. Let me simplify for the one-dimensional case. This means your formal group has the shape $F(x,y)\in\mathfrak o[[x,y]]$ satisfying conditions that you evidently know, and where $\mathfrak o$ is probably the ring of integers in a finite extension of $\Bbb Q_p$ for some $p$.
As Alex says, to even have a Tate module, you need your formal group to be $p$-divisible, and for formal group laws of dimension one, this means that the $[p]$-endomorphism has a unit coefficient somewhere, the first such necessarily appearing in degree $p^h$ for some $h\ge1$. This number is called the
height of $F$, as again you probably already know.
The important thing is that the first unit coefficient of $[p^n](x)$ will be in degree $p^{nh}$, and this means that according to Weierstrass Preparation, the roots of $[p^n](x)$ in the maximal ideal of the integers of the algebraic closure of $\Bbb Q_p$ are the same as the roots of a polynomial $P_n(x)\in\mathfrak o[x]$, where $\deg(P_n)=p^{nh}$. The roots are all of multiplicity one, as is easily seen, and this means that $[p^n]$ has kernel of order $p^{nh}$. The kernels fit together in just the way you know from the theory of elliptic curves, and you get a Tate module that’s a free $\Bbb Z_p$-module of rank $h$.
I've personally only seen this notion defined when $\mathbb{G}$ is $p$-divisible. This means, that if we're working over $A$ (some Noetherian local ring with residue characteristic $p$, e.g. $\overline{\mathbb{F}_p}$ or $\mathbb{Z}_p$) then the map
$$[p^n]^\ast:A[[T_1,\ldots,T_n]]\to A[[T_1,\ldots,T_n]]$$
is finite and free (i.e. makes the RHS into a finite free module over the left) for all $n$ (equiv. just for $n=1$).
So, now suppose that $\mathbb{G}$ is such a formal group. Then, define for all $n\geqslant 1$
$$\mathbb{G}[p^n]=\text{Spec}\left(\frac{A[[ T_1,\ldots,T_n]]}{([p^n]^\ast(T_1),\ldots,[p^n]^\ast(T_n))}\right)$$
which, one shows (although not too difficultly) is a finite flat group scheme over $A$ of rank $p^{nh}$ (if $h$ is the height of your formal group over the residue field of $A$).
These $\mathbb{G}[p^n]$ fit together, in the obvious way, to create a $p$-divisble group denoted $\mathbb{G}[p^\infty]$ of height $h$. This notion is important since Tate showed, in his famous article on $p$-divisible groups, that $\mathbb{G}\mapsto \mathbb{G}[p^\infty]$ is an equivalence between connected $p$-divisible groups over $A$ and $p$-divisible formal groups over $A$ which, as already mentioned, preserves height.
So with all this being said let's now assume that $A=k$ is a field. Then, one defines
$$T_p\mathbb{G}:= T_p\mathbb{G}[p^\infty]:=\varprojlim \mathbb{G}[p^n](\overline{k})$$
as per usual. |
Newform invariants
Coefficients of the \(q\)-expansion are expressed in terms of a basis \(1,\beta_1,\beta_2,\beta_3\) for the coefficient ring described below. We also show the integral \(q\)-expansion of the trace form.
Basis of coefficient ring in terms of a root \(\nu\) of \(x^{4}\mathstrut -\mathstrut \) \(6\) \(x^{2}\mathstrut +\mathstrut \) \(4\):
\(\beta_{0}\) \(=\) \( 1 \) \(\beta_{1}\) \(=\) \( \nu \) \(\beta_{2}\) \(=\) \((\)\( \nu^{3} - 4 \nu \)\()/2\) \(\beta_{3}\) \(=\) \( \nu^{2} - 3 \)
\(1\) \(=\) \(\beta_0\) \(\nu\) \(=\) \(\beta_{1}\) \(\nu^{2}\) \(=\) \(\beta_{3}\mathstrut +\mathstrut \) \(3\) \(\nu^{3}\) \(=\) \(2\) \(\beta_{2}\mathstrut +\mathstrut \) \(4\) \(\beta_{1}\)
For each embedding \(\iota_m\) of the coefficient field, the values \(\iota_m(a_n)\) are shown below.
For more information on an embedded modular form you can click on its label.
\( p \) Sign \(3\) \(-1\) \(223\) \(-1\)
This newform can be constructed as the intersection of the kernels of the following linear operators acting on \(S_{2}^{\mathrm{new}}(\Gamma_0(6021))\):
\(T_{2}^{2} \) \(\mathstrut -\mathstrut 2 \) \(T_{5}^{4} \) \(\mathstrut -\mathstrut 6 T_{5}^{2} \) \(\mathstrut +\mathstrut 4 \) |
Thanks for providing more info. In general, stackoverflow is here to answer specific questions more of the "why does this code error out?" or "Why is my view always black even though I'm...." variety than questions about approach or code review. I think you are asking for help with handing the XML data that was sent to you, but you haven't supplied that code. Saying this is not very helpful to you, so I'm going to post an "answer" that can continue the dialog outside of comments. — tobinjim32 secs ago
Can a c function return an array? so that if i defined a function sayint sum(int A[5][5],int B[5][5]){int z[5][5],i,j;for(i=0;i<5;i++){for(j=0;j<5;j++){z[i][j]=A[i][j]+B[i][j];}}return z;}am i allowed to return z.?
http://www.codechef.com/APRIL15/problems/BROKPHONMy code is giving the correct o/p. But it's not accepted. Pls correct my code.#include<iostream>using namespace std;int main(){int t;cin>>t;while(t--){long int count = 0, N, *msg;cin>>N;if(N>=...
I've got an exercise for training my abilities to write algorithms when i were in job's interview.I got a description of exercise:A non-empty zero-indexed array A consisting of N integers is given. A pit in this array is any triplet of integers (P, Q, R) such that:0 ≤ P < Q < R < N;sequenc...
I was practicing some C and decided to write this simple command line utility for stripping leading and trailing white-space characters.The code:#include <ctype.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#define LINE_LENGTH 8096#define HELP_MESSAGE "Usage: trim [FILE1, [...
Generally using this amount of pointers in C++ code is frowned upon, and hints that the design of the tree classes/interfaces is not ideal. If you want us to look at your entire binary tree code, I suggest posting it in codereview.stackexchange.com — zennehoy8 secs ago
like full Documentation of my project. I hate even thinking about it, doesn't feel as rewarding as actually writing the code... but I know that it must be done, and now I have this project and about 4 others that I need to document as well.
Unit tests are not meant to describe what a method does, they're meant to describe what a particular workflow of a unit should yield. Note that I used unit, not method since a unit can span multiple methods.I will reiterate this from a recent post of mine on CodeReview: if you phrase the name o...
Every year we ask our users to tell us a little about themselves. This year we asked our users to tell us a lot.
For 2 weeks in February 2015, we ran a 45 question survey. We asked where you live, what programming languages & frameworks you use, how much money you make, how much coffee you drink, and whether you prefer tabs or spaces when writing code. More than 26,000 of you responded, making this year’s survey quite possibly the most authoritative developer survey ever conducted.
I am thinking that you could shorten up your private boolean methods a little bit by including the early return into the return conditional, so you would take thisprivate bool HasExplicitCallStatement(VBAParser.ECS_MemberProcedureCallContext context){if (context == null){re...
> 26,086 people from 157 countries participated in our 45-question survey. 6,800 identified as full-stack developers, 1,900 as mobile developers, 1,200 as front-end developers, 2 as farmers, and 12,000 as something else.
For a problem description, see other question.Imagine I would write this code at an interview. What would you say?Time complexity: \$O(n)\$Space complexity: \$O(n)\$Auxiliary space complexity: \$O(1)\$public class FindDeepestPit {public static void main(String[] args) {in...
@skiwi and @JeroenVannevel I don't think it has to do with being a bad student in a majority of cases, a lot of people need to work while they go to school these days, which means they would need to take less classes at a time in order to give each class the amount of study time required to perform well.
I've passed all but one programming-related course. The only ones I have left are economics and math classes. It's particularly hard to motivate yourself for some stupid bookkeeping stuff when you have all the knowledge needed to start working
I don't have time for a full review (at the moment) but the code below is O(N) and just looks for the maximal bounds of each pit:public static int findDeepestPitDepth( int[] heights ){int leftPosition = 0;int bottomPosition = 0;int rightPosition = 0;int maxDepth = 0;final int len...
See the previous iteration: A trivial command line utility for trimming whitespace from lines in CNow my code looks like this:#include <ctype.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#define HELP_MESSAGE "Usage: trim [-h] [-v] [FILE1, [FILE2, [...]]]\n" \...
I am trying to make my wordpressblogg responsive. When Im trying to set the sizes to 100% the posts gets really thin even though it says that the element is 360px wide. I don't get what I'm doing wrong here.Here is my style.css@media only screen and (max-width:1100px){.datebox, #sidebar {...
I'm trying to figure out which namespaces (not assemblies) are predefined in C#. By predefined, I mean built-in to the language and that, to use a predefined namespace, we don't have to reference any assemblies. For instance, I've created the following program.class Program{static void M...
@rolfl yup, I've seen it. I didn't quite understand what you were doing, the point of the inflection variable especially confused me. It seems like I didn't miss much though as it doesn't produce correct results.
I have a list of error messages:def errorMessages = ["Line : 1 Invoice does not foot Reported""Line : 2 Could not parse INVOICE_DATE value""Line 3 : Could not parse ADJUSTMENT_AMOUNT value""Line 4 : MATH ERROR"...
I adapted pseudocode for an optimized binary search tree algorithm from Cormen to run some sample data, but there is a logic error somewhere that is preventing correct output in the root list.My output is a two-dimensional list of nothing but zeros because t always ends up as inf. I've checked ...
I'm looking to pull all the youtube links from a string of text and was wondering how does this look?if (preg_match('/(https?:\/\/)?(www\.)?(youtube\.com)\/[^\s]*watch\?v=([a-zA-Z0-9_-]+)[^\s]*/gim', $this->content, $matches)) {}Obviously this doesn't take into account the youtu.be links a...
I am running a test load using Telerik Test Studio using 100 users at once entering search filters on a page that calls a stored procedure.My application that calls the stored procedure is an asp.net mvc 5 application and it returns "Timeout expired. The timeout period elapsed prior to completio...
This is my first attempt to a serious Node.js application, I've tried to follow the async style of Node at my best, but I have several doubts and I am sure that this code can be massively improved.The idea behind this little service is to get a minified stacktrace and return a 'translated' stac...
I have a code snippet which i wish to improve to increase my programs FPS, but i do not know how. I know the problem, which is the fact my counter mechanism used to delay attacks is causing lag. I just dont know how to change my code to minimize this lag. I am looking for helpful criticism and co...
I am not asking for a code review. I provided a working solution to demonstrate that I have tried to solve the problem, but, as I stated, my solution is a hack relying on overriding styles later on in the stylesheet. I am seeking other methods entirely, not a review of my own. — forTruce46 secs ago
The exercise 1.3 of the book Structure and Interpretation of Computer Programs asks the following:Exercise 1.3. Define a procedure that takes three numbers as arguments and returns the sum of the squares of the two larger numbers.My answer is this:#![feature(core)]use std::cmp;fn su...
^^^ That is the most interesting tidbit in the survey
> Upon closer examination of the data, a trend emerges: Developers increasingly prefer spaces as they gain experience. Stack Overflow reputation correlates with a preference for spaces, too: users who have 10,000 rep or more prefer spaces to tabs at a ratio of 3 to 1.
The task was to build a simple parallel server in erlang that prints what it receives through a socket and alerts when the socket is closed.-module(server).-export([start/0]).start() ->{ok, ListenSocket} = gen_tcp:listen(8211,[binary,{packet,4},{active, once}]),spawn(fun() -> acce...
The first if group can be slightly simplified using the else if:if (ageForPeople[placeValueHolder] <= 0){// Trigger error.}else if (ageForPeople[placeValueHolder] <= 18){agesForCensusGroups[0]++;}else if (ageForPeople[placeValueHolder] <= 30){agesForCensusGroups[1]++;}e...
When you get this all working correctly, I'd strongly suggest posting it to CodeReview.SE so that people can work help point out these small things that aren't that small and you can understand all the minor gotchas and ways to implement what you are working on better. — MichaelT1 min ago
Hmm, I prefer spaces, but prefer pressing tab. Also, I like backspace to clear a tab's worth of spaces at a time.
I like Xcode.
Also, I am slightly below average age for my country.
I told you guys the op wouldn't like my answer, @Mat'sMug
I can't believe you'd compare this to the ease of #import statements! :) ...that said, this is an extremely helpful guide, and it does look like this is the best we're going to get. Practically, it means that I'm going to write everything first and then chop it up into modules near the end of the project, which isn't that bad a way of doing things. Thank you very much, this will be really helpful. — Le Mot Juiced2 mins ago
"didn't wanted the question to become too long" I consider 'too long' to be when the site software approaches the point where it hits a char limit for posts. I experienced that on the File Browser GUI thread. That code was pushing 650 LOC. ;) — Andrew Thompson41 secs ago
Am trying to create a new Map <String, List<String>> headErrors with selective elements from another Map <String, List<String>> invoiceErrorLinesinvoiceErrorLines = ['1660277':['Line : 1 Invoice does not foot Reported', 'Line : 2 MATH ERROR'],'1660278':['Line : 5 Invoice d...
I'm working on this small program for practice and would like to know how I could improve it from here. I'm still a relative beginner with C++ and would like to know if there is a better why to design this program.Things I want to do:implement a while loop so that the program will continue t...
In the code below, it only builds successfully if I comment out the code on <Line 11>, the constructor for class Course.It also builds successfully if don't comment-out <Line 11> but instead I change <Line 18> to a pointer, as in list<Student> *roster. But then I am having trouble inserting ne...
I have created a Library Management system. Can anyone please look at if see if there's any bad practices or if anything could be done better.Main.cpp#include <stdlib.h>#include <iostream>#include <fstream>#include<iomanip>#include<stdio.h>#include <string.h>#include "Book.h"#include ...
I wrote a LISP REPL using parboiled2:Here's the AST that I used:sealed trait SExprsealed trait Atom extends SExprcase class Comb(sExprs: List[SExpr]) extends SExprcase class Number(x: Int) extends Atomcase class Ident(x: String) extends AtomMy main method, evaluate, has the type sig...
So I'm extremely amateur when it comes to C++ in general, but I wanted to make a little "Pyramid Escape" text based game. I'm not done whatsoever but before I continue I want to see what I could possible do differently.//// main.cpp// TextFiles//// Created by Neal Carico on 3/12/15.// ...
That is a keen observation, @SimonAndréForsberg. You're probably mistaken on a couple of things though (as well as me). As I look at it, I can't actually say for sure that the lookup table is faster. We're talking about 256 or so bytes in memory that will be cached very close to the core (L1 cache), but, 4 if/else statements may be faster than 1 L1 lookup. It's a margin call. I would guess that all the systems (if/else, 4-value table, and all-value table) will be about as fast as each other in this situation.... Code Readability is probably the important thing, and I am not sure mine is best. — rolfl ♦55 secs ago
@Simon - it will likely require a benchmark to prove which system is fastest.
This code finds expectation and standard deviation of sum(x) of 6 digited base-49 numbers with all digits distinct.The expectation is sum(y) of sums(x) upon the total no of sums(x),i.e. e=(x_1+x_2+..x_n)/n whereas standard deviation is squareroot of sum of squares of sums(x) minus the expec...
Please format your code so it is easy to read. That wouldn't only be helpful for us to help you but also for you to find those kind of errors easily. The checkbox construct especially is something I, as a code reviewer, would just throw in the bin. — b.enoit.be7 secs ago
@RubberDuck i develop VBA myself, and my comment was not supposed to transfer any negative message towards VBA development. I was just pointing out, that there is no error in the code, and that it is better placed in code review as you mentioned, because here i would expect more questions where things are not working at all or have errors. So from that perspective the code is fine and it surely can be optimized - but not here :) — hexerei software18 secs ago
This code finds expectation and standard deviation of sum(x) of 6 digited base-49 numbers with all digits distinct.The expectation is: \$\mu={\rm E[X]}=\frac1n\sum_{k=0}^{n}x_i=\frac{x_1+x_2+\ldots+x_n}{n}\$ whereas standard deviation is: \${\rm \sigma[X]}=\sqrt{{\rm E[(X-\mu)^2]}}=\sqrt{\f...
Welcome to Stack Overflow! According to your question, your script works. It's not clear what you are asking for help with. Stack Overflow is great when you have a specific question, but it's hard for us to provide help with "how can I make this better" questions. If you're looking for a code review, try codereview.stackexchange.com — Dave Bacher1 min ago
interface#import "DXYBaseViewControllerHD.h"// Displays a table view for its UI.//// Unlike `UITableViewController`, this controller's `view` is not the table// view itself. This means it is possible to insert content above the table// view.@interface DXYBaseTableViewControllerHD : DXYBa...
I wrote a sub to remove the blank entries in a row without shifting the cells around but it seems unnecessarily clunky and I'd like to get some advice on how to improve it.Public Sub removeBlankEntriesFromRow(inputRow As Range, pasteLocation As String)'Removes blank entries from inputRow a...
Related:What criteria do you use to upvote a question?What makes a good question?I feel that it is currently too easy to copy some code that you have written and put it up for review here on Code Review. My opinion is that if you want a review of your code, then you should also be able to ... |
OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.
The simplest somewhat-plausible way to handle #1 is to suppose that the value to you of your material assets is roughly proportional to the
logarithm of their total value or, equivalently, that the value to you of your "next" dollar is roughly inversely proportional to your current net wealth. (There's some evidence that it actually grows a bit slower than that, but it'll do).
The way I'll handle #2 is to suppose that the probability that you pay up when I win \$$X$ in the game is roughly proportional to $\frac a{a+X}$ where $a$ is a fairly large number; so when $X$ is small you almost certainly pay and when $X$ is very large you almost certainly don't. (We should expect $a$ to be of the same order of magnitude as your net wealth in dollars; it's the size of payoff at which I think it's equally likely that you will or won't pay up.)
Then, if my net wealth before playing the game is \$$w$ and your pay-or-don't parameter is \$$a$, the "correct" price \$$p$ is such that
$$\frac12\log\frac{w-p}w+\sum_{k=0}^{\infty}\frac1{2^{k+2}}\frac{a}{a+2^k}\log\frac{w-p+2^k}w=0.$$
This isn't the sort of thing we should expect to be able to solve analytically; so far as I know there is no nice closed form for that sum. But we can do it numerically. Here are some specific results. (I cannot guarantee that I haven't messed up the calculations, though the numbers seem plausible enough to me.)
Suppose your $a$ parameter is (\$)1000000, so that your probability of paying up has reduced to 1/2 by the time you owe me a megabuck. And suppose my own net wealth happens also to be one megabuck. Then I should be willing to pay about \$4.87. Suppose my net wealth remains at \$1M but now $a$ is \$$10^9$; even once you owe me a billion dollars you might well pay up. Then the amount I'm willing to pay increases to about ... \$5.46. Suppose you remain a billionaire but I am much poorer, having only \$1000 to my name. Then the amount I'm willing to pay goes down to \$2.98. (It goes down because the poorer I am, the faster the value-to-me of a dollar decreases as I get more.) Suppose we're both poor, so I'm at \$1000 and your $a$ parameter is the same. Then I am willing to pay about \$2.39. Suppose I am a billionaire and you have infinitely much money. Then I am willing to pay about \$12.80.
So I'm fairly comfortable answering the original question as follows:
I am willing to pay somewhere between about \$2 and about \$13, depending on how wealthy we both are and how much I trust you to pay up even if doing so is painful for you. |
Tagged: determinant Problem 548
An $n\times n$ matrix $A$ is said to be
invertible if there exists an $n\times n$ matrix $B$ such that $AB=I$, and $BA=I$,
where $I$ is the $n\times n$ identity matrix.
If such a matrix $B$ exists, then it is known to be unique and called the
inverse matrix of $A$, denoted by $A^{-1}$.
In this problem, we prove that if $B$ satisfies the first condition, then it automatically satisfies the second condition.
So if we know $AB=I$, then we can conclude that $B=A^{-1}$.
Let $A$ and $B$ be $n\times n$ matrices.
Suppose that we have $AB=I$, where $I$ is the $n \times n$ identity matrix.
Prove that $BA=I$, and hence $A^{-1}=B$.Add to solve later
Problem 452
Let $A$ be an $n\times n$ complex matrix.
Let $S$ be an invertible matrix. (a) If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix. (b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.
Add to solve later
(c) Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$. Problem 438
Determine whether each of the following statements is True or False.
(a) If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$. (b) If the characteristic polynomial of an $n \times n$ matrix $A$ is \[p(\lambda)=(\lambda-1)^n+2,\] then $A$ is invertible. (c) If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible. (d) If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$. (e) If $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ with corresponding eigenvalue $\lambda_1$, and if $\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_2$, then $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$.
(
Stanford University, Linear Algebra Exam Problem) Read solution Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$.
Add to solve later
(b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. Problem 391 (a) Is the matrix $A=\begin{bmatrix} 1 & 2\\ 0& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 1& 2 \end{bmatrix}$? (b) Is the matrix $A=\begin{bmatrix} 0 & 1\\ 5& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ 4& 3 \end{bmatrix}$? (c) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 0& 2 \end{bmatrix}$?
Add to solve later
(d) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ -1& 4 \end{bmatrix}$? Problem 389 (a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$. (b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$. (c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of $A$?
(
Harvard University, Linear Algebra Exam Problem) Problem 374
Let \[A=\begin{bmatrix}
a_0 & a_1 & \dots & a_{n-2} &a_{n-1} \\ a_{n-1} & a_0 & \dots & a_{n-3} & a_{n-2} \\ a_{n-2} & a_{n-1} & \dots & a_{n-4} & a_{n-3} \\ \vdots & \vdots & \dots & \vdots & \vdots \\ a_{2} & a_3 & \dots & a_{0} & a_{1}\\ a_{1} & a_2 & \dots & a_{n-1} & a_{0} \end{bmatrix}\] be a complex $n \times n$ matrix. Such a matrix is called circulant matrix. Then prove that the determinant of the circulant matrix $A$ is given by \[\det(A)=\prod_{k=0}^{n-1}(a_0+a_1\zeta^k+a_2 \zeta^{2k}+\cdots+a_{n-1}\zeta^{k(n-1)}),\] where $\zeta=e^{2 \pi i/n}$ is a primitive $n$-th root of unity. Problem 363 (a) Find all the eigenvalues and eigenvectors of the matrix \[A=\begin{bmatrix} 3 & -2\\ 6& -4 \end{bmatrix}.\]
Add to solve later
(b) Let \[A=\begin{bmatrix} 1 & 0 & 3 \\ 4 &5 &6 \\ 7 & 0 & 9 \end{bmatrix} \text{ and } B=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}.\] Then find the value of \[\det(A^2B^{-1}A^{-2}B^2).\] (For part (b) without computation, you may assume that $A$ and $B$ are invertible matrices.) Problem 338
Each of the following sets are not a subspace of the specified vector space. For each set, give a reason why it is not a subspace.
(1) \[S_1=\left \{\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \R^3 \quad \middle | \quad x_1\geq 0 \,\right \}\] in the vector space $\R^3$. (2)\[S_2=\left \{\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \R^3 \quad \middle | \quad x_1-4x_2+5x_3=2 \,\right \}\] in the vector space $\R^3$.
(3)\[S_3=\left \{\, \begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2 \quad \middle | \quad y=x^2 \quad \,\right \}\] in the vector space $\R^2$.
(4)Let $P_4$ be the vector space of all polynomials of degree $4$ or less with real coefficients.
\[S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}\] in the vector space $P_4$.
(5)\[S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}\] in the vector space $P_4$. (6)Let $M_{2 \times 2}$ be the vector space of all $2\times 2$ real matrices.
\[S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\} \] in the vector space $M_{2\times 2}$.
(7)\[S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\} \] in the vector space $M_{2\times 2}$.
(
Linear Algebra Exam Problem, the Ohio State University) (8)Let $C[-1, 1]$ be the vector space of all real continuous functions defined on the interval $[a, b]$.
\[S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\} \] in the vector space $C[-2, 2]$.
(9)\[S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$. (10)Let $C^2[a, b]$ be the vector space of all real-valued functions $f(x)$ defined on $[a, b]$, where $f(x), f'(x)$, and $f^{\prime\prime}(x)$ are continuous on $[a, b]$. Here $f'(x), f^{\prime\prime}(x)$ are the first and second derivative of $f(x)$.
\[S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$.
