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Let $T: \R^n \to \R^m$ be a linear transformation.Suppose that the nullity of $T$ is zero.
If $\{\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k\}$ is a linearly independent subset of $\R^n$, then show that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.
Let $A$ be the matrix given by\[A=\begin{bmatrix}-2 & 0 & 1 \\-5 & 3 & a \\4 & -2 & -1\end{bmatrix}\]for some variable $a$. Find all values of $a$ which will guarantee that $A$ has eigenvalues $0$, $3$, and $-3$.
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.
Define two functions $T:\R^{2}\to\R^{2}$ and $S:\R^{2}\to\R^{2}$ by\[T\left(\begin{bmatrix}x \\ y\end{bmatrix}\right)=\begin{bmatrix}2x+y \\ 0\end{bmatrix},\;S\left(\begin{bmatrix}x \\ y\end{bmatrix}\right)=\begin{bmatrix}x+y \\ xy\end{bmatrix}.\]Determine whether $T$, $S$, and the composite $S\circ T$ are linear transformations.
Let\[\mathbf{v}_{1}=\begin{bmatrix}1 \\ 1\end{bmatrix},\;\mathbf{v}_{2}=\begin{bmatrix}1 \\ -1\end{bmatrix}.\]Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?
Let $A$ be an $m \times n$ matrix.Suppose that the nullspace of $A$ is a plane in $\R^3$ and the range is spanned by a nonzero vector $\mathbf{v}$ in $\R^5$. Determine $m$ and $n$. Also, find the rank and nullity of $A$.
Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\]still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample.
For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]
For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]
(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?
(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.
(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.
(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]
(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?
(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$. |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\]which matrix do you get?(a) $A$(b) $C^{-1}A^{-1}BC^{-1}AC^2$(c) $B$(d) $C^2$(e) $C^{-1}BC$(f) $C$
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 $Q$.
(The Ohio State University, Linear Algebra Midterm)
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 $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
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 $W$.
(The Ohio State University, Linear Algebra Midterm)
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 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
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)
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 1 and contains the first three problems.Check out Part 2 and Part 3 for the rest of the exam problems.
Problem 1. Determine all possibilities for the number of solutions of each of the systems of linear equations described below.
(a) A consistent system of $5$ equations in $3$ unknowns and the rank of the system is $1$.
(b) A homogeneous system of $5$ equations in $4$ unknowns and it has a solution $x_1=1$, $x_2=2$, $x_3=3$, $x_4=4$.
Problem 2. Consider the homogeneous system of linear equations whose coefficient matrix is given by the following matrix $A$. Find the vector form for the general solution of the system.\[A=\begin{bmatrix}1 & 0 & -1 & -2 \\2 &1 & -2 & -7 \\3 & 0 & -3 & -6 \\0 & 1 & 0 & -3\end{bmatrix}.\]
Problem 3. Let $A$ be the following invertible matrix.\[A=\begin{bmatrix}-1 & 2 & 3 & 4 & 5\\6 & -7 & 8& 9& 10\\11 & 12 & -13 & 14 & 15\\16 & 17 & 18& -19 & 20\\21 & 22 & 23 & 24 & -25\end{bmatrix}\]Let $I$ be the $5\times 5$ identity matrix and let $B$ be a $5\times 5$ matrix.Suppose that $ABA^{-1}=I$.Then determine the matrix $B$.
(Linear Algebra Midterm Exam 1, the Ohio State University) |
Recently I was watching this interview of Andrew Wiles where a secondary school teacher asked this question:
How do you teach square roots?
He doesn't answer the question so I'd like to ask it here?
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As he says, this is a convention, and you just have to make a choice and stick with it. The usual convention (which makes the square root a function) is to take the positive branch.
The teacher seems a bit confused. He seems to think that there is a contradiction between saying $\sqrt{9} = 3$ on one hand, and on the other hand writing $x^2=9 \implies x=\pm3$ on the other.
If we stick to the convention that the square root takes only positive values, then there are two sensible ways to approach the quadratic equation.
One is to recognize that $\sqrt{x^2} = |x|$, not $x$. So $x^2 =9 \implies |x|=3 \implies x = \pm 3$.
The other way is to factor $x^2=9 \implies x^2-9=0 \implies (x-3)(x+3) = 0 \implies x = \pm3$.
In either case, the $\pm$ does not need to be part of the definition of the square root, and including it makes life more complicated because then square root is not a (single valued) function.
EDIT:
As mentioned in the comments, it is worth making the distinction between square roots and principle square roots. We can define $y$ to be a square root of $x$ if and only if $y^2=x$. Then every positive number $x$ has two square roots, $0$ has only $0$ as its square root, and negative numbers have no real square roots.
The relation $y^2=x$ is not a function of $x$, but if we throw out all the pairs with negative $y$ values we do obtain a function, called the "principle square roots function". This takes a nonnegative real number $x$ as input and returns its nonnegative square root as output.
Another observation: this choice is nice for many reasons, not the least of which is that it is consistent with the definition $x^a = e^{a \log(x)}$ which can be used to define any positive power function (assuming the exponential and logarithm have been defined).
This might help to clarify my remarks above the edit.
Original Answer
I first define that a square root of a number $a$ is a number $x$ such that $x^2 = a$. I then give some examples like: What are the square roots of 4? $2$ and $-2$ because $2^2=4$ and $(-2)^2=4$. I then state that positive real numbers have two square roots, one positive and one negative.
So what does $\sqrt{a}$ mean? There are two choices, right?
Define that the symbol $\sqrt{a}$ always denotes the positive square root of $a$.
Then I given a couple of examples: The square roots of 4 are $\sqrt{4}=2$ and $-\sqrt{4}=-2$. The square roots of 2 are $\sqrt{2}$ and $-\sqrt{2}$.
$n$-th roots are defined in a similar way, as solutions of $x^n=a$. Example: $-2$ is cube root of $-8$ because $(-2)^3=-8$ To entice students, I like to mention that there are actually three cube roots of $-8$: $-2$ and two complex roots (which they will see in their future)
So what does $\sqrt[3]{a}$ mean? There are three choices?
Answer: The symbol $\sqrt[3]{a}$ always denotes the real $3$-rd root of $a$.
Example: $\sqrt[3]{-8} = -2$
The symbol $\sqrt[n]{a}$ denotes the real $n$-th root of $a$, unless there are two real $n$-th roots, in which case pick the positive $n$-th root. Alternatively, break it up according to whether $n$ is odd or even.
Added in response to a question in the comments:
The formulas $(\sqrt{x})^2 = x$ and $\sqrt{x^2} = |x|$ and often confuse students. Here's how I approach it.
Question to students: $(\sqrt{x})^2$ = ?
By definition, $\sqrt{x}$ is the non-negative number such that $(\sqrt{x})^2 = x$.
So $(\sqrt{x})^2 = x$.
Example: $(\sqrt{5})^2 = 5$
Question to students: $\sqrt{x^2}$ = ?
Some students will say $\sqrt{x^2}$ = x$.
Let's test this formula.
Example: $\sqrt{2^2} = \sqrt{4} = 2$. That works.
Example: $\sqrt{(-2)^2}$ = ?. Since $(-2)^2 = 4$, we have $\sqrt{(-2)^2} = \sqrt{4} = 2$. But the suggested formula $\sqrt{x^2} = x$ says that $\sqrt{(-2)^2} = -2$. The formula doesn't work.
Some student will probably now suggest $\sqrt{x^2} = |x|$. If not, you should suggest it.
Now I explain why it works.
By definition, $\sqrt{x^2}$ is the non-negative number such that $(\sqrt{x^2})^2 = x^2$.
So either $\sqrt{x^2} = x$ (because $x^2 = x^2$) or $\sqrt{x^2} = -x$ (because $(-x)^2 = x^2$). How to choose? By definition, pick the non-negative one. If $x \geq 0$, then $\sqrt{x^2} = x$. If $x < 0$, then $\sqrt{x^2} = -x$.
Let's write this as a piecewise function:
$$ \sqrt{x^2} = \left\{ \begin{array}{ll} x & \text{if } x \geq 0\\ -x & \text{if } x < 0\\ \end{array} \right. $$
Does this function seems familiar? This exactly the formula for $|x|$. Therefore $$ \sqrt{x^2} = |x|. $$
NOTE: I am assuming here I have introduced the piecewise formula for the absolute value $|x|$.
The answer in the video was a bit unsatisfying for you, and I understand why. You expect a firm answer.
IRL, when I get this question, I need to quickly determine the level the student is at in her studies. When square roots (or nth roots for that matter) are first discussed, we are only concerned about the positive value. After all, my square with an area of 4 square units has a side that’s a positive number.
But then the students move on the the square root as part of the quadratic equation solving, and we must be mindful of both positive and negative roots.
At some point, the 4th root of 16 has 4 solutions, as the +/-2i is also a root.
It’s for the adult math teacher to understand what level the class is at, and whether that level contains the negative or imaginary root as a solution. |
Question:
Is there any sense of uniqueness in Yang-Mills gauge field theories?
Details:
Let's say we are after the most general Lagrangian Quantum Field Theory of (possibly self-interacting) $N$ spin $j=1$ particles (and matter). Yang-Mills' construction is based on the following:
Pick a compact semi-simple Lie Group $G$ with $\dim G=N$, and introduce $N$ vector fields $A_\mu^a$, $a=1,\dots,N$. Then $$ F^a_{\mu\nu}\equiv 2\partial_{[\mu}A_{\nu]}^a+gf^{abc}A_\mu^b A_\nu^c $$
The Lagrangian is given by $$ \mathcal L=-\frac12\text{tr}(F^2)+\mathcal L_\mathrm{matter}(\psi,\nabla\psi)+\text{gauge-fixing} $$ where $\nabla\psi\equiv\partial\psi-ig T^a A^a$.
My question is about how unique this procedure is. For example, some questions that come to mind:
Is $-\frac12\text{tr}(F^2)$ the most general Lagrangian $\mathcal L=\mathcal L(A^a_\mu)$ that leads to a consistent theory? or can we add new self interactions, and new free terms, without spoiling unitarity, covariance, or renormalisability?
Is minimal coupling $\partial\to \nabla$ the most general introduction of interactions with the matter fields? or can we add non-minimal interactions without spoiling unitarity, covariance, or renormalisability?
In short: does Yang-Mills' construction lead to the most general Lagrangian that can accommodate the interactions of these spin $j=1$ particles consistently? This construction has many different ingredients, some of which can be motivated through geometric considerations, but I've never seen any claim about
uniqueness. |
Prove or disprove:
Let $\rho : \mathbb{N} \rightarrow \mathbb{N}$ injective. Let $(a_{n})_{n \in \mathbb{N}}$ be a sequence.
(i) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ absolutely converges then $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ also converges absolutely.
(ii) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ converges then $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ also converges.
So my understanding is that a series converges if the infinite sum of the series is the limit? It converges absolutely if $\displaystyle\sum\limits_{n=1}^\infty |a_{n}|$ converges. The way I am reading the $a_{\rho(n)}$ 's is that they denote some sort of permutation or rearrangement of the original series. Aside from that though, I am absolutely lost when it comes to approaching this problem... In general am I looking for an $\epsilon >0$ which is greater than $a_{\rho(1)}+ \dots +a_{\rho(n)}$ ? If so, how do I begin trying to find it? |
EDIT updated (improved) description of phase detection circuit
There are two principles used in these systems.
The first is the time-of-flight principle. As you noted, if you wanted to get down to 3 mm accuracy, you need timing resolution of 20 ps (20, not 10, because you would be timing the round trip of the light). That's challenging - certainly not the realm of cheap consumer electronics. The problem is not only the need to detect a fast edge - you have to detect the actual reflected pulse, and not every other bit of noise around. Signal averaging would be your friend: sending a train of pulses and timing their average round trip time helps. This immediately suggests that continuous modulation would probably work better - it has an inherent filtering characteristic.
That leads to the second way to get an accurate measurement: by comparing the phase of the emitted and returned signal. If you modulate your laser at a modest 300 MHz, the wave length of one complete "wave" is 1 m; to measure a change in distance of 3 mm (6 mm round trip), it is sufficient to detect a phase shift of $\frac{6}{1000}\times 2\pi$. This is quite trivial with a circuit that squares the transmitted and reflected wave, then takes the XOR of the two signals and averages the result. Such a circuit will give minimum voltage when the two signals are exactly in phase, and maximum voltage when they are exactly out of phase; and the voltage will be very linear with phase shift. You then add a second circuit that detects whether signal 2 is high when signal 1 has a rising edge: that will distinguish whether signal 1 or signal 2 is leading.
Putting the output of the logic gates into a low pass filter (resistor and capacitor) and feeding it into a low speed 12 bit ADC is sufficient to determine the phase with high accuracy. There are ready made circuits that can do this for you - for example, the AD8302
The only problem with the phase method is that you will find the distance modulo half the wavelength; to resolve this, you use multiple frequencies. There is only a single distance that has the right wavelength for all frequencies.
A possible variation of this uses a sweeping frequency source, and detects the zero crossings of the phase - that is, every time the phase detector output is zero (perfectly in phase) you record the modulation frequency at which this occurred. This can easily be done very accurately - and has the advantage that "detecting zero phase" doesn't even require an accurate ADC. A wise man taught me many years ago that "the only thing you can measure accurately is zero". The distance would correspond to the round trip time of the lowest frequency which has a zero crossing - but you don't necessarily know what that frequency is (you may not be able to go that low). However, each subsequent zero crossing will correspond to the same increase in frequency - so if you measure the $\Delta f$ between zero crossings for a number of crossings, you get an accurate measure of the distance.
Note that a technique like that requires very little compute power, and most of the processing is the result of very simple signal averaging in analog electronics.
You can read for example US patent application US20070127009 for some details on how these things are implemented.
A variation of the above is actually the basis of an incredibly sensitive instrument called the lock-in amplifier. The principle of a lock-in amplifier is that you know there is a weak signal at a known frequency, but with unknown phase (which is the case for us when we look at the reflected signal of a modulated laser). Now you take the input signal, and put it through an IQ detector: that is, you multiply it by two signals of the same frequency, but in quadrature (90° phase shift). And then you average the output over many cycles. Something interesting happens when you do that: the circuit acts, in effect, as a phase sensitive bandpass filter, and the longer you wait (the more cycles' output you average over), the narrower the filter becomes. Because you have both the I and the Q signals (with their phase shift), you get both amplitude and phase information - with the ability to recover a tiny signal on top of a hug amount of noise, which is exactly the scenario you will often have with a laser range finder. See for example the wiki article.
The quadrature detection becomes quite trivial when you use a clock at twice the modulation frequency, and put two dividers on it: one that triggers on the positive edge, and one that triggers on the negative edge. A couple of (fast, electronic) analog switches and a simple RC circuit complete the project. You can now sweep the driving frequency and watch the phase on the two outputs "wrap" - and every time it makes a full circle, you have increased the frequency by an amount $\Delta f = \frac{c}{2d}$ where $c$ is the speed of light, and $d$ is the distance to the target. Which has turned a very hard measurement into a really easy one. |
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Step 7: Divide does not depend on the choice of the weighting matrix. the coefficient is always positive. The heights were originally given rounded to the nearest inch averages rather than measurements on individual women.When this assumption is violated the regressors videos smaller than 600MB.
constant) we have a quadratic model in the second regressor. of linear unbiased estimators, which is quite restrictive. It is customary to split this assumption into two parts: Homoscedasticity: E[ εi2 | X ] = σ2, Coefficient Privacy policy.tests such as for example Wald test or LR test should be used.
In fact, you'll find the formula on the AP statistics minimum-variance mean-unbiased estimation when the errors have finite variances. This approach allows for more natural study |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
Wigner has many friends, and they can disagree:
Daniela Frauchiger and Renato Renner: Quantum Theory Cannot Consistently Describe the Use of Itself. Lidia del Rio: Journal club: Frauchiger-Renner No-Go theorem for Single-World Interpretations of Quantum Theory: "In this talk I will go over the recent paper by Daniela Frauchiger and Renato Renner, "Single-World Interpretations of Quantum Theory Cannot Be Self-Consistent" (arXiv:1604.07422)...
...The paper introduces an extended Wigner's friend thought experiment, which makes use of Hardy's paradox to show that agents will necessarily reach contradictory conclusions-unless they take into account that they themselves may be in a superposition, and that their subjective experience of observing an outcome is not the whole story. Frauchiger and Renner then put this experiment in context within a general framework to analyse physical theories. This leads to a theorem saying that a theory cannot be simultaneously (1) compliant with quantum theory, including at the macroscopic level, (2) single-world, and (3) self-consistent across different agents. In this talk I will (1) describe the experiment and its immediate consequences, (2) quickly review how different interpretations react to it, (3) explain the framework and theorem in more detail...
Scott Aaronson: It’s Hard to Think When Someone Hadamards Your Brain: "So: a bunch of people asked for my reaction to the new Nature Communications paper by Daniela Frauchiger and Renato Renner, provocatively titled 'Quantum Theory Cannot Consistently Describe the Use of Itself'...
...While I found their paper thought-provoking, I reject the contention that there’s any new problem with QM’s logical consistency: for reasons I’ll explain, I think there’s only the same quantum weirdness that (to put it mildly) we’ve known about for quite some time.... The new argument rests on just three assumptions... QM works, measurements have definite outcomes, and the “transitivity of knowledge”... if you reject the argument, then you must reject at least one of the three assumptions; and how different interpretations (Copenhagen, Many-Worlds, Bohmian mechanics, etc.) make different choices about what to reject. But I reject an assumption that Frauchiger and Renner never formalize. That assumption is, basically: “it makes sense to chain together statements that involve superposed agents measuring each other’s brains in different incompatible bases, as if the statements still referred to a world where these measurements weren’t being done”....
The argument can be understood as simply the “Wigner’s-friendification” of Hardy’s Paradox.... Take Hardy’s paradox from 1992, and promote its entangled qubits to the status of conscious observers who are in superpositions over thinking different thoughts. Having talked to Renner about it, I don’t think he fully endorses the preceding statement. But since I fully endorse it, let me explain the two ingredients that I think are getting combined here—starting with Hardy’s paradox, which I confess I didn’t know (despite knowing Lucien Hardy himself!) before the Frauchiger-Renner paper forced me to learn it.... If you strip away the words and look only at the actual setup, it seems to me that Frauchiger and Renner’s contribution is basically to combine Hardy’s paradox with the earlier Wigner’s friend paradox. They thereby create something that doesn’t involve counterfactuals quite as obviously as Hardy’s paradox does, and so requires a new discussion...
My attempt to understand: We have Alice, Bob, Wigner, Wigner's Friend, and Wigner's Other Friend...
Prepare the state $ |\psi\rangle $:
$ |\psi\rangle = \frac{|↑↑\rangle + |↑↓\rangle + |↓↑\rangle}{\sqrt{3}} $
Alice measures the first qubit; Bob measures the second.
Conditional on Alice's qubit being in state |↑〉:
$ |Bob\rangle = \frac{|↑\rangle + |↓\rangle}{\sqrt{2}} = |→\rangle $
so Bob can never see $ |←\rangle = \frac{|↑\rangle - |↓\rangle}{\sqrt{2}} $.
Conditioned on Bob’s qubit being in the state |↑〉:
$ |Alice\rangle = \frac{|↑\rangle + |↓\rangle}{\sqrt{2}} = |→\rangle> $
so Alice can never see $ |←\rangle = \frac{|↑\rangle - |↓\rangle}{\sqrt{2}} $.
Since there is no $ |↓↓\rangle> $ component in the state, either Bob or Alice must be |↑〉. Therefore either Alice or Bob must be $ |→\rangle $ Therefore we can never see $ |←←\rangle $ when Alice and Bob measure. Yet we do, 1/12 of the time.
So now we Wignerize it:
Alice measures ↑↓. Wigner observes, writes "I have observed Alice's measurement and drawn the appropriate conclusions about whether Bob can see $ |←\rangle $", and leaves that message behind. Wigner departs beyond an event horizon, leaving no possibility of decohering with the rest of the universe. A quantum eraser is applied to Alice and her particle. Bob measures ↑↓. Wigner's friend observes, writes "I have observed Bob's measurement and drawn the appropriate conclusions about whether Alice can see $ |←\rangle $", and leaves that message behind. Wigner's friend departs beyond an event horizon, leaving no possibility of decohering with the rest of the universe. A quantum eraser is applied to Bob and his particle. Off beyond their event respective event horizons, either Wigner is in the state $ \frac{|↑↑\rangle + |↑↓\rangle}{\sqrt{2}} $ or Wigner's friend is in the state $ \frac{|↑↑\rangle + ||↓↑\rangle}{\sqrt{3}} $, and so either Wigner or Wigner's friend knowsBob or Alice, respectively, cannot have seen $ |←\rangle $ . You, knowing what one of Wigner and Wigner's friend knowsbeyond their respective event horizons, use the "consistent reasoning" heuristic to conclude that $ |←← \rangle $ is impossible. Alice and Bob then measure →←. Wigner's other friend looks, and 1/12 of the time sees a fully decohered $ |←← \rangle $. I think I have got it! Apropos of "Wigner Has Many Friends"... Wigner has many friends, and they can disagree... Renato Renner and Scott Aaronson: It’s hard to think when someone Hadamards your brain Daniela Frauchiger and Renato Renner: Quantum theory cannot consistently describe the use of itself http://nbviewer.jupyter.org/github/braddelong/weblog-support/blob/master/2018-09-27%20Aaronson%2C%20del%20Rio%2C%20Frauchiger%2C%20Renner-%20Quantum%20Theory%20Cannot%20Consistently%20Describe%20the%20Use%20of%20Itself.ipynb
#quantummechanicswignerhasmanyfriends
#shouldread |
Interested in the following function:$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$where $\pi(n)$ is the prime counting function.When $s=2$ the sum becomes the following:$$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...
Consider a random binary string where each bit can be set to 1 with probability $p$.Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...
The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.
Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?
Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...
@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.
Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.
Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).
Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...
I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.
On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?
suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable
Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}]$ .
Can you give some hint?
My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$
If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.
I have a bilinear functional that is bounded from below
I try to approximate the minimum by a ansatz-function that is a linear combination
of any independent functions of the proper function space
I now obtain an expression that is bilinear in the coeffcients
using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)
I get a set of $n$ equations with the $n$ the number of coefficients
a set of n linear homogeneus equations in the $n$ coefficients
Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists
This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.
Avoiding the neccessity to solve for the coefficients.
I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.
I wonder if there is something deeper in the background, or so to say a more very general principle.
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert. Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!! |
The Annals of Applied Probability Ann. Appl. Probab. Volume 13, Number 2 (2003), 475-489. Harmonic moments and large deviation rates for supercritical branching processes Abstract
Let $ \{Z_{n}, n \ge 1 \}$ be a single type supercritical Galton--Watson process with mean $EZ_{1} \equiv m$, initiated by a single ancestor. This paper studies the large deviation behavior of the sequence $\{R_n \equiv \frac{Z_{n+1}}{Z_n}\dvtx n \ge 1 \}$ and establishes a "phase transition" in rates depending on whether $r$, the maximal number of moments possessed by the offspring distribution, is less than, equal to or greater than the Schröder constant $\alpha$. This is done via a careful analysis of the harmonic moments of $Z_n$.
Article information Source Ann. Appl. Probab., Volume 13, Number 2 (2003), 475-489. Dates First available in Project Euclid: 18 April 2003 Permanent link to this document https://projecteuclid.org/euclid.aoap/1050689589 Digital Object Identifier doi:10.1214/aoap/1050689589 Mathematical Reviews number (MathSciNet) MR1970272 Zentralblatt MATH identifier 1032.60081 Citation
Ney, P. E.; Vidyashankar, A. N. Harmonic moments and large deviation rates for supercritical branching processes. Ann. Appl. Probab. 13 (2003), no. 2, 475--489. doi:10.1214/aoap/1050689589. https://projecteuclid.org/euclid.aoap/1050689589 |
Let $X$ be a normal complex projective variety, $T\subseteq X$ a subvariety of dimension $m$ (possibly $T=X$ ) and $a:X\rightarrow A$ a morphism to an abelian variety such that $\text{Pic}^{0}(A)$ injects into $\text{Pic}^{0}(T)$ ; let $L$ be a line bundle on $X$ and $\unicode[STIX]{x1D6FC}\in \text{Pic}^{0}(A)$ a general element.
We introduce two new ingredients for the study of linear systems on $X$ . First of all, we show the existence of a factorization of the map $a$ , called the eventual map of $L$ on $T$ , which controls the behavior of the linear systems $|L\otimes \unicode[STIX]{x1D6FC}|_{|T}$ , asymptotically with respect to the pullbacks to the connected étale covers $X^{(d)}\rightarrow X$ induced by the $d$ -th multiplication map of $A$ .
Second, we define the so-called continuous rank function $x\mapsto h_{a}^{0}(X_{|T},L+xM)$ , where $M$ is the pullback of an ample divisor of $A$ . This function extends to a continuous function of $x\in \mathbb{R}$ , which is differentiable except possibly at countably many points; when $X=T$ we compute the left derivative explicitly.
As an application, we give quick short proofs of a wide range of new Clifford–Severi inequalities, i.e., geographical bounds of the form
$$\begin{eqnarray}\displaystyle \text{vol}_{X|T}(L)\geqslant C(m)h_{a}^{0}(X_{|T},L), & & \displaystyle \nonumber\end{eqnarray}$$ where $C(m)={\mathcal{O}}(m!)$ depends on several geometrical properties of $X$ , $L$ or $a$ . |
The character $\chi_R : G \to \mathbb{C}$ of a representation $R$ is defined by $\chi_R(U) = Tr_R(U)$, namely by taking the trace of $U$ in the representation $R$. See for example Appendix A of Aharony et al.. Then the equation you write seems reasonable: presumably there are $N_f$ hypermultiplets, each with fundamental and anti-fundamental fields that give $Tr(U^m)$ and $Tr(U^{-m})$respectively in the sum over $R$. The trace is taken of the matrix $U$ because the fields are in the fundamental.
The factor of 2 should come from the details of which representations exactly are being summed over, the field content of the hypermultiplets, etc., which I didn't try to follow.
Now, regarding the distribution of eigenvalues that appears in (C.5), let me give you a physical explanation of where this comes from. They consider the thermodynamics of a theory on a sphere, so the topology is that of a sphere times $S^1$ -- the Euclidean compact time. In this case there is a zero-mode of the gauge field, which comes from the fact that you can turn on a constant field in the thermal direction. This mode cannot be removed by a gauge transformation (in general), so in a path integral you need to integrate over it. This is explained for example in section 4.1 of Aharony et al. You can use a gauge transformation to diagonalize this zero-mode matrix, so you are left with the discrete eigenvalues of the matrix to integrate over. You can also show that these eigenvalues live on a circle, because you can shift them by $2\pi$ (in some normalization) using a gauge transformation, so you need to integrate them over a circle. In the large $N$ limit you have an infinite number of such eigenvalues, but they are still restricted to live on a circle. So you need to describe them using a density, and that is what they call the distribution $\rho$.
Edit: adding explanation of (C.6)
To derive (C.6), consider first the sum
$$ \sum_i \cos(n \alpha_i) = N \int_0^{2\pi} \rho(\theta) \cos(n\theta) $$
where $\rho(\theta) = \frac{1}{N} \sum_i \delta(\theta - \alpha_i)$. Notice that $\rho$ has the correct normalization.
When taking $N \to \infty$ the delta functions in $\rho$ will become very dense, and we will be able to approximate $\rho$ by a smooth function, whose value at $\theta$ depends on the density of the delta functions across the small interval $[\theta,\theta+\epsilon]$. This will be a good approximation because the function $\cos(n\theta)$ that we are integrating over will not vary much in these intervals, so averaging it instead of sampling it will not change the result by much. When we reach the limit $N=\infty$ this will no longer be an approximation, because the delta functions will become continuous.
So this explains the derivation of the second term in (C.6). As for the first term, the idea is similar except that you have two integrals over $\theta,\theta'$ corresponding to the $i,j$ sums. The contributions from the $i=j$ terms ($\theta=\theta'$ in the integral) are subleading in the large $N$ limit: they scale as $N$ while the rest of it scales as $N^2$, so they are neglected.
Now, replace $\cos(n(\theta-\theta')) = \cos(n\theta) \cos(n\theta') + \sin(n\theta) \sin(n \theta')$. Notice that in the path integral over $\rho$ you may consider only eigenvalue distributions that are symmetric under $\theta \to -\theta$, because the original integrand is invariant under this. This is explicitly mentioned for example in Schnitzer, who does the same computation. So this means that $\int d\theta \rho \sin(n\theta) = 0$. I think it should be clear from this point. |
The power spectral density describes the density of power in a stationary random process $X(t)$ per unit of frequency. By the Wiener-Khinchin theorem, it can be calculated as follows for a wide-sense stationary random process:
$$S_{xx}(f) = \int_{-\infty}^{\infty} r_{xx}(\tau) e^{-j2\pi f \tau} d\tau$$
where $r_{xx}(\tau)$ is the autocorrelation function of the process $X(t)$:
$$r_{xx}(\tau) = \mathbb{E}\left(X(t)X(t - \tau)\right)$$
This is only valid for a wide-sense stationary process because its autocorrelation function is only a function of the time lag $\tau$ and not the absolute time $t$; stated differently, this means that its second-order statistics don't change as a function of time.
With that said, if you have a sufficiently-detailed and accurate statistical model for your signal, then you can calculate its power spectral density using the relationship above. As an example, this can be used to calculate the power spectral density of communications signals, given the statistics of the information symbols carried by the signal and any pulse shaping employed during transmission.
In most practical situations, this level of information is not available, however, and one must resort to
estimating a given signal's power spectral density. One very straightforward approach is to take the squared magnitude of its Fourier transform (or, perhaps, the squared magnitude of several short-time Fourier transforms and average them) as the estimate of the PSD. However, assuming that the signal you're observing contains some stochastic component (which is often the case), this is again just an estimate of what the true underlying PSD is based upon a single realization (i.e. a single observation) of the random process. Whether the power spectrum that you calculate bears any meaningful resemblance to the actual PSD of the process is situation-dependent.
As this previous post notes, there are many methods for PSD estimation; which is most suitable depends upon the character of the random process, any
a priori information that you might have, and what features of the signal you're most interested in. |
Category: Ring theory Problem 624
Let $R$ and $R’$ be commutative rings and let $f:R\to R’$ be a ring homomorphism.
Let $I$ and $I’$ be ideals of $R$ and $R’$, respectively. (a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$. (b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$
Add to solve later
(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$ Problem 618
Let $R$ be a commutative ring with $1$ such that every element $x$ in $R$ is idempotent, that is, $x^2=x$. (Such a ring is called a
Boolean ring.) (a) Prove that $x^n=x$ for any positive integer $n$.
Add to solve later
(b) Prove that $R$ does not have a nonzero nilpotent element. Problem 543
Let $R$ be a ring with $1$.
Suppose that $a, b$ are elements in $R$ such that \[ab=1 \text{ and } ba\neq 1.\] (a) Prove that $1-ba$ is idempotent. (b) Prove that $b^n(1-ba)$ is nilpotent for each positive integer $n$.
Add to solve later
(c) Prove that the ring $R$ has infinitely many nilpotent elements. |
Tagged: subspace Problem 709
Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[ \mathbf{v}_{1}= \begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{2}= \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix} ,\;\mathbf{v}_{3}= \begin{bmatrix} 1 \\ 5 \\ -1 \\ 5 \end{bmatrix} ,\;\mathbf{v}_{4}= \begin{bmatrix} 1 \\ 1 \\ 4 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{5}= \begin{bmatrix} 2 \\ 7 \\ 0 \\ 2 \end{bmatrix} .\] Find a basis for the span $\Span(S)$. Problem 706
Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\] still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later
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 $W$.
Add to solve later
(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 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 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 |
First of all I want to mention that this is homework, so don't spoil it and reveal all the answer. just some guidenss :)
Let $H$ be a Hilbert space, $T:H\rightarrow H$ a bounded linear operator for which there exists an $r>0$, such that for all $x\in H: \|Tx\|\geq r\|x\|$.
Define $B:H\rightarrow H$, a bilinear form, by $B(x,y)=\langle Tx,y\rangle$. Prove or disprove that $B$ is coercive.
What I did so far:
my intuition says it is false. I managed to prove that $B$ is bounded I realised that $T$ is an injection, therefore if $H$ is finite-dimensional $T$ is invertible, and my guts say that it is trueiff $T$ is invertible (use the Lax-Milgram theorem and the fact that any inner-product is coercive). Therefore I tried looking into infinite dimentional Hilbert spaces, with a non invertible $T$ that would answer to the conditions yet make the bilinear form not coercive. I have't even found a $T$ that follows the conditions (let alone disporoves the claim).
I would appriciate help and assistence in order to make my mind more organized.
Thanks! |
If $x_n$ is a Cauchy sequence, then uniform continuity of $f$ allows us to conclude that the sequence $f(x_n)$ is also Cauchy.
The sequences $a_n=a+\frac{1}{n}$ and $b_n =b-\frac{1}{n}$ are Cauchy, hence so are the sequences $f(a_n)$, $f(b_n)$. Since $\mathbb{R}$ is complete, these sequences converge to some numbers $f_a, f_b$ respectively. Define the function $\overline{f}:[a,b] \to \mathbb{R}$ by $\overline{f}(a) = f_a$, $\overline{f}(b) = f_b$ and $\overline{f}(x) = f(x)$ for $x \in (a,b)$. Clearly $\overline{f}$ is continuous in $(a,b)$, it only remains to show continuity at $a,b$.
