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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 $V$?
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 $\Fun(S, V)$.
Let $\mathrm{P}_3$ denote the set of polynomials of degree $3$ or less with real coefficients. Consider the ordered basis\[B = \left\{ 1+x , 1+x^2 , x – x^2 + 2x^3 , 1 – x – x^2 \right\}.\]Write the coordinate vector for the polynomial $f(x) = -3 + 2x^3$ in terms of the basis $B$.
Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$ matrices. Define the linear transformation $T : V \rightarrow W$ by\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a+b & 2d \\ 2b – d & -3c \\ 2b – c & -3a \end{bmatrix}.\]
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,\[T (f) (x) = x f(x).\]
Prove that $T$ is a linear transformation, and find its range and nullspace.
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
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 $W$.
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.
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)
Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\]be a subset in $C[-1, 1]$.
(a) Prove that $V$ is a subspace of $C[-1, 1]$.
(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.
(c) Prove that\[B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}\]is a basis for $V$. |
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!! |
Let us define a predicate
$$T(i,k_1,...,k_n)\in\{True, False\}$$Where $T(i,k_1,...,k_n)$ means "using the $i$ first values of $S$, we can find $n$ disjoint subsets with sums $k_1, ... k_n$". The answer you are looking for is $T(|S|,K,..,K)$ where $K$ appears $n$ times. We have the following recurrence formula : $$T(i,k_1,...,k_n) = \left\{\begin{matrix} T(i-1,k_1,...,k_n) & \text{// we don't use value $x_i$} \\ \vee T(i-1,k_1-x_i,...,k_n) \text{ if $x_i \geq k_1$} &\text{// we put value $x_i$ in set 1}\\ \vdots \\ \vee T(i-1,k_1,...,k_n-x_i) \text{ if $x_i \geq k_n$} &\text{// we put value $x_i$ in set n}\\ \end{matrix}\right.$$This formula is a big boolean "or", I don't know how I could make it look better. We also the following initialisation : $\forall i, T(i,0,...,0) = True$. You can see it as a big $n+1$-dimensional array, where the first dimension has length $|S|$, and the others have length $K$, which gives $O(K^n|S|)$ values. In that array, all the values can be computed from the other values in time $O(n)$ (the big "or" is over $n+1$ values). Therefore, the worst case complexity is $O(nK^n|S|)$.
About implementation, it might be easier to do with recursively with memorization. However, you're gonna blow up the stack really fast, so you might want to think of something iterative.
To get the actual values in each set, run a backtracking algorithm once you have that truth array. |
The aim of this test case is to validate the following functions:
The simulation results of SimScale were compared to the numerical results presented in [Roark]. The meshes used in (A) and (B) were created with the parametrized-tetrahedralization-tool on the SimScale platform. The meshes used in (C) and (D) were locally created with Salome.
The shaft has a radius r = 0.1 m and a length of l = 0.5 m.
Tool Type : CalculiX/Code_Aster Analysis Type : Static Mesh and Element types :
Case Mesh type Number of nodes Element type (A) linear tetrahedral 12940 3D isoparametric (B) quadratic tetrahedral 94919 3D isoparametric (C) linear hexahedral 10325 3D isoparametric (D) quadratic hexahedral 40935 3D isoparametric
Material:
Constraints:
Loads:
\[\begin{equation}\label{ref1}
J = \frac{1}{2} \pi r^4 = 1.57 \cdot 10^{-4} m^4 \end{equation}\]\[\begin{equation}\label{ref2} \tau_{max} = \frac {T r} {J} = 31.847 N/mm^2 \end{equation}\]\[\begin{equation}\label{ref3} \theta = \frac {\tau_{max} l} {G r} = 1.9904 \cdot 10^{-4} rad \end{equation}\]
Important
Comparison of the maximum shear stress τmax and the angle of twist θ obtained with SimScale and the results derived from [Roark].
Case Tool Type [Roark] SimScale Error (A) CalculiX 31.847 30.212 5.13% (A) Code_Aster 31.847 30.212 5.13% (B) CalculiX 31.847 31.842 0.02% (B) Code_Aster 31.847 31.838 0.03% (C) Code_Aster 31.847 32.066 -0.69% (D) Code_Aster 31.847 31.879 -0.10%
Case Tool Type [Roark] SimScale Error (A) CalculiX 0.0019904 0.001953 1.88% (A) Code_Aster 0.0019904 0.001969 1.08% (B) CalculiX 0.0019904 0.001969 1.08% (B) Code_Aster 0.0019904 0.001989 0.07% (C) Code_Aster 0.0019904 0.002004 -0.68% (D) Code_Aster 0.0019904 0.00199 0.02%
[Roark] (1, 2, 3, 4) (2011)”Roark’s Formulas For Stress And Strain, Eighth Edition”, W. C. Young, R. G. Budynas, A. M. Sadegh |
Category: Group Theory
Group Theory Problems and Solutions.
Popular posts in Group Theory are:
Problem 625
Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$.
(a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.
Add to solve later
(b) Prove that a group cannot be written as the union of two proper subgroups. Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$.
Add to solve later
(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. Problem 613
Let $m$ and $n$ be positive integers such that $m \mid n$.
(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined. (b) Prove that $\phi$ is a group homomorphism. (c) Prove that $\phi$ is surjective.
Add to solve later
(d) Determine the group structure of the kernel of $\phi$. If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 497
Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively. Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.
Also determine whether the statement is true if $G$ is a non-abelian group.Add to solve later |
1. How much time will it take for an amount of Rs. 900 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
A. 2 years
B. 3 years
C. 1 year
D. 4 years
Here is the answer and explanation
Answer : Option A
Explanation :
P = Rs.900
SI = Rs.81 T = ? R = 4.5% $MF#%\text{T= }\dfrac{100 ×\text{SI}}{\text{PR}} = \dfrac{100 × 81}{900 × 4.5} = 2 \text{ years}$MF#%
2. Arun took a loan of Rs. 1400 with simple interest for as many years as the rate of interest. If he paid Rs.686 as interest at the end of the loan period, what was the rate of interest?
A. 8%
B. 6%
C. 4%
D. 7%
Here is the answer and explanation
Answer : Option D
Explanation :
Let rate = R%
Then, Time, T = R years P = Rs.1400 SI = Rs.686 $MF#%\begin{align}&\text{SI= }\dfrac{\text{PRT}}{100} \\ \\ &\Rightarrow \text{686 = }\dfrac{\text{1400 × R × R}}{100} \\ \\ &\Rightarrow 686 = 14 \text{ R}^2 \\ \\ &\Rightarrow 49 = \text{R}^2 \\ \\ &\Rightarrow \text{R} = 7\end{align}$MF#% i.e.,Rate of Interest was 7%
3. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is :
A. Rs. 700
B. Rs. 690
C. Rs. 650
D. Rs. 698
Here is the answer and explanation
Answer : Option D
Explanation :
Simple Interest (SI) for 1 year = 854-815 = 39
Simple Interest (SI) for 3 years = 39 × 3 = 117 Principal = 815 - 117 = Rs.698
4. A sum fetched a total simple interest of Rs. 929.20 at the rate of 8 p.c.p.a. in 5 years. What is the sum?
A. Rs. 2323
B. Rs. 1223
C. Rs. 2563
D. Rs. 2353
Here is the answer and explanation
Answer : Option A
Explanation :
SI = Rs.929.20
P = ? T = 5 years R = 8% $MF#%\text{P = }\dfrac{100 \times \text{SI}}{\text{RT}}=\dfrac{100 \times 929.20}{8 \times 5}\text{ = Rs.2323}$MF#%
5. Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
A. Rs. 6400
B. Rs. 7200
C. Rs. 6500
D. Rs. 7500
Here is the answer and explanation
Answer : Option A
Explanation :
Let the investment in scheme A be Rs.x
and the investment in scheme B be Rs.(13900 - x) $MF#%\begin{align}&\text{We know that }\text{SI = }\dfrac{\text{PRT}}{100}\\\\ &\text{Simple Interest for Rs.x in 2 years at 14% p.a. = }\dfrac{x \times 14 \times 2}{100} = \dfrac{28x}{100}\\\\ &\text{Simple Interest for Rs.(13900 - x) in 2 years at 11% p.a. = }\dfrac{(13900 - x) \times 11 \times 2}{100} = \dfrac{22(13900 - x)}{100}\\\\ &\text{Total interest = Rs.3508}\\\\ &\dfrac{28x}{100} + \dfrac{22(13900 - x)}{100} = 3508\\\\ &28x + 305800 -22x = 350800\\\\ &6x = 45000\\\\ &x = \dfrac{45000}{6} = 7500\end{align}$MF#% Investment in scheme B = 13900 - 7500 = Rs.6400
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Let $X_1,X_2,...,X_n$ be a random sample from a uniform distribution on $(\mu-\sqrt 3\sigma,\mu+\sqrt3\sigma)$.
Here the unknown parameters are two, namely $\mu$ and $\sigma$, which are the population mean and standard deviation.
Find the point estimator of $\mu$ and $\sigma$.
I have tried to do that by the Method-of-Moments(MOM). The procedure is,
$M\prime_j=\mu\prime_j(\theta_1,\theta_2,...,\theta_k); j=1,2,...,k$
where $M\prime_j$ is the $j^{th}$ sample moment about zero & $M\prime_j=\frac{1}{n}\sum_{i=1}^n X_i^j$
& $\mu\prime_j$ is the $j^{th}$ moment about zero ,ie, $j^{th}$ raw moment.
Now,
$M\prime_1=\mu\prime_1=\mu\prime_1(\mu,\sigma)=\mu$
And
$M\prime_1=\frac{1}{n}\sum_{i=1}^n X_i=\bar X$
Again,
$M\prime_2=\mu\prime_2=\mu\prime_2(\mu,\sigma)=\sigma^2+\mu^2$
$\Rightarrow M\prime_2=\sigma^2+\mu^2$
$\Rightarrow \sigma^2=M\prime_2-\mu^2$
$\Rightarrow \sigma=\sqrt{\frac{1}{n}\sum_{i=1}^n(X_i-\bar X^2)}$
Hence Method-of-Moment estimators are $\bar X$ for $\mu$
and $\sqrt{\frac{1}{n}\sum_{i=1}^n(X_i-\bar X^2)}$ for $\sigma$.
But the procedure seems illogical to me for the following reason:
$\bullet$ I haven't considered the pdf of uniform density. so this procedure is also applicable for normal density. Then where is the difference?
What is the correct process of finding the point estimators for the above situation? |
The short answer is yes: Survey Monkey ignores exactly how you obtained your sample. Survey Monkey is not smart enough to assume that what you have gathered
isn't a convenience sample, but virtually every Survey Monkey survey is a convenience sample. This creates massive discrepancy in exactly what you're estimating which no amount of sheer sampling can/will eliminate. On one hand you could define a population (and associations therein) you would obtain from a SRS. On the other, you could define a population defined by your non-random sampling, the associations there you can estimate (and the power rules hold for such values). It's up to you as a researcher to discuss the discrepancy and let the reader decide exactly how valid the non-random sample could be in approximating a real trend.
As a point, there are inconsistent uses of the term bias. In probability theory, the bias of an estimator is defined by $\mbox{Bias}_n = \theta - \hat{\theta}_n$. However an estimator can be biased, but consistent, so that bias "vanishes" in large samples, such as the bias of maximum likelihood estimates of the standard deviation of normally distributed RVs. i.e. $\hat{\theta} \rightarrow_p \theta$. Estimators which don't have vanishing bias, (e.g. $\hat{\theta} \not\to_p \theta$) are called
inconsistent in probability theory. Study design experts (like epidemiologists) have picked up a bad habit of calling inconsistency "bias". In this case, it's selection bias or volunteer bias. It's certainly a form of bias, but inconsistency implies that no amount of sampling will ever correct the issue.
In order to estimate population level associations from convenience sample data, you would have to correctly identify the sampling probability mechanism and use inverse probability weighting in all of your estimates. In very rare situations does this make sense. Identifying such a mechanism is next to impossible in practice. A time that it can be done is in a cohort of individuals with previous information who are approached to fill out a survey. Nonresponse probability can be estimated as a function of that previous information, e.g. age, sex, SES, ... Weighting gives you a chance to extrapolate what results would have been in the non-responder population. Census is a good example of the involvement of inverse probability weighting for such analyses. |
Well there should be a distinction between types and terms somehow. If we had a "plus 10" function it wouldn't make any sense to apply it to a type. What is a type plus 10? So we need some kind of type system that differentiates between types and terms. This much is unavoidable. I suspect that this is the brunt of your issue.
But lets try and consolidate things a bit.
$$\begin{align}t, u ::=&~x && \text{(variable)} \\ &|~\lambda x : A. t && \text{(term abstraction)} \\ &|~t~u && \text{(application)} \\ &|~\lambda \alpha : Type. t && \text{(type abstraction)}\\\end{align}$$
where Type is a new keyword. Then applications can be type checked to ensure that only terms that have types of the form $\forall a. t$ can have types applied to them and that terms of the form $\lambda x : Type. t$ have type $\forall x. t$
This is however just a rebranding of System-F! All I have done is unified the syntax a bit. This does manage to consolidate application but you might then ask what happens when we say "Type is a Type too!" so that we can have only one kind of abstraction as well. So you might try and define things like this
$$\begin{align}A, B ::=&~\alpha && \text{(type variable)} \\ &|~A \rightarrow B && \text{(function type)} \\ &|~\forall \alpha. B && \text{(universal quantification)}\\ &|~Type && \text{(type of types)} \\\end{align}$$
and then try something like this for terms
$$\begin{align}t, u ::=&~x && \text{(variable)} \\ &|~\lambda x : A. t && \text{(abstraction)} \\ &|~t~u && \text{(application)} \\\end{align}$$
Now I have introduced some terms I didn't want to! like $\lambda x : Type \to Type. t$. We don't even have the proper machinery to use something like that! So it doesn't seem like we can consolidate much further than our first attempt lest we introduce functions on types which would take us outside of system F! Infact the above move takes us outside of system-$F\omega$ as well! Consider the term $\lambda x : \forall a. a \to Type. t$. That's a dependent type!
If we morph the system into something closer to dependent types we eventually can drop the distinction if we are very carful. Such a system would not be System-F however and thus the distinction is needed as long as we are in System-F. |
This is a problem that I encountered in my research and have no clues to fully resolve it. Basically, I need large (or moderate) deviation bounds on the difference between an order statistic of independent and identically distributed (i.i.d.) random variables on the compact interval $\left[ 0,1\right] $ and the expectation of this order statistic.
Let $X_{1}% ,...,X_{n}$, $n\in\mathbb{N}$ be i.i.d. and uniformly distributed on $\left[ 0,1\right] $. Let their order statistics be $X_{\left( 1\right) }\leq X_{\left( 2\right) }\leq...\leq X_{\left( n\right) }$, where we can ignore the zero probability event that any two order statistics are equal. Let $E$ and $V$ denote respectively the expectation operator and variance operator. Then each $X_{\left( r\right) }$ follows a Beta distribution, $$ E\left[ X_{\left( r\right) }\right] =\frac{r}{n+1},r=1,...,n $$ and $$ V\left[ X_{\left( r\right) }\right] =\frac{r\left( n-r+1\right) }{\left( n+1\right) ^{2}\left( n+2\right) },r=1,...,n $$
This implies two things:
(a) when $r=o\left( n\right) $ where the small $o$ notation means that $\lim_{n\rightarrow\infty}\frac{r}{n}=0$, for any $\varepsilon>0,$ \begin{equation} P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}\right\vert >\varepsilon\right) \leq\frac{o\left( 1\right) }{n \varepsilon^{2}},\label{eq1}% \end{equation} where $o\left( 1\right) $ denotes a nonnegative sequence that converges to $0$ as $n\rightarrow\infty$; in this case, $\varepsilon=n^{-\alpha}$ with $0 \le \alpha < \frac{1}{2}$ can be set, such that \begin{equation} P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}\right\vert >\frac {1}{n^{\alpha}}\right) \le \frac{o(1)}{n^{1-2\alpha}}\rightarrow0,\label{eq3}% \end{equation} where the $o\left( 1\right) $ can be set to be no smaller in order than rate $\frac{r}{n}$.
(b) when $r=O\left( n\right) $ where the big $O$ notation here means that $\liminf_{n\rightarrow\infty}\frac{r}{n}>0$, for any $\varepsilon>0$, \begin{equation} P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}\right\vert >\varepsilon\right) \leq C\frac{1}{\left( n+2\right) \varepsilon^{2}% }\label{eq2}% \end{equation} from some constant $C\leq2$; in this case $\varepsilon=o\left( \sqrt {n}\right) $ can be set such that \begin{equation} P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}\right\vert >\frac {1}{n^{\alpha}}\right) \leq\frac{2}{n^{1-2\alpha}}\rightarrow0\label{eq4}% \end{equation} for any $0\leq\alpha<\frac{1}{2}$.
Observation:A simple conclusion from the above discussion is that, regardless of the valueof $r$, we have that $X_{\left( r\right) }$ for each $r$ converges to$E\left[ X_{\left( r\right) }\right] $ in probability as $n\rightarrow\infty$. Further, we know the rate of convergence is $n^{-\alpha}$ for any $0 \le \alpha < 1/2$. My question is " are the deviation bounds given abovethe best?" Very likely NOT. Question:Let $a_{n,r}$ be a positive sequencethat depends on $n$ and $r$ such that$$\lim_{n\rightarrow\infty}a_{n,r}=0$$for each $r=1,...,n$, what is the best result available on$$\beta_{n,r} \ge P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}%\right\vert >a_{n,r}\right) $$where $\beta_{n,r} \to 0$ as $n \to \infty$? By this I mean, what is the bound $\beta_{n,r}$ corresponding to the sequence $a_{n,r}$ that converges to $0$ at relatively and possibly the fastest speed?
Any pointers or hints would be greatly appreciated! Thanks!
Update 1: (see update 3 below)
Thanks for Henry's comment. I found this: https://projecteuclid.org/euclid.ecp/1465263184, Concentration inequalities for order statistics. But this paper is mainly about concentration of order statistics of i.i.d. standard Gaussian random variables. The second paragraph in the Introduction of this paper quotes without a proof a general concentration of measure phenomenon for i.i.d. standard Gaussian random variables.
If someone can point out to me a reference on how this result is obtained, that will be great. Since I guess I can reverse engineer this result from its proof to get a result for i.i.d. standard uniform random variables. Update 3 (April 14, 2017):
By Theorem 2 of Chung 1949 "An estimate concerning the Kolmogoroff limit distribution, Trans. Amer. Math. Soc. 67: 36–50", we see $$ P\left( \limsup\limits_{n\rightarrow\infty}\dfrac{n\sup\limits_{t\in\mathbb{R}% }\left\vert \mathbb{S}_{n}\left( t\right) -S_{\ast}\left( t\right) \right\vert }{\left( 2^{-1}n\log_{\left( 2\right) }n\right) ^{1/2}}=1\right) =1 $$ for any continuous CDF $S_{\ast}$ on $\mathbb{R}$ with $\mathbb{S}_{n}\ $ as its empirical CDF (ECDF), where $\log_{(s)}$ means the natural logarithm composed by itself $s$ times. Therefore, these order statistics converges at a rate as per the iterated logrithm.
In other words, when $n$ is very large, the classic location of each $X_{(r)}$ is $r/n$, with asymptotic deviation upper bounded by$$\tag{1}\frac{\sqrt{2 \log_{(2)}n}}{\sqrt{n}}$$Basically, we know where these order statistics are.
However, this raises an interesting question as follows. Let $u_1 = X_{(1)}$, $u_2 = X_{(2)}-X_{(1)}$, $\ldots$, $u_k = X_{(k)}-X_{(k-1)}$, $\ldots$ , $u_n = X_{(n)}-X_{(n-1)}$, and $u_{n+1}=1-X_{(n)}$ be the uniform spacings. Further, let$$u^{\ast} = \max_{1 \le k \le n+1} u_{k}$$be the maximal uniform spacing. Then by Devroye (1981, 1982) "Laws of the iterated logarithm for order statistics of uniform spacings" and "A log log law for maximal uniform spacings", we know$$P\left( \limsup\limits_{n\rightarrow\infty}\frac{n{u}^{\ast}-\log n}{2\log_{\left( 2\right) }n}=1\right) =1,$$
Namely, the maximal spacing is asymptotically no larger than$$\tag{2}\frac{\log n + 2 \log_{(2)} n}{n}$$Devroye (1981, 1982) also provide the law of the iterative logarithm for the minimal uniform spacing $u_{\ast} = \min_{1 \le k \le n+1} u_k$. Specifically,$$P\left( \liminf\limits_{n\rightarrow\infty}\left( n {u}_{\ast}-\log n +\log_{\left( 3\right) }n\right) =-\log2\right) =1$$ This means that the miminal spacing asympotitcally is no smaller than$$\tag{3}\frac{-\log 2 + \log n - \log_{(3)}n}{n}$$With this piece of information, we can approximately locate all $X_{(r)}$, by first locating $X_{(n)}$, then $X_{(n-1)}$ and so on. Question #2:
I am curious on the following: Look at the largest order statistic $X_{(n)}$ and the smallest $X_{(1)}$. Then each of them has two different rates of convergence, $(1)$ and $(2)$, respectively to $0$ and $1$. Since both rates are correct, we see that the extreme statistics converge much faster than non-extreme ones.
Why is the maximal uniform spacing so small in magnitude compared to the maximal oscillation in the empirical distribution? Which rate of convergence would you use to identify the locations of uinform order statistics? (I am aware of one quick and intuitive answer, which is "on average the individual spacing should be around $n^{-1}$ and for the empirical distribution central limit theorem plays a role to give $n^{-1/2}$". But this does not seems to be somewhat convincing.) Question #3 (April 16, 2017):
For the convergence of the order statistics to their classic locations, the first rate is based on deviation of empirical distribution, whereas the second based on uniform spacing.