(11)Let $S_{11}$ be the set of real polynomials of degree exactly $k$, where $k \geq 1$ is an integer, in the vector space $P_k$. (12)Let $V$ be a vector space and $W \subset V$ a vector subspace. Define the subset $S_{12}$ to be the complementof $W$,
\[ V \setminus W = \{ \mathbf{v} \in V \mid \mathbf{v} \not\in W \}.\] Add to solve later |
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Now showing items 1-10 of 18
J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ... |
Oscillations and Waves Force Law and Energy in Simple Harmonic Motion The rate of change of velocity gives acceleration . The acceleration of the particle in SHM a = − Aw 2sin wt
where A = Amplitude, w = Angular frequency, t = time.
If 'F' is force acting on SHM at a displacement 'Y' from mean position F = −mw 2y (w = Angular frequency) Using Newtons second law of motion F = −mw 2y and f = −Kx ⇒ \tt w = \sqrt{\frac{K}{M}} The work done to displace simple harmonic oscillator is stored in the form of potential energy U = \frac{1}{2} mw^{2} x^{2} = \frac{1}{2} mw^{2} \sin^{2} wt At mean position KE is maximum and PE is minimum at extreme position KE is minimum and PE is maximum. But total energy is constant. Force Law of Simple Harmonic Motion View the Topic in this video From 0:19 To 5:36 View the Topic in this video From 0:18 To 12:40
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Both velocity and acceleration of a body executing simple harmonic motion are periodic functions. Velocity amplitude is
v max = ωA and acceleration amplitude is a= ω max 2A. |
Three - dimensional Geometry Direction Cosines and Direction Ratios of a Line If l, m, n are the direction cosines of a line, then l 2+ m 2+ n 2= 1. i.e., cos 2α + cos 2β + cos 2γ = 1 i.e., sin 2α + sin 2β + sin 2γ = 2 If A(a 1, b 1, c 1) B(a 2, b 2, c 2), then direction ratios of a line joining points A and B is = (a 2− a 1, b 2− b 1, c 2− c 1) If a, b, c are the direction ratios of the line then its direction cosines are \left(\frac{\pm a}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{\pm b}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{\pm c}{\sqrt{a^{2}+b^{2}+c^{2}}}\right) If l, m, n are the direction cosines of the line then its direction ratios are = (kl, km, kn) ∀ k ∈ R View the Topic in this video From 22:50 To 31:18
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1. If l, m, n are the direction cosines of a line, then l
2 + m 2 + n 2 = 1.
2. Direction cosines of a line joining two points P(x
1, y 1, z 1) and Q(x 2, y 2, z 2) are \frac{x_{2}-x_{1}}{PQ},\frac{y_{2}-y_{1}}{PQ},\frac{z_{2}-z_{1}}{PQ} where PQ=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}
3. If l, m, n are the direction cosines and a, b, c are the direction ratios of a line then l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}};m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}};n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}} |
One shouldn't expect to have a "good" formula for the local isometric embeddings of a constant negative curvature surface in
Euclidean $\mathbb{R}^3$. This is due to a little theorem proved by David Hilbert around 1901:
Theorem There does not exist a smooth immersion of the hyperbolic plane into Euclidean 3 space.
The theorem has been further studied in the years following. In 1961 Efimov showed that any
complete surface with curvature strictly bounded above (that is to say, if there exists a negative number $K_0 < 0$ such that the Gaussian curvature is always strictly less than $K_0$) cannot admit a smooth (twice continuously differentiable) isometric immersion into Euclidean three space.
That is to say, if you try to "extend" any surface in Euclidean 3 space that satisfies constant negative curvature, you are guaranteed to hit a singularity. In particular, you cannnot expect the surface to be described by $F(x,y,z) = 0$ where $F(x,y,z)$ has a nice algebraic expression (say, polynomial) and has smooth level sets.
Typically the image one usually use to illustrate the notion of
negative (but not constant) curvature is the graph $$ z = x^2 - y^2 $$which produces a classical saddle, or the catenoid whose Gaussian curvature, while everywhere negative, is not constant. (Though it has constant [in fact everywhere vanishing] mean curvature.)
Lastly, however, despite the above, it is possible to embed "patches" of hyperbolic plane into Euclidean 3 space. There are many ways of doing so (one can search for the term
pseudosphere; though some people use the same term for the hyperboloid/de Sitter spaces embedded in higher dimensional Minkowski space), but one of the more well-known is the tractricoid. (See Wiki entry here.) Parametrically in cylindrical coordinates $(z,r,\theta)$ the surface can be described by:
$$ \mathbb{R}_+\times\mathbb{S}^1 \ni (t,\omega) \mapsto \left(z=\frac{1}{\cosh t},r=t-\tanh t , \theta = \omega\right) $$
and has constant negative curvature. |
Dual Nature of Matter and Radiation Electron Emission and Experimental Study of Photoelectric Effect Photoelectric effect supports the quantum theory of light. When light of suitable frequency with certain wavelength falls on some alkali metals, electrons are released from the metal. This effect is known as PHOTO ELECTRIC EFFECT. Photoelectric effect is interaction of energy with matter THRESHOLD FREQUENCY (ν 0), is the minimum frequency of the incident radiation below which no photoelectric emission takes place. THRESHOLD WAVE LENGTH (λ 0), is the maximum wave length of the incident radiation above which no photo electric emission takes place. WORK FUNCTION (W) is the minimum energy required to remove a free electron from the metal \tt W=h\nu_{0}=\frac{hc}{\lambda_{0}} The minimum retarding potential to be applied to the collector to just stop the electrons of maximum K.E reaching the anode is called STOPPING POTENTIAL The maximum K.E. emitted from of the electron is equal to the product of stopping potential, V sand electron charge, e. k max= eV s The stopping potential, (V s) depends on (a) nature of the metal or work function (b) frequency or wavelength of the incident radiation V sstopping potential is independent of intensity of the incident radiation. View the Topic in this video From 00:08 To 19:03
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1. The most energetic photoelectrons, with the maximum kinetic energy (K
max), so that K max = e V 0
2. Photon Counts Emitted by a Source Per Second
E = hv = \frac{hc}{\lambda} |
I'll keep this short. (I assume you are just asking generally and not about a specific application. Writing a book on all uses of the zener is beyond the scope here.)
The
zener voltage is probably better seen as the "break down voltage" for the reverse-biased zener diode. If you impress a higher voltage on it, it will collapse from the applied voltage and allow huge currents to flow if there isn't something else to limit those currents.
So if you take the 6V2 and hook up a \$10\:\text{V}\$ power supply to it, you will pretty much destroy the zener.
However, if you put a \$680\:\Omega\$ resistor in series with the zener and that same \$10\:\text{V}\$ power supply, then at first there will be no current (for all intents) and the voltage drop across the resistor starts out at \$0\:\text{V}\$ (for the tiniest moment.) This impresses the entire \$10\:\text{V}\$ across the zener, which immediately begins to collapse and start a current flowing. The current rapidly rises and as it does the voltage drop across the \$680\:\Omega\$ resistor increases, thereby reducing the voltage difference across the zener (good thing.) Eventually, the whole process stabilizes when the voltage drop across this resistor is about \$3.8\:\text{V}\$, leaving the desired \$6.2\:\text{V}\$ across the zener itself. At this point, the zener stops increasing the current and just allows the impressed voltage across it to remain stable at this value.
Different zeners will be designed to reach this stable point at different voltages. It is your job as a designer to make sure that the current that results in the zener is the rated value (approximately.) In your datasheet example, this current is \$5\:\text{mA}\$. So, with the \$680\:\Omega\$ resistor I mentioned, we can expect about \$\frac{3.8\:\text{V}}{680\:\Omega}\approx 5.6\:\text{mA}\$. And this is close enough to the spec that you can expect about the right voltage across the 6V2 zener.
From this discussion you have your answer about \$I_Z\$.
Note also that this datasheet includes maximums and minimums for the zener voltage. This means that you cannot actually expect a precise \$6.2\:\text{V}\$ from the 6V2, but instead \$6.2\pm 0.4\:\text{V}\$. This is over the range of parts you might find in a box, or in a bunch of different boxes of them bought at different times. They are telling you that you cannot expect too much accuracy from these devices.
The value of \$Z_Z\$ can be used to estimate the worst case variation of the voltage across the zener, if you know the current variations. So let's continue with the 6V2 with \$Z_Z=10\:\Omega\$. We just computed an estimate of \$5.6\:\text{mA}\$ using a \$680\:\Omega\$ resistor and assuming an exact zener voltage (that we now know we can't be entirely sure of.)
Let's see where that takes us. The zener voltage for the 6V2 should be \$6.2\pm 0.4\:\text{V}\$. Assuming a 1% resistor of \$680\:\Omega\$, we may have a current ranging from \$\frac{10\:\text{V}-6.6\:\text{V}}{680\:\Omega+1\%}\approx 4.95\:\text{mA}\$ to \$\frac{10\:\text{V}-5.8\:\text{V}}{680\:\Omega-1\%}\approx 6.24\:\text{mA}\$. A difference of about \$1.25\:\text{mA}\$. While we don't know the exact voltage for some specific zener here, we can still estimate that there will be an additional variation of about \$1.25\:\text{mA}\cdot 10\:\Omega\approx 12.5\:\text{mV}\$ due to \$Z_Z\$.
This is actually not so important here, though. It's just mathy number twisting, really. Where it becomes important is instead when you add a circuit that uses the zener voltage. Often, this is an emitter follower BJT. (See this question: Explain the logic of a 12 V to 9 V conversion.) The base of the BJT will require some base current and this base current will vary depending on the load requirements.
So the point here is that a designer can estimate the load current variation for some larger circuit that uses the zener. And from this load current variation estimate a base current variation. And from this base current variation and \$Z_Z\$ estimate how much the zener voltage will vary due to the load current variation.
This may be important (or not.) But it gives you a starting point to estimate how bad it might be once you calibrate your circuit and start applying a realistic load, now.
The value of \$I_R\$ includes a referenced voltage in the table. This basically helps you to understand how much
prior leakage you can expect from the zener diode if the impressed reverse voltage is LESS than the face value. So if you have the 6V2, you can see that they specify it for \$4\:\text{V}\$, which is well below their minimum of \$5.8\:\text{V}\$. (But it is also as big as possible, short of that, so that the leakage current will be a "worst case" scenario.) So if you didn't use a \$10\:\text{V}\$ power supply but instead applied \$4\:\text{V}\$, then this value of \$I_R\$ is the worst you would expect to see (about \$3\:\mu\text{A}\$.) This would produce only about \$2\:\text{mV}\$ across the \$680\:\Omega\$ resistor, in the example case I've been discussing here. But there are other circumstances where this leakage might be more important to know. |
Given two uncorrelated strategies, each with a Sharpe ratio of 1, what is the of Sharpe ratio of the ensemble?
If we assume that by ensemble you mean an equally weighted portfolio of the two. We can express that portfolio as $$P = \frac{1}{2}x + \frac{1}{2}y$$ and the sharpe ratio of $P$, $S(P)$, will be $$\frac{\frac{1}{2}\mu_x + \frac{1}{2}\mu_y - r_f}{\sigma_{\frac{1}{2}x + \frac{1}{2}y}}$$ becuase $x$ and $y$ are uncorellated, this reduces to $$\frac{\mu_x + \mu_y - 2r_f}{\sqrt{\sigma_x^2 + \sigma_y^2}}$$ becuase the sharpe ratios $$S(x)=\frac{\mu_x - r_f}{\sigma_x}=S(y)=\frac{\mu_y - r_f}{\sigma_y} = 1$$ we get $$\mu_x - r_f = \sigma_x \\\mu_y - r_f = \sigma_y $$ thus $$\mu_x + \mu_y - 2r_f = \sqrt{\sigma_x^2} + \sqrt{\sigma_y^2} $$ and $$S(P) = \frac{\sqrt{\sigma_x^2} + \sqrt{\sigma_y^2}}{\sqrt{\sigma_x^2 + \sigma_y^2}}$$ What can you say about this ratio? How does it relate to Jensen's inequality? what happens if they are perfectly correlated?
Of course, it depends on the weights of your 'ensemble'. The optimal combination will have the following Sharpe ratio:
$$ S_{opt} = \sqrt{S_1^2+S_2^2} $$
i.e. $S_{opt} = \sqrt{2} \approx 1.414$ in you example
Proof:Let $x$ be the expectation, and $V$ the covariance matrix of a vector of assets. The Sharpe ratio of a portfolio with weights $w$ is defined by $S_w=\frac{x^Tw}{\sqrt{w^TVw}}$.
First, we transform the problem in a simpler one:
It follows that if $w_1$ has an optimal Sharpe ratio $S^*$, which is always positive, then $a \: w_1$ has the same Sharpe ratio for any positive real number $a$. Setting $a=1/x^Tw_1$, shows that there exists a portfolio $w$ with optimal Sharpe ratio and $x^Tw=1$.
Now, we can find $S^*$ by maximizing $S_w$ subject to $x^Tw=1$, i.e. minimize $w^TVw$ subject to $x^Tw=1$. Using one Lagrange multiplyer $\lambda$ gives the following conditions: $$ \nabla_w(w^TVw+\lambda x^Tw)=2 Vw + \lambda x\stackrel{!}{=}0 $$ $$ x^Tw=1$$ The solution is $w=\frac{V^{-1}x}{w^TV^{-1}w}$ and the optimal Sharpe ratio is thus $$ S^*=\sqrt{x^TV^{-1}x}$$
Application to your case: Two uncorrelated assets with volas $\sigma_1$ and $\sigma_2$ i.e. $V^{-1}=\left(\begin{array} c\sigma_1^{-2}& 0\\0&\sigma_2^{-2}\end{array}\right)$, and Sharpe ratios $S_i=x_i/\sigma_i$ gives the above result. |
I'm facing problems with this book and it is the only book that discusses localization in depth. The results that I'm getting makes no sense. I've read a lot of papers, majority of them copy the localization algorithm from this book. My question here is why $\bar{\mu}$ and $\bar{\Sigma}$ are being changed every iteration?? I'm using them to get the predicted measurements in lines 11- 13, so they should be fixed.
9. for all observed features $z^{i} = [r^{i} \ \phi^{i} \ s^{i}]^{T} $ do
10. $j = c^{i}$
11. $q = (m_{x} - \bar{\mu}_{x})^{2} + (m_{y} - \bar{\mu}_{y})^{2}$
12. $\hat{z}^{i} = \begin{bmatrix} \sqrt{q} \\ atan2(m_{y} - \bar{\mu}_{y}, m_{x} - \bar{\mu}_{x} ) - \bar{\mu}_{\theta} \\ m_{s} \\ \end{bmatrix}$
13. $ \hat{H}^{i} = \begin{bmatrix} h_{11} & h_{12} & h_{13} \\ h_{21} & h_{22} & h_{23} \\ h_{31} & h_{32} & h_{33} \\ \end{bmatrix} $
14. $\hat{S}^{i} = H^{i} \bar{\Sigma} [H^{i}]^{T} + Q $
15. $K^{i} = \bar{\Sigma} [H^{i}]^{T} [S^{i}]^{-1}$
16. $\bar{\mu} = \bar{\mu} + K^{i}(z^{i}-\hat{z}^{i}) $
17. $\bar{\Sigma} = (I - K^{i} H^{i}) \bar{\Sigma} $
18. endfor
19. $\mu = \bar{\mu}$
20. $\Sigma = \bar{\Sigma}$
Please suggest me other books that discuss EKF localization in depth. |
You not only can, but also must treat symbols for units by the ordinary rules of algebra, since unit symbols are mathematical entities and not abbreviations.
The
value of a quantity is expressed as the product of a number and a unit. That number is called the numerical value of the quantity expressed in this unit.
This relation may be expressed in the form
$$Q = \left\{ Q \right\} \cdot \left[ Q \right]$$
where $Q$ is the symbol for the quantity, $\left[ Q \right]$ is the symbol for the unit, and $\left\{ Q \right\}$ is the symbol for the numerical value of the quantity $Q$ expressed in the unit $\left[ Q \right]$.
For example, the mass of a sample is
$$m = 100\ \mathrm g$$
Here, $m$ is the symbol for the quantity mass, $\mathrm g$ is the symbol for the unit gram (a unit of mass), and $100$ is the numerical value of the mass expressed in grams. Thus, the value of the mass is $100\ \mathrm g$.
It is important to distinguish between the quantity $Q$ itself and the numerical value $\left\{ Q \right\}$ of the quantity expressed in a particular unit $\left[ Q \right]$. The value of a particular quantity $Q$ is independent of the choice of unit $\left[ Q \right]$, although the numerical value $\left\{ Q \right\}$ will be different for different units.
For example, changing the unit for the mass in the previous example from the gram to the kilogram, which is $10^3$ times the gram, leads to a numerical value which is $10^{-3}$ the numerical value of the mass expressed in grams, whereas the value of the mass stays the same.
$$m = 100\ \mathrm g = 0.100\ \mathrm{kg}$$
Since symbols for units are mathematical entities, both the numerical value and the unit may be treated by the ordinary rules of algebra. For example, the equation $m = 100\ \mathrm g$ may equally be written $$m/\mathrm g = 100$$
It is often convenient to label the axes of a graph in this way, so that the tick marks are labelled only with numbers. The quotient of a quantity and a unit may also be used in this way for the heading of a column in a table, so that the entries in the table are all simply numbers.
Performing the mathematical operations of quantities is called
quantity calculus. Quantities are multiplied and divided by one another according to the rules of algebra, resulting in new quantities.
The quotient of two quantities, $Q_1$ and $Q_2$, satisfies the relation$$\begin{align}\frac{Q_1}{Q_2} &= \frac{ \left\{ Q_1 \right\} \cdot \left[ Q_1 \right] }{ \left\{ Q_2 \right\} \cdot \left[ Q_2 \right] } \\[6pt]&= \frac{ \left\{ Q_1 \right\} }{ \left\{ Q_2 \right\} } \cdot \frac{ \left[ Q_1 \right] }{ \left[ Q_2 \right] }\end{align}$$Thus, the quotient $\left\{ Q_1 \right\}/\left\{ Q_2 \right\}$ is the numerical value $\left\{ Q_1/Q_2 \right\}$ of the quantity $Q_1/Q_2$, and the quotient $\left[ Q_1 \right]/\left[ Q_2 \right]$ is the unit $\left[ Q_1/Q_2 \right]$ of the quantity $Q_1/Q_2$.
For example, assuming a volume of $V = 0.127\ \mathrm{l}$, the density $\rho$ of the above-mentioned sample is$$\begin{align}\rho &= \frac{m}{V} \\[6pt]&= \frac{ 0.100\ \mathrm{kg} }{ 0.127\ \mathrm{l} } \\[6pt]&= \frac{ 0.100 }{ 0.127 } \cdot \frac{ \mathrm{kg} }{ \mathrm{l} } \\[6pt]&= 0.79\ \mathrm{kg/l}\end{align}$$
Similarly, the product of two quantities, $Q_1$ and $Q_2$, satisfies the relation$$\begin{align}Q_1 \cdot Q_2 &= \left( \left\{ Q_1 \right\} \cdot \left[ Q_1 \right] \right) \cdot \left( \left\{ Q_2 \right\} \cdot \left[ Q_2 \right] \right) \\[6pt]&= \left\{ Q_1 \right\}\left\{ Q_2 \right\} \cdot \left[ Q_1 \right] \left[ Q_2 \right]\end{align}$$
Thus, the product $\left\{ Q_1 \right\}\left\{ Q_2 \right\}$ is the numerical value $\left\{ Q_1Q_2 \right\}$ of the quantity $Q_1Q_2$, and the product $\left[ Q_1 \right]\left[ Q_2 \right]$ is the unit $\left[ Q_1Q_2 \right]$ of the quantity $Q_1Q_2$.
For example, considering the standard acceleration of free fall $g_\mathrm n = 9.80665\ \mathrm{m/s^2}$, the weight $F_\mathrm g$ of the above-mentioned sample is$$\begin{align}F_\mathrm g &= m \cdot g_\mathrm n \\[6pt]&= 0.100\ \mathrm{kg} \times 9.80665\ \frac{\mathrm m}{\mathrm{s^2}} \\[6pt]&= 0.100 \times 9.80665 \times \mathrm{kg} \cdot \frac{\mathrm m}{\mathrm{s^2}} \\[6pt]&= 0.98\ \frac{\mathrm {kg\ m}}{\mathrm{s^2}} \\[6pt]&= 0.98\ \mathrm{N}\end{align}$$
In forming products and quotients of unit symbols, the normal rules of algebraic multiplication or division apply.
For example, the expansion work $W$ at constant pressure $p = 100\,000\ \mathrm{Pa} = 100\,000\ \mathrm{kg\ m^{-1}\ s^{-2}}$ associated with a volume change of $\Delta V = 0.5\ \mathrm{m^3}$ is
$$\begin{align}W &= p \cdot \Delta V \\[6pt]&= 100\,000\ \frac{\mathrm{kg}}{\mathrm{m\ s^{2}}} \times 0.5\ \mathrm{m^3}\\[6pt]&= 50\,000\ \frac{\mathrm{kg}}{\mathrm{m\ s^{2}}}\cdot\mathrm{m^3}\\[6pt]&= 50\,000\ \frac{\mathrm{kg\ m^2}}{\mathrm{s^{2}}}\\[6pt]&= 50\,000\ \mathrm J\end{align}$$
Two or more quantities cannot be added or subtracted unless they belong to the same kind. The expression shall be written as the sum or difference of expressions for the quantities$$l=12\ \mathrm m-7\ \mathrm m$$or parentheses shall be used to combine the numerical values, placing the common unit symbol after the complete numerical value$$l=\left(12-7\right)\ \mathrm m$$but it is not permissible to write$$l=12-7\ \mathrm m\quad\color{red}{\small\text{(wrong!)}}$$For the same reason, quantities on each side of an equal sign in an equation must be of the same kind$$\begin{align}m_\text{total} &= m_1+m_2 \\1.8\ \mathrm{kg} &= 1.5\ \mathrm{kg}+0.3\ \mathrm{kg}\end{align}$$However, quantities of the same kind do not necessarily have the same unit.$$250\ \mathrm g = 0.250\ \mathrm{kg}$$$$10\ \mathrm{m/s} = 36\ \mathrm{km/h}$$Anyway, quantities on each side of an equal sign in an equation must not be of different kinds.
$$1\ \mathrm{mol} = 22.414\ \mathrm l\quad\color{red}{\small\text{(wrong!)}}$$ |
I have been fascinated by a very intriguing question - Can lasers push objects up?
I have done the below math to find out
Lets say we have a $1000~\text{mW}$ laser and we would like to lift an object of weight $100~\text{g}$.
By definition:
$1~\text{W} = 1 \frac{~\text{J}}{~\text{s}}$
That means the laser is emitting $1~\text{J}$ of energy per second.
On the other hand energy required to lift an object off the ground is given by $m \cdot g \cdot h$.
Putting in the number and lets say we want to solve for
$0.1~\text{kg} \cdot 9.8 \frac{~\text{m}}{~\text{s}^{2}} \cdot h = 1~\text{J}$
So, $h \approx 1~\text{m}$.
You see, if we had a $1000~\text{mW}$ laser we could lift an object of $100~\text{g}$ weight up to 1 meter in one second.
I can't see anything wrong with the above math. If this is correct, can anyone tell me then why on Earth we use heavy rockets to send objects into space? |
Algebraic Notation
The following algebra notation can be entered in Show My Work boxes.
Note In addition to the keyboard shortcuts listed in this topic, some symbols can be typed using the keyboard shortcuts for your operating system; for example, you can press ALT + 0247 on Windows to type ÷.
Notation Keyboard Button Notes Variables Type variables exactly as specified in the question. Variables are automatically displayed in italics. Lowercase Greek letter \ letter_name For example, type \alpha, \pi, or \theta. Uppercase Greek letter \ Letter_name For example, type \Delta or \Omega. |
@HarryGindi So the $n$-simplices of $N(D^{op})$ are $Hom_{sCat}(\mathfrak{C}[n],D^{op})$. Are you using the fact that the whole simplicial set is the mapping simplicial object between cosimplicial simplicial categories, and taking the constant cosimplicial simplicial category in the right coordinate?