Suppose $x_n\in [a,b]$, and $x_n \to a$. We have $|\overline{f}(x_n) - \overline{f}_a| \leq |\overline{f}(x_n) - \overline{f}(a_n)| + | \overline{f}(a_n) - \overline{f}_a|$. Let $\epsilon >0$, then by uniform continuity, there exists $\delta>0$ such that if $|x-y|< \delta$, with $x,y \in (a,b)$, then $|f(x)-f(y)| < \epsilon$. Choose $n$ large enough such that $|x_n-a_n| < \delta$, and $| f(a_n) - f_a| < \epsilon$. If $x_n = a$, then $|\overline{f}(x_n) - \overline{f}_a| = 0$, otherwise we have $|\overline{f}(x_n) - \overline{f}_a| \leq |f(x_n) - f(a_n)| + | f(a_n) - f_a| < 2 \epsilon$. Consequently $\overline{f}(x_n) \to \overline{f}_a$, hence $\overline{f}$ is continuous at $a$. Similarly for $b$.
For the second case, take $f(x) = \frac{1}{x-a}$. Then $f$ is continuous on $(a,b)$, but the domain cannot be extended to $[a,b]$ while keeping $f$ continuous, and $\mathbb{R}$ valued. To prove this, take the sequence $a_n$ above, then $f(a_n) = n$, and clearly $\lim_n f(a_n) = \infty$. If $f$ could be continuously extended to $\overline{f}$, then $\overline{f}(a) \in \mathbb{R}$, which would be a contradiction. |
I assume that your definition of abelian category includes additive and I define additive categories as follows:An
Ab-category $\mathscr C$ is additive if there is a zero object $0 \in \mathscr C$ and products $X \times Y$ exist for every pair $X, Y \in \mathscr C$.
Denote products $X \times Y$ by $X \oplus Y$. Now prove the following theorem:In an additive category $\mathscr C$, there exist unique morphisms $$X \xrightarrow{i_1} X \oplus Y \xleftarrow{i_2} Y$$such that $p_1 i_1 = 1_{X}$, $p_2 i_2 = 1_{Y}$, $p_1i_2 = 0$, $p_2i_1 = 0$ and $i_1p_1 + i_2p_2 = 1_{X \oplus Y}$, where $$X \xleftarrow{p_1} X \oplus Y \xrightarrow{p_2} Y$$are the projection maps of the product $X \oplus Y$.
Then we get the following corollary: Suppose maps $i_1, i_2$ are constructed as above. Then $(X \oplus Y, i_1, i_2)$ is the coproduct of $X$ and $Y$. Thus, sums and products are the same in an additive category.
Finally, prove that if $g$ and $h$ are isomorphisms then $\operatorname{Ker} f \cong \operatorname{Ker}hfg$ and similarly for cokernels.
Now your result is easy:$$\begin{align}\operatorname{Im} \left( \bigoplus_{1\le i\le n} \varphi_i \right) &\cong \operatorname{Ker} \left(\bigoplus_{1\le i\le n} \operatorname{Coker} \varphi_i \right) \\&\cong \operatorname{Ker} \left(\prod_{1\le i\le n} \operatorname{Coker} \varphi_i \right) \\&\cong \prod_{1\le i\le n} \operatorname{Ker}\operatorname{Coker} \varphi_i \\&\cong \bigoplus_{1\le i\le n} \operatorname{Ker}\operatorname{Coker} \varphi_i \\&\cong \bigoplus_{1\le i\le n} \operatorname{Im} \varphi_i.\end{align}$$ |
I am tasked with the proof that $x =0$ is a unique stationary point and a minimum of $\cos(x) + \cosh(x) = 2\sum_{k=0}^{\infty} \frac{x^{4k}}{(4k)!}$.
What the markscheme does, concerning the "minimum" and "uniqueness" part is incomprehensible to me:
Given that the derivative is the series for $\sinh(x) - \sin(x)$ which is odd and strictly positive for all $x > 0$, we have proven that the point is unique. The fact that the point is a min can be proven in two ways: either by the fact that the series for $\cos(x) + \cosh(x)$ is positive for all $x \neq 0$ or by finding the fourth derivative which equals 2 at $x=0$, hence its a local minimum.
I have no idea why $\sinh(x) - \sin(x)$ being odd and strictly positive for all $x > 0$ helps me prove that the point is unique. Similarly, I dont understand the issue about the fourth derivative or the actual series being positive for all x not equal to 0 implying that the point is a min. Can someone please clarify this for me? |
Let $K$ be the forward exchange rate determined at time $t$ for maturity $T$. Then the payoff at time $T$ is given by $S_T-K$, which has zero value at time $t$. Let $Q$ and $Q^f$ be the respective domestic and foreign risk-neutral measures, and $E^Q$ and $E^{Q^f}$ be the corresponding expectation operators. Moreover, let $B^d_T = e^{\int_0^t r^d_sds}$ and $B^f_T = e^{\int_0^t r^f_sds}$ be the respective domestic and foreign money market account values at time $t$. Then, \begin{align*}B^d_tE^Q\left(\frac{S_T-K}{B^d_T} \mid \mathcal{F}_t\right) = 0.\end{align*}That is,\begin{align*}K = \frac{B^d_tE^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)}{E^Q\left(\frac{B^d_t}{B^d_T} \mid \mathcal{F}_t\right)}= \frac{B^d_tE^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)}{P^d(t, T)}. \tag{1}\end{align*}Let $Q^T$ be the domestic $T$-forward measure and $E^{Q^T}$ be the corresponding expectation operator. Then\begin{align*}\frac{dQ}{dQ^T}\big|_{[t, T]} = \frac{B^d_T}{B^d_tP^d(t, T)}.\end{align*}From $(1)$,\begin{align*}K &= \frac{B^d_t E^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)}{P^d(t, T)}\\&=E^{Q^T}(S_T \mid \mathcal{F}_t).\end{align*}That is,\begin{align*}E^{Q^T}(S_T \mid \mathcal{F}_t) = F_t(T).\end{align*}On the other hand, we note that\begin{align*}\frac{dQ}{dQ^f}\big|_{[t, T]} = \frac{B^d_TB^f_t S_t}{B^f_T B^d_tS_T}.\end{align*}Then\begin{align*}E^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right) &= E^{Q^f}\left(\frac{B^d_TB^f_t S_t}{B^f_T B^d_tS_T}\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)\\&=\frac{S_t}{B^d_t}E^{Q^f}\left(\frac{B^f_t}{B^f_T} \mid \mathcal{F}_t\right)\\&=\frac{S_t}{B^d_t}P^f(t, T).\end{align*}Moreover, from $(1)$,\begin{align*}K &= \frac{B^d_t E^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)}{P^d(t, T)}\\&=S_t \frac{P^f(t, T)}{P^d(t, T)}.\end{align*}That is,\begin{align*}E^{Q^T}(S_T \mid \mathcal{F}_t) = F_t(T) = S_t \frac{P^f(t, T)}{P^d(t, T)}.\tag{2}\end{align*}
However, we note that, with stochastic interest rates, generally,\begin{align*}E^{Q}(S_T \mid \mathcal{F}_t) \ne F_t(T).\end{align*}
$$$$As OP has already pointed out, Fromula $(2)$ can also be shown by no arbitrage argument. Specifically, at time $t$, while entering a forward contract with forward exchange rate $F_t(T)$, we borrow one unit domestic currency (which can be used to buy $\frac{1}{P^d(t, T)}$ units domestic zero-coupon bond with maturity $T$), convert into $\frac{1}{S_t}$ units foreign currency, and buy $\frac{1}{S_t P^f(t, T)}$ units foreign zero-coupon bond with maturity $T$. The net value of this trading strategy is zero.
At maturity $T$, in domestic currency, the above trading strategy has value $$F_t(T)\frac{1}{S_tP^f(t, T)}-\frac{1}{P^d(t, T)},$$ which should also have zero value. That is,\begin{align*}F_t(T) &=S_t \frac{P^f(t, T)}{P^d(t, T)}.\end{align*} |
Electronic Devices Intrinsic Semiconductor, Extrinsic Semiconductor, p-n Junction A semiconductor in pure form is called INTRINSIC SEMI CONDUCTOR Ex: Si, Ge,.. At any temperature the number of holes in valance band and the number of free electrons in conduction band are equal (n e= n h). For intrinsic semiconductors, the fermi energy level lies exactly at the middle of the forbidden band. Ie = neAVe (current due to electrons) Ih = neAV h(current due to holes) The total current at any instant in semi-conductor is I = I e+ I h HALF WAVE RECTIFIER converts half cycle of applied A.C into DC When a half wave rectifier is used to convert 'n' Hz A.C into D.C., then the number of pulses per second present in the rectified voltage is 'n' only. HALF WAVE RECTIFIER \tt EFFICIENCY\left(\eta\right)=\frac{DC\ Power\ output}{AC\ Power\ input} \tt EFFICIENCY\left(\eta\right)=\left[\frac{0.406\times R_{L}}{r_{F}+R_{L}}\right]\times 100\% n max= 40.6% for an ideal diode. Rectification is the process of converting AC to DC FULL WAVE RECTIFIER converts the whole cycle of applied input A.C signal into D.C signal. When a full wave rectifier is used to convert ‘n’ Hz A.C into D.C, then the no of pulses per second present in the rectified voltage is 2n Efficiency of full wave Rectifier, \tt Efficiency\left(\eta\right)=\frac{DC\ Power\ output}{AC\ Power\ output} \tt EFFICIENCY\left(\eta\right)=\left[\frac{0.812\times R_{L}}{r_{F}+R_{L}}\right]\times 100\%=81.2\% n max= 81.2% for an ideal diode. With the increase of temperature, the ratio \tt \frac{n_e}{n_h} for n-type decreases, but \tt \left(\frac{n_e}{n_h}>1\right) As the temperature increases, the ratio \tt \frac{n_e}{n_h} for P-type increases, but \tt \left(\frac{n_e}{n_h}<1\right) View the Topic in this video From 00:16 To 14:34
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1. Forward resistance of diode Junction = \frac{\Delta V}{\Delta I}
2. Peak value of current is halfwave Rectifier I_{m} = \frac{V_{m}}{r_{f} + R_{L}}
3. Peak value of current is I_{m} = \frac{V_{m}}{r_{f} + R_{L}} in full wave Rectifier.
4. Dynamic output resistance R_{o} = \left[\frac{\Delta V_{CB}}{\Delta i_{c}}\right]_{i_{e} = {\tt constant}}
5. Dynamic input resistance R_{i} = \left[\frac{\Delta V_{BE}}{\Delta I_{B}}\right]_{V_{CE} \rightarrow {\tt constant}} |
Vectors can be defined in multiple ways depending on the context where it is utilized. A vector has both magnitude and direction that is shown over directed line segment where length denotes the magnitude of vector and the arrow indicates the direction from tail to head.
Two vectors are similar if they have same magnitude and direction. The magnitude or direction of a vector with respect to the position doesn’t change. But if you stretch or move the vector from head or tail then both magnitude and direction will change.
In other words, the vector is a quantity having both magnitude and direction. There are scalar quantities that only have magnitude and given a vector measurement. A vector is not important in mathematics only but physics too like aeronautical space, space, traveling guide etc. Pilots use vector quantities while sitting on the plane and taking it to the other direction safely.
Once you are sure on the definition of vector and its usage then next important step is to study the vectors representation. They are represented in the form of a ray and written either in lowercase or upper case. Generally, a single vector is represented in both forms –uppercase and lowercase alphabets. If the vector is written in the form like AB then A is the tail and B is the head.
Vectors are divided into two major categories – one is the dot product, and the other is a cross product. A list of basic formula is available for both the categories to solve the geometrical transformation in 2 dimensions and 3 dimensions. These formulas are frequently used in physics and mathematics. Further, they are widely acceptable for analytical or coordinate geometry problems.
Formula of Magnitude of a Vector
Magnitude of a vector when end point is origin. Let
x and y are the components of the vector,
\[\large \left|v\right|=\sqrt{x^{2}+y^{2}}\]
Magnitude of a vector when starting points are $(x_{1}$, $y_{1})$ end points are $(x_{2}$, $y_{2})$,
\[\large \left|v\right|=\sqrt{\left(x_{2}+x_{1}\right)^{2}+\left(y_{2}+y_{1}\right)^{2}}\]
The formula of resultant vector is given as:
\[\large \overrightarrow{R}=\sqrt{\overrightarrow{x^{2}}+\overrightarrow{y^{2}}}\]
Vector Projection formula is given below:
\[\large proj_{b}\,a=\frac{\vec{a}\cdot\vec{b}}{\left|\vec{b}\right|^{2}}\;\vec{b}\]
The Scalar projection formula defines the length of given vector projection and is given below:
\[\large proj_{b}\,a=\frac{\vec{a}\cdot\vec{b}}{\left|\vec{a}\right|}\]
Unit Vector Formula is given by
\[\large \widehat{V}=\frac{v}{\left|v\right|}\]
formula of direction is
\[\LARGE \theta =\tan^{-1}\frac{y}{x}\]
Parts in vectors are taken as the angles that are directed towards the coordinate axes. Take an example, if some vector is directed at northwest then its parts would be westward vector and the northward vector. So, vectors are generalized into two parts mostly where names could be different but the concept is same.
With the study of old geometry books, you would know about the evolution of vectors in algebra and how is it beneficial for students. Vectors were initially named as the algebra of segments and directed to displacements. Let us see some of the benefits why students should learn Vectors in school and during higher studies too.
Vectors are important in both physics and mathematics and it was discovered to make the geometry transformations easier. It signifies that quick insights can be gained into Geometry and taken an important part of linear algebra. The popular application of vectors includes – particle mechanics, fluid mechanics, planar description, trajectories calculation, 3D motion etc.
The other area where vectors are used is electromagnetism, analytical geometry, and the coordinate geometry etc. With a clear understanding of Vectors, students not only progress in their career but clear various competitive exams too. |
Tagged: subspace Problem 709
Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[ \mathbf{v}_{1}= \begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{2}= \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix} ,\;\mathbf{v}_{3}= \begin{bmatrix} 1 \\ 5 \\ -1 \\ 5 \end{bmatrix} ,\;\mathbf{v}_{4}= \begin{bmatrix} 1 \\ 1 \\ 4 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{5}= \begin{bmatrix} 2 \\ 7 \\ 0 \\ 2 \end{bmatrix} .\] Find a basis for the span $\Span(S)$. Problem 706
Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\] still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later
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 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 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 |
I'm new to classical mechanics and learned a new cool thing, second type equations of Lagrangian Mechanics. So, I was just testing if it really works or not. So, I made a question myself to test it, but I think that I've some problem here.
First, assume the $x$ coordinate of the ball of mass $m$. Initially, it's at $x=0$, under the influence of Gravity it should fall (assuming the setup is in such a way that Gravity is acting from downwards.)
So, the cool equation says that :
$$\displaystyle{\boxed{\boxed{ \dfrac{d}{dt} \left ( \dfrac{\partial L}{\partial {x'} } \right) - \dfrac{\partial L}{\partial x} = 0 }} \quad \dots \quad (*)}$$ Where, $L = K.E. - P.E. = ( mg \sin \theta x) - (mg \sin \theta h)\quad \dots \quad (1)$
However using the Euler-Lagrange equation (*) with the Lagrangian (1) does not give me a sensible answer, so my question is where is my mistake?
So, I've calculated K.E. here as: $K = \frac 1 2 m v^2 = \frac 1 2 m (2(g \sin \theta) x) \quad \dots \quad (\text{As }v^2 = u^2 + 2as)$
and $P = mgh = mg\sin \theta(h - x )$
The issue might be to do with using $v^2=2as$ in the kinetic energy (as suggested in comments) but I don't see why this is improper.
So, Why I can't substitute the Velocity equations in that Lagrangian Equation (That Kinetic energy part). |
I am considering the Zimm model for polymer dynamics, and have come across a question
Find an expression for the time it takes for the polymer to diffuse a distance equal to its contour length $L=Nb$, if the drag coefficient for the polymer is $\gamma = N\beta b$ where the polymer consists of $N$ segments of Kuhn length $b$.
My thoughts on this question were:
For Fickian diffusion $\langle R^2 \rangle = 2Dt$ So if we plug in $D=\frac{k_BT}{\gamma}$ and $R=L$, we should get the right answer?
For some reason, I am not convinced that "diffusing a distance $L$" translates to $\langle R^2 \rangle = \langle L^2 \rangle$.
I'm not sure if this should be obvious or not. Unfortunately, I do not have answers to compare with. |
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 |
Introduction
The relationship between the maximum obtainable directivity and aperture of an antenna is described by:
$$ D = \frac{4\pi A_e}{\lambda^2} $$
Where $D$ is the directivity, $A_e$ the effective aperture and $\lambda$ the used wavelength. The derivation can be found in the following document: pdf.
However, if the antenna does not scan as a sphere but in $x, y$ but not $z$. Following the same path as the above-mentioned document the equation becomes:
$$ D = \frac{2\pi A_e}{\lambda^2} $$
The aperture of a lossless isotropic antenna is then also $A_e = \frac{\lambda²}{2 \pi}$ instead of $\frac{\lambda²}{4 \pi}$
Problem
I cannot manage to use that equation for a microphone array using delay-and-sum beamforming.
For example, if I have an array of 25 microphones spaced as a grid and that my setup has the following properties:
Smallest distance $d$ between them is $0.01$ meters. Amount of orientations $o$ that it looks at is $64$. Frequency $f$ to be detected is 8 kHz.
The wavelength $\lambda = \frac{v}{f} = \frac{343}{8000} = 0.0429 $
The aperture $A$ is $A = \frac{\lambda^2}{precision} = \frac{0.0429^2}{\frac{2\pi}{64}} = 0.0187 $
Thus:
$$ D = \frac{2\pi A_e}{\lambda^2} = \frac{2\pi A}{\lambda^2} = \frac{2 \pi \cdot 0.0184}{0.0429^2} = 62 $$
I rounded the numbers for the simplicity of formulating my problem, if you do not round then you get $64$ instead of $62$ which is the number of orientations that it looks at furthermore the results cannot be true because:
It does not involve the number of microphones used. It does not involve the smallest spacing between microphones. I probably made a mistake somewhere. How can I calculate the maximum obtainable directivity for an array of sensors? |
Consider one period $T=2\pi/\omega$ and suppose that the random time $t$ at which you measure the displacement $x$ of oscillator is uniformly distributed over the period of oscillation: $$f(t)=\frac{1}{T},\quad t\in[0,T)$$
By symmetry the p.d.f. $g(x)$ for displacement $x$ is an odd function i.e. $g(x)=g(-x)$ for $x\in[-a,a]$, so it suffices to work out $g(x)$ for $x\in[0,a]$. A given value of $x$, except the endpoints $x=\pm a$, is obtained at two distinct time instants in $[0,T)$. For $x\in[0,a)$ let the two time instants $t_1,t_2,$ be such that $\omega t_1=\sin^{-1}(x/a)\in[0,\pi/2]$ and $\omega t_2=\sin^{-1}(x/a)\in[\pi/2,\pi]$. Then we have:\begin{align}g(x)dx&=f(t_1)dt+f(t_2)dt\\g(x)&=\frac{2}{T}\frac{dt}{dx}=\frac{1}{\pi\sqrt{a^2-x^2}},\quad x\in[-a,a]\end{align}in which I have also included the endpoints $x=\pm a$ because the probability measure over a finite set of points is zero anyway. You may verify that $g(x)$ is correctly normalised.
Now let us consider an ensemble of oscillators. The oscillators have random amplitude $a$ whose p.d.f. is $q(a)$. Let $h(x)$ be the corresponding p.d.f. for displacement $x$; $h(x)$ is supposed given and we want to find $q(a)$. The p.d.f. $g(x)$ found previously I shall write as $g(x|a)$ which is to be read as the p.d.f. for displacement
given that the oscillator has amplitude $a$. From Bayes theorem we have: $$h(x)=\int_0^\infty da~g(x|a)q(a)$$ The integral needs to be inverted somehow to obtain $q(a)$ and here is a trick which I think will work. Consider the transformation $\eta=a^2$. Then since $a\geq 0$ there is one-to-one correspondence between $\eta$ and $a$, which allows us to relate $q(a)$ to p.d.f. of $\eta$: $$q(a)=q^*(\eta)\frac{d\eta}{da}=2a~q^*(\eta)$$ Therefore $$h(x)=\int_0^\infty d\eta~\frac{1}{\pi\sqrt{\eta-x^2}}\times q^*(\eta)$$
We now similarly transform $x$. By symmetry we have $h(x)=h(-x)$, so let us limit ourselves to $x\in[0,\infty)$. Transforming to $X=x^2$ then gives: $$h(x)=h^*(X)\frac{dX}{dx}=2x~h^*(X)=2\sqrt{X}~h^*(X)$$ Therefore $$2\sqrt{X}~h^*(X)=\int_0^\infty d\eta~\frac{1}{i\pi\sqrt{X-\eta}}\times q^*(\eta)$$
We have a convolution integral above. Applying Laplace transform $\mathbf{L}\equiv\int_0^\infty ds~\exp(-sX)$ gives: $$ q^*(\eta)=\mathbf{L}^{-1}\left[ \frac{\mathbf{L}\left[ 2\sqrt{X}~h^*(X) \right]}{\mathbf{L}\left[ \frac{1}{i\pi\sqrt{X-\eta}} \right]} \right]$$ from which $q(a)$ may be obtained. |
Question
So recently I was thinking about this: How many scalars are available in $4$ dimensions in General Relativity (without being redundant)? For example, with metric we can construct the following scalar:
$$ g^{\mu \nu} g_{\mu \nu} = 4 $$ is the same as: $$ (g^{\mu \nu} \otimes g^{\rho \kappa}) \cdot (g_{\mu \nu} \otimes g_{\rho \kappa} ) = 16 $$
We also have scalars like curvature, torsion, inner product of the riemann tensor with itself, etc.
Motivation
My motivation for doing so is as follows: GR is currently through (rank $2$ symmetric) tensors formulated as:$$ R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$ Hence any solution of the above automatically satisfies:
$$ (R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu}) (R^{\mu \nu} - \frac{1}{2} R g^{\mu \nu}) = \bigg(\frac{8 \pi G}{c^4}\bigg)^2 T_{\mu \nu} T^{\mu \nu} $$
But note the later equation is written purely in invariant observables. I was wondering if General Relativity could also be written purely in terms of observables? If not how many short are we? Can the remaining variables be expressed as something invariant and not a scalar (not sure if it would be a tensor either)? |
Can matter be described as the result of the curvature of space, rather than the curvature of space being the result of matter, and energy being the cause of the curvature of space?
Maybe one day.
This idea, at least it's mathematical genesis seems to have begun with Riemann and later Clifford. In 1870 Clifford (a very good mathematician), building upon Riemann gave a lecture stating:
1) That small portions of space are in fact analogous to little hills on a surface which is on the average flat namely that the ordinary laws of geometry are not valid.
2) That this property of being curved or distorted is continually being passed on from one portion of space to another after the matter of a wave.
3) That this variation of curvature of space is what really happens in that phenomena we call the motion of matter, whether ponderable or etherial.
4)That in the physical world nothing else takes place but this variation, subject (possibly) to the law of continuity.
This was the type of thinking that Led Einstein to consider space and time as dynamic entities and develop General Relativity(using Riemann's then developed geometry). Many others have sought to describe more of the universe than gravity through this type of program.
As of now, it's a no; however some progress has been made.
Because Electromagnetism curves spacetime like matter does, it has been shown (Rainich, Misner, Wheeler) that just the "footprints" left on spacetime by electromagnetic fields are enough to reconstruct the fields themselves (up do a "duality" rotation). This of course only holds for classical electrodynamics. This was called "Geometrodynamics" by Wheeler (who was famous for coining other phrases like blackhole and wormhole as well).
Wheeler showed analytically that a properly constructed ball of gravitational and electromagnetic radiations would possess the properties of a massive object (he called this a geon) and he applied similar topological ideas such that electric charges can appear to exist when there is in fact none (a trick of spacetime geometry).
To go further, one would need to be able to describe the gauge fields of the standard model in terms of geometry and properly quantize it. Whether these things can be found withing the topology of general relativity's spacetime, well that jury is still out.
The equations involved here are highly nonlinear, especially when you have the feedback effect of describing say the electromagnetic field through geometry which in turn effects the geometry again.
Anyway, when I read your question I am literally reading a book in front of my face:
"The Geometrodynamics of Gauge Fields. On the geometry of Yang-Mills fields and Gravitational gauge theories" Eckehard W. Mielke
I'll update this when I'm done maybe, it's an excellent book by the way.
PS: I had the pleasure of seeing Kip Thorne Lecture this (last?) year and he seems a large proponent of geometrodynamics. As one of the founders of LIGO (we're FINALLY detecting gravitational waves!!! still can't believe it) there's a potential to test such theories in the foreseeable future.
The answer is
yes !
In General Relativity, there's a not very well known theorem called the
that states that any metric in 4D spacetime (including matter) could be represented (or Campbell-Magaard theorem embeded) as a metric in empty5D spacetime (Ricci flat spacetime, wich implies pure geometry) :
There are several simple solutions to Einstein's equation in 5D vacuum ($R_{AB}^{(5)} = 0$) that represent matter in 4D ($R_{\mu \nu}^{(4)} \ne 0$). This could be interpreted as matter made of pure geometry, in higher dimensions spacetimes.
Here's a very simple example. Consider the following metric in 5D spacetime ($\theta$ is a cyclic coordinate in the fifth dimension) :\begin{equation}\tag{1}ds_{(5)}^2 = dt^2 - a^2(t) (dx^2 + dy^2 + dz^2) - b^2(t) \, d\theta^2.\end{equation}Substitute this metric into the 5D Einstein's equation without any matter :\begin{equation}\tag{2}R_{AB}^{(5)} = 0.\end{equation}Then you get as a non-trivial solution these two scale factors :\begin{align}\tag{3}a(t) &= \alpha \, t^{1/2}, & b(t) &= \beta \, t^{- 1/2}.\end{align}This is the same as
pure radiation in an homogeneous 4D spacetime :\begin{equation}\tag{4}ds_{(4)}^2 = dt^2 - a^2(t) (dx^2 + dy^2 + dz^2).\end{equation}With\begin{equation}\tag{5}R_{\mu \nu}^{(4)} - \frac{1}{2} \, g_{\mu \nu}^{(4)} \, R^{(4)} = -\, \kappa \, T_{\mu \nu}^{(4)},\end{equation}and $T_{\mu \nu}^{(4)}$ describing a perfect fluid of incoherent radiation ($p_{rad} = \frac{1}{3} \, \rho_{rad}$):
\begin{equation}\tag{6} T_{\mu \nu}^{(4)} = (\rho_{rad} + p_{rad}) \, u_{\mu}^{(4)} \, u_{\nu}^{(4)} - g_{\mu \nu}^{(4)} \, p_{rad}. \end{equation}
This is the subject of the
, and is extremely fascinating! For more on the Campbell-Magaard theorem and the induced-matter theory : induced matter hypothesis |
Electricalis the branch of physics dealing with electricity, electronics and electromagnetism. Electrical formulas play a great role in finding the parameter value in any electrical circuits. Most commonly used electrical formulas are formulas related to voltage, current, power, resistance etc. Volt is a unit of electrical potential or motive force – potential is required to send one ampere of current through one ohm of resistance. Watt is a unit of electrical energy or power – one watt is the product of one ampere and one volt – one ampere of current flowing under the force of one volt gives one watt of energy
Below are given some commonly used Electrical formulas which may be helpful for you.
Quantity Formula Unit Charge Q = C $\times$ V Coulomb (C) Capacitance
C = $\frac{Q}{V}$
Farad (F) Inductance
V
L = – L $\frac{di}{dt}$ Henry (L or H) Voltage V = I R Volt (V) Current
I = $\frac{V}{R}$
Ampere (A) Resistance
R = $\frac{V}{I}$
ohm ($\omega$) Power P = VI Watt (W) Conductance
G = $\frac{1}{R}$
mho ($mho$) Impedance Z 2 = R 2 + (x L – x c) 2 ohm ($\omega$) Resonant Frequency f = $\frac{1}{2 \pi \sqrt{LC}}$ Hertz (Hz) Electrical Formulashelps us to calculate the parameters related to electricals in any electrical components. Electrical Problems Solved Examples Question 1:A wire carrying a current of 4 Amperes is having resistance of 5 $\omega$. Calculate the potential difference across its ends. Solution:
Given: Current I = 4 A,
Resistance R = 5 $\omega$
The Potential difference is given by V = IR
= 4 A $\times$ 5 $\omega$
= 20 V.
Question 2:Calculate the charge across the capacitor 5mF and the voltage applied is 25 V. Solution:
Given: Capacitance of the capacitor C = 5 mF,
Voltage applied V = 25 V,
The Charge across the capacitor is given by Q = CV
= 5 mF $\times$ 25 V
= 125 $\times$ 10
-3C
= 0.125 C. |
2019-10-09 06:01
HiRadMat: A facility beyond the realms of materials testing / Harden, Fiona (CERN) ; Bouvard, Aymeric (CERN) ; Charitonidis, Nikolaos (CERN) ; Kadi, Yacine (CERN)/HiRadMat experiments and facility support teams The ever-expanding requirements of high-power targets and accelerator equipment has highlighted the need for facilities capable of accommodating experiments with a diverse range of objectives. HiRadMat, a High Radiation to Materials testing facility at CERN has, throughout operation, established itself as a global user facility capable of going beyond its initial design goals. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPRB085 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPRB085 Registo detalhado - Registos similares 2019-10-09 06:01
Commissioning results of the tertiary beam lines for the CERN neutrino platform project / Rosenthal, Marcel (CERN) ; Booth, Alexander (U. Sussex (main) ; Fermilab) ; Charitonidis, Nikolaos (CERN) ; Chatzidaki, Panagiota (Natl. Tech. U., Athens ; Kirchhoff Inst. Phys. ; CERN) ; Karyotakis, Yannis (Annecy, LAPP) ; Nowak, Elzbieta (CERN ; AGH-UST, Cracow) ; Ortega Ruiz, Inaki (CERN) ; Sala, Paola (INFN, Milan ; CERN) For many decades the CERN North Area facility at the Super Proton Synchrotron (SPS) has delivered secondary beams to various fixed target experiments and test beams. In 2018, two new tertiary extensions of the existing beam lines, designated “H2-VLE” and “H4-VLE”, have been constructed and successfully commissioned. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW064 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW064 Registo detalhado - Registos similares 2019-10-09 06:00
The "Physics Beyond Colliders" projects for the CERN M2 beam / Banerjee, Dipanwita (CERN ; Illinois U., Urbana (main)) ; Bernhard, Johannes (CERN) ; Brugger, Markus (CERN) ; Charitonidis, Nikolaos (CERN) ; Cholak, Serhii (Taras Shevchenko U.) ; D'Alessandro, Gian Luigi (Royal Holloway, U. of London) ; Gatignon, Laurent (CERN) ; Gerbershagen, Alexander (CERN) ; Montbarbon, Eva (CERN) ; Rae, Bastien (CERN) et al. Physics Beyond Colliders is an exploratory study aimed at exploiting the full scientific potential of CERN’s accelerator complex up to 2040 and its scientific infrastructure through projects complementary to the existing and possible future colliders. Within the Conventional Beam Working Group (CBWG), several projects for the M2 beam line in the CERN North Area were proposed, such as a successor for the COMPASS experiment, a muon programme for NA64 dark sector physics, and the MuonE proposal aiming at investigating the hadronic contribution to the vacuum polarisation. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW063 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW063 Registo detalhado - Registos similares 2019-10-09 06:00
The K12 beamline for the KLEVER experiment / Van Dijk, Maarten (CERN) ; Banerjee, Dipanwita (CERN) ; Bernhard, Johannes (CERN) ; Brugger, Markus (CERN) ; Charitonidis, Nikolaos (CERN) ; D'Alessandro, Gian Luigi (CERN) ; Doble, Niels (CERN) ; Gatignon, Laurent (CERN) ; Gerbershagen, Alexander (CERN) ; Montbarbon, Eva (CERN) et al. The KLEVER experiment is proposed to run in the CERN ECN3 underground cavern from 2026 onward. The goal of the experiment is to measure ${\rm{BR}}(K_L \rightarrow \pi^0v\bar{v})$, which could yield information about potential new physics, by itself and in combination with the measurement of ${\rm{BR}}(K^+ \rightarrow \pi^+v\bar{v})$ of NA62. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW061 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW061 Registo detalhado - Registos similares 2019-09-21 06:01
Beam impact experiment of 440 GeV/p protons on superconducting wires and tapes in a cryogenic environment / Will, Andreas (KIT, Karlsruhe ; CERN) ; Bastian, Yan (CERN) ; Bernhard, Axel (KIT, Karlsruhe) ; Bonura, Marco (U. Geneva (main)) ; Bordini, Bernardo (CERN) ; Bortot, Lorenzo (CERN) ; Favre, Mathieu (CERN) ; Lindstrom, Bjorn (CERN) ; Mentink, Matthijs (CERN) ; Monteuuis, Arnaud (CERN) et al. The superconducting magnets used in high energy particle accelerators such as CERN’s LHC can be impacted by the circulating beam in case of specific failure cases. This leads to interaction of the beam particles with the magnet components, like the superconducting coils, directly or via secondary particle showers. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPTS066 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPTS066 Registo detalhado - Registos similares 2019-09-20 08:41
Performance study for the photon measurements of the upgraded LHCf calorimeters with Gd$_2$SiO$_5$ (GSO) scintillators / Makino, Y (Nagoya U., ISEE) ; Tiberio, A (INFN, Florence ; U. Florence (main)) ; Adriani, O (INFN, Florence ; U. Florence (main)) ; Berti, E (INFN, Florence ; U. Florence (main)) ; Bonechi, L (INFN, Florence) ; Bongi, M (INFN, Florence ; U. Florence (main)) ; Caccia, Z (INFN, Catania) ; D'Alessandro, R (INFN, Florence ; U. Florence (main)) ; Del Prete, M (INFN, Florence ; U. Florence (main)) ; Detti, S (INFN, Florence) et al. The Large Hadron Collider forward (LHCf) experiment was motivated to understand the hadronic interaction processes relevant to cosmic-ray air shower development. We have developed radiation-hard detectors with the use of Gd$_2$SiO$_5$ (GSO) scintillators for proton-proton $\sqrt{s} = 13$ TeV collisions. [...] 2017 - 22 p. - Published in : JINST 12 (2017) P03023 Registo detalhado - Registos similares 2019-04-09 06:05
The new CGEM Inner Tracker and the new TIGER ASIC for the BES III Experiment / Marcello, Simonetta (INFN, Turin ; Turin U.) ; Alexeev, Maxim (INFN, Turin ; Turin U.) ; Amoroso, Antonio (INFN, Turin ; Turin U.) ; Baldini Ferroli, Rinaldo (Frascati ; Beijing, Inst. High Energy Phys.) ; Bertani, Monica (Frascati) ; Bettoni, Diego (INFN, Ferrara) ; Bianchi, Fabrizio Umberto (INFN, Turin ; Turin U.) ; Calcaterra, Alessandro (Frascati) ; Canale, N (INFN, Ferrara) ; Capodiferro, Manlio (Frascati ; INFN, Rome) et al. A new detector exploiting the technology of Gas Electron Multipliers is under construction to replace the innermost drift chamber of BESIII experiment, since its efficiency is compromised owing the high luminosity of Beijing Electron Positron Collider. The new inner tracker with a cylindrical shape will deploy several new features. [...] SISSA, 2018 - 4 p. - Published in : PoS EPS-HEP2017 (2017) 505 Fulltext: PDF; External link: PoS server In : 2017 European Physical Society Conference on High Energy Physics, Venice, Italy, 05 - 12 Jul 2017, pp.505 Registo detalhado - Registos similares 2019-04-09 06:05
CaloCube: a new homogenous calorimeter with high-granularity for precise measurements of high-energy cosmic rays in space / Bigongiari, Gabriele (INFN, Pisa)/Calocube The direct observation of high-energy cosmic rays, up to the PeV region, will depend on highly performing calorimeters, and the physics performance will be primarily determined by their acceptance and energy resolution.Thus, it is fundamental to optimize their geometrical design, granularity, and absorption depth, with respect to the total mass of the apparatus, probably the most important constraints for a space mission. Furthermore, a calorimeter based space experiment can provide not only flux measurements but also energy spectra and particle identification to overcome some of the limitations of ground-based experiments. [...] SISSA, 2018 - 5 p. - Published in : PoS EPS-HEP2017 (2017) 481 Fulltext: PDF; External link: PoS server In : 2017 European Physical Society Conference on High Energy Physics, Venice, Italy, 05 - 12 Jul 2017, pp.481 Registo detalhado - Registos similares 2019-03-30 06:08 Registo detalhado - Registos similares 2019-03-30 06:08 Registo detalhado - Registos similares |
This is problem 2.8.3 from Miller's
Quantum Mechanics For Scientists And Engineers. I'm getting stuck when I try to figure out the wave equation on the right-hand side of the barrier. The original problem is:
Graph the (relative) probability density as a function of distance for an electron wave of energy 1.5 eV incident from the left on a barrier of height 1 eV. Continue your graph far enough in distance on both sides of the barrier to show the characteristic behavior of this probability density.