This prompts me to think:
is there a rate of convergence for uniform order statistics that is better than $(1)$ if we know that $F(t)=t$, i.e., is the rate in Chung's 1949 paper optimal for convergence of empirical distribution of uniform random variables? (Probably not since that result was obtained for the space of functions of totally bounded variation.) Update (April 21, 2017): answer to question #3The convergence rate of uniform order statistics to their expecations obtained from $(1)$ is NOT optimal for order statisic $X_{(k)}$ for which $k$ or $n-k$ is bounded. However, it is optimal for $X_{(k)}$ for which $k= c_n n$ with $$1 > \limsup c_n \ge \liminf c_n >0.$$ In other words, close to the edges 0 or 1, the optimal reate of convergence is proportional to $$\frac{\log n}{n},$$ whereas in other parts of the compact interval $[0,1]$ the optimal rate of convergence is $$\frac{\sqrt{2 \log_{(2)}n}}{\sqrt{n}}.$$ This can be seen from: Lemma A2.1 on page 148 of "asymptotic expansions for the power of distribution free tests in the one-sample problem" by Albers, Bickel and Zwet. |
Let $X$ be a smooth complex projective manifold, and $L$ an ample line bundle. By Bertini's Theorem, for every integer $q$ big enough there exists an open dense subset $U_q\subset |qL|$ such that every divisor $D$ in $U_q$ is smooth.
Warm up question: is the complement of $U_q$ always a divisor?
We can define a bigger open subset $V_q\subset |qL|$ as $$ V_q:=\{D\in |qL| \; \textrm{s. t.} \; (X,\frac{1}{q}D) \; \textrm{ is klt} \} $$
My question is: what is the dimension of the complement of $V_q$ ? (or at least can we bound its asymptotic in $q$ ? e.g. is it upper-bounded by $aq^{\dim X}$ with $a$ a constant which is strictly smaller than the volume of $L$? ) |
I have a set of equations to integrate something in time $t$. At each time step I compute a scalar field $\phi(t)$ and a potential $V(\phi)$. I should also control the conservation of energy with an equation:
$E=\frac{1}{2}\left(\dot\phi^2+V(\phi)\right)$
Or numerically:
$E=\frac{1}{2}\left(\left(\frac{\phi_{n}-\phi_{n-1}}{t_{n}-t_{n-1}}\right)^2+V(\phi)\right)$
My question is: is it $V(\phi_{n})$ or $V(\phi_{n-1})$ ? |
I am using the leave one out cross-validation technique to evaluate my model. If the prediction on the test sample is right the output is 1 otherwise 0. So I have a array of N samples with 0's and 1's at the end of it. I then average these values to get the average prediction accuracy and calculate the Standard deviation. I am getting the Average as 0.6 but the standard deviation is 0.5 (which is large). But the Mean + Standard deviation is more than the range , is that normal or is it wrong ? I have read that leave one out tends to have high variance because of high correlation between models. The second question is there a significance test I can do on the cross validation results to evaluate them?
The big standard deviation is completely. In fact, is is completely determined by the mean prediction accuracy: Your values are either 0 or 1 (with 60% being 1), so the standard deviation is $\sqrt{0.4\cdot(0-0.6)^2 + 0.6\cdot(1-0.6)^2} \approx 0.49$.
For why the variance of the prediction error rate is higher for leave-one-out CV compared to 10-fold CV, see this older answer.
Regarding your second question, I would use a permutation test: Completely shuffle the mapping between factors and labels of your training data, train a new model on it and calculate its mean prediction accuracy -- this estimates the accuracy that you get by chance[1]. Repeat this procedure several times to get a distribution of the chance prediction accuracy. Now compare your actual prediction accuracy (with unshuffled labels) to that distribution -- your $p$ value is the percentage of chance accuracies that are better than your actual prediction accuracy.
If you have few data points, you should do the permutation test with all possible permutations. Otherwise, you need enough repetitions to make sure that (the complete confidence interval of) the $p$ value is below your significance level. I don't have good rules of thumb for the general case; the relevant Wikipedia article links to this paper.
[1] For a balanced two-class problem, this chance level typically should be about 0.5; but it will decrease with the number of classes and increase if some classes are more frequent in your training data than others. |
Since the other answer is more a wave of the hand than an answer one would submit to your instructor I thought a proper, economical, write-up might be welcome. (Not that there is anything wrong with "hand-waving"we all do it: the intention is "Here is the idea -- You write it up yourself now.")
Also the other poster was less than pleased by the discussion and asked (one presumes) for details.
Problem. Let $f:[a,b]\to \mathbb R$ have the property that the variation $V_c^b f \leq M$ for all $a<c<b$. Show that $f$ has bounded
variation on $[a,b]$ and that $$V_a^b f \leq M + |f(a+)-f(a)|.$$ In
particular, if $f$ is continuous on the right at $a$, then $V_a^b f
\leq M$.
First observe that, for any $a<x<b$ $$ |f(b) -f(x)| \leq V_x^b f \leq M.$$Hence $|f|$ is bounded on $(a,b]$ by $M+|f(b)|$.
Now take any subdivision$a=a_0<a_1 <a_2< \dots < a_n=b$ and any $a<t<a_1$. Observe that$$ \sum_{i=1}^n |f(a_{i+1})-f(a_i)| \leq |f(t)-f(a)| + |f(a_1)-f(t)| + \dots +|f(a_n)-f(a_{n-1})|$$$$ \leq |f(t)-f(a)| + M .$$This is true for all such $t$ and $f$ is bounded, so$$ \sum_{i=1}^n |f(a_{i+1})-f(a_i)| \leq \liminf_{t\to a+}|f(t)-f(a)| + M< \infty.$$
Since this is true for all such subdivisions of $[a,b]$ we have $$ V_a^b f \leq \liminf_{t\to a+}|f(t)-f(a)| + M$$and so $f$ has bounded variation on $[a,b]$.
All functions of bounded variation are regulated (i.e. they have finite right and left hand limits at each point) andso $$\liminf_{t\to a+}|f(t)-f(a)| = \lim_{t\to a+}|f(t)-f(a)| = |f(a+)-f(a)|$$completing the proof. |
$u$ substitution doesnt work. I don't see any connection with the Weierstrass substitution either. integration by parts results in a infinite integral series.
this integral does not have a solution in terms of elementary functions. But, you can solve it using series methods. For instance, since
$$ \sin x = \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$
$$ \text{then} \; \; \sin (x^2) = \sum \frac{(-1)^n x^{4n+2}}{(2n+1)!}$$
$$ \int \sin (x^2) = \int \sum \frac{(-1)^n x^{4n+2}}{(2n+1)!} = \sum \frac{(-1)^n x^{4n+3}}{(2n+1)!(4n+3)} + K $$
Mathematica returns:
$$\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} x\right)$$
So, unless you consider Fresnel Sine to be an elementary function, that explains your troubles. |
M 3: a new muon missing momentum experiment to probe (g – 2) μ and dark matter at Fermilab Abstract
Here, new light, weakly-coupled particles are commonly invoked to address the persistent $$\sim 4\sigma$$ anomaly in $$(g-2)_\mu$$ and serve as mediators between dark and visible matter. If such particles couple predominantly to heavier generations and decay invisibly, much of their best-motivated parameter space is inaccessible with existing experimental techniques. In this paper, we present a new fixed-target, missing-momentum search strategy to probe invisibly decaying particles that couple preferentially to muons. In our setup, a relativistic muon beam impinges on a thick active target. The signal consists of events in which a muon loses a large fraction of its incident momentum inside the target without initiating any detectable electromagnetic or hadronic activity in downstream veto systems. We propose a two-phase experiment, M$^3$ (Muon Missing Momentum), based at Fermilab. Phase 1 with $$\sim 10^{10}$$ muons on target can test the remaining parameter space for which light invisibly-decaying particles can resolve the $$(g-2)_\mu$$ anomaly, while Phase 2 with $$\sim 10^{13}$$ muons on target can test much of the predictive parameter space over which sub-GeV dark matter achieves freeze-out via muon-philic forces, including gauged $$U(1)_{L_\mu - L_\tau}$$.
Authors: Princeton Univ., Princeton, NJ (United States) Fermi National Accelerator Lab. (FNAL), Batavia, IL (United States) Publication Date: Research Org.: Fermi National Accelerator Lab. (FNAL), Batavia, IL (United States) Sponsoring Org.: USDOE Office of Science (SC), High Energy Physics (HEP) (SC-25) OSTI Identifier: 1439466 Report Number(s): arXiv:1804.03144; FERMILAB-PUB-18-087-A Journal ID: ISSN 1029-8479; 1667037; TRN: US1900618 Grant/Contract Number: AC02-07CH11359 Resource Type: Journal Article: Accepted Manuscript Journal Name: Journal of High Energy Physics (Online) Additional Journal Information: Journal Volume: 2018; Journal Issue: 9; Journal ID: ISSN 1029-8479 Publisher: Springer Berlin Country of Publication: United States Language: English Subject: 79 ASTRONOMY AND ASTROPHYSICS; 46 INSTRUMENTATION RELATED TO NUCLEAR SCIENCE AND TECHNOLOGY; 72 PHYSICS OF ELEMENTARY PARTICLES AND FIELDS; Fixed target experiments Citation Formats
Kahn, Yonatan, Krnjaic, Gordan, Tran, Nhan, and Whitbeck, Andrew.
M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab. United States: N. p., 2018. Web. doi:10.1007/JHEP09(2018)153.
Kahn, Yonatan, Krnjaic, Gordan, Tran, Nhan, & Whitbeck, Andrew.
M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab. United States. doi:10.1007/JHEP09(2018)153.
Kahn, Yonatan, Krnjaic, Gordan, Tran, Nhan, and Whitbeck, Andrew. Wed . "M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab". United States. doi:10.1007/JHEP09(2018)153. https://www.osti.gov/servlets/purl/1439466.
@article{osti_1439466,
title = {M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab}, author = {Kahn, Yonatan and Krnjaic, Gordan and Tran, Nhan and Whitbeck, Andrew}, abstractNote = {Here, new light, weakly-coupled particles are commonly invoked to address the persistent $\sim 4\sigma$ anomaly in $(g-2)_\mu$ and serve as mediators between dark and visible matter. If such particles couple predominantly to heavier generations and decay invisibly, much of their best-motivated parameter space is inaccessible with existing experimental techniques. In this paper, we present a new fixed-target, missing-momentum search strategy to probe invisibly decaying particles that couple preferentially to muons. In our setup, a relativistic muon beam impinges on a thick active target. The signal consists of events in which a muon loses a large fraction of its incident momentum inside the target without initiating any detectable electromagnetic or hadronic activity in downstream veto systems. We propose a two-phase experiment, M$^3$ (Muon Missing Momentum), based at Fermilab. Phase 1 with $\sim 10^{10}$ muons on target can test the remaining parameter space for which light invisibly-decaying particles can resolve the $(g-2)_\mu$ anomaly, while Phase 2 with $\sim 10^{13}$ muons on target can test much of the predictive parameter space over which sub-GeV dark matter achieves freeze-out via muon-philic forces, including gauged $U(1)_{L_\mu - L_\tau}$.}, doi = {10.1007/JHEP09(2018)153}, journal = {Journal of High Energy Physics (Online)}, issn = {1029-8479}, number = 9, volume = 2018, place = {United States}, year = {2018}, month = {9} } Citation information provided by Web of Science
Web of Science
Figures / Tables: left) and vector ( right) forces that couple predominantly to muons. In both cases, a relativistic muon beam is incident on a fixed target and scatters coherently off a nucleus to produce the new particle as initial- ormore » |
I am struggling with this question:
Prove or give a counterexample: If $f : X \to Y$ is a continuous mapping from a compact metric space $X$, then $f$ is uniformly continuous on $X$.
Thanks for your help in advance.
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I am struggling with this question:
Prove or give a counterexample: If $f : X \to Y$ is a continuous mapping from a compact metric space $X$, then $f$ is uniformly continuous on $X$.
Thanks for your help in advance.
The answer is yes, if $f$ is continuous on a compact space then it is uniformly continuous:
Let $f: X \to Y$ be continuous, let $\varepsilon > 0$ and let $X$ be a compact metric space. Because $f$ is continuous, for every $x$ in $X$ you can find a $\delta_x$ such that $f(B(\delta_x, x)) \subset B({\varepsilon\over 2}, f(x))$. The balls $\{B(\delta_x, x)\}_{x \in X}$ form an open cover of $X$. So do the balls $\{B(\frac{\delta_x}{2}, x)\}_{x \in X}$. Since $X$ is compact you can find a finite subcover $\{B(\frac{\delta_{x_i}}{2}, x_i)\}_{i=1}^n$. (You will see in a second why we are choosing the radii to be half only.)
Now let $\delta_{x_i}' = {\delta_{x_i}\over 2}$.
You want to choose a distance $\delta$ such that for any two $x,y$ they lie in the same $B(\delta_{x_i}', x_i)$ if their distance is less than $\delta$.
How do you do that?
Note that now that you have finitely many $\delta_{x_i}'$ you can take the minimum over all of them: $\min_i \delta_{x_i}'$. Consider two points $x$ and $y$. Surely $x$ lies in one of the $B(\delta_{x_i}', x_i) $ since they cover the whole space and hence $x$ also lies in $B(\delta_{x_i}', x_i)$ for some $i$.
Now we want $y$ to also lie in $B(\delta_{x_i}', x_i)$. And this is where it comes in handy that we chose a subcover with radii divided by two:
If you pick $\delta : = \min_i \delta_{x_i}'$ (i.e. $\delta = \frac{\delta_{x_i}}{2}$ for some $i$) then $y$ will also lie in $B(\delta_{x_i}, x_i)$:
$d(x_i, y) \leq d(x_i, x) + d(x,y) < \frac{\delta_{x_i}}{2} + \min_k \delta_{x_k} \leq \frac{\delta_{x_i}}{2} + \frac{\delta_{x_i}}{2} = \delta_{x_i}$.
Hope this helps.
Let $(X, d)$ be a compact metric space, and $(Y, \rho)$ be a metric space. Suppose $f : X \to Y$ is continuous. We want to show that it is uniformly continuous.
Let $\epsilon > 0$. We want to find $\delta > 0$ such that $d(x,y) < \delta \implies \rho(f(x), f(y))< \epsilon$.
Ok, well since $f$ is continuous at each $x \in X$, then there is some $\delta_{x} > 0$ so that $f(B(x, \delta_{x})) \subseteq B(f(x), \frac{\epsilon}{2})$.
Now, $\{B(x, \frac{\delta_{x}}{2})\}_{x \in X}$ is an open cover of $X$, so there is a finite subcover $\{B(x_{i}, \frac{\delta_{x_{i}}}{2})\}_{i =1}^{n}$.
If we take $\delta := \min_{i} (\frac{\delta_{x_{i}}}{2})$, then we claim $d(x,y) < \delta \implies \rho(f(x), f(y)) < \epsilon$. Why?
Well, suppose $d(x,y) < \delta$. Since $x \in B(x_{i}, \frac{\delta_{x_{i}}}{2})$ for some $i$, we get $y \in B(x_{i}, \delta_{x_{i}})$. Why? $d(y, x_{i}) \leq d(y,x) + d(x,x_{i}) < \frac{\delta_{x_{i}}}{2} + \frac{\delta_{x_{i}}}{2} = \delta_{x_{i}}$.
Ok, finally, if $d(x,y) < \delta$, then we claim $\rho(f(x), f(y)) < \epsilon$. This is because $\rho(f(x), f(y)) \leq \rho(f(x), f(x_{i})) + \rho(f(x_{i}), f(y)) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. |
I have a question. If $X$ and $Y$ are Banach spaces, we have to prove that a compact linear operator is completely continuous. A mapping $T \colon X \to Y$ is called completely continuous, if it maps a weakly convergent sequence in $X$ to a strongly convergent sequence in $Y$ , i.e., $x_n\underset{n\to +\infty}\rightharpoonup x$ implies $\lVert Tx_n- Tx\rVert_Y\to 0$.
Since it's a homework question I will just give some steps.
By linearity, we can assume that $x=0$. We have to show that for each subsequence of $\{Tx_n\}$, we can extract a further subsequence which converges to $0$ in norm in $Y$. A weakly converging sequence is bounded. $T$ maps bounded sets to sets with a compact closure.
Once the second steps is shown, we can conclude. Indeed, assume that $Tx_n$ doesn't converge to $0$. Then we are able to find $\delta>0$ and $A$ an infinite subset of the natural numbers such that $\lVert Tx_k\rVert_Y\geq\delta$ for each element of $A$. We can consider it as a subsequence, and we can't extract a further subsequence which converges to $0$, a contradiction. |
I am an adult software developer who is trying to do a math reboot. I am working through the exercises in the following book.
Ayres, Frank , Jr. and Elliott Mendelson. 2013.
Schaum's Outlines Calculus Sixth Edition (1,105 fully solved problems, 30 problem-solving videos online). New York: McGraw Hill. ISBN 978-0-07-179553-1.
Most of the precalculus problems are easy, but every chapter seems to have a problem that is significantly harder than the others. The following problem struck me as especially hard. Frankly, it seems completely out of depth compared to the other problems. Clearly I have a gap in my knowledge.
Chapter 4 circles, problem 24.
Let $\mathscr{C}_1$ and $\mathscr{C}_2$ be two intersecting circles determined by the equations $x^2+y^2+A_1x+B_1y+C_1=0$ and $x^2+y^2+A_2x+B_2y+C_2=0$. For any number $k \ne -1$, show that $$ x^2+y^2+A_1x+B_1y+C_1+k(x^2+y^2+A_2x+B_2y+C_2)=0 $$ is the equation of a circle through the intersection of $\mathscr{C}_1$ and $\mathscr{C}_2$. Show, conversely, that every such circle may be represented by such an equation for a suitable $k$.
My attempt.
I spent weeks with SageMath and WolframAlpha figuring out that the intersecting points $P$ and $Q$ of $\mathscr{C}_1$ and $\mathscr{C}_2$ given $A_1$, $B_1$, $C_1$, $A_2$, $B_2$ and $C_2$ are as follows.
$$ P_x = \frac{ (A_1 - A_2)(C_2 - C_1) - (B_1 - B_2) \bigg(\frac{A_2 B_1 - A_1 B_2}{2} - \sqrt{ \frac{(A_2 B_1 - A_1 B_2)^2}{4} + (A_1 - A_2)(A_2 C_1 - A_1 C_2) + (B_1 - B_2)(B_2 C_1 - B_1 C_2) - (C_1 - C_2)^2 }\bigg)}{ ( A_1 - A_2 )^2 + ( B_1 - B_2 )^2 } \\ P_y = \frac{ (B_1 - B_2)(C_2 - C_1) - (A_1 - A_2)\bigg(\frac{A_1 B_2 - A_2 B_1}{2} + \sqrt{ \frac{(A_2 B_1 - A_1 B_2)^2}{4} + (A_1 - A_2) (A_2 C_1 - A_1 C_2) + (B_1 - B_2) (B_2 C_1 - B_1 C_2) - (C_1 - C_2)^2 }\bigg)}{ ( A_1 - A_2 )^2 + ( B_1 - B_2 )^2 } \\ Q_x = \frac{ (A_1 - A_2)(C_2 - C_1) - (B_1 - B_2)\bigg(\frac{A_2 B_1 - A_1 B_2}{2} + \sqrt{ \frac{(A_2 B_1 - A_1 B_2)^2}{4} + (A_1 - A_2)(A_2 C_1 - A_1 C_2) + (B_1 - B_2)(B_2 C_1 - B_1 C_2) - (C_1 - C_2)^2 }\bigg)}{ ( A_1 - A_2 )^2 + ( B_1 - B_2 )^2 } \\ Q_y = \frac{ (B_1 - B_2) (C_2 - C_1) - (A_1 - A_2) \bigg(\frac{A_1 B_2 - A_2 B_1}{2} - \sqrt{ \frac{(A_2 B_1 - A_1 B_2)^2}{4} + (A_1 - A_2) (A_2 C_1 - A_1 C_2) + (B_1 - B_2) (B_2 C_1 - B_1 C_2) - (C_1 - C_2)^2 }\bigg)}{ ( A_1 - A_2 )^2 + ( B_1 - B_2 )^2 } $$
I then found that the intersecting points of $\mathscr{C}_1$ or $\mathscr{C}_2$ and $x^2+y^2+A_1x+B_1y+C_1+k(x^2+y^2+A_2x+B_2y+C_2)=0$ are the same as the above. $k$ was "magically" simplified out. I am convinced that the assertion in the question is correct, but I do not feel that I properly solved the problem.
I have three questions.
What does a proper proof that solves the problem look like? What are effective ways to approach problems like this? What is this precalculus warm up problem trying to teach me as I move towards calculus? |
I am given that $a_n \ge 0$ and $\sum a_n$ converges.
I need to show that $\sum \frac{\sqrt{a_n}}{n}$ converges also.
After a long time, I came up with a solution by re-ordering the series into:
$\sum_{a_n < \frac{1}{n^2}} \frac{\sqrt{a_n}}{n} + \sum_{a_n \ge \frac{1}{n^2}} \frac{\sqrt{a_n}}{n}$ which is bounded by $\sum\frac{1}{n^2} + \sum{a_n}$
I was wondering if there is a more basic solution because I don't want to be using ad-hoc methods on an exam. |
This question already has an answer here:
I am desperately trying to solve the following problem, and would really appreciate help!
Suppose $R$ and $Q$ are two quantum systems with the same Hilbert space $\mathcal{H}$ with $\dim(\mathcal{H})=N$. Let $|i_R\rangle$ and $|i_Q\rangle$ be orthonormal basis sets for $R$ and $Q$. Let $A$ be an operator on $R$ and $B$ an operator on $Q$. Define $|m\rangle := \sum_{i=1}^N |i_R\rangle\otimes |i_Q\rangle$. Show that $$ tr(A^\dagger B)=\langle m|A\otimes B| m\rangle $$
where the multiplication on the left hand side is of matrices, and it is understood that the matrix elements of $A$ are taken with respect to the basis $|i_R\rangle$ and those for $B$ with respect to the basis $|i_Q\rangle$.