I guess I'm just very confused about how you're saying anything about the entire simplicial set if you're not producing it, in one go, as the mapping space between two cosimplicial objects. But whatever, I dunno. I'm having a very bad day with this junk lol.
It just seems like this argument is all about the sets of n-simplices. Which is the trivial part.
lol no i mean, i'm following it by context actually
so for the record i really do think that the simplicial set you're getting can be written as coming from the simplicial enrichment on cosimplicial objects, where you take a constant cosimplicial simplicial category on one side
@user1732 haha thanks! we had no idea if that'd actually find its way to the internet...
@JonathanBeardsley any quillen equivalence determines an adjoint equivalence of quasicategories. (and any equivalence can be upgraded to an adjoint (equivalence)). i'm not sure what you mean by "Quillen equivalences induce equivalences after (co)fibrant replacement" though, i feel like that statement is mixing category-levels
@JonathanBeardsley if nothing else, this follows from the fact that \frakC is a left quillen equivalence so creates weak equivalences among cofibrant objects (and all objects are cofibrant, in particular quasicategories are). i guess also you need to know the fact (proved in HTT) that the three definitions of "hom-sset" introduced in chapter 1 are all weakly equivalent to the one you get via \frakC
@IlaRossi i would imagine that this is in goerss--jardine? ultimately, this is just coming from the fact that homotopy groups are defined to be maps in (from spheres), and you only are "supposed" to map into things that are fibrant -- which in this case means kan complexes
@JonathanBeardsley earlier than this, i'm pretty sure it was proved by dwyer--kan in one of their papers around '80 and '81
@HarryGindi i don't know if i would say that "most" relative categories are fibrant. it was proved by lennart meier that model categories are Barwick--Kan fibrant (iirc without any further adjectives necessary)
@JonathanBeardsley what?! i really liked that picture! i wonder why they removed it
@HarryGindi i don't know about general PDEs, but certainly D-modules are relevant in the homotopical world
@HarryGindi oh interesting, thomason-fibrancy of W is a necessary condition for BK-fibrancy of (R,W)?
i also find the thomason model structure mysterious. i set up a less mysterious (and pretty straightforward) analog for $\infty$-categories in the fappendix here: arxiv.org/pdf/1510.03525.pdf
as for the grothendieck construction computing hocolims, i think the more fundamental thing is that the grothendieck construction itself is a lax colimit. combining this with the fact that ($\infty$-)groupoid completion is a left adjoint, you immediately get that $|Gr(F)|$ is the colimit of $B \xrightarrow{F} Cat \xrightarrow{|-|} Spaces$
@JonathanBeardsley If you want to go that route, I guess you still have to prove that ^op_s and ^op_Delta both lie in the unique nonidentity component of Aut(N(Qcat)) and Aut(N(sCat)) whatever nerve you mean in this particular case (the B-K relative nerve has the advantage here bc sCat is not a simplicial model cat)
I think the direct proof has a lot of advantages here, since it gives a point-set on-the-nose isomorphism
Yeah, definitely, but I'd like to stay and work with Cisinski on the Ph.D if possible, but I'm trying to keep options open
not put all my eggs in one basket, as it were
I mean, I'm open to coming back to the US too, but I don't have any ideas for advisors here who are interested in higher straightening/higher Yoneda, which I am convinced is the big open problem for infinity, n-cats
Gaitsgory and Rozenblyum, I guess, but I think they're more interested in applications of those ideas vs actually getting a hold of them in full generality
@JonathanBeardsley Don't sweat it. As it was mentioned I have now mod superpowers, so s/he can do very little to upset me. Since you're the room owner, let me know if I can be of any assistance here with the moderation (moderators on SE have network-wide chat moderating powers, but this is not my turf, so to speak).
There are two "opposite" functors:$$ op_\Delta\colon sSet\to sSet$$and$$op_s\colon sCat\to sCat.$$The first takes a simplicial set to its opposite simplicial set by precomposing with the opposite of a functor $\Delta\to \Delta$ which is the identity on objects and takes a morphism $\langle k...
@JonathanBeardsley Yeah, I worked out a little proof sketch of the lemma on a notepad
It's enough to show everything works for generating cofaces and codegeneracies
the codegeneracies are free, the 0 and nth cofaces are free
all of those can be done treating frak{C} as a black box
the only slightly complicated thing is keeping track of the inner generated cofaces, but if you use my description of frak{C} or the one Joyal uses in the quasicategories vs simplicial categories paper, the combinatorics are completely explicit for codimension 1 face inclusions
the maps on vertices are obvious, and the maps on homs are just appropriate inclusions of cubes on the {0} face of the cube wrt the axis corresponding to the omitted inner vertex
In general, each Δ[1] factor in Hom(i,j) corresponds exactly to a vertex k with i<k<j, so omitting k gives inclusion onto the 'bottom' face wrt that axis, i.e. Δ[1]^{k-i-1} x {0} x Δ[j-k-1] (I'd call this the top, but I seem to draw my cubical diagrams in the reversed orientation).
> Thus, using appropriate tags one can increase ones chances that users competent to answer the question, or just interested in it, will notice the question in the first place. Conversely, using only very specialized tags (which likely almost nobody specifically favorited, subscribed to, etc) or worse just newly created tags, one might miss a chance to give visibility to ones question.
I am not sure to which extent this effect is noticeable on smaller sites (such as MathOverflow) but probably it's good to follow the recommendations given in the FAQ. (And MO is likely to grow a bit more in the future, so then it can become more important.) And also some smaller tags have enough followers.
You are asking posts far away from areas I am familiar with, so I am not really sure which top-level tags would be a good fit for your questions - otherwise I would edit/retag the posts myself. (Other than possibility to ping you somewhere in chat, the reason why I posted this in this room is that users of this room are likely more familiar with the topics you're interested in and probably they would be able to suggest suitable tags.)
I just wanted to mention this, in case it helps you when asking question here. (Although it seems that you're doing fine.)
@MartinSleziak even I was not sure what other tags are appropriate to add.. I will see other questions similar to this, see what tags they have added and will add if I get to see any relevant tags.. thanks for your suggestion.. it is very reasonable,.
You don't need to put only one tag, you can put up to five. In general it is recommended to put a very general tag (usually an "arxiv" tag) to indicate broadly which sector of math your question is in, and then more specific tags
I would say that the topics of the US Talbot, as with the European Talbot, are heavily influenced by the organizers. If you look at who the organizers were/are for the US Talbot I think you will find many homotopy theorists among them. |
Throughout many fields of science, one finds quantities which behave (or are claimed to behave) according to a
power-law distribution. That is, one quantity of interest, y, scales as another number x raised to some exponent:
[tex] y \propto x^{-\alpha}.[/tex]
Power-law distributions made it big in complex systems when it was discovered (or rather re-discovered) that a simple procedure for growing a network, called “preferential attachment,” yields networks in which the probability of finding a node with exactly
k other nodes connected to it falls off as k to some exponent:
[tex]p(k) \propto k^{-\gamma}.[/tex]
The constant γ is typically found to be between 2 and 3. Now, from my parenthetical remarks, the Gentle Reader may have gathered that the story is not quite a simple one. There are, indeed, many complications and subtleties, one of which is an issue which might sound straightforward: how do we know a power-law distribution when we see one? Can we just plot our data on a log-log graph and see if it falls on a straight line? Well, as Eric and I are fond of saying, “You can hide a multitude of sins on a log-log graph.”
Power-law distributions occur in many situations of scientific interest and have significant consequences for our understanding of natural and man-made phenomena. Unfortunately, the empirical detection and characterization of power laws is made difficult by the large fluctuations that occur in the tail of the distribution. In particular, standard methods such as least-squares fitting are known to produce systematically biased estimates of parameters for power-law distributions and should not be used in most circumstances. Here we describe statistical techniques for making accurate parameter estimates for power-law data, based on maximum likelihood methods and the Kolmogorov-Smirnov statistic. We also show how to tell whether the data follow a power-law distribution at all, defining quantitative measures that indicate when the power law is a reasonable fit to the data and when it is not. We demonstrate these methods by applying them to twenty-four real-world data sets from a range of different disciplines. Each of the data sets has been conjectured previously to follow a power-law distribution. In some cases we find these conjectures to be consistent with the data while in others the power law is ruled out.
After going over the theory involved, the authors look at twenty-four different real-world data sets which have been claimed to follow a power-law distribution. They compute
p-values based on the power-law model to judge whether or not a power law is actually a good description of that data. For most of the twenty-four datasets, the power-law description holds up pretty well. In one case — the distribution of word frequencies in Moby Dick — the power-law description trumps all others, while in other cases, fits to other curves may still be plausible, but power laws certainly aren’t ruled out. However, in seven of the twenty-four datasets, the authors’ calculations show that a power-law model is just no good!
In particular, the distributions for the HTTP connections, earthquakes, web links, fires, wealth, web hits, and the metabolic network cannot plausibly be considered to follow a power law; the probability of getting a fit as poor as that observed purely by chance is very small in each case and one would have to be unreasonably optimistic to see power-law behavior in any of these data sets. (For two data sets — the HTTP connections and wealth distribution — the power law, while not a good fit, is nonetheless better than the alternatives, implying that these data sets are not well-characterized by any of the functional forms considered here.)
Furthermore, they find that it is in general difficult to tell a power-law distribution apart from another kind known as a
log-normal, whose probability density function looks like this:
[tex]p(x;\mu,\sigma) = \frac{1}{\sqrt{2\pi}} \frac{e^{-(\ln x – \mu)^2/(2\sigma^2)}}{x \sigma},[/tex]
where μ is the mean and σ the standard deviation of the
logarithm of x. The authors note that
the log-normal is not ruled out for any of our data sets, save the HTTP connections. In every case it is a plausible alternative and in a few it is strongly favored. In fact, we find that it is in general extremely difficult to tell the difference between a log-normal and true power-law behavior. Indeed over realistic ranges of
xthe two distributions are very closely equal, so it appears unlikely that any test would be able to tell them apart unless we have an extremely large data set. Thus one must again rely on physical intuition to draw any final conclusions. Otherwise, the best that one can say is that the data do not rule out the power law, but that other distributions are as good or better a fit to the data.
Because this is the age of Web 3.1, Aaron Clauset has a blog entry devoted to discussing this paper, here. He and his co-authors have even made their code publicly available — if you can use MATLAB and R, it’s all yours!
Applications to the technological singularity are left as an exercise to the interested reader. |
Electromagnetic Induction Inductance
An inductor is a device for storing energy in a magnetic field.an inductor is generally called as inductance.
Self induction : The phenomena of production of induced emf in a circuit due to change in current flowing in its own, is called self – induction.
Coefficient of self induction
(Φ ) = LI Mutual induction: The phenomena of production of induced emf in a circuit due to the change in magnetic flux in its neighboring circuit is called mutual induction.
Coefficient of mutual induction
( Φ ) = MI View the Topic in this video From 25:08 To 59:21
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1. By Faraday's second law, induced emf e = -N\frac{d \phi}{dt}, which gives e = -L \frac{di}{dt}; if \frac{di}{dt} = 1 amp/sec then |e| = L.
2. According to Faraday's second law emf induce in secondary is e_{2} = - N_{2} \frac{d \phi_{2}}{dt}; e_{2} = -M \frac{di_{1}}{dt}
3. Relation between M, L
1 and L 2 : For two magnetically coupled coils M = k\sqrt{L_{1}L_{2}}
4. Series: If two coils of self-inductances L
1 and L 2 having mutual inductance M are in series and are far each other, so that the mutual induction between them is negligible, then net self inductance L s = L 1 + L 2
5. Parallel : If two coils of self-inductances L
1 and L 2 having mutual inductance are connected in parallel and are far from each other, then net inductance L is \frac{1}{L_{P}} = \frac{1}{L_{1}} + \frac{1}{L_{2}} |
Chapters
Chapter 2: Functions
Chapter 3: Binary Operations
Chapter 4: Inverse Trigonometric Functions
Chapter 5: Algebra of Matrices
Chapter 6: Determinants
Chapter 7: Adjoint and Inverse of a Matrix
Chapter 8: Solution of Simultaneous Linear Equations
Chapter 9: Continuity
Chapter 10: Differentiability
Chapter 11: Differentiation
Chapter 12: Higher Order Derivatives
Chapter 13: Derivative as a Rate Measurer
Chapter 14: Differentials, Errors and Approximations
Chapter 15: Mean Value Theorems
Chapter 16: Tangents and Normals
Chapter 17: Increasing and Decreasing Functions
Chapter 18: Maxima and Minima
Chapter 19: Indefinite Integrals
Chapter 20: Definite Integrals
Chapter 21: Areas of Bounded Regions
Chapter 22: Differential Equations
Chapter 23: Algebra of Vectors
Chapter 24: Scalar Or Dot Product
Chapter 25: Vector or Cross Product
Chapter 26: Scalar Triple Product
Chapter 27: Direction Cosines and Direction Ratios
Chapter 28: Straight Line in Space
Chapter 29: The Plane
Chapter 30: Linear programming
Chapter 31: Probability
Chapter 32: Mean and Variance of a Random Variable
Chapter 33: Binomial Distribution
RD Sharma Mathematics Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) Chapter 14: Differentials, Errors and Approximations Chapter 14: Differentials, Errors and Approximations Exercise 14.1 solutions [Pages 9 - 12]
If y = sin x and x changes from π/2 to 22/14, what is the approximate change in y ?
For the function
y = x 2, if x = 10 and ∆x = 0.1. Find ∆ y ?
The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume ?
A circular metal plate expends under heating so that its radius increases by
k%. Find the approximate increase in the area of the plate, if the radius of the plate before heating is 10 cm.
Find the percentage error in calculating the surface area of a cubical box if an error of 1% is made in measuring the lengths of edges of the cube ?
If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere ?
The pressure
p and the volume v of a gas are connected by the relation pv 1.4 = const. Find the percentage error in p corresponding to a decrease of 1/2% in v .
The height of a cone increases by
k%, its semi-vertical angle remaining the same. What is the approximate percentage increase (i) in total surface area, and (ii) in the volume, assuming that k is small ?
Show that the relative error in computing the volume of a sphere, due to an error in measuring the radius, is approximately equal to three times the relative error in the radius ?
1 Using differential, find the approximate value of the following:
\[\sqrt{25 . 02}\]
Using differential, find the approximate value of the following: \[\left( 0 . 009 \right)^\frac{1}{3}\]
Using differential, find the approximate value of the following: \[\left( 0 . 007 \right)^\frac{1}{3}\]
Using differential, find the approximate value of the \[\sqrt{401}\] ?
Using differential, find the approximate value of the \[\left( 15 \right)^\frac{1}{4}\] ?
Using differential, find the approximate value of the \[\left( 255 \right)^\frac{1}{4}\] ?
Using differential, find the approximate value of the \[\frac{1}{(2 . 002 )^2}\] ?
Using differential, find the approximate value of the log
4.04, it being given that log e 104 = 0.6021 and log 10 e= 0.4343 ?
Using differential, find the approximate value of the log
10.02, it being given that log e 10 = 2.3026 ? e
Using differential, find the approximate value of the log
10 10.1, it being given that log 10 e = 0.4343 ?
Using differentials, find the approximate values of the cos 61°, it being given that sin60° = 0.86603 and 1° = 0.01745 radian ?
Using differential, find the approximate value of the \[\frac{1}{\sqrt{25 . 1}}\] ?
Using differential, find the approximate value of the \[\sin\left( \frac{22}{14} \right)\] ?
Using differential, find the approximate value of the \[\cos\left( \frac{11\pi}{36} \right)\] ?
Using differential, find the approximate value of the \[\left( 80 \right)^\frac{1}{4}\] ?
Using differential, find the approximate value of the \[\left( 29 \right)^\frac{1}{3}\] ?
Using differential, find the approximate value of the \[\left( 66 \right)^\frac{1}{3}\] ?
Using differential, find the approximate value of the \[\sqrt{26}\] ?
Using differential, find the approximate value of the \[\sqrt{37}\] ?
Using differential, find the approximate value of the \[\sqrt{0 . 48}\] ?
Using differential, find the approximate value of the \[\left( 82 \right)^\frac{1}{4}\] ?
Using differential, find the approximate value of the \[\left( \frac{17}{81} \right)^\frac{1}{4}\] ?
Using differential, find the approximate value of the \[\left( 33 \right)^\frac{1}{5}\] ?
Using differential, find the approximate value of the \[\sqrt{36 . 6}\] ?
Using differential, find the approximate value of the \[{25}^\frac{1}{3}\] ?
Using differential, find the approximate value of the \[\sqrt{49 . 5}\] ?
Using differential, find the approximate value of the \[\left( 3 . 968 \right)^\frac{3}{2}\] ?
Using differential, find the approximate value of the \[\left( 1 . 999 \right)^5\] ?
Using differential, find the approximate value of the \[\sqrt{0 . 082}\] ?
Find the approximate value of f (2.01), where f (
x) = 4 x 2 + 5 x + 2 ?
Find the approximate value of f (5.001), where f (x) = x
3 − 7x 2 + 15 ?
Find the approximate value of log
10 1005, given that log 10 e = 0.4343 ?
If the radius of a sphere is measured as 9 cm with an error of 0.03 m, find the approximate error in calculating its surface area ?
Find the approximate change in the surface area of a cube of side
x metres caused by decreasing the side by 1% ?
If the radius of a sphere is measured as 7 m with an error of 0.02 m, find the approximate error in calculating its volume ?
Find the approximate change in the value
V of a cube of side x metres caused by increasing the side by 1% ? Chapter 14: Differentials, Errors and Approximations solutions [Page 12]
For the function
y = x 2, if x = 10 and ∆ x = 0.1. Find ∆ y.
If
y = log e x, then find ∆ y when x = 3 and ∆ x = 0.03 ?
If the relative error in measuring the radius of a circular plane is α, find the relative error in measuring its area ?
If the percentage error in the radius of a sphere is α, find the percentage error in its volume ?
A piece of ice is in the form of a cube melts so that the percentage error in the edge of cube is
a, then find the percentage error in its volume ? Chapter 14: Differentials, Errors and Approximations solutions [Page 13]
If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is
1%
2%
3%
4%
If there is an error of a% in measuring the edge of a cube, then percentage error in its surface is
2
a%
\[\frac{a}{2} \%\]
3
a%
none of these
If an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is
k%
3k%
2k%
k/3%
The height of a cylinder is equal to the radius. If an error of α % is made in the height, then percentage error in its volume is
α %
2α %
3α %
none of these
While measuring the side of an equilateral triangle an error of
k % is made, the percentage error in its area is k%
2
k%
\[\frac{k}{2}\%\]
3
k%
If log
4 = 1.3868, then log e 4.01 = e
1.3968
1.3898
1.3893
none of these
A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is
12000 π mm
3
800 π mm
3
80000 π mm
3
120 π mm
3
If the ratio of base radius and height of a cone is 1 : 2 and percentage error in radius is λ %, then the error in its volume is
λ %
2 λ %
3 λ %
none of these
The pressure
P and volume V of a gas are connected by the relation PV 1 /4 = constant. The percentage increase in the pressure corresponding to a deminition of 1/2 % in the volume is
\[\frac{1}{2} \%\]
\[\frac{1}{4} \%\]
\[\frac{1}{8} \%\]
none of these
If y = x
n then the ratio of relative errors in y and x is
1 : 1
2 : 1
1 : n
n : 1
The approximate value of (33)
1/5 is
2.0125
2.1
2.01
none of these
The circumference of a circle is measured as 28 cm with an error of 0.01 cm. The percentage error in the area is
\[\frac{1}{14}\]
0.01
\[\frac{1}{7}\]
none of these
Chapter 14: Differentials, Errors and Approximations RD Sharma Mathematics Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) Textbook solutions for Class 12 RD Sharma solutions for Class 12 Mathematics chapter 14 - Differentials, Errors and Approximations
RD Sharma solutions for Class 12 Maths chapter 14 (Differentials, Errors and Approximations) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics for Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com are providing such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 12 Mathematics chapter 14 Differentials, Errors and Approximations are Maximum and Minimum Values of a Function in a Closed Interval, Maxima and Minima, Simple Problems on Applications of Derivatives, Graph of Maxima and Minima, Approximations, Tangents and Normals, Increasing and Decreasing Functions, Rate of Change of Bodies Or Quantities, Introduction to Applications of Derivatives.
Using RD Sharma Class 12 solutions Differentials, Errors and Approximations exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 12 prefer RD Sharma Textbook Solutions to score more in exam.
Get the free view of chapter 14 Differentials, Errors and Approximations Class 12 extra questions for Maths and can use Shaalaa.com to keep it handy for your exam preparation |
U. Biccari, V. Hernández-Santamaría Null controllability of a nonlocal heat equation with integral kernel, DOI: Abstract: We consider a linear nonlocal heat equation in a bounded domain $\Omega\subset\mathbb{R}^d$ with Dirichlet boundary conditions. The non-locality is given by the presence of an integral kernel. We analyze the problem of controllability when the control acts on a…
Hernández-Santamaría V., Lazar M., Zuazua E. Greedy optimal control for elliptic problems and itsapplication to turnpike problems DOI: 10.1007/s00211-018-1005-z Abstract: We adapt and apply greedy methods to approximate in an efficient way the optimalcontrols for parameterized elliptic control problems. Our results yield an optimal approximation procedure that, in particular, performs better than simply sampling theparameter-space…
Hernández-Santamaría V., de Teresa L., Poznyak A. Corrigendum and addendum to “Hierarchic control for a coupled parabolic system” , DOI: 10.4171/PM/1998 Abstract: In [2] we used three controls for a system of two coupled parabolic equations. We defined three functionals to be minimized and a hierarchy on the controls obtaining from the optimality condition a…
Hernández-Santamaría V., de Teresa L., Boyer, F. Insensitizing controls for a semilinear parabolic equation: a numerical approach , DOI: Abstract: In this paper, we study the insensitizing control problem in the discrete setting of finite-differences. We prove the existence of a control that insensitizes the norm of the observed solution of a 1-D semidiscrete parabolic…
Hernández-Santamaría V., de Teresa L. Robust Stackelberg controllability for linear and semilinear heat equations , DOI: Abstract: In this paper we study a Stackelberg–Nash strategy to control systems of coupled linear parabolic partial differential equations. We assume that we can act on the system through several controls called followers, intended to solve a Nash multi-objective…
Hernández-Santamaría V., de Teresa L. Some remarks on the hierarchic control for coupled parabolic PDEs , DOI: To appear Abstract: In this paper, we study Stackelberg-Nash strategies to control a system of two coupled parabolic equations. We assume that we act in the system by means of a hierarchy of controls. First, controls in each…
U. Biccari, V. Hernández-Santamaría The Poisson equation from non-local to local, Electronic Journal of Differential Equations, Vol. 2018 (2018), No. 145, pp. 1-13. DOI: arXiv:1801.09470 Abstract: We analyze the limit behavior as $s\to 1^-$ of the solution to the fractional Poisson equation $\fl{s}{u_s}=f_s$, $x\in\Omega$ with homogeneous Dirichlet boundary conditions $u_s\equiv 0$, $x\in\Omega^c$. We show that…
Hernández-Santamaría V., de Teresa L., Poznyak A. Hierarchic control for a coupled parabolic system , DOI: 10.4171/PM/1979 Abstract: In this paper we study a Stackelberg–Nash strategy to control systems of coupled linear parabolic partial differential equations. We assume that we can act on the system through several controls called followers, intended to solve a Nash…
Hernández-Santamaría V., Lazar M., Zuazua E. , DOI: 10.1007/s00211-018-1005-z Abstract: In this paper, we deal with the approximation of optimal controls for parameter-dependent elliptic and 7 parabolic equations. We adapt well-known results on greedy algorithms to approximate in an efficient 8 way the optimal controls for parameterized elliptic control problems. Our results yield an optimal…
U. Biccari, V. Hernández-Santamaría Null controllability of a nonlocal heat equation with integral kernel, DOI: Abstract: We consider a linear nonlocal heat equation in a bounded domain $\Omega\subset\mathbb{R}^d$ with Dirichlet boundary conditions. The non-locality is given by the presence of an integral kernel. We analyze the problem of controllability when the control acts on a… |
Let's have a look at your grammar:
$\qquad \begin{align}X &\to aE \mid IXE \mid (X)E \\E &\to IE \mid BXE \mid \varepsilon \\I &\to \text{++} \mid \text{--} \\B &\to \text{+} \mid \text{-} \mid \varepsilon \end{align}$
Note that $X$ does not need left-factoring: all rules have disjoint FIRST sets¹.If you want to make this obvious, you can drop $I$ and inline it:
$\qquad \begin{align}X &\to aE \mid \text{++}XE \mid \text{--}XE \mid (X)E \\E &\to \text{++}E \mid \text{--}E \mid BXE \mid \varepsilon \\B &\to \text{+} \mid \text{-} \mid \varepsilon \end{align}$
Similarly, we can inline $B$:
$\qquad \begin{align}X &\to aE \mid \text{++}XE \mid \text{--}XE \mid (X)E \\E &\to \text{++}E \mid \text{--}E \mid \text{+}XE \mid \text{-}XE \mid XE \mid \varepsilon \end{align}$
Now we see that we actually have to do left-factoring on $E$: we have obvious conflicts, and we get additional conflicts via $XE$. So, let's inline $X$ once at $XE$:
$\qquad \begin{align}X &\to aE \mid \text{++}XE \mid \text{--}XE \mid (X)E \\E &\to \text{++}E \mid \text{--}E \mid \text{+}XE \mid \text{-}XE \mid aEE \mid \text{++}XEE \mid \text{--}XEE \mid (X)EE \mid \varepsilon \end{align}$
And now we can left-factor as easily as in your example:
$\qquad \begin{align}X &\to aE \mid \text{++}XE \mid \text{--}XE \mid (X)E \\E &\to \text{+}P \mid \text{-}M \mid aEE \mid (X)EE \mid \varepsilon \\P &\to \text{+}E \mid XE \mid \text{+}XEE \\M &\to \text{-}E \mid XE \mid \text{-}XEE\end{align}$
By now we can see that we are not getting anywhere: by factoring away $\text{+}$ or $\text{-}$ from the alternatives, we dig up another $X$ which again has both $\text{+}$ and $\text{-}$ in its FIRST set.