In the text, Miller solves a similar problem where the energy of the incident particle is less than the energy of the barrier. However, in this case, since the particle's energy is greater than the energy of the barrier, I'm using the same form of the wave function on either side of the barrier. Namely,
$$ \psi_{left}(x) = C \; exp(ik_{left}x)+D \; exp(-ik_{left}x) $$ $$ \psi_{right}(x) = G \; exp(ik_{right}x) $$
where the potential is zero for $x \lt 0$ and 1 eV for $x \ge 0.$ Substituting the above into Schrodinger's equation produces $k_{left}=\sqrt{\frac{2m(E-V_{left})}{\hbar^2}}$ and $k_{right}=\sqrt{\frac{2m(E-V_{right})}{\hbar^2}}.$ Using the boundary conditions that $ \psi_{left}(0)=\psi_{right}(0) $ and $\psi_{left}^\prime(0)=\psi_{right}^\prime(0),$ we obtain,
$$ C+D=G $$ $$ ik_{left}C-ik_{left}D=ik_{right}G. $$
From this, we can solve for $D$ and $G$ to obtain,
$$ D=\frac{k_{left}-k_{right}}{k_{left}+k_{right}}C $$ $$ G=\frac{2k_{left}}{k_{left}+k_{right}}C. $$
Substituting back into the wave functions gives,
$$ \psi_{left}(x) = C \; exp(ik_{left}x)+\frac{k_{left}-k_{right}}{k_{left}+k_{right}}C \; exp(-ik_{left}x) $$ $$ \psi_{right}(x) = \frac{2k_{left}}{k_{left}+k_{right}}C \; exp(ik_{right}x). $$
A this point, I feel like I've done something wrong, since $|\psi_{left}|^2$ is a sinusoidal standing wave, but $|\psi_{right}|^2$ is constant. Have I made a mistake? Will an incident electron with energy greater than that of an infinitely long potential barrier have a probability distribution that is sinusoidal before the barrier and constant after the barrier?
Also, although the original problem uses the word "relative", I'm still troubled by the fact that even in a potential well situation (which is different from the statement of the problem) the above solution would lead to a contradiction. Namely, a finite and constant probability distribution outside of the well would not tend to zero at plus infinity. Can someone either confirm this reasoning, or perhaps point out where I made a mistake?
Please note that I've tagged this question as homework, although it does not pertain to any actual homework assignment. I'm just reading on my own. |
Let us denote this set of edges conatined in some MST by $M\subseteq E$ and let $w: \mathcal{P} E \to \mathbb{N}_0$ denote the weight function, i.e. for some set of edges $S\subseteq E$, $w(S)$ denotes the sum of the weights of the edges in $S$. I'm using the variables from http://en.wikipedia.org/wiki/Kruskal%27s_algorithm#Pseudocode.
You do not need to compute all possible MSTs wit h$e$ as an edge. You only need one MST and another (as small as possible) spanning tree containing $e$:
To check whether some $e\in E(G)$ is in $M$, you can do the following: compute just any MST $T$. Then start kruskal again: But as the initial set of edges in the tree don't use $A:=\emptyset$ but $A:=\{e\}$. Starting with that, compute a spanning tree $T_e$. Of course, it is $T_e$. Now we have:$$ e \in M \Leftrightarrow T_e\text{ is a MST} \Leftrightarrow w(T_e) = w(T).$$
If you want to check it for multiple candidates $e_1,\ldots,e_n$ you need to compute $T$ only once and you can abort computing $T_e$ if the weight of the tree you build already exceeds $w(T)$.
So computing the whole $M$ takes $\mathcal{O}(E^2 \cdot \log E)$. I'm not sure whether this is the optimal complexity for this problem. |
I was writing lecture notes for a class and wanted to provide the students with a version of the lecture notes that they can fill in. In particular, I wanted to have some examples, where I can show the question, but not the derivation. So I created a new environment called ‘example’. I also added in a counter so that the examples would be numbered and can be referenced (with the usual \ref{}).
I had to include these packages:
\usepackage{comment} \usepackage{ifthen} \usepackage{color}
The counter is defined by:
\newcounter{example} \def\theexample{\thechapter-\arabic{example}}
so example is the counter, but the label will reference things by chapter number and example.
To choose whether to show examples:
\newboolean{showexamples} \setboolean{showexamples}{false} % set this to true or false
The environment is as follows:
\ifthenelse{\boolean{showexamples}}{% \newenvironment{example}[3] {\refstepcounter{example} \clearpage\vspace{1ex}\hrule\vspace*{1ex}\noindent EXAMPLE \theexample: #2\\ #3 \\} {\vspace{1ex}\hrule\vspace{1ex}} } {% \newenvironment{example}[3] {\refstepcounter{example} \clearpage\vspace{1ex}\hrule\vspace*{1ex}\noindent EXAMPLE \theexample: #2\\ #3 \vspace*{\dimexpr#1}\begingroup\color{white}} {\endgroup\vspace{1ex}\hrule\vspace{1ex}} }
It takes 3 arguments: a padding size, a description of the example, and a figure to place for the example (optional). If one chooses to show the example, it shows all the body in the example environment. If one chooses to hide the example, it only shows the title, figure, some blank space, and some additional blank space (determined by the padding). The way the body is ‘hidden’ is by setting the text color to white.
I also defined the following environment for a captioned figure:
\newenvironment{capfig}[2]{\begin{figure}[!h]\center\includegraphics[width=0.5\textwidth]{#1}\caption{#2}\end{figure}}{}
it takes 2 arguments (the filename with the figure and the caption).
Here is an example use:
\begin{example}{0pt}{The Compound Pendulum is a typical example of a simple system that is awkward to describe using vectorial analysis, but straightforward using analytic mechanics.}{\capfig{figures/CompoundPenduluum.png}{Vectorial analysis of the compound pendulum}} In order to describe the motion of the two masses as a function of time, we start by defining an origin, and label the coordinates of mass $i$ with $x_i$ and $y_i$. We wish to find the functions $x_i(t)$ and $y_i(t)$ given initial conditions on the positions and velocities of the masses. We have 6 unknowns (the accelerations in $x$ and $y$ of the two masses, and the two string tensions). \begin{align*} m_1\vec{a}_1&=\vec{T}_1+\vec{T}_2+m_1\vec{g}\\ m_2\vec{a}_2&=\vec{T}_2+m_2\vec{g}\\ \end{align*} Constraints from in-extensible strings: \begin{align*} m_1g\cos{\theta}+T_2\cos{\phi}\cos{\theta}&=T_1\\ m_2g\cos{\phi}&=T_2\\ \end{align*} The angles are given by: \begin{align*} \cos{\theta}&=\frac{x_1}{L_1}\\ \cos{\phi}&=\frac{x_2-x_1}{L_2}\\ \end{align*} \label{ex:vectorCompPend} \end{example} |
Can anyone help with the following (slightly weird) random walk question? I have a random walk starting at $X_0 = 1 = S_0$ where the steps $X_i$ are independent uniform random variates in $[-1,1]$. The entire walk is conditioned on two hypotheses:
The walk terminates after $n$ steps at $0$ (so $S_n = X_0 + X_1 + \cdots + X_n = 0$). Each partial sum $S_k \geq \frac{X_k - 1}{2}$. Alternatively, the weighted partial sum $2 X_0 + 2 X_1 + 2 X_2 + \cdots + 2 X_{k-1} + X_k \geq -1$.
I'm interested in the variable $M_n = \max_{0 \leq i \leq n} S_i$.
What's the distribution of $M_n$? Can one find upper or lower bounds on $P[M_n < a]$?
This came up in the same project as Distribution of maximum of random walk conditioned to stay positive, which is certainly more standard. For this one, I completely don't know whether this is standard or difficult.
I've looked up some standard stuff (eg. on sequential sampling) where you have a boundary condition given by absorbing boundaries at $0$ and $a$, but the weighted sum seems to make things harder. Again, I'd be very happy to learn that this is a standard thing with a good reference, or for advice as well as complete solutions. Any thoughts? |
I know all of the following are used for showing the absolute value:
$\lvert x \rvert$
$| x |$
$\mid x \mid$
I myself think that there is another one
\abs; but I can't use this last one.
What is the difference between them?
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I know all of the following are used for showing the absolute value:
$\lvert x \rvert$
$| x |$
$\mid x \mid$
I myself think that there is another one
\abs; but I can't use this last one.
What is the difference between them?
You ask for the differences and here's an analysis for them.
First let's get rid of the last case:
\mid is a relation symbol like
= or
< and TeX adds spaces around it that disqualify the command from being used for the absolute value.
The other two seem good, but the first one is better as it doesn't need precautions. The fact is that
| is considered an ordinary symbol, whereas
\lvert is an opening (like
[) and
\rvert is a closing (like
]).
Let's compare some cases.
\documentclass{article}\usepackage{amsmath}\begin{document}$\begin{array}{lllll}\text{good} & \lvert x\rvert\le 1 & \lvert-1\rvert=1 & \lvert\sin x\rvert & \log\lvert x-1\rvert \\[3ex]\text{so and so} & |x|\leq 1 & |-1|=1 & |\sin x| & \log|x-1| \\[3ex]\text{wrong} & \mid x\mid \leq 1 & \mid -1\mid =1 & \mid \sin x\mid & \log\mid x-1\mid\end{array}$\end{document}
You can see in the picture that
\lvert and
\rvert produces the right spacing in all cases, whereas
|...| doesn't. Not to speak about
\mid.
Using
| would force you to input
|{-1}||{\sin x}|\log\!|x-1|
You can
define
\abs as suggested in answers to Absolute Value Symbols but I recommend
to follow the suggestion of making not
\left and
\right automatic, because it will most of the time choose a wrong size. |
First, observe that it suffices to show $$\sum_{i=1}^n w_ia_i \ge \left( \sum_{i=1}^n a_i^p\right)^{1/p}$$for all sequences satisfying your conditions and $a_1=1$.
Now let $(w_i)$ and $(a_i)$ be sequences satisfying the non-negativity and monotonicity properties.
Let $n\ge 1$. Define $$f_n (a_n) = \left(\sum_{i=1}^n w_ia_i\right)^p - \sum_{i=1}^n a_i^p.$$Let us study the monotonicity of $a_n\mapsto f(a_n)$. It holds$$\begin{split}f_n'(a_n)& = p \left(\sum_{i=1}^n w_ia_i\right)^{p-1}w_n - p a_n^{p-1}\\& \ge p \left(n w_na_n\right)^{p-1}w_n - p a_n^{p-1} \\&= pa_n^{p-1}(n^{p-1} w_n^p-1).\end{split}$$This implies that $a_n\mapsto f(a_n)$ is monotonically increasing if $$w_n \ge n^{-(p-1)/p}.$$Now, we will show that the choice $$w_n:=\frac1{n^{\frac{p-1}p}}$$is sufficient to prove the inequality.
We do this by induction with respect to $n$. For $n=1$ the inequality is trivially true as $w_1=1$. Suppose the inequality holds for some $n\ge 1$ and all feasible sequences.
Now let $a_1 \ge a_2 \ge \dots \ge a_{n+1}\ge0 $ be given. Consider again the function $f_{n+1}$ from above with the values of $a_1\dots a_n$ being fixed.From the induction hypothesis, we obtain that $f_{n+1}(0)\ge0$.Since by our choice of $w_i$ the function $a \mapsto f_{n+1}(a)$ is monotonically increasing, this implies $f_{n+1}(a_{n+1}) \ge0$, which is equivalent to$$\sum_{i=1}^{n+1} w_ia_i \ge \left( \sum_{i=1}^{n+1} a_i^p\right)^{1/p}.$$And the claim is proven.
If $p=1$ there cannot exists a sequence $(w_i)$ satisfying all conditions. If $a_i=1$ for all $i$, then the inequality is equivalent to$$0 \le \left(\sum_{i=1}^n w_i \right)-n = \sum_{i=1}^n (w_i-1)$$for all $n$. If $w_i>0$ and $\lim w_i=0$, then the right-hand side tends to $-\infty$ for $n\to \infty$. |
Consider a two period OLG model where each young agent recieves an endowment of $w$ units of the single commodity good in the 1st period of his/her life, and nothing in the second period.
Each period the government consumes the fraction $\bar{g}$ of the total endowment. $\bar{g}$ is i.i.d with mean $\bar{g}$. The government's expenditures are financed by lump-sum taxes, money and one-period discount bonds. $M_{t}$ and $B_{t}$ are the supplies of money and nominal bonds at the end of period $t$. The government's budget constraint is then:
$\frac{B_{t-1}}{p_{t}} + \bar{g}w = \tau_{y}(t)+\tau_{o}(t) + \frac{M_{t}-M_{t-1}}{p_{t}} + \frac{B_{t}}{(1+i_{t})p_{t}}$
$\tau_{i}(t)$ is the lump-sum tax where $i$ is either the old or young generation.
My only problem with this budget constraint is the $\frac{B_{t-1}}{p_{t}}$ component. I understand that it is the current real obligation but I don't understand why there is no interest component. I am self-studying OLG models so my reasoning might be flawed but here is how I see this constraint:
The RHS is the government's revenue in period $t$, which consists of tax revenue from the young and the old, seigniorage and new bond issuance. The LHS is the expenditure of the government in period $t$. It can spend the income on consumption ($\bar{g}w$), to pay off the existing debt ( $\frac{B_{t-1}}{p_{t}}$) but what about the interest on the existing debt?
I am used to seeing government constraints in the form of:
$g_{t} + r_{t-1}b_{t-1} = t_{t}+(b_{t}-b_{t-1}) + h_{t} - \frac{h_{t-1}}{1 + \pi _{t}}$
This is almost identical to the author's budget constraint only that it includes an interest component. What am I missing here?
For reference, the budget constraint was obtained from Aiyagari and Gertler (1985) |
Keeping Your Perspective
Just keep in mind what's being asked. The question is asking about the phase of the current, relative to the phase of the voltage. For a parallel circuit, you want to look at the admittance/susceptance/conductance of each device, which is the inverse of their impedance/reactance/resistance. Just a cursory glance at your values and you can see that the three admittances look like this:
$$\frac{1}{6\:\text{k}\Omega}+\frac{1}{-4\:\text{k}\Omega\:j}+\frac{1}{2\:\text{k}\Omega\:j}$$
It's quick to see that the inductance's susceptance dominates this expression.
Here's a quick manipulation of the above expression:
$$\begin{align*}\frac{1}{6\:\text{k}\Omega}+\frac{1}{-4\:\text{k}\Omega\:j}+\frac{1}{2\:\text{k}\Omega\:j}&=\frac{1}{6\:\text{k}\Omega}+\frac{1}{-4\:\text{k}\Omega\:j}\cdot\frac{1.5\:j}{1.5\:j}+\frac{1}{2\:\text{k}\Omega\:j}\cdot\frac{3\:j}{3\:j}\\\\&=\frac{1}{6\:\text{k}\Omega}+\frac{1.5\:j}{6\:\text{k}\Omega}+\frac{-3\:j}{6\:\text{k}\Omega}\\\\&=\frac{1-1.5\:j}{6\:\text{k}\Omega}\end{align*}$$
At this point it is helpful to remember the following picture:
To remain consistent with the
, the Impedance Triangle 's angles rotate in the opposite direction around the circle. (The hypotenuse of the left triangle is the Admittance Triangle and the hypotenuse of the right triangle is the Impedance .) Admittance Getting the Answer
From the above expression which contains a negative susceptance, the resulting angle will be positive. (See the triangle on the right, instead of the left.) The divisor isn't important for this calculation since it scales both factors in the numerator equally. So all you need recognize is that you have a ratio of \$\frac{\text{susceptance}}{\text{conductance}}=\frac{-1.5}{1}=-1.5\$. Since the admittance angle rotates in the opposite direction to the usual mathematical standard, you negate this result.
So the answer is \$\theta=\operatorname{ATAN}\left[-\left(-1.5\right)\right]= 0.982793723\:\text{rad}\$. The angle is positive, as expected, and not negative (given that the inductance dominates the admittances here.) |
Performing Topology Optimization with the Density Method
Engineers are given significant freedom in their pursuit of lightweight structural components in airplanes and space applications, so it makes sense to use methods that can exploit this freedom, making topology optimization a popular choice in the early design phase. This method often requires regularization and special interpolation functions to get meaningful designs, which can be a nuisance to both new and experienced simulation users. To simplify the solution of topology optimization problems, the COMSOL® software contains a density topology feature.
About the Density Method for Topology Optimization
As the name suggests, topology optimization is a method that has the ability to come up with new and better topologies for an engineering structure given an objective function and set of constraints. The method comes up with these new topologies by introducing a set of design variables that describe the presence, or absence, of material within the design space. These variables are defined either within every element of the mesh or on every node point of the mesh. Changing these design variables thus becomes analogous to changing the topology. This means that holes in the structure can appear, disappear, and merge as well as that boundaries can take on arbitrary shapes. In addition, the control parameters are somewhat automatically defined and tied to the discretization.
As of COMSOL Multiphysics® software version 5.4, the add-on Optimization Module includes a density topology feature to improve the usability of topology optimization. The feature is designed to be used as a density method (Ref. 3), meaning that the control parameters change a material parameter through an interpolation function. Interpolation functions for solid and fluid mechanics are built into the feature and used in example models throughout the Application Library in COMSOL Multiphysics.
A bracket geometry is topology optimized, leaving only 50% of the material, which contributes the most to the stiffness. The printed bracket geometry.
The density method involves the definition of a control variable field, \theta_c, which is bounded between 0 and 1. In solid mechanics, \theta_c=1 corresponds to the material from which the structure is to be built, while \theta_c=0 corresponds to a very soft material. By default, the void Young’s modulus is 0.1% of the solid Young’s modulus. In fluid mechanics, convention dictates that \theta_c=1 corresponds to fluid, while \theta_c=0 is a (slightly) permeable material with an inverse permeability factor, \alpha; i.e., a damping term is added to the Navier-Stokes equation:
The damping term is 0 in fluid domains, while a large value is used in solid domains. These different values give a good approximation of the no-slip boundary condition on the interface between the domains.
An Introduction to the Density Model Feature
The
Density Model feature supports regularization via a Helmholtz equation (Ref. 1). This introduces a minimum length scale using the filter radius, R_\mathrm{min}:
Here, \theta_c is the raw control variable, which is modified by the optimizer, and \theta_f is the filtered variable. The mesh edge size is the default value for the filter radius. While this works well in terms of regularizing the optimization problem, it’s important to set a fixed length (larger than the mesh edge size) to get mesh-independent results.
Top: The equation for the Helmholtz filter can be solved analytically for a 1D Heaviside function. Bottom: This plot is taken from the MBB beam optimization model. It shows the raw control variables to the left and the filtered version to the right.
The Helmholtz filter gives rise to significant grayscale, which does not have a clear physical interpretation. The grayscale can be reduced by applying a smooth step function in what is referred to as
projection in topology optimization. Projection reduces grayscale, but it also makes it more difficult for the optimizer to converge. The density topology feature supports projection based on the hyperbolic tangent function, and the amount of projection can be controlled with the projection steepness, \beta.
Here, \theta_{\beta} is the projection point.
Plot showing the filtered field to the left and the projected field to the right.
Projection makes it possible to avoid grayscale, but grayscale can still appear if the optimization problem favors it. If the same interpolation function is used for the mass and the stiffness, grayscale is optimal in volume-constrained minimum compliance problems. It is thus common to use interpolation functions that cause intermediate values to be associated with little stiffness relative to their cost (compared to the fully solid value). You can think of this as a
penalization of intermediate values for the material volume factor, and the Density Model interface (shown below) supports two such interpolation schemes for solid mechanics: solid isotropic material with penalization (SIMP) and rational approximation of material properties method (RAMP) interpolation. Darcy interpolation is provided for fluid mechanics. The interpolated variable is called the penalized material volume factor, \theta_p, and is used for interpolating the material parameters, e.g., for SIMP interpolation, the p_\textsc{simp} exponent can be increased to reduce the stiffness of intermediate values, so that grayscale becomes less favorable.
\theta_p %26= \theta_\mathrm{min}(1-\theta_\mathrm{min})\theta^{p_\textsc{simp}}\\
E_p %26= E\theta_p
\end{align}
Here, E is the Young’s modulus of the solid material and E_p is the penalized Young’s modulus to be used throughout all optimized domains.
The Density Model feature is available under Topology Optimization in Component > Definitions . The mesh edge length is taken as the default filter radius and it works well, but it has to be replaced with a fixed value in order to produce mesh-independent results.
The penalized Young’s modulus can be defined as a domain variable, or (as in the case of the bracket model) it can be defined directly in the materials.
Topology optimization with the density method involves varying the Young’s modulus spatially. In this case, it is achieved by going to the material properties and multiplying the solid Young’s modulus with the penalized material volume factor, dtopo1.theta_p.
In summary, the density topology feature adds four variables. The filtered material volume factor is defined implicitly using a dependent variable.
Symbol Description Equation \theta_c Control material volume factor 0\leq\theta_c\leq1 \theta_f Filtered material volume factor \theta_f = R_\mathrm{min}^2\mathbf{\nabla}^2\theta_f + \theta_c \theta Material volume factor \theta = \frac{\tanh(\beta(\theta_f-\theta_{\beta}))+\tanh(\beta\theta_{\beta})}{\tanh(\beta(1-\theta_{\beta}))+\tanh(\beta\theta_{\beta})} \theta_p Penalized material volume factor \theta_p = \theta_\mathrm{min}+(1-\theta_\mathrm{min})\theta^{p_\textsc{simp}} or \theta_p = \frac{q_\mathrm{Darcy}(1-\theta)}{q_\mathrm{Darcy}+\theta}
When the filtering is disabled, the filtered variable becomes undefined and the projection instead uses the control material volume factor directly. If the projection is disabled, the material volume factor still exists, but it becomes identical to the projection input.
Applying Continuation to Avoid Local Minima
When the topology is not too complicated, the default values of the density topology feature work well. This is the case for the MBB beam optimization and topology optimized hook models. If the optimal design is more complicated (such as for the bracket example shown at the top of this post), there might be many local minima. To avoid these minima, you can use continuation in the SIMP exponent and the projection slope. This can be achieved by modifying the initial value expression in the density topology feature and adding a
Parametric Sweep feature, as shown below. As a result, the solver ramps over the specified parameters, using the optimum from the previous case as the initial value for the next optimization step. That is, it starts with a small SIMP exponent and projection slope and then continues to higher values. It is possible to apply continuation by combining a parametric sweep with a study reference. See the Bracket — Topology Optimization tutorial model for details. Objectives and Constraints in Topology Optimization
If the geometry is optimized for a single load case (as shown below to the left), the resulting design will be optimal with respect to that load case. This can seem obvious, but often designers make assumptions about symmetries and the design topology. Unless these assumptions are formalized as constraints, they will not be respected. Therefore, the design shown to the right below uses eight load cases (two load groups times four constraint groups).
Left: The bracket geometry is optimized for a single load case, resulting in an asymmetric design with two loosely connected halves. Right: The bracket geometry with eight load cases.
Designers often have several objectives that need to be weighted. To make an informed decision about these objectives, a designer can trace the Pareto optimal front using several optimizations with different weights.
The Pareto optimal front for the bracket geometry can be traced by varying the weight in a parametric sweep. Animation of the topology optimized bracket. (Download the glTF™ file from the Application Gallery in GLB-file format to rotate the geometry yourself.) Exporting and Importing Topology Optimization Results
It is possible to analyze the result of a topology optimized design with respect to stress concentration and buckling without remeshing. However, if you want to be completely sure that the void phase does not play a role, you can eliminate it by exporting and importing the resulting design, as shown below. The details of this procedure are discussed in a previous blog post.
The contour (left) for the topology optimized MBB beam design is exported and imported as an interpolation curve (right). Next Steps
To learn more about the built-in tools and features for solving optimization problems, check out the Optimization Module product page by clicking the button below.
Further Resources Try using the density feature for topology optimization with these example models: Read more about topology optimization on the COMSOL Blog: References B.S. Lazarov and O. Sigmund, “Filters in topology optimization based on Helmholtz‐type differential equations,” International Journal for Numerical Methods in Engineering, vol. 86, no. 6, pp. 765–781, 2011. F. Wang, B.S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Structural and Multidisciplinary Optimization, vol. 43, pp. 767–784, 2011. M.P. Bendsøe, “Optimal shape design as a material distribution problem,” Structural Optimization, vol. 1, pp. 193–202, 1989.
glTF and the glTF logo are trademarks of the Khronos Group Inc. Comments (0) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
Set of natural numbers: \(\mathbb{N}\)
Set of integers: \(\mathbb{Z}\) Set of rational numbers: \(\mathbb{Q}\) Set of real numbers: \(\mathbb{R}\) Bases: \(a\), \(b\) Powers (exponents): \(n\), \(m\), \(r\), \({r_n}\), \(\beta\), \(u\), \(v\)
Set of integers: \(\mathbb{Z}\)
Set of rational numbers: \(\mathbb{Q}\)
Set of real numbers: \(\mathbb{R}\)
Bases: \(a\), \(b\)
Powers (exponents): \(n\), \(m\), \(r\), \({r_n}\), \(\beta\), \(u\), \(v\)
Natural numbers: \(n\), \(m\), \(q\)
Integers: \(p\) Rational numbers: \(r\), \({r_n}\) Irrational numbers: \(\beta\) Real numbers: \(u\), \(v\)
Integers: \(p\)
Rational numbers: \(r\), \({r_n}\)
Irrational numbers: \(\beta\)
Real numbers: \(u\), \(v\)
The power of a real number \(a\) with a natural exponent \(n\) is defined as \({a^n} = \underbrace {a \cdot a \ldots a}_{n\text{ times}},\) where \(a \in \mathbb{R},\) \(n \in \mathbb{N}\). Multiplication of powers with the same base \({a^n}{a^m} = {a^{n + m}}\) Division of powers with the same base \({a^n}/{a^m} = {a^{n – m}}\) \(\left( {n \gt m} \right)\) Power of a product \({\left( {ab} \right)^n} = {a^n}{b^n}\) Power of a quotient \({\left( {\large\frac{a}{b}}\normalsize \right)^n} = {\large\frac{{{a^n}}}{{{b^n}}}\normalsize}\;\) \(\left( {b \ne 0} \right)\) Raising a power to another power \({\left( {{a^n}} \right)^m} = {a^{nm}}\) \({0^n} = 0\) \({1^n} = 1\) \({a^1} = a\) Raising a negative number to an even power \({\left( { – a} \right)^{2n}} = {a^{2n}}\;\) \(\left( {a \gt 0} \right)\) Raising a negative number to an odd power \({\left( { – a} \right)^{2n + 1}} = – {a^{2n + 1}}\;\) \(\left( {a \gt 0} \right)\) Zero power \({a^0} = 1\;\;\left( {a \ne 0} \right)\) The expression \({0^0}\) is not defined. Negative power \({a^{ – r}} = 1/{a^r},\) where \(r \in \mathbb{Q},\) \(a \ne 0\). Properties of powers with rational exponents The power of a positive real number \(a\) with a rational exponent \(p/q\) is defined as \({a^{p/q}} = \sqrt[\large q\normalsize]{{{a^p}}},\) where \(a \ge 0,\) \(p \in \mathbb{Z},\) \(q \in \mathbb{N}\). Definition of power with an irrational exponent \(\beta:\) \({a^\beta } = \lim\limits_{{r_n} \to \beta } {a^{{r_n}}},\) where \({r_n}\) is an arbitrary sequence of rational numbers converging to the exponent \(\beta\). For any real exponents \(u\), \(v\) under condition of \(a \gt 0\) and \(b \gt 0\) the following operations with powers are valid: \({a^u}{a^v} = {a^{u + v}},\) \({\left( {{a^u}} \right)^v} = {a^{uv}},\) \({a^{ – u}} = 1/{a^u},\) \(\large\frac{{{a^u}}}{{{a^v}}}\normalsize = {a^{u – v}},\) \({\left( {ab} \right)^u} = {a^u}{b^u},\) \({\left( {\large\frac{a}{b}}\normalsize \right)^u} = \large\frac{{{a^u}}}{{{b^u}}}\normalsize .\) |
No.
Angular momentum is a manifestation of linear momentum at a distance. For a single particle of small mass $m_i$ and velocity $\boldsymbol{v}_i$ the linear and angular momentum (about the origin) is
$$ \begin{aligned} \boldsymbol{p}_i & = m_i \boldsymbol{v}_i \\ \boldsymbol{L}_i & = \boldsymbol{r}_i \times \boldsymbol{p}_i\end{aligned} $$
where $\boldsymbol{r}_i$ is the position vector of the particle from the origin.
Now in the context of rigid bodies we split the velocity of each particle into the velocity of the center of mass, and a rotation about the center of mass. Without loss of generality we can move the origin at the center of mass to simplify the equations.
The particle speed is thus $ \boldsymbol{v}_i = \boldsymbol{v}_{C} + \boldsymbol{\omega} \times \boldsymbol{r}_i$.