What I have is the following:
LHS: $$ tr(A^\dagger B)=\sum_{n=1}^N \langle n_Q|\left( \sum_{i,j,k=1}^N a^*_{ji}b_{jk} |i_Q\rangle \langle k_R| \right)|n_R\rangle $$
Since $Q$ and $R$ are over the same Hilbert space, we can disregard the fact that one set of basis vectors is in system $Q$ and the other one in system $R$. This yields $$ tr(A^\dagger B)=\sum_{i,j=1}^N a^*_{ji} b_{ji} $$
RHS: $$ \begin{align*} \langle m'|(A\otimes B)|m\rangle &= \sum_{e,f=1}^N \Bigg( \langle e_R| \left( \sum_{k,l=1}^N a_{kl} |k_R\rangle \langle l_R\right) |f_R\rangle \cdot \langle e_Q|\left(\sum_{i,j=1}^N b_{ij}|i_Q\rangle\langle j_Q| \right)|f_Q\rangle\Bigg)\\ &=\sum_{e,f,k,l,i,j=1}^N a_{kl} b_{ij}\langle e_R |k_R \rangle \langle l_R|f_R \rangle \langle e_Q|i_Q \rangle\langle j_Q|f_Q \rangle\\ &=\sum_{i,j=1}^N a_{ij} b_{ij} \end{align*} $$
As you can see, the two expressions are
almost the same. Only the first number on the LHS evaluation is a complex conjugate and on the RHS it's not.
Any ideas where I went wrong?
Thanks!
P.S: Full disclosure: I posted this question on PhysicsForum.com as well. |
No, the equation reflects the fact that energy is conserved. If a particle has an energy, E, then in the absence of any interactions the energy will remain E. The distribution of E between KE and PE will vary- that's suggested by the left hand side of the equation which shows a potential varying over space.
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Let me not answer this question but only provide you with a hint. You are already close to what you are looking for.I don't think that your approach is flawed. There are just a few gaps that you have overlooked. You are right in concluding that the work $W$ should depend on the mass $m$ and the velocity $\vec{v}$. However, you know that $W$ is a scalar. ...
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First, you should understand what "conservation" means in a physics sense. Conservation of a quantity means it is neither created nor destroyed in a time-limited process (we're not talking about eternity past or future). It is merely moved around to other systems/objects or (for energy) changes forms. The means that if a quantity (like momentum or total ...
The mistake you are making is saying the second equation is due to "conservation of energy". The equation only addresses conservation of kinetic energy. Kinetic energy is not conserved. But total energy is conserved. The kinetic energy that is lost is dissipated primarily as heat because the collision is inelastic.Hope this helps.
What you are talking about is an inelastic collision, which is defined by loss of kinetic energy (converted into e.g. heat or sound, as you had before the edit). Check this wikipedia page for more info.On a side note, if you have two expressions that tell you different things, this means that at least one of them is wrong. You don't just choose one of them ... |
Physical World and Measurements Errors in Measurements and Accuracy and precision of measuring instruments Uncertainty in measurement of a physical quantity is called the error in measurement. The difference between the measured value and true value as per standard method without mistakes is called error. Accuracy refers to how closely a measured value agrees with the true values. Precision refers to what limit or resolution the given physical quantity can be measured. The errors arise due to external conditions like charges in environment changes in temperature pressure, humidity etc. In the measurement of a physical quantity the arithmetic mean of all readings which is found to be very close to the most accurate reading is to be taken as true value of the quantities. Absolute errors, mean absolute error, relative error, percentage error. View the Topic in this video From 0:36 To 1:02:01<>
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1. The absolute error in measured values is given by
Δa n = a m − a n
2. Mean Absolute Error: The arithmetic mean of the magnitude of absolute errors in all the measurements is called mean absolute error.
\tt \overline{\Delta a}=\frac{\mid\Delta a_{1}\mid+\mid\Delta a_{2}\mid+...+\mid\Delta a_{n}\mid}{n}
3. Relative error = \tt\frac{Mean\ absolute\ error}{True \ value}=\frac{\overline{\Delta a}}{a_{m}}
4. Percentage error = \tt\ \frac{\overline{\Delta a}}{a_{m}}\times100\%
5. Error in Addition or Subtraction: Maximum absolute error in their addition or subtraction
Δ x = ± (Δ a + Δ b)
6. Error in Multiplication or Division: Maximum relative error \tt \frac{\Delta x}{x}=\pm\left(\frac{\Delta a}{a}+\frac{\Delta b}{b} \right)
7. Error in Raised to a Power: Maximum error \tt \frac{\Delta Z}{Z}=p\left(\frac{\Delta A}{A}\right)+q\left(\frac{\Delta B}{B}\right)+r\left(\frac{\Delta C}{C}\right) |
10.7 The optimal reconciliation approach
Optimal forecast reconciliation will occur if we can find the \(\bm{G}\) matrix which minimises the forecast error of the set of coherent forecasts. We present here a simplified summary of the approach. More details are provided in Wickramasuriya et al. (2019).
Suppose we generate coherent forecasts using Equation (10.6), repeated here for convenience: \[ \tilde{\bm{y}}_h=\bm{S}\bm{G}\hat{\bm{y}}_h. \]
First we want to make sure we have unbiased forecasts. If the base forecasts \(\hat{\bm{y}}_h\) are unbiased, then the coherent forecasts \(\tilde{\bm{y}}_h\) will be unbiased provided \(\bm{S}\bm{G}\bm{S}=\bm{S}\) (Hyndman et al., 2011). This provides a constraint on the matrix \(\bm{G}\). Interestingly, no top-down method satisfies this constraint, so all top-down methods are biased.
Next we need to find the error in our forecasts. Wickramasuriya et al. (2019) show that the variance-covariance matrix of the \(h\)-step-ahead coherent forecast errors is given by \[\begin{equation*} \bm{V}_h = \text{Var}[\bm{y}_{T+h}-\tilde{\bm{y}}_h]=\bm{S}\bm{G}\bm{W}_h\bm{G}'\bm{S}' \end{equation*}\] where \(\bm{W}_h=\text{Var}[(\bm{y}_{T+h}-\hat{\bm{y}}_h)]\) is the variance-covariance matrix of the corresponding base forecast errors.
The objective is to find a matrix \(\bm{G}\) that minimises the error variances of the coherent forecasts. These error variances are on the diagonal of the matrix \(\bm{V}_h\), and so the sum of all the error variances is given by the trace of the matrix \(\bm{V}_h\). Wickramasuriya et al. (2019) show that the matrix \(\bm{G}\) which minimises the trace of \(\bm{V}_h\) such that \(\bm{S}\bm{G}\bm{S}=\bm{S}\), is given by \[ \bm{G}=(\bm{S}'\bm{W}_h^{-1}\bm{S})^{-1}\bm{S}'\bm{W}_h^{-1}. \] Therefore, the optimal reconciled forecasts are given by \[\begin{equation} \tag{10.8} \tilde{\bm{y}}_h=\bm{S}(\bm{S}'\bm{W}_h^{-1}\bm{S})^{-1}\bm{S}'\bm{W}_h^{-1}\hat{\bm{y}}_h. \end{equation}\] We refer to this as the “MinT” (or Minimum Trace) estimator.
To use this in practice, we need to estimate \(\bm{W}_h\), the forecast error variance of the \(h\)-step-ahead base forecasts. This can be difficult, and so we provide four simplifying approximations which have been shown to work well in both simulations and in practice.
Set \(\bm{W}_h=k_h\bm{I}\) for all \(h\), where \(k_{h} > 0\).
20This is the most simplifying assumption to make, and means that \(\bm{G}\) is independent of the data, providing substantial computational savings. The disadvantage, however, is that this specification does not account for the differences in scale between the levels of the structure, or for relationships between series. This approach is implemented in the
forecast()function by setting
method = "comb"and
weights = "ols".
The weights here are referred to as OLS (ordinary least squares) because setting \(\bm{W}_h=k_h\bm{I}\) in (10.8) gives the least squares estimator we introduced in Section 5.7 with \(\bm{X}=\bm{S}\) and \(\bm{y}=\hat{\bm{y}}\).
Set \(\bm{W}_{h} = k_{h}\text{diag}(\hat{\bm{W}}_{1})\) for all \(h\), where \(k_{h} > 0\), \[ \hat{\bm{W}}_{1} = \frac{1}{T}\sum_{t=1}^{T}\bm{e}_{t}\bm{e}_{t}', \] and \(\bm{e}_{t}\) is an \(n\)-dimensional vector of residuals of the models that generated the base forecasts stacked in the same order as the data. The approach is implemented by setting
method = "comb"and
weights = "wls".
This specification scales the base forecasts using the variance of the residuals and it is therefore referred to as the WLS (weighted least squares) estimator using
variance scaling.
Set \(\bm{W}_{h}=k_{h}\bm{\Lambda}\) for all \(h\), where \(k_{h} > 0\), \(\bm{\Lambda}=\text{diag}(\bm{S}\bm{1})\), and \(\bm{1}\) is a unit vector of dimension \(n\). This specification assumes that the bottom-level base forecast errors each have variance \(k_{h}\) and are uncorrelated between nodes. Hence each element of the diagonal \(\bm{\Lambda}\) matrix contains the number of forecast error variances contributing to each node. This estimator only depends on the structure of the aggregations, and not on the actual data. It is therefore referred to as
structural scaling. Applying the structural scaling specification is particularly useful in cases where residuals are not available, and so variance scaling cannot be applied; for example, in cases where the base forecasts are generated by judgmental forecasting (Chapter 4). This approach is implemented by setting
method="comb"and
weights = "nseries".
Set \(\bm{W}_h = k_h \bm{W}_1\) for all \(h\), where \(k_h>0\). Here we only assume that the error covariance matrices are proportional to each other, and we directly estimate the full one-step covariance matrix \(\bm{W}_1\). The most obvious and simple way would be to use the sample covariance. This is implemented by setting
method = "comb",
weights = "mint", and
covariance = "sam".
However, for cases where the number of bottom-level series \(m\) is large compared to the length of the series \(T\), this is not a good estimator. Instead we use a shrinkage estimator which shrinks the sample covariance to a diagonal matrix. This is implemented by setting
method = "comb",
weights = "mint", and
covariance = "shr".
In summary, unlike any other existing approach, the optimal reconciliation forecasts are generated using all the information available within a hierarchical or a grouped structure. This is important, as particular aggregation levels or groupings may reveal features of the data that are of interest to the user and are important to be modelled. These features may be completely hidden or not easily identifiable at other levels.
For example, consider the Australian tourism data introduced in Section 10.1, where the hierarchical structure followed the geographic division of a country into states and zones. Some coastal areas will be largely summer destinations, while some mountain regions may be winter destinations. These differences will be smoothed at the country level due to aggregation.
Example: Forecasting Australian prison population
We compute the forecasts for the Australian prison population, described in Section 10.2. Using the default arguments for the
forecast() function, we compute coherent forecasts by the optimal reconciliation approach with the WLS estimator using variance scaling.
To obtain forecasts for each level of aggregation, we can use the
aggts() function. For example, to calculate forecasts for the overall total prison population, and for the one-factor groupings (State, Gender and Legal Status), we use:
A simple plot is obtained using
A nicer plot is available using the following code. The results are shown in Figure 10.8. The vertical line marks the start of the forecast period.
prisonfc <- ts(rbind(groups, fcsts), start=start(groups), frequency=4)p1 <- autoplot(prisonfc[,"Total"]) + ggtitle("Australian prison population") + xlab("Year") + ylab("Total number of prisoners ('000)") + geom_vline(xintercept=2017)cols <- sample(scales::hue_pal(h=c(15,375), c=100,l=65,h.start=0,direction = 1)(NCOL(groups)))p2 <- as_tibble(prisonfc[,-1]) %>% gather(Series) %>% mutate(Date = rep(time(prisonfc), NCOL(prisonfc)-1), Group = str_extract(Series, "([A-Za-z ]*)")) %>% ggplot(aes(x=Date, y=value, group=Series, colour=Series)) + geom_line() + xlab("Year") + ylab("Number of prisoners ('000)") + scale_colour_manual(values = cols) + facet_grid(. ~ Group, scales="free_y") + scale_x_continuous(breaks=seq(2006,2018,by=2)) + theme(axis.text.x = element_text(angle=90, hjust=1)) + geom_vline(xintercept=2017)gridExtra::grid.arrange(p1, p2, ncol=1)
Similar code was used to produce Figure 10.9. The left panel plots the coherent forecasts for interactions between states and gender. The right panel shows forecasts for the bottom-level series.
The
accuracy() command is useful for evaluating the forecast accuracy across hierarchical or grouped structures. The following table summarises the accuracy of the bottom-up and the optimal reconciliation approaches, forecasting 2015 Q1 to 2016 Q4 as a test period.
The results show that the optimal reconciliation approach generates more accurate forecasts especially for the top level. In general, we find that as the optimal reconciliation approach uses information from all levels in the structure, it generates more accurate coherent forecasts than the other traditional alternatives which use limited information.
MAPE MASE MAPE MASE Total 5.32 1.84 3.08 1.06 State 7.59 1.88 7.62 1.85 Legal status 6.40 1.76 4.32 1.14 Gender 8.62 2.68 8.72 2.74 Bottom 15.82 2.23 15.25 2.16 All series 12.41 2.16 12.02 2.08 Bibliography
Hyndman, R. J., Ahmed, R. A., Athanasopoulos, G., & Shang, H. L. (2011). Optimal combination forecasts for hierarchical time series.
Computational Statistics and Data Analysis, 55(9), 2579–2589. https://robjhyndman.com/publications/hierarchical/
Wickramasuriya, S. L., Athanasopoulos, G., & Hyndman, R. J. (2019). Optimal forecast reconciliation for hierarchical and grouped time series through trace minimization.
J American Statistical Association, 114(526), 804–819. https://robjhyndman.com/publications/mint/ |
My question is how to find the easiest way to find the number of non-negative integer solutions to $$x+2y+4z=400$$ I know that I can use generating functions, and think of it as partitioning $400$ with $x$ $1$'s, $y$ $2$'s, and $z$ $4$'s. The overall generating function is: $$(1+x+x^2 + \cdots x^{400})(1+x^2+x^4+\cdots x^{200})(1+x^4+x^8+\cdots x^{100})$$ And then from this I have to calculate the coefficient of $x^{400}$, which I don't know how to do. If there's an easier way to do, I'd love to know.
My question is how to find the easiest way to find the number of non-negative integer solutions to $$x+2y+4z=400$$
I think the following way is easy (I'm not sure if it's the easiest, though).
Since $x+2y+4z=400$, $x$ has to be even. So, setting $x=2m$ gives you $$2m+2y+4z=400\Rightarrow m+y+2z=200.$$ Since $m+y$ has to be even, setting $m+y=2k$ gives you $$2k+2z=200\Rightarrow k+z=100.$$
There are $101$ pairs for $(k,z)$ such that $k+z=100$. For each $k$ such that $m+y=2k$, there are $2k+1$ pairs for $(m,y)$.
Hence, the answer is $$\sum_{k=0}^{100}(2k+1)=1+\sum_{k=1}^{100}(2k+1)=1+2\cdot \frac{100\cdot 101}{2}+100=10201.$$
You were on the right track, but instead of truncating the factors, just consider the coefficient of $x^{400}$ in: $$(1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^4+x^8+x^{12}+\ldots)=\frac{1}{(1-x)(1-x^2)(1-x^4)},\tag{1}$$ then write the RHS of $1$ as a sum of terms like $\frac{A}{(1-\xi x)^k}$, with $\xi\in\{1,-1,i,-i\}$, and exploit the identities: $$\frac{1}{1-\xi x}=\sum_{k=0}^{+\infty}(\xi x)^k,$$ $$\frac{1}{(1-\xi x)^2}=\sum_{k=0}^{+\infty}(k+1)(\xi x)^k,$$ $$\frac{1}{(1-\xi x)^3}=\sum_{k=0}^{+\infty}\frac{(k+2)(k+1)}{2}(\xi x)^k$$ to recover the final expression:
$$[x^n]\frac{1}{(1-x)(1-x^2)(1-x^4)}=\frac{2n^2+14n+21}{32}+\frac{(7+2n) (-1)^n}{32}+\frac{1}{8} \cos\left(\frac{n \pi }{2}\right)+\frac{1}{8} \sin\left(\frac{n \pi }{2}\right)$$
that if $8\mid n$ becomes:
$$[x^n]\frac{1}{(1-x)(1-x^2)(1-x^4)}=\frac{1}{16}(n+4)^2.$$ |
How many digits are there in the number $200^{2010}$? I have tried to re-write it as $(2\cdot 100)^{2010} = (2\cdot 10^2)^{2010} = 2^{2010} \cdot 10^{4020} = 1024^{201} \cdot 10^{4020} = 1.024^{201} \cdot 10^{4623}$. But how do I write $1.024^{201}$ as a power of base 10? I would like to solve the problem without logarithms or a calculator.
$\log 200^{2010}=2010\times\log 200=2010\times(2+0.3010)=4625.01$
$\therefore$ $4625\leq \log 200^{2010} <4626$
$\therefore$ $10^{4625}\leq 200^{2010}<10^{4626}$
$\therefore$ $4626$ digits
Think about some very simple cases $10$ has $2$ digits as does $99$, but $100$ has $3$ digits.$1\le \log_{10} 99 \lt 2$. A few explorations should convince you that the number of decimal digits of a positive integer $N$ is $\lfloor \log_{10} N\rfloor +1$.
You can prove this by writing $N=a\times 10^k$ where $1\le a\lt 10$ and $k$ is an integer and taking logs to base 10.
Now $\log_{10} (200^{2010})=2010 \log_{10}(200)=2010 \cdot(2+\log_{10}2)$ so you just need to have a sufficiently accurate estimate of $\log_{10}2$.
$1024^{201} \approx (10^3)^{201}=10^{603}$ is a rough estimate for that term.
We can do slightly better by writing $1024=10^3 + 24$, so
$1024^{201} = (10^3 + 24)^{201} = 10^{603} + 201\cdot24\cdot10^{600} + \tbinom {201}2\cdot24^2\cdot10^{597} + \cdots$
After just a few terms, they should be small enough so as not to add digits, and you can probably work that out without too much trouble.
This is kind of clunky, but it does completely avoid logarithms, if that's important.
You cannot really solve it without logarithms since this is exactly what logarithm is – it measures how many digits you need to write a number in a certain base. The number of decimal digits is $⌊\log_{10}(n)⌋ + 1$. $\log(200^{2010}) = 2010 * \log(200) = 2010 * \log(2 * 100) = 2010 * (\log(2) + 2) = 4625.07…$. So the exact number of digits is 4626. If you don't want to use calculator, you can estimate $\log_{10}(2)$ by another method. |
The key concepts with confidence intervals are
coverage, correctness and accuracy. Coverage
The
coverage or confidence level should be explained first. It is the percentage of times the random interval is expected to include the true value of the parameter.
The way this is best shown is to take a probability statement about a pivotal statistic and show how the statement is inverted to get a confidence interval. An example might be getting a confidence interval for the mean $\mu$ of a normal distribution when the variance is known to be $1$. Let the sample be denoted $X_i$, $i=1,2,\ldots,n$. The students will know from undergraduate courses or have been taught earlier in this particular graduate course that the sample mean $X_b=\sum X_i/n$ is normal with mean $\mu$ and variance $1/n$. Then the pivotal quantity is$$Z= \sqrt{n} (X_b - \mu)$$ and $Z$ has a $N(0,1)$ distribution. So of course $\Pr(|Z| \le 1.96) = 0.95$ (from a table of the standard normal distribution). You do the inversion to show that this probability is the same as $\Pr(X_b-1.96/\sqrt{n} \le \mu \le X_b+1.96/\sqrt{n})$. Then the random interval $[X_b-1.96/\sqrt{n}, X_b+1.96/\sqrt{n}]$ is a prescription for a 95% confidence interval for $\mu$.
It should be clear that if you repeated an experiment many times where each time you randomly select $n$ observations from an $N(\mu, 1)$ distribution, then in close to 95% of the cases the interval will contain $\mu$ and of course it also means that in the remaining approximate 5% of the cases $\mu$ will lie outside the interval. This is how I would explain an exact 95% confidence interval.
You could certainly use some other simple example such as estimating the rate parameter for an exponential distribution. The idea is to construct a pivotal quantity whose distribution is known and is independent of any unknown parameters.
To explain the relationship between coverage and confidence, you can just point out that if you substituted $1.645$ for $1.96$ in the original probability statement for $Z$ you would get a probability of $0.90$ and hence by replacing $1.96$ by $1.645$ in the confidence interval prescription you would get a 90% confidence interval. This also illustrates how lowering the coverage tightens the width of the interval.
Accuracy
The other two important concepts Efron calls accuracy and correctness. I like using that terminology. We looked at examples of exact confidence intervals. They are
accurate in the sense that the nominal coverage of 95% is the exact coverage probability. But sometimes it is convenient to use asymptotic theory. Instead of using the exact distribution for the pivotal quantity, we compute a distribution that it will converge to as the sample size $n$ goes to infinity. Using this asymptotic distribution, the coverage of an advertised 95% confidence interval will not be exact for given values of $n$. But if the approximation is good we can say that the approximate confidence interval is reasonably accurate.
(This is important in the bootstrap literature for confidence intervals because bootstrap confidence intervals are never exact and in some situations certain bootstrap variants (
e.g. the BCa method) give more accurate intervals than others. The bootstrap theory on order of accuracy was developed by Peter Hall and others who defined accuracy by the rate the interval approaches the advertised confidence level as $n$ goes to infinity. The results involve the use of Edgeworth expansions and can be found detailed in Hall's book The Bootstrap and Edgeworth Expansion.) Correctness
Last of all I would discuss
correctness. For many problems there are several ways to construct confidence intervals with exact or asymptotic coverage 95%. How do we choose between them? Well they will have different average lengths. An exact confidence interval that has the shortest expected length is called correct and is the optimal one to choose. When they exist and an efficient estimator of the parameter exists, correct confidence intervals can be constructed by choosing an efficient estimate for the parameter. |
what is the product of delta function with itself ? what is the dot product with itself ?
closed as off-topic by José Carlos Santos, user21820, Brahadeesh, Lord Shark the Unknown, Did Sep 18 '18 at 16:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
" This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, user21820, Brahadeesh, Lord Shark the Unknown, Did
A distribution is actually a linear functional on the space of compactly supported infinitely differentiable functions (the so called "test functions"). A function $f$ is compactly supported if $\overline{\{x : f(x) \neq 0\}}$ is compact (the overline denotes the closure).