So let's have a look at your language. Via
$\qquad \displaystyle X \Rightarrow aE \Rightarrow^* aI^n E \Rightarrow aI^nBXE$
and
$\qquad \displaystyle X \Rightarrow aE \Rightarrow^* aI^n E \Rightarrow aI^nIE$
you have
arbitrarily long prefixes of the form $+^+$ which end differently, semantic-wise: an LL(1) parser can not decide whether any given (next) $\text{+}$ belongsto a pair -- which would mean choosing alternative $IE$ -- or comes alone -- which would mean choosing $BXE$.
In consequence, it doesn't look like your
language can be expressed by any LL(1) grammar, so trying to convert yours into one is futile.
It's even worse: as $BXE \Rightarrow BIXEE \Rightarrow^* BI^n X E^n E$, you can not decide to chose $BXE$ with
any finite look-ahead. This is not a formal proof, but it strongly suggests that your language is not even LL.
If you think about what you are doing -- mixing Polish notation with unary operators -- it is not very surprising that parsing should be hard. Basically, you have to count from the left
and from the right to identify even a single $B$-$\text{+}$ in a long chain of $\text{+}$. If I think of multiple $B$-$\text{+}$ in a chain, I'm not even sure the language (with two semantically different but syntactically equal $\text{+}$) can be parsed deterministically (without backtracking) at all. That would be the sets of terminals that can come first in derivations of a non-terminal/rule alternative. |
In operator formalism, for example a 2-point time-ordered Green's function is defined as
$\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{op}=\theta(x_1-x_2)\phi(x_1)\phi(x_2)+\theta(x_2-x_1)\phi(x_2)\phi(x_1),$
where the subscript "op" refers to operator formalism. Now if one is to take a time derivative of it, the result will be $\frac{\partial}{\partial x_1^0}\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{op}=\langle\mathcal{T}{\frac{\partial \phi(x_1)}{\partial x_1^0}}\phi(x_2)\rangle_{op}+\delta(x_1^0-x_2^0)[\phi(x_1),\phi(x_2)]$, the delta function comes from differentiating the theta functions. This means time derivative does not commute with time ordering.
If we consider path integral formalism, the time-ordered Green's function is defined as
$\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{pi}=\int\mathcal{D}\phi\phi(x_1)\phi(x_2)e^{iS(\phi)}$.
Of course $\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{op}=\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{pi},$ as is proved in any QFT textbook. However in path integral case time derivative commutes with time ordering, because we don't have anything like a theta function thus $\frac{\partial}{\partial x_1^0}\int\mathcal{D}\phi\phi(x_1)\phi(x_2)e^{iS(\phi)}=\int\mathcal{D}\phi\frac{\partial}{\partial x_1^0}\phi(x_1)\phi(x_2)e^{iS(\phi)}$. I did a bit googling and found out that for the path integral case the time-ordered product is called "$\mathcal{T^*}$ product" and operator case just "$\mathcal{T}$ product".
I am not that interested in what is causing the difference(still explanations on this are welcomed), because I can already vaguely see it's due to some sort of ambiguity in defining the product of fields at equal time. The question that interests me is, which is the right one to use when calculating Feynman diagrams?
I did find a case where both give the same result, i.e. scalar QED(c.f. Itzykson & Zuber, section 6-1-4), but is it always the case? If these two formulations are not effectively equivalent, then it seems every time we write down something like $\langle\partial_0\phi\cdots\rangle$, we have to specify whether it's in the sense of the path integral definition or operator definition.
EDIT: As much as I enjoy user1504's answer, after thinking and reading a bit more I don't think analytic continuation is all the mystery. In Peskin&Schroeder chap 9.6 they manage to use path integral to get a result equivalent to operator approach, without any reference to analytic continuation. It goes like this : Consider a T-product for free KG field $\langle T\{\phi(x)\phi(x_1)\}\rangle=\int\mathcal{D}\phi\phi(x)\phi(x_1)e^{iS(\phi)}$. Apply Dyson-Schwinger equation, we get $\int\mathcal{D}\phi(\partial^2+m^2)\phi(x)\phi(x_1)e^{iS}=-i\delta^4(x-x_1)$, then they just assume the $\partial^2$ commute with path integration(which is already weird according to our discussion) and they conclude
$(\partial^2+m^2)\int\mathcal{D}\phi\phi(x)\phi(x_1)e^{iS}=(\partial^2+m^2)\langle T\{\phi(x)\phi(x_1)\}\rangle=-i\delta^4(x-x_1)$.
This is just the right result given by operator approach, in which $\delta(x^0-x_1^0)$ comes from $\theta$ function. Given my limited knowledge on the issue, this consistency looks almost a miracle to me. What is so wicked behind these maths?
Response to @drake:If $a$ is a positive infinitesimal, then $\int \dot A(t) B(t) \,e^{iS}\equiv\int D\phi\, {A(t+a)-A(t)\over a}B(t)\,e^{iS}=\frac{1}{a}\langle T\{A(t+a)B(t)\}\rangle-\frac{1}{a}\langle A(t)B(t)\rangle$, notice the second term has an ordering ambiguity from path integral(say $A=\dot{\phi},B=\phi$), and we can make it in any order we want by choosing an appropriate time discretization, c.f. Ron Maimon's post cited by drake. Keeping this in mind we proceed:
$\frac{1}{a}\langle T\{A(t+a)B(t)\}\rangle-\frac{1}{a}\langle A(t)B(t)\rangle\\=\frac{1}{a}\theta(a)\langle A(t+a)B(t)\rangle+\frac{1}{a}\theta(-a)\langle B(t)A(t+a)\rangle-\frac{1}{a}\langle A(t)B(t)\rangle\\=\frac{1}{a}\theta(a)\langle A(t+a)B(t)\rangle+\frac{1}{a}[1-\theta(a)]\langle B(t)A(t+a)\rangle-\frac{1}{a}\langle A(t)B(t)\rangle\\=\frac{\theta(a)}{a}\langle [A(t+a),B(t)]\rangle+\frac{1}{a}[\langle B(t)A(t+a)\rangle-\langle A(t)B(t)\rangle]$
Now taking advantage of ordering ambiguity of the last term to make it $\langle B(t)A(t)\rangle$(this amounts to defining A using backward discretization, say $A=\dot{\phi}(t)=\frac{\phi(t+\epsilon^-)-\phi(t)}{\epsilon^-}$), then the finally:
$\frac{\theta(a)}{a}\langle [A(t+a),B(t)]\rangle+\frac{1}{a}\langle B(t)[A(t+a)-A(t)\rangle]\to \frac{1}{2a}\langle [A(t),B(t)]\rangle+\langle B(t)\dot{A}(t)\rangle$.(Here again a very dubious step, to get $\frac{1}{2a}$ we need to assume $\theta(a\to 0^+)=\theta(0)=\frac{1}{2}$, but this is really not true because $\theta$ is discontinuous)
However on the other hand, since $a$ was defined to be a postive infinitesimal, at the very beginning we could've written $\frac{1}{a}\langle T\{A(t+a)B(t)\}\rangle-\frac{1}{a}\langle A(t)B(t)\rangle=\frac{1}{a}\langle A(t+a)B(t)\rangle-\frac{1}{a}\langle A(t)B(t)\rangle$, then all the above derivation doesn't work. I'm sure there are more paradoxes if we keep doing these manipulations. |
For a 6DoF robot with all revolute joints the Jacobian is given by: $$ \mathbf{J} = \begin{bmatrix} \hat{z_0} \times (\vec{o_6}-\vec{o_0}) & \ldots & \hat{z_5} \times (\vec{o_6}-\vec{o_5})\\ \hat{z_0} & \ldots & \hat{z_5} \end{bmatrix} $$ where $z_i$ is the unit z axis of joint $i+1$(using DH params), $o_i$ is the origin of the coordinate frame connected to joint $i+1$, and $o_6$ is the origin of the end effector. The jacobian matrix is the relationship between the Cartesian velocity vector and the joint velocity vector: $$ \dot{\mathbf{X}}= \begin{bmatrix} \dot{x}\\ \dot{y}\\ \dot{z}\\ \dot{r_x}\\ \dot{r_y}\\ \dot{r_z} \end{bmatrix} = \mathbf{J} \begin{bmatrix} \dot{\theta_1}\\ \dot{\theta_2}\\ \dot{\theta_3}\\ \dot{\theta_4}\\ \dot{\theta_5}\\ \dot{\theta_6}\\ \end{bmatrix} = \mathbf{J}\dot{\mathbf{\Theta}} $$
Here is a singularity position of a Staubli TX90XL 6DoF robot:
$$ \mathbf{J} = \begin{bmatrix} -50 & -425 & -750 & 0 & -100 & 0\\ 612.92 & 0 & 0 & 0 & 0 & 0\\ 0 & -562.92 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 1 & 0\\ 1 & 0 & 0 & -1 & 0 & -1 \end{bmatrix} $$
You can easily see that the 4th row corresponding to $\dot{r_x}$ is all zeros, which is exactly the lost degree of freedom in this position.
However, other cases are not so straightforward.
$$ \mathbf{J} = \begin{bmatrix} -50 & -324.52 & -649.52 & 0 & -86.603 & 0\\ 987.92 & 0 & 0 & 0 & 0 & 0\\ 0 & -937.92 & -375 & 0 & -50 & 0\\ 0 & 0 & 0 & 0.5 & 0 & 0.5\\ 0 & 1 & 1 & 0 & 1 & 0\\ 1 & 0 & 0 & -0.866 & 0 & -0.866 \end{bmatrix} $$
Here you can clearly see that joint 4 and joint 6 are aligned because the 4th and 6th columns are the same. But it's not clear which Cartesian degree of freedom is lost (it should be a rotation about the end effector's x axis in red).
Even less straightforward are singularities at workspace limits.
$$ \mathbf{J} = \begin{bmatrix} -50 & 650 & 325 & 0 & 0 & 0\\ 1275.8 & 0 & 0 & 50 & 0 & 0\\ 0 & -1225.8 & -662.92 & 0 & -100 & 0\\ 0 & 0 & 0 & 0.86603 & 0 & 1\\ 0 & 1 & 1 & 0 & 1 & 0\\ 1 & 0 & 0 & 0.5 & 0 & 0 \end{bmatrix} $$
In this case, the robot is able to rotate $\dot{-r_y}$ but not $\dot{+r_y}$. There are no rows full of zeros, or equal columns, or any clear linearly dependent columns/rows.
Is there a way to determine which degrees of freedom are lost by looking at the jacobian? |
The Annals of Statistics Ann. Statist. Volume 6, Number 1 (1978), 132-141. Bound on the Classification Error for Discriminating Between Multivariate Populations with Specified Means and Covariance Matrices Abstract
Let $\mathscr{F}_1, \mathscr{F}_2$ be two families of $p$-variate distribution functions with specified means $\mathbf{\mu}_i (i = 1,2)$ and nonsingular covariance matrices $\Sigma_i$, and let $\pi_i$ be the prior probability assigned to $\mathscr{F}_i$ for $i = 1, 2$. The objective is to discriminate whether an observation $\mathbf{x}$ is from a distribution $F_1 \in \mathscr{F}_1$ or $F_2 \in \mathscr{F}_2$. Given a pair $F = (F_1, F_2)$ the error probability for classification rule $\phi$ is denoted by $e(\phi, F)$. In this paper the values of $\sup_F \inf_\phi e(\phi, F)$ and $\inf_\phi \sup_F e(\phi, F)$ are found and conditions for the existence of a saddle point of $e(\phi, F)$ are given. Also a saddle point is found when it exists. When $\phi$ is restricted to linear classification rules the same problems are considered. The mathematical programming method for finding a saddle point is also outlined.
Article information Source Ann. Statist., Volume 6, Number 1 (1978), 132-141. Dates First available in Project Euclid: 12 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aos/1176344072 Digital Object Identifier doi:10.1214/aos/1176344072 Mathematical Reviews number (MathSciNet) MR468036 Zentralblatt MATH identifier 0377.62032 JSTOR links.jstor.org Citation
Isii, K.; Taga, Y. Bound on the Classification Error for Discriminating Between Multivariate Populations with Specified Means and Covariance Matrices. Ann. Statist. 6 (1978), no. 1, 132--141. doi:10.1214/aos/1176344072. https://projecteuclid.org/euclid.aos/1176344072 |
Post's Correspondence Problem is known to be undecidable. A variant of PCP, namely PCP with partially commutative alphabets is also known to be undecidable. Is the following variant also known to be undecidable?
PCP with commutative alphabet. Given and alphabet $\Sigma = \{a_1, a_2, \ldots, a_K\}$ and $2N$ sequences $\gamma_1, \gamma_2, \ldots, \gamma_N, \delta_1, \delta_2, \ldots, \delta_N \in \Sigma^*$.Determine if there is a sequence of integers $\{i_j\}_{j=1}^M$ such that for each $1 \leq j \leq M$, $i_j \in \{1, 2, \ldots, N\}$ and for each $1 \leq m \leq K$,
$\#(a_m, \gamma_{i_1} \gamma_{i_2} \ldots \gamma{i_M}) = \#(a_m, \delta_{i_1} \delta_{i_2} \ldots \delta_{i_M})$ where $\#(a, w)$ denotes the number of occurences of the symbol $a$ in the string $w$.
Notice why I used the term "commutative" for the alphabet: the order of the symbols in the strings does not matter; only the counts need to be equal.
In general, is any variant of (unbounded) PCP with commutative alphabet known to be decidable or undecidable? |
The first thing to note is that each of the slits produces a diffraction pattern the width of which is controlled by the width of the slit and the wavelength of the light.
The amount of light travelling from a slit in a particular direction is controlled by the diffraction pattern due to a single slit. The light waves from each of the slits superpose (interfere) and produce an interference pattern. The intensity of the fringes produced by the interference of light from the slits is modulated by the diffraction pattern produced by each of the slits. That is why the intensity of the interference fringes deceases as the order of the fringes increases.
So here is the modulated interference pattern for one slit, two slits, three slits and five slits with all slits the same width and with the same slit separation.
Note the modulation of light intensity of the interference fringes by the diffraction envelope.
Also note that the separation of the principal maximum for the 2, 3 and 5 slit arrangement is the same. The spacing of the principal maxima is controlled by the separation of the slits $d$ and the wavelength of the light $\lambda$The condition for the $n^{\text{th}}$ principal maximum is $n\lambda = d \sin \theta_n$. You would have met this equation when studying the diffraction grating but it is the same equation for any number $N$ of slits provided that you are dealing with the principal maxima.
When two slit interference is studied the angle $\theta$ is small (< 0.1 radian or < 5$^\circ$) and so the approximation $\sin \theta \approx \theta$ is a good one.
So the condition for a maximum becomes $n\lambda = d \theta_n$ which results in the fringes appearing to be equally spaced.
When using the diffraction grating because the slit separation is small compared with that of the normal 2 slit arrangement the angles at which there are maxima are large.
So the small angle approximation cannot be made and the fringes are not equally spaced.
The other striking thing about the patterns for 2,3 and 5 slits is that the principal maxima get narrower as the number of slits increases and there are also in between the principal maxima much less intense subsidiary maxima.
What is shown by the next diagram is that as well as the principal maxima getting narrower they at the same time get brighter.
What is happening is that as the number of slits is increased the amount of light coming through the slits is increased and at the same time the light is being channelled into a smaller angular width (the fringe width).
Ignoring the diffraction envelope for 2 slits the intensity of a principal maximum $I_2 \propto (2A)^2$ where $A$ is the amplitude of a wave from a single slit. For 3 slits $I_3 \propto (3A)^2$ and for five slits $I_5 \propto (5A)^2$.
So in a diffraction grating set up if the number of number of slits being used is reduced, say half the grating is covered up with black paper, the interference pattern would become less bright and the width of the principal maxima would increase.
Your three images are not comparing like for like.
For example the double slit pattern in the middle would appear to have slits which are much narrower than the slit used for the single slit pattern. The reason for this inference is that the width of the diffraction envelope modulation of intensity is much broader in the second diagram than the first. The last image of the pattern from a diffraction grating probably shows a much greater angular range than for the middle image because it shows the unequal spacing of the fringes. It also shows that probably the width of the slits in the diffraction grating are much smaller than those in the two slit arrangement because there seems to be hardly any evidence of diffraction envelope modulation of intensity over a very wide angular range for the diffraction grating picture.
Although all the intensity graphs can be derived mathematically it is perhaps more informative to use phasor diagram to explain what is happening.
To make the analysis easier I have ignored the effect of the diffraction envelope.
For three slits you have the superposition of waves from three coherent sources each of amplitude $A$.
When $\theta = 0^\circ$ the three wave then the phase difference between the waves is zero and so when they overlap they produce a resulting amplitude for a principal maximum of $3A$. This is the $n=0$ fringe.
The same thing happens when the phase difference is $360^\circ$ which is a path difference of $\lambda$. This again results in a principal maximum of amplitude of $3A$. This is the $n = \pm 1$ fringe.
When the phase difference is $180^\circ$ which is a path difference of $\frac \lambda 2$, there is a secondary maximum of amplitude $A$.
For phase differences of $120^\circ$ and $240^\circ$ which correspond to path differences of $\frac {\lambda}{3}$ and $\frac {2\lambda}{3}$ the resultant amplitude is zero. There is a minimum in those positions.
So in the space between adjacent maxima for 2 slits there are now two minima and a secondary maximum. Thus the width of the principal maxima must have decreased.
Imagine how narrow and bright the principal maxima are for a diffraction grating if there are 5000 slits being used.
Finally. The separation of the principal maxima is controlled by the separation of the slits, the wavelength of the light and the order of the fringes whereas the width and intensity of the principal maxima is controlled by the number of slits. |
Natural numbers: \(n\), \(k\)
Real numbers: \(x\)
Real numbers: \(x\)
Factorial of \(n\): \(n!\)
Gamma function: \(\Gamma \left( x \right)\)
Gamma function: \(\Gamma \left( x \right)\)
The factorial of a non-negative integer \(n\) is called the product of all positive integers less than or equal to the number \(n.\) The factorial of \(n\) is denoted as \(n!\) \(n! =\) \( 1 \cdot 2 \cdot 3 \ldots \left( {n – 1} \right) \cdot n\) factorial of zero by definition is equal to\(1:\) \(0! = 1\) Factorials of the numbers \(1-10\) Recursive formula \(\left( {n + 1} \right)! = n! \cdot \left( {n + 1} \right)\) Extension of the factorial function to non-negative real numbers The factorial of a non-negative real number \(x\) is expressed through the gamma function by the formula \(x! = \Gamma \left( {x + 1} \right),\) which allows to calculate the factorial of any real numbers \(x \ge 0\). Rate of increase The factorial function increases faster than the exponential function. The inequality \(n! \gt \exp \left( n \right)\) holds for all \(n \ge 6\). When \(n \ge 1,\) the following relation is valid: \(n \le n! \le {n^n}.\) Stirling formula For large \(n\) the approximating factorial value can be determined by the Stirling formula: \(n! \approx\) \( {n^n}\sqrt {2\pi n} \,\exp \left( { – n} \right) \cdot\) \(\Big[ {1 + {\large\frac{1}{{12n}}\normalsize} + {\large\frac{1}{{288{n^2}}}\normalsize} }-\) \({ {\large\frac{{139}}{{51840{n^3}}}\normalsize} – \ldots } \Big]\). Taking into account only the first term in the expansion, this formula takes the form \(n! \approx {n^n}\sqrt {2\pi n} \,\exp \left( { – n} \right)\). Double factorial The double factorial is the product of all odd integers from \(1\) to some odd positive integer \(n\). The double factorial is denoted by \(n!!\) \(\left( {2k + 1} \right)!! =\) \( 1 \cdot 3 \cdot 5 \cdots \left( {2k – 1} \right) \cdot\) \( \left( {2k + 1} \right).\) Sometimes, the double factorial is considered for even integers. We can define it as \(\left( {2k} \right)!! =\) \( 2 \cdot 4 \cdot 6 \cdots \left( {2k – 2} \right) \cdot 2k\) It is implied here that \(0!! = 1.\) |
Category: Group Theory
Group Theory Problems and Solutions.
Popular posts in Group Theory are:
Problem 625
Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$.
(a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.
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(b) Prove that a group cannot be written as the union of two proper subgroups. Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$.
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(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. Problem 613
Let $m$ and $n$ be positive integers such that $m \mid n$.
(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined. (b) Prove that $\phi$ is a group homomorphism. (c) Prove that $\phi$ is surjective.
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(d) Determine the group structure of the kernel of $\phi$. If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 497
Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively. Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.
Also determine whether the statement is true if $G$ is a non-abelian group.Add to solve later |
Cauchy and Heine Definitions of Limit
Let \(f\left( x \right)\) be a function that is defined on an open interval \(X\) containing \(x = a\). (The value \(f\left( a \right)\) need not be defined.)
The number \(L\) is called the limit of function \(f\left( x \right)\) as \(x \to a\) if and only if, for every \(\varepsilon \gt 0\) there exists \(\delta \gt 0\) such that
\[\left| {f\left( x \right) – L} \right| \lt \varepsilon ,\]
whenever
\[0 \lt \left| {x – a} \right| \lt \delta .\]
This definition is known as \(\varepsilon-\delta-\) or Cauchy definition for limit.
There’s also the Heine definition of the limit of a function, which states that a function \(f\left( x \right)\) has a limit \(L\) at \(x = a\), if for every sequence \(\left\{ {{x_n}} \right\}\), which has a limit at \(a,\) the sequence \(f\left( {{x_n}} \right)\) has a limit \(L.\) The Heine and Cauchy definitions of limit of a function are equivalent.
One-Sided Limits
Let \(\lim\limits_{x \to a – 0} \) denote the limit as \(x\) goes toward \(a\) by taking on values of \(x\) such that \(x \lt a\). The corresponding limit \(\lim\limits_{x \to a – 0} f\left( x \right)\) is called the left-hand limit of \(f\left( x \right)\) at the point \(x = a\).
Similarly, let \(\lim\limits_{x \to a + 0} \) denote the limit as \(x\) goes toward \(a\) by taking on values of \(x\) such that \(x \gt a\). The corresponding limit \(\lim\limits_{x \to a + 0} f\left( x \right)\) is called the right-hand limit of \(f\left( x \right)\) at \(x = a\).
Note that the \(2\)-sided limit \(\lim\limits_{x \to a} f\left( x \right)\) exists only if both one-sided limits exist and are equal to each other, that is \(\lim\limits_{x \to a – 0}f\left( x \right) \) \(= \lim\limits_{x \to a + 0}f\left( x \right) \). In this case,
\[{\lim\limits_{x \to a}f\left( x \right) = \lim\limits_{x \to a – 0}f\left( x \right)} ={ \lim\limits_{x \to a + 0}f\left( x \right).}\]
Solved Problems
Click a problem to see the solution.