The total linear and angular momentum are now
$$ \require{cancel} \begin{aligned} \boldsymbol{p} = \sum_i \boldsymbol{p}_i & = \sum_i m_i (\boldsymbol{v}_{C}+ \boldsymbol{\omega} \times \boldsymbol{r}_i) = m\, \boldsymbol{v}_C + \boldsymbol{\omega} \times \cancel{\left(\sum_i m_i \boldsymbol{r}_i \right)} \\ \boldsymbol{L}_C = \sum_i \boldsymbol{L}_i & = \sum_i \boldsymbol{r}_i \times m_i (\boldsymbol{v}_{C}- \boldsymbol{r}_i \times \boldsymbol{\omega}) = \cancel{\left(\sum_i m_i \boldsymbol{r}_i \right)} \times \boldsymbol{v_C} - \sum_i m_i\, \boldsymbol{r}_i \times (\boldsymbol{r}_i \times \boldsymbol{\omega}) \\\end{aligned} $$
Where $m=\sum_i m_i$ and $\sum_i m_i \boldsymbol{r}_i = 0$ from the definition of the Center of Mass.
And that is how we defined the mass moment of inertia tensor about the origin $\boldsymbol{L}_C = \mathrm{I}_C \boldsymbol{\omega}$ or
$$ \mathrm{I}_C = \sum_i \left(-m_i [\boldsymbol{r}_i \times][\boldsymbol{r}_i \times] \right) = \sum_i m_i \left| \matrix{y_i^2+z_i^2 & -x_i y_i & -z_i x_i \\ -x_i y_i & x_i^2+z_i^2 & -y_i z_i \\ -z_i x_i & -y_i z_i & x_i^2+y_i^2} \right| $$
Thus by definition angular momentum as the moment of momentum (linear momentum at a distance) we derive the equations of motion at the center of mass
$$\left. \begin{aligned} \boldsymbol{p} & = m\, \boldsymbol{v}_C \\ \boldsymbol{L}_C & = \mathrm{I}_C \boldsymbol{\omega}\end{aligned}\, \right\} \begin{aligned} \boldsymbol{F}_{\rm net} & = \frac{\rm d}{{\rm d}t} \boldsymbol{p} = m\, \boldsymbol{a}_C \\ \boldsymbol{\tau}_{\rm net} & = \frac{\rm d}{{\rm d}t} \boldsymbol{L}_C = \mathrm{I}_C \boldsymbol{\alpha} + \boldsymbol{\omega} \times \boldsymbol{L}_C\end{aligned} $$
Now let's look at the relationship between linear velocity $\boldsymbol v$ and rotational velocity $\boldsymbol \omega$, as well as between angular momentum $\boldsymbol{L}$ and linear momentum $\boldsymbol p$ and finally between torque $\boldsymbol \tau$ and force $\boldsymbol F$:
$$\begin{array}{r|c|c|l} \mbox{quantity} & \mbox{direction vector} & \mbox{moment vector} & \mbox{transformation law} \\ \hline \mbox{motion} & \boldsymbol{\omega} & \boldsymbol{v} = \boldsymbol{r} \times \boldsymbol{\omega} & \boldsymbol{v}_A = \boldsymbol{v}_B + (\boldsymbol{r}_A - \boldsymbol{r}_B) \times \boldsymbol{\omega} \\ \hline \mbox{momentum} & \boldsymbol{p} & \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p} & \boldsymbol{L}_A = \boldsymbol{L}_B + (\boldsymbol{r}_A - \boldsymbol{r}_B) \times \boldsymbol{p} \\ \hline \mbox{loading} & \boldsymbol{F} & \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} & \boldsymbol{\tau}_A = \boldsymbol{\tau}_B + (\boldsymbol{r}_A - \boldsymbol{r}_B) \times \boldsymbol{F} \\ \end{array}$$
Which leads to the following definitions:
Linear velocity is the moment of rotation. Angular momentum is the moment of momentum.
Torque
is the moment of force.
Together a . direction vector and a moment vector describe a line in space
This line is often called the
, or axis of rotation of force and finally the line of action for momentum. axis of percussion
Please
read this answer to get more details on how forces or impulses affect various rigid bodies. |
Consider a plane curve defined by the equation \(y = f\left( x \right).\) Suppose that the tangent line is drawn to the curve at a point \(M\left( {x,y} \right).\) The tangent forms an angle \(\alpha\) with the horizontal axis (Figure \(1\text{).}\)
At the displacement \(\Delta s\) along the arc of the curve, the point \(M\) moves to the point \({M_1}.\) The position of the tangent line also changes: the angle of inclination of the tangent to the positive \(x-\text{axis}\) at the point \({M_1}\) will be \(\alpha + \Delta\alpha.\) Thus, as the point moves by the distance \(\Delta s,\) the tangent rotates by the angle \(\Delta\alpha.\) (The angle \(\alpha\) is supposed to be increasing when rotating counterclockwise.)
The absolute value of the ratio \(\large\frac{{\Delta \alpha }}{{\Delta s}}\normalsize\) is called the mean curvature of the arc \(M{M_1}.\) In the limit as \(\Delta s \to 0,\) we obtain the curvature of the curve at the point \(M:\)
\[K = \lim\limits_{\Delta s \to 0} \left| {\frac{{\Delta \alpha }}{{\Delta s}}} \right|.\]
From this definition it follows that the curvature at a point of a curve characterizes the speed of rotation of the tangent of the curve at this point.
For a plane curve given by the equation \(y = f\left( x \right),\) the curvature at a point \(M\left( {x,y} \right)\) is expressed in terms of the first and second derivatives of the function \(f\left( x \right)\) by the formula
\[K = \frac{{\left| {y^{\prime\prime}\left( x \right)} \right|}}{{{{\left[ {1 + {{\left( {y’\left( x \right)} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}.\]
If a curve is defined in parametric form by the equations \(x = x\left( t \right),\) \(y = y\left( t \right),\) then its curvature at any point \(M\left( {x,y} \right)\) is given by
\[K = \frac{{\left| {x’y^{\prime\prime} – y’x^{\prime\prime}} \right|}}{{{{\left[ {{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}.\]
If a curve is given by the polar equation \(r = r\left( \theta \right),\) the curvature is calculated by the formula
\[K = \frac{{\left| {{r^2} + 2{{\left( {r’} \right)}^2} – rr^{\prime\prime}} \right|}}{{{{\left[ {{r^2} + {{\left( {r’} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}.\]
The radius of curvature of a curve at a point \(M\left( {x,y} \right)\) is called the inverse of the curvature \(K\) of the curve at this point:
\[R = \frac{1}{K}.\]
Hence for plane curves given by the explicit equation \(y = f\left( x \right),\) the radius of curvature at a point \(M\left( {x,y} \right)\) is given by the following expression:
\[R = \frac{{{{\left[ {1 + {{\left( {y’\left( x \right)} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}{{\left| {y^{\prime\prime}\left( x \right)} \right|}}.\]
Solved Problems
Click a problem to see the solution.
Example 1Calculate the curvature of the ellipse Example 2Find the curvature and radius of curvature of the parabola \(y = {x^2}\) at the origin. Example 3Find the curvature and radius of curvature of the curve \(y = \cos mx\) at a maximum point. Example 4Calculate the curvature and radius of curvature of the graph of the function \(y = \sqrt x \) at \(x = 1.\) Example 5Consider the curve given by the equation \({y^2} + {x^3} = 0.\) Find its curvature at the point \(\left( { – 1,1} \right).\) Example 6Find the curvature of the cardioid \(r = a\left( {1 + \cos \theta } \right)\) at \(\theta = 0.\) Example 7Find the radius of curvature of the cycloid Example 8Determine the curvature of the curve \(y = \arctan x\) at \(x = 0\) and at infinity. Example 9Determine the least radius of curvature of the exponential function \(y = {e^x}.\) Example 10Find the least radius of curvature of the cubic parabola \(y = {x^3}.\) Example 1.Calculate the curvature of the ellipse
Solution.
Obviously, it suffices to find the curvature of the ellipse at points \(A\left( {a,0} \right)\) and \(B\left( {0,b} \right)\) (Figure \(2\)), because due to the symmetry of the curve, the curvature at the two opposite vertices of the ellipse will be the same.
To calculate the curvature, it is convenient to pass from the canonical equation of the ellipse to the equation in parametric form:
\[x = a\cos t,\;\;\;y = b\sin t,\]
where \(t\) is a parameter. The parameter has the value \(t = 0\) at the point \(A\left( {a,0} \right)\) and is equal to \(t = \large\frac{\pi }{2}\normalsize\) at the point \(B\left( {0,b} \right).\)
Find the first and second derivatives:
\[ {x’ = {x’_t} = {\left( {a\cos t} \right)^\prime } = – a\sin t,}\;\;\;\kern-0.3pt {x^{\prime\prime} = {x^{\prime\prime}_{tt}} = {\left( { – a\sin t} \right)^\prime } }={ – a\cos t;} \]
\[ {y’ = {y’_t} = {\left( {b\sin t} \right)^\prime } = b\cos t,}\;\;\;\kern-0.3pt {x^{\prime\prime} = {x^{\prime\prime}_{tt}} = {\left( { b\cos t} \right)^\prime } = – b\sin t.} \]
The curvature of a parametrically defined curve is expressed by the formulas
\[K = \frac{{\left| {x’y^{\prime\prime} – y’x^{\prime\prime}} \right|}}{{{{\left[ {{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}.\]
Substituting the above derivatives, we get:
\[
{K } = {\frac{{\left| {ab\,{{\sin }^2}t + ab\,{{\cos }^2}t} \right|}}{{{{\left( {{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t} \right)}^{\large\frac{3}{2}\normalsize}}}} } = {\frac{{\left| {ab\left( {{{\sin }^2}t + {{\cos }^2}t} \right)} \right|}}{{{{\left( {{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t} \right)}^{\large\frac{3}{2}\normalsize}}}} } = {\frac{{ab}}{{{{\left( {{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t} \right)}^{\large\frac{3}{2}\normalsize}}}}.} \]
Now we calculate the values of the curvature at the vertices \(A\left( {a,0} \right)\) and \(B\left( {0,b} \right):\)
\[
{K\left( A \right) = K\left( {t = 0} \right) } = {\frac{{ab}}{{{{\left( {{a^2}{{\sin }^2}0 + {b^2}{{\cos }^2}0} \right)}^{\large\frac{3}{2}\normalsize}}}} } = {\frac{{ab}}{{{{\left( {{b^2}} \right)}^{\large\frac{3}{2}\normalsize}}}} } = {\frac{{ab}}{{{b^3}}} } = {\frac{a}{{{b^2}}};} \]
\[
{K\left( B \right) = K\left( {t = \frac{\pi }{2}} \right) } = {\frac{{ab}}{{{{\left( {{a^2}{{\sin }^2}\frac{\pi }{2} + {b^2}{{\cos }^2}\frac{\pi }{2}} \right)}^{\large\frac{3}{2}\normalsize}}}} } = {\frac{{ab}}{{{{\left( {{a^2}} \right)}^{\large\frac{3}{2}\normalsize}}}} } = {\frac{{ab}}{{{a^3}}} } = {\frac{b}{{{a^2}}}.} \] |
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 |
A triangle on a flat plane is described by its angles and side lengths, and you don’t need all of the angles and side lengths to work out everything about the triangle. (This is the same as last time.) However, this time, the triangle will not necessarily have a right angle. This is where more trigonometry comes in. Break out your trig again, people.
I am going to skip over a bit of the problem (that of parsing the input), but I think what’s left is more than enough to make an interesting problem.
Thinking back to high school, triangles are defined by six pieces of information: three sides and three angles:
But really, you only need any three pieces of those (with the exception of the three angles, with just that you have no sense of scale). More specifically, there are a set of relationships between the sides and angles in a triangle such that you can derive all six:
Sum of angles Rule of sines Rule of cosines
\alpha + \beta + \gamma = \pi
\frac{a}{sin\ \alpha} = \frac{b}{sin\ \beta} = \frac{c}{sin\ \gamma}
a^2 = b^2 + c^2 - 2bc\ cos\ \alpha
Sounds good, right? But how do we turn that into code?
First, we need some structure to work with. Perhaps a structure that represents the eventual information about the triangle that we are going to solve
2:
; Represent a triangle as three angles and three sides; Angles can be in either degrees or radians so long as all three are the same; Any value is either numeric or #f if it is currently unknown(struct triangle (∠ɑ ∠β ∠γ a b c) #:transparent)
Fair enough.
Next, how do we know when we’re done? Since we’re representing both angles and sides as numbers, we just need all six fields to be numeric. We could have a giant
and, or we could play with a few neat functions. Specifically
struct->:
; If all six fields are numeric, we've solved the triangle(define (solved? t) (andmap number? (cdr (vector->list (struct->vector t)))))
(If you’d like to follow along, the full code is here: triangle-trouble.rkt)
We need that extra
cdr in there since the first field is struct id (
triangle).
Luckily though, that should be all of the framework we need for the moment. Next, we just need a solver.
At first glance, this might not look like a particularly recursive problem. In recursion, you need to be able to identify one or more base cases (in this case: the triangle is solved) and then break the problem into small steps, each of which is ‘smaller’ than the original problem. So if we can figure out how to at least make some progress / solve one additional side or angle, that should be good enough, right?
On first glance, the easiest solution would be to work out all of the possible cases. Perhaps we know a, b, and ∠ɑ. Perhaps ∠ɑ, ∠β, and c. But that’s a lot of cases. Specifically:
We can do better than that.
Let’s start with the framework:
; Given some of the sides/angles of a triangle try to solve for the rest(define (solve t) (let loop ([t t]) (match t ; We have all three sides and angles, return [(? solved?) t] ; Two angles, solve for the third ... ; Sine rule ... ; Cosine rule ... ; Reorder ...)))
We’re using
match because
match is awesome. More specifically because
match is great at pattern matching against structs. For example, take the first case–two angles, solving for the third:
...; Two angles, solve for the third[(triangle (? number? ∠ɑ) (? number? ∠β) #f a b c) (loop (triangle ∠ɑ ∠β (- pi ∠ɑ ∠β) a b c))]...
Here we’re matching against the six values that make up a triangle. For the first two clauses, we have the form
(? number? ∠ɑ). The
? signifies that we are matching against a predicate,
number?. The last part is the name we’re binding to that value. After that, we’re matching a literal
#f (an unspecified third angle). Finally, we have the three sides. As they are, they will match any values, either numeric or
#f.
So this case matches if and only if the first two angles are solved and the third is not, ignoring the sides for the moment. If this pattern doesn’t match, we’ll move on to the next (we’ll deal with the case that it’s ∠ɑ or ∠β that we’re missing in a bit). Then, all we have to do is calculate the new value of ∠γ. Since Racket’s trig functions work in radians
3 4, we’ll subtract from 180° = π.
Straight forward. Now how about the sine rule. Well, it turns out that there are two cases we can use this for. Since we’re relating matched sides and angles, we can take a matching side and angle along with
either a second side or angle to solve for the other. More specifically, we can solve: ∠ɑ, a, ∠β → b or ∠ɑ, a, b → ∠β. In either case, we need to rearrange the equation slightly: ∠ɑ, a, ∠β → b = \frac{a}{sin\ \alpha} b = \frac{a}{sin\ \alpha}\ sin\ \beta
And another:
∠ɑ, a, ∠β → b \frac{sin\ \beta}{b} = \frac{sin\ \alpha}{a} sin\ \beta = b \frac{sin\ \alpha}{a} \beta = arcsin \bigl( b \frac{sin\ \alpha}{a} \bigr)
Turning that into Racket and we have:
...; Sine rule 1: Matching side/angle + angle, solve for missing side[(triangle (? number? ∠ɑ) (? number? ∠β) ∠γ (? number? a) #f c) (loop (triangle ∠ɑ ∠β ∠γ a (/ (* a (sin ∠β)) (sin ∠ɑ)) c))]; Sine rule 2: Matching side/angle + side, solve for missing angle[(triangle (? number? ∠ɑ) #f ∠γ (? number? a) (? number? b) c) (loop (triangle ∠ɑ (asin (/ (* b (sin ∠ɑ)) a)) ∠γ a b c))]...
After that, we have the law of cosines. From here, we can work with two sides and the mismatched angle or directly with three sides:
∠ɑ, b, c → a a^2 = b^2 + c^2 - 2bc\ cos\ \alpha a = \sqrt{b^2 + c^2 - 2bc\ cos\ \alpha}
Second one:
a, b, c → ∠ɑ b^2 + c^2 - 2bc\ cos\ \alpha = a^2 -2bc\ cos\ \alpha = a^2 - b^2 - c^2 cos\ \alpha = -\frac{a^2 - b^2 - c^2}{2bc} \alpha = arccos\bigl(-\frac{a^2 - b^2 - c^2}{2bc}\bigr)
In Racket:
...; Cosine rule 1: Angle and the other two sides, solve for third side[(triangle (? number? ∠ɑ) ∠β ∠γ #f (? number? b) (? number? c)) (loop (triangle ∠ɑ ∠β ∠γ (sqrt (+ (sqr b) (sqr c) (- (* b c (cos ∠ɑ)))))))]; Cosine rule 2: Three sides, solve for one angle[(triangle #f ∠β ∠γ (? number? a) (? number? b) (? number? c)) (loop (triangle (acos (/ (+ (sqr b) (sqr c) (- (sqr a))) (* 2 b c))) ∠β ∠γ a b c))]...
Those are all of the hard cases. But as I said, we still have to deal with reordering. Here’s where one concession that I’m making compared to the original problem comes into play: I do not care what order the final sides / angles are in, just so long as the triangles are equivalent. So we’re perfectly fine to rotate the sides and angles, just so long as we do it to both:
...; Try another ordering[(triangle ∠ɑ ∠β ∠γ a b c) (loop (triangle ∠β ∠γ ∠ɑ b c a))]
This won’t
quite work though. Can you see why?
Well what happens if the variables are just ordered badly. Right now, it’s possible that none of the cases will trigger in any of these three orderings. What happens then?
Well, right now we’ll keep looping forever.
Oops.
There are a few ways that we can deal with this. We could keep a counter. If we’ve tried three rotations without moving forward, we’re done. But even more elegant (in my mind), we could keep a list of partial solutions we’ve already tried. Even better, that will let us backtrack. So if we get to a case we’ve already seen (that didn’t work), back up naturally using the structure of recursion. All together, it will look something like this:
; Given some of the sides/angles of a triangle try to solve for the rest(define (solve t) (define tried '()) (let loop ([t t]) (set! tried (cons t tried)) (match t ; We have all three sides and angles, return [(? solved?) t] ; We've already tried this solution, backtrack [(? (curryr member (cdr tried))) #f] ...)))
We could have done without the explicit mutation, either using a
parameter to make the dynamic scope more obvious or by passing the
tried variable around in the loop. Personally though, I think this is probably the ‘cleanest’ solution.
So plug in the rules from above and we’re good to go, right?
Well, not quite. There are still a few ordering issues. Sometimes the sides and angles just aren’t lined up 5 rules that we have. So the last thing we need is a second ordering rule:
...; Try another ordering[(triangle ∠ɑ ∠β ∠γ a b c) (or (loop (triangle ∠β ∠γ ∠ɑ b c a)) (loop (triangle ∠β ∠ɑ ∠γ b a c)))]
This is actually pretty elegant for two reasons: Because we return
#f when we backtrack, that will go to the second case if and only if the first doesn’t find a solution. This works because anything not exactly
#f in Racket is true and
or returns the first non-
#f argument (if one exists).
And that’s it. It can already solve a good number of problems:
> (solve (triangle #f #f #f 1 1 (sqrt 2)))(triangle 0.785 0.785 1.571 1 1 1.414)> (solve (triangle (/ pi 2) #f #f (sqrt 2) 1 #f))(triangle 0.785 0.785 1.571 1 1.000 1.414)> (solve (triangle (/ pi 3) (/ pi 3) #f #f #f 3))(triangle 1.0472 1.0472 1.0472 2.999 2.999 3)
We do have a fair few rounding errors creeping in, but unfortunately that’s just part of the price we have to pay when working with real (analog) numbers in a digital computer.
And that’s pretty much it. Or it would be, if not for one little extra feature I wanted to throw in. Remember how I said that because Racket’s trig functions deal in radians, so too would our triangles? Well, just in case, here are a few conversion functions (since
radians->degrees and
degrees->radians are both built in if you’re using the full
racket module):
; Convert all angles in a triangle from radians to degrees(define (triangle-radians->degrees t) (define (r->d ∠) (and ∠ (radians->degrees ∠))) (match t [(triangle ∠ɑ ∠β ∠γ a b c) (triangle (r->d ∠ɑ) (r->d ∠β) (r->d ∠γ) a b c)] [#f #f])); Convert all angles in a triangle from degrees to radians(define (triangle-degrees->radians t) (define (d->r ∠) (and ∠ (degrees->radians ∠))) (match t [(triangle ∠ɑ ∠β ∠γ a b c) (triangle (d->r ∠ɑ) (d->r ∠β) (d->r ∠γ) a b c)] [#f #f])); Given some of the sides/angles of a triangle try to solve for the rest(define (solve/radians t) (define tried '()) (let loop ([t t]) ...)); Given some of the sides/angles of a triangle try to solve for the rest(define (solve/degrees t) (triangle-radians->degrees (solve/radians (triangle-degrees->radians t))))
Now you can solve the examples given in the problem:
> (solve/degrees (triangle 39.0 56.0 #f 2.45912 #f #f))(triangle 56.0 85.0 39.0 3.240 3.893 2.459)> (solve/degrees (triangle 43.0 #f 70.0 #f #f 7))(triangle 43.0 67.0 70.0 5.080 6.857 7)
And that’s it. You can find the full code on GitHub here: triangle-trouble.rkt
Bonus round: 1) Without actually running it, can you figure out what will happen if you give it an impossible triangle? Such as three angles or two angles already too large?
Bonus bonus round:2) What about cases with three sides that
should be impossible but technically have a solution 5… Such as a=1, b=2, c=4. |
Hastie and Tibshirani mention in section 4.3.2 of their book that in the linear regression setting, the least squares approach is in fact a special case of maximum likelihood. How can we prove this result?
PS: Spare no mathematical details.
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Hastie and Tibshirani mention in section 4.3.2 of their book that in the linear regression setting, the least squares approach is in fact a special case of maximum likelihood. How can we prove this result?
PS: Spare no mathematical details.
The linear regression model
$Y = X\beta + \epsilon$, where $\epsilon \sim N(0,I\sigma^2)$
$Y \in \mathbb{R}^{n}$, $X \in \mathbb{R}^{n \times p}$ and $\beta \in \mathbb{R}^{p}$
Note that our model error (residual) is ${\bf \epsilon = Y - X\beta}$. Our goal is to find a vector of $\beta$s that minimize the $L_2$ norm squared of this error.
Least Squares
Given data $(x_1,y_1),...,(x_n,y_n)$ where each $x_{i}$ is $p$ dimensional, we seek to find:
$$\widehat{\beta}_{LS} = {\underset \beta {\text{argmin}}} ||{\bf \epsilon}||^2 = {\underset \beta {\text{argmin}}} ||{\bf Y - X\beta}||^2 = {\underset \beta {\text{argmin}}} \sum_{i=1}^{n} ( y_i - x_{i}\beta)^2 $$
Maximum Likelihood
Using the model above, we can set up the likelihood of the data given the parameters $\beta$ as:
$$L(Y|X,\beta) = \prod_{i=1}^{n} f(y_i|x_i,\beta) $$
where $f(y_i|x_i,\beta)$ is the pdf of a normal distribution with mean 0 and variance $\sigma^2$. Plugging it in:
$$L(Y|X,\beta) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(y_i - x_i\beta)^2}{2\sigma^2}}$$
Now generally when dealing with likelihoods its mathematically easier to take the log before continuing (products become sums, exponentials go away), so let's do that.
$$\log L(Y|X,\beta) = \sum_{i=1}^{n} \log(\frac{1}{\sqrt{2\pi\sigma^2}}) -\frac{(y_i - x_i\beta)^2}{2\sigma^2}$$
Since we want the maximum likelihood estimate, we want to find the maximum of the equation above, with respect to $\beta$. The first term doesn't impact our estimate of $\beta$, so we can ignore it:
$$ \widehat{\beta}_{MLE} = {\underset \beta {\text{argmax}}} \sum_{i=1}^{n} -\frac{(y_i - x_i\beta)^2}{2\sigma^2}$$
Note that the denominator is a constant with respect to $\beta$. Finally, notice that there is a negative sign in front of the sum. So finding the maximum of a negative number is like finding the minimum of it without the negative. In other words:
$$ \widehat{\beta}_{MLE} = {\underset \beta {\text{argmin}}} \sum_{i=1}^{n} (y_i - x_i\beta)^2 = \widehat{\beta}_{LS}$$
Recall that for this to work, we had to make certain model assumptions (normality of error terms, 0 mean, constant variance). This makes least squares equivalent to MLE under certain conditions. See here and here for more discussion.
For completeness, note that the solution can be written as:
$${\bf \beta = (X^TX)^{-1}X^Ty} $$ |
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 |
There is actually a stronger result; A problem is in the class $\mathrm{FPTAS}$ if it has an
fptas 1: an $\varepsilon$-approximation running in time bounded by $(n+\frac{1}{\varepsilon})^{\mathcal{O}(1)}$ (i.e. polynomial in both the size and the approximation factor). There's a more general class $\mathrm{EPTAS}$ which relaxes the time bound to $f(\frac{1}{\varepsilon})\cdot n^{\mathcal{O}(1)}$ - essentially an $\mathrm{FPT}$-like running time with respect to the approximation factor.
Clearly $\mathrm{FPTAS}$ is a subset of $\mathrm{EPTAS}$, and it turns out that $\mathrm{EPTAS}$ is a subset of $\mathrm{FPT}$ in the following sense:
Theorem If an NPO problem $\Pi$ has an eptas, then $\Pi$ parameterized by the cost of the solution is fixed-parameter tractable.
The theorem and proof is given in Flum & Grohe [1] as Theorem 1.32 (pp. 23-24), and they attribute it to Bazgan [2], which puts it two years before Cai & Chen's weaker result (but in a French technical report).
I'll give a sketch of the proof, because I think it's a nice proof of the theorem. For simplicity I'll do the minimization version, just mentally do the appropriate inversions for maximization.
Proof. Let $A$ be the eptas for $\Pi$, then we can construct a parameterized algorithm $A'$ for $\Pi$ parameterized by the solution cost $k$ as follows: given input $(x,k)$, we run $A$ on input $x$ where we set $\varepsilon := \frac{1}{k+1}$ (i.e. we choose the approximation ratio bound $1+\frac{1}{k+1}$). Let $y$ be the solution, $\text{cost}(x,y)$ be the cost of $y$ and $r(x,y)$ be the actual approximation ratio of $y$ to $\text{opt}(x)$, i.e. $\text{cost}(x,y) = r(x,y)\cdot \text{opt}(x)$.
If $\text{cost}(x,y) \leq k$, then accept, as clearly $\text{opt}(x) \leq \text{cost}(x,y) \leq k$. If $\text{cost}(x,y) > k$, reject as $r(x,y) \leq 1+\frac{1}{k+1}$ as $A$ is an
eptas and
$$\text{opt}(x) = \frac{\text{cost}(x,y)}{r(x,y)} \geq \frac{k+1}{1+\frac{1}{k+1}} > k$$
Of course you get the running time bound for $A'$ simply from $A$ being an
eptas. $\Box$
Of course as Pål points out, parameterized hardness results imply the non-existence of any
eptas unless there is some collapse, but there are problems in $\mathrm{FPT}$ with no eptas (or even ptas), so $\mathrm{EPTAS}$ is a strict subset of $\mathrm{FPT}$ (in the sense of the theorem). Footnotes: An fptas (equivalently eptas or ptas) is an approximation scheme with the running time bounded as described above. The class $\mathrm{FPTAS}$ (equiv. $\mathrm{EPTAS}$, $\mathrm{PTAS}$) is the set of problems in $\mathrm{NPO}$ that have such a scheme.
[1]: J. Flum and M. Grohe,
Parameterized Complexity Theory, Springer, 2006. [2]: C. Bazgan. Schémas d’approximation et complexité paramétrée, Rapport deDEA, Université Paris Sud, 1995. |
Answer
The judge should rule in favor of the police officer.
Work Step by Step
We use the equation for the velocity considering the Doppler effect: $v = \frac{c\Delta f}{2f} = \frac{(10.8\times10^8 \ km/h)(15.6\times10^3 \ Hz)}{2(70\times10^9 \ Hz)} = 120 \ km/h$ The judge should rule in favor of the police officer, for this is the same as the recorded speed. |
Mechanical Properties of Solids Application of Elastic Behavior of Material Railway tracks and bridges declared unsafe after long use due to Elastic fatigue. If the material breaking after Elastic limit it is called brittle material. Brittle material is used to make Glass. If the material breaking point and Elastic limit are far it is called ductile material. Ductile material is used to make thin wires. Elastic property of a material increase with addition of impurity. Slow cooling of material after heating called Annealing. For invar steel Elasticity is independent of temperature. Springs are made of steel because it is more Elastic Young's modulus of a material does not depend up on Dimensions of body but depends upon nature of material. Young's modulus (Y) of a material is the ratio of linear stress to linear strain. Shear modulus (n) is defined as the ratio of shear stress to shear strain. Bulk modulus (k) is defined as the ratio of Bulk stress to bulk strain. Y. nk are having same units = N/m 2 When a wire is stretched young's modulus, When twisted rigidity modulus involved. View the Topic in this video From 01:10 To 58:27
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1. For a beam with circular cross-section depression is given by \tt \delta=\frac{WL^{3}}{12\pi r^{4}Y}
2. Torque required to produce a unit twist in a solid shaft
\tau_{solid}=\frac{\pi\eta\ r^{4}}{2l}
3. Torque required to produce a unit twist in a hollow shaft
\tau_{hollow}=\frac{\pi\eta(r_2^4-r_1^4)}{2l} |
You have the right idea! There's a very general construction for public-key encryption called KEM/DEM that works as follows to encrypt a message $m$:
Key encapsulation mechanism, KEM: Generate a key $k$ and an encapsulation $y$ using a public key. Data encapsulation mechanism, DEM: Use $k$ to authenticate and encrypt $m$ giving an authenticated ciphertext $c$. Transmit the encapsulation $y$ along with the ciphertext $c$.
The recipient, with the secret key, can recover $k$ from $y$ and then decrypt $c$.
For RSA, you can make a KEM as follows with a public key $n$:
Pick an integer $x$ between $0$ and $n$ uniformly at random. Compute $y = x^3 \bmod n$. Compute $k = H(x)$, where $H$ is (say) SHA-256.
The recipient uses secret knowledge of the solution $d$ to $3d \equiv 1 \pmod{\lambda(n)}$, where $\lambda(n) = \operatorname{lcm}(p - 1, q - 1)$ if $n = pq$, to recover $x = y^d \bmod n$, and then computes the same $k = H(x)$.
The security requirement for a DEM is modest. An authenticated cipher like AES-GCM or NaCl crypto_secretbox_xsalsa20poly1305, with nonce set to zero, will do just fine. (I recommend AES-256 if you must use AES, to avoid multi-target attacks; I recommend crypto_secretbox_xsalsa20poly1305 over AES-GCM, to avoid side channel attacks on AES and GHASH in fast software implementations.)
ECIES is an example of the KEM/DEM structure—for a public key $A$ on an elliptic curve with standard base point $B$, you pick a scalar $t$ uniformly at random, compute $T = [t]B$, derive the key $k = H([t]A)$, and use $T$ as the encapsulation of the secret key $k$ which you use in an authenticated cipher to encrypt the message; the recipient knows the secret $a$ such that $A = [a]B$, and given $T$ recovers $H([a]T) = H([a\cdot t] B) = H([t\cdot a]B) = H([t]A) = k$ to decrypt the authenticated ciphertext.
You can also generate $k$ independently and shoe-horn it into an encapsulation like RSAES-OAEP. But, while this is probably the most common way to do RSA encryption out of inertia, it is more complicated than necessary. |
In the derivation of the minimum shift keying in-phase and quadrature components in Simon-Haykin Communication Systems 4th ed (p.387-390) we start off with the definition of the signal to be
$\ s(t) $= $\sqrt {2E_b/T_b}$ $\cos(2\pi f_c t+\theta(t))$
where $\theta(t) $ = $\theta(0) $ $\ \pm $ $\ \frac {\pi}{2T_b} t $
$\theta(0) $ is the phase of previous bit and $\ 0 <= t <=T_b$
but in the subsequent steps : for the in-phase component,
$\ s_I $=$\ \pm $ $\sqrt {2E_b/T_b}$ $\cos( \frac {\pi}{2T_b} t )$ and $\ -T_b <= t <=T_b$
for the quadrature-phase component,
$\ s_Q $=$\ \pm $ $\sqrt {2Eb/Tb}$ $\sin( \frac {\pi}{2T_b} t )$ and $\ 0 <= t <= 2T_b$
How did this change of the limits of t change from $\ 0 <= t <=T_b$ to $\ -T_b <= t <=T_b$ or $\ 0 <= t <= 2T_b$ ? How are we allowed to extend the definition like that?
Thanks in advance. |
I am struggling with an equivalence of categories.