The $\delta$-distribution is a linear functional such that for all $\phi \in C_c^\infty(\mathbb{R}^n)$ we have that $\langle \delta, \phi \rangle = \phi(0)$.
When you want to compute the product of distributions the problem is that you don't have a property which you would really like to have, that is associativity. So for distributions $\alpha$, $\beta$ and $\gamma$ we usually have that $(\alpha \cdot \beta) \cdot \gamma \neq \alpha \cdot (\beta \cdot \gamma)$. Wikipedia gives an example. However this does not really turn out to be a problem in applications. What we do have is convolution.
When we want to do convolution we prefer a smaller class of distributions (for example because on the smaller class the Fourier transform of a distribution in this class is again a distribution in this class). This actually has a rougher class of test-functions, as test functions here we take the Schwartz functions, that are the smooth functions of which the function itself and all its derivatives are rapidly decreasing. $f$ is said to be rapidly decreasing if there are constants $M_n$ such that $|f(x)| \leq M_N |x|^{-N}$ as $x \to \infty$ for $N = 1,2,3,\ldots$.
To begin defining the convolution we first define what the convolution of Schwartz-function is with a tempered distribution. Let $f$ be our tempered distribution, then we can show that the following definition actually makes sense: $$\langle \phi * f, \psi \rangle := \langle f, \tilde{\phi} * \psi \rangle$$ where $\tilde{\phi}(x) = \phi(-x)$. Note that the RHS is well-defined. Convolution is a nice thing, we can see that if we start with a tempered distribution and convolute it with a test function, the result will be smooth. Now, $L_1 * L_2$ is the unique distribution $L$ with the property that $L * \phi = L_1 * (L_2 * \phi)$. We can show that this is commutative.
Fine, now note that $\delta * \phi(x) = \phi(x - y)|_{y = 0} = \phi(x)$. So we see that $\delta * \delta = \delta$.
If you want me to comment on the dot product of distributions, you first would have to explain what you mean with that.
So far for this short digression on distributions.
EDIT: Okay, you want to compute $\delta^2$. Let $\phi_n$ be an approximation to the identity and let it converge to $\delta$ in the sense of distributions, but $\phi_n^2$ does not converge at all since the integral against a test function that does not vanish at the origin blows up as $n \to \infty$.
Here is a heuristic to suggest that it will be difficult to define the square of the delta function.
The Fourier transform has the property that it takes the convolution of two functions to the product of their Fourier transforms, and vice versa, ie, it takes the product of two functions to their convolution.
Remember that the Fourier transform of the delta function is the constant function($=1$). Now suppose that $\delta^2$ exists. Then its Fourier transform would be the convolution of two constant functions. Such a convolution would shoot to infinity at every point. Even the theory of distributions can't handle such kind of stuff. So how would you imagine the inverse Fourier transform of this? How would it make sense?
This answer is primarily to expand on this comment. From that comment and the following, it seems to me draks thinks of the delta as a function, and from the title, it seems the OP also does. Or at least, this was true at the time of posting the comment and the question respectively.
Eradicate this misconception from your minds at once, if it is there. The delta is not a function, although it is sometimes called "Delta function".
Let me give you a bit of background, a little timeline of my relationship with the Delta.
I first heard of it from my father, a Physics professor and physicist, who introduced it to me as a function equalling 0 outside 0 and infinity in 0. Such a function seemed abstruse to me, but I had other worries up my mind so I didn't bother investigating. This is how Dirac originally thought of the Delta when introducing it, but, as we shall see, this definition is useless because it doesn't yield the one most used identity involving this "function"; Then I had Measure theory, and voilà a Dirac Delta again, this time a measure, which gives a set measure 0 if 0 is not in it, and 1 if 0 is in. More precisely, $\delta_0$ is a measure on $\mathbb{R}$, and if $A\subseteq\mathbb{R}$, then $\delta_0(A)=0$ if $0\not\in A$, and 1 otherwise. Actually, I was introduced to uncountably many Deltas, one for each $x\in\mathbb{R}$. $\delta_x$, for $x\in\mathbb{R}$, was a measure on the real line, giving measure 0 to a set $A\subseteq\mathbb{R}$ with $x\not\in A$, and 1 to a set containing $x$; Then I had Physics 2 and Quantum Mechanics, and this Delta popped up as a function, and I was like, WTF! It's a measure, not a function! Both courses didsay it was a distribution, and not a function, so I was like, what in the world isa distribution? But both courses, when using it, always treated it like a function; Then I had Mathematical Physics, including a part of Distribution theory, and I finally was like, oh OK, thatis what a distribution is! The measure and the distribution are close relatives, since the distribution is nothing but the integral with respect to the measure of the function this distribution is given as an argument.
In both settings, it is
a priori meaningless to multiply two deltas. Well, one could make a product measure, but that would just be another delta on a Cartesian product, no need for special attention. In the distribution setting, we have what this answer says, which gives us an answer as to what the product might be defined as, and what problems we might run into.
So what is the product of deltas? And what is the comment's statement all about?
The answer to the first question is: there is
no product of deltas. Or rather, to multiply distributions you need convolutions, and those need some restrictions to be associative.
The second question can be answered as follows. That statement is a formal abbreviations. You will typically use that inside a double integral like: $$\int_{\mathbb{R}}f(\xi)\int_{\mathbb{R}}\delta(\xi-x)\delta(x-\eta)dxd\xi,$$ which with the formal statement reduces to $f(\eta)$. I have seen such integrals in Quantum Mechanics, IIRC. I remember some kind of spectral theorem for some kind of operators where there was a part of the spectrum, the discrete spectrum, which yielded an orthonormal system of eigenvectors, and the continuous spectrum somehow yielded deltas, but I will come back here to clarify after searching what I have of those lessons for details.
Edit:$\newcommand{\braket}[1]{\left|#1\right\rangle}\newcommand{\xbraket}[1]{|#1\rangle}$I have sifted a bit, and found the following:
Spectral theorem Given a self-adjoint operator $A$, the set of eigenvectors $\braket{n}$ of $A$ can be completed with a family of distributions $\braket{a}$, indicised by a continuous parameter $a$, which satisfy: \begin{align*} A\braket{n}={}&a_n\braket{n} && \braket{n}\in H, \\ A\braket{a}={}&a\braket{a} && \braket{a}\text{ distribution}, \end{align*} in such a way as to form a "generalized" basis of $H$, in the sense that
allthe vectors of $H$ can be written as an infinite linear combination: $$\braket{\psi}=\sum c_n\braket{n}+\int da\,c(a)\braket{a}.$$ The set of eigenvalues (proper and generalized) of $A$ is called the spectrumof $A$ and is a subset of $\mathbb{R}$.
What happens to the Parseval identity? Naturally: $$\langle\psi,\psi\rangle=\sum|c_n|^2+\int da\,|c(a)|^2.$$ So this "basis" is orthonormal in the sense that the eigenvectors are, the distributions have as product a family of deltas, or: $$\langle a,a'\rangle=\delta(a-a'),$$ and multiplying the eigenvectors by the distributions also yields a nice big 0.
The famous identity I mentioned in the timeline above and then forgot to expand upon is actually what
defines the delta, or at least what the QM teacher used to define it:$$\int_{\mathbb{R}}f(x)\delta(x-x_0)=f(x_0),$$for any function $f:\mathbb{R}\to\mathbb{R}$ and $x_0\in\mathbb{R}$. If the $\delta$ were a function, it would have to be zero outside 0, but I'm sure you know all too well that altering the value of a function in a single point doesn't alter the integral, and the integral in the identity above would be an integral of a function that is 0 save for a point, so it would be 0, and if $f(x_0)\neq0$ the identity wouldn't hold.
Notice how this formal statement is much like an analogous statement for Kronecker deltas: $$\sum_n\delta_{nm}\delta_{nl}=\delta_{ml}.$$ Imagine taking this to the continuum: the sums become integrals, and what can $\delta_{nm}$ become if not $\delta(n-m)$? So the statement is just a formal analog of the true statement with Kronecker Deltas when going into the continuum. Of course, distributionally it makes no sense, nor in terms of measure.
I have no idea how integrals with two deltas may be useful, and I have found none in my sifting. I will sift more, and perhaps Google, and if I find anything interesting, I'll be back.
Update:$\newcommand{\lbar}{\overline}\newcommand{\pa}[1]{\left(#1\right)}$I decided I'd just stop the sifting and concentrate on my exams. I googled though, and found this.
Another argument I thought up myself in favor of the statement is the following. Let $\phi$ be a functions. It is pretty natural to say:$$\phi=\int_{\mathbb{R}}\phi(a)\delta(x-a)da,$$since for any $x$ this yields $\phi(x)$. Now what happens to the $L^2$-norm?$$N:=\|\phi\|_{L^2}^2=\int_{\mathbb{R}}\lbar{\phi(x)}\phi(x)dx=\int_{\mathbb{R}}\lbar{\int_{\mathbb{R}}\phi(a')\delta(x-a')da'}\cdot\pa{\int_{\mathbb{R}}\phi(a)\delta(x-a)da}dx.$$The complex conjugation can be brought inside the first integral. Now to a physicist integrals that don't swap are
evil, and we surely don't want any evil around, so we assume we can reorder the three integrals the way we want, and get:$$N=\int_{\mathbb{R}}da\,\phi(a)\cdot\pa{\int_{\mathbb{R}}da'\,\lbar{\phi(a')}\cdot\pa{\int_{\mathbb{R}}dx\,\delta(x-a)\delta(x-a')}}.$$Suppose the formal statement holds. Then the innermost integral yields $\delta(a-a')$, and the second innermost one yields $\lbar{\phi(a)}$, which then combines with $\phi(a)$ outside it to form $|\phi(a)|^2$, which integrated gives the $L^2$ norm of $\phi$, squared. If the statement doesn't hold, it seems unreasonable to think we can still get the squared norm out of that mess. So the statement must hold, otherwise the integrals won't swap.
I think the answer provided by Jonas is correct, but Jonas assumed "product" meant convolution as this is normally the only binary operation that usually make sense with the Dirac delta function.
So if by product you mean convolution then: δ∗δ=δ
If by "product" you meant point-wise multiplication, then the answer is: Undefined.
The usual approach is to treat δ as the limit of some nascent delta function. See Delta Function for examples of these nascent delta functions. So if you multiply two nascent δ functions and then take the limit, the result will vary depending on the particular pair of nascent δ functions selected. In most cases the integral of the point-wise product of two nascent δ functions is zero as the limit is taken, in some cases the integral is 1, in other cases the value of the integral tends to infinity as the limit of nascent δ is taken.
For example If you multiply the rectangle nascent δ with the nascent δ that is a triangle pulse, then integral of the resulting cubic tends to zero as the interval tends to 0. If you multiply the sinc version of the nascent δ with itself, then the integral is equal to the frequency of the sinc function. In this case the integral tends to infinity as the frequency of the sinc function tends to infinity.
Because of this, the original answer given (The question makes no sense) is the best because the term "product" in its common meaning (e.g. point-wise multiplication of two functions) leads to contradictory results for the limit process. Using the term "product" when you mean convolution (the only operation that produces meaningful results) is at best confusing. |
This is an extension question of my previous post.
Say I have an equation like $y = 7 \cos(0.96(x-3)) + 11$.
How would I find the sine equivalent that lines exactly with it? I thought that $\sin$ and $\cos$ differ only by a phase shift of $\displaystyle -\frac{\pi}{2}$, when do I need to use reflections?
Thanks |
I am trying to prove the following bound using induction (where N>D):
$\sum_{i=0}^{D}\binom{N}{i} \leq N^{D} + 1$
I would appreciate any help on how to prove this. Thanks.
Currently I have looked at low values of D (base cases) but I don't understand how to formulate an induction step without using any prior information on the final result.
D=0: $\sum_{i=0}^{0}\binom{N}{i} \leq N^{0} + 1 \Leftrightarrow 1 \leq 2$
D=1: $\sum_{i=0}^{1}\binom{N}{i} \leq N^{1} + 1 \Leftrightarrow N + 1 \leq N + 1$
D=2: $\sum_{i=0}^{2}\binom{N}{i} \leq N^{2} + 1 \Leftrightarrow \frac{N^{2}}{2} + \frac{N}{2} + 1 \leq \frac{N^{2}}{2} + \frac{N^{2}}{2} + 1 \leq N^{2} + 1$
Maybe a first step in solving this is:
$\sum_{i=0}^{D}\binom{N}{i} = \sum_{i=0}^{D-1}\binom{N}{i} + \binom{N}{D}$ ?? |
All this is easier with a theory of causality. For example, let's use here the Structural Causal Models (which includes the Potential Outcomes) approach.
A Structural Causal Model (SCM) is triplet $M = \langle V, U, F\rangle$ where $U$ is a set of exogeneous variables, $V$ a set of endogenous variables and $F$ is a set of structural equations that determines the values of each endogenous variable. The structural equations are in the sense of
assignments not equalities. For example, consider the simple structural equation $Y = X^2$. This is meant to be read $Y\leftarrow X^2$, in the sense that if I experimentally set $X=2$ then this causally determines the value of $Y = 4$ but experimentally setting $Y = 4$ does nothing to $X$. The asymmetry is important/fundamental in causality: rain causes the floor to be wet, but making the floor wet does not cause rain.
So our causal model can be thought as functional relationships among variables and we are considering these relationships as autonomous. You can think of it as an idealized representation of the real world, where the variables $V$, the endogenous variables, are what we choose to model, and the variables $U$ are the aspects we chose to ignore. Since we chose not to model the $U$, what we usually do is to represent our ignorance about $U$ with a probability distribution $P(U)$ over the domain of $U$, giving us a probabilistic SCM which is pair $\langle M, P(U) \rangle$. Notice this means that causal relationships are ultimately functional relationships, therefore causal relationships may or may not translate to specific probabilistic dependencies. Finally, every causal model can be associated with a directed (acyclic) graph $G(M)$.
Hence, one way to simulate from a probabilistic causal model is by specifying: (i) the endogenous variables $V$ you are going to model; (ii) the exogenous variables $U$ which are usually the "disturbances", along with their joint probability distribution; and, (iii) the (causal) structural relationships among the variables. It might be easier to start this process qualitatively by first drawing the causal DAG with the main features that you want to illustrate and then add the details of the simulation (functional forms) later.
To see how this can be easily done in practice, let's simulate a simple causal model that illustrates simpson's paradox (for more see Pearl). Suppose our model $M$ is given by the following causal DAG, where the variables in parenthesis are "unobserved" and each variable has an associated exogenous disturbance $U$ which is omitted for convenience:
More specifically we will assume the following structural equations $F$:
$$\begin{aligned}W_1 &= U_{W_1}\\W_2 &= U_{W_2}\\Z &= W_1 + W_2 + U_{Z}\\X &= W_1 + U_{x}\\Y &= X+ 10W_2+ U_{y}\\\end{aligned}$$
Finally, assume all disturbances in $U$ are independent standard normal random variables. Now it's easy to simulate from our causal model. In
R for instance:
rm(list = ls())
set.seed(1)
n <- 1e5
w1 <- rnorm(n)
w2 <- rnorm(n)
z <- w1 + w2 + rnorm(n)
x <- w1 + rnorm(n)
y <- x + 10*w2 + rnorm(n)
This example is interesting because if you run the regression $Y \sim X$ you get $1$:
lm(y ~ x)
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
0.01036 1.00081
But if you further "control" for $Z$, which is a pre-treatment variable correlated with both $X$ and $Y$ --- and some people still erroneously would say it's a confounder --- you will actually see a sign reversal of the estimate and get $-1$:
lm(y ~ x + z)
Call:
lm(formula = y ~ x + z)
Coefficients:
(Intercept) x z
0.00845 -1.01127 4.00041
In this example, since we simulated the data, we know the true causal effect is $1$ which is captured by the first regression. But you can only know that if you know the true causal structure. There's nothing in the data itself that tells you which one is the correct answer. Hence if you simulate this and give to a researcher only the variables $x$, $y$ and $z$ he can't tell the right answer just from looking at the correlations. If you want further play/simulate causal models with multi-stage Simpson's paradox reversals, you can check it here.
Simulating correlations/dependencies
To simulate correlations/dependencies you can take a similar approach. You can simply create a causal model that gives you the correlations/dependencies you want (adding latent variables if needed), simulate as above and the resulting data will have the desired correlations/dependencies. To make it easier, you can start by drawing the causal DAG (bayesian network) and read from the graph if the desired conditional dependencies/independencies are implied by your model. After that you might think of specific functional forms to get other quantitative aspects that you want. Notice that several models with different causal interpretations can give you the same correlations. |
Electrochemistry Specific conductivity, Equivalent conductivity Specific conductance (K): (i) In C.G.S system: The conducting power of all the ions that are present in 1 cm 3of a solution is called specific conductance. (ii) In M.K.S: The conducting power of all the ions that are present in 1 m 3of a solution is called specific conductance. In case of conductance (G) volume is not restricted but in case of specific conductance (K) volume of solution is restricted to 1 cm 3 Effect of dilution: (i) Upon dilution number of ions increases therefore conductance increases (ii) Upon dilution number of ions present in unit volume decreases therefore specific conductance decreases Conductance depends upon following factors (a) Greater the number of ions higher the conductance. (b) When number of ions are same greater the charge of the ion higher is the conductance. (c) When number of ions and charge are equal smaller the size of the ion in aqueous solution greater is the conductivity. (d) Conductance order Cs +> Rb +> K +> Na +> Li +→ Big size of Li +ion in aqueous solution due to "solvation" Equivalent conductance (ΛThe conducting power of all the ions in 1 gram equivalent of electrolyte present in a solution of any volume is called equivalent conductance. equi): Relationship between Λ equi& K: Case(i): 1 GEq of an electrolyte is present in 1 cm 3of solution. ∴ Λ c= K Case (ii): 1 GEq of an electrolyte is present in V cm 3of solution. ∴ Λ equi= K × V Where V is volume of electrolytic solution in cm 3which contain 1 GEq of electrolyte Case (iii): Units for Λ c: ohm −1cm 2gram equi −1(C.G.S) ohm −1m 2gram equi −1(M.K.S) Effect of dilution:Up on dilution equivalent conductance increases (because here increase in volume is more than decrease in K) Relationship between equivalent conductance and normality: (i) In C.G.S system \wedge_{equi}=\frac{K\times 100}{N} Units: ohm −1cm 2gram equi −1 (ii) In M.K.S system \wedge_{equi}=\frac{K}{1000\times N}
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1.
Specific Conductivity (K): It is the reciprocal of specific resistance. \tt k=\frac{1}{\rho}=\frac{l}{R.a}=G\times\frac{l}{a}=G\times cell\ constant \left(G^\star\right) (\tt \frac{l}{a} = cell constant)
2. Equivalent Conductivity (∧
eq) \tt \wedge_{eq}=\frac{k\times1000}{N}
where, N = Normality. |
Given a collection of sets $S_i$ of disjoint subsets $sub_i$ of a set $X$, find a set $A$ of disjoint subsets $asub$ such that each one of these subsets is subset or equal to at most one subset in each of $S_i$, i.e., for
$S_1$ = $\{sub_11,sub_12,sub_13\}$, where $sub_11$, $sub_12$, $sub_13$ are disjoint,
$S_2$ = $\{sub_21,sub_22\}$, where $sub_21$, $sub_22$ are disjoint
Then find $A$ = $\{asub1,asub2,asub3\}$, where $asub1$, $asub2$ are disjoint
and
$asub1$ can be a subset of only one of $sub_11$ or $sub_12$ or $sub_13$ but not more than one of them. Same goes for $asub2$ and $asub3$.
Similarly
$asub1$ can be a subset of only of either $sub_21$ or $sub_22$ or but not both of them. Same goes for $asub2$ and $asub3$.
More formally,
$\forall sub_i \in S_i,$ and given a subset $asub \in A,$
If $Asub=\{sub_i\mid asub\subseteq sub_i\}$
$then$ $|Asub|=1$
$Example 1:$
$X = \{e1,e2,e3,e4,e5\}$
$S_1 = \{\{e1,e2\},\{e3,e4\},\{e5\}\}$
$S_2 = \{\{e1\},\{e2,e3,e4\},\{e5\}\}$
$S_3 = \{\{e1,e2,e3,e4\},\{e5\}\}$
$A= \{\{e1\},\{e2\},\{e3,e4\},\{e5\}\}$
$Example 2:$
$X = \{e1,e2,e3,e4,e5,e6,e7,e8,e9,e10,e11,e12\}$ $S_1 = \{\{e1,e2,e11,e12\},\{e3,e4,e5,e6\},\{e7,e8,e9,e10\}\}$ $S_2 = \{\{e1,e2,e3\},\{e4,e5,e6,e7,e8\},\{e9,e10,e11\},\{e12\}\}$
$A= \{\{e1,e2\},\{e3\},\{e4,e5,e6\},\{e7,e8\},\{e9,e10\},\{e11\},\{e12\}\}$
Preferably the algorithm for finding $A$ should take optimal time.
P.S. Forgive me if the mathematical description of the problem is not formal enough or if there are mistakes in it, I am not a mathematician. |
The famous formula of Heron connects the sides, $a,b,c$ of a triangle with the area, $A$ of the triangle, i.e.
$$ A = \sqrt{s(s-a)(s-b)(s-c)}$$
1995
The famous formula of Heron connects the sides, $a,b,c$ of a triangle with the area, $A$ of the triangle, i.e.
$$ A = \sqrt{s(s-a)(s-b)(s-c)}$$
It is impossible to overestimate our debt to the ancient Greeks in a wide range of subjects including mathematics.