Example 1Using the \(\varepsilon-\delta-\) definition of limit, show that \(\lim\limits_{x \to 3} \left( {3x – 2} \right) = 7.\) Example 2Using the \(\varepsilon-\delta-\) definition of limit, show that \(\lim\limits_{x \to 2} {x^2} = 4\). Example 3Using the \(\varepsilon-\delta-\) definition of limit, find the number \(\delta\) that corresponds to the \(\varepsilon\) given with the following limit: Example 4Prove that \(\lim\limits_{x \to \infty } {\large\frac{{x + 1}}{x}\normalsize} = 1\). Example 5Prove that \(\lim\limits_{x \to \infty } {\large\frac{{2x – 3}}{{x + 1}}\normalsize} = 2\). Example 1.Using the \(\varepsilon-\delta-\) definition of limit, show that \(\lim\limits_{x \to 3} \left( {3x – 2} \right) = 7.\)
Solution.
Let \(\varepsilon \gt 0\) be an arbitrary positive number. Choose \(\delta = {\large\frac{\varepsilon }{3}\normalsize}\). We see that if
\[0 \lt \left| {x – 3} \right| \lt \delta, \]
then
\[{\left| {f\left( x \right) – L} \right| = \left| {\left( {3x – 2} \right) – 7} \right|} ={ \left| {3x – 9} \right| }
={ 3\left| {x – 3} \right| \lt 3\delta } = {3 \cdot \frac{\varepsilon }{3} = \varepsilon .} \]
Thus, by Cauchy definition, the limit is proved.
Example 2.Using the \(\varepsilon-\delta-\) definition of limit, show that \(\lim\limits_{x \to 2} {x^2} = 4\).
Solution.
For convenience, we will suppose that \(\delta = 1,\) i.e.
\[\left| {x – 2} \right| \lt 1.\]
Let \(\varepsilon \gt 0\) be an arbitrary number. Then we can write the following inequality:
\[{\left| {{x^2} – 4} \right| \lt \varepsilon ,\;\;}\Rightarrow
{\left| {x – 2} \right|\left| {x + 2} \right| \lt \varepsilon ,\;\;}\Rightarrow {\left| {x – 2} \right|\left( {x + 2} \right) \lt \varepsilon .} \]
Since the maximum value of \(x\) is \(3\) (as we supposed above), we obtain
\[{5\left| {x – 2} \right| \lt \varepsilon \;\;(\text{if } \left| {x – 2} \right| \lt 1),\;\;}\kern-0.3pt
{\text{or}\;\left| {x – 2} \right| \lt \frac{\varepsilon }{2}.} \]
Then for any \(\varepsilon \gt 0\) we can choose the number \(\delta\) such that
\[\delta = \min \left( {\frac{\varepsilon }{2},1} \right).\]
As a result, the inequalities in the definition of limit will be satisfied. Therefore, the given limit is proved. |
Tagged: normal subgroup If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 470
Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$. Then prove that $H$ is a normal subgroup of $G$.
(
Michigan State University, Abstract Algebra Qualifying Exam) Problem 332
Let $G=\GL(n, \R)$ be the
general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices. Consider the subset of $G$ defined by \[\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.\] Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$. The subgroup $\SL(n,\R)$ is called special linear group |
What do the curly brackets mean in this context?
$$f(x) = 2 \cdot x \cdot 1_{x>0}$$
Is this a condition?
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The $\Bbb 1_{x>0}$ is the indicator function $^1$ that indicates if $x$ is greater than $0$ or not. $$1_{x>0}=\begin{cases} 1, ~~x>0\\0, ~~x \le 0\end{cases}$$
This makes your function into,
$$f(x)=\begin{cases}2x,~~x>0\\0, ~~x \le0 \end{cases}$$
Footnotes:
$^1$(also called the Characteristic function, denoted by $\chi_A$ is the function that takes $1$ when $x \in A$ and $0$ otherwise.) |
1) Let us work in units where the speed-of-light $c=1$ is one.
In Ref. 1 is derived the radial geodesic equation for a particle in the equatorial plane
$$\tag{7.47} (\frac{dr}{d\lambda})^2+2V(r)~=~E^2, $$
with potential
$$ \tag{7.48} 2V(r)~:=~(1-\frac{r_s}{r})((\frac{L}{r})^2+\epsilon). $$
Here $\epsilon=0$ for a massless particle and $\epsilon=1$ for a massive particle. The energy $E$ and angular momentum $L$ are constants of motion (which reflect Killing-symmetries of the Schwarzschild metric); $\lambda$ is the affine parameter of the geodesic; and $r_s\equiv\frac{2GM}{c^2}$ is the Schwarzschild-radius. (More precisely, in the massive case $\epsilon=1$, the quantities $E$ and $L$ are specific quantities, i.e. quantities per unit rest mass; and $\lambda$ is proper time.)
2) By differentiating eq. (7.47) wrt. $\lambda$, we find that the condition for a circular orbit
$$r(\lambda)~\equiv~ r_{*} \qquad\Rightarrow\qquad \frac{dr}{d\lambda}~\equiv~0$$
is
$$\tag{1}V'(r_{*})~=~0\qquad\Leftrightarrow\qquad \frac{2r_{*}}{r_s}~=~3+\epsilon(\frac{r_{*}}{L})^2.$$
3) Let us next investigate an incoming particle, which has non-constant radial coordinate $\lambda\mapsto r(\lambda)$, and that is precisely on the critical border between being captured and not being captured by the black hole. It would have a radial turning point $\frac{dr}{d\lambda}=0$ precisely at the radius $r=r_{*}$, so that
$$\tag{2} 2V(r_{*})~=~E^2\qquad\Leftrightarrow\qquad (1-\frac{r_s}{r_{*}})((\frac{L}{r_{*}})^2+\epsilon)~=~E^2.$$
4)
The massless case $\epsilon=0$. Eq. (1) yields
$$\tag{3}r_{*}~=~\frac{3}{2}r_s.$$
Plugging eq. (3) into eq. (2) then yields the ratio
$$\tag{4} \frac{L}{E}~=~\frac{3}{2}\sqrt{3}r_s. $$
We next use that $L$ and $E$ are constants of motion, so that we can easily identify them at spacial infinity $r=\infty$, where special relativistic formulas apply. The critical impact parameter $b$ is precisely this ratio
$$\tag{5} b~=~\frac{L}{p}~=~\frac{L}{E}~\stackrel{(4)}{=}~\underline{\underline{\frac{3}{2}\sqrt{3}r_s}}. $$
5)
The non-relativistic case $v_{\infty}\ll 1$. The specific energy $E\approx 1$ consists mostly of rest energy. Solving eqs. (1) and (2) then leads to a unique solution
$\tag{6}r_{*}~\approx~ 2r_s~\approx~ L.$
The critical impact parameter $b$ becomes
$$\tag{5} b~=~\frac{L}{v_{\infty}}~\approx~\underline{\underline{2r_s\frac{c}{v_{\infty}}}}, $$
cf. Ref. 2. The cross section is $\sigma=\pi b^2$.
References:
S. Carroll,
Lecture Notes on General Relativity, Chapter 7, p.172-179. The pdf file is available from his website.
V.P. Frolov and I.D. Novikov,
Black Hole Physics: Basic Concepts and New Developments, p.48. |
Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$.
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(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. Problem 613
Let $m$ and $n$ be positive integers such that $m \mid n$.
(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined. (b) Prove that $\phi$ is a group homomorphism. (c) Prove that $\phi$ is surjective.
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(d) Determine the group structure of the kernel of $\phi$. Problem 612
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$. (a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
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(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$. Problem 611
An $n\times n$ matrix $A$ is called
orthogonal if $A^{\trans}A=I$. Let $V$ be the vector space of all real $2\times 2$ matrices.
Consider the subset
\[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$. Problem 607
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.
Let \[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\] where \begin{align*} p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\ p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3. \end{align*} (a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$. (b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 606
Let $V$ be a vector space and $B$ be a basis for $V$.
Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$. Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form
\[\begin{bmatrix} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) What is the dimension of $V$? (b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 605
Let $T:\R^2 \to \R^3$ be a linear transformation such that
\[T\left(\, \begin{bmatrix} 3 \\ 2 \end{bmatrix} \,\right) =\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \text{ and } T\left(\, \begin{bmatrix} 4\\ 3 \end{bmatrix} \,\right) =\begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix}.\] (a) Find the matrix representation of $T$ (with respect to the standard basis for $\R^2$). (b) Determine the rank and nullity of $T$.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 604
Let
\[A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) Find a basis for the null space $\calN(A)$. (b) Find a basis of the range $\calR(A)$. (c) Find a basis of the row space for $A$.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 603
Let $C[-2\pi, 2\pi]$ be the vector space of all continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the functions \[f(x)=\sin^2(x) \text{ and } g(x)=\cos^2(x)\] in $C[-2\pi, 2\pi]$.
Prove or disprove that the functions $f(x)$ and $g(x)$ are linearly independent.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 601
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let \[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\] (a) Show that $W$ is a subspace of $V$. (b) Find a basis of $W$. (c) Find the dimension of $W$.
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(The Ohio State University, Linear Algebra Midterm) Read solution |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
Correlation coefficients in mathematics are used to measure the relationship between two variables. However, there are different types of Correlation coefficients but the most common type is Pearson’s correlation that is immensely popular in linear regression.
Most of the times, when we are discussing Correlation coefficients then it is all about Pearson’s until any specific type is not mentioned in the problem. With the help of Correlation coefficients formulas, you can check how strong is the relationship among data. The value is returned between -1 to 1 generally where,
\[\large r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}\]
Here,
n= Quantity of information.
Σx = Total of the first variable value. Σy = Total of the second variable value. Σxy =Sum of the product of first & second value. Σx 2 = Sum of the squares of the first value. Σy 2 = Sum of the squares of the second value.
Linear correlation is used to find the relationship among two variables in a population. With the help of Formula, you can find how two variables are connected together and the value will always be calculated between -1 and 1. For a strong relationship, the value is 1. If value comes closer to -1 then the relationship is negative and in case of a zero, there is no relationship exists between data given.
Where, n is the number of observations, x
i and y i are the variables.
To measure the relative variability, the coefficient of variation (CV) formula is used. This is taken as the ratio of standard deviation to the mean. The most common application of CV is the comparison of the result of two surveys or tests. Researchers and scientists are already familiar with the application of CV in different conditions. There could be conducted multiple tests till the time you are not sure of the final outcome.
\[\ Coefficient\;of\;Variation\;Formula = \frac{Standard\;Deviation}{Mean}\]
As per sample and population data type, the formula for standard deviation may vary –
\[\ Sample\;Standard\;Deviation=\frac{\sqrt{\sum_{i=1}^{n}(X_{i}-\overline{X})^{2}}}{n-1}\]
\[\ Population\;Standard\;Deviation=\frac{\sqrt{\sum_{i=1}^{n}(X_{i}-\overline{X})^{2}}}{n}\]
Where,
x i = Terms given in the data $\overline{x}$ = Mean n = Total number of terms.
There are different states of correlation Formula, one of the most popular types if Pearson’s correlation coefficient formula. For the basic tests, it is used frequently. With the help of Pearson’s Formula, you can quickly identify the dependent or independent variables. For example, how high-calorie diet and diabetes are correlated, researchers generally analyze the associated data before they reach the final conclusion.
\[\large r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}\]
Where,
r = Pearson correlation coefficient x = Values in the first set of data y = Values in the second set of data n = Total number of values.
Pearson’s correlation coefficient formula has a multiple of apps in real-life too. For example, if you want to test a portion of population growth for some product then this technique can be used. It is frequently used for a variety of research techniques, sample analysis, and higher studies in colleges. With the basic understanding of the concept, you can apply the Formula for real-life complex problems later during your practical work. If you will check online there are numerous studies where correlation is shown between the two variables. Like, how carbohydrates increased weight and low carbohydrate content in your diet will trigger the weight loss. There are more studies that can be easily found online and used in our daily practices too.
\[\large r = \frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{\left [ n\sum x^{2}-(\sum x)^{2} \right ]\left [ n\sum y^{2}-(\sum y)^{2} \right ]}}\]
Where,
r = Correlation coefficient x = Values in first set of data y = Values in second set of data n = Total number of values. |
Can I be a pedant and say that if the question states that $\langle \alpha \vert A \vert \alpha \rangle = 0$ for every vector $\lvert \alpha \rangle$, that means that $A$ is everywhere defined, so there are no domain issues?
Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of:We can always detect uniform motion with respect to a medium by a positive result to a Michelson...
Hmm, it seems we cannot just superimpose gravitational waves to create standing waves
The above search is inspired by last night dream, which took place in an alternate version of my 3rd year undergrad GR course. The lecturer talks about a weird equation in general relativity that has a huge summation symbol, and then talked about gravitational waves emitting from a body. After that lecture, I then asked the lecturer whether gravitational standing waves are possible, as a imagine the hypothetical scenario of placing a node at the end of the vertical white line
[The Cube] Regarding The Cube, I am thinking about an energy level diagram like this
where the infinitely degenerate level is the lowest energy level when the environment is also taken account of
The idea is that if the possible relaxations between energy levels is restricted so that to relax from an excited state, the bottleneck must be passed, then we have a very high entropy high energy system confined in a compact volume
Therefore, as energy is pumped into the system, the lack of direct relaxation pathways to the ground state plus the huge degeneracy at higher energy levels should result in a lot of possible configurations to give the same high energy, thus effectively create an entropy trap to minimise heat loss to surroundings
@Kaumudi.H there is also an addon that allows Office 2003 to read (but not save) files from later versions of Office, and you probably want this too. The installer for this should also be in \Stuff (but probably isn't if I forgot to include the SP3 installer).
Hi @EmilioPisanty, it's great that you want to help me clear out confusions. I think we have a misunderstanding here. When you say "if you really want to "understand"", I've thought you were mentioning at my questions directly to the close voter, not the question in meta. When you mention about my original post, you think that it's a hopeless mess of confusion? Why? Except being off-topic, it seems clear to understand, doesn't it?
Physics.stackexchange currently uses 2.7.1 with the config TeX-AMS_HTML-full which is affected by a visual glitch on both desktop and mobile version of Safari under latest OS, \vec{x} results in the arrow displayed too far to the right (issue #1737). This has been fixed in 2.7.2. Thanks.
I have never used the app for this site, but if you ask a question on a mobile phone, there is no homework guidance box, as there is on the full site, due to screen size limitations.I think it's a safe asssumption that many students are using their phone to place their homework questions, in wh...
@0ßelö7 I don't really care for the functional analytic technicalities in this case - of course this statement needs some additional assumption to hold rigorously in the infinite-dimensional case, but I'm 99% that that's not what the OP wants to know (and, judging from the comments and other failed attempts, the "simple" version of the statement seems to confuse enough people already :P)
Why were the SI unit prefixes, i.e.\begin{align}\mathrm{giga} && 10^9 \\\mathrm{mega} && 10^6 \\\mathrm{kilo} && 10^3 \\\mathrm{milli} && 10^{-3} \\\mathrm{micro} && 10^{-6} \\\mathrm{nano} && 10^{-9}\end{align}chosen to be a multiple power of 3?Edit: Although this questio...
the major challenge is how to restrict the possible relaxation pathways so that in order to relax back to the ground state, at least one lower rotational level has to be passed, thus creating the bottleneck shown above
If two vectors $\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}$ and$\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}$, have angle $\theta$ between them then the dot product (scalar product) of $\vec{A}$ and $\vec{B}$ is$$\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos \theta$$$$\vec{A}\cdot\...
@ACuriousMind I want to give a talk on my GR work first. That can be hand-wavey. But I also want to present my program for Sobolev spaces and elliptic regularity, which is reasonably original. But the devil is in the details there.
@CooperCape I'm afraid not, you're still just asking us to check whether or not what you wrote there is correct - such questions are not a good fit for the site, since the potentially correct answer "Yes, that's right" is too short to even submit as an answer |
Motivation:
Many interesting
irrational numbers (or numbers believed to be irrational) appear as answers to natural questions in mathematics. Famous examples are $e$, $\pi$, $\log 2$, $\zeta(3)$ etc. Many more such numbers are described for example in the wonderful book "Mathematical Constants" by Steven R. Finch. The question:
I am interested in theorems where a "special"
rational number makes a surprising appearance as an answer to a natural question. By a special rational number I mean one with a large denominator (and preferably also a large numerator, to rule out numbers which are simply the reciprocals of large integers, but I'll consider exceptions to this rule). Please provide examples.
For illustration, here are a couple of nice examples I'm aware of:
The average geodesic distance between two random points of the Sierpinski gasket of unit side lengths is equal to $\frac{466}{885}$. This is also equivalent to a natural discrete math fact about the analysis of algorithms, namely that the average number of moves in the Tower of Hanoi game with $n$ disks connecting a randomly chosen initial state to a randomly chosen terminal state with a shortest number of moves, is asymptotically equal to $\frac{466}{885}\times 2^n$. See here and here for more information.
The answer to the title question of the recent paper ""The density of primes dividing a term in the Somos-5 sequence" by Davis, Kotsonis and Rouse is $\frac{5087}{10752}$.
Rules:
1) I won't try to define how large the denominator and numerator need to be to for the rational number to qualify as "special". A good answer will maximize the ratio of the number's information theoretic content to the information theoretic content of the statement of the question it answers. (E.g., a number like 34/57 may qualify if the question it answers is simple enough.) Really simple fractions like $3/4$, $22/7$ obviously do not qualify.
2) The question the number answers needs to be natural. Again, it's impossible to define what this means, but avoid answers in the style of "what is the rational number with smallest denominator solving the Diophantine equation [some arbitrary-sounding, unmotivated equation]".
Edit: a lot of great answers so far, thanks everyone. To clarify my question a bit, while all the answers posted so far represent very beautiful mathematics and some (like Richard Stanley's and Max Alekseyev's answers) are truly astonishing, my favorite type of answers involve questions that are conceptual in nature (e.g., longest increasing subsequences, tower of Hanoi, Markov spectrum, critical exponents in percolation) rather than purely computational (e.g., compute some integral or infinite series) and to which the answer is an exotic rational number. (Note that someone edited my original question changing "exotic" to "special"; that is fine, but "exotic" does a better job of signaling that numbers like 1/4 and 2 are not really what I had in mind. That is, 2 is indeed quite a special number, but I doubt anyone would consider it exotic.) |
Following the terminology of Rosenblatt, I will say that a bounded function $f:\mathbb Z\rightarrow\mathbb R$ has a unique mean value if for every pair of finitely additive translation invariant probability measures $\lambda_1,\lambda_2$ on $\mathbb Z$, one has $\int fd\lambda_1=\int fd\lambda_2$.
I will say that $f$ has Fubini's property if for all finitely additive probability measures $\mu,\nu$ on $\mathbb Z$, one has $\int\int f(x+y)d\mu d\nu=\int\int f(x+y)d\nu d\mu$.
Question: Is it true that the two properties above are equivalent?
It is obvious that if $f$ has Fubini's property, then it has also a unique mean value, but the converse is not clear to me. I don't have a real evidence why the two properties should be the same.. let's say, that I am interested in studying the relation between these two properties and I was not able to find a function with a unique mean value which does not have Fubini's property. |
@JosephWright Well, we still need table notes etc. But just being able to selectably switch off parts of the parsing one does not need... For example, if a user specifies format 2.4, does the parser even need to look for e syntax, or ()'s?
@daleif What I am doing to speed things up is to store the data in a dedicated format rather than a property list. The latter makes sense for units (open ended) but not so much for numbers (rigid format).
@JosephWright I want to know about either the bibliography environment or \DeclareFieldFormat. From the documentation I see no reason not to treat these commands as usual, though they seem to behave in a slightly different way than I anticipated it. I have an example here which globally sets a box, which is typeset outside of the bibliography environment afterwards. This doesn't seem to typeset anything. :-( So I'm confused about the inner workings of biblatex (even though the source seems....
well, the source seems to reinforce my thought that biblatex simply doesn't do anything fancy). Judging from the source the package just has a lot of options, and that's about the only reason for the large amount of lines in biblatex1.sty...
Consider the following MWE to be previewed in the build in PDF previewer in Firefox\documentclass[handout]{beamer}\usepackage{pgfpages}\pgfpagesuselayout{8 on 1}[a4paper,border shrink=4mm]\begin{document}\begin{frame}\[\bigcup_n \sum_n\]\[\underbrace{aaaaaa}_{bbb}\]\end{frame}\end{d...
@Paulo Finally there's a good synth/keyboard that knows what organ stops are! youtube.com/watch?v=jv9JLTMsOCE Now I only need to see if I stay here or move elsewhere. If I move, I'll buy this there almost for sure.
@JosephWright most likely that I'm for a full str module ... but I need a little more reading and backlog clearing first ... and have my last day at HP tomorrow so need to clean out a lot of stuff today .. and that does have a deadline now
@yo' that's not the issue. with the laptop I lose access to the company network and anythign I need from there during the next two months, such as email address of payroll etc etc needs to be 100% collected first
@yo' I'm sorry I explain too bad in english :) I mean, if the rule was use \tl_use:N to retrieve the content's of a token list (so it's not optional, which is actually seen in many places). And then we wouldn't have to \noexpand them in such contexts.
@JosephWright \foo:V \l_some_tl or \exp_args:NV \foo \l_some_tl isn't that confusing.
@Manuel As I say, you'd still have a difference between say \exp_after:wN \foo \dim_use:N \l_my_dim and \exp_after:wN \foo \tl_use:N \l_my_tl: only the first case would work
@Manuel I've wondered if one would use registers at all if you were starting today: with \numexpr, etc., you could do everything with macros and avoid any need for \<thing>_new:N (i.e. soft typing). There are then performance questions, termination issues and primitive cases to worry about, but I suspect in principle it's doable.
@Manuel Like I say, one can speculate for a long time on these things. @FrankMittelbach and @DavidCarlisle can I am sure tell you lots of other good/interesting ideas that have been explored/mentioned/imagined over time.
@Manuel The big issue for me is delivery: we have to make some decisions and go forward even if we therefore cut off interesting other things
@Manuel Perhaps I should knock up a set of data structures using just macros, for a bit of fun [and a set that are all protected :-)]
@JosephWright I'm just exploring things myself “for fun”. I don't mean as serious suggestions, and as you say you already thought of everything. It's just that I'm getting at those points myself so I ask for opinions :)
@Manuel I guess I'd favour (slightly) the current set up even if starting today as it's normally \exp_not:V that applies in an expansion context when using tl data. That would be true whether they are protected or not. Certainly there is no big technical reason either way in my mind: it's primarily historical (expl3 pre-dates LaTeX2e and so e-TeX!)
@JosephWright tex being a macro language means macros expand without being prefixed by \tl_use. \protected would affect expansion contexts but not use "in the wild" I don't see any way of having a macro that by default doesn't expand.
@JosephWright it has series of footnotes for different types of footnotey thing, quick eye over the code I think by default it has 10 of them but duplicates for minipages as latex footnotes do the mpfoot... ones don't need to be real inserts but it probably simplifies the code if they are. So that's 20 inserts and more if the user declares a new footnote series
@JosephWright I was thinking while writing the mail so not tried it yet that given that the new \newinsert takes from the float list I could define \reserveinserts to add that number of "classic" insert registers to the float list where later \newinsert will find them, would need a few checks but should only be a line or two of code.
@PauloCereda But what about the for loop from the command line? I guess that's more what I was asking about. Say that I wanted to call arara from inside of a for loop on the command line and pass the index of the for loop to arara as the jobname. Is there a way of doing that? |
As this Wikipedia article describes, we can define probability density functions (pdfs) for discrete random variables using Dirac delta functions, which is called "generalized pdf".
I considered the following pdf and tried to calculate its entropy (differential entropy): $$ f(t) = \sum\limits_{i = 1}^n {{p_i}\delta (t - {x_i})} $$
I obtained $-\infty$ by solving the integral (actually because $\int\delta(0)log(\delta(0))dt$ terms appeared). But this result seems odd to me, as I found the entropy is always $-\infty$ regardless of the parameters $n$ and $\{p_i\}$. I think it is not intuitive, since if we compare it to the definition of entropy for discrete distributions, we expect that the entropy gets higher as the distribution gets closer to uniform.