Let $\mathbf{Mon}$ be the category of monoids, and let $\mathbf{Cat}$ be the category whose objects are all categories with exactly one object. The morphisms from object $A$ to $B$ in $\mathbf{Cat}$ are all functors from $A$ to $B.$ I want to show that $\mathbf{Mon}$ and $\mathbf{Cat}$ are equivalent categories. Please help. Thanks in advance.
I am struggling with an equivalence of categories.
A monoid $(M,*,1)$ can be seen as a category with only one object $\bullet$, the same way a group is a groupoid with a single object. The composition of morphisms corresponds to the multiplication $"*"$ of elements and the rule $f\circ(g∘h)=(f∘g)∘h$ means that multiplication is associative, while the single identity arrow $1_\bullet$ is the unit of $*$ since $1∘f=f∘1=f$ for any $f\in M$.
A functor $F:M\to N$ between two categories with a single object is uniquely determined by its arrow function, and since there is only one hom-set, any function is allowed, as long as it respect the composition of arrows and the identity, i.e. $F(1_∙)=1_∙$ and $F(g∘f)=F(g)∘F(f)$. But these are just the properties that we require from a homomorphism between monoids. Hence, a functor between two monoids, considered as categories with a single object, is basically the same as a homomorphism, and vice versa. |
A linear homogeneous system of \(n\) differential equations with constant coefficients has the form:
\[ {\mathbf{X}’\left( t \right) = A\mathbf{X}\left( t \right),\;\;}\kern-0.3pt {\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{x_1}\left( t \right)}\\ {{x_2}\left( t \right)}\\ \vdots \\ {{x_n}\left( t \right)} \end{array}} \right],\;\;}\kern-0.3pt {A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}\\ {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}\\ \cdots & \cdots & \cdots & \cdots \\ {{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}} \end{array}} \right].} \]
Here \(\mathbf{X}\left( t \right)\) is an \(n\)-dimensional vector, \(A\) is a square matrix with constant coefficients of size \(n \times n.\)
Next, we describe a general algorithm for solving this system and consider specific cases where the solution is constructed by the method of undetermined coefficients.
We seek a solution of the given equation in the form of vector functions
\[\mathbf{X}\left( t \right) = {e^{\lambda t}}\mathbf{V},\]
where \(\lambda\) is the eigenvalue of \(A,\) and \(\mathbf{V}\) is the eigenvector associated with the eigenvalue.
The eigenvalues \({\lambda _i}\) are found from the auxiliary equation
\[\det \left( {A – \lambda I} \right) = 0,\]
where \(I\) is the identity matrix.
Since some of the roots \({\lambda _i}\) can be multiple, in the general case of an \(n\)th order system, the auxiliary equation has the form:
\[{{\left( { – 1} \right)^n}{\left( {\lambda – {\lambda _1}} \right)^{{k_1}}}{\left( {\lambda – {\lambda _2}} \right)^{{k_2}}} \cdots}\kern0pt{ {\left( {\lambda – {\lambda _m}} \right)^{{k_m}}} }={ 0.}\]
Here the following condition holds:
\[{k_1} + {k_2} + \cdots + {k_m} = n.\]
The power \({k_i}\) of the factor \(\left( {\lambda – {\lambda _i}} \right)\) is called the algebraic multiplicity of the eigenvalue \({\lambda _i}.\)
For each eigenvalue \({\lambda _i},\) we can find the eigenvector (or more eigenvectors in the case of a multiple \(\left.{\lambda _i}\right)\) using the formula
\[\left( {A – {\lambda _i}I} \right){\mathbf{V}_i} = \mathbf{0}.\]
The number of eigenvectors associated with the eigenvalue \({\lambda _i}\) is called the geometric multiplicity of \({\lambda _i}\) (we denote it by \({s_i}\)). Thus, the eigenvalue \({\lambda _i}\) is characterized by two quantities: the algebraic multiplicity \({k_i}\) and geometric multiplicity \({s_i}.\) The following relationship holds:
\[0 \lt {s_i} \le {k_i},\]
i.e., the geometric multiplicity \({s_i}\) (or the number of eigenvectors) does not exceed the algebraic multiplicity \({k_i}\) of the eigenvalue \({\lambda _i}.\)
A fundamental system of solutions and, hence, the general solution of the system strongly depend on the algebraic and geometric multiplicity of the eigenvalues \({\lambda _i}.\) In the simplest case \({s_i} = {k_i} = 1,\) when the eigenvalues \({\lambda _i}\) of the matrix \(A\) are distinct and each number \({\lambda _i}\) is associated with one eigenvector \({\mathbf{V}_i},\) the fundamental system of solutions consists of the functions
\[{{e^{{\lambda _1}t}}{\mathbf{V}_1},\;}\kern-0.3pt{{e^{{\lambda _2}t}}{\mathbf{V}_2}, \;\ldots,\;}\kern-0.3pt{ {e^{{\lambda _n}t}}{\mathbf{V}_n}.}\]
In this case, the general solution is written as
\[ {\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{x_1}\left( t \right)}\\ {{x_2}\left( t \right)}\\ \vdots \\ {{x_n}\left( t \right)} \end{array}} \right] } = {{C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} }+{ {C_2}{e^{{\lambda _2}t}}{\mathbf{V}_2} + \cdots } + {{C_n}{e^{{\lambda _n}t}}{\mathbf{V}_n} } = {\sum\limits_{i = 1}^n {{C_i}{e^{{\lambda _i}t}}{\mathbf{V}_i}} ,} \]
where \({C_i}\) are arbitrary constants.
Let’s discuss the case of complex roots of the characteristic equation. If all the coefficients in the equations are real numbers, the complex roots will be “born” in the form of pairs of complex conjugate numbers \(\alpha \pm i\beta .\) To construct a solution that is associated with such a pair, it is enough to take one number, for example, \(\alpha + i\beta\) and find for it the eigenvector \(\mathbf{V},\) which may also have complex coordinates. Then the solution will be presented by the complex-valued vector function \({e^{\left( {\alpha + i\beta } \right)t}}\mathbf{V}\left( t \right).\) The exponential function can be expanded by Euler’s formula:
\[
{{e^{\left( {\alpha + i\beta } \right)t}} }={ {e^{\alpha t}}{e^{i\beta t}} } = {{e^{\alpha t}}\left( {\cos \beta t + i\sin \beta t} \right).} \]
As a result, the part of the general solution corresponding to the pair of eigenvalues \(\alpha \pm i\beta,\) will be presented in the form of
\[ {\mathbf{X}\left( t \right) \text{ = }}\kern0pt {{ {e^{\alpha t}}\left( {\cos \beta t + i\sin \beta t} \right) \cdot}}\kern0pt{{ \left( {{\mathbf{V}_\text{Re}} + i{\mathbf{V}_\text{Im}}} \right) }} = {{{e^{\alpha t}}\left[ {\cos \left( {\beta t} \right){\mathbf{V}_\text{Re}} }\right.}-{\left.{ \sin \left( {\beta t} \right){\mathbf{V}_\text{Im}}} \right] }} + {{i{e^{\alpha t}}\left[ {\cos \left( {\beta t} \right){\mathbf{V}_\text{Im}} }\right.}+{\left.{ \sin \left( {\beta t} \right){\mathbf{V}_\text{Re}}} \right] }} = {{\mathbf{X}^{\left( 1 \right)}}\left( t \right) + i{\mathbf{X}^{\left( 2 \right)}}\left( t \right),} \]
where \(\mathbf{V} =\) \( {\mathbf{V}_\text{Re}} + i{\mathbf{V}_\text{Im}}\) is the complex eigenvector. The vector functions \({\mathbf{X}^{\left( 1 \right)}}\) and \({\mathbf{X}^{\left( 2 \right)}}\) in the real and imaginary parts of the resulting expression form two linearly independent real solutions.
As it can be seen, the solution for a pair of complex conjugate eigenvalues is constructed in the same manner as for the real eigenvalues. It’s only necessary to clearly distinguish the real and imaginary parts of the vector function at the end of the transformations.
Now consider the case of multiple roots \({\lambda _i}.\) For simplicity, we assume them to be real numbers. Here again, the solution process is split into two scenarios.
If the algebraic multiplicity \({k_i}\) and geometric multiplicity \({s_i}\) of an eigenvalue \({\lambda _i}\) coincide \(\left( {{k_i} = {s_i} \gt 1} \right),\) there exist \({k_i}\) eigenvectors for this value \({\lambda _i}\). As a result, the eigenvalue \({\lambda _i}\) will be associated with \({k_i}\) linearly independent solutions of the form
\[{{e^{{\lambda _i}t}}\mathbf{V}_i^{\left( 1 \right)},\;}\kern-0.3pt{{e^{{\lambda _i}t}}\mathbf{V}_i^{\left( 2 \right)},\; \ldots ,\;}\kern-0.3pt{{e^{{\lambda _i}t}}\mathbf{V}_i^{\left( {{k_i}} \right)}.}\]
In this case, the system of \(n\) equations will have total \(n\) eigenvectors forming a fundamental system of solutions. Examples of such systems are given on the web page Method of Eigenvalues and Eigenvectors.
The most interesting is the case of multiple \({\lambda _i}\) when the geometric multiplicity \({s_i}\) is less than the algebraic multiplicity \({k_i}.\) This means that we have only \({s_i}\) \(\left( {{s_i} \lt {k_i}} \right)\) eigenvectors associated with \({\lambda _i}.\) The number of eigenvectors \({s_i}\) is given by the formula
\[{{s_i} }={ n – \text{rank}\left( {A – {\lambda _i}I} \right),}\]
where \(\text{rank}\left( {A – {\lambda _i}I} \right)\) means the rank of the matrix \({A – {\lambda _i}I}\) with substituted value of \({\lambda _i}.\)
The solution corresponding to \({\lambda _i}\) can be sought as the product of a polynomial of degree \({k_i} – {s_i}\) by the exponential function \({e^{{\lambda _i}t}}:\)
\[ {{\mathbf{X}_i}\left( t \right) = {\mathbf{P}_{{k_i} – {s_i}}}\left( t \right){e^{{\lambda _i}t}},\;\;}\kern-0.3pt{ \text{where}}\;\; {{\mathbf{P}_{{k_i} – {s_i}}}\left( t \right) }={ {\mathbf{A}_0} + {\mathbf{A}_1}t + \cdots }+{ {\mathbf{A}_{{k_i} – {s_i}}}{t^{{k_i} – {s_i}}}.} \]
Here \({\mathbf{P}_{{k_i} – {s_i}}}\left( t \right)\) is a vector polynomial, i.e., each of the \(n\) coordinates corresponds to a polynomial of degree \({{k_i} – {s_i}}\) with some coefficients to be determined.
In fact, the method of undetermined coefficients is needed only in the case of multiple roots \({\lambda _i},\) when the number of linearly independent eigenvectors is less than the algebraic multiplicity of the root \({\lambda _i}.\)
To find the vectors \({\mathbf{A}_0},\) \({\mathbf{A}_1}, \ldots ,\) \({\mathbf{A}_{{k_i} – {s_i}}}\) for each such eigenvalue \({\lambda _i},\) one should substitute the vector function \({\mathbf{X}_i}\left( t \right)\) in the original system of equations. Equating the coefficients of the terms with the same power in the left and right sides of each equation, we obtain an algebraic system of equations for the unknown vectors \({\mathbf{A}_0},\) \({\mathbf{A}_1}, \ldots ,\) \({\mathbf{A}_{{k_i} – {s_i}}}.\)
This method for constructing the general solution of a system of differential equations is sometimes referred to as the Euler method.
Solved Problems
Click a problem to see the solution. |
Linear models describe a continuous response variable as a function of one or more predictor variables. They can help you understand and predict the behavior of complex systems or analyze experimental, financial, and biological data.
Linear regression is a statistical method used to create a linear model. The model describes the relationship between a dependent variable \(y\) (also called the response) as a function of one or more independent variables \(X_i\) (called the predictors). The general equation for a linear model is:
\[y = \beta_0 + \sum \ \beta_i X_i + \epsilon_i\]
where \(\beta\) represents linear parameter estimates to be computed and \(\epsilon\) represents the error terms.
There are several types of linear regression:
Simple linear regression:models using only one predictor Multiple linear regression:models using multiple predictors Multivariate linear regression:models for multiple response variables Generate predictions Compare linear model fits Plot residuals Evaluate goodness-of-fit Detect outliers
To create a linear model that fits curves and surfaces to your data, see Curve Fitting Toolbox. To create linear models of dynamic systems from measured input-output data, see System Identification Toolbox. To create a linear model for control system design from a nonlinear Simulink model, see Simulink Control Design. |
I think you're confusing a few issues. Firstly, $F$ isn't an 'algorithm' in and of itself; it's a function. You can ask if $F$ is polynomially bounded, and whether there's a polynomial-time algorithm to compute $F$, but those two are different statements. I'm assuming that you're after polynomial boundedness - that is, that there's some polynomial $p(n)$ such that $F\in O(p(n))$ , but clarifying the question would be immensely useful.
The first thing to note is that the inner summation is trivial, and in fact the presence of 'b' as a lower summation limit is somewhat misleading; your formula works out to $F(n) = \Sigma_{i=a}^{f(a,n)}\left(g(a,b,i,n)-b\right)$ and it's easy to absorb the $-b$ term into the definition of $g$. Now, if $g$ is polynomially bounded
in all its arguments, then its maximum over all $i$ in some polynomial range is also polynomially bounded as a function of $a$, $b$, and $n$; this is essentially the statement that polynomials are closed under composition, that is that if $p(n)$ is a polynomial and $q(n)$ is a polynomial, then $r(n)=p(q(n))$ is also a polynomial (and this extends to the multivariate case). So the sum (as a function of $a$, $b$, and $n$) is bounded by $f(a,n)$ times the maximum value of $g$ over the specified range, and thus is polynomially bounded.
Note that you need polynomial bounding in all the parameters for this argument to go through, and that 'polynomially computable' doesn't imply 'polynomially bounded'. For instance, ignoring the parameters $a$ and $b$ for a moment, it's easy to find $g(n,i) = n+2^i$ in time polynomial in $n$ and $i$, and it's even polynomially bounded as a function of $n$ for fixed values of $i$ - but then for something as simple as $f(n) = n$ the resulting sum isn't polynomially bounded (it grows exponentially with $n$).
Added on the clarification: with the clarified description, yes, it follows almost trivially that $F(n)$ is polynomial-time computable if $f$ and $g$ are polynomially bounded, for essentially the same reasons; seen from an algorithmic perspective, a polynomial-in-$n$ number of steps are executed a polynomial-in-$n$ number of times and by the composition theorem mentioned above the result is still polynomially bounded as a function of $n$. That's not necessarily useful, though, because for a problem of this nature what one would ideally like is to have a subpolynomially-bounded algorithm - for instance, things like multiplication can be easily done this way (choose $a=b=1$, $f(n)=n$, $g(n)=c$ to compute $F(n)=cn$) but one isn't usually interested in linear-time algorithms for multiplication! Ideally what you'd like is a polylogarithmic algorithm - something that can compute these sums in time $O(\log(n)^d)$ for some $d$ - and that's a much harder question, but I believe (without a specific example) that this can't be computed in polylogarithmic time; for instance, it shouldn't be hard to construct a reduction from this to something like the sum-of-square-roots problem. |
Chemical Kinetics Rate of a chemical reaction Rate of reaction: The change in concentration of a reactant or a product in unit time is called rate of reaction. Its unit is mole L −1 sec −1 (or) mol dm −3 sec −1 for solutions and kPa sec −1 or Nm −2 sec −1 for gases.
For example aA → bB
(i) Average rate of reaction = \tt \frac{-1}{a}\frac{\Delta \left[A\right]}{\Delta t}=\frac{1}{b}\frac{\Delta \left[B\right]}{\Delta t} (ii) \tt -\frac{\Delta \left[A\right]}{\Delta t}\ and\ +\frac{\Delta \left[B\right]}{\Delta t} are average rate of consumption of A and average rate of production of B respectively. (iii) −Δ[A] is decrease in concentration of A and +Δ[B] is increase in concentration of B in time taken Δt.
Instantaneous rate of reaction is the change in concentration of a reactant or a product at a given time.
It can be identified graphically only For example w.r.t "A" Instantaneous rate of reaction \tt \lim_{\Delta t \rightarrow 0}\frac{\Delta \left[A\right]}{\Delta t}=-\frac{dx}{dt} dx is very small change of concentration in very small change of time dt.
The kinetics of reaction can be followed (i.e., order, rate constant etc, can be established) by measuring a property which changes with time.
eg: (i) Total pressure in a gaseous reaction. (ii) Volume of a reagent (Acidic Basic, oxidising or reducing agent) (iii) Volume of a gaseous mixture (V) (iv) Optical rotation (r) For a reaction A 0 → nB
t = 0 c 0 t = t c − x nx t = ∞ 0 nc
For any measurable property X proportional to the concentration of reaction mixture at various times, in terms of
(i) x 0 and x \tt k=\frac{1}{t}\ ln\ \frac{x_0}{x_0-x} (ii) x 0 and x t \tt k=\frac{1}{t}\ ln\ \frac{x_0\left(n-1\right)}{nx_0-x_t} (iii) x ∞ and x t \tt k=\frac{1}{t}\ ln\ \frac{\left(n-1\right)x_\infty}{n\left(x_\infty-x_t\right)} (iv) x 0, x t and x ∞ \tt k=\frac{1}{t}\ ln\ \left(\frac{x_\infty-x_0}{X_\infty-x_t}\right) x = amount of reacted in time t. x 0 = measured property at t = 0 x t = measured property at t = t x ∞ = measured property at t = ∞ Part1: View the Topic in this Video from 1:10 to 50:09 Part2: View the Topic in this Video from 33:00 to 39:06
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1. For the reaction aA + bB → cC + dD
Rate of disappearance of a reactant is negative \tt -\frac{d\left[A\right]}{dt} = Rate of disappearance of A \tt -\frac{d\left[B\right]}{dt}= Rate of disappearance of B Rate of formation of a product is positive \tt \frac{d\left[C\right]}{dt}= Rate of formation of C \tt \frac{d\left[D\right]}{dt}= Rate of formation of D
2. In terms of stoichiometric coefficient rate may be expressed as
\tt \frac{dx}{dt}=-\frac{1}{a}\frac{d\left[A\right]}{dt}=-\frac{1}{b}\frac{d\left[B\right]}{dt}=\frac{1}{c}\frac{d\left[C\right]}{dt}=\frac{1}{d}\frac{d\left[D\right]}{dt}
3. Rate in atm time
-1 = Rate in mole L -1 time -1 × RT |
Conditional Probability is the probability of one event occurrence having the same relationship with other events too. For example –
Look at these events carefully having a direct relationship to each other. If it will be raining then you are not supposed to go outside in the case. The event B will happen when A has occurred already. If A will not happen then B will also not occur automatically. This probability is written P(B|A), the notation for the probability of B given A.
When A and B events are not related to each other somehow then they are simply written as P(B), P(A).From the above definition, the conditional probability could be calculated quickly and it is given by the formula –
When B is given by A, then conditional probability is,
\[\ P(B|A)= \frac{P(A\cap B)}{P(A)}\]
When A is given by B, then the conditional probability is
\[\ P(A|B)= \frac{P(A\cap B)}{P(B)}\]
The formula can be applied successfully only if the value of P(A) is greater than zero. The conditional probability could have many applications in the real-life like finance sector, insurance, or politics etc. The election process is an example of conditional probability where one can win only if the opponent fails.
The binomial probability is simply thought of as the probability of success or failure outcomes during an experiment or survey which are related somehow. This is also named as the binomial distribution with chances of two possible outcomes. Take an example of the coin tossed in the air has only two outcomes i.e. Head or Tail. At the same time, when a student appears for an exam, there are two possibilities again i.e. Pass or Fail. The binomial distribution should satisfy the three criteria, as given below –
\[\ P(X) = C_{x}^{n} P^{x} q^{n-x}\]
Where,
n = Total number of trials x = Total number of successful trials p = probability of success in a single trial q = probability of failure in a single trial = 1-p
A probability distribution tells you about the probability of the event occurrence and it can be used for such complex systems like the success rate of a drug during the cancer treatment. It can be represented with the help of a table in mathematics. The sum of probability distributions should always be 100 percent or 1 in the decimal.
\[\large p(x)=\frac{1}{\sqrt{2\pi \sigma^{2}}}\;e^{\frac{(x-\mu)^{2}}{2\sigma^{2}}}\]
Where,
μ = Mean σ = Standard Distribution. If mean(μ) = 0 and standard deviation(σ) = 1, then this distribution is known to be normal distribution. x = Normal random variable.
\[\large p(x)=\frac{n!}{r!(n-r)!}\cdot p^{r}(1-p)^{n-1}=C(n,r)\cdot p^{r}(1-p)^{n-r}\]
Where,
n = Total number of events r = Total number of successful events. p = Probability of success on a single trial. nC r = \[\ \frac{n!}{r!(n – r)!}\] 1 – p = Probability of failure.
The other name for empirical probability is experimental probability to calculate the probability of an experiment and a certain result too. It is usually required during the survey when the experiment is conducted over 100 people or more and give educational data accordingly.
\[\ P(E)=\frac{Number\;of\;times\;event\;occurs}{Total\;number\;of\;times\;experiment\;performed}\]
\[\ P(E)=\frac{f}{n}\]
P(E) = Empirical Probability |
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D-meson nuclear modification factor and elliptic flow measurements in Pb–Pb collisions at $\sqrt {s_{NN}}$ = 5.02TeV with ALICE at the LHC
(Elsevier, 2017-11)
ALICE measured the nuclear modification factor ($R_{AA}$) and elliptic flow ($\nu_{2}$) of D mesons ($D^{0}$, $D^{+}$, $D^{⁎+}$ and $D^{s+}$) in semi-central Pb–Pb collisions at $\sqrt{s_{NN}} =5.02$ TeV. The increased ...
ALICE measurement of the $J/\psi$ nuclear modification factor at mid-rapidity in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV
(Elsevier, 2017-11)
ALICE at the LHC provides unique capabilities to study charmonium production at low transverse momenta ( p T ). At central rapidity, ( |y|<0.8 ), ALICE can reconstruct J/ ψ via their decay into two electrons down to zero ...
Net-baryon fluctuations measured with ALICE at the CERN LHC
(Elsevier, 2017-11)
First experimental results are presented on event-by-event net-proton fluctuation measurements in Pb- Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV, recorded by the ALICE detector at the CERN LHC. The ALICE detector is well ... |
Suppose that $\mu(n)$ is a negligible function, which means that for every $c>0$ there is some $N$ such that for all $n>N$ it holds that $\mu(n)\leq n^{-c}$.
Now, imagine that some encryption scheme, signature scheme, or some cryptographic primitive in general has an "error" (distinguishing error, forging error, etc.) of $\mu(n)$. Imagine that now I want to set up my primitive so that this error is no larger than $2^{-80}$.If I had the
explicit expression for $\mu(n)$ (which is likely to be the case if I read the security proof of the given primitive) then I would solve the inequality $\mu(n)\leq 2^{-80}$ for $n$.In practice (even though this is not necessarily the case), negligible functions look pretty much like exponentials $2^{-n}$, so we know that this $n$ will be polynomial in $80$ and therefore it won't be too large.
Now think of the general case.Suppose I want to find some $n$ such that $\mu(n)\leq 2^{-\kappa}$ for some $\kappa$, let's denote the minimum of such $n$'s by $n(\kappa)$.We know that such value must exist simply because $\mu$ tends to zero as $n$ approaches infinity. Moreover, since $\mu$ is negligible, intuition on the definition says that $n(\kappa)$ shouldn't be too large in terms of $\kappa$ (e.g. if $\mu(n) = 2^{-n}$ then $n(\kappa) = \kappa$).
My question is: how do we actually prove this?
Here's my question, extracted from the motivation above:
Let $\mu$ be a negligible function, and let $n(\kappa)$ be the smallest $n$ such that $\mu(n)\leq 2^{-\kappa}$. Prove that $n(\kappa)$ does not grow too fast as $\kappa\to\infty$.
Part of the question requires to specify precisely what is meant by "too fast". I would love some statement like "$n(\kappa)$ is polynomial in $\kappa$", but I'm not sure how to prove this.
I believe this property is important as this is one of the reasons why we choose negligible functions (besides being closed on additions and multiplication by polynomials): it allows us to get small error probabilities without increasing the security parameter too much.
Any insights on this are welcome. |
The Eilenberg-MacLane spaces $K(G,q)$ are readily generalized to study cohomology with local coefficients.The generalized Eilenberg-MacLane space $K_{\pi}(G,q)$ are spaces with only two nnvanishing homotopy groups, one of them the fundamental group is $\pi$, the other is $G$ in dimension $q$. And the fundamental group $\pi$ acts on $G$ by representation $\rho:\pi\rightarrow Aut(G)$, where $Aut(G)$ is the group of automorphisms of $G$.
It has been given the construction of Eilenberg-MacLane space K(G,n) in the category of CW complex in All Hathcer's book 'ALGEBRAIC TOPOLOGY'. In Samuel Gitler's article 'Cohomology Operations with Local Coefficients' . The generalized Eilenberg-MacLane space $K_{\pi}(G,q)$ are constructed in the category of Kan Complex, but I have no idea about Kan complex and I have not learned them before. Is it possible to reference me some books or articles which talk about $K_{\pi}(G,q)$ in the category of CW complex?
And I have some idea about construction in the follow. But I am not sure if I am right.
Let $K(\pi,1)$ be the base space and homomorphism $\rho:\pi\rightarrow Aut(G)$ be the representation. The homomorphism $\rho$ can induce representation $\rho_{*}:\pi\rightarrow Homeo(K(G,q);K(G,q))$, where $Homeo(K(G,q);K(G,q))$ is the group of homeomorphisms of $K(G,q)$ itself. Then we can construct a fiber bundle $K(G,q)\rightarrow E\rightarrow K(\pi,1)$ by the representation $\rho_{*}$.
I want know if the total space $E$ is the generalized Eilenberg-MacLane space $K_{\pi}(G,q)$ we need? |
Dear community,
In light of the recent work of DeBacker/Reeder on the depth zero local Langlands correspondence, I was wondering if there is an attempt to "geometrize" the depth zero local Langlands correspondence.
In particular, in Teruyoshi Yoshida's thesis, one can see a glimpse of this for $GL(n,F)$, where $F$ is a $p$-adic field. Namely : Suppose $k$ is the residue field of $F$. Let $w$ be the cyclic permutation $(1 \ 2 \ 3 \ ... \ n)$ in the Weyl group $S_n$ of $GL(n,F)$. Let $\widetilde{Y_w}$ be the Deligne-Lusztig variety associated to $w$, and denote by $H^*(\widetilde{Y_w})$ the alternating sum of the cohomologies $H_c^i(\widetilde{Y_w}, \overline{\mathbb{Q}_{\ell}})$. Let $T_w(k) = k_n^*$ be the elliptic torus in $GL(n,k)$, where $k_n$ is the degree $n$ extension of $k$.
Since $T_w(k)$ and $GL(n,k)$ act on cohomology, $H^*(\widetilde{Y_w})$ is an element of the Grothendieck group of $GL(n,k) \times T_w(k)$-modules. There is a canonical surjection $I_F \rightarrow k_n^* = T_w(k)$, where $I_F$ is the inertia subgroup of the Weil group of $F$. Therefore, we may pull back the $GL(n,k) \times T_w(k)$ action on $H^*(\widetilde{Y_w})$ to an action of $GL(n,k) \times I_F$.
By Deligne-Lusztig Theory, as $GL(n,k) \times T_w(k)$-representations,
$$H_c^{n-1}(\widetilde{Y_w}, \overline{\mathbb{Q}_{\ell}})^{cusp} = $$
$$\displaystyle\sum_{\theta \in C} \pi_{\theta} \otimes \theta$$
where $C$ denotes the set of all characters of $k_n^*$ that don't factor through the norm map $k_n^* \rightarrow k_m^*$ for any integer $m$ such that $m \neq n$ and $m$ divides $n$, and where cusp denotes "cuspidal part". Here, $\pi_{\theta}$ is the irreducible cuspidal representation of $GL(n,k)$ associated to the torus $T_w(k)$ and the character $\theta$ of $T_w(k)$.
One of Yoshida's main theorems is that in this decomposition $$\displaystyle\sum_{\theta \in C} \pi_{\theta} \otimes \theta,$$ the correspondence $\theta \leftrightarrow \pi_{\theta}$ is indeed the depth zero local Langlands correspondence for $GL(n,k)$, ''up to twisting'' (this twist is unimportant for my question), by comparing with Harris-Taylor.
So my question is : Has anyone tried to generalize this to more general groups, but still working only in depth zero local Langlands? One could try to do this, and then compare to the recent work of DeBacker/Reeder (they write down a fairly general depth zero local Langlands correspondence). In other words, has anyone tried to realize depth zero local Langlands in the cohomology of Deligne-Lusztig varieties outside of the case $GL(n,F)$, which Yoshida did?
A priori the above idea for $GL(n)$ won't work on the nose for other reductive groups since the tori that arise in other reductive groups vary considerably, but something similar might. One would possibly want to try to pull back the action of $T_w(k)$ on cohomology to the inertia group $I_F$ in a more general setting, where now $T_w(k)$ is a more general torus in a more general reductive group. Then, one could compare to DeBacker/Reeder.
I took a look at the case of unramified $U(3)$, and it seems that things will work quite nicely.
My other question is : It might turn out that what I'm proposing is an easy check if one understands DeBacker/Reeder and Deligne-Lusztig enough to write this down in general. If so, then is my original question even interesting? It would basically say that Deligne-Lusztig theory is very naturally compatible with local Langlands correspondence, but the hard work is really in DeBacker/Reeder and Deligne-Lusztig, and putting everything together might not be difficult. Is the original question interesting regardless of whether or not it is difficult to answer?
Sincerely,
Moshe Adrian |
I understand that the speed of light can be derived from Maxwell's equations, giving $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$
I furthermore understand how the principle of invariance of laws w.r.t. inertial reference frames gives rise to special relativity in order to preserve the above equation.
I am aware that gluons are theoretically massless and also travel at $c$.
I am also aware that the speed of light is considered to be the "speed of massless particles" or "the speed of information", but I'll get to that in a moment.
My question is:
why should the speed of gluons by given by the electric and magnetic constants $\mu_0$ and $\varepsilon_0$? This connection seems sensible in the case of the photon, an electromagnetic particle, but why should this apply to the gluon as well?
I reject the "all massless particles" and "speed of information" answers as an explanation because they
don't actually explain anything -- the situation is just as mysterious after these "answers" are given as before.
If "all massless particles" is really the answer, then $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ needs to explain how $\mu_0$ and $\varepsilon_0$ are derived from $c$, not the other way round. This is two new mysteries: firstly, how do we obtain $\mu_0$ and $\varepsilon_0$ from $c$ in a philosophically sound manner, and secondly, why should the classical derivation coincidentally obtain the same answer?
If "speed of information" is the answer, we need both to supply a sensible fundamental definition of "information" and furthermore show that photons and gluons actually satisfy that definition. Then we still have the problem in the bullet point above.
Can anyone shed some light (ha ha) on this?
How can we present these results such that the speed of the gluon is naturally given by electronic and magnetic constants, or how do we derive $\mu_0$ and $\varepsilon_0$ from some common concept? |
I think it'd be nice to have a whole set of possible messages, each associated with some equation which has zero on the right hand side. It'd look something like this:
"404; Did you just [insert operation that yields zero*]? Because there's nothing here! (Insert relevant equation**)"
*Examples; (**Corresponding equations):
1. symmetrize the electromagnetic field strength tensor \(\bigl(F^{(\mu\nu)}=0\bigr)\)
2. take a covariant derivative of the metric \(\bigl(\nabla_\rho g_{\mu\nu}=0\bigr)\)
3. calculate a lightlike interval \(\bigl(ds^2=0\bigr)\)
4. vary the action of the internet \(\bigl(\delta S=0\bigr)\)
5. take the d'Alembertian of a massless field \(\bigl(\square\phi=0\bigr)\)
6. check the Bianchi identities \(\bigl(\nabla_{[\mu}F_{\nu\sigma]}=0\bigr)\ \text{or}\ \bigl(\nabla_{[\mu}R_{\nu\sigma]}=0\bigr)\)
feel free to add other/better ones if you'd like |
also, if you are in the US, the next time anything important publishing-related comes up, you can let your representatives know that you care about this and that you think the existing situation is appalling
@heather well, there's a spectrum
so, there's things like New Journal of Physics and Physical Review X
which are the open-access branch of existing academic-society publishers
As far as the intensity of a single-photon goes, the relevant quantity is calculated as usual from the energy density as $I=uc$, where $c$ is the speed of light, and the energy density$$u=\frac{\hbar\omega}{V}$$is given by the photon energy $\hbar \omega$ (normally no bigger than a few eV) di...
Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." — tparker3 mins ago
> A representative of RELX Group, the official name of Elsevier since 2015, told me that it and other publishers “serve the research community by doing things that they need that they either cannot, or do not do on their own, and charge a fair price for that service”
for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @EmilioPisanty
> for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing)
@0celo7 but the bosses are forced because they must continue purchasing journals to keep up the copyright, and they want their employees to publish in journals they own, and journals that are considered high-impact factor, which is a term basically created by the journals.