School textbooks often give the impression that Mathematics has all been worked out. In fact, nothing could be further from the truth.
Peter Kanka, of James Cook Boys Technology High School, provided the following proof that $$\sin(18) = \frac{\sqrt{5} - 1}{4}.$$
Q.1. Let $n$ be a positive integer. If the polynomial
$$(x+1)(x+2)(x+3) \cdots (x+n)$$
is expanded (a) find the sum of all the coefficients; (b) find the sum of the coefficients of odd powers of $x$.
Q.949 (a) We have a collection of numbers, each of which is either zero or one. |
A particle moves along the x-axis so that at time t its position is given by $x(t) = t^3-6t^2+9t+11$ during what time intervals is the particle moving to the left? so I know that we need the velocity for that and we can get that after taking the derivative but I don't know what to do after that the velocity would than be $v(t) = 3t^2-12t+9$ how could I find the intervals
Fix $c\in\{0,1,\dots\}$, let $K\geq c$ be an integer, and define $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$.I believe I have numerically discovered that$$\sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n \quad \text{ as } K\to\infty$$but cannot ...
So, the whole discussion is about some polynomial $p(A)$, for $A$ an $n\times n$ matrix with entries in $\mathbf{C}$, and eigenvalues $\lambda_1,\ldots, \lambda_k$.
Anyways, part (a) is talking about proving that $p(\lambda_1),\ldots, p(\lambda_k)$ are eigenvalues of $p(A)$. That's basically routine computation. No problem there. The next bit is to compute the dimension of the eigenspaces $E(p(A), p(\lambda_i))$.
Seems like this bit follows from the same argument. An eigenvector for $A$ is an eigenvector for $p(A)$, so the rest seems to follow.
Finally, the last part is to find the characteristic polynomial of $p(A)$. I guess this means in terms of the characteristic polynomial of $A$.
Well, we do know what the eigenvalues are...
The so-called Spectral Mapping Theorem tells us that the eigenvalues of $p(A)$ are exactly the $p(\lambda_i)$.
Usually, by the time you start talking about complex numbers you consider the real numbers as a subset of them, since a and b are real in a + bi. But you could define it that way and call it a "standard form" like ax + by = c for linear equations :-) @Riker
"a + bi where a and b are integers" Complex numbers a + bi where a and b are integers are called Gaussian integers.
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Does anyone know if $T: V \to R^n$ is an inner product space isomorphism if $T(v) = (v)_S$, where $S$ is a basis for $V$? My book isn't saying so explicitly, but there was a theorem saying that an inner product isomorphism exists, and another theorem kind of suggesting that it should work.
@TobiasKildetoft Sorry, I meant that they should be equal (accidently sent this before writing my answer. Writing it now)
Isn't there this theorem saying that if $v,w \in V$ ($V$ being an inner product space), then $||v|| = ||(v)_S||$? (where the left norm is defined as the norm in $V$ and the right norm is the euclidean norm) I thought that this would somehow result from isomorphism
@AlessandroCodenotti Actually, such a $f$ in fact needs to be surjective. Take any $y \in Y$; the maximal ideal of $k[Y]$ corresponding to that is $(Y_1 - y_1, \cdots, Y_n - y_n)$. The ideal corresponding to the subvariety $f^{-1}(y) \subset X$ in $k[X]$ is then nothing but $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$. If this is empty, weak Nullstellensatz kicks in to say that there are $g_1, \cdots, g_n \in k[X]$ such that $\sum_i (f^* Y_i - y_i)g_i = 1$.
Well, better to say that $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$ is the trivial ideal I guess. Hmm, I'm stuck again
O(n) acts transitively on S^(n-1) with stabilizer at a point O(n-1)
For any transitive G action on a set X with stabilizer H, G/H $\cong$ X set theoretically. In this case, as the action is a smooth action by a Lie group, you can prove this set-theoretic bijection gives a diffeomorphism |
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 |
Mechanical Properties of Fluids Bernoullis principle and its applications An ideal fluid is that which incompressible, irrotational non viscous called ideal fluid. Bernoullis theorem states that the sum of the pressure energy kinetic energy and potential energy at any point in steady flow calculated per unit mass is constant. Bernoullies theorem represents law of conservation of energy. When flow is horizontal, height is same and hence sum of pressure head and velocity head is constant. When the speed of flow is zero every where Bernoullies theorem changes as P 1− P 2= ρg(h 2−h 1) The velocity of efflux of a liquid through an orifice is equal to that of the velocity acquired by a freely falling body from a height which is equal to that of the liquid level from orifice. Horizontal range is maximum when the orifice is at the middle. The plane motion of a spinning ball gets charged due to an effect called magnus effect. Atomiser, paint gun and bunsen burner work on the basis of Bernoullies theorem. Verturimeter is an ideal device of measuring rate of flow of liquid through pipe. Reynold number is the ratio of inertial force to viscous force. Blowing of roofs in heavy wind, Aeroplanes Aerodynamic lift magnus effect of spinning ball are applications of Bernoullies theorem. Bernoulli's principle and its applications View the Topic in this video From 0:40 To 22:35
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1.
Bernoulli's theorem: If an ideal fluid is flowing in streamlined flow, then total energy, i.e., sum of pressure energy, kinetic energy and potential energy per unit volume of the liquid remains constant at every cross-section of the tube. \tt \frac{p}{\rho g}+\frac{v^{2}}{2g}+h=constant
2.
Toricelli's theorem: Velocity of efflux, v = \sqrt{2gh}
3. Venturimeter is a device to measure the flow speed of incompressible fluid.
Rate of flow of liquid, v = \tt \sqrt{\frac{2\rho_{m} gh}{\rho}}\left[\left(\frac{A}{a}\right)^2-1\right]^{-1/2} |
Problem 270
Let
\[A=\begin{bmatrix} 4 & 1\\ 3& 2 \end{bmatrix}\] and consider the following subset $V$ of the 2-dimensional vector space $\R^2$. \[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\] (a) Prove that the subset $V$ is a subspace of $\R^2$.
Add to solve later
(b) Find a basis for $V$ and determine the dimension of $V$. |
Non-autonomous Schrödinger-Poisson system in $\mathbb{R}^{3}$
1.
School of Mathematics and Statistics, Shandong University of Technology Zibo 255049, China
2.
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, China
3.
Department of Applied Mathematics, National University of Kaohsiung, Kaohsiung 811, Taiwan
4.
School of Mathematical and Statistical Sciences, University of Texas-Rio Grande Valley, Edinburg, Texas 78539, USA
$\left\{ {\begin{array}{*{20}{l}} { - \Delta u + u + \lambda K\left( x \right)\phi u = a\left( x \right){{\left| u \right|}^{p - 2}}u}&{{\text{in }}{\mathbb{R}^3},} \\ { - \Delta \phi = K\left( x \right){u^2}}&{{\text{in }}{\mathbb{R}^3},} \end{array}} \right.$
$\lambda >0$
$2 < p \le 4$
$K\left( x\right) $
$a\left( x\right) $
$\mathbb{R}^{3}$
$% \lim_{\left\vert x\right\vert \rightarrow \infty }K\left( x\right) = K_{\infty }\geq 0$
$\lim_{\left\vert x\right\vert \rightarrow \infty }a\left( x\right) = a_{\infty }>0$
$\lambda$
$p$
$3.18 \approx \frac{{1 + \sqrt {73} }}{3} < P \le 4$ Keywords:Positive solution, Schrödinger-Poisson system, variational method, ground state, fibering maps, Sobolev embedding theorem, concentration-compactness principle. Mathematics Subject Classification:Primary: 35J20, 35J61, 35A01; Secondary: 35B40. Citation:Juntao Sun, Tsung-Fang Wu, Zhaosheng Feng. Non-autonomous Schrödinger-Poisson system in $\mathbb{R}^{3}$. Discrete & Continuous Dynamical Systems - A, 2018, 38 (4) : 1889-1933. doi: 10.3934/dcds.2018077
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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 |
The title of Sloane's A001037 is: Number of degree-$n$ irreducible polynomials over $GF(2)$; number of $n$-bead necklaces with beads of 2 colors when turning over is not allowed and with primitive period $n$; number of binary Lyndon words of length $n$.
The first few terms of the sequence are (for $n=1,2,...$ ) $2,1,2,3,6,9,...$
The formula for the sequence is $\frac{1}{n}\sum_{d|n}\mu(\frac{n}{d})\cdot 2^d$.
I am familiar with the derivation given by Wilf in Generatingfunctiontology on page 62. This derivation explains why the formula enumerates binary Lyndon words and equivalently the "$n$ bead necklaces" statement in the title.
I know the 2 irreducible polynomials of degree 1 are $x$ and $x+1$. The degree 2 polynomial is $x^2+x+1$. The degree 3 polynomials are $x^3+x^2+1$ and $x^3+x+1$. The degree 4 polynomials are $x^4+x+1$, $x^4+x^3+x^2+x+1$ and $x^4+x^3+1$.
The binary Lyndon words are: $a(1)=2=\#\{"0","1"\}$, $a(2)=1=\#\{"01"\}$, $a(3)=2=\#\{"001","011"\}$, $a(4)=3=\#\{"0001","0011","0111"\}$
I would like to know if there is an easy correspondence between these objects or if there is some explanation as to why the formula counts the irreducible polynomial over $GF(2)$. |
The example I'm trying to understand is:
$ \hat{S}_{x} \begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{pmatrix} = 1/2 \begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{pmatrix} $
My interpretation of this is that the vector shows you the probabilities of a particle being spin up or spin down if you square them.
And I've been told that $ \hat{S}_{x} $ gives you the spin as an eigenvalue, but how? Since its 50:50 of getting -1/2 and 1/2. $ \hat{S}_{x} $ has only given you one of them.
Is it that $ \hat{S}_{x} $ only measures the magnitude of spin in the x direction?This post imported from StackExchange Physics at 2014-03-22 16:56 (UCT), posted by SE-user 9k9 |
Below is my solution to a problem from a text book. I do not have confidence that my solution is right. I feel like I am missing something. Am I?
Thanks, Bob Problem: An airline finds that $5$ percent of the persons making reservations on a certain flight will not show up for that flight. If the airline sells $160$ seats tickets for a fight with only $155$ seats, what is the probability that a seat will be available for every person holding a reservation and planning on flying. Answer: First realize that we have a binomial distribution with $n = 160$, $p = 0.95$ and $q = 0.05$. We are going to approximate that with a normal distribution. \begin{eqnarray*} u &=& np = 160(0.95) = 152 \\ \sigma^2 &=& npq = 0.95(0.05)(160) =7.6 \\ \sigma &=& 2.75681 \\ \end{eqnarray*} Observe that $155$ is $1.08821$ standard deviations above the mean. We then run the following command in R: pnorm(1.08821) and got $0.8617488$. We conclude the probability that all the passengers will have seats is $0.8617488$. The book gets $0.8980$. We will now do the problem again using Yates's correction. This time we ask what is the probability that we have $155.5$ passengers or less. Now we are $1.26991$ standard deviations above the mean. We then run the following command in R: pnorm(1.26991) and got $0.0.89794$ which matches the answer in the book.
Below is my solution to a problem from a text book. I do not have confidence that my solution is right. I feel like I am missing something. Am I?
There is a precise answer, you do not need to estimate. Note that using normal distribution instead of the binomial one is an approximation which is very accurate under some conditions, of course, but still, not better that the precise answer. Although the problem did not state this, I guess we may assume that customers decide to show up or not independently from each other. This is not realistic: if I book a flight with my wife for our honeymoon, then we both show up or we both don't, normally. Anyway, without the independence assumption, we have nothing.
So as you correctly put, $p=0.95$, $n=160$, and then the number $X$ of people showing up has distribution $X\sim Binom(160, 0.95)$.
The question is $P(n\leq 155)$. To simplify the calculation, we should compute the complement probability: $P(n> 155)= P(n=156) + P(n=157) + P(n=158) + P(n=159) + P(n=160)= \binom{160}{156}\cdot 0.95^{156}\cdot 0.05^4 + \binom{160}{157}\cdot 0.95^{157}\cdot 0.05^3 + \binom{160}{158}\cdot 0.95^{158}\cdot 0.05^2 + \binom{160}{159}\cdot 0.95^{159}\cdot 0.05^1 + \binom{160}{160}\cdot 0.95^{160}\cdot 0.05^0$.
Let's try this using standard notation: The number of people who show for the flight is $X \sim \mathsf{Binom}(n = 160, p = .95).$
Thus, on average, the number who show is $$\mu = E(X) = np = 160(0.95) = 152.$$ So if the average number always show up, there will always be room for everyone.
However, the number actually showing up for any one flight is variable: specifically, $\sigma^2 = Var(X) = np(1-p) = 7.68$ and $$\sigma = SD(X) = \sqrt{np(1-p)} = 2.75681.$$ Roughly speaking, it isn't unlikely for a binomial random variable to be as much as $2\sigma$ on either side of the mean. So the number of people actually showing for a particular flight might be between 146 and 175, which means sometimes 155 seats won't be enough to accommodate everyone who shows.
The exact probability that everyone showing gets a seat is $P(X \le 155) = 0.9061.$ This probability can be computed exactly: using the PDF formula for the binomial distribution (as shown by @APongracz), or using software (as commented by @JMoravitz)--either a statistical calculator or software program such as R. (The result from R is shown below.)
pbinom(155, 160, .95)## 0.9061461
The answer can also be approximated using the normal distribution $\mathsf{Norm}(\mu = 152, \sigma = 2.7568).$ This can be done directly in software or by standardizing and using printed normal tables. (Roughly as in the deleted answer; using normal tables may require some rounding and thus give a slightly different answer.)
pnorm(155.5, 152, 2.75681)## 0.8978835
By any of the methods, the probability of being able to accommodate everyone who shows is about 0.90.$ (On about 10% of these flights there might be announcements asking if some people with reservations are willing to take the next flight in exchange for some sort of 'bribe'.)
In the figure below, the exact binomial probability is the sum of the heights of the vertical bars to the left of the vertical broken line, and the normal approximation is the area under the normal curve to the left of that line. The normal curve does not exactly match the binomial bars, so you can only expect about two places of accuracy from the normal approximation. |
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 $\Fun(S, V)$. |
Binary Dihedral Groups¶
AUTHORS:
Travis Scrimshaw (2016-02): initial version class
sage.groups.matrix_gps.binary_dihedral.
BinaryDihedralGroup(
n)¶
The binary dihedral group \(BD_n\) of order \(4n\).
Let \(n\) be a positive integer. The binary dihedral group \(BD_n\) is a finite group of order \(4n\), and can be considered as the matrix group generated by\[\begin{split}g_1 = \begin{pmatrix} \zeta_{2n} & 0 \\ 0 & \zeta_{2n}^{-1} \end{pmatrix}, \qquad\qquad g_2 = \begin{pmatrix} 0 & \zeta_4 \\ \zeta_4 & 0 \end{pmatrix},\end{split}\]
where \(\zeta_k = e^{2\pi i / k}\) is the primitive \(k\)-th root of unity. Furthermore, \(BD_n\) admits the following presentation (note that there is a typo in [Sun2010]):\[BD_n = \langle x, y, z | x^2 = y^2 = z^n = x y z \rangle.\]
(The \(x\), \(y\) and \(z\) in this presentations correspond to the \(g_2\), \(g_2 g_1^{-1}\) and \(g_1\) in the matrix group avatar.)
REFERENCES:
cardinality()¶
Return the order of
self, which is \(4n\).
EXAMPLES:
sage: G = groups.matrix.BinaryDihedral(3) sage: G.order() 12
order()¶
Return the order of
self, which is \(4n\).
EXAMPLES:
sage: G = groups.matrix.BinaryDihedral(3) sage: G.order() 12 |
The article "Construction of a selfadjoint, strictly positive transfer matrix for euclidean lattice gauge theories" (Lüscher 1977), about lattice QCD, says the following:
> The fermion Hilbert space $\mathscr{H}_F$ is the Fock space built from an operator spinor field $\hat{\chi}_n$ which satisfies the usual canonical anticommutation relations:
> $$\{\hat{\chi}_{n,\alpha},\hat{\chi}^\dagger_{m,\beta} \}=\delta_{n,m} \delta_{\alpha,\beta}\qquad[...]\tag{11}$$ > The field operator $\hat{\psi}$ acts in $\mathscr{H}_F \otimes \mathscr{H}_G$.
(This is the hilbert space of fermions tensor product the hilbert space of the gauge fields.)
> It does _not_ have a canonical anticommutator, but:
> $$\{\hat{\psi}_{n,\alpha},\hat{\psi}^\dagger_{m,\beta} \}= B^{-1}_{n\alpha,m\beta}\qquad [...]\tag{12}$$ > The matrix $B_{n\alpha,m\beta}$ depends on the gauge field and is given by > $$B_{n\alpha,m\beta} = \delta_{n,m}\delta_{\alpha,\beta} - K \sum_{j=1,2,3} U(n,j)_{\alpha,\beta}\delta_{n+\hat{j},m}+U(m,j)^{\dagger}_{\alpha,\beta}\delta_{m+\hat{j},n}\tag{13}$$
Then the article says in equation (14) that the relation between the field $\psi$ and the canonical field $\chi$ is
$$\psi_{n\alpha} = \sum_{m\beta} (B^{-1/2})_{n\alpha,m\beta} \chi_{m,\beta}$$ where $n$ and $m$ are the lattice sites, $U$ are the links variables (the gauge fields), $K$ is the Wilson hopping parameter, and the action is supposed to be the improved Wilson action: $$S_F= \sum_{n} \biggl\{\bar{\psi}(n)\psi(n)-K \sum_{j=1,2,3}\bar{\psi}(n) U(n,j)(1-\gamma_j)\psi(n+\hat{j})+\bar{\psi}(n+\hat{j})U(n,j)^{\dagger}(1+\gamma_j)\psi(n) \biggr\}$$
How can these equations be justified? I have tried for days (without any success) to come up with an explanation nor I have found any demonstration from other sources for these assertions.
Especially on why the anticommutator of the fields is $B^{-1}$. |
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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 ... |
In fluid kinematics I can't understand the meaning of these terms : vorticity and circulation.
Can somebody give me a description of these terms so that a lay person can understand them easily?
In fluid kinematics I can't understand the meaning of these terms : vorticity and circulation.
There are fundamental types of motion (or deformation) for a fluid element: translation, rotation, linear strain and shear strain. Usually all these types of motion occur concurrently which makes the analysis of fluid dynamics somehow difficult.
One can express the rate of translation vector mathematically by the velocity vector $\overrightarrow{V}$: $$\overrightarrow{V} = u\overrightarrow{i} + v\overrightarrow{j} + w\overrightarrow{k}$$
When it comes to express the rate of rotation of a fluid element it becomes quite challenging, Why? because a fluid element translates and deforms as it rotates, imagine an initially rectangular fluid element that starts to rotate while each line of the rectangular having a different angular velocity than the other. You can check White's book for the complete derivation but we can express the rotation vector $\overrightarrow{\omega}$ for now as follows:
$$\overrightarrow{\omega} = \frac{1}{2} (\frac{\partial w}{\partial y} - \frac{\partial v}{\partial z})\overrightarrow{i} + \frac{1}{2} (\frac{\partial u}{\partial z} - \frac{\partial w}{\partial x})\overrightarrow{j} + \frac{1}{2} (\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y})\overrightarrow{k}$$
Putting it simply as half the curl of the velocity vector (just a mathematical manipulation, no fear): $$\overrightarrow{\omega} = \frac{1}{2}\overrightarrow{\nabla}\times \overrightarrow{V}$$
Now, let us define a vector which is called
Vorticity vector which is twice the angular velocity (we just got rid of the silly $\frac{1}{2}$) and we might call it $\xi $:$$\overrightarrow{\xi} = \overrightarrow{\nabla}\times \overrightarrow{V}$$
Okay, enough with the math. What does it mean?
For an arbitary point in a flow field:
Any fluid element (particle) that occupy that point having a non-zero vorticity, that point is called rotational. Vice versa, Any fluid element (particle) that occupy that point having a zero vorticity, that point is called irrotationalwhich means particle is not rotating.
Flow from A to B is rotational (has voriticity) while flow from A to C is irrotational (has no vorticity).
You can find many examples for rotational flows such as in wake regions behind blunt bodies and flow through turbomachines.
According to this lecture:
• Circulation and vorticity are the two primary measures of rotation in a fluid.
• Circulation, which is a scalar integral quantity, is a macroscopic measure of rotation for a finite area of the fluid.
• Vorticity, however, is a vector field that gives a microscopic measure of the rotation at any point in the fluid. |
Let $(K,w)$ be a henselian field such that the residue field $k=k(w)$ is an algebraic extension of a finite field $\mathbb{F}_p$. Let $\ell\neq p$ be a prime with $\mu_\ell\subset K$.
Now, in the paper I'm reading, it states:
Let $G_\ell$ be a pro-$\ell$ Sylow group of $G_K$. By the ramification theory for general valuations (see e.g.
O. Endler, Valuation Theory, Springer, 1972, §20) we have $G_\ell\cong \mathbb{Z}_\ell(1)^r \rtimes G_k(\ell)$ where the action is defined via the cyclotomic character with $r=\dim_{\mathbb{F}_\ell}(\Gamma_w/\ell\Gamma_w)$ and $G_k(\ell)$ being the pro-$\ell$ Sylow group of $G_k$ hence either $\cong \mathbb{Z}_\ell$ or $\cong 0$.
I've looked into "Valuation Theory", §20 by O. Endler and other books on the subject ramification theory, but the only thing I've found is how the ramification group is the pro-$p$ Sylow group of the inertia group, which, of course, is something entirely different, right?
Anyway, I'm wondering why this assertion holds and if there are any better references?
Update: I've found the same assertion in the paper On Grothendieck's Conjecture of Birational Anabelian Geometry by Florian Pop, Ann. Math. 138 (1994), p.155, 1.6 (2). But there's no reference. |
$R$ is finite commutative ring without zero divisors which has at least two elements. How to show that $R$ is field?