Have I made a mistake in solving the integral? Or the entropy is really independent of distribution parameters in this case? |
2018-09-11 04:29
Proprieties of FBK UFSDs after neutron and proton irradiation up to $6*10^{15}$ neq/cm$^2$ / Mazza, S.M. (UC, Santa Cruz, Inst. Part. Phys.) ; Estrada, E. (UC, Santa Cruz, Inst. Part. Phys.) ; Galloway, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; Gee, C. (UC, Santa Cruz, Inst. Part. Phys.) ; Goto, A. (UC, Santa Cruz, Inst. Part. Phys.) ; Luce, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; McKinney-Martinez, F. (UC, Santa Cruz, Inst. Part. Phys.) ; Rodriguez, R. (UC, Santa Cruz, Inst. Part. Phys.) ; Sadrozinski, H.F.-W. (UC, Santa Cruz, Inst. Part. Phys.) ; Seiden, A. (UC, Santa Cruz, Inst. Part. Phys.) et al. The properties of 60-{\mu}m thick Ultra-Fast Silicon Detectors (UFSD) detectors manufactured by Fondazione Bruno Kessler (FBK), Trento (Italy) were tested before and after irradiation with minimum ionizing particles (MIPs) from a 90Sr \b{eta}-source . [...] arXiv:1804.05449. - 13 p. Preprint - Full text Подробная запись - Подобные записи 2018-08-25 06:58
Charge-collection efficiency of heavily irradiated silicon diodes operated with an increased free-carrier concentration and under forward bias / Mandić, I (Ljubljana U. ; Stefan Inst., Ljubljana) ; Cindro, V (Ljubljana U. ; Stefan Inst., Ljubljana) ; Kramberger, G (Ljubljana U. ; Stefan Inst., Ljubljana) ; Mikuž, M (Ljubljana U. ; Stefan Inst., Ljubljana) ; Zavrtanik, M (Ljubljana U. ; Stefan Inst., Ljubljana) The charge-collection efficiency of Si pad diodes irradiated with neutrons up to $8 \times 10^{15} \ \rm{n} \ cm^{-2}$ was measured using a $^{90}$Sr source at temperatures from -180 to -30°C. The measurements were made with diodes under forward and reverse bias. [...] 2004 - 12 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 533 (2004) 442-453 Подробная запись - Подобные записи 2018-08-23 11:31 Подробная запись - Подобные записи 2018-08-23 11:31
Effect of electron injection on defect reactions in irradiated silicon containing boron, carbon, and oxygen / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Yakushevich, H S (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) Comparative studies employing Deep Level Transient Spectroscopy and C-V measurements have been performed on recombination-enhanced reactions between defects of interstitial type in boron doped silicon diodes irradiated with alpha-particles. It has been shown that self-interstitial related defects which are immobile even at room temperatures can be activated by very low forward currents at liquid nitrogen temperatures. [...] 2018 - 7 p. - Published in : J. Appl. Phys. 123 (2018) 161576 Подробная запись - Подобные записи 2018-08-23 11:31 Подробная запись - Подобные записи 2018-08-23 11:31
Characterization of magnetic Czochralski silicon radiation detectors / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) Silicon wafers grown by the Magnetic Czochralski (MCZ) method have been processed in form of pad diodes at Instituto de Microelectrònica de Barcelona (IMB-CNM) facilities. The n-type MCZ wafers were manufactured by Okmetic OYJ and they have a nominal resistivity of $1 \rm{k} \Omega cm$. [...] 2005 - 9 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 548 (2005) 355-363 Подробная запись - Подобные записи 2018-08-23 11:31
Silicon detectors: From radiation hard devices operating beyond LHC conditions to characterization of primary fourfold coordinated vacancy defects / Lazanu, I (Bucharest U.) ; Lazanu, S (Bucharest, Nat. Inst. Mat. Sci.) The physics potential at future hadron colliders as LHC and its upgrades in energy and luminosity Super-LHC and Very-LHC respectively, as well as the requirements for detectors in the conditions of possible scenarios for radiation environments are discussed in this contribution.Silicon detectors will be used extensively in experiments at these new facilities where they will be exposed to high fluences of fast hadrons. The principal obstacle to long-time operation arises from bulk displacement damage in silicon, which acts as an irreversible process in the in the material and conduces to the increase of the leakage current of the detector, decreases the satisfactory Signal/Noise ratio, and increases the effective carrier concentration. [...] 2005 - 9 p. - Published in : Rom. Rep. Phys.: 57 (2005) , no. 3, pp. 342-348 External link: RORPE Подробная запись - Подобные записи 2018-08-22 06:27
Numerical simulation of radiation damage effects in p-type and n-type FZ silicon detectors / Petasecca, M (Perugia U. ; INFN, Perugia) ; Moscatelli, F (Perugia U. ; INFN, Perugia ; IMM, Bologna) ; Passeri, D (Perugia U. ; INFN, Perugia) ; Pignatel, G U (Perugia U. ; INFN, Perugia) In the framework of the CERN-RD50 Collaboration, the adoption of p-type substrates has been proposed as a suitable mean to improve the radiation hardness of silicon detectors up to fluencies of $1 \times 10^{16} \rm{n}/cm^2$. In this work two numerical simulation models will be presented for p-type and n-type silicon detectors, respectively. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 2971-2976 Подробная запись - Подобные записи 2018-08-22 06:27
Technology development of p-type microstrip detectors with radiation hard p-spray isolation / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) ; Díez, S (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) A technology for the fabrication of p-type microstrip silicon radiation detectors using p-spray implant isolation has been developed at CNM-IMB. The p-spray isolation has been optimized in order to withstand a gamma irradiation dose up to 50 Mrad (Si), which represents the ionization radiation dose expected in the middle region of the SCT-Atlas detector of the future Super-LHC during 10 years of operation. [...] 2006 - 6 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 566 (2006) 360-365 Подробная запись - Подобные записи 2018-08-22 06:27
Defect characterization in silicon particle detectors irradiated with Li ions / Scaringella, M (INFN, Florence ; U. Florence (main)) ; Menichelli, D (INFN, Florence ; U. Florence (main)) ; Candelori, A (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Bruzzi, M (INFN, Florence ; U. Florence (main)) High Energy Physics experiments at future very high luminosity colliders will require ultra radiation-hard silicon detectors that can withstand fast hadron fluences up to $10^{16}$ cm$^{-2}$. In order to test the detectors radiation hardness in this fluence range, long irradiation times are required at the currently available proton irradiation facilities. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 589-594 Подробная запись - Подобные записи |
A number $n$ is called
insipid if the groups having a core-free maximal subgroup of index $n$ are exactly $A_n$ and $S_n$. There is an OEIS enter for these numbers: A102842. There are exactly $486$ insipid numbers less than $1000$. Question: Are there infinitely many insipid numbers?
Let $\iota(r)$ be the number of insipid numbers less than $r$. The following plot (from OEIS) leads to:
Bonus question: Is it true that $\lim_{r \to \infty}r/\iota(r)=2$? |
Edit:
Hilbert Derived the EFE before Einstein, but using the EH Action as a postulate. Einstein took the EFE as a postulate.
I think the matter is only the formulation of General Relativity; the postulates made.
Hilbert's postulate was elegantly simple. It was only that
$$\mathcal L_G=\lambda R$$
I.e. that the Lagrangian Density of Gravity was proportional to the Ricci Scalar. From then on, he applied the Hamilton's principle to this's equation and found that the field equations for gravity are $$G_{\mu\nu}=\kappa T_{\mu\nu}$$ And then, he showed how this is a better theory of gravity etc. etc.
Einstein's postulate was not so elegant or simple, in my opinion. However, many hold that it was simpler. His postulate was the field equation itself! I.e that
$$G_{\mu\nu}=\kappa T_{\mu\nu}$$
The reason why people hold this to be very elegant is that $\nabla^\mu G_{\mu\nu}=0$ and $\nabla^\mu T_{\mu\nu}=0$.
So, it is no different than whether Einstein or Minkowski should get the credit. Minkowski did it the elegant way.
Similarly, Hilbert did it the elegant way.
Edit:
That Einstein's postulate was the EFE has its reference in Einstein's original GR paper "On the Foundations of the General Theory of Relativity". As for Hilbert's postulate being the EH Action, I think it was from some book... Can't remember. Note:
P.S. The fact that Einstein's postulate itself was the EFE means that he first wrote down the EFE. The fact that Hilbert's postulate itself was the EH Action means that he first wrote down the EH Action, (that is, 5 days before Einstein wrote down the EFE).
P.P.S. By "wrote down", I mean "published", who knows wheter Einstein/Hilbert actually wrote down the action/FE, or forced someone else to write it for them? Who knows...
P.P.P.S. Another important point of this answer is that there is no controversy, since it is just the formulation that matters, like Einstein&Minkowski, Newton&Leibniz, Heisenberg&Schrodinger&Feynman&somerandompersonwhodiscoveredthevariationalformulation&..., etc. Because the OP seems to think there's a big controversy... But there's not, at least there shouldn't be. |
Tagged: invertible matrix Problem 583
Consider the $2\times 2$ complex matrix
\[A=\begin{bmatrix} a & b-a\\ 0& b \end{bmatrix}.\] (a) Find the eigenvalues of $A$. (b) For each eigenvalue of $A$, determine the eigenvectors. (c) Diagonalize the matrix $A$.
Add to solve later
(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$. Problem 582
A square matrix $A$ is called
nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.
Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample. Problem 562
An $n\times n$ matrix $A$ is called
nonsingular if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$. Using the definition of a nonsingular matrix, prove the following statements. (a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular. (b) Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then: The matrix $B$ is nonsingular. The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.) Problem 552
For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix.
Add to solve later
(a) $A=\begin{bmatrix} 1 & 3 & -2 \\ 2 &3 &0 \\ 0 & 1 & -1 \end{bmatrix}$ (b) $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$. Problem 548
An $n\times n$ matrix $A$ is said to be
invertible if there exists an $n\times n$ matrix $B$ such that $AB=I$, and $BA=I$,
where $I$ is the $n\times n$ identity matrix.
If such a matrix $B$ exists, then it is known to be unique and called the
inverse matrix of $A$, denoted by $A^{-1}$.
In this problem, we prove that if $B$ satisfies the first condition, then it automatically satisfies the second condition.
So if we know $AB=I$, then we can conclude that $B=A^{-1}$.
Let $A$ and $B$ be $n\times n$ matrices.
Suppose that we have $AB=I$, where $I$ is the $n \times n$ identity matrix.
Prove that $BA=I$, and hence $A^{-1}=B$.Add to solve later
Problem 546
Let $A$ be an $n\times n$ matrix.
The $(i, j)$
cofactor $C_{ij}$ of $A$ is defined to be \[C_{ij}=(-1)^{ij}\det(M_{ij}),\] where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column.
Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$.
The matrix $\Adj(A)$ is called the adjoint matrix of $A$.
When $A$ is invertible, then its inverse can be obtained by the formula
For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula.
(a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$. (b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$. Problem 506
Let $A$ be an $n\times n$ invertible matrix. Then prove the transpose $A^{\trans}$ is also invertible and that the inverse matrix of the transpose $A^{\trans}$ is the transpose of the inverse matrix $A^{-1}$.
Namely, show that \[(A^{\trans})^{-1}=(A^{-1})^{\trans}.\] Problem 500
10 questions about nonsingular matrices, invertible matrices, and linearly independent vectors.
The quiz is designed to test your understanding of the basic properties of these topics.
You can take the quiz as many times as you like.
The solutions will be given after completing all the 10 problems.
Click the View question button to see the solutions. Problem 452
Let $A$ be an $n\times n$ complex matrix.
Let $S$ be an invertible matrix. (a) If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix. (b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.
Add to solve later
(c) Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$. Problem 438
Determine whether each of the following statements is True or False.
(a) If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$. (b) If the characteristic polynomial of an $n \times n$ matrix $A$ is \[p(\lambda)=(\lambda-1)^n+2,\] then $A$ is invertible. (c) If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible. (d) If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$. (e) If $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ with corresponding eigenvalue $\lambda_1$, and if $\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_2$, then $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$.
(
Stanford University, Linear Algebra Exam Problem) Read solution |
MathJaxExamples
See how math markup works with SimpleMathJax extension.
You can now add TeX formulas to a wiki page and have the rendered beautifully. Simple put the TeX code between and opening <math> and closing </math> tag and the page will render the TeX.
Here are some examples taken from TeX examples:
Enter the text:
<math>E=mc^2</math>
And the page will render it as:
[math]E=mc^2[/math]
Here are the Lorenz equations (view the source of this page to see the wiki code):
[math]\begin{align} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{align}[/math] |
I thought of this question while studying inelastic collisions in special relativity, where kinetic energy is converted into mass-energy.
I was wondering if it's possible to formalize a version of special relativity where mass conservation still holds. Basically I'm imagining that, in this hypothetical theory, the loss of kinetic energy in an inelastic collision would be accounted for in the same way as in classical kinematics: By assuming that the kinetic energy associated with center-of-mass-motion is converted into kinetic energy associated with disorganized, relative motion
about the center of mass - that is, thermal energy.
Of course, this theory would be empirically wrong. But here is my question: Would it be wrong simply because Nature "decided" not to do things that way, or is there a compelling theoretical reason why we ought to doubt it? For example, would such a theory contradict Einstein's postulates of special relativity in some way? Or, perhaps, would it violate certain symmetries?
Thanks.
Edit: To make my question more concrete, here's an example of some calculations in this hypothetical theory, per Ismasou's request:
Consider a collision between two objects, of masses $m_1$ and $m_2$. Suppose these two masses "stick together" upon collision.
Given mass conservation, the total mass of the resulting object is just $m_1 + m_2$. Now, let's restrict our attention to the inertial frame where $m_2$ is at rest, and $m_1$ is moving at some initial speed $v_i$ (assume this is a 1-dimensional problem). What is the final speed $v_f$ of the composite mass?
Well, supposing that relativistic 3-momentum conservation still holds, we know that the initial and final relativistic momentum $p$ is given by:
$$p=\frac{1}{\sqrt{1-v_i^2/c^2}} m_1 v_i=\frac{1}{\sqrt{1-v_f^2/c^2}} (m_1+m_2) v_f$$
If my calculations are right, solving this equation for $v_f$ yields:
$$v_f = \frac{1}{\sqrt{1-\left[1-\left(\frac{m_1}{m_1+m_2}\right)^2 \right]\frac{v_i^2}{c^2}}} \frac{m_1}{m_1+m_2} v_i$$
Notice that in the limit of low speeds, the square root factor approaches 1, and we recover the result obtained in a classical inelastic collision problem.
Now, here, it is not obvious that relativistic energy $E=\gamma m c^2$ is conserved - it seems like we've had to give up energy conservation to maintain mass conservation. However, in this theory, relativistic energy might be conserved in the same way that kinetic energy is conserved in a classical inelastic collision: By postulating that the kinetic energy which moved $m_1$ forward has simply been dispersed into the many disorganized motions of the composite mass about its center of mass (thermal energy). |
Examples of Differential Operators
Differential operators are a generalization of the operation of differentiation.
The simplest differential operator \(D\) acting on a function \(y,\) “returns” the first derivative of this function:
\[Dy\left( x \right) = y’\left( x \right).\]
Double \(D\) allows to obtain the second derivative of the function \(y\left( x \right):\)
\[{{D^2}y\left( x \right) = D\left( {Dy\left( x \right)} \right) }={ Dy’\left( x \right) }={ y^{\prime\prime}\left( x \right).}\]
Similarly, the \(n\)th power of \(D\) leads to the \(n\)th derivative:
\[{D^n}y\left( x \right) = {y^{\left( n \right)}}\left( x \right).\]
Here we assume that the function \(y\left( x \right)\) is \(n\) times differentiable and defined on the set of real numbers. The function \(y\left( x \right)\) itself can take complex values.
Differential operators may be more complicated depending on the form of differential expression.
For example, the nabla differential operator often appears in vector analysis. It is defined as
\[\nabla = \frac{\partial }{{\partial x}}\mathbf{i} + \frac{\partial }{{\partial y}}\mathbf{j} + \frac{\partial }{{\partial z}}\mathbf{k},\]
where \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are the unit vectors along the coordinate axes \(x,\) \(y,\) \(z.\)
As a result of acting of the operator \(\nabla\) on a scalar field \(F,\) we obtain the gradient of the field \(F:\)
\[{\nabla F }={ \frac{{\partial F}}{{\partial x}}\mathbf{i} + \frac{{\partial F}}{{\partial y}}\mathbf{j} }+{ \frac{{\partial F}}{{\partial z}}\mathbf{k}.}\]
The gradient vector always points in the direction of greatest increase of the function \(F,\) and its length indicates the rate of increase of the function in this direction.
The scalar product of vector \(\nabla\) and the vector field \(\mathbf{V}\) is known as the divergence of the vector \(\mathbf{V}:\)
\[{\nabla \cdot \mathbf{V} = \text{div}\,\mathbf{V} }={ \frac{{\partial {V_x}}}{{\partial x}} + \frac{{\partial {V_y}}}{{\partial y}} + \frac{{\partial {V_z}}}{{\partial z}}.}\]
The vector product of vectors \(\nabla\) and \(\mathbf{V}\) gives the curl of the vector \(\mathbf{V}:\)
\[ {\nabla \times \mathbf{V} = \text{rot}\,\mathbf{V} } = {\left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{V_x}}&{{V_y}}&{{V_z}} \end{array}} \right|.} \]
The scalar product of \(\nabla \cdot \nabla = {\nabla ^2}\) corresponds to a scalar differential operator, called the Laplace operator or Laplacian. It is also denoted by the symbol \(\Delta:\)
\[{\Delta = {\nabla ^2} }={ \frac{{{\partial ^2}}}{{\partial {x^2}}} + \frac{{{\partial ^2}}}{{\partial {y^2}}} + \frac{{{\partial ^2}}}{{\partial {z^2}}}.}\]
The introduction of differential operators allows to investigate differential equations in terms of operator theory and functional analysis. This generalized approach turns out powerful and effective. In particular, considering application to higher order linear differential equations, we obtain a compact way of writing equations, and in some cases, the possibility of a quick solution.
Differential Operator \(L\left( D \right)\)
Consider the linear differential equation of the \(n\)th order:
\[ {{y^{\left( n \right)}}\left( x \right) }+{ {a_1}\left( x \right){y^{\left( {n – 1} \right)}}\left( x \right) + \cdots } + {{a_{n – 1}}\left( x \right)y’\left( x \right) }+{ {a_n}\left( x \right)y\left( x \right) }={ f\left( x \right).} \]
Using the differential operator \(D,\) this equation can be written as
\[L\left( D \right)y\left( x \right) = f\left( x \right),\]
where \(L\left( D \right)\) is the differential polynomial equal to
\[{L\left( D \right) }={ {D^n} + {a_1}\left( x \right){D^{n – 1}} + \cdots }+{ {a_{n – 1}}\left( x \right)D }+{ {a_n}\left( x \right).}\]
In other words, the operator \(L\left( D \right)\) is an algebraic polynomial, in which the differential operator \(D\) plays the role of a variable.
Let us consider some properties of the operator \(L\left( D \right).\)
The operator \(L\left( D \right)\) is linear:\[ {L\left( D \right)\left[ {{C_1}{y_1}\left( x \right) + {C_2}{y_2}\left( x \right)} \right] } = {{C_1}L\left( D \right){y_1}\left( x \right) }+{ {C_2}L\left( D \right){y_2}\left( x \right).} \]In the case of several operators \(L\left( D \right),\) \(M\left( D \right)\) and \(N\left( D \right)\) (the degree of the differential polynomials can be different), the following properties also hold: Commutative law of addition:\[{L\left( D \right) + M\left( D \right) }={ M\left( D \right) + L\left( D \right).}\] Associative law of addition:\[ {\left[ {L\left( D \right) + M\left( D \right)} \right] + N\left( D \right) } = {L\left( D \right) + \left[ {M\left( D \right) + N\left( D \right)} \right].} \]For two operators \(L\left( D \right)\) and \(M\left( D \right)\), one can also define the multiplication operation: Commutative law of multiplication:\[{L\left( D \right) \cdot M\left( D \right) }={ M\left( D \right) \cdot L\left( D \right)}\] Associative law of multiplication:\[ {{\left[ {L\left( D \right) \cdot M\left( D \right)} \right] \cdot N }\kern0pt{\left( D \right) }} = {{L\left( D \right) \cdot}\kern0pt{ \left[ {M\left( D \right) \cdot N\left( D \right)} \right]}} \] Distributive law of multiplication over addition:\[ {{L\left( D \right) \cdot}\kern0pt{ \left[ {M\left( D \right) + N\left( D \right)} \right] }} = {L\left( D \right) \cdot M\left( D \right) }+{ L\left( D \right) \cdot N\left( D \right)} \]We also mention another useful property of the operator \(D:\) \({D^m}{D^n} = {D^{m + n}}.\)
For such operators, conditions \(4-6\) are satisfied:
As it can be seen, the differential operators \(L\left( D \right)\) with constant coefficients have the same properties as ordinary algebraic polynomials. Consequently, as well as algebraic polynomials, we can multiply, factor or divide differential operators \(L\left( D \right)\) with constant coefficients. These properties are used in the operator method of solution of differential equations.
Solved Problems
Click a problem to see the solution. |
Mass and Static Moments of a Solid
Suppose we have a solid occupying a region \(U.\) Its volume density at a point \(M\left( {x,y,z} \right)\) is given by the function \(\rho\left( {x,y,z} \right).\) Then the mass of the solid \(m\) is expressed through the triple integral as
\[m = \iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} .\]
The static moments of the solid about the coordinate planes \(Oxy, Oxz, Oyz\) are given by the formulas
\[
{{M_{xy}} = \int\limits_U {z\rho \left( {x,y,z} \right)dxdydz} ,\;\;}\kern-0,3pt {{M_{yz}} = \int\limits_U {x\rho \left( {x,y,z} \right)dxdydz} ,\;\;}\kern-0,3pt {{M_{xz}} = \int\limits_U {y\rho \left( {x,y,z} \right)dxdydz} .} \]
The coordinates of the center of gravity of the solid are described by the expressions:
\[
{{\bar x = \frac{{{M_{yz}}}}{m} }={ \frac{{\iiint\limits_U {x\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }},\;\;}}\kern-0.3pt {{\bar y = \frac{{{M_{xz}}}}{m} }={ \frac{{\iiint\limits_U {y\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }},\;\;}}\kern-0.3pt {{\bar z = \frac{{{M_{xy}}}}{m} }={ \frac{{\iiint\limits_U {z\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }}.}} \]
If a solid is homogeneous with density \({\rho \left( {x,y,z} \right)} = 1\) for all points \({M\left( {x,y,z} \right)}\) in the region \(U,\) then the center of gravity of the solid is determined only by the shape of the solid and is called the centroid.
Moments of Inertia of a Solid
The moments of inertia of a solid about the coordinate planes \(Oxy, Oxz, Oyz\) are given by
\[
{{{I_{xy}} }={ \iiint\limits_U {{z^2}\rho \left( {x,y,z} \right)dxdydz} ,\;\;}}\kern-0.3pt {{{I_{yz}} }={ \iiint\limits_U {{x^2}\rho \left( {x,y,z} \right)dxdydz} ,\;\;}}\kern-0.3pt {{{I_{xz}} }={ \iiint\limits_U {{y^2}\rho \left( {x,y,z} \right)dxdydz},}} \]
and the moments of inertia of a solid about the coordinate axes \(Ox, Oy, Oz\) are expressed by the formulas
\[
{{{I_x} = \iiint\limits_U {\left( {{y^2} + {z^2}} \right)\cdot}}\kern0pt{{\rho \left( {x,y,z} \right)dxdydz} ,\;\;}}\kern-0.3pt {{{I_y} = \iiint\limits_U {\left( {{x^2} + {z^2}} \right)\cdot}}\kern0pt{{\rho \left( {x,y,z} \right)dxdydz} ,\;\;}}\kern-0.3pt {{{I_z} = \iiint\limits_U {\left( {{x^2} + {y^2}} \right)\cdot}}\kern0pt{{\rho \left( {x,y,z} \right)dxdydz} .\;\;}}\kern-0.3pt \]
As it can be seen, the following properties are valid:
\[
{{I_x} = {I_{xy}} + {I_{xz}},\;\;}\kern-0.3pt {{I_y} = {I_{xy}} + {I_{yz}},\;\;}\kern-0.3pt {{I_z} = {I_{xz}} + {I_{yz}}.} \]
The moment of inertia about the origin is called the integral
\[{I_0} = \iiint\limits_U {\left( {{x^2} + {y^2} + {z^2}} \right)}\cdot\kern0pt{\rho \left( {x,y,z} \right)dxdydz}.\]
The moment of inertia about the origin can be expressed through the moments of inertia about the coordinate planes as follows:
\[{I_0} = {I_{xy}} + {I_{yz}} + {I_{xz}}.\]
Tensor of Inertia
Using the \(6\) numbers considered above: \({I_x},{I_y},{I_z},{I_{xy}},{I_{xz}},{I_{yz}},\) we can construct the so-called matrix of inertia or the tensor of inertia of the solid:
\[{I }={ \left( {\begin{array}{*{20}{c}} {{I_x}}&{ – {I_{xy}}}&{ – {I_{xz}}}\\ { – {I_{xy}}}&{{I_y}}&{ – {I_{yz}}}\\ { – {I_{xz}}}&{ – {I_{yz}}}&{{I_z}} \end{array}} \right).}\]
This tensor is symmetric and, hence, it can be transformed to a diagonal view by choosing the appropriate coordinate axes \(Ox’, Oy’, Oz’.\) The values of the diagonal elements (after transforming the tensor to a diagonal form) are called the main moments of inertia, and the indicated directions of the axes are called the eigenvalues or the principal axes of inertia of the body.