@BalarkaSen I think one can cheat a little. I'm trying to solve $\Delta u=f$. In coordinates that's $$\frac{1}{\sqrt g}\partial_i(g^{ij}\partial_j u)=f.$$ Buuuuut if I write that as $$\partial_i(g^{ij}\partial_j u)=\sqrt g f,$$ I think it can work...
@BalarkaSen Plan: 1. Use functional analytic techniques on global Sobolev spaces to get a weak solution. 2. Make sure the weak solution satisfies weak boundary conditions. 3. Cut up the function into local pieces that lie in local Sobolev spaces. 4. Make sure this cutting gives nice boundary conditions. 5. Show that the local Sobolev spaces can be taken to be Euclidean ones. 6. Apply Euclidean regularity theory. 7. Patch together solutions while maintaining the boundary conditions.
Alternative Plan: 1. Read Vol 1 of Hormander. 2. Read Vol 2 of Hormander. 3. Read Vol 3 of Hormander. 4. Read the classic papers by Atiyah, Grubb, and Seeley.
I am mostly joking. I don't actually believe in revolution as a plan of making the power dynamic between the various classes and economies better; I think of it as a want of a historical change. Personally I'm mostly opposed to the idea.
@EmilioPisanty I have absolutely no idea where the name comes from, and "Killing" doesn't mean anything in modern German, so really, no idea. Googling its etymology is impossible, all I get are "killing in the name", "Kill Bill" and similar English results...
Wilhelm Karl Joseph Killing (10 May 1847 – 11 February 1923) was a German mathematician who made important contributions to the theories of Lie algebras, Lie groups, and non-Euclidean geometry.Killing studied at the University of Münster and later wrote his dissertation under Karl Weierstrass and Ernst Kummer at Berlin in 1872. He taught in gymnasia (secondary schools) from 1868 to 1872. He became a professor at the seminary college Collegium Hosianum in Braunsberg (now Braniewo). He took holy orders in order to take his teaching position. He became rector of the college and chair of the town...
@EmilioPisanty Apparently, it's an evolution of ~ "Focko-ing(en)", where Focko was the name of the guy who founded the city, and -ing(en) is a common suffix for places. Which...explains nothing, I admit. |
Whenever there is a battery connected in a circuit we assume that the positive terminal of the battery is at a higher potential then the negative terminal by convention. Also by convention + charge flows from positive to negative terminal .The driving force here is the potential difference.
Correct
Electrons flow in the opposite direction. The driving force here is the force exerted by the electric field.
Correct.
But don't be confused, the driving force here is the same as before. The electric potential difference $V$ corresponds to an electric field $E=\frac{dV}{dx}=\frac{V}{L}$ (the
change or gradient of the potential is the field), which causes this electric force $F=Eq$ (where $q$ here is the negative electron charge, and thus the force is opposite towards the higher potential).
Electric field direction is from positive to negative and if a particle travels in its direction , it loses potential energy and vice versa.(since I learned that electric field lines point in the direction of decreasing electric potential)
Correct, if you to your last line add: "...
for a positive charge".
As electrons move they constantly gain potential energy,but they also gain kinetic energy as they are being accelerated by E .So isn't it violating conservation of energy?
No. See the comment for point 3. The electrons are as well moving in the direction of decreasing potential energy.
Even though electrons are moving towards
higher potential $V$ they are still moving towards lower potential , because $U=Vq$. $q$ is negative for electrons and thus the potential energy $U$ here changed sign and is lower than at the other pole. energy
A note: Remember that electrons moving one way corresponds to positive charge moving the other way. So what causes positive charge to flow is exactly the same mechanism that makes negative charge flow, just looked upon up-side-down if you will.
For an electron entering a resistor Initial KE+Initial PE=Final KE +Final PE+Heat dissipated. But since Initial KE=Final KE. Therefore Initial PE should be greater than Final PE but according to point 3 this cannot happen.So what is happening?
You forget here that there is much heat loss in an electric circuit. The complete energy conservation equation says:
$$K_{initial} + U_{initial}=K_{final}+U_{final}+W_{loss}$$
$W_{loss}$ is any energy loss and for a circuit this is heat. Heat generated and lost to the ambient air in resistances in components and wires etc.
Note: The energy that is converted into heat comes from the loss in potential energy - the kinectic energy of the electron remains the same - so $W_{loss}=U_{initial}-U_{final}=-\Delta U$
Also why do we say that current everywhere in a circuit is the same,even before and after entering a resistance?
This is Kirchhoff's current law:
$$\sum i = 0\implies i_{in}=i_{out}$$
It simply states that charge does not accumulate anywhere. All charge entering at any point must also leave at the same rate, so no charge is build up.
Current in must equal current out.
This is intuitively logical:
There cannot leave more charge every second than what enters every second at a point, because where does the charge that leaves come from then. So current out cannot be higher than current in. And if more current enters than what leaves, then charge will build up infinitly. Soon you will have a huge amount of charge here. This huge chunk of negative charge will for sure repel any further incoming electrons and stop the current. So if you do have a current running, then you know that no charge is build up anywhere, and what enters must equal what leaves. |
Let's say a have a correlation matrix $\Omega$ for 25 assets which I use to generate a Monte-Carlo simulation. Let's assume that $\Omega$ is valid (i.e positive-semi-definite, etc...) and estimated empirically with market data.
Now assume that I want to add a risk factor $r$, but I only know the correlation $\rho$ of that risk factor to a given asset $m$. As $r$ is not easily observable, I can't include it in the empirical estimation. Plus, traders can't mark the correlation with the remaining 24 asset without making the correlation matrix invalid (i.e not positive-definite anymore).
I was actually thinking about:
generating $N$ correlated standard normal numbers for the 25 assets for which correlation is known yielding a set $Z$ of size $N \times 25$. Taking from $Z$ the random numbers corresponding to asset $m$, let's denote them $Z_m$, which is a vector of $N$ standard normal random numbers Correlating a new set of indepentand standard normal numbers $X$ with $Z_m$ yielding $Y$, a vector of size $N$ "Appending" $Y$ to $Z$.
Somebody suggested me to do something different:
Extend $\Omega$ adding $r$ Setting $\Omega_{r,m}=\rho$ For each asset $i$ in $\Omega$ such that $i \neq r,m$: Set $\Omega_{r,i}=\Omega_{m,i} \cdot \Omega_{r,m}$ Correlate $Z$ as mentioned in the first point above.
Theoretically, is one of these methods "more right than the other"?
Is there another common approach to solving this kind of problem? |
Every time when two rays intersect or half-lines projecting the common endpoints then the corner points of angles are named as vertices or angles of the rays are named as sides. Angle is also termed as the measurement of a turn between any two lines. The unit of an angle is degree or radian. Further, angles could be divided into multiple categories like double-angle formula, half angle formula, compound angle, or interior angle etc. The angle formula in mathematics is given as below –
When multiple angles are expanded then it will make double angles and take the sum of different angles then again apply the double angle formula.
\[\ sin(A+B)=sinA\;cosB+cosA\;sinB\]
\[\ sin(A-B)=sinA\;cosB-cosA\;sinB\]
\[\ cos(A+B)=cosA\;cosB-sinA\;sinB\]
\[\ cos(A-B)=cosA\;cosB+sinA\;sinB\]
\[\ sin\alpha +sin\beta =2sin\frac{\alpha +\beta }{2}cos\frac{\alpha -\beta }{2}\]
\[\ sin\alpha -sin\beta =2sin\frac{\alpha -\beta }{2}cos\frac{\alpha +\beta }{2}\]
\[\ cos\alpha +cos\beta =2cos\frac{\alpha +\beta }{2}cos\frac{\alpha -\beta }{2}\]
\[\ cos\alpha -cos\beta =-2sin\frac{\alpha +\beta }{2}sin\frac{\alpha -\beta }{2}\]
\[\ sin2\alpha =2\;sin\alpha\;cos\alpha\]
\[\ cos2\alpha =cos^{2}\alpha -sin^{2}\alpha = 2cos^{2}\alpha -1=1-2sin^{2}\alpha\]
\[\ tan2\alpha =\frac{2tan\alpha }{1-tan^{2}\alpha }\]
In case of special identities where sum and differences of sine and cosine functions are calculated, it would be termed as double angle identities or half angle identities. Any double angle when divided by two, the half-angle formula can be derived as given below.
\[\ Sine\;of\;a\;Half\;Angle = \sin \frac{a}{2} = \pm \sqrt{\frac{(1- \cos a)}{2}}\]
\[\ Cosine\;of\;a\;Half\;Angle = \cos \frac{a}{2} = \pm \sqrt{\frac{(1+ \cos a)}{2}}\]
\[\ Tangent\;of\;a\;Half\;Angle = \tan \left ( \frac{a}{2} \right ) = \frac{1 – \cos a}{\sin a} = \frac{\sin a}{1 + \cos a}\]
First Trigonometric expression is an example of double angle formula and the second equation is an example of half-angle formula. In the same way, their multiple halves – angle formulas can be derived for multiple trigonometric functions one by one.
The trigonometric functions for multiple angles are named as multiple angle formula. For double or triple angles formulas, there would come multiple angle formulas ahead. The popular Trigonometric functions are Sine, Cosine, Tangent etc.
The sin formula for multiple angle is:
\[\large sin \theta = \sum_{k=0}^{n}\;cos^{k}\theta \; Sin^{n-k}\theta\; Sin\left [\frac{1}{2}\left(n-k\right)\right]\pi\]
Where n=1,2,3,……
General formulas are,
\[\large sin^{2}\theta =2 \times cos\,\theta \; sin\,\theta\]
\[\large sin^{3}\theta =3 \times cos^{2}\,\theta \; sin\, \theta \; sin^{3}\,\theta\]
The multiple angle’s Cosine formula is given below:
\[\large Cos\;n\, \theta =\sum_{k=0}^{n}cos^{k}\theta \,sin^{n-k}\theta \;cos\left [\frac{1}{2}\left(n-k\right)\pi\right]\]
Where n = 1,2,3
The general formula goes as:
\[\large cos^{2}\, \theta =cos^{2}\, \theta – sin^{2}\, \theta\]
\[\large cos^{3}\, \theta =cos^{3}\, \theta – cos\, \theta \; sin^{2}\, \theta\]
Tangent Multiple Angles formula
\[\large Tan\;n\theta = \frac{sin\;n\theta}{cos\;n\theta}\]
The three sides for a right-angle triangle in mathematics are given as Perpendicular, Base, and the Hypotenuse. The largest side that is opposite to the right angle will be termed as the Hypotenuse. To find a particular side of a Triangle, we should know the other two sides of the Triangle. And the formula is given as –
\[\large Hypotenuse^{2}=(Adjacent\;Side)^{2}+(Opposite\;Side)^{2}\]
The other popular name for right angle formula is the Pythagorean theorem and a right angle is an angle that exactly measures 90-degree. This is the most used formula is mathematics and should be clearly understood by students when preparing for higher studies or competitive exams. There is a special notation in mathematics for the right-angle and it is given by a small square between two sides. Let us understand through figure how it looks alike –
The right angles could be seen at multiple places in our daily life. For example, every rectangular or square object you see around you is a right angle. One of the most common places forthe right angle is a triangle. If there are no right-angles, then Trigonometry existence is not possible in this case. All Trigonometry concepts are based on the right-angle formulas only. Also, the right-angle formula has multiple applications in real-life too.
For example, when you want to calculate the distance up to the slope or you wanted to measure the height of a hill, only right-angle triangle formulas are useful. In the same way, there are just the endless applications for right-angle formula in mathematics.
There are three popular steps for side angle side formulas. These are –
\[\ Area=\frac{ab\;Sin\,C}{2}\] |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\]which matrix do you get?(a) $A$(b) $C^{-1}A^{-1}BC^{-1}AC^2$(c) $B$(d) $C^2$(e) $C^{-1}BC$(f) $C$
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 $Q$.
(The Ohio State University, Linear Algebra Midterm)
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 $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
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 $W$.
(The Ohio State University, Linear Algebra Midterm)
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 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
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) |
This is one of the problem in our past comprehensive exams. I don't mind getting full solution.
Suppose $f$ is a bounded, measurable function on $[0,1]$, $\epsilon>0,$ and for all $x>\epsilon\,$ one has
$$0=\int_0^1 f(s )\exp(-xs)ds$$
Show that $f=0$ almost everywhere.
Someone gave me a hint to solve the problem using Urysohn's lemma. I am not totally comfortable with that lemma. I have a hunch that we can prove this along the line of Fourier analysis. I am not that sure on this approach either. I don't even know how to get started. |
Notes on General Equilibrium
Tags:
Table of contents
I took these notes while preparing for the placement exam for 14.123, the third quarter of the first-year “theory sequence” at MIT. It covers eseentially Part 4 of the this OCW course.
Private Ownership EconomiesA private ownership economy \(\mathcal{E}\) is a tuple \begin{equation} \label{eq:1} \mathcal{E} = (\{ (X_i, \succeq_i) \}_{i=1}^I, \{Y_j\}_{j=1}^J, \{(\omega_i, \theta_{i1}, \dots, \theta_{iJ})\}_{i=1}^I) \end{equation}
where \(\theta_{ij} \in [0,1]\) and \(\sum_i \theta_{ij} = 1\) for each \(j = 1, \dots, J\). The interpretations are:
\(X_i\) and \(\succeq_i\) are agent \(i\)’s consumption set and preferences over consumption. \(Y_j\) is the production set of firm \(j\). \(\omega_i\) is agent \(i\)’s endownment. \(\theta_{ij}\) is the shares of firm \(j\) held by agent \(i\).
It is assumed that \(X_i, Y_j \subseteq \mathbf{R}^L\), where \(L\) is the number of commodities. In addition, assume that \(\omega_i \in X_i\).
Feasible Allocations and Pareto OptimalityAn allocation is a vector \((x_1, \dots, x_I, y_1, \dots, y_J)\) such that \(x_i \in X_i\) and \(y_j \in Y_j\). A feasible allocation is an allocation \((x_1, \dots, x_I, y_1, \dots, y_J)\) such that \begin{equation} \label{eq:2} \sum_i x_i = \sum_i \omega_i + \sum_j y_j = \bar \omega + \sum_j y_j \end{equation}
where \(\bar \omega = \sum_i \omega_i\).A feasible allocation \((x_1, \dots, x_I, y_1, \dots, y_J)\) is Pareto Optimal if there is no other feasible allocation \((x'_1, \dots, x'_I, y'_1, \dots, y'_J)\) such that \begin{equation} \label{eq:3} x'_i \succeq x_i \quad \text{for every $i$ and strict for at least one $i$.} \end{equation}
Walrasian Equilibrium
An allocation \((x^{*}, y^{*}) \in \prod X_i \times \prod Y_j\) and a price vector \(p = (p_1, \dots, p_L)\) constitute a Walrasian equilibrium if
\(y_j^{*}\) maximizes profits for firm \(j\) \begin{equation} \label{eq:4} p \cdot y_j \leq p \cdot y_j^{*} \quad \text{for every $y_j \in Y_j$} \end{equation} \(x_i^{*}\) maximizes utility for agent \(i\) within her budget set in that \begin{equation} \label{eq:5} x_i \succ x_i^{*} \implies x_i \notin B(p, \omega_i, y_j^{*}) := \big\{ x_i \in X_i : p \cdot x_i \leq p \cdot \omega_i + \sum_j \theta_{ij} p \cdot y_j^{*} \big\} \end{equation} The allocation is feasible \begin{equation} \label{eq:6} \sum x_i^{*} = \bar \omega + \sum_j y_j^{*}. \end{equation} Walrasian Equilibrium with TransfersAn allocation \((x^{*}, y^{*}) \in \prod X_i \times \prod Y_j\) and a price vector \(p = (p_1, \dots, p_L)\) constitute a Walrasian equilibrium with transfers if there is some \((w_1,\dots,w_I)\) such that the above conditions hold, except \eqref{eq:5} is replaced with \begin{equation} \label{eq:7} x_i^{*} \quad \text{is maximal in} \quad \{x_i \in X_i : p \cdot x_{i} \leq w_i\} \end{equation} A price quasi-equilibrium with transfers means that \eqref{eq:7} is replaced with \begin{equation} \label{eq:8} x_i \succ x_i^{*} \implies p \cdot x_i \geq w_i. \end{equation} In particular, any Walrasian equilibrium is a Walrasian equilibrium with transfers with \begin{equation} \label{eq:9} w_i = p \cdot \omega_i + \sum_j \theta_{ij} p \cdot y_j^{*} \end{equation} First Welfare Theorem
If preferences are
locally non-satiated
and if \((x^{*}, y^{*}, p)\) is a price equilibrium with transfers, then the allocation \((x^{*}, y^{*})\) is Pareto optimal.
Second Welfare Theorem
If preferences are
convex locally non-satiated
and production sets \(Y_j\) are
convex
then for every Pareto optimal allocation \((x^{*}, y^{*})\), there is a price vector \(p\) such that \((x^{*}, y^{*}, p)\) is a price quasi-equilibrium with transfers. If there is a cheaper consumption bundle \(x'_i\) for agent \(i\) in that \(p \cdot x'_i < w_i\), then \eqref{eq:7} holds for agent \(i\). |
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Energy Transfer Between Squaraine Polymer Sections: From helix to zig-zag and All the Way Back (2015)
Joint experimental and theoretical study of the absorption spectra of squaraine polymers in solution provide evidence that two different conformations are present in solution: a helix and a zig-zag structure. This unique situation allows investigating ultrafast energy transfer processes between different structural segments within a single polymer chain in solution. The understanding of the underlying dynamics is of fundamental importance for the development of novel materials for light-harvesting and optoelectronic applications. We combine here femtosecond transient absorption spectroscopy with time-resolved 2D electronic spectroscopy showing that ultrafast energy transfer within the squaraine polymer chains proceeds from initially excited helix segments to zig-zag segments or vice versa, depending on the solvent as well as on the excitation wavenumber. These observations contrast other conjugated polymers such as MEH-PPV where much slower intrachain energy transfer was reported. The reason for the very fast energy transfer in squaraine polymers is most likely a close matching of the density of states between donor and acceptor polymer segments because of very small reorganization energy in these cyanine-like chromophores.
We have investigated the photodynamics of \(\beta\)-D-glucose employing our field-induced surface hopping method (FISH), which allows us to simulate the coupled electron-nuclear dynamics, including explicitly nonadiabatic effects and light-induced excitation. Our results reveal that from the initially populated S\(_{1}\) and S\(_{2}\) states, glucose returns nonradiatively to the ground state within about 200 fs. This takes place mainly via conical intersections (CIs) whose geometries in most cases involve the elongation of a single O-H bond, while in some instances ring-opening due to dissociation of a C-O bond is observed. Experimentally, excitation to a distinct excited electronic state is improbable due to the presence of a dense manifold of states bearing similar oscillator strengths. Our FISH simulations explicitly including a UV laser pulse of 6.43 eV photon energy reveals that after initial excitation the population is almost equally spread over several close-lying electronic states. This is followed by a fast nonradiative decay on the time scale of 100-200 fs, with the final return to the ground state proceeding via the S\(_{1}\) state through the same types of CIs as observed in the field-free simulations.
Excitonic Properties of Ordered Metal Nanocluster Arrays: 2D Silver Clusters at Multiporphyrin Templates (2016)
The design of ordered arrays of metal nanoclusters such as for example 2D cluster organic frameworks might open a new route towards the development of materials with tailored optical properties. Such systems could serve as plasmonically enhanced light-harvesting materials, sensors or catalysts. We present here a theoretical approach for the simulation of the optical properties of ordered arrays of metal clusters that is based on the ab initio parametrized Frenkel exciton model. We demonstrate that small atomically precise silver clusters can be assembled in one- and two-dimensional arrays on suitably designed porphyrin templates exhibiting remarkable optical properties. By employing explicit TDDFT calculations on smaller homologs, we show that the intrinsic optical properties of metal clusters are largely preserved but undergo J- and H-type excitonic coupling that results in controllable splitting of their excited states. Furthermore, ab initio parameterized Frenkel exciton model calculations allow us to predict an energetic splitting of up to 0.77 eV in extended two-dimensional square arrays and 0.79 eV in tilted square aggregates containing up to 25 cluster-porphyrin subunits.
We present a joint experimental and computational study of the nonradiative deactivation of the benzyl radical, C\(_7\)H\(_7\) after UV excitation. Femtosecond time-resolved photoelectron imaging was applied to investigate the photodynamics of the radical. The experiments were accompanied by excited state dynamics simulations using surface hopping. Benzyl has been excited at 265 nm into the D-band (\(\pi\pi^*\)) and the dynamics was probed using probe wavelengths of 398 nm or 798 nm. With 398 nm probe a single time constant of around 70-80 fs was observed. When the dynamics was probed at 798 nm, a second time constant \(\tau_2\)=1.5 ps was visible. It is assigned to further non-radiative deactivation to the lower-lying D\(_1\)/D\(_2\) states.
We introduce a general theoretical approach for the simulation of photochemical dynamics under the influence of circularly polarized light to explore the possibility of generating enantiomeric enrichment through polarized-light-selective photochemistry. The method is applied to the simulation of the photolysis of alanine, a prototype chiral amino acid. We show that a systematic enantiomeric enrichment can be obtained depending on the helicity of the circularly polarized light that induces the excited-state photochemistry of alanine. By analyzing the patterns of the photoinduced fragmentation of alanine we find an inducible enantiomeric enrichment up to 1.7%, which is also in good correspondence to the experimental findings. Our method is generally applicable to complex systems and might serve to systematically explore the photochemical origin of homochirality.
Molecules containing multiple bonds between atoms—most often in the form of olefins—are ubiquitous in nature, commerce, and science, and as such have a huge impact on everyday life. Given their prominence, over the last few decades, frequent attempts have been made to perturb the structure and reactivity of multiply-bound species through bending and twisting. However, only modest success has been achieved in the quest to completely twist double bonds in order to homolytically cleave the associated π bond. Here, we present the isolation of double-bond-containing species based on boron, as well as their fully twisted diradical congeners, by the incorporation of attached groups with different electronic properties. The compounds comprise a structurally authenticated set of diamagnetic multiply-bound and diradical singly-bound congeners of the same class of compound.
Molecules containing multiple bonds between atoms—most often in the form of olefins—are ubiquitous in nature, commerce, and science, and as such have a huge impact on everyday life. Given their prominence, over the last few decades, frequent attempts have been made to perturb the structure and reactivity of multiply-bound species through bending and twisting. However, only modest success has been achieved in the quest to completely twist double bonds in order to homolytically cleave the associated π bond. Here, we present the isolation of double-bond-containing species based on boron, as well as their fully twisted diradical congeners, by the incorporation of attached groups with different electronic properties. The compounds comprise a structurally authenticated set of diamagnetic multiply-bound and diradical singly-bound congeners of the same class of compound. |
This vignette describes the concept of neighborhood-inclusion, its connection with network centrality and gives some example use cases with the
netrankr package. The partial ranking induced by neighborhood-inclusion can be used to assess partial centrality or compute probabilistic centrality.
In an undirected graph \(G=(V,E)\), the
neighborhood of a node \(u \in V\) is defined as \[N(u)=\lbrace w : \lbrace u,w \rbrace \in E \rbrace\] and its closed neighborhood as \(N[v]=N(v) \cup \lbrace v \rbrace\). If the neighborhood of a node \(u\) is a subset of the closed neighborhood of a node \(v\), \(N(u)\subseteq N[v]\), we speak of neighborhood inclusion. This concept defines a dominance relation among nodes in a network. We say that \(u\) is dominated by \(v\) if \(N(u)\subseteq N[v]\). Neighborhood-inclusion induces a partial ranking on the vertices of a network. That is, (usually) some (if not most!) pairs of vertices are incomparable, such that neither \(N(u)\subseteq N[v]\) nor \(N(v)\subseteq N[u]\) holds. There is, however, a special graph class where all pairs are comparable (found in this vignette).
The importance of neighborhood-inclusion is given by the following result:
\[ N(u)\subseteq N[v] \implies c(u)\leq c(v), \] where \(c\) is a centrality index defined on special path algebras. These include many of the well known measures like closeness (and variants), betweenness (and variants) as well as many walk-based indices (eigenvector and subgraph centrality, total communicability,…).
Very informally, if \(u\) is dominated by \(v\), then u is less central than \(v\) no matter which centrality index is used, that fulfill the requirement. While this is the key result, this short description leaves out many theoretical considerations. These and more can be found in
Schoch, David & Brandes, Ulrik. (2016). Re-conceptualizing centrality in social networks.
European Journal of Appplied Mathematics, 27(6), 971–985. (link)
netrankr Package
library(netrankr)library(igraph)set.seed(1886) #for reproducibility
We work with the following simple graph.
g <- graph.empty(n=11,directed = FALSE)g <- add_edges(g,c(1,11,2,4,3,5,3,11,4,8,5,9,5,11,6,7,6,8, 6,10,6,11,7,9,7,10,7,11,8,9,8,10,9,10))V(g)$name <- 1:11plot(g, vertex.color="black",vertex.label.color="white", vertex.size=16,vertex.label.cex=0.75, edge.color="black", margin=0,asp=0.5)
We can compare neighborhoods manually with the
neighborhood function of the
igraph package. Note the
mindist parameter to distinguish between open and closed neighborhood.
u <- 3v <- 5Nu <- neighborhood(g,order=1,nodes=u,mindist = 1)[[1]] #N(u) Nv <- neighborhood(g,order=1,nodes=v,mindist = 0)[[1]] #N[v] Nu
## + 2/11 vertices, named, from 666e02d:## [1] 5 11
Nv
## + 4/11 vertices, named, from 666e02d:## [1] 5 3 9 11
Although it is obvious that
Nu is a subset of
Nv, we can verify it with
R.
all(Nu%in%Nv)
## [1] TRUE
Checking all pairs of nodes can efficiently be done with the
neighborhood_inclusion() function from the
netrankr package.
P <- neighborhood_inclusion(g)P
## 1 2 3 4 5 6 7 8 9 10 11## 1 0 0 1 0 1 1 1 0 0 0 1## 2 0 0 0 1 0 0 0 1 0 0 0## 3 0 0 0 0 1 0 0 0 0 0 1## 4 0 0 0 0 0 0 0 0 0 0 0## 5 0 0 0 0 0 0 0 0 0 0 0## 6 0 0 0 0 0 0 0 0 0 0 0## 7 0 0 0 0 0 0 0 0 0 0 0## 8 0 0 0 0 0 0 0 0 0 0 0## 9 0 0 0 0 0 0 0 0 0 0 0## 10 0 0 0 0 0 0 0 0 0 0 0## 11 0 0 0 0 0 0 0 0 0 0 0
If an entry
P[u,v] is equal to one, we have \(N(u)\subseteq N[v]\).
The function
dominance_graph() can alternatively be used to visualize the neighborhood inclusion as a directed graph.
g.dom <- dominance_graph(P)plot(g.dom, vertex.color="black",vertex.label.color="white", vertex.size=16, vertex.label.cex=0.75, edge.color="black", edge.arrow.size=0.5,margin=0,asp=0.5)
We start by calculating some standard measures of centrality found in the
ìgraph package for our example network. Note that the
netrankr package also implements a great variety of indices, but they need further specifications described in this vignette.
cent.df <- data.frame( vertex=1:11, degree=degree(g), betweenness=betweenness(g), closeness=closeness(g), eigenvector=eigen_centrality(g)$vector, subgraph=subgraph_centrality(g))#rounding for better readabilitycent.df.rounded <- round(cent.df,4) cent.df.rounded
## vertex degree betweenness closeness eigenvector subgraph## 1 1 1 0.0000 0.0370 0.2260 1.8251## 2 2 1 0.0000 0.0294 0.0646 1.5954## 3 3 2 0.0000 0.0400 0.3786 3.1486## 4 4 2 9.0000 0.0400 0.2415 2.4231## 5 5 3 3.8333 0.0500 0.5709 4.3871## 6 6 4 9.8333 0.0588 0.9847 7.8073## 7 7 4 2.6667 0.0526 1.0000 7.9394## 8 8 4 16.3333 0.0556 0.8386 6.6728## 9 9 4 7.3333 0.0556 0.9114 7.0327## 10 10 4 1.3333 0.0526 0.9986 8.2421## 11 11 5 14.6667 0.0556 0.8450 7.3896
Notice how for each centrality index, different vertices are considered to be the most central node. The most central from degree to subgraph centrality are \(11\), \(8\), \(6\), \(7\) and \(10\). Note that only
undominated vertices can achieve the highest score for any reasonable index. As soon as a vertex is dominated by at least one other, it will always be ranked below the dominator. We can check for undominated vertices simply by forming the row Sums in
P.
which(rowSums(P)==0)
## 4 5 6 7 8 9 10 11 ## 4 5 6 7 8 9 10 11
8 nodes are undominated in the graph. It is thus entirely possible to find indices that would also rank \(4, 5\) and \(9\) on top.
Besides the top ranked nodes, we can check if the entire partial ranking
P is preserved in each centrality ranking. If there exists a pair \(u\) and \(v\) and index \(c()\) such that \(N(u)\subseteq N[v]\) but \(c(v)>c(u)\), we do not consider \(c\) to be a valid index.
In our example, we considered vertex \(3\) and \(5\), where \(3\) was dominated by \(5\). It is easy to verify that all centrality scores of \(5\) are in fact greater than the ones of \(3\) by inspecting the respective rows in the table. To check all pairs, we use the function
is_preserved. The function takes a partial ranking, as induced by neighborhood inclusion, and a score vector of a centrality index and checks if
P[i,j]==1 & scores[i]>scores[j] is
FALSE for all pairs
i and
j.
apply(cent.df[,2:6],2,function(x) is_preserved(P,x))
## degree betweenness closeness eigenvector subgraph ## TRUE TRUE TRUE TRUE TRUE
All considered indices preserve the neighborhood inclusion preorder on the example network.
NOTE: Preserving neighborhood inclusion on one network does not guarantee that an index preserves it on all networks. For more details refer to the paper cited in the first section. |
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I would like to know some resources where I can learn all these different physics concepts that are on Brilliant. If you know any good books, websites, etc. just write them in the comments.
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MIT OCW Walter Lewin lectures are the best! The scholar version is filled with testing and all!! Freedman and young or Halliday Resnick and walker can serve as text books!! But Walter Lewin is the best!
University physics by freedman and young
Yo can refer "concept of physics" by HC Verma
I would try to get hold of classical electromagnetism in a nutshell and the Feynman lectures.
To fall in love with Physics, watch the entire series by Walter Lewin on Mechanics, Electricity and Magnetism on the official MIT channel on youtube.
flippingphysics.com
YES! ME TOO! It makes no sense!
How are you so good at electricity and magnetism? I only managed to get to lvl 2 because of some basic concepts and multiple choice questions.
Well... I took the pre-test and got Level 4 by guessing, and then (feeling cocky) proceeded to lose that level to a Level 2 problem. I don't even know what E+M is! :D
@Finn Hulse – Really, I started learning this year.
For books I'd say you should read "Lectures on Physics by R.P.Feynman " ,or "Fundamentals of physics by Resnick,Halliday and Walker "
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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 |
Structure of Atom Quantum Numbers and Electronic Configuration of Elements
Quantum Numbers : a. Principal quantum number (n) b. Azimuthal quantum number (l) c. Magnetic quantum number (m) d. Spin quantum number (s) a. Principal quantum number (n) r_{n} = 0.54\frac{n^{2}}{Z} Indicates size of the orbit. Indicates energy of electron b. Azimuthal quantum number (l) Denoted by letter ' l ' and can have value from 0,1 to (n − 1) l = 0 → s orbital → spherical l = 1 → p orbital → dumb-bell l = 2 → d orbital → double dumb-bell l = 3 → f orbital → fore fold c. Magnetic quantum number (m) can have values −l to +l total values (2l + 1) d. Spin quantum number (s) It can have values s = +½ and s = −½
Exchange Energy :
Exchange Energy =
nC 2 =\frac{n!}{(n - 2)! \ 2!} n = no.of unpaired electrons Cr = 3d 5 4s 1
n = 6
6C 2 = 15
Magnetic \ moment (\mu) = \sqrt{n(n + 2)} BM
n μ=\sqrt{n(n + 2)} BM 1 \sqrt{3}=1.732 BM 2 \sqrt{8}=2.823 BM 3 \sqrt{15}=3.89 BM 4 \sqrt{24}=4.88 BM 5 \sqrt{35}=5.89 BM
Electronic configuration :
nl
x method moeller diagram :
1s < 2s < 2p <3s <3p < 4s < 3d < 4p < 3s < 4d < 5p < 6s < 4f < < 5d < 6p < 7s < 5f < 6d < 7p < 8s
Part1: View the Topic in this Video from 37:35 to 50:55 Part2: View the Topic in this Video from 1:17 to 18:35
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1. The number of electrons in subshell = 2(2
l + 1). 2. Number of orbitals in main energy level = n 2. 3. Maximum number of electrons in nth shell = 2n 24. The orbital angular momentum of an electron = \tt \sqrt{l\left(l + 1\right)}\frac{h}{2\pi}
5. Spin angular momentum = \tt \sqrt{s\left(s + 1\right)}\frac{h}{2\pi}
6. Magnetic moment = \tt \sqrt{n\left(n + 2\right)} B.M. (Bohr magneton) of n unpaired e
- |
Let's first recall the normal construction for Shamir's Secret Sharing:
We work in a finite field $\mathbb{K}$. This will usually be $\mathbb{F}_{256}$ (the binary field with 256 elements), so that the sharing can be conveniently done on a per-byte basis. We want an $m$-out-of-$n$ scheme, so the finite field must have at least $n+1$ elements. For each value to share, a random polynomial $P \in \mathbb{K}[X]$ of degree at most $m-1$: we write $P = \sum_{i=1}^{m-1} p_i X^i$ where the $p_i$ are chosen randomly and uniformly in $\mathbb{K}$, except for $p_0$, which we set to the secret value to share.