I'm just starting with abstract algebra and I'd really appreciate if someone could explain it to me.
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$R$ is finite commutative ring without zero divisors which has at least two elements. How to show that $R$ is field?
I'm just starting with abstract algebra and I'd really appreciate if someone could explain it to me.
It follows from the pigeon-hole principle.
First, we show that $R$ has an identity. (Sometimes the existence of an identity is included in the definition of a ring, but you do not need it here.)
For each $0 \ne a \in R$, consider the map $$ \phi_{a} : R \to R, \qquad x \mapsto a x. $$ Because $a$ is not a zero-divisor, the map $\phi_{a}$ is injective, thus surjective. So there is $e \in R$ such that $$a = \phi_{a}(e) = e a.\tag{1}$$ Again because the map $\phi_{a}$ is surjective, every $b \in R$ is of the form $$b = \phi_{a}(x_{b}) = a x_{b}$$ for some $x_{b}$, so multiplying (1) by $x_{b}$ you find $$ b = e b $$ for all $b \in R$, so $e$ is the identity.
Now use once more the fact that $\phi_{a}$ is surjective, to show that there is $c \in R$ such that $$a c = \phi_{a}(c) = e,$$ so $c$ is an inverse of $a$, $a$ being an arbitrary non-zero element.
This is a classic exercise. It's more precisely stated as a finite integral domain with more than one element is a field.
Here's a hint. Let $a \in R$ be non-zero and consider the map $\phi: R \rightarrow R$ given by $\phi(r)=ra$. What can you say about $R$? Is it injective? What's in the image?
Another possibility: Let $x\in R$, $x\neq 0$. Since $R$ is finite, there must be some repetition in the infinite list $x^0, x^1, x^2, x^3, x^4,\ldots$. Hence there are exponents $i < j$ with $x^i = x^j$. So $x^i(x^{j-i} - 1) = 0$. Since $R$ has no zero divisors, we get $x^{j-i} - 1 = 0$ ($x^i \neq 0$ since $x\neq 0$ and $R$ has no zero divisors). So $x\cdot x^{j-i-1} = 1$, showing that $x$ is invertible. |
We know that given a multiplicative function $f$ for which the series $\sum_{n=1}^\infty f(n)$ converges
absolutely then so does the Euler product $\prod_{p}\sum_{k=0}^\infty f(p^k)$, but does the reverse hold (at least up to conditional convergence)?
We know that given a multiplicative function $f$ for which the series $\sum_{n=1}^\infty f(n)$ converges
Suppose $f$ is multiplicative and $\prod_p \sum_k |f(p^k)|$ converges, i.e. there is $L$ such that for every $\epsilon > 0$ there exist $K,P$ such that $\left|L - \prod_{p \le P'} \sum_{k \le K'} |f(p^k)|\right| < \epsilon$ whenever $K' > K$ and $P' > P$. Now $$\prod_{p \le P_1} \sum_{k \le K_1} |f(p^k)| \le \sum_{n \le N} |f(n)| \le \prod_{p \le P_2} \sum_{k \le K_2} |f(p^k)|$$ where $P_1$ and $K_1$ are such that all positive integers $\prod_{p \le P_1} p^{k(p)}$ with all $k(p) < K_1$ are at most $N$, while $P_2$ and $K_2$ are such that all $n \le N$ are of the form $\prod_{p \le P_2} p^{k(p)}$ with all $k(p) \le K_2$. We conclude that the sum converges absolutely. |
Why Casey Mulligan is wrong about everything Matthew Martin8/15/2013 12:11:00 PM
Tweetable
I won't dispute Mulligan's claims over the existence of these "taxes" and their incentive effects (though someone should). Suppose he's right about that, here's a simpleton argument why he's wrong about everything else:
Consider a standard microeconomic model. Household utility is thus: $$U=ln\left(c\right)+ln\left(1-L\right)$$ with a budget constraint of $$c\leq \left(1-\tau\right)wL+\theta$$ Maximizing utility, the household chooses labor supply $L$ according to $$L=\frac{1}{2}-\frac{\theta}{2\left(1-\tau\right)w}$$ Note that the expression $\left(1-\tau\right)w$ is just a way of writing the worker's after-tax wage rate (lets call the after tax wage $w_{at}\equiv \left(1-\tau\right)w$ , and that the worker really doesn't care at all about the tax rate per se. That is, the worker is indifferent between being paid $w$ at a tax rate of $\tau$ versus being paid $w_{at}$ with no taxes. Furthermore, note that the after-tax wage nearly falls out of this equation--when $\theta=0$, taxes have no effects on the employment supply decision (ok, implicit assumptions of the maximization procedure required an interior point--at the boundary point of a 100 percent tax rate, this labor supply equation does not hold).
What is $\theta$? It refers to the profits the household gets from ownership of firms. To the extent that firms are perfectly competitive, $\theta=0$. The point is that $\theta$ is generally small--most people don't have huge streams of income other than their wages.
Why do I bring up such a simplistic model when you could easily reverse the result with a more complicated one? Because it appears to be basically right:
The result holds.
My point is this: analytically, lower effective marginal taxes is identical to higher wage rates, so rising wages due to productivity growth should have identical effects to cutting taxes. So why hasn't the labor supply exploded?
Economists often fall victim of incentive-worship, whereby an economist considers only the effect of prices on substitution, and not the effect of prices on wealth. Mulligan's whole argument presumes that the substitution effect--whereby lower after-tax wages decrease labor supply--is always much larger than the wealth effect--whereby lower after-tax wages induce people to work more to make up the difference. In reality, what we see is that for the most part the two effects balance out, so that doubling wages had very little effect on labor supply--in fact it decreased.
So lets revisit Mulligan's hypothesis. Let's suppose that the ACA imposes massive new hidden taxes, amounting to 50 percent of your disposable income. That implies that the average weekly hours will be at 1960s levels--5 hours a week more than we are currently working!
Thus, for the most part, substitution effect=wealth effect. This has to be a fairly robust result, because labor supply is always bounded--there are only 24 hours in a day. And, we know of a utility function of which this is true. It's called cobb-douglas. Perhaps Casey Mulligan has heard of it? |
First note that in the special case $\kappa = \omega_1$, countably closed implies countably directed-closed.
Answer 1: No. In any model of CH, there is a countably closed, $\omega_2$-c.c. forcing that is inequivalent to $\mathbb P = \text{Add}(\omega_1,\omega_2)$.
Proof: Let $\mathbb Q$ be Jensen's partial order for adding a Kurepa tree. Conditions are $(t,b)$, where $t$ is a countable tree of successor height $\alpha$, $b$ is a countable subset of $\mathcal P(\omega_1)$, and for all $x \in b$, $x \cap \alpha \subseteq t$. Ordering is $(t_1,b_1) \leq (t_0,b_0)$ when $t_1$ end-extends $t_0$ and $b_0 \subseteq b_1$. It is not hard to show that $\mathbb Q$ is countably closed and $\omega_2$-c.c. under CH. Any two conditions with the same first coordinate are compatible. The union of the first coordinates in any generic is a Kurepa tree $T$ by a density argument.
Note that every subset of $\omega_1$ added by $\mathbb P$ is added by some $\omega_1$-sized regular suborder. If $\mathbb Q$ were forcing equivalent to $\mathbb P$, then there would be some $\omega_1$-sized regular subalgebra $\mathbb R$ of $ro(\mathbb Q)$ that adds the generic Kurepa tree $T$. $\mathbb Q$ forces an $\omega_2$-size subset of $\mathcal P(\omega_1)^V$ to be contained in the set of branches of $T$. Thus suppose $G * H$ is $\mathbb R * \mathbb Q / \mathbb R$-generic. But in $V[G]$ we can compute whether $(p,q) \in H$, since we just determine whether $p \subseteq T$ and whether all $x \in q$ are branches of $T$. Thus $V[G * H] = V[G]$, and $\mathbb Q$ must have a dense set of size $\omega_1$. But this is false: If $\{ (t_i,b_i) : i < \omega_1 \} \subseteq \mathbb Q$, then pick $x \notin \bigcup b_i$. No $(t_i,b_i)$ can force $x$ to be a branch of $T$ since we can extend $t$ to rule out $x$ as a branch.
Note that $\mathbb Q$ doesn't have infima to countable chains since we have a lot of choices for the "top level" of the countable tree. But...
Answer 2: Not necessarily, even for $\omega_1$-closed with infima. Let $V$ be a model of GCH, and let $G \subseteq \text{Col}(\omega_1,\omega_2)$ be generic. $\omega_3^V = \omega_2^{V[G]}$. I claim that in $V[G]$, $\mathbb P = \text{Col}(\omega_1,<\omega_2)^{V[G]}$ and $\mathbb Q = \text{Col}(\omega_2,<\omega_3)^{V}$ are both countably closed with infima and both $\omega_2$-c.c., but are not forcing equivalent.
The closure follows from the fact that $V$ and $V[G]$ have the same countable sequences. The $\omega_2$-c.c. for $\mathbb Q$ follows from the fact that $\mathbb Q \times \text{Col}(\omega_1,\omega_2)$ is $\omega_3$-c.c. in $V$.
Next, one can prove a similar fact to Mohammad's answer. In $V[G]$,
(1) $\Vdash_{\mathbb P} \exists A \in [\omega_2]^{\omega_2}$ such that $A$ does not contain any $\omega_1$-sized set from $V[G]$.
(2) $\Vdash_{\mathbb Q} \exists A \in [\omega_2]^{\omega_2}$ such that for all $X \in [\omega_2]^{\omega_2} \cap V[G]$, there is $y \in [X]^{\omega_1} \cap V[G]$ such that $y \cap A = \emptyset$.
The justification for
(1) is the same as for Mohammad's answer. To see (2), let $\dot A$ be a name for a code of the generic $H$ for $\mathbb Q$. In $V$, let $\dot B$ be a name for a code for $G \times H$. If $X \in ([\omega_2]^{\omega_2})^{V[G]}$, let $\dot X$ be a name for it in $V$. Let $p \in \text{Col}(\omega_1,\omega_2)$ be arbitrary, and let $\{ p_i : i < \omega_3 \} \subseteq \text{Col}(\omega_1,\omega_2) \restriction p$ be such that $p_i$ decides the $i^{th}$ member of $\dot X$. Let $p'$ be such that $p_i = p'$ for $\omega_3$ many $i$, giving a large set $Y$ such that $p' \Vdash \check Y \subseteq \dot X$. Let $q \in \mathbb Q$ be arbitrary. Interpreting $Y$ as a subset of $\mathbb Q$, find $Y' \subseteq Y$ of size $\omega_3$ such that $q \nleq y$ for all $y \in Y'$. For each such $y$, choose $q_y \leq q$ such that $q_y \perp y$. Now form a $\Delta$-system among the $q_y$, and let $z$ be an infimum of a set $S$ of $\omega_1$ many of them. Then $z$ forces that the generic $H$ misses $y$ for $q_y \in S$. As $p$ and $q$ were arbitrary, we are done.
Likewise, we can show that $\mathbb P$ forces the negation of the the property under the forcing sign in
(2), and vice versa. Some details are in section 2.1.2 of my thesis. I may post more later. |
What is the difference between the three terms below?
percentile quantile quartile
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0 quartile = 0 quantile = 0 percentile
1 quartile = 0.25 quantile = 25 percentile
2 quartile = .5 quantile = 50 percentile (median)
3 quartile = .75 quantile = 75 percentile
4 quartile = 1 quantile = 100 percentile
Percentiles go from $0$ to $100$.
Quartiles go from $1$ to $4$ (or $0$ to $4$).
Quantiles can go from anything to anything.
Percentiles and quartiles are examples of quantiles.
In order to define these terms rigorously,it is helpful to first define the
quantile functionwhich is also known as the inverse cumulative distribution function.Recall that for a random variable $X$,the cumulative distribution function $F_X$ is defined by the equation$$F_X(x) := \Pr(X \le x).$$The quantile function is defined by the equation$$Q(p)\,=\,\inf\left\{ x\in \mathbb{R} : p \le F(x) \right\}.$$
Now that we have got these definitions out of the way, we can define the terms:
percentile: a measure used in statistics indicatingthe value below which a given percentage of observationsin a group of observations fall.
Example: the 20th percentile of $X$ is the value $Q_X(0.20)$
quantile: values taken from regular intervalsof the quantile function of a random variable.For instance, for some integer $k \geq 2$,the $k$-quartiles are defined as the valuesi.e. $Q_X(j/k)$ for $j = 1, 2, \ldots, k - 1$.
Example: the 5-quantiles of $X$ are the values $Q_X(0.2), Q_X(0.4), Q_X(0.6), Q_X(0.8)$
It may be helpful for you to work out an example of what these definitions mean when say $X \sim U[0,100]$, i.e. $X$ is uniformly distributed from 0 to 100.
References from Wikipedia:
From wiki page: https://en.wikipedia.org/wiki/Quantile
Some q-quantiles have special names:
The only 2-quantile is called the median The 3-quantiles are called tertiles or terciles → T The 4-quantiles are called quartiles → Q The 5-quantiles are called quintiles → QU The 6-quantiles are called sextiles → S The 8-quantiles are called octiles → O (as added by @NickCox - now on wiki page also) The 10-quantiles are called deciles → D The 12-quantiles are called duodeciles → Dd The 20-quantiles are called vigintiles → V The 100-quantiles are called percentiles → P The 1000-quantiles are called permilles → Pr
The difference between
quantile,
quartile and
percentile becomes obvious.
Percentile : The percent of population which lies below that value Quantile : The cut points dividing the range of probability distribution into continuous intervals with equal probability There are q-1 of q quantiles one of each k satisfying 0 < k < q Quartile : Quartile is a special case of quantile, quartiles cut the data set into four equal parts i.e. q=4 for quantiles so we have First quartile Q1, second quartile Q2(Median) and third quartile Q3
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A slightly magic proof.
Lemma: If $a,b$ are relatively prime positive integers, then for any integer $n$ there is solution to $ax+by=n$ with $x,y$ integers and $0\leq x\leq b-1.$ (So, possibly with $y<0$.)
Proof: We know we can find integers $x_0,y_0$ so that $ax_0+by_0=n.$
By the division algorithm, we can find integers $q,r$ so that $x_0=bq+r$ with $0\leq r\leq b-1.$ Then:
$$ar + b(y_0+aq)=a(r+bq)+by=ax_0+by_0=n$$
So there is a solution $(x,y)=(r,y_0+aq)$ with $0\leq r=x\leq b-1.$
Theorem: If $a,b$ are relatively prime positive integers, then for each integer $n$, there is a non-negative integer solution to at least on of the equations $ax+by=n$ or to $ax+by=ab-a-b-n.$
(There is a stronger statement, which is that for each $n$, exactly one of those two equations has a solution.)
In your case, this would mean for each $n$, one of $3x+5y=n$ and $3x+5y=7-n$ has a non-negative solution.
Proof: Solve $ax_0+by_0=n$ with integers $x_0,y_0$ and $0\leq x_0\leq b-1.$
If $y_0\geq 0$ we are done.
If $y_0<0$ then:
$$ab-a-b-n = a(b-1)-a-b-(ax_0+by_0)=a(b-1-x_0)+b(-1-y_0)$$
Since $0\leq b-1-x_0$ and $0\leq -1-y_0$ we either have non-negative solutions ot $ax+by=n$ or to $ax+b=ab-a-b-n.$
Now, if $N>ab-a-b$, or $N\geq (a-1)(b-1)$, then there there can't be non-negative solutions to $ax+by=ab-a-b-N<0,$ so there must be a solution to $ax+by=N.$
The stronger version of the theorem is that, for any integer $n,$
exactly one of the equations $$\begin{align}ax+by&=n\\ax+by&=ab-a-b-n\end{align}$$
has a non-negative solution.
You can show this by showing that $ax+by=ab-a-b$ has no non-negative integer solution $(x,y)$. This is because if both the equations have non-negative solutions, you could add them to get a non-negative solution for $ax+by=ab-a-b.$
If $ax+by=ab-a-b$ then $a(x+1)+b(y+1)=ab.$ So $b\mid a(x+1)$ and $a\mid b(y+1).$ But since $a,b$ are relatively prime, we get that $b\mid x+1$ and $a\mid y+1.$
But if $x,y\geq 0,$ $x\geq b-1$ and $y\geq a-1$, and thus $ax+by\geq 2ab-a-b>ab-a-b,$ reaching a contradiction.
This lets us count the number of non-negative integers $n$ for which there is no solution:
$$\frac{(a-1)(b-1)}{2}$$
A generating function approach is to ask which coefficients of:
$$f(x)=\frac{1}{1-x^a}\frac{1}{1-x^b}$$
are $0$. We can write:
$$f(x)=\frac{\left(1+x^a+\cdots+x^{a(b-1)}\right)\left(1+x^b+\cdots+x^{b(a-1)}\right)}{\left(1-x^{ab}\right)^2}$$
The denominator contributes only additional multiples of $ab$ to the exponents, and since we know that all sufficiently large values have a solution, this means that the exponents in the numerator must cover all the values modulo $ab$, which means that they must all have coefficient $1$. If $S$ is the set of positive values $n$ for which no $ax+by=n$, we can write the numerator as:
$$\sum_{k=0}^{ab-1} x^k -\left(1-x^{ab}\right)\sum_{k\in S} x^k$$
So, one way to enumerate the numbers in $S$ is:
$$S=\{ax+by-ab\mid 1\leq x\leq b-1, 1\leq y\leq a-1, ax+by\geq ab\}$$
We can see exactly half the pairs $x,y$ have $ax+by>ab$ beacuse if $ax+by>ab$ then $a(b-x)+b(a-y)=2ab-(ax+by)<ab$. (Not that there is no $x,y$ with $ax+by=ab$ since $x\leq b-1,y\leq a-1$.) So again we get that there must be $\frac{(a-1)(b-1)}{2}$ elements of $S$.
We can count $S$ another way and get the result: $$\sum_{x=1}^{b-1}\left\lfloor \frac{ax}{b}\right\rfloor=\frac{(a-1)(b-1)}{2}.$$
Now, our generating function is:$$f(x)=\frac{\frac{x^{ab}-1}{x-1}+\left(x^{ab}-1\right)\sum_{k\in S} x^k}{(1-x^{ab})^2}=\frac{1}{(x-1)(x^{ab}-1)}-\frac{\sum_{k\in S} x^k}{1-x^{ab}}$$
The coefficient of $x^n$ in the left part is $1+\left\lfloor \frac{n}{ab}\right\rfloor$ the coefficient of $x^n$ in the right term is $1$ if $n$ is congruent to some $s\in S$ modulo $ab$. So the number of non-negative solutions to $ax+by=n$ is:
$$\left\lfloor \frac{n}{ab}\right\rfloor+\begin{cases}0&n\equiv s\pmod{ab} \text{ for some } s\in S\\1&\text{otherwise}\end{cases}$$
The case when $b=2a-1$ gives us an easy way to list $S$:
$$S=\{i+aj\mid 1\leq i\leq a-1, 0\leq j\leq 2(a-i-1)\}$$
This gives, for $a=3,b=5$ that $S=\{1,4,7,2\}.$
For $a=5,b=9$ this gives $S=\{1,6,11,16,21,26,31,2,7,12,17,22,3,8,13,4\}.$
More generally, if $a<b$, we can list, for each $i\in\{1,\dots,a-1\}$ the values $i+aj$ until we get one divisible by $b$.
For example, $a=3,b=10$ gives:
$$\{1,1+3,1+6,2,2+3,2+6,2+9,2+12,2+15\}$$
If $a<b$ and $ax\equiv -1\pmod {b}$ for $1\leq x\leq b-1$, then you get:
$$S=\{i+bj\mid 1\leq i\leq a-1, 0\leq j<(ix\bmod b)\}$$
If $b=ax+1$ then $0\leq j< xi$. For example, when $a=4,b=13,x=3,$ you get:$$S=\{1,1+4,1+4\cdot 2,\\2,2+4,\dots,2+4\cdot 5,\\3,3+4,\dots,3+4\cdot 8\}$$
If $b=ay-1$ for $y>1,$ then $x = b-y$ and for $i$ you get $0\leq j< b-iy.$
This formula also lets you count $S$ again, since the number of elements for $i$ is $ix\bmod b$ and the number of elements for $a-i$ is $(a-i)x\bmod b$, and $xi+(a-i)x=ax\equiv -1\pmod b$. Since $(xi\bmod b)+((a-i)x\bmod b)\leq 2b-2$, we must have the total for these $i,a-i$ to be $b-1$, so the average over all $i$ must be $\frac{b-1}{2}$, and we get, again, that $|S|=(a-1)\frac{(b-1)}{2}.$ |
I am trying to find
$$\int_{0}^{\infty} {\frac{\ln(x)}{(x+1)^3}}dx$$
Using the residue formula.
I am mainly having a hard time finding a contour that works, since we must include the pole of order three at $x=-1$.
I have tried a partial, indented circle that goes from $\theta = 0$ to $4\pi/3$ and a $3/4$ circle as well, but in both cases, if $\gamma_3(t) = te^{i\theta}, t \in (R,0)$ for a fixed $\theta$ in the third quadrant, we are left with the real part of the integral, as well as a complex integral that is just as difficult to solve.
The residue I calculated is:
$$res_{-1}f(z) = \lim_{z \rightarrow -1}\frac{1}{3!} \frac{d^2}{dz^2} (z+1)^3 \frac{\ln{z}}{(1+z)^3}$$ $$=\frac{1}{2}* \frac{-1}{-1^2} = \frac{-1}{2}$$
So that means the entire complex integral should be $2\pi i* \frac{-1}{2} = -\pi i$...
I had another idea for a contour that catches the pole: we have
$$\gamma_1(t) = t, t \in (\epsilon, R)$$ $$\gamma_2(t) = Re^{it}, t\in (0, \pi/2)$$ $$\gamma_3(t) = e^{it} -1, t \in (\pi, 2\pi - \epsilon)$$ $$\gamma_4(t) = $$ the little tiny circle to complete the contour.