If a body rotates about an axis which does not coincide with a principal axis of inertia, it will experience vibrations at the high rotation speeds. Therefore, when designing such devices it is necessary the axis of rotation to be coinciding with one of the principal axes of inertia. For example, when replacing car tires, it’s often necessary to balance the wheels by attaching small lead weights to ensure the coincidence of the rotation axis with the principal axis of inertia and to eliminate vibration.
Gravitational Potential and Attraction Force
The Newton potential of a body at a point \(P\left( {x,y,z} \right)\) is called the integral
\[{u\left( {x,y,z} \right) \text{ = }}\kern0pt{ \iiint\limits_U {\rho \left( {\xi ,\eta ,\zeta } \right)\frac{{d\xi d\eta d\zeta }}{r}} ,}\]
where \({\rho \left( {\xi ,\eta ,\zeta } \right)}\) is the density of the body and
\[{r \text{ = }}\kern0pt{\sqrt {{{\left( {\xi – x} \right)}^2} + {{\left( {\eta – y} \right)}^2} + {{\left( {\zeta – z} \right)}^2}} }\]
The integration is performed over the whole volume of the body. Knowing the potential, one can calculate the force of attraction of the material point of mass \(m\) and the distributed body with the density \({\rho \left( {\xi ,\eta ,\zeta } \right)}\) by the formula
\[\mathbf{F} = – Gm\,\mathbf{\text{grad}}\,u,\]
where \(G\) is the gravitational constant.
Solved Problems
Click a problem to see the solution.
Example 1Find the centroid of a homogeneous half-ball of radius \(R.\) Example 2Determine the mass and coordinates of the center of gravity of the unit cube with the density \(\rho \left( {x,y,z} \right) =\) \( x + 2y + 3z\) (Figure \(2\)). Example 3Find the mass of a ball of radius \(R\) whose density \(\gamma\) is proportional to the squared distance from the center. Example 4Find the moment of inertia of a right circular homogeneous cone about its axis. The cone has base radius \(R,\) height \(H\) and the total mass \(m\) (Figure \(3\)). Example 5With what force does a homogeneous ball of mass \(M\) attract a material point of mass \(m,\) located at distance \(a\) from the center of the ball \(\left( {a \gt R} \right)?\) Example 6Suppose that a planet has a radius \(R\) and its density is expressed by the formula Example 1.Find the centroid of a homogeneous half-ball of radius \(R.\)
Solution.
We introduce the system of coordinates in such a way that the half-ball is located at \(z \ge 0\) and centered at the origin (Figure \(1\text{).}\)
Using this system of coordinates, we find the centroid (the center of gravity) of the solid. Obviously, by symmetry,
\[\bar x = \bar y = 0.\]
Calculate the coordinate \(\bar z\) of the centroid by the formula
\[
{\bar z = \frac{{{M_{xy}}}}{m} } = {\frac{{\iiint\limits_U {z\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }}.} \]
Since the half-ball is homogeneous, we set \(\rho \left( {x,y,z} \right) = {\rho _0}.\) Then
\[\require{cancel}
{\bar z = \frac{{{\bcancel{\rho _0}}\iiint\limits_U {zdxdydz} }}{{{\bcancel{\rho _0}}\iiint\limits_U {dxdydz} }} } = {\frac{{\iiint\limits_U {zdxdydz} }}{{\iiint\limits_U {dxdydz} }} } = {\frac{{\iiint\limits_U {zdxdydz} }}{V}.} \]
The symbol \(V\) in the denominator denotes the volume of the solid, which is equal to
\[{V = \frac{1}{2}\left( {\frac{4}{3}\pi {R^3}} \right) }={ \frac{2}{3}\pi {R^3}.}\]
It remains to compute the triple integral \({\iiint\limits_U {zdxdydz} }.\) For this, we pass to spherical coordinates. In this case, the radial coordinate is denoted by \(r\) in order not to be confused with the density \(\rho.\) As a result, we have
\[
{\iiint\limits_U {zdxdydz} } = {\iiint\limits_{U’} {r\cos \theta {r^2}\sin \theta drd\varphi d\theta } } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\cos \theta \sin \theta d\theta } } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\sin \theta d\left( {\sin \theta } \right)} } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \cdot}\kern0pt{ \left[ {\left. {\left( {\frac{{{{\sin }^2}\theta }}{2}} \right)} \right|_{\theta = 0}^{\theta = \large\frac{\pi }{2}\normalsize}} \right] } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \cdot \frac{1}{2} = \frac{1}{2}\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_0^R} \right] } = {\frac{{{R^4}}}{8}\int\limits_0^{2\pi } {d\varphi } } = {\frac{{{R^4}}}{8} \cdot 2\pi } = {\frac{{\pi {R^4}}}{4}.} \]
Thus, the coordinate \(\bar z\) of the center of gravity is
\[
{\bar z = \frac{{\iiint\limits_U {zdxdydz} }}{V} } = {\frac{{\frac{1}{4}\pi {R^4}}}{{\frac{2}{4}\pi {R^3}}} }={ \frac{{3R}}{8}.} \] |
When dealing with several numbers and long equations, it's common to make careless arithmetic mistakes that give the wrong answer. I was wondering if anyone had tips to catch these mistakes, or even better avoid them more often. (aside from the obvious checking your work- that's a must)
There are certain quick methods called
sanity checks which will catch most (but not all) arithmetic errors. One common one is to replace each number with the sum of its digits, which is the "casting out nines" method mentioned in Robert Israel's answer. To check a computation, say $567\times 894=506,898$, we replace $567$ with $5+6+7=18$ and $894$ with $8+9+4=21$, and then replace each of these with the sum of their digits to get $9$ and $3$ (in general we keep doing this until we get down to $1$ digit), while on the other side we get $5+0+6+8+9+8=36$ and then $3+6=9$. We then check that $9\times 3=9$ after casting out nines on both sides, and so our answer is probably right (though not necessarily). This method is called "casting out nines" because it ensures that whatever answer you are checking differs from the actual answer by a multiple of nine (hopefully $0\times 9$).
However, this method has a serious drawback: if the answer you are checking is correct except for having the digits switched around (a relatively common error) the method will not catch the error. A remedy for this is to use "casting out elevens" where you take the
alternating sums of the digits instead of the sums, such that the last digit is always added rather than subtracted. In our previous example, this becomes $5-6+7=6$, $8-9+4=3$ and $5+0-6+8-9+8=-4$. Here we have to be a little careful: we want to take the equation $6\times 3-(-4)=0$, cast out elevens (take the alternating sum) and verify that the the resulting equation holds, which in this case it does. We move everything to one side so that we don't have to work with numbers of different signs ($6\times 3=-4$ is true $\bmod 11$, which is what matters, but it is not obvious how to cast out elevens to show this). This ensures that whatever answer you are checking differs from the actual answer by a multiple of eleven (hopefully $0\times 11$), hence the name. Edit: These methods can both be made rigrous with modular arithmetic. The first simple checks that an equation holds $\bmod 9$, and adding the digits comes from the fact that $$\begin{eqnarray}d_n\cdots d_1d_0 &=& \sum\limits_{i=1}^n 10^id_i\\&\equiv& \sum\limits_{i=1}^n 1^id_i (\bmod 9)\\&=&\sum\limits_{i=1}^n d_i\end{eqnarray}$$while the second checks that an equation holds $\bmod 11$, and the alternating sum of the digits comes from the fact that $$\begin{eqnarray}d_n\cdots d_1d_0 &=& \sum\limits_{i=1}^n 10^id_i\\&\equiv& \sum\limits_{i=1}^n (-1)^id_i (\bmod 11)\end{eqnarray}$$
Credit where credit is due: I believe I read about this years ago in a question to Dr. Math from an elementary school teacher who had been teaching the method and wanted to know how it worked.
In my experience, the best way to avoid computational errors is to avoid computation. Develop general algorithms for whatever quantity that you are looking for and then proceed to "plug and chug" as the last step. Mathematics requires precision, however, and you often cannot avoid having to comb over your work tediously.
I have found the best way to avoid these types of computational mistakes is:
To have
extremelyneat and clear handwriting.
To effectively use the space on the page to organize the work in a logical manner.
Use a pencil. Never cross things out, but erase them instead.
Always take the time to think things through slowly and carefully.
Keep your desk very neat and well organized.
I believe that if you write things down in a clear way, then you will think in a clear way; and if you write in a sloppy way, you will think in a sloppy way.
In my opinion, for research, use a computer. Your brain is just never going to be perfect.
For exams, though, go through the common computations lots of times before an exam, for instance if you have a linear algebra exam, it is wise to compute matrix inverses a few times (say like 20-40 times depending on your brain's capacity to compute) to be at ease with the algorithmic details and be able to focus on the numbers more easily during computation.
But then again, even if you've practiced for a week, a month later, I've already lost the habit of computing the thing in question and start abusing my brain like nuts to compute...
Hope that helps!
One simple tool that catches many arithmetic mistakes is "casting out nines". See http://en.wikipedia.org/wiki/Casting_out_nines
One way to check arithmetic calculations in a ring is to map the computation homomorphically into rings where calculation is easier. For example, many rings have parity, i.e. have $\mathbb Z/2$ as an image, and mapping the arithmetic mod $2$ yields a simple parity check that often catches errors. Casting nines is another example of a modular arithmetic check (which also works for fractions whose denominator is coprime to $9$). More generally one can
verify equalities using a sufficient number of modular checks, by employing CRT (Chinese Remainder). For polynomial rings one can similarly apply evaluation maps as checks. Again, with enough evaluations, one can verify equalities (which here is CRT = Lagrange interpolation).
Like CRT, such factorizations or decompositions of an algebraic structure into simpler structures is a powerful problem-solving technique, applicable not only to checking arithmetic, but also quite generally. It is the algebraists way to
divide-and-conquer. When combined with a little logic this yields even greater power. One nontrivial example is the model-theoretic proof of Jacobson's theorem, that rings satisfying the identity $\rm\:x^m = x\:$ are commutative. This proceeds by a certain type of ring factorization, which reduces the problem to the (subdirectly) irreducible rings satisfying the identity. These turn out to be certain finite fields, which are commutative, as desired. In a sense, the proof works by exploiting the fact that the statement need only be checked on a certain set of simpler cases (finite fields), where the verification is much easier. Thus this can be seen as a grand generalization of the ideas employed in the more elementary cases above.
One technique that's under-hyped is to simply redo the calculation on a separate sheet of paper
without looking at your prior calculations. You are still vulnerable to committing the same error but it'll catch frivolous one's quite well.
You should also check out
Vedic maths
You can always use all kinds of computer applications, but those only help you if you have them around.
I suppose that you're talking about making calculations on paper or in head, and I would have one suggestion here, which proved to be more than excellent for me.
It's quite simple: concentration, concentration, concentration.
You have to think about what you're computing, and nothing else. It may sound stupid, but you will be suprised about your own mental power if you stay in absolute focus.
Good luck!
As mentioned in other answers computers are very good at arithmetic. If such devices are not available try to calculate the answer in several different ways or at least do the steps in a different order. If you always get the same answer it is likely to be correct. If the answers are different investigate the source of the difference. Casting out nines will not catch transposition errors but it will catch carrying errors.
After writing the above I recalled this graphical method of multiplication: Multiplication by counting points .
I use checksums for memory operations.
A simple example is whenever my wife gives me a long verbal list of groceries to buy, I count the total. If a big number say > 20 items, then split into 3 categories. If I pretend a simple sum is easy to remember for short term memory, then I can accumulate or count down until the process ( in this case; simple remember, search and fetch) as long as I remember the total number of items, I rarely forget one. For fun I may remember the word I can spell with the 1st letter of each item. This gives more redundancy and checksum info in an easy way to remember.
The challenge is to count brackets, variables & transforms and visualize the process as your checksums.
I often apply a simple intuitive check at the end of my calculations to make sure that at the coarsest level the answer makes sense.
For example:
Am I expecting the answer to be a larger or smaller number than the input values? by how many orders of magnitude? Should the answer be positive or negative? Does it makes sense that the answer is between 0 and 1 (or greater than 100, etc...)?
While these checks are not as sensitive as some of the other examples, they can catch a lot of careless mistakes and have the advantage of forcing one to think about and understand what the calculations are doing and what type of answer should make sense.
When doing algebraic computations, for example when finding the partial fraction decomposition of a certain expression, something I find helpful is to substitute a certain number and see if the two expressions match.
For example to "verify" $\frac{1}{(1+x)(5+x)}=\frac{1}{4(x+1)}-\frac{1}{4(x+5)}$, you may want to substitute in some value of x, say $x=9$, and see that both sides become $1/140$. |
To clarify, I'm asking about series where $n$th term depends only on $n$th prime number. From the famous result (by Euler, I think) we have:
$$\sum_{n=1}^\infty \ln \left(1-\frac{1}{p_n^s} \right)=-\ln \zeta(s)$$
As a particular case:
$$\sum_{n=1}^\infty \ln \left(1-\frac{1}{p_n^2} \right)=-\ln \frac{\pi^2}{6}$$
What about the general series:
$$\sum_{n=1}^\infty f(p_n)$$
Does it have a nontrivial closed form for some other function $f$?
In general, I'm also interested in infinite products (they can be easily written as series, as I did for the titular one), nested radicals, continued fractions, etc.
Edit
Thanks to the great advice of @user1952009, I looked up Dirichlet beta function, which has the prime expansion:
$$\sum_{n=2}^\infty \ln \left(1-\frac{(-1)^{(p_n-1)/2}}{p_n^s} \right)=-\ln \beta(s)$$
Among it's special values we have:
$$\sum_{n=2}^\infty \ln \left(1-\frac{(-1)^{(p_n-1)/2}}{p_n} \right)=-\ln \frac{\pi}{4}$$
$$\sum_{n=2}^\infty \ln \left(1-\frac{(-1)^{(p_n-1)/2}}{p_n^2} \right)=-\ln G$$
(Catalan's constant)
$$\sum_{n=2}^\infty \ln \left(1-\frac{(-1)^{(p_n-1)/2}}{p_n^3} \right)=-\ln \frac{\pi^3}{32}$$
Etc. |
Suppose I have the following Cobb-Douglas function $$U(x,y) = x^\alpha y^{1-\alpha} = 1$$ where $\alpha \in [0,1]$.
$$MRS = -\frac{U_x}{U_y} = - \frac{\alpha}{1-\alpha} \frac{y}{x} $$ $$\frac{\partial MRS}{\partial \alpha} = -\frac{1}{(1-\alpha)^2}\frac{y}{x}$$
So suppose I have the following set of graphs:
Here I just picked some values. Red is $\alpha = \frac{1}{4}$, blue is $\alpha = \frac{1}{2}$, black is $\alpha = \frac{3}{4}$
I understand how the steepness and flatness of the different curves change as I vary $\alpha$. But I am confused as to what $\frac{\partial MRS}{\partial \alpha} $ tells me about the graph. In particular, $\frac{\partial MRS}{\partial \alpha} $ is dependent on $\alpha$. So even though I know $\frac{\partial MRS}{\partial \alpha}$ tells me how much $MRS$ changes as I change alpha....changing $\alpha$ changes $\frac{\partial MRS}{\partial \alpha}$...so I am very confused!
My Question
What is $\frac{\partial MRS}{\partial \alpha}$ telling me about the graph? Since $\frac{\partial MRS}{\partial \alpha}$ is dependent on $\alpha$, how does this affect things? |
History and origin
According to Robert D Cousins$^{1}$ and Tommaso Dorigo$^{2}$, the origin of the $5\sigma$ threshold origin lies in the early particle physics work of the 60s when numerous histograms of scattering experiments were investigated and searched for peaks/bumps that might indicate some newly discovered particle. The threshold is a rough rule to account for the multiple comparisons that are being made.
Both authors refer to a 1968 article from Rosenfeld$^3$, which dealt with the question whether or not there are far out mesons and baryons, for which several $4 \sigma$ effects where measured. The article answered the question negatively by arguing that the number of published claims corresponds to the statistically expected number of fluctuations. Along with several calculations supporting this argument the article promoted the use of the $5\sigma$ level:
Rosenfeld: "Before we go on to survey far-out mass spectra where bumps have been reported in
$(K\pi\pi)_{3/2},(\pi \rho)^{--}$ we should first decide what threshold of significance to demand in 1968. I want to show you that although experimentalists should probably note $3\sigma$-effects, theoreticians and phenomenologists would do better to wait till the effect reaches $>4\sigma$."
and later in the paper (emphasis is mine)
Rosenfeld: "Then to repeat my warning at the beginning of this section; we are generating at least 100 000 potential bumps per year, and should expect several $4\sigma$ and hundreds of $3\sigma$ fluctuations. What are the implications? To the theoretician or phenomenologist the moral is simple;
wait for $5\sigma$ effects."
Tommaso seems to be careful in stating that it started with the Rosenfeld article
Tommaso: "However, we should note that the article was written in 1968, but the strict criterion of five standard deviations for discovery claims was not adopted in the seventies and eighties. For instance, no such thing as a five-sigma criterion was used for the discovery of the W and Z bosons, which earned Rubbia and Van der Meer the Nobel Prize in physics in 1984."
But in the 80s the use of $5\sigma$ was spread out. For instance, the astronomer Steve Schneider$^4$ mentions in 1989 that it is something being taught (emphasize mine in the quote below):
Schneider: "Frequently, 'levels of confidence' of 95% or 99% are quoted for apparently discrepant data, but this amounts to only two or three statistical sigmas.
I was taught not to believe anything less than five sigma, which if you think about it is an absurdly stringent requirement --- something like a 99.9999% confidence level. But of course, such a limit is used because the actual size of sigma is almost never known. There are just too many free variables in astronomy that we can't control or don't know about."
Yet, in the field of particle physics many publications where still based on $4\sigma$ discrepancies up till the late 90s. This only changed into $5\sigma$ at the beginnning of the 21th century. It is probably prescribed as a guidline for publications around 2003 (see the prologue in Franklin's book Shifting Standards$^5$)
Franklin: By 2003 the 5-standard-deviation criterion for "observation of" seems to have been in effect
...
A member of the BaBar collaboration recalls that about this time the 5-sigma criterion was issued as a guideline by the editors of the
Physical Review Letters Modern use
Currently, the $5\sigma$ threshold is a textbook standard. For instance, it occurs as a standard article on physics.org$^6$ or in some of Glen Cowan's works, such as the statistics section of the Review of Particle Physics from the particle data group$^7$ (albeit with several critical sidenotes)
Glen Cowan:
Often in HEP, the level of significance where an effect is said to qualify as a discovery is $Z = 5$, i.e., a $5\sigma$ effect, corresponding to a p-value of $2.87 \times 10^{−7}$ . One’s actual degree of belief that a new process is present, however, will depend in general on other factors as well, such as the plausibility of the new signal hypothesis and the degree to which it can describe the data, one’s confidence in the model that led to the observed p-value, and possible corrections for multiple observations out of which one focuses on the smallest p-value obtained (the “look-elsewhere effect”).
The use of the $5\sigma$ level is now ascribed to 4 reasons:
History based on practice one found that $5\sigma$ is a good threshold. (exotic stuff seems to happen randomly, even between $3\sigma$ to $4\sigma$, like recently the 750 GeV diphoton excess) The look elsewhere effect (or the multiple comparisons). Either because multiple hypotheses are tested, or because experiments are performed many times, people adjust for this (very roughly) by adjusting the bound to $5\sigma$. This relates to the history argument. Systematic effects and uncertainty in $\sigma$ often the uncertainty of the experiment outcome is not well known. The $\sigma$ is derived, but the derivation includes weak assumptions such as the absence of systematic effects, or the possibility to ignore them. Increasing the threshold seems to be a way to sort of a protect against these events. (This is a bit strange though. The computed $\sigma$ has no relation to the size of systematic effects and the logic breaks down, an example is the "discovery" of superluminal neutrino's which was reported to be having a $6\sigma$ significance.) Extraordinary claims require extraordinary evidence Scientific results are reported in a frequentist way, for instance using confidence intervals or p-values. But, they are often interpreted in a Bayesian way. The $5\sigma$ level is claimed to account for this.
Currently several criticisms have been written about the $5\sigma$ threshold by Louis Lyons${^{8,}}$$^9$, and also the earlier mentioned articles by Robert D Cousins$^{1}$ and Tommaso Dorigo$^{2}$ provide critique.
Other Fields
It is interesting to note that many other scientific fields do not have similar thresholds or do not, somehow, deal with the issue. I imagine this makes a bit sense in the case of experiments with humans where it is very costly (or impossible) to extend an experiment that gave a .05 or .01 significance.
The result of not taking these effects into account is that over half of the published results may be wrong or at least are not reproducible (This has been argued for the case of psychology by Monya Baker $^{10}$, and I believe there are many others that made similar arguments. I personaly think that the situation may be even worse in nutritional science). And now, people from other fields than physics are thinking about how they should deal with this issue (the case of medicine/pharmacology$^{11}$).
Cousins, R. D. (2017). The Jeffreys–Lindley paradox and discovery criteria in high energy physics. Synthese, 194(2), 395-432. arxiv link
Dorigo, T. (2013) Demystifying The Five-Sigma Criterion, from science20.com 2019-03-07
Rosenfeld, A. H. (1968). Are there any far-out mesons or baryons? web-source: escholarship
Burbidge, G., Roberts, M., Schneider, S., Sharp, N., & Tifft, W. (1990, November). Panel discussion: Redshift related problems. In NASA Conference Publication (Vol. 3098, p. 462). link to photocopy on harvard.edu
Franklin, A. (2013). Shifting standards: Experiments in particle physics in the twentieth century. University of Pittsburgh Press.
What does the 5 sigma mean? from physics.org 2019-03-07
Beringer, J., Arguin, J. F., Barnett, R. M., Copic, K., Dahl, O., Groom, D. E., ... & Yao, W. M. (2012). Review of particle physics. Physical Review D-Particles, Fields, Gravitation and Cosmology, 86(1), 010001. (section 36.2.2. Significance tests, page 394, link aps.org )
Lyons, L. (2013). Discovering the Significance of 5 sigma. arXiv preprint arXiv:1310.1284. arxiv link
Lyons, L. (2014). Statistical Issues in Searches for New Physics. arXiv preprint arxiv link
Baker, M. (2015). Over half of psychology studies fail reproducibility test. Nature News. from nature.com 2019-03-07
Horton, R. (2015). Offline: what is medicine's 5 sigma?. The Lancet, 385(9976), 1380. from thelancet.com 2019-03-07 |
Introduction
Built at the Jet Propulsion Laboratory by an Investigation Definition Team (IDT) headed by John Trauger, WFPC2 was the replacement for the first Wide Field and Planetary Camera (WF/PC-1) and includes built-in corrections for the spherical aberration of the HST Optical Telescope Assembly (OTA). The WFPC2 was installed in HST during the First Servicing Mission in December 1993.