Thus, $P(0) = p_0$ is the secret value. The "shares" are $P(v_1)$, $P(v_2)$... $P(v_n)$, where the $v_j$ values are $n$ distinct non-zero elements of $\mathbb{K}$, that are chosen arbitrarily through a fixed, public convention (exactly which convention is used does not matter).
To retrieve the secret out of $m$ shares, we really rebuild the polynomial $P$, using Lagrange's polynomials. Let $N_i \in \mathbb{K}[X]$ be the following polynomial:$$ N_i = \left(\prod_{1\leq j\leq n, j\neq i} (X - v_j) \right) $$It is easily seen that $N_i(v_j) = 0$ for any $j \neq i$, but $N_i(v_i) \neq 0$. We thus define the Lagrange polynomial $L_i = N_i / N_i(v_i)$; it has value $1$ on $v_i$, and $0$ on all other $v_j$.
Then, $P$ is reconstructed as a linear combination of the Lagrange polynomials. If we have shares $P(v_1)$, $P(v_2)$,... $P(v_m)$, then we can write:$$ P = \sum_{i=1}^{m} P(v_i) L_i $$Therefore, the secret is:$$p_0 = P(0) = \sum_{i=1}^{m} P(v_i) L_i(0)$$
In a practical implementation, the coefficients $L_i(0)$ will be precomputed, and the reconstruction needs not bother with actually rebuilding the polynomial $P$; it just recomputes the secret $p_0$ as a linear combination of the shares $p_i$.
Now, let's examine the problem at hand: you have $k$ predefined "shares" and want to use them in a secret sharing. To do that, you simply apply the reconstruction above, and use fresh random values when needed. In other words, you first declare that the polynomial $P$ really exists, and that the $k$ predefined shares are $P(v_1)$, $P(v_2)$,... $P(v_k)$. You then generate randomly and uniformly $m-k-1$ extra shares, which you call $P(v_{k+1})$, $P(v_{k+2})$,... $P(v_{m-1})$. Finally, you define that your secret value is $P(0)$. At that point, you have the evaluation of $P$ on exactly $m$ values ($0$, and the non-zero elements $v_1$ to $v_{m-1}$). You then define new Lagrange polynomials for a set of values that include $0$:\begin{eqnarray*}N'_0 &=& \prod_{1\leq j\leq m-1} (X - v_j) \\N'_i &=& X \prod_{1\leq j\leq m-1, j\neq i} (X - v_j) \\L'_0 &=& N'_0 / N'_0(0) \\L'_i &=& N'_i / N'_i(0) \\\end{eqnarray*}At that point, you can find $P$:$$ P = P(0) L'_0 + \sum_{i=1}^{m-1} P(v_i) L'_i $$
What happened here is that you rebuilt a polynomial $P$ that matches your predefined shares
and the target secret value; you also generated new random shares in order to have enough value to make a unique candidate for $P$. Now that you have the polynomial $P$, you can compute as many extra shares $P(v_i)$ (for $i\geq m$) as you need.
Some extra notes:
If $k < m$, you can still choose the secret value, as shown above. It does not need to be "something random". If you do not want to choose the shared secret, you can simply create $m-k$ extra shares (instead of $m-k-1$) and use the normal Lagrange interpolation to get $P$, at which point you can obtain $P(0)$ as the shared secret. The computations always work, for all input values (that's how the scheme is secure, indeed). This would also be the situation if you started from exactly $m$ predefined shares: if you want a threshold of $m$ and already have $m$ shares, then there is no room for choice of the secret.
In all of the above, I used $P(0)$ for the secret, but that's purely conventional. $0$ is not special here. The gist of Lagrange interpolation is that, for any set of values $P(v_i)$ on $m$ different field elements $v_i$, there is a unique matching polynomial $P$ of degree less than $m$ (i.e. a polynomial that can be represented as $m$ coefficients). |
Let's prove that $latex e^{\pi} > \pi^e$ without a calculator. If you haven't seen this before, give it a try before you read any further. Consider the function $latex f(x) = x/\ln(x)$ on the interval $latex (1,\infty)$. This function shoots off to positive infinity as $latex x$ tends towards either endpoint of the domain. Let's … Continue reading e^pi > pi^e
Consider the problem of finding the following limit:$latex \lim_{x \rightarrow 0} x^x&s=2$It's actually not too bad. We can write$latex x^x = e^{ x \ln x}&s=2$and bring the limit into the exponent (as exponentiating is continuous) to get that$latex \lim_{x \rightarrow 0} x^x = e^{\lim_{x \rightarrow 0} x \ln x}&s=2$From here, all we need to do … Continue reading Limits and Towers
Feel up to seeing a bit of mathematical magic? See here.
http://www.youtube.com/watch?v=g0qbNksZLgo Just me doing my 3MT.
At the end of my undergraduate degree, I remember thinking "what better way to mark such an event than to write and record a maths rap?" Well, I did end up recording such a thing, but it was pretty woeful. However, a couple of months ago, I plucked up the zest to do it again. … Continue reading Don’t Mess with the Mathematician
Integrals can tell us quite a lot. For those of you who are so disgustingly bored that you've found your way onto my blog, you should have a go at evaluating exactly the following integral:$latex \int_0^1 \frac{x^4 (1-x)^4}{1+x^2} dx$It might take some effort, but it's well worth it. Of course, Wolfram Mathematica could do it … Continue reading Late night integration
I thought it would be a good idea to relay some of the advice given by Sir Michael Francis Atiyah during his talk on Tuesday. Sir Michael Atiyah during his talk at #hlf13 Picture: HLFF @Bernhard Kreutzer Always ask yourself questions. Atiyah says that one of the secrets of his success is to always be curious. … Continue reading Advice to a Young Mathematician |
Heisenberg's uncertainty principle tells us that there is a lower limit on the accuracy with which we can determine both the momentum and the position of a particle simultaneously.
$\Delta$$x$ $\Delta$$p$ $\ge$ $\frac{\hbar}{2}$
Suppose we decide to determine the position with 100 % accuracy thus remaining completely uncertain about the momentum.After measurement we can represent such a system by a delta function, which represents a state with definite position.At least this works theoretically.
Is this possible practically? I mean is there an experimental set up for which we can make $\Delta$$x$=$0$ and $\Delta$$p$ = $\infty$ rather than having $\Delta$$x$ as small finite value close to $0$ and $\Delta$$p$ as some large but finite huge value.
For more clarity on what I mean,consider Heisenberg's Microscope.We cannot determine the position of the electron with complete. If we try to use high wavelength light we get a fuzzy image and if we use a low wavelength light, we disturb the electron and its exact position cannot be determined. Is there some way in this experiment of some other experiment where we can arrange the experiment setup in such a way that we are forced to get $\Delta$$x$ =$0$ and $\Delta$$p$=$\infty$ ?(or are we forced to have $\Delta$$x$ as some non-zero finite quantity as long we are doing the experiment practically?) |
Could someone please tell me if I am correct and if I am not, tell me where I went wrong?
Using the laws of logic prove that $ [\neg q \land (p \rightarrow q)] \rightarrow \neg p$ is a tautology.
First I used the Implication law $(p \rightarrow q) \equiv (\neg p \vee q)$ to show that $$[\neg q \land (p \rightarrow q)] \equiv [\neg q \land (\neg p \vee q)]$$
Then I "factored" (?) the $\neg$ out and had $$ \neg [q \vee (p \land \neg q)] $$
And since $(p \land \neg q)$ denotes to "$p$ but not $q4" then I assumed I could leave $q$ out, leaving me with $$ ¬[q∨(p)] $$
Which is
$$ \equiv (\neg q \land \neg p) $$
Which says "not $q$ and not $p$".
Since it is "not $p$", does that mean that
$$ \equiv (\neg q \land \neg p) \rightarrow \neg p $$
and prove that $ [\neg q \land (p \rightarrow q)] \rightarrow \neg p$ is a tautology? |
The existence and nonexistence results of ground state nodal solutions for a Kirchhoff type problem
1.
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
2.
School of History Culture and Ethnology, Southwest University, Chongqing 400715, China
$ -\left( a+b\int_{\Omega }{|}\nabla u{{|}.{2}}dx \right)\vartriangle u=\lambda u+|u{{|}.{2}}u,\ \ u\in H_{0}.{1}(\Omega ), $ [, X.H. Tang and B.T. Cheng (2016) 21] [. 22] Mathematics Subject Classification:Primary: 35J20; Secondary: 35J65. Citation:Xiao-Jing Zhong, Chun-Lei Tang. The existence and nonexistence results of ground state nodal solutions for a Kirchhoff type problem. Communications on Pure & Applied Analysis, 2017, 16 (2) : 611-628. doi: 10.3934/cpaa.2017030
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X. H. Tang and B. T. Cheng,
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Sign changing solutions of superlinear Schrödinger equations,
[3]
T. Bartsch and T. Weth,
Three nodal solutions of singular perturbed elliptic equations on domains without topology,
[4] [5] [6]
K. J. Brown and T. F. Wu,
A semilinear elliptic system involving nonlinear boundary condition and sign-changing weight function,
[7]
K. J. Brown and Y. P. Zhang,
The Nehari manifold for a semilinear elliptic equation with a sign-changing weight function,
[8]
Y. B. Deng, S. J. Peng and W. Shuai,
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Caisheng Chen, Qing Yuan.
Existence of solution to $p-$Kirchhoff type problem in
$\mathbb{R}^N$ via Nehari manifold.
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[3]
Hongyu Ye.
Positive high energy solution for Kirchhoff equation in $\mathbb{R}^{3}$ with superlinear nonlinearities via Nehari-Pohožaev manifold.
[4]
Xianhua Tang, Sitong Chen.
Ground state solutions of Nehari-Pohozaev type for Schrödinger-Poisson problems with general potentials.
[5]
Sitong Chen, Junping Shi, Xianhua Tang.
Ground state solutions of Nehari-Pohozaev type for the planar Schrödinger-Poisson system with general nonlinearity.
[6]
Patrizia Pucci, Mingqi Xiang, Binlin Zhang.
A diffusion problem of Kirchhoff type involving the nonlocal fractional
[7]
Zhiqing Liu, Zhong Bo Fang.
Global solvability and general decay of a transmission problem for kirchhoff-type wave equations with nonlinear damping and delay term.
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Nemat Nyamoradi, Kaimin Teng.
Existence of solutions for a Kirchhoff-type-nonlocal operators of elliptic type.
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The Nehari manifold for a fractional Laplacian equation involving critical nonlinearities.
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Middle East Technical University General Seminars
Approximation of Exit Probabilities of Constrained Random Walks
Ali Devin Sezer
METU, Turkey
Özet : Let $X$ be the constrained random walk on $\mathbb{Z}_+^2$ having increments $(1,0)$, $(-1,1)$, and $(0,-1)$ with probabilities $\lambda$, $\mu_1$, and~$\mu_2$ representing the lengths of two tandem queues. $X$ is assumed stable and $\mu_1 \neq \mu_2$. Let $\tau_n$ be the first time when the sum of the components of $X$ equals $n$. Let $Y$ be the constrained random walk on $\mathbb{Z} \times \mathbb{Z}_+$ having increments $(-1,0)$, $(1,1)$, and $(0,-1)$ with probabilities $\lambda$, $\mu_1$, and $\mu_2$. Let $\tau$ be the first time that the components of $Y$ are equal to each other. We prove that $P_{n-x_n(1),x_n(2)}(\tau < \infty)$ approximates $p_{x_n}(\tau_n < \tau_0)$ with relative error {\em exponentially decaying} in $n$ for $x_n = \lfloor nx \rfloor,\, x \in \mathbb{R}^2,\, 0 < x(1) + x(2) < 1$, $x(1) >0$. An affine transformation moving the origin to the point $(n,0)$ and letting $n\rightarrow \infty$ connects the $X$ and $Y$ processes. We use a linearm combination of basis functions constructed from single and conjugate points on a characteristic surface associated with $X$ to derive a simple expression for$P_y(\tau < \infty)$ in terms of the ratios $\lambda/\mu_i.$ The proof that the relative error decays exponentially in $n$ uses an upper bound on the error probability and a lower bound on $p_n$ obtained via sub and super solutions of a related Hamilton-Jacobi-Bellman equation. We carry out a similar analysis also for the constrained random walk with increments $(1,0)$, $(-1,0)$, $(0,-1)$ and $(0,1)$ representing the lengths of two queues in parallel. Although the main ideas generalize from the tandem case there are also significant differences. We provide a comparison of these two cases.
Tarih : 18.04.2019 Saat : 15:40 Yer : Gündüz İkeda Seminar Room Dil : English |
At the core of MIDI is a representation of music as discrete note events, each of those having a static pitch. This is perfect for representing music as played on keyboard instruments. You can convert any frequency corresponding to a note on the tempered scale into a MIDI note number, using:
$69 + 12 \times \log_2 \frac{frequency}{440}$
Under the assumption that the MIDI receiver is calibrated for A4 = 440 Hz.
This representation is alright for piano music, but the problem is how to represent pitches which are not mapped to the tempered scale (non-western music, non musical sounds), and how to represent pitch variations over the duration of a note (glissando, vibrato).
This is done in MIDI by using "pitch bend messages" which instruct the synthesizer to shift the pitch of the currently played note by a small interval. Most synthesizers are calibrated by default for +/- 2 semitones over the course of the pitch bend message range (0 .. 16383). 8192 corresponds to no pitch bending - the emitted pitch is exactly that of the note value. The mapping between the pitch bend value and the frequency shifting ratio is given by:
$\frac{f_{emitted\_note}}{f_{note\_message}} = 2 ^ {\frac{pitchbend - 8192}{4096 \times 12}}$
You can thus get the frequency of a note played by a syntheszier from the following formula:
$440 \times 2^{\frac{note - 69}{12.0} + \frac{pitchbend - 8192}{4096 \times 12}}$
Where
note is the 7-bits MIDI note number of the last received Note On message ; and pitchbend is the 14-bits value of the last received Pitch bend message. A synthesizer starts with its pitch bend register set to 8192, and this value is also reset during the reception of a "Reset all controllers" message.
Let us take the following example. You want to express a flute trill with the following frequency trajectory: 500 Hz, 510 Hz, 500 Hz, as MIDI messages.
The base note number is:
$round(69 + 12 \times \log_2(500 / 440)) = 71$.
So you send a "note on" message with note# equal to 71. This is equivalent to a pitch of:
$440 \times 2 ^ {(71 - 69) / 12} = 493.88$
Which is the nearest pitch on the tempered scale. You need to send a pitch bend message to raise the pitch by a factor of:
$\frac{500}{493.88} = 1.0124$
And get your 500 Hz. The corresponding pitch bend value is:
$round(8192 + 4096 \times 12 \times log_2 1.0124) = 9065$
To get your 510 Hz, the pitch bend value would be:
$round(8192 + 4096 \times 12 \times log_2 \frac{510}{493.88}) = 10469$
So your sequence of MIDI messages for 500, 510, 500 Hz would be:
NOTE 71 PITCH BEND 9065 ... PITCH BEND 10469 ... PITCH BEND 9065
You can think of the MIDI note number as the "integral" part of the pitch ; and the pitch bend as a redundant "fractional" part of the pitch. |
The obvious choice for Alex's [vampire Dr. Alex Schwartz's, PHANG's] research area, given the Oxford topology group's interests, appears to be Topological Quantum Field Theory, or TQFT. But what is a TQFT? And what does it have to do with category theory?...
In honor of Halloween, I would like to discuss mathematics associated with one of my favorite fictional mathematicians: Dr. Alex Schwartz, from Charles Stross's Hugo-award-nominated series
The Laundry Files. Alex is from the north of England and "suffers from an inordinate case of impostor syndrome." Like many people with math backgrounds, he has a job outside academia: at the beginning of The Rhesus Chart, he is working in London as a "quant," developing new financial models for the research arm of a major investment bank. He is also a vampire.
There's a long literary tradition associating certain kinds of geometry with horror. It began with the astonishingly influential (and astonishingly racist) writer H.P. Lovecraft's references to non-Euclidean geometry and continues in more light-hearted references to twisting spaces and Escher's art. In Stross's fiction, this association is literal: sufficiently complex algorithmic or mathematical calculations can open gateways to other worlds and the unspeakable horrors that lurk within. Alex's multidimensional financial models create just such a gateway.
My favorite game, though, is trying to work out the topic of Alex's dissertation. What do we know about Alex's mathematical background?
Armed with this information, one can try to identify Alex's research area, and perhaps even his advisor. There are multiple people in the Oxford topology group working on problems in algebraic topology inspired by theoretical physics. Category theory is a meta-subject, focused on common structures that appear in different branches of mathematics, so perhaps it's not surprising for physics-inspired topology problems to cross into category theory. (Oxford also has a very active mathematical physics group incorporating some people who literally work on string theory, but Alex's interest in physics seems to be limited to the math problems it produces.) The obvious choice for Alex's research area, given the Oxford topology group's interests, appears to be Topological Quantum Field Theory, or TQFT. But what is a TQFT? And what does it have to do with category theory?
Let's start with a very simple story motivated by string theory. Suppose we have a little tiny loop. Because this is a string theory story, this loop represents a fundamental component of the universe; perhaps a physicist in a lab would identify it as a light particle like a photon, or something more exotic, like a neutrino. After a little while, our loop splits into two loops. In the physicist's version of events, one particle has become two. We can illustrate this story with a surface, the aptly named pair of pants:
At the top of the picture we have our starting loop, and at the bottom of the picture we have the final two loops. Intermediate slices of the surface correspond to intermediate moments of time.
Now, even if the fundamental components of the universe do act like tiny loops--a matter which is a subject of much debate--we can't observe these loops directly. Not only is there no microscope powerful enough, it is impossible to build a microscope powerful enough. Instead, we associate to each loop a
quantum state, representing quantities that someone could measure in a lab. Geometrically, we can think of these numbers as a vector, a little arrow with its tail at the origin and its head at the coordinates given by our list of numbers. If you're a quantum physics enthusiast, you may recognize that this vector describes the probability that a physicist will make particular measurements. At the moment, though, we're more interested in the shape of our pair of pants. Shapes like this will give us the topology in Topological Quantum Field Theory.
Suppose we have two loops that, after a while, join to form a single loop:
We can combine the pair of pants shape with the upside-down pair of pants shape to form a new shape, corresponding to the story of a loop that splits into two, then recombines into a single loop:
Next, we'd like to recast our stories about loops and strings in the language of category theory. To specify a category in the sense of category theory, you start by specifying the
objects you are studying. For example, I might have a category where my objects are collections of miniature animals:
Categories also come with
morphisms, which are ways of transforming the objects in the category. You can think of morphisms as particularly nice maps or functions. Each morphism has a source (sometimes called a "domain") and a target. In the miniature animal collection category, I have morphisms that consist of "adding an animal":
If the target of one morphism matches the source of another morphism, we can compose the two morphisms. For example, I can add a frog after adding a chicken:
But I can't compose my "adding a chicken" morphism with itself: I only own one toy chicken, so an animal collection with a chicken in it cannot be the source of an adding a chicken morphism.
In our stories about strings, the objects are one or more loops, and the morphisms are the surfaces that connect them. These connecting surfaces are called
cobordisms. The word "cobordism" combines "bord-" from border with "-ism" from morphism; we say "cobordisms" rather than "bordisms" because we are reversing the operation of taking the border of a surface.
It's worth pointing out that there can be multiple different cobordisms relating objects. For instance, one cobordism from a single loop to a single loop is a cylinder:
But we have already seen another cobordism starting and ending with one loop that has a hole in the middle:
Our category of cobordisms has some extra structure, which comes from the geometric operation of combining loops. Mathematically, the procedure of combining multiple loops without gluing them together is called the
disjoint union. (It's "disjoint" because they are not joined!) The nice thing about disjoint union is that it's compatible with our morphisms, the cobordisms, as well as our objects. For example, we can combine the pair of pants with the hole-in-the-middle cobordism to make a cobordism taking two loops to three loops:
In this context, one can think of the disjoint union operation as being like multiplication. There's even an identity element: taking the disjoint union with the empty set gives us the same cobordism we started with. Mathematically, adding this multiplication-like operation to our category makes it a
monoidal category.
Mathematicians don't necessarily stop at 1-dimensional loops and 2-dimensional cobordisms. You can also make a monoidal category out of smooth
d-dimensional shapes (strictly speaking, d-manifolds) that form the boundaries of d+1-dimensional cobordisms. The loops and surface cobordisms category has two advantages, though: it matches the stories from string theory, and it's easy to draw examples.
Our story about strings also included vectors, associated with quantum states. Let's use some of these ingredients to build a different kind of monoidal category. There is certainly a category whose objects are vectors, but we want to do something a little fancier: our objects will be
vector spaces. For example, we could take the two-dimensional vector space consisting of all of the vectors with their tails at the origin and their heads at a point in a plane. In physics terms, each vector space represents all of the possible quantum states that a string might be in at a particular moment in time.
The morphisms in our category will be
linear transformations. These include operations such as stretching, rotating, or flipping all the vectors in a vector space. In terms of equations, each linear transformation corresponds to multiplying by a matrix. For example, we can include a two-dimensional vector space as the $xy$-plane in a three-dimensional vector space by multiplying by an appropriate matrix: $$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \\ 0 \end{pmatrix}$$
We have objects and morphisms now. For a monoidal category, we also need a "multiplication" operation. This means we need a way to combine two vector spaces
that might have different dimensions in order to make a new vector space. We will use an operation called the tensor product of vectors, which is usually written $\otimes$. Tensor product is a really multiplication-heavy multiplication: we will multiply every single coordinate of the first vector by every coordinate of the second vector, then line up the resulting numbers into a new vector. Thus, if $\vec{u}$ has $m$ coordinates and $\vec{v}$ has $n$ coordinates, the tensor product $\vec{u} \otimes \vec{v}$ has $mn$ coordinates. We can organize the computation of $\vec{u} \otimes \vec{v}$ by computing the matrix $\vec{u} \vec{v}^T$ and then listing its entries in a vector of length $mn$.
As an example, let's compute $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \otimes \begin{pmatrix} 4\\ 5 \end{pmatrix}$.
$$ \begin{align} \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \begin{pmatrix} 4\\ 5 \end{pmatrix}^T &= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \begin{pmatrix} 4 & 5 \end{pmatrix} \\ &= \begin{pmatrix} 4 & 5 \\ 8 & 10 \\ 12 & 15\end{pmatrix} . \end{align} $$
Thus,
$$\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \otimes \begin{pmatrix} 4\\ 5 \end{pmatrix} = \begin{pmatrix} 4 \\ 5 \\ 8 \\ 10 \\ 12 \\ 15 \end{pmatrix}.$$
Now, if we have an $m$-dimensional vector space $V$ and an $n$-dimensional vector space $W$, the tensor product of vector spaces $V \otimes W$ is just all the possible tensor products of a vector from $V$ and a vector from $W$. The new vector space $V \otimes W$ is $mn$-dimensional. It's also possible to take the tensor product of two linear transformations, by finding the output of each linear transformation and then taking the tensor product of those outputs.
Now that we have two different monoidal categories, we want a way to relate them. When mathematicians want to compare two categories, they usually look for a
functor. Functors assign each object in one category to an object in the other category, and assign each morphism to a corresponding morphism. For example, I could create a category from my "collections of miniature animals" category to a "sets of names" category. Perhaps the bear is named Emmy, the green turtle is named Logo, and the blue turtle is named Pig. Then my functor takes this collection:
and yields {Emmy, Logo, Pig}. The functor takes the "adding an animal" morphism to the "adding a name" morphism. For example, adding a chicken corresponds to the transformation from {Emmy, Logo, Pig} to {Emmy, Logo, Pig, Henrietta Swan}.
We now have all of the language we need to describe topological quantum field theory. A Topological Quantum Field Theory, or TQFT, is a functor from the category of cobordisms of loops (or, more generally, from the category of cobordisms of $d$-dimensional manifolds) to the category of vector spaces that takes disjoint union to tensor product. In other words, a TQFT is a rule that tells you how to associate a vector space to a loop and a linear transformation to a cobordism. Disjoint unions on the cobordism side give you tensor products on the vector space side.
From a mathematical perspective, the interesting thing about a TQFT is that it relates very different styles of mathematics. Cobordisms are easy to visualize, at least in low dimensions, but the geometric information associated with them can grow very complicated very quickly. In contrast, working with tensor products can be algebraically intensive, but it's a very straightforward kind of labor. In particular, though it's hard to teach a computer to understand the boundaries of surfaces, it's very easy to make a computer manipulate matrices for you.
We've seen that the study of TQFTs combines topology (from cobordisms), algebra (from tensor products), category theory (to describe the functors), and even potentially some coding (to make a computer work out examples). This is excellent preparation for the modern, mathematically inclined vampire! |
Title Minimization of functionals of the gradient by Baire's theorem Publication Type Journal Article Year of Publication 2000 Authors Zagatti, S Journal SIAM J. Control Optim. 38 (2000) 384-399 Abstract
We give sufficient conditions for the existence of solutions of the minimum problem $$ {\mathcal{P}}_{u_0}: \qquad \hbox{Minimize}\quad \int_\Omega g(Du(x))dx, \quad u\in u_0 + W_0^{1,p}(\Omega,{\mathbb{R}}), $$ based on the structure of the epigraph of the lower convex envelope of g, which is assumed be lower semicontinuous and to grow at infinity faster than the power p with p larger than the dimension of the space. No convexity conditions are required on g, and no assumptions are made on the boundary datum $u_0\in W_0^{1,p}(\Omega,\mathbb{R})$.
URL http://hdl.handle.net/1963/3511 DOI 10.1137/S0363012998335206 Minimization of functionals of the gradient by Baire's theorem
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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 ... |
The purpose of this assignment is to get students hands on practice with fundamental concepts from vector math and also to introduce them to Javascript and the glmatrix library (version 3.1), and particularly the vec3 class. Students will implement a series of mini algorithms using vector concepts that are commonly used in larger geometry systems. These are real world primitive operations that pop up all the time in graphics, robotics, and physics simulations, to name a few of the many application areas. For a large number students, this will serve as an introduction to translating mathematics into code, which is an extremely useful but sometimes underrepresented skill.
This assignment is split up into two parts to keep everyone on track. You will have five days to complete each part. It is highly recommended that you move onto the second part as soon as you finish the first part, as the second part is conceptually trickier.
Getting Started
To get started, check out or download the code in the two github repositories for each part at
https://github.com/Ursinus-CS476-F2019/Mini1_GeometricPrimitives_Part1 https://github.com/Ursinus-CS476-F2019/Mini1_GeometricPrimitives_Part2
. All of the
GeomPrimitives.js
.html files are GUI front end debugging tools for you to check to make sure your code is working properly, as will be described in the individual tasks. I have included a minified version of the glmatrix library which contains the
vec3 class for manipulating 3D vectors, as well as a minified version of the plot.ly library for 3D plotting (which I used to help create some GUI visualizations to help you debug...no need to worry about it unless you're interested).
Collaboration
This is an individual assignment, so all code must be completely your own. However, feel free to post questions on Microsoft Teams and to work with each other on the conceptual math part. Also, if you need help debugging, it is fine to show each other Javascript code that you develop for that purpose (just not code that you write for the actual implementations in
GeomPrimitives.js). If you are unsure if something is allowed, please ask.
Tips Be sure to carefully reference the documentation of the vec3 class as you go along. Be sure to allocate new vector objects with
vec3.create()or
vec3.fromValues()before storing results in them or using them in any way.
Since this is our first Javascript assignment, don't forget to hit
F12to bring up the powerful debugging suite, including the console and the ability to set breakpoints. Use logging in the console with
console.log("Test strings" + x + whatever)to output intermediate values to the debug console.
Ask questions on Microsoft Teams as you are going along and get stuck. Javascript is very "loose" with declarations and types, so there are many things that can happen that wouldn't happen in other languages. One of the most common errors I have personally made is accidentally re-using variable names. This is particularly bad when the original variable was an object and the new variable is a numerical value, and then I try to use the numerical value as if it were an object. This will commonly manifest itself as some kind of "undefined" variable if you log it to the console. to Canvas when you are finished. The first version will contain implementations of tasks 1-4 (
GeometricPrimitives.js
due Wednesday 9/4 at 11:59PM), and the second version will contain implementations of tasks 5, 6, and 7 ( due Friday 9/13). Please also submit a file with both submissions with the following information each time:
README.txt
Your name Approximately how many hours it took you to finish this assignment ( I will not judge you for this at all...I am simply using it to gauge if the assignments are too easy or hard) Your overall impression of the assignment. Did you love it, hate it, or were you neutral? One word answers are fine, but if you have any suggestions for the future let me know. Any other concerns that you have. For instance, if you have a bug that you were unable to solve but you made progress, write that here. The more you articulate the problem the more partial credit you will receive (fine to leave this blank)
Study this code carefully and post on Microsoft Teams if you have any questions
Part 1 (Due Wednesday 9/4 at 11:59PM)vec3 class. You will write on function for parallel projection and one function for perpendicular projection of vectors (what's left over after parallel projection) Code to write: and
projVector(u, v)
in
projPerpVector(u, v)
GeomPrimitives.js
Gui Tester:
projection.html
Tips: Write the code for parallel projection first, then see if you can call that function to help with perpendicular projection. a, band c, compute the angle in degreesbetween band cwith respect to a, as shown in the picture below Code to write:Fill in in
getAngle(a, b, c)
GeomPrimitives.js
Gui Tester:
angle.html
Alternatively, you can use the dot product and the fact that
\[ \sin(\theta) = \sqrt{ 1 - \cos^2(\theta) } \](this is simply the pythagorean theorem on unit triangles) Code to write:Fill in in
getTriangleArea(a, b, c)
GeomPrimitives.js
Gui Tester:
area.html
Tips: Test this with some examples you know in the plane (z=0) where you can apply the formula (1/2*base*height) to make sure it makes sense a, b, and c, determine if the point dis above or below the plane determined by the triangle , with the plane normal determined by the right hand rule traversing the points counter-clockwise. Return +1 if the point is above, -1 if the point is below, or 0 of the point lies on the plane. /\abc
In the picture below, point
dis above the plane determined by abcin counter-clockwise order using the right hand rule: Code to write:Fill in in
getAboveOrBelow(a, b, c, d)
GeomPrimitives.js
Gui Tester:
aboveOrBelow.html
Tips: Hint: Use the cross product between two vectors to determine the plane normal, and verify conceptually with the right hand rule to make sure it points in the direction consistent with counter-clockwise specification of abcbefore you write your code. Then use the sign of the dot product between that normal and some other vector that involves d
This part of the assignment will get you to warm up with some very important tasks that you will re-implement in GLSL when you go to do your ray tracer. Click here to visit the repository that has the skeleton code for this part.
Note: Unit tests have shipped with this part of the assignment. Some of the default outputs actually make it look like you're passing, but be careful; the default output is actually what you're supposed to do for some of the corner cases.
Given a vector described by \[\vec{p_0} + t\vec{v} \] and a sphere defined by a center vector
c and a radius
r, determine the intersection point of the ray and the sphere, if it exists (see API comments for more details on return types and corner cases). Code to write: Fill in in rayIntersectSphere(p0, v, c, r)
GeomPrimitives.js
Gui Tester: for 2D only (you should look at the unit tests for 3D cases). A screenshot of the GUI on a working implementation is shown below: rayCircle.html Tips: Hint: The
vec3.scaleAndAddfunction in
vec3may save you time when you go to perform an operation
p0 + tv
Given three vertices
on a triangle and another point
a, b, c
, determine the 3 barycentric coordinates of
p
with respect to
p
(see API comments for more details on return types and corner cases).
a, b, c Code to write: Fill in in getBarycentricCoords(a, b, c, p)
GeomPrimitives.js
Gui Tester: for 2D only (you should look at the unit tests for 3D cases). A screenshot of the GUI on a working implementation is shown below: Barycentric.html Tips: Hint: Use the area ratio method, it is by far the easiest technique, and it relies on code you already wrote to compute the area of a 3D triangle! (See Figure 2.38 in section 2.7.1 of the textbook)
Given a vector described by \[\vec{p_0} + t\vec{v} \] and a triangle determined by three vertices
a, b, c, determine the intersection point of the ray and the triangle, if it exists (see API comments for more details on return types and corner cases).