But this one, for $\gamma_3$ gives us a weird integral:
$$\int_{\pi}^{2\pi - \epsilon}\frac{\ln{e^{it} - 1}}{(e^{it} - 1 +1)^3}(ie^{it})$$
Which looks a little better, but I'm not sure how to hande the log function in this case, and it doesn't look promising. Any help would be appreciated. |
I am trying to do a double integral over the following region in polar coordinates:
I know that the limits of integration are:
$$\theta = -\frac{\pi}{2} \quad \to \quad \theta = \frac{\pi}{2} \\ r = 0 \quad \to \quad r = \cos \theta$$
However, I don't understand how $r = 0 \quad \to \quad r = \cos \theta$ works. Cosine is a function (not just a relation) meaning that it has only one value of $r$ for every value of $\theta$. However, it seems like the graph $r = \cos \theta$ has two values of $r$ for every value of $\theta$. Why does $r = \cos \theta$ produce a circle? |
Functions An online exercise on function notation, inverse functions and composite functions.
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Level 1 - Describe function machines using function notation.
Level 2 - Evaluate the given functions.
Level 3 - Solve the equations given in function notation.
Level 4 - Find the inverse of the given functions.
Level 5 - Simplify the composite functions.
Level 6 - Mixed questions.
Exam Style questions are in the style of GCSE or IB/A-level exam paper questions and worked solutions are available for Transum subscribers.
The following notes are intended to be a reminder or revision of the concepts and are not intended to be a substitute for a teacher or good textbook.
Function notation is quite different to the algebraic notation you have learnt involving brackets. \(f(x)\) does not mean the value of f multiplied by the value of x. In this case f is the name of the function and you would read \(f(x) = x^2\) as "f of x equals x squared".
In terms of function machines, if the input is \(x\) then the output is \(f(x)\).
Example
\(x \to \)\( + 3 \)\( \to \)\( \times 4 \)\( \to f(x)\)
In this case 3 is added to \(x\) and then the result is multiplied by 4 to give \(f(x)\)
\( (x+3) \times 4 = f(x) \)
\( f(x) = 4(x+3) \)
Example
if \(f(x)=x^2 + 3\) calculate the value of \(f(6)\)
This means replace the \(x\) with a 6 in the given function to obtain the result.
\(f(6) = 6^2+3\)
\(f(6) = 39\)
Example
\(f(x)=3(x+7) \) find \(x\) if \(f(x) = 30\)
\(3(x+7)=30\)
\(x+7 = 10\)
\(x = 3\)
The inverse of a function, written as \(f^{-1}(x) \) can be thought of as a way to 'undo' the function. If the function is written as a function machine, the inverse can be thought of as working backwards with the output becomming the input and the input becoming the output.
Example
\( f(x) = 4(x+3) \)
\(x \to \)\( + 3 \)\( \to \)\( \times 4 \)\( \to f(x)\)
\( f^{-1}(x) \leftarrow \)\( - 3 \)\( \leftarrow \)\( \div 4 \)\( \leftarrow x \)
\( f^{-1}(x) = \frac{x}{4} - 3 \)
A quicker way of finding the inverse of \(f(x)\) is to replace the \(f(x)\) with \(x\) on the left side of the equals sign and replace the \(x\) with \( f^{-1}(x) \) on the right side of the equals sign. Then rearrange the equation to make \( f^{-1}(x) \) the subject.
A composite function contains two functions combined into a single function. One function is applied to the result of the other function. You should evaluate the function closest to \(x\) first.
Example
if \(f(x)=2x+7\) and \(g(x)=5x^2\) find \(fg(3)\)
\(g(3) = 5 \times 3^2\)
\(g(3) = 5 \times 9\)
\(g(3) = 45\)
\(f(45) = 2 \times 45 + 7\)
\(f(45) = 97\)
so \( fg(3) = 97\)
Example
if \(f(x)=x+2\) and \(g(x)=3x^2\) find \(gf(x)\)
\( gf(x) = 3(x+2)^2\)
\( gf(x) = 3(x^2+4x+4) \)
\( gf(x) = 3x^2+12x+12 \)
Example
Find \(f(x-2)\) if \(f(x)=5x^2+3\)
\(f(x-2) =5(x-2)^2+3\)
\(f(x-2) =5(x^2-4x+4)+3\)
\(f(x-2) =5x^2-20x+20+3\)
\(f(x-2) =5x^2-20x+23\)
TI-nSpire:
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Can you provide a direct (not based on the neighborhood definition of pretopologies) proof for pretopological spaces (expressed as closure operators) that a function $f$ from a topological space $\mu$ to a topological space $\nu$ is continuous (in the customary topology sense) that is preimage of every open set is open if and only if it is continuous in the pretopology sense that is when $f[\operatorname{cl}_{\mu}A] \subseteq\operatorname{cl}_{\nu} f[A]$ for every subset $A$ of space $\mu$?
Given topological spaces $X,Y$ and a function $f: X \to Y$, I'll show that the two forms of continuity are equivalent. Note that I'll be using the overline $\overline{\phantom{A}}$ to denote the closure operator in both topological spaces.
Assume that $f : X \to Y$ is continuous "in the customary topological sense". Given $A \subseteq X$, note that $\overline{f[A]}$ is closed in $Y$, so $Y \setminus \overline{f[A]}$ is open in $Y$, and therefore $f^{-1} [ Y \setminus \overline{f[A]}] = X \setminus f^{-1} [ \overline{ f[A] } ]$ is open in $X$. It follows that $f^{-1} [ \overline{ f[A] } ]$ is closed in $X$. Note, too, that $$A \subseteq f^{-1} [ f [ A ] ] \subseteq f^{-1} [ \overline{ f[A] } ],$$ and so it follows that $$\overline{A} \subseteq f^{-1} [ \overline{ f[A] } ].$$ We now have that $$f [ \overline{A} ] \subseteq f [ f^{-1} [ \overline{ f[A] } ] ] \subseteq \overline{f[A]},$$ as desired.
Assume that $f : X \to Y$ is continuous "in the pretopological sense". Given an open $V \subseteq Y$, we want to show that $f^{-1} [ V ]$ is open, or, equivalently that $X \setminus f^{-1} [ V ] = f^{-1} [ Y \setminus V ]$ is closed. That is, we want to show that $\overline{ f^{-1} [ Y \setminus V ] } = f^{-1} [ Y \setminus V ]$. By properties of closure, we already know that $f^{-1} [ Y \setminus V ] \subseteq \overline{ f^{-1} [ Y \setminus V ] }$, and so we need only show the reverse inclusion.
By pretopological continuity, $f[ \overline{ f^{-1} [ Y \setminus V ] } ] \subseteq \overline{ f[ f^{-1} [ Y \setminus V ] ] } \subseteq \overline{ Y \setminus V} = Y \setminus V$ (recall that $Y \setminus V$ is closed). Therefore $$f^{-1} [ Y \setminus V ] \supseteq f^{-1} [ f [ \overline{ f^{-1} [ Y \setminus V ] } ] ] \supseteq \overline{ f^{-1} [ Y \setminus V ] },$$ as desired.
(Note that in the above I am making use of such facts as $f^{-1} [ f [ A ] ] \supseteq A$ and $f [ f^{-1} [ B ] ] \subseteq B$.) |
On Monday, Celestalon kicked off the official Alpha Theorycrafting season by posting a Theorycrafting Discussion thread on the forums. And he was kind enough to toss a meaty chunk of information our way about
Resolve, the replacement for Vengeance.
Resolve: Increases your healing and absorption done to yourself, based on Stamina and damage taken (before avoidance and mitigation) in the last 10 sec.
In today’s post, I want to go over the mathy details about how Resolve works, how it differs from Vengeance, and how it may (or may not) fix some of the problems we’ve discussed in previous blog posts.
Mathemagic
Celestalon broke the formula up into two components: one from stamina and one from damage taken. But for completeness, I’m going to bolt them together into one formula for resolve $R$:
$$ R =\frac{\rm Stamina}{250~\alpha} + 0.25\sum_i \frac{D_i}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t_i )}{10} \right )$$
where $D_i$ is an individual damage event that occurred $\Delta t_i$ seconds ago, and $\alpha$ is a level-dependent constant, with $\alpha(100)=261$. The sum is carried out over all damaging events that have happened in the last 10 seconds.
The first term in the equation is the stamina-based contribution, which is always active, even when out of combat. There’s a helpful buff in-game to alert you to this:
My premade character has 1294 character sheet stamina, which after dividing by 250 and $\alpha(90)=67$, gives me 0.07725, or about 7.725% Resolve. It’s not clear at this point whether the tooltip is misleadingly rounding down to 7% (i.e. using floor instead of round) or whether Resolve is only affected by the stamina from gear. The Alpha servers went down as I was attempting to test this, so we’ll have to revisit it later. We’ve already been told that this will update dynamically with stamina buffs, so having Power Word: Fortitude buffed on you mid-combat will raise your Resolve.
Once you’re in combat and taking damage, the second term makes a contribution:
I’ve left this term in roughly the form Celestalon gave, even though it can obviously be simplified considerably by combining all of the constants, because this form does a better job of illustrating the behavior of the mechanic. Let’s ignore the sum for now, and just consider an isolated damage event that does $D$ damage:
$$0.25\times\frac{D}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t )}{10} \right )$$
The 0.25 just moderates the amount of Resolve you get from damaging attacks. It’s a constant multiplicative factor that they will likely tweak to achieve the desired balance between baseline (stamina-based) Resolve and dynamic (damage-based) Resolve.
The factor of $D/{\rm MaxHealth}$ means that we’re normalizing the damage by our max health. So if we have 1000 health and take an attack that deals 1000 damage (remember, this is before mitigation), this term gives us a factor of 1. Avoided auto-attacks also count here, though instead of performing an actual damage roll the game just uses the mean value of the boss’s auto-attack damage. Again, nothing particularly complicated here, it just makes Resolve depend on the percentage of your health the attack would have removed rather than the raw damage amount. Also note that we’ve been told that dynamic health effects from temporary multipliers (e.g. Last Stand) aren’t included here, so we’re not punished for using temporary health-increasing cooldowns.
The term in parentheses is the most important part, though. In the instant the attack lands, $\Delta t=0$ and the term in parentheses evaluates to $2(10-0)/10 = 2.$ So that attack dealing 1000 damage to our 1000-health tank would give $0.25\times 1 \times 2 = 0.5,$ or 50% Resolve.
However, one second later, $\Delta t = 1$, so the term in parentheses is only $2(10-1)/10 = 1.8$, and the amount of resolve it grants is reduced to 45%. The amount of Resolve granted continues to linearly decrease as time passes, and by the time ten seconds have elapsed it’s reduced to zero. Each attack is treated independently, so to get our total Resolve from all damage taken we just have to add up the Resolve granted by every attack we’ve taken, hence the sum in my equation.
You may note that the time-average of the term in parentheses is 1, which is how we get the advertised “averages to ~Damage/MaxHealth” that Celestalon mentioned in the post. In that regard, he’s specifically referring to just the part within the sum, not the constant factor of 0.25 outside of it. So in total, your average Resolve contribution from damage is 25% of Damage/MaxHealth.
Comparing to Vengeance
Mathematically speaking, there’s a world of difference between Resolve and Vengeance. First and foremost is the part we already knew: Resolve doesn’t grant any offensive benefit. We’ve talked about that a lot before, though, so it’s not territory worth re-treading.
Even in the defensive component though, there are major differences. Vengeance’s difference equation, if solved analytically, gives solutions that are exponentials. In other words, provided you were continuously taking damage (such that it didn’t fall off entirely), Vengeance would decay and adjust to your new damage intake rather smoothly. It also meant that damage taken at the very beginning of an encounter was still contributing some amount of Vengeance at the very end, again, assuming there was no interruption. And since it was only recalculated on a damage event, you could play some tricks with it, like taking a giant attack that gave you millions of Vengeance and then riding that wave for 20 seconds while your co-tank takes the boss.
Resolve does away with all of that. It flat-out says “look, the only thing that matters is the last 10 seconds.” The calculation doesn’t rely on a difference equation, meaning that when recalculating, it doesn’t care what your Resolve was at any time previously. And it forces a recalculation at fixed intervals, not just when you take damage. As a result, it’s much harder to game than Vengeance was.
Celestalon’s post also outlines a few other significant differences:
No more ramp-up mechanism No taunt-transfer mechanism Resolve persists through shapeshifts Resolve only affects self-healing and self-absorbs
The lack of ramp-up and taunt-transfer mechanisms may at first seem like a problem. But in practice, I don’t think we’ll miss either of them. Both of these effects served offensive (i.e. threat) and defensive purposes, and it’s pretty clear that the offensive purposes are made irrelevant by definition here since Resolve won’t affect DPS/threat. The defensive purpose they served was to make sure you had
some Vengeance to counter the boss’s first few hits, since Vengeance had a relatively slow ramp-up time but the boss’s attacks did not.
However, Resolve ramps up a
lot faster than Vengeance does. Again, this is in part thanks to the fact that it isn’t governed by a difference equation. The other part is because it only cares about the last ten seconds.
To give you a visual representation of that, here’s a plot showing both Vengeance and Resolve for a player being attacked by a boss. The tank has 100 health and the boss swings for 30 raw damage every 1.5 seconds. Vengeance is shown in arbitrary units here since we’re not interested in the exact magnitude of the effect, just in its dynamic properties. I’ve also ignored the baseline (stamina-based) contribution to Resolve for the same reason.
As a final note, while the blog post says that Resolve is recalculated every second, it seemed like it was updating closer to every half-second when I fooled with it on alpha, so these plots use 0.5-second update intervals. Changing to 1-second intervals doesn’t significantly change the results (they just look a little more fragmented).
The plot very clearly shows the 50% ramp-up mechanism and slow decay-like behavior of Vengeance. Note that while the ramp-up mechanism gets you to 50% of Vengeance’s overall value at the first hit (at t=2.5 seconds), Resolve hits this mark as soon as the second hit lands (at 4.0 seconds) despite not having
any ramp-up mechanism.
Resolve also hits its steady-state value much more quickly than Vengeance does. By definition, Resolve gets there after about 10 seconds of combat (t=12.5 seconds). But with Vengeance, it takes upwards of 30-40 seconds to even approach the steady-state value thanks to the decay effect (again, a result of the difference equation used to calculate Vengeance). Since most fights involve tank swaps more frequently than this, it meant that you were consistently getting stronger the longer you tanked a boss. This in turn helped encourage the sort of “solo-tank things that should not be solo-tanked” behavior we saw in Mists.
This plot assumes a boss who does exactly 30 damage per swing, but in real encounters the boss’s damage varies. Both Vengeance and Resolve adapt to mimic that change in the tank’s damage intake, but as you could guess, Resolve adapts much more quickly. If we allow the boss to hit for a random amount between 20 and 40 damage:
You can certainly see the similar changes in both curves, but Resolve reacts quickly to each change while Vengeance changes rather slowly.
One thing you’ve probably noticed by now is that the Resolve plot looks very jagged (in physics, we might call this a “sawtooth wave”). This happens because of the linear decay built into the formula. It peaks in the instant you take the attack – or more accurately, in the instant that Resolve is recalculated after that attack. But then every time it’s recalculated it linearly decreases by a fixed percent. If the boss swings in 1.5-second intervals, then Resolve will zig-zag between its max value and 85% of its max value in the manner shown.
The more frequently the boss attacks, the smoother that zig-zag becomes; conversely, a boss with a long swing timer will cause a larger variation in Resolve. This is apparent if we adjust the boss’s swing timer in either direction:
It’s worth noting that every plot here has a new randomly-generated sequence of attacks, so don’t be surprised that the plots don’t have the same profile as the original. The key difference is the size of the zig-zag on the Resolve curve.
I’ve also run simulations where the boss’ base damage is 50 rather than 30, but apart from the y-axis having large numbers there’s no real difference:
Note that even a raw damage of 50% is pretty conservative for a boss – heroic bosses in Siege have frequently had raw damages that were larger than the player’s health. But it’s not clear if that will still be the case with the new tanking and healing paradigm that’s been unveiled for Warlords.
If we make the assumption that raw damage will be lower, then these rough estimates give us an idea of how large an effect Resolve will be. If we guess at a 5%-10% baseline value from stamina, these plots suggest that Resolve will end up being anywhere from a 50% to 200% modifier on our healing. In other words, it has the potential to double or triple our healing output with the current tuning numbers. Of course, it’s anyone’s guess as to whether those numbers are even remotely close to what they’ll end up being by the end of beta.
Is It Fixed Yet?
If you look back over our old blog posts, the vast majority of our criticisms of Vengeance had to do with its tie-in to damage output. Those have obviously been addressed, which leaves me worrying that I’ll have nothing to rant about for the next two or three years.
But regarding everything else, I think Resolve stands a fair chance of addressing our concerns. One of the major issues with Vengeance was the sheer magnitude of the effect – you could go from having 50k AP to 600k AP on certain bosses, which meant your abilities got up to 10x more effective. Even though that’s an extreme case, I regularly noted having over 300k AP during progression bosses, a factor of around 6x improvement. Resolve looks like it’ll tamp down on that some. Reasonable bosses are unlikely to grant a multiplier larger than 2x, which will be easier to balance around.
It hasn’t been mentioned specifically in Celestalon’s post, but I think it’s a reasonable guess that they will continue to disable Resolve gains from damage that could be avoided through better play (i.e. intentionally “standing in the bad”). If so, there will be little (if any) incentive to take excess damage to get more Resolve. Our sheer AP scaling on certain effects created situations where this was a net survivability gain with Vengeance, but the lower multiplier should make that impossible with Resolve.
While I still don’t think it needs to affect anything other than active mitigation abilities, the fact that it’s a multiplier affecting everything equally rather than a flat AP boost should make it easier to keep talents with different AP coefficients balanced (Eternal Flame and Sacred Shield, specifically). And we already know that Eternal Flame is losing its Bastion of Glory interaction, another change which will facilitate making both talents acceptable choices.
All in all, I think it’s a really good system, if slightly less transparent. It’s too soon to tell whether we’ll see any unexpected problems, of course, but the mechanic doesn’t have any glaring issues that stand out upon first examination (unlike Vengeance). I still have a few lingering concerns about steady-state threat stability between tanks (ironically, due to the
removal of Vengeance), but that is the sort of thing which will become apparent fairly quickly during beta testing, and at any rate shouldn’t reflect on the performance of Resolve. |
L # 1
Show that
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Last edited by krassi_holmz (2006-03-09 02:44:53)
IPBLE: Increasing Performance By Lowering Expectations.
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It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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L # 2
If
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Let
log x = x' log y = y' log z = z'. Then:
x'+y'+z'=0.
Rewriting in terms of x' gives:
IPBLE: Increasing Performance By Lowering Expectations.
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Well done, krassi_holmz!
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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L # 3
If x²y³=a and log (x/y)=b, then what is the value of (logx)/(logy)?
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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loga=2logx+3logy
b=logx-logy loga+3b=5logx loga-2b=3logy+2logy=5logy logx/logy=(loga+3b)/(loga-2b). Last edited by krassi_holmz (2006-03-10 20:06:29)
IPBLE: Increasing Performance By Lowering Expectations.
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Very well done, krassi_holmz!
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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L # 4
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It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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You are not supposed to use a calculator or log tables for L # 4. Try again!
Last edited by JaneFairfax (2009-01-04 23:40:20)
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No, I didn't
I remember
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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You still used a calculator / log table in the past to get those figures (or someone else did and showed them to you). I say again:
no calculators or log tables to be used (directly or indirectly) at all!! Last edited by JaneFairfax (2009-01-06 00:30:04)
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log a = 2log x + 3log y
b = log x log y
log a + 3 b = 5log x
loga - 2b = 3logy + 2logy = 5logy
logx / logy = (loga+3b) / (loga-2b)
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Hi ganesh
for L # 1 since log(a)= 1 / log(b), log(a)=1 b a a we have 1/log(abc)+1/log(abc)+1/log(abc)= a b c log(a)+log(b)+log(c)= log(abc)=1 abc abc abc abc Best Regards Riad Zaidan
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Hi ganesh
for L # 2 I think that the following proof is easier: Assume Log(x)/(b-c)=Log(y)/(c-a)=Log(z)/(a-b)=t So Log(x)=t(b-c),Log(y)=t(c-a) , Log(z)=t(a-b) So Log(x)+Log(y)+Log(z)=tb-tc+tc-ta+ta-tb=0 So Log(xyz)=0 so xyz=1 Q.E.D Best Regards Riad Zaidan
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Gentleman,
Thanks for the proofs.
Regards.
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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log_2(16) = \log_2 \left ( \frac{64}{4} \right ) = \log_2(64) - \log_2(4) = 6 - 2 = 4, \,
log_2(\sqrt[3]4) = \frac {1}{3} \log_2 (4) = \frac {2}{3}. \,
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L # 4
I don't want a method that will rely on defining certain functions, taking derivatives,
noting concavity, etc.
Change of base:
Each side is positive, and multiplying by the positive denominator
keeps whatever direction of the alleged inequality the same direction:
On the right-hand side, the first factor is equal to a positive number less than 1,
while the second factor is equal to a positive number greater than 1. These facts are by inspection combined with the nature of exponents/logarithms.
Because of (log A)B = B(log A) = log(A^B), I may turn this into:
I need to show that
Then
Then 1 (on the left-hand side) will be greater than the value on the
right-hand side, and the truth of the original inequality will be established.
I want to show
Raise a base of 3 to each side:
Each side is positive, and I can square each side:
-----------------------------------------------------------------------------------
Then I want to show that when 2 is raised to a number equal to
(or less than) 1.5, then it is less than 3.
Each side is positive, and I can square each side:
Last edited by reconsideryouranswer (2011-05-27 20:05:01)
Signature line:
I wish a had a more interesting signature line.
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Hi reconsideryouranswer,
This problem was posted by JaneFairfax. I think it would be appropriate she verify the solution.