Early IDT report of the WFPC2 on-orbit performance: Trauger et al. (1994, ApJ, 435, L3) A more detailed assessment of its capabilities: Holtzman et al. (1995, PASP, 107, page 156 and page 1065).
The WFPC2 was used to obtain high resolution images of astronomical objects over a relatively wide field of view and a broad range of wavelengths (1150 to 11,000 Å). WFPC2 was installed during the first HST Servicing Mission in 1993 and removed during Servicing Mission 4 in 2009. WFPC2 data can be found on the MAST Archive.
ISRs Filter WFPC2 ISRs Listing Results 2010-04: The Dependence of WFPC2 Charge Transfer Efficiency on Background Illumination 2010-01: WFPC2 Standard Star CTE Optical Configuration
While it was in operation, the WFPC2 field of view was located at the center of the HST focal plane. The central portion of the f/24 beam coming from the OTA would be intercepted by a steerable pick-off mirror attached to the WFPC2 and diverted through an open port entry into the instrument. The beam would then pass through a shutter and interposable filters. An assembly of 12 filter wheels contained a total of 48 spectral elements and polarizers. The light would then fall onto a shallow-angle, four-faceted pyramid, located at the aberrated OTA focus. Each face of the pyramid was a concave spherical surface, dividing the OTA image of the sky into four parts. After leaving the pyramid, each quarter of the full field of view would then be relayed by an optically flat mirror to a Cassegrain relay that would form a second field image on a charge-coupled device (CCD) of 800 x 800 pixels. Each of these four detectors were housed in a cell sealed by a MgF2 window, which is figured to serve as a field flattener.
The aberrated HST wavefront was corrected by introducing an equal but opposite error in each of the four Cassegrain relays. An image of the HST primary mirror would then be formed on the secondary mirrors in the Cassegrain relays. The spherical aberration from the telescope's primary mirror would be corrected on these secondary mirrors, which were extremely aspheric; the resulting point spread function was quite close to that originally expected for WF/PC-1.
Field of View
The U2,U3 axes were defined by the "nominal" Optical Telescope Assembly (OTA) axis, which was near the center of the WFPC2 FOV. The readout direction was marked with an arrow near the start of the first row in each CCD; note that it rotated 90 degrees between successive chips. The x,y arrows mark the coordinate axes for any POS TARG commands that may have been specified in the proposal. An optional special requirement in HST observing proposals, places the target an offset of POS TARG (in arcsec) from the specified aperture.
Camera Configurations
Camera Pixels Field of View Scale f/ratio PC (PC1) 800 x 800 36" x 36" 0.0455" per pixel 28.3 WF2, 3, 4 800 x 800 80" x 80" 0.0996" per pixel
12.9
A Note about HST File Formats
Data from WFPC2 are made available to observers as files in Multi-Extension FITS (MEF) format, which is directly readable by most PyRAF/IRAF/STSDAS tasks. All WFPC2 data are now available in either waivered FITS or MEF formats. The user may specify either format when retrieving that data from the HDA. WFPC2 data, in either Generic Edited Information Set (GEIS) or MEF formats, can be fully processed with STSDAS tasks.
The figure below provides a physical representation of the typical data format.
Resources Charge Traps
There are about 30 pixels in WFPC2 that are "charge traps" which do not transfer charge efficiently during readout, producing artifacts that are often quite noticeable. Typically, charge is delayed into successive pixels, producing a streak above the defective pixel. In the worst cases, the entire column above the pixel can be rendered useless. On blank sky, these traps will tend to produce a dark streak. However, when a bright object or cosmic ray is read through them, a bright streak will be produced. Here, we show streaks (a) in the background sky, and (b) stellar images produced by charge traps in the WFPC2. Individual traps have been cataloged and their identifying numbers are shown.
Warm Pixels and Annealing
Decontaminations (anneals), during which the instrument is warmed up to about 22
o C for a period of six hours, were performed about once per month. These procedures are required in order to remove the UV-blocking contaminants which gradually build-up on the CCD windows (thereby restoring the UV throughput) as well as fix warm pixels. Examples of warm pixels are presented in the figure below.
Calibration
Procedure Estimated Accuracy Notes Bias subtraction 0.1 DN rms Unless bias jump is present Dark subtraction 0.1 DN/hr rms Error larger for warm pixels; absolute error uncertain because of dark glow Flat fielding <1% rms large scale Visible, near UV 0.3% rms small scale Visible, near UV ~10% F160BW; however, significant noise reduction achieved with use of correction flats
Relative Photometry
Procedure Estimated Accuracy Notes Residuals in CTE correction < 3% for the majority (~90%) of cases up to 1-% for extreme cases (e.g., very low backgrounds) Long vs. short anomaly (uncorrected) < 5% Magnitude errors <1% for well-exposed stars but may be larger for fainter stars. Some studies have failed to confirm the effect. (see Chapter 5 of IHB for more details) Aperture correction 4% rms focus dependence (1 pixel aperture) Can (should) be determined from data <1% focus dependence (> 5 pixel) Can (should) be determined from data 1-2% field dependence (1 pixel aperture) Can (should) be determined from data Contamination correction 3% rms max (28 days after decon) (F160BW) 1% rms max (28 days after decon) (filters bluer than F555W) Background determination 0.1 DN/pixel (background > 10 DN/pixel) May be difficult to exceed, regardless of image S/N Pixel centering < 1%
Absolute Photometry
Precedure Estimated Accuracy Sensitivity < 2% rms for standard photometric filters 2% rms for broad and intermediate filters in visible < 5% rms for narrow-band filters in visible 2-8% rms for UV filters
Astrometry
Procedure Estimated Accuracy Notes Relative 0.005" rms (after geometric and 34th-row corrections) Same chip 0.1" (estimated) Across chips Absolute 1" rms (estimated)
Photometric Systems Used for WFPC2 Data
The WFPC2 flight system is defined so that stars of color zero in the Johnson-Cousins UBVRI system have color zero between any pair of WFPC2 filters and have the same magnitude in V and F555W. This system was established by Holtzman et al. (1995b)
The zeropoints in the WFPC2 synthetic system, as defined in Holtzman et al. (1995b), are determined so that the magnitude of Vega, when observed through the appropriate WFPC2 filter, would be identical to the magnitude Vega has in the closest equivalent filter in the Johnson-Cousins system.
\(m_{AB} = -48.60-2.5\log f_\nu \)
\(m_{ST} = -21.10-2.5\log f_\lambda\)
Photometric Corrections
A number of corrections must be made to WFPC2 data to obtain the best possible photometry. Some of these, such as the corrections for UV throughput variability, are time dependent, and others, such as the correction for the geometric distortion of WFPC2 optics, are position dependent. Finally, some general corrections, such as the aperture correction, are needed as part of the analysis process. Here we provide examples of factors affecting photometric corrections.
Cool Down on April 23, 1994
PSF Variations
34th Row Defect
Gain Variation
Pixel Centering
Possible Variation in Methane Quad Filter Transmission
Polarimetry
WFPC2 has a polarizer filter which can be used for wide-field polarimetric imaging from about 200 through 700 nm. This filter is a
quad, meaning that it consists of four panes, each with the polarization angle oriented in a different direction, in steps of 45 o. The panes are aligned with the edges of the pyramid, thus each pane corresponds to a chip. However, because the filters are at some distance from the focal plane, there is significant vignetting and cross-talk at the edges of each chip. The area free from vignetting and cross-talk is about 60" square in each WF chip, and 15" square in the PC. It is also possible to use the polarizer in a partially rotated.
Accurate calibration of WFPC2 polarimetric data is rather complex, due to the design of both the polarizer filter and the instrument itself. WFPC2 has an aluminized pick-off mirror with a 47° angle of incidence, which rotates the polarization angle of the incoming light, as well as introducing a spurious polarization of up to 5%. Thus, both the HST roll angle and the polarization angle must be taken into account. In addition, the polarizer coating on the filter has significant transmission of the perpendicular component, with a strong wavelength dependence.
Astrometry
Astrometry with WFPC2 means primarily
relative astrometry. The high angular resolution and sensitivity of WFPC2 makes it possible, in principle, to measure precise positions of faint features with respect to other reference points in the WFPC2 field of view. On the other hand, the absolute astrometry that can be obtained from WFPC2 images is limited by the positions of the guide stars, usually known to about 0.5" rms in each coordinate, and by the transformation between the FGS and the WFPC2, which introduces errors of order of 0.1"
Because WFPC2 consists of four physically separate detectors, it is necessary to define a coordinate system that includes all four detectors. For convenience, sky coordinates (right ascension and declination) are often used; in this case, they must be computed and carried to a precision of a few mas, in order to maintain the precision with which the relative positions and scales of the WFPC2 detectors are known. It is important to remember that the coordinates are
not known with this accuracy. The absolute accuracy of the positions obtained from WFPC2 images is typically 0.5" rms in each coordinate and is limited primarily by the accuracy of the guide star positions. |
Meaning of Denominator
The divisor of a fraction is called the denominator.
A Fraction is one part when a whole is divided into equal parts. Mathematically its is shown by the division of two numbers. Let’s go through some examples and you recall all the divisibility rules to make sure you understand the formulas easily.
For example:
1/4 is one part out of the four equal parts created from that one whole thing.
A fraction consists of two numbers. Symbolically it is represented by two numbers separated by a horizontal line. The number above the line is numerator, and the number below the line is the denominator.
For Example:
1 ⟶ Numerator 4⟶ Denominator Numerator and Denominator
The denominator indicates the number of equal parts in which the whole thing has to be divided.
The numerator indicates the number of divisions selected out of the total number of equal parts.
What are Numerator and Denominator?
This would be better explained with the help of an Example.
3/4 is a fraction in which the denominator 4 represents that 4 equal divisions have to be made.
3 parts selected out of 4 equal parts created out from 1 circle can be represented as 3/4.
Diagrammatic representation of ¾ is as follows:
The diagram clearly shows three equal parts taken out when the whole circle is divided into four equal parts.
Least Common Denominator(LCD)
The least common denominator of two or more non-zero denominators is the smallest whole number that is divisible by each of the denominators.
Methods to find LCD 1.Multiply both the denominators (when the denominators have no common multiple)
For Example:
There are two fractions as follows:-
⅓ and ⅕
3*5 = 15
Multiply both the fractions with the product (15) with the top as well as bottom:
⅓ * 15/15 = 5/15
⅕ * 15/15 = 3/15
Thus, we have a common denominator in both our fractions.
2.List the multiples of both denominators:
We have the two fractions: ⅓ and ⅙
3 = 3,6,9,12,18,21,24,27,30.
6=6,12,18,24,30,36,42,48,60.
The smallest common multiple that can be seen is 6.
So that is the LCD.
Rationalize the Denominator
The denominator of a fraction needs to be rationalized when it is an irrational number so that further calculations can be made easily on the fraction.
Irrational denominator includes the root numbers.
How to Rationalize The Denominator Example 1. Monomial Denominator
\(\frac{1}{\sqrt{3}}\)
To remove the radical multiply the numerator and denominator by the \(\sqrt{3}\)
\(\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}\)
By multiplying the top and bottom with the, we have created the smallest possible perfect square in the denominator and removed the radical as per our requirement.
Example 2. Binomial Denominator:
When there are two terms in the denominator that make it irrational.
\(\frac{3}{5 + \sqrt{3}}\)
In this case, multiply the numerator and denominator with the conjugate of the denominator.
The conjugate means the same denominator but with the opposite sign.
The conjugate of \(5 + \sqrt{3}\)
To rationalize:
\(\frac{3}{5 + \sqrt{3}} \times \frac{5 – \sqrt{3}}{5 – \sqrt{3}}\)
\(\frac{15 – 3\sqrt {3}}{5^{2} – (\sqrt{3})^{2}} = \frac{15 – 3\sqrt {3}}{28}\)
This was all about denominator and how to rationalize the fraction. Learn more about ratio and proportion, percentage, fractions etc. by visiting our site BYJU’S.
‘ |
A cone is a popular geometrical shape with a flat surface that tapers to appoint on another side. Cone can be of different types but we will focus on right circular cone throughout this article. This is the cone with a flat surface tapers to appoint that is angled at 90-degree from the midpoint of a circle.
In other words, the cone is the 3-dimensional structure with a circular base, a set of segments of lines that is connecting all the points together on the base with a common focused point i.e. apex. Thus, a cone can be seen as the set of non-congruent discs that are circular and stacked over one another where the radius of the adjacent disc would remain the constant.
A cone can also be taken as the triangle that is rotated along one of its vertices. There are a predefined set of basic cone formulas that are used to calculate its curved area, surface area, the volume of a cone, total surface area etc.
The curved surface area of a cone is the multiplication of pi, slant height, and the radius. Further, the surface area of a cone is given as the sum of the base and curved surface area. In mathematics, there is a special formula to figure it out –
\[\large Surface\;Area\;of\;cone=\pi r \left (s+r \right )\]
Where, r is the radius of cone. h is the height of cone. s is the slant height of the cone.
Where r is the radius, s is the slant height and value of π is constant = 3.14
For a better understanding of cone formulas, you must have good knowledge of basic terminologies like radius, height, slant height etc. Radius is the distance from the centre to the edge of the circle till the end. Height is the distance from the centre point to the top of the cone. The slant height is the length from the tip of the cone to the edge of the cone. Pi (π) is a special term that is used along with the circles whose value is constant i.e. 3.14. wherever, you see the symbol (π) in mathematics, put the value 3.14 against it to solve an equation.
There is used a special formula to find the volume of the cone. The volume gives you an idea of how much space will be taken inside the cone. The final answer would be express in terms of cubic units. In mathematics, the volume of a cone formula is given as –
\[\large Vomule\;of\;cone=\frac {1}{3}\pi r^{2}h\]
Where,
r is the radius of cone. h is the height of cone.
Where r is the radius, s is the slant height and value of π is constant = 3.14
The curved surface area of a cone is the multiplication of pi, slant height, and the radius. In mathematics, the volume of a cone formula is given as –
\[\large Curved\;Surface\;Area\;of\;cone=\pi rs\]
Where,
r is the radius of cone. h is the height of cone. s is the slant height of the cone.
Where r is the radius, l is the slant height and value of π is constant = 3.14
The formula to compute the total surface area of a cone in mathematics is given as –
\[\large Total\;Surface\;Area\;of\;Right\;circular\;cone=\pi r(r+ \sqrt {h^{2}+r^{2}}) \]
Where, r is the radius of cone. h is the height of cone.
Where r is the radius, l is the slant height and value of π is constant = 3.14. to compute the slant height of cone, you can apply Pythagoras theorem, if you know the height and the radius. |
After a while, you just get tired. An honest science blogger can only handle so much science jargon thrown around without meaning, only a limited amount of Choprawoo and quantum flapdoodle. How long can anyone with integrity, curiosity and a dose of genuine knowledge endure the trumpeting that, say, the brain’s limited ability to recover after injury is evidence for some quantum spirit? The brain is living flesh, made of living cells: by the same so-called logic, the scabbed knees of childhood are all evidence of quantum skin.
After a while, a deep reserve of psyche cries out, “Enough! If my freedom means aught, I must
stop responding to these charlatans and move beyond. I must send a message of my own, a message which is not a reaction but an expression unto itself. I must sing the quantum genuine!”
In my case, this means another post on supersymmetric quantum mechanics.
RECAP
Last time, we deduced some interesting properties of Hamiltonians which can be factored into operators and adjoints:
[tex]H_1 = A^\dag A,\ H_2 = AA^\dag.[/tex]
We observed that [tex]H_1[/tex] and [tex]H_2[/tex] are
isospectral. That is, while the forms of their eigenfunctions may be different, the eigenvalues associated with those functions are the same; or, in physical terms, the wavefunctions have different shapes, but the energies match. The only exception is the ground state: if [tex]H_1[/tex] has a zero-energy ground state, then [tex]H_2[/tex] will not. Furthermore, the operator [tex]A[/tex] maps eigenstates of [tex]H_1[/tex] into those of [tex]H_2[/tex], and the operator [tex]A^\dag[/tex] maps eigenstates in the reverse direction: We can sketch the full set of [tex]H_1[/tex] eigenstates as a ladder with, potentially, an infinite number of rungs. The eigenstates of [tex]H_2[/tex] sit right alongside them; the bottom “rung” is missing, as we mentioned earlier.
That “missing rung” has an important implication: instead of mapping the ground state of [tex]H_1[/tex] into a state of [tex]H_2[/tex], the operator [tex]A[/tex] annihilates it.
[tex]A\ket{\psi_0^{(1)}} = 0.[/tex]
We could, therefore, solve for the wavefunction [tex]\psi_0^{(1)}(x)[/tex] by solving a first-order, linear differential equation — a simpler task than tackling the full, second-order Schrödinger Equation.
Now, we’ll make a qualitative statement which we’ll firm up in a moment. If the Hamiltonian [tex]H_2[/tex] were “roughly similar” to [tex]H_1[/tex], then we should be able to “pull the same trick again,” writing it as
[tex]H_2 = A_2^\dag A_2,[/tex]
and inventing a third Hamiltonian,
[tex]H_3 = A_2 A_2^\dag,[/tex]
which will be isospectral with [tex]H_2[/tex] except for a ground state satisfying
[tex]A_2\ket{\psi_0^{(2)}} = 0.[/tex]
Then, why not, we do the same thing a whole bunch of times over again, building a whole hierarchy of Hamiltonia stretching off beyond the margin of our page.
Note that the “bottom rung” in each “ladder” satisfies a first-order differential equation, and that we can build any state of [tex]H_1[/tex] by stacking up a sequence of [tex]A^\dag[/tex] operators. Start with a state annihilated by some [tex]A_n[/tex] and then apply
[tex]A^\dag_{n-1}A^\dag_{n-2}\cdots A^\dag_{1}[/tex]
to get an eigenstate of [tex]H_1[/tex] there on the left edge. This is reminiscent of the way we solve the harmonic oscillator using creation and annihilation operators, with the difference that instead of building
up, we’re working in from the side. DEFINITION OF SHAPE INVARIANCE
To see this in action, we need to solidify the notion of rough resemblance we invoked above. Suppose we have a potential [tex]V_1[/tex] associated with some system of interest. If we compute the superpotential [tex]W(x)[/tex] and find that the partner potential [tex]V_2[/tex] has a similar shape to [tex]V_1[/tex], we can calculate the energy states of [tex]V_1[/tex] in a remarkably straightforward way. In this context, “similar shape” means that if [tex]V_1[/tex] depends upon a set of parameters, we can reach [tex]V_2[/tex] by substituting different values of those parameters into [tex]V_1[/tex]. More mathematically, two partner potentials [tex]V_1[/tex] and [tex]V_2[/tex] are said to be
shape invariant if they satisfy
[tex]V_2(x;a_1) = V_1(x;a_2) + R(a_1)[/tex]
where [tex]a_1[/tex] and [tex]a_2[/tex] are sets of parameters, and [tex]a_2[/tex] is uniquely determined by [tex]a_1[/tex] (that is, [tex]a_2 = f(a_1)[/tex] for some function [tex]f[/tex]). The residual term [tex]R(a_1)[/tex] does not depend upon the coordinate [tex]x[/tex]. We shall see that if [tex]H_1[/tex] has a shape-invariant partner potential in [tex]H_2[/tex], the entire spectrum of [tex]H_1[/tex] can be computed through operator manipulations.
In the material which follows, we’ll work in “natural units,” to save ourselves from carrying around excess notational baggage. Our operators will therefore take the form
[tex]A = \partial_x + W(x)[/tex]
and
[tex]A^\dag = -\partial_x + W(x).[/tex]
HIERARCHY OF HAMILTONIANS
We are usually interested in examining the physics of a particular system, for which we have written an [tex]H_1[/tex]. (We are restricted to work in one dimension; however, many higher-dimensional problems can be attacked through separation of variables. When studying the hydrogen atom, the radial dependence may be factored out of the Coulomb Hamiltonian to give a pseudo-one-dimensional potential; when studying the Landau effect, a convenient choice of gauge shows that along one axis, electrons behave as if in a harmonic oscillator potential.) As mentioned above, we can slide the energy scale up and down, so for convenience we fix the [tex]n = 0[/tex] eigenstate of [tex]H_1[/tex] to have zero energy:
[tex]E_0^{(1)}(a_1) = 0.[/tex]
We can solve for the ground state knowing the superpotential to show that
[tex]\psi_0^{(1)}(x;a_1) = C\exp\left(-\int^x W(y;a_1)dy\right)[/tex]
with [tex]C[/tex] being a constant of normalization. (Having the freedom to choose this constant makes the lower limit of the integral arbitrary.)
To investigate the spectrum of [tex]H_1[/tex], we construct the series of Hamiltonia [tex]\{H_s\}[/tex], where
s ranges over the positive integers:
[tex]H_s = -\partial_x^2 + V_1(x;a_s) + \sum_{k=1}^{s-1}R(a_k),[/tex]
in which we have defined
[tex]a_s = f^{s-1}(a_1).[/tex]
Here, [tex]f^{s-1}(a_1)[/tex] denotes applying the transformation [tex]f[/tex] to the parameter set [tex]a_1[/tex] a total of [tex]s-1[/tex] times. Each time, we get a new, uniquely determined set of parameters. With this definition, it is easy to see that [tex]H_s[/tex] and [tex]H_{s+1}[/tex] are SUSY partners:
[tex]H_{s+1} = -\partial_x^2 + V_1(x;a_{s+1}) + \sum_{k=1}^{s}R(a_k),[/tex]
which is also equal to
[tex] -\partial_x^2 + V_2(x;a_s) + \sum_{k=1}^{s-1}R(a_k).[/tex]
Because [tex]H_s[/tex] and [tex]H_{s+1}[/tex] are SUSY partners, they share the same eigenenergies, save for the ground state, whose energy is just given by the sum of the residuals,
[tex]E_0^{(s)} = \sum_{k=1}^{s-1}R(a_k).[/tex]
The SUSY between [tex]H_1[/tex] and [tex]H_2[/tex] means that they are isospectral; the [tex]n=1,2,3,\ldots[/tex] eigenstates of [tex]H_1[/tex] have the same energies as the [tex]n = 0,1,2\ldots[/tex] states of [tex]H_2[/tex]. The same holds true for [tex]H_2[/tex] and [tex]H_3[/tex], [tex]H_3[/tex] and [tex]H_4[/tex], [tex]H_{17}[/tex] with [tex]H_{18}[/tex] and so on until we run out of bound states. (For a problem like the hydrogen atom, we never do: [tex]n[/tex] ranges upward to infinity.) We “lose” one state, the lowest-energy ground state, every time we go from [tex]H_s[/tex] to [tex]H_{s+1}[/tex]. Following the degeneracies through and liberally applying the transitive law, we see that the
n-th eigenstate of [tex]H_1[/tex] is degenerate with the ground state of [tex]H_n[/tex]. Therefore, we can recover the entire spectrum of [tex]H_1[/tex] from the equation above.
We can get the eigenfunctions as well as the eigenvalues by this approach. Up to a normalization constant,
[tex]\psi_n^{(s+1)} \propto A_s \psi_{n+1}^{(s)},[/tex]
so knowing the eigenfunctions of [tex]H_1[/tex] would allow us to compute those of [tex]H_2[/tex]. The proportionality works in reverse, too:
[tex]\psi_n^{(s)} \propto A^\dag_{s+1}\psi_{n-1}^{(s+1)}.[/tex]
Therefore, if we knew the ground-state wavefunction of [tex]H_n[/tex], we could apply a sequence of operators [tex]A^\dag_n,A^\dag_{n-1},\ldots[/tex] until we reached a state in the spectrum of [tex]H_1[/tex]. But, we
do know the ground state of [tex]H_n[/tex]: it is that state which is annihilated by the operator [tex]A_n[/tex]. Remember that [tex]A_n[/tex] involves a single first-order derivative [tex]\partial_x[/tex], so solving [tex]A_n \psi = 0[/tex] only involves solving a first-order differential equation. In this way, for any [tex]H_1[/tex] which has a shape-invariant potential (SIP), we can exactly solve for its eigenvalues and eigenfunctions.
As we noted earlier, this is reminiscent of the operator-based solution of the harmonic oscillator. With that in mind, the interested reader can for an exercise see what happens when the superpotential takes the simple form
[tex]W(x) = x.[/tex]
READING Fred Cooper, Avinash Khare, Uday Sukhatme, “Supersymmetry and Quantum Mechanics” Phys.Rept. 251(1995): 267–385. Available as arXiv:hep-th/9405029. SUSY QM SERIES |
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