Code to write: Fill in in rayIntersectTriangle(p0, v, a, b, c)
GeomPrimitives.js
Gui Tester:
rayTriangle.html. A screenshot of the GUI on a working implementation is shown below:
Tips: Hint: Figure out the intersection point on the plane spanned by the triangle first (using "implicit form type 2" that we talked about in class), and then use a technique very similar to computing barycentric coordinates to figure out of that point is actually in the triangle. Hint: The
vec3.scaleAndAddfunction in
vec3may save you time when you go to perform an operation
p0 + tv |
Are there any normal interpretations of $(=, <)$ with countable domain that are elementary equivalent but not isomorphic to $\mathbb{Q}$? To $\mathbb{Q}^+ = \mathbb{Q} \cap [0; +\infty)$?
First, I know that $(=, <)$ on $\mathbb{Q}$ allows quantifier elimination, so, intuitively, the only expressible closed formulas are those that express the axioms of $=$ and $<$ and the properties of $\mathbb{Q}$ (namely, having no minimum and maximum elements and being dense). So any other elementary equivalent normal interpretation should have a linear order, no min/max elements and be dense. So, the claim basically boils down to the question: is there a set $M$ having these properties but not isomorphic to $\mathbb{Q}$?
I tried proving that there is none by contradiction by assuming that for every bijection $\alpha : \mathbb{Q} \rightarrow M$ there exist $x, y \in \mathbb{Q} : x < y \land \alpha(x) > \alpha(y)$, but I couldn't come to any and got a bit stuck.
I also tried proving that there is none by taking some bijection $\alpha$ and then trying to build a "fixed" bijection $\alpha'$ that preserves the order, but it gets messy too.
I haven't paid as much attention to $\mathbb{Q}^+$ yet, but I seem to be able to prove that it also allows quantifier elimination after constant $0$ is added to the signature, so, intuitively, after this it should be the same as with $\mathbb{Q}$. |
This question already has an answer here:
A vector potential $A(x, t)$ satisfies the wave equation $\left(\Delta-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right) A(x, t)=0$.
It is written in my physics textbook that the general solution of the wave function is superposition of $e^{(1)}a_k e^{i(k\cdot x-\omega t)}$.
How to prove there are no other solutions? |
Galois Group of $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is cumbersome to computing. Is easy to find the all four possible candidates but is cumberstone to show that they are automorphisms:
For multiplication is too long showing that
$f((a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6}))=f(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})f(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6})$
For some non trivial candidate since the left hand side expands in $16$ terms. What about $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$? there are $7$ non trivial candidates and multiplication expands in $64$ terms!
How to do it in a tricker form? Edit(Let me be more precise in my question):
I know that if $f\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$, then $f(\sqrt{2})=\pm\sqrt{2}$, $f(\sqrt{3})=\pm\sqrt{3}$ so there are $4$ candidates for $\mathbb{Q}$-Automorphisms. Candidates are given by:
$f_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$
$f_2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$
$f_3(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}$
$f_4(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}$
But I need to check that they are $\mathbb{Q}$-Automorphisms in $\mathbb Q(\sqrt{2},\sqrt{3})$. All candidates $f_i$ fixes $\mathbb{Q}$ and is obvious that all preserves sums, $f_1$ obviously preserves product also since is the identity, but is cumberstone to show that $f_i(xy)=f_i(x)_if(y)$ to conclude $f_i\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$. |
I've wondered about this too, but never pursued it before. Here is an educated guess based on dimensional analysis.
A pressure has the same units as an energy density:$$\mathrm{\frac N{m^2} = \frac{N\,m}{m^3} = \frac J{m^3}} $$
Electron capture takes kind of 1 MeV of energy, so maybe the transition becomes allowed once the energy density is something like 1 MeV/atom. Let's test this idea out: I vaguely remember that coal (or something) turns to diamond somewhere above one gigapascal:\begin{align}\mathrm{1 \,GPa }&= \mathrm{10^9\frac J{m^3} \cdot \frac{1\,eV}{1.6\times10^{-19} \,J} \cdot \left(\frac{1\,m}{10^9\,nm}\right)^3 }\\&\mathrm{= \frac 58\times10^{1}\,\frac{eV}{nm^3} }\\ &\mathrm{= 6 \,\frac{eV}{nm^3}}\end{align}Diamond is pure carbon, density 3.5 g/cm$^3$, so the volume of one atom is\begin{align}V &= \mathrm{ \frac{1\,cm^3}{3.5\,g} \cdot \frac{12\,g}{1\,mole} \cdot \frac{1 \,mole}{6\times10^{23}\,atoms}\cdot \left(\frac{10^9\,nm}{100\,cm}\right)^3} \\&\mathrm{ = \frac 2{3.5} \times10^{-2} \frac{nm^3}{atom} }\\&\mathrm{ = 0.006 \frac{nm^3}{atom}}\end{align}Graphite (the initial state) has a density closer to 2 g/cm$^3$ so the volume of each atom is about 0.0035 nm$^3$.
This suggests that the transition from carbon to diamond at constant pressure takes an energy $E = P\Delta V \approx 0.015$ eV. This corresponds pretty closely to a tabulation of formation enthalpies which includes diamond:creating diamond requires about 1.9 kJ/mole, or about 20 meV/atom. Shockingly close! I must have made a set of mistakes that cancel each other out :-)
Assuming this method is sound, the energy for electron capture (and other nuclear transitions) is generally about 1 MeV. So if your block of carbon kept the same density 0.006 nm$^3$/atom, and electron that disappears frees up 1/6 of the atom's volume, you'd need$$P = \frac E{\Delta V} = \mathrm{{10^6\,eV} \cdot \frac{6\,GPa\,nm^3}{1\,eV} \cdot \frac{6\, electrons/atom}{0.006\,nm^3/atom} = 6\times10^{9}\,GPa}$$Here's some company that thinks 15,000 bar = 1.5 GPa is a lot of pressure, so we have the sanity check that this estimate $P$ for the degeneracy pressure is "astronomical" (it's sixty trillion atmospheres).
I'm very interested to see if you get an answer from someone who isn't winging it.
Two years after I wrote this answer, Rob Jeffries adds in a comment
Neutronisation threshold in Carbon nuclei is 13.9 MeV. This is reached by degenerate electrons at pressure of about $10^{19}$ GPa.
So I guessed the neutronization energy to within a factor of ten, but my pressure is kind of embarrassingly off. But I explicitly made the assumption that the density of the diamond wouldn't change between ordinary conditions and the onset of neutron capture --- which of course isn't what happens.In nature, the "overcoming" of electron degeneracy pressure occurs only in white dwarfs and degenerate stellar cores, where the density is as high as $10^7\rm\,g/cm^3$.
If you take my estimate $10^{10}\rm\,GPa$ above, increase the energy by a factor of ten, and decrease the volume per "atom" by a factor of $10^7$, you get within a factor of ten of RobJeffries's statement of the upper limit of electron degeneracy pressure in white dwarfs. |
Limits and Derivatives Derivatives Derivative of a function:Let y = f(x) be a function defined on the interval [a, b]. Let for a small increment δx in x, the corresponding increment in the value of ‘y’ be δy. Then \tt \frac{dy}{dx}=\lim_{\delta x \rightarrow 0}\ \frac{\delta y}{\delta x}= \lim_{\delta x \rightarrow 0}\ \frac{f\left(x+\delta x\right)-f\left(x\right)}{\delta x} \tt \frac{dy}{dx} represents geometrically the slope of the tangent at the point (x,y) on the curve y = f(x). Derivatives of Algebraic functions: \frac{d}{dx}(x^{n})=nx^{n-1} \frac{d}{dx}(x)=1 \frac{d}{dx}(x^{2})=2x \frac{d}{dx}(x^{3})=3x^{2} \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}} \frac{d}{dx}\left(\frac{1}{x}\right)=\frac{-1}{x^{2}} \frac{d}{dx}\left(\frac{1}{x^{2}}\right)=\frac{-2}{x^{3}} \frac{d}{dx}\left(\frac{1}{x^{3}}\right)=\frac{-3}{x^{4}} \frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right)=\frac{-1}{2x^{3/2}} \tt \frac{d}{dx}\ f\left(x\right)\cdot g\left(x\right)=f\left(x\right)\frac{d}{dx}\ g\left(x\right)+g\left(x\right)\cdot\frac{d}{dx}\ f\left(x\right) \tt \frac{d}{dx}\ \frac{f\left(x\right)}{g\left(x\right)}=\frac{g\left(x\right)\cdot\frac{d}{dx}\ f\left(x\right)-f\left(x\right)\cdot\frac{d}{dx}\ g\left(x\right)}{\left[g\left(x\right)\right]^{2}} Derivative of Trigonometric functions: \frac{d}{dx}(\sin \ x)=\cos x \frac{d}{dx}(\cos \ x)=\sin x \frac{d}{dx}(\tan \ x)=\sec^{2} x \frac{d}{dx}(\cot \ x)= -cosec^{3} x \frac{d}{dx}(\sec \ x)=\sec x \ \tan x \frac{d}{dx}(cosec \ x)=-cosec x \ \cot x Part1: View the Topic in this video From 00:14 To 11:45 Part2: View the Topic in this video From 00:13 To 06:32 Part3: View the Topic in this video From 00:12 To 09:12 Part4: View the Topic in this video From 00:12 To 13:52
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1. f'(x)=\frac{d}{dx}f(x)=\lim_{\delta x \rightarrow 0}\frac{f(x+\delta x)-f(x)}{\delta x}
2. \frac{d}{dx}(x^n)=nx^{n-1},x\in R, n \in R
3. (i) \frac{d}{dx}\left\{cf(x)\right\}=c \ \frac{d}{dx} f(x), where c is a constant.
(ii) \frac{d}{dx}\left\{f(x)\pm g(x)\right\}= \frac{d}{dx} f(x)\pm\frac{d}{dx}g(x) (sum and difference rule)
(iii) \frac{d}{dx}\left\{f(x) g(x)\right\}= f(x) \ \frac{d}{dx} \ g(x)+g(x) \ \frac{d}{dx} \ f(x) (product rule)
4. (i) \frac{d}{dx}\left\{\frac{f(x)}{g(x)}\right\}= \frac{g(x) \ \frac{d}{dx} \ f(x)- f(x) \ \frac{d}{dx} \ g(x)}{\left\{g(x)^2\right\}} (quotient rule)
(ii) If \frac{d}{dx}f(x)=\phi(x), then \frac{d}{dx}f(ax+b)=a\phi(ax+b)
(iii) Differentiation of a constant function is zero i.e., \frac{d}{dx}(c)=0.
5. \frac{d}{dx}(\sin x)=\cos x
6. \frac{d}{dx}(\cos x)=-\sin x
7. \frac{d}{dx}(\tan x)=\sec^{2} x,x\neq(2n+1)\frac{\pi}{2}, n\in I
8. \frac{d}{dx}(\cot x)=-cosec^{2} x,x\neq n \pi, n\in I
9. \frac{d}{dx}(\sec x)=\sec x \tan x, x \neq(2n+1)\frac{\pi}{2}, n\in I
10. \frac{d}{dx}(cosec \ x)=- cosec \ x \cot x, x \neq n \pi, n \in I |
It is known that a morphism of schemes $f\colon X \to S$ is smooth at a point $x \in X$ if and only if there is an open neighborhood $U$ of $x$ and an étale map $g \colon U \to \mathbb A^n_S$ such that $g \circ p=f_{|U}$, where $p \colon \mathbb A^n_S \to S$ is the natural projection.
I'm looking for a similar characterization for semistable curves $f \colon X \to S$. I'm interested in the case $S=Spec(k)$, with $k$ a field, and in the case $S=Spec(V)$, with $V$ a discrete valuation ring, where now $X$ is generically smooth.
In particular my question is: in the second case it is true that we can find $\lbrace Spec(R_i)\rbrace _{i \in I}$, an affine open covering of $X$, such that for each $i$, there is an étale map $V[x,y]/(xy-\pi) \to R_i$, where $\pi$ is a uniformizer of $V$?
Thanks.
Ricky |
It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced.
Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit.
@Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form.
A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts
I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it.
Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis.
Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)?
No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet.
@MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it.
Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow.
@QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary.
@Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer.
@QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits...
@QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right.
OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ...
So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study?
> I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago
In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a...
@MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really.
When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.?
@tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...)
@MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences). |
It's also a (failed) replicator. If you look closely you can see the puffer being generated again, and even again before crashing into the oscillators.Gamedziner wrote: That's actually a spacefiller.
Current rule interest: B2ce3-ir4a5y/S2-c3-y
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x = 4, y = 4, rule = B2i34ceiknry8/S23-a4city3o$o2bo$o2bo$b3o!
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x = 4, y = 4, rule = B2i34ceiknry8/S23-a4cityb3o$o2bo$o2bo$3o!#C [[ TRACKBOX 14/36 1/36 0 0 ]]
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x = 3, y = 3, rule = B2n34eiqrtz5eijnr8/S23-a4city2o$b2o$bo!
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x = 4, y = 3, rule = B2in3aeikn4city8/S23-a4city2bo$b3o$o2bo!
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x = 4, y = 4, rule = B2in3aceir4city8/S23-a4cityb2o$2o$b3o$2bo!
An awesome gun firing cool spaceships:
It's class F. I've edited the OP.GUYTU6J wrote:EDIT2:How about this?
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x = 4, y = 3, rule = B2in3aeikn4city8/S23-a4city 2bo$b3o$o2bo!
965808 is period 336 (max = 207085118608).
Code: Select all
x = 5, y = 4, rule = B2k3aeiqy4city8/S23-a4cityb3o$bo2bo$bob2o$o!
Code: Select all
x = 3, y = 3, rule = B2k3aeinr4city7c/S23-a4citybo$3o$2bo!
Code: Select all
x = 4, y = 4, rule = B2k3aeinr4city7e/S23-a4city3o$o2bo$o2bo$b3o!
An awesome gun firing cool spaceships:
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x = 4, y = 4, rule = B2ik3aeikr4ceijqry78/S23-a4city78b3o$o2bo$o$2o!
An awesome gun firing cool spaceships:
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x = 7, y = 2, rule = B3-y5a/S234c5ek3ob3o$bo3bo!
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x = 31, y = 27, rule = B3-y5a/S234c5ekbo3bo19bo3bo$3ob3o17b3ob3o24$3ob3o$bo3bo!
-Terry Pratchett
BlinkerSpawn Posts:1906 Joined:November 8th, 2014, 8:48 pm Location:Getting a snacker from R-Bee's It seems to follow OEIS A189007 (1, 4, 8, 16, 16, 32, 32, 64, 32, ...), so it does not appear to be in any existing class.muzik wrote:Possibly a class r:
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$
http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
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x = 1, y = 1, rule = B1e/S04eo!
Do these count as replicators, or spacefillers?
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x = 1, y = 1, rule = B12cn4c/S0o!
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x = 1, y = 1, rule = B1c2c4c/So!
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x = 1, y = 1, rule = B12ci3ci4c6i8/S02i8o!
Spacefillers.
965808 is period 336 (max = 207085118608).
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
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x = 1, y = 1, rule = B1e2e/S2co!
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x = 1, y = 1, rule = B1e2e/S2c3i4w5yo!
These also don't have the "dying edge replicators" muzik talked about
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
Code: Select all
x = 3, y = 7, rule = B3-cny4cekryz5c6in7e8/S2-cn3-aeky4eit5ackn6ei7e83o$obo$obo2$obo$obo$3o!
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x = 4, y = 4, rule = B2kn3-ckny4irt5r8/S2aek3ijnqr4ib3o$o2bo$o2bo$3o!
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x = 7, y = 8, rule = B2i3-ekny4z5r7/S2-cn3-ace4eiz5ejknq6i2bo$b3o$2obo3$3bob2o$3b3o$4bo!
That rule has a second replicator!
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x = 5, y = 4, rule = B2i3-ekny4z5r7/S2-cn3-ace4eiz5ejknq6i2ob2o$o3bo$b3o$2bo!
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x = 4, y = 3, rule = B2in3aijr4eq5j6c/S2-in3ijnqr4i5cnr6kb3o$bobo$2obo!
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x = 26, y = 17, rule = B2in3aijr4eq5j6c/S2-in3ijnqr4i5cnr6kb3o$bobo$2obo4$23b3o$23bobo$22b2obo6$9b3o$9bobo$8b2obo!
Things to work on:
- Find a (7,1)c/8 ship in a Non-totalistic rule
- Finish a rule with ships with period >= f_e_0(n) (in progress)
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
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x = 4, y = 4, rule = B2kn3-ckny4irt5r8/S2aek3ijnqr4ib3o$o2bo$o2bo$3o!
Are there any other polygons it can trace out? I doubt so, but if so, those with 4n sides seem to be most likely.
LaundryPizza03 Posts:457 Joined:December 15th, 2017, 12:05 am Location:Unidentified location "https://en.wikipedia.org/wiki/Texas"
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x = 3, y = 3, rule = B34j5k6en/S235e6c2o$b2o$2bo!
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x = 4, y = 2, rule = B2n3aijnr4jrz5r6-e/S2-in3knr4aeijknq5-r8b2o$4o!
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x = 19, y = 19, rule = B3-q4z5y/S234k5j12b2o$12b2o3$b5o$o3bo3b2o$o$bobo4$9b2o$3b2o3b2o7b2o$7b2o8b2o4$5b2o$5b2o!
Could Class S replicator hasslers be used for signal circuitry?
LaundryPizza03 Posts:457 Joined:December 15th, 2017, 12:05 am Location:Unidentified location "https://en.wikipedia.org/wiki/Texas" LaundryPizza03 Posts:457 Joined:December 15th, 2017, 12:05 am Location:Unidentified location "https://en.wikipedia.org/wiki/Texas"
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x = 3, y = 3, rule = B2k3578/S3-e456k83o$3o$obo!
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x = 3, y = 3, rule = B2a/S3n4y2o$obo$b2o!
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x = 3, y = 2, rule = B2a3ir5aik7e/S2k4zobo$3o!
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x = 3, y = 2, rule = B2a3ir4i5aik7e/S2k3y4zobo$3o!
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x = 3, y = 2, rule = B2a3ir4ik5aik7e/S2k3y4zobo$3o!
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x = 3, y = 2, rule = B1e2i3a4i5eiq6i/S2a3eiq4inrbo$3o!
Technically it would be a 1.5849D replicator though?
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x = 11, y = 11, rule = B2ce3y4e5y6c/S1c2i3ciy4ct5ey6i7e85bo$4bobo$5bo$5bo$bo7bo$ob2o3b2obo$bo7bo$5bo$5bo$4bobo$5bo!
That's interesting — it's a replicator with infinite growth and finite copies.muzik wrote:Hm.
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x = 11, y = 11, rule = B2ce3y4e5y6c/S1c2i3ciy4ct5ey6i7e8 5bo$4bobo$5bo$5bo$bo7bo$ob2o3b2obo$bo7bo$5bo$5bo$4bobo$5bo!
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$
http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce |
Let $k$ be a field (of possibly positive characteristic), let $U_n$ denote the space of all $n \times n$ unipotent upper triangular matrices over $k$, and let $G$ be an algebraic subgroup of $U_n$ (hence a unipotent algebraic group itself). Then each $X \in \text{Lie}(G)$ (thought of as a member of $\text{Lie}(U_n)$, i.e. a strictly upper triangular $n \times n$ matrix) is nilpotent, so it makes sense to define
$\text{exp}(X) = 1 + X + X^2/2! + \dots + X^{n-1}/(n-1)!$
(This definition makes sense even in characteristic $p > 0$ so long as $p \geq n$, i.e. so that $p$ never divides $1!, 2!, \dots, (n-1)!$). We can also define, for $g \in G$, since $g-1$ is nilpotent,
$\log(g) = (g-1) - (g-1)^2/2 + (g-1)^3/3 - \dots \pm (g-1)^{n-1}/(n-1) $
Obviously $\text{exp}$ and $\log$ define maps from $\text{Lie}(U_n)$ to $U_n$ and back to $\text{Lie}(U_n)$, and are bijective, being inverses of one another.
My Question: If $g \in G$, is $\log(g) \in \text{Lie}(G)$? Or, equivalently, for $X \in \text{Lie}(G)$, is $\text{exp}(X) \in G$?
I feel like there should be an obvious proof of this, but I don't see it. If $G$ were a Lie group, the Lie algebra of $G$ would often just be defined to be all $X$ such that $e^{tX} \in G$ for all $t \in \mathbb{R}$, and so for Lie groups $\text{exp}$ maps from $\text{Lie}(G)$ to $G$ simply by definition. In the algebraic group context this definition no longer makes sense generally, and even when it does, is not used in the literature (so far as I've seen), so I tried using each of the following equivalent definitions of $\text{Lie}(G)$, with no success:
$\text{Lie}(G) = \text{Dist}_1^+(G)$ (distributions of order no greater than $1$ without constant term)
$\text{Lie}(G) = $ the subspace of $\text{Lie}(U_n) = \text{Dist}_1^+(U_n)$ which kills $I = (\text{defining polynomials of $G$})$
$\text{Lie}(G) = \{M \in \text{Lie}(U_n): 1 + \tau M \in G(k[\tau]) \}$ where $\tau^2 = 0$
$\text{Lie}(G) = \{M \in \text{Lie}(U_n) : 1 + \tau M \text{ satisfies the defining polynomials of } G \}$, again where $\tau^2 = 0$
$\text{Lie}(G) = $ left invariant derivations on the Hopf algebra of $G$
It is certainly believable on it's face; we have that $\text{Lie}(G) \stackrel{\text{exp}}{\longrightarrow} U_n \stackrel{\log}{\longrightarrow} \text{Lie}(G)$ composes to the identity, similarly for $G \stackrel{\log}{\longrightarrow} \text{Lie}(U_n) \stackrel{\text{exp}}{\longrightarrow} G$, but I don't see why in the meantime that $\log(G) \subset \text{Lie}(G)$ or that $\text{exp}(\text{Lie}(G)) \subset G$.
If it makes a difference, I'm actually only interested in the case where the defining polynomials of $G$ have integer (perhaps mod $p$) coefficients.
Thanks in advance for any help.
EDIT: Here's a more basic question, one which might help answer the above.
Suppose $k = \mathbb{R}$. Then $G$ is also a Lie group, and it is customary to define
$\text{Lie}(G) = \{ X \in \text{Lie}(U_n): e^{tX} \in G \text{ for all } t \in \mathbb{R} \}$
Can someone explain, or point me to a reference explaining, why this definition is equivalent to any of the above definitions for $\text{Lie}(G)$ as an algebraic group? |
RNA-seq is a widely used technique allowing sensitive differential gene expression analysis. The typical RNA-seq experiment involves the preparation of mRNA samples, fragmentation of the mRNA molecules, reverse transcription to cDNA, and the conversion of the sample into a molecular library for sequencing. The sequencing output consists of millions of reads generated from the cDNA fragments. The reads are then aligned to a reference genome or transcriptome, in order to determine the qualitative and quantitative composition of the cDNA library. The ultimate goal is the estimation of the relative abundance of the underlying genomic features, i.e. the genomic regions that can appear represented in the sample, such as exons or genes.
Along with the relative expression of the genomic feature, which is exactly the target measurement in RNA-seq experiments, read counts depend on some confounding factors. Namely, read counts for each genomic feature depend heavily on the total number of obtained reads (sequencing depth) and the feature's length, as well as transcriptome and nucleotide composition. The dependence on feature length may be surprising to you, as you might expect the method to produce a number of reads proportional to the number of cDNA molecules in the sample. The key is the fragmentation step referred above. Sequencing requires that the molecules are broken down into smaller pieces. Longer transcripts will naturally yield a higher number of fragments and, thus, a higher read count.
In this post I will go over different RNA-seq expression units and the basic math behind them, as well as their implementation in the popular gene expression analysis Bioconductor package edgeR.
RNA-seq experiments generate relative, rather than absolute, measurements. In order for any units to be comparable between samples or between experiments, we need to perform inter-sample normalization. While feature expression comparisons within samples need to account for all factors referred in the second paragraph above, comparisons between samples need to be corrected for differences in sequencing depth and transcriptome composition only. Attributes like feature length and GC content (nucleotide composition) do not vary and thus cancel out.
How differences in sequencing depth affect read count is self explanatory. The impact of transcriptome composition, on the other hand, can be understood using the following example. Given some feature present in two RNA-seq samples with the same number of total reads (sequencing depth), if one sample has a larger transcriptome, a smaller proportion of the reads will pertain to that feature. Similarly, for samples with the same qualitative transcriptome and sequenced at the same depth, if a subset of other shared features is more highly expressed in one sample it will make up a larger proportion of the total read pool, leaving a lower proportion of the reads mapping to the test feature.
For this reason, final counts of aligned reads (known as feature summarization data) need to be corrected for such biases. The typical analysis of feature summarization data will drop reads for features with counts below a certain threshold, calculate sample normalization factors, and scale library sizes (sum of reads in each sample) accordingly. This will address the biases brought about by differences in sequencing depth and transcriptome composition. In the specific case of edgeR, an empirical approach based on the trimmed mean of M values (TMM) method is used, implemented in the function
After sample normalization, expression units are chosen for inter-sample and within-sample differential feature expression analysis. I will go over the most popular units next.
The simplest RNA-seq feature expression unit reports normalized counts, or the number of reads that align to a particular feature after correcting for sequencing depth and transcriptome composition bias. Normalized counts are the most popular unit among differential expression analysis methods (including edgeR). However, feature length normalization is skipped, with the important consequence that within-sample differential feature expression analysis is not possible.
This unit is known as
counts per million reads mapped (CPM). In its basic form, for each feature \(i\), \(CPM\) is the count of sequenced fragments mapping to the feature (the random variable I am calling \(r_i\) here) scaled by the total number of reads (\(R\)) times one million (to bring it up to a more convenient number).
$$ \text{CPM}_i = \dfrac{r_i}{\dfrac{R}{10^6}} = \dfrac{r_i}{R}\cdot 10^6 $$ with
\( R = \sum_{j=1}^n r_j \)
and
\( n = \) number of represented features
If we drop the multiplication by one million, \( CPM_i \) becomes simply the number of reads mapping to feature \( i \) relative to the total number of fragments in the sample and thus has the properties of relative counts:
\( 0 \leq \dfrac{r_i}{R} \leq 1 \) and \( \sum_{i=1}^n \dfrac{r_i}{R} = 1 \)
edgeR includes function
> edgeR::cpm.defaultfunction (x, lib.size = NULL, log = FALSE, prior.count = 0.25, ...){ x <- as.matrix(x) if (is.null(lib.size)) lib.size <- colSums(x) if (log) { prior.count.scaled <- lib.size/mean(lib.size) * prior.count lib.size <- lib.size + 2 * prior.count.scaled } lib.size <- 1e-06 * lib.size if (log) log2(t((t(x) + prior.count.scaled)/lib.size)) else t(t(x)/lib.size)}
The key part of their implementation can be written as
Of course this basic implementation ignores sample normalization factors and thus fails to account for differences in transcriptome composition. As mentioned in the previous section, that is accomplished by scaling library sizes according to sample normalization factors (calculated in edgeR using
> edgeR::cpm.DGEListfunction (x, normalized.lib.sizes = TRUE, log = FALSE, prior.count = 0.25, ...){ lib.size <- x$samples$lib.size if (normalized.lib.sizes) lib.size <- lib.size * x$samples$norm.factors cpm.default(x$counts, lib.size = lib.size, log = log, prior.count = prior.count)}
If (normalization factor-scaled) CPM is appropriate for inter-sample comparison, within-sample comparisons must account for differences in feature length. The most commonly used unit in this case is
reads per kilobase of transcript per million reads mapped (RPKM), introduced with one of the original RNA-seq papers (Mortazavi et al. Nature Methods, 2008). The alternative unit FPKM is nothing more than a generalization of RPKM to deal with the fact that a single sequenced cDNA fragment can actually produce more than one read, depending on the sequencing technology. This currently refers to paired-end sequencing experiments, in which a read pair, rather than a single read, corresponds to a cDNA fragment. The word 'reads' in RPKM is replaced by 'fragments' in FPKM. RPKM (or FPKM) is defined formally for each feature \(i\) as the count (the random variable \( r_i \)) scaled by the feature's length (\( l_i \)) times one thousand (to kilobase) and further scaled by the total number of reads (\( R \)) times one million (again, in order to bring it up to a more convenient number). $$ \text{RPKM}_i = \dfrac{r_i}{ \left(\dfrac{l_i}{10^3}\right) \left( \dfrac{R}{10^6} \right)} = \dfrac{r_i}{l_i R} \cdot 10^9 $$
edgeR includes function
> edgeR::rpkm.defaultfunction (x, lib.size = NULL, log = FALSE, prior.count = 0.25, ...){ y <- cpm.default(x = x, lib.size = lib.size, log = log, prior.count = prior.count, ...) gene.length.kb <- gene.length/1000 if (log) y - log2(gene.length.kb) else y/gene.length.kb}
However, the RPKM is inconsistent across samples (Wagner
et al. Theory Biosci., 2012). The reason is explained by the following insight: RPKM normalizes by the total number of reads and this is not a measure of total transcript number. This relationship depends on the length distribution of transcripts, which is variable between samples. In other words, the relationship between the total number of reads (\( R \) in the equations) and the actual total number of cDNA fragments sampled depends on the feature length distribution. For this reason, this normalization factor is variable across samples, hence the statement that opens this paragraph: the resulting unit is inconsistent across samples. A clear manifestation of the problem is the fact that the sum of all RPKM values will vary from sample to sample.
Inter-feature differential expression analysis within samples will still work. Running the risk of weakening the underlying message of this article, I will even go as far as stating that even inter-sample RPKM comparisons will, in most cases, be ok. In practical terms, since we typically look at thousands of features and it is reasonable to assume that a majority of them will not be differentially expressed between samples, we are probably not that far off when using RPKM. We can probably trust most conclusions drawn in such terms in the literature. The inconsistency is still real and is there, though, so it needs fixing.
A possible fix emerges from the explanation of RPKM's inconsistency above. Transcript length distributions can vary between samples and that introduces an inconsistency in RPKM calculation. So, instead of dividing by the total number of sequenced reads (\( R \)) for a given sample, we can instead divide by the total number of length-normalized reads. This unit,
transcripts per million fragments sequenced (TPM), was introduced originally in Li et al. Bioinformatics, 2010 and further supported in Wagner et al. Theory Biosci., 2012. Compare the TPM equation with CPM and RPKM's, re-written here for direct comparison. For simplicity, sample normalization factor scaling of library size is excluded. $$ \text{CPM}_i = \dfrac{r_i}{\sum_{j=1}^n r_j} \cdot 10^6 $$ $$ \text{RPKM}_i = \dfrac{r_i}{\sum_{j=1}^n r_j} \cdot \dfrac{1}{\dfrac{l_i}{10^3}} \cdot 10^6 $$ $$ \text{TPM}_i = \dfrac{r_i}{\sum_{j=1}^n \dfrac{r_j}{l_j}} \cdot \dfrac{1}{l_i} \cdot 10^6 $$
The interpretation is that \( TPM_i \) represents the number of transcripts of feature \(i \) found in a total number of one million transcripts. Interestingly, we can easily convert RPKM values to TPM by simply dividing each feature's RPKM by the sum of the RPKM values of all features and multiplying by one million. In fact, TPM is really just RPKM scaled by a constant to correct the sum of all values to 1 million. $$ \begin{aligned} \text{TPM}_i &= \dfrac{\text{RPKM}_i}{\sum_{j=1}^n \text{RPKM}_j } \cdot 10^6 \end{aligned} $$
Armed with this information, we can convert RPKM to TPM in two different ways: from pre-calculated RPKM, by diving by the sum of RPKM values, or directly from the normalized counts. Below I have written some example R code to calculate TPM starting from RPKM values computed using edgeR's
# Start from feature summarization data (as an edgeR DGEList object)# Filter-out low expression features, perform sample normalization and calculate RPKMy <- edgeR::calcNormFactors(y)y <- edgeR::rpkm(y)# Calculate TPM from RPKMtpm_from_rpkm <- function(x) { rpkm.sum <- colSums(x) return(t(t(x) / (1e-06 * rpkm.sum)))}tpm <- tpm_from_rpkm(y)
And the following is a simple implementation of TPM computation starting from edgeR-normalized counts.
# Start from feature summarization data (as an edgeR DGEList object)# Filter-out low expression features and perform sample normalizationy <- edgeR::calcNormFactors(y)# Calculate TPM from normalized countscalc_tpm <- function(x, gene.length) { x <- as.matrix(x) len.norm.lib.size <- colSums(x / gene.length) return((t(t(x) / len.norm.lib.size) * 1e06) / gene.length)}tpm_man <- calc_tpm(y, gene.length = y$genes$Length)
TPM accounts for the lengths of all transcripts found in the sample and thus brings us one step closer to a good solution. This unit is more stable across samples than RPKM. However, the abundances of all transcripts will still change between samples, meaning that the denominator of the TPM equation (sum of length-normalized read counts) is still variable. This way, TPM is
not directly comparable between experiments either.
So if no unit is perfect, how are we supposed to analyse differential feature expression? Maybe a better unit will be developed at some point, but in the mean time CPM is an appropriate unit for inter-sample comparison. When comparing feature expression within samples, TPM should be used instead of RPKM/FPKM.
If you want to dive deeper in the subject of why TPM is a better unit than RPKM/FPKM, besides the papers I cited above I would recommend two very informative blog posts published in Bits of DNA (by Prof. Lior Pachter) and The Farago (by Prof. Pachter's student Harold Pimentel). |
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