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Hi all,
I saw this post today and saw the probs on log. Well, they are not bad, they are good. But you can also try these problems here by me (Credit: to a book):
http://www.mathisfunforum.com/viewtopic … 93#p399193
Practice makes a man perfect.
There is no substitute to hard work All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam
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JaneFairfax, here is a basic proof of L4:
For all real a > 1, y = a^x is a strictly increasing function.
log(base 2)3 versus log(base 3)5
2*log(base 2)3 versus 2*log(base 3)5
log(base 2)9 versus log(base 3)25
2^3 = 8 < 9
2^(> 3) = 9
3^3 = 27 < 25
3^(< 3) = 25
So, the left-hand side is greater than the right-hand side, because
Its logarithm is a larger number.
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On Monday, Celestalon kicked off the official Alpha Theorycrafting season by posting a Theorycrafting Discussion thread on the forums. And he was kind enough to toss a meaty chunk of information our way about
Resolve, the replacement for Vengeance.
Resolve: Increases your healing and absorption done to yourself, based on Stamina and damage taken (before avoidance and mitigation) in the last 10 sec.
In today’s post, I want to go over the mathy details about how Resolve works, how it differs from Vengeance, and how it may (or may not) fix some of the problems we’ve discussed in previous blog posts.
Mathemagic
Celestalon broke the formula up into two components: one from stamina and one from damage taken. But for completeness, I’m going to bolt them together into one formula for resolve $R$:
$$ R =\frac{\rm Stamina}{250~\alpha} + 0.25\sum_i \frac{D_i}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t_i )}{10} \right )$$
where $D_i$ is an individual damage event that occurred $\Delta t_i$ seconds ago, and $\alpha$ is a level-dependent constant, with $\alpha(100)=261$. The sum is carried out over all damaging events that have happened in the last 10 seconds.
The first term in the equation is the stamina-based contribution, which is always active, even when out of combat. There’s a helpful buff in-game to alert you to this:
My premade character has 1294 character sheet stamina, which after dividing by 250 and $\alpha(90)=67$, gives me 0.07725, or about 7.725% Resolve. It’s not clear at this point whether the tooltip is misleadingly rounding down to 7% (i.e. using floor instead of round) or whether Resolve is only affected by the stamina from gear. The Alpha servers went down as I was attempting to test this, so we’ll have to revisit it later. We’ve already been told that this will update dynamically with stamina buffs, so having Power Word: Fortitude buffed on you mid-combat will raise your Resolve.
Once you’re in combat and taking damage, the second term makes a contribution:
I’ve left this term in roughly the form Celestalon gave, even though it can obviously be simplified considerably by combining all of the constants, because this form does a better job of illustrating the behavior of the mechanic. Let’s ignore the sum for now, and just consider an isolated damage event that does $D$ damage:
$$0.25\times\frac{D}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t )}{10} \right )$$
The 0.25 just moderates the amount of Resolve you get from damaging attacks. It’s a constant multiplicative factor that they will likely tweak to achieve the desired balance between baseline (stamina-based) Resolve and dynamic (damage-based) Resolve.
The factor of $D/{\rm MaxHealth}$ means that we’re normalizing the damage by our max health. So if we have 1000 health and take an attack that deals 1000 damage (remember, this is before mitigation), this term gives us a factor of 1. Avoided auto-attacks also count here, though instead of performing an actual damage roll the game just uses the mean value of the boss’s auto-attack damage. Again, nothing particularly complicated here, it just makes Resolve depend on the percentage of your health the attack would have removed rather than the raw damage amount. Also note that we’ve been told that dynamic health effects from temporary multipliers (e.g. Last Stand) aren’t included here, so we’re not punished for using temporary health-increasing cooldowns.
The term in parentheses is the most important part, though. In the instant the attack lands, $\Delta t=0$ and the term in parentheses evaluates to $2(10-0)/10 = 2.$ So that attack dealing 1000 damage to our 1000-health tank would give $0.25\times 1 \times 2 = 0.5,$ or 50% Resolve.
However, one second later, $\Delta t = 1$, so the term in parentheses is only $2(10-1)/10 = 1.8$, and the amount of resolve it grants is reduced to 45%. The amount of Resolve granted continues to linearly decrease as time passes, and by the time ten seconds have elapsed it’s reduced to zero. Each attack is treated independently, so to get our total Resolve from all damage taken we just have to add up the Resolve granted by every attack we’ve taken, hence the sum in my equation.
You may note that the time-average of the term in parentheses is 1, which is how we get the advertised “averages to ~Damage/MaxHealth” that Celestalon mentioned in the post. In that regard, he’s specifically referring to just the part within the sum, not the constant factor of 0.25 outside of it. So in total, your average Resolve contribution from damage is 25% of Damage/MaxHealth.
Comparing to Vengeance
Mathematically speaking, there’s a world of difference between Resolve and Vengeance. First and foremost is the part we already knew: Resolve doesn’t grant any offensive benefit. We’ve talked about that a lot before, though, so it’s not territory worth re-treading.
Even in the defensive component though, there are major differences. Vengeance’s difference equation, if solved analytically, gives solutions that are exponentials. In other words, provided you were continuously taking damage (such that it didn’t fall off entirely), Vengeance would decay and adjust to your new damage intake rather smoothly. It also meant that damage taken at the very beginning of an encounter was still contributing some amount of Vengeance at the very end, again, assuming there was no interruption. And since it was only recalculated on a damage event, you could play some tricks with it, like taking a giant attack that gave you millions of Vengeance and then riding that wave for 20 seconds while your co-tank takes the boss.
Resolve does away with all of that. It flat-out says “look, the only thing that matters is the last 10 seconds.” The calculation doesn’t rely on a difference equation, meaning that when recalculating, it doesn’t care what your Resolve was at any time previously. And it forces a recalculation at fixed intervals, not just when you take damage. As a result, it’s much harder to game than Vengeance was.
Celestalon’s post also outlines a few other significant differences:
No more ramp-up mechanism No taunt-transfer mechanism Resolve persists through shapeshifts Resolve only affects self-healing and self-absorbs
The lack of ramp-up and taunt-transfer mechanisms may at first seem like a problem. But in practice, I don’t think we’ll miss either of them. Both of these effects served offensive (i.e. threat) and defensive purposes, and it’s pretty clear that the offensive purposes are made irrelevant by definition here since Resolve won’t affect DPS/threat. The defensive purpose they served was to make sure you had
some Vengeance to counter the boss’s first few hits, since Vengeance had a relatively slow ramp-up time but the boss’s attacks did not.
However, Resolve ramps up a
lot faster than Vengeance does. Again, this is in part thanks to the fact that it isn’t governed by a difference equation. The other part is because it only cares about the last ten seconds.
To give you a visual representation of that, here’s a plot showing both Vengeance and Resolve for a player being attacked by a boss. The tank has 100 health and the boss swings for 30 raw damage every 1.5 seconds. Vengeance is shown in arbitrary units here since we’re not interested in the exact magnitude of the effect, just in its dynamic properties. I’ve also ignored the baseline (stamina-based) contribution to Resolve for the same reason.
As a final note, while the blog post says that Resolve is recalculated every second, it seemed like it was updating closer to every half-second when I fooled with it on alpha, so these plots use 0.5-second update intervals. Changing to 1-second intervals doesn’t significantly change the results (they just look a little more fragmented).
The plot very clearly shows the 50% ramp-up mechanism and slow decay-like behavior of Vengeance. Note that while the ramp-up mechanism gets you to 50% of Vengeance’s overall value at the first hit (at t=2.5 seconds), Resolve hits this mark as soon as the second hit lands (at 4.0 seconds) despite not having
any ramp-up mechanism.
Resolve also hits its steady-state value much more quickly than Vengeance does. By definition, Resolve gets there after about 10 seconds of combat (t=12.5 seconds). But with Vengeance, it takes upwards of 30-40 seconds to even approach the steady-state value thanks to the decay effect (again, a result of the difference equation used to calculate Vengeance). Since most fights involve tank swaps more frequently than this, it meant that you were consistently getting stronger the longer you tanked a boss. This in turn helped encourage the sort of “solo-tank things that should not be solo-tanked” behavior we saw in Mists.
This plot assumes a boss who does exactly 30 damage per swing, but in real encounters the boss’s damage varies. Both Vengeance and Resolve adapt to mimic that change in the tank’s damage intake, but as you could guess, Resolve adapts much more quickly. If we allow the boss to hit for a random amount between 20 and 40 damage:
You can certainly see the similar changes in both curves, but Resolve reacts quickly to each change while Vengeance changes rather slowly.
One thing you’ve probably noticed by now is that the Resolve plot looks very jagged (in physics, we might call this a “sawtooth wave”). This happens because of the linear decay built into the formula. It peaks in the instant you take the attack – or more accurately, in the instant that Resolve is recalculated after that attack. But then every time it’s recalculated it linearly decreases by a fixed percent. If the boss swings in 1.5-second intervals, then Resolve will zig-zag between its max value and 85% of its max value in the manner shown.
The more frequently the boss attacks, the smoother that zig-zag becomes; conversely, a boss with a long swing timer will cause a larger variation in Resolve. This is apparent if we adjust the boss’s swing timer in either direction:
It’s worth noting that every plot here has a new randomly-generated sequence of attacks, so don’t be surprised that the plots don’t have the same profile as the original. The key difference is the size of the zig-zag on the Resolve curve.
I’ve also run simulations where the boss’ base damage is 50 rather than 30, but apart from the y-axis having large numbers there’s no real difference:
Note that even a raw damage of 50% is pretty conservative for a boss – heroic bosses in Siege have frequently had raw damages that were larger than the player’s health. But it’s not clear if that will still be the case with the new tanking and healing paradigm that’s been unveiled for Warlords.
If we make the assumption that raw damage will be lower, then these rough estimates give us an idea of how large an effect Resolve will be. If we guess at a 5%-10% baseline value from stamina, these plots suggest that Resolve will end up being anywhere from a 50% to 200% modifier on our healing. In other words, it has the potential to double or triple our healing output with the current tuning numbers. Of course, it’s anyone’s guess as to whether those numbers are even remotely close to what they’ll end up being by the end of beta.
Is It Fixed Yet?
If you look back over our old blog posts, the vast majority of our criticisms of Vengeance had to do with its tie-in to damage output. Those have obviously been addressed, which leaves me worrying that I’ll have nothing to rant about for the next two or three years.
But regarding everything else, I think Resolve stands a fair chance of addressing our concerns. One of the major issues with Vengeance was the sheer magnitude of the effect – you could go from having 50k AP to 600k AP on certain bosses, which meant your abilities got up to 10x more effective. Even though that’s an extreme case, I regularly noted having over 300k AP during progression bosses, a factor of around 6x improvement. Resolve looks like it’ll tamp down on that some. Reasonable bosses are unlikely to grant a multiplier larger than 2x, which will be easier to balance around.
It hasn’t been mentioned specifically in Celestalon’s post, but I think it’s a reasonable guess that they will continue to disable Resolve gains from damage that could be avoided through better play (i.e. intentionally “standing in the bad”). If so, there will be little (if any) incentive to take excess damage to get more Resolve. Our sheer AP scaling on certain effects created situations where this was a net survivability gain with Vengeance, but the lower multiplier should make that impossible with Resolve.
While I still don’t think it needs to affect anything other than active mitigation abilities, the fact that it’s a multiplier affecting everything equally rather than a flat AP boost should make it easier to keep talents with different AP coefficients balanced (Eternal Flame and Sacred Shield, specifically). And we already know that Eternal Flame is losing its Bastion of Glory interaction, another change which will facilitate making both talents acceptable choices.
All in all, I think it’s a really good system, if slightly less transparent. It’s too soon to tell whether we’ll see any unexpected problems, of course, but the mechanic doesn’t have any glaring issues that stand out upon first examination (unlike Vengeance). I still have a few lingering concerns about steady-state threat stability between tanks (ironically, due to the
removal of Vengeance), but that is the sort of thing which will become apparent fairly quickly during beta testing, and at any rate shouldn’t reflect on the performance of Resolve. |
"Abstract: We prove, once and for all, that people who don't use superspace are really out of it. This includes QCDers, who always either wave their hands or gamble with lettuce (Monte Zuma calculations). Besides, all nonsupersymmetric theories have divergences which lead to problems with things like renormalons, instantons, anomalons, and other phenomenons. Also, they can't hide from gravity forever."
Can a gravitational field possess momentum? A gravitational wave can certainly possess momentum just like a light wave has momentum, but we generally think of a gravitational field as a static object, like an electrostatic field.
"You have to understand that compared to other professions such as programming or engineering, ethical standards in academia are in the gutter. I have worked with many different kinds of people in my life, in the U.S. and in Japan. I have only encountered one group more corrupt than academic scientists: the mafia members who ran Las Vegas hotels where I used to install computer equipment. "
so Ive got a small bottle that I filled up with salt. I put it on the scale and it's mass is 83g. I've also got a jup of water that has 500g of water. I put the bottle in the jug and it sank to the bottom. I have to figure out how much salt to take out of the bottle such that the weight force of the bottle equals the buoyancy force.
For the buoyancy do I: density of water * volume of water displaced * gravity acceleration?
so: mass of bottle * gravity = volume of water displaced * density of water * gravity?
@EmilioPisanty The measurement operators than I suggested in the comments of the post are fine but I additionally would like to control the width of the Poisson Distribution (much like we can do for the normal distribution using variance). Do you know that this can be achieved while still maintaining the completeness condition $$\int A^{\dagger}_{C}A_CdC = 1$$?
As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including:ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room.An altern...
You're always welcome to ask. One of the reasons I hang around in the chat room is because I'm happy to answer this sort of question. Obviously I'm sometimes busy doing other stuff, but if I have the spare time I'm always happy to answer.
Though as it happens I have to go now - lunch time! :-)
@JohnRennie It's possible to do it using the energy method. Just we need to carefully write down the potential function which is $U(r)=\frac{1}{2}\frac{mg}{R}r^2$ with zero point at the center of the earth.
Anonymous
Also I don't particularly like this SHM problem because it causes a lot of misconceptions. The motion is SHM only under particular conditions :P
I see with concern the close queue has not shrunk considerably in the last week and is still at 73 items. This may be an effect of increased traffic but not increased reviewing or something else, I'm not sure
Not sure about that, but the converse is certainly false :P
Derrida has received a lot of criticism from the experts on the fields he tried to comment on
I personally do not know much about postmodernist philosophy, so I shall not comment on it myself
I do have strong affirmative opinions on textual interpretation, made disjoint from authoritorial intent, however, which is a central part of Deconstruction theory. But I think that dates back to Heidegger.
I can see why a man of that generation would be leaned towards that idea. I do too. |
The exact value of the Kolmogorov complexity depends on the language chosen to represent strings. This language has to be Turing complete, so representing all strings as themselves isn't an option.By the pigeonhole principle, if there is at least one string of length at most $n$ whose representation is shorter than itself, then there is also at least one ...
The central concept here is Kolmogorov complexity, and more specifically compressibility. To get a intuitive feeling of compressibility, consider two strings $A \in \mathbb{B}^*$ and $B \in \mathbb{B}^*$, where $\mathbb{B} = \{ 0,1 \}$. Let$A = 1010$ $1010$ $1010$ $1010$, and$B = 1011$ $0110$ $0111$ $1001$.Note that $|A| = |B| = 16$. How could ...
I guess one possible answer to your question is this: Take a pseudorandom number generator $G$. Try to chose a generator which has some powerful attacks against it: a random number generator attack for $G$ is (for our purposes), an algorithm $A$ which, when given an imput string $s$, determines a seed $A(s)$, such that $G(A(s))=s$. Then approximate the KC of ...
I'll assume that the paper you're referring to is Veness et al. (2011) - A Monte-Carlo AIXI Approximation. The paper (and the AIXI model more generally) is rather technical, and so it is difficult to talk about without using much math. I'll do my best to explain informally here.The question is in two parts, so I'll address each separately.ModelsThe ...
To my knowledge, this is not one of the "classical" approaches used to characterize regular languages.This approach is discussed in "A New Approach to Formal Language Theory byKolmogorov Complexity", by Ming Li and Paul M.B. Vitanyi (see section 3.1).They give several examples where one can use the statement you mentioned instead of using the pumping ...
Fancy answerWe need to show that the set$$R = \{x \in \{0,1\}^* \mid C(x) < |x|\}$$is c.e. (allow me to use the new terminology). The defining condition for $R$ is equivalent to$$\exists n, m \in \mathbb{N} \,.\, T(n,0,m) \land U(m) = x \land |n| < |x| \tag{1}$$where $T$ is Kleene's predicate and $U$ the associated output function.In words, the ...
For any given problem, you can create a programming language where a program to encode the solution to that problem is a single character. (cf. HQ9+). Kolmogorov complexity is language dependent. The answer to your question about which problems cause a blowup will depend heavily upon whichever "standard formal language" you choose.There are some ...
It is impossible to give a specific example of a string with high Kolmogorov complexity. If I was to give an example in this answer, then the following piece of code would retrieve it:wget http://cs.stackexchange.com/q/9721(plus some O(1) post-processing).A string of high Kolmogorov complexity is as elusive as a random number:There is in fact a ...
Any probability distribution. If you have a computable probability distribution that gives your data probability $p(x)$, then by the Kraft inequality, there's a computable compressor that compresses it in $-\log p(x)$ bits (round up if you object to fractional bits). This means pretty much any generative machine learning algorithm can be used.This is why ...
We want to show $\overline{R_c}\in RE$.$\overline{R_c}=\left\{x|\exists M\hspace{1mm} s.t. \hspace{1mm} M(\epsilon)=x,|\langle M\rangle|<|x| \right\}$, i.e. $x$ is not a Kolmogorov-random string if there exists a Turing machine which outputs $x$ (say, when initialized with blank input), with description $\langle M\rangle$ shorter than $|x|$.Given $x$, ...
No. This is basically Chaitin's incompleteness theorem.Roughly, the theorem says that there exists a concrete constant $C$ (which is a function of your consistent set of axioms) for which no fixed string can be proven to have Kolmogorov complexity larger than $C$. The core idea is that if you could do it for every string, then you can write a program of ...
I apologize in advance for that I give way too many details, but I'm about to contradict people.About $K(x)≤K'(x)+c$The fact that $K_1(x)≤K_2(x)+c$ usually comes from an interpreter of the description language #2 into the description language #1 and not from a translation from programs of #2 into programs of #1.For example $K_\mathtt{C}(x)≤K_\mathtt{...
A result of Shannon's states that there exists a sequence of functions $f_n\colon\{0,1\}^n\rightarrow \{0,1\}$ that is so that the problem $P(n)$ of computing $f_n(x)$ for $x\in\{0,1\}^n$ requires at least $\Theta(\frac{2^n}{n})$ boolean operations (i.e. the circuit complexity of computing $f_n(x)$ is at least $\Theta(\frac{2^n}{n})$).This theorem is not ...
The problem is in your poor definition of "algorithmically random number" as applied to irrational numbers. In particular:has the same length (number of bits) as the number itself.has no meaning if the number is of unbounded length.Your Wikipedia link gives better definitions, which don't have this problem. For example (and paraphrasing formatting):...
The description of a string considered here is an input to some universal Turing machine. You can think of it as a C program. The string hello world does not, by itself, form a C program, but the following one does: int main(int argc, char *argv[]) { printf("hello world"); }. As you can see, the overhead is constant but not zero.
Should Kc be restated as being resource based and not solely program size based?No. If you change the definition like that, you get a different concept, and the different concepts deserves its own different name. If you want to examine that concept, you should give it its own name. Kolmogorov complexity is a standard term with an accepting meaning; ...
The standard notion of pseudorandomness is about a process. You can say that the process (the pseudorandom generator) is pseudorandom, or not. The notion of pseudorandomness of a single string is not defined; that's not something you can talk about.Kolmogorov randomness is a property of a bit-string. You can say that a particular bit-string (sequence) ...
This is a classical question in the foundations of Kolmogorov complexity: does the programming language matter for the sake of defining minimal description length? It depends on your model, but in general the answer is "no" or "almost no".Suppose that $P_1,P_2$ are two Turing complete programming languages. In particular, in $P_1$ you can write an ...
No, there is no general algorithm to compute a close approximation to the Kolmogorov complexity of the sequence $1,2,\dots,n$. Any candidate algorithm you come up with will have some inputs where it gives a bad answer (a poor approximation to the correct answer).Denote by $[n]$ the binary encoding of the natural number $n$, and let $[[n]] = [1],[2],\ldots,...
grammar coding is a less-frequently used version of a compression algorithm and can be taken as a "rough" estimate of Kolmogorov complexity. grammar coding is not as commonly used as a compression algorithm as other more common approaches maybe mainly because it doesnt improve much on compression from eg Lempel-Ziv on text based-corpuses, but it may do well ...
What Li and Vitanyi mean by a reasonable programming language is one that is Turing-complete. This means that you can simulate it on a Turing machine and you can simulate a Turing machine on it, so it has the same expressive power as a Turing machine. This gives you a kind of class of languages. I can prove that C is equivalent to a Turing machine, and that ...
For every $n$ there is a "busy beaver" machine (i.e., a Turing machine on the tape alphabet $\{0,1\}$ run on the empty tape) which outputs $1^n$ using $O(\log n/\log\log n)$ states, which is asymptotically optimal (up to constants).Here is how this machine works. For every $C$, we will construct such a machine having $O(\log n/C) + O(C2^C)$ states. The ...
Collatz conjecture:The following program always halts:void function( ArbitraryInteger input){while( input > 1){if(input % 2 == 0)input /= 2;elseinput = (input*3) + 1;}// Halt here}Slight variation (still a conjecture, because it's based on a result from Collatz's one):...
That's right. You can run all machines in parallel and you'll find shorter and shorter descriptions, but at some point there will always be a shorter program still running, for which you can't prove that it won't halt with $x$ as output.The probability you describe (of a program not halting, assuming some distribution on inputs) is the complement to ...
Like Kaveh said, you can view a decision problem as a language consisting of the descriptions of all the "yes" instances. Then the Kolmogorov complexity of the language is, as you said, the length of the smallest program that produces all those descriptions.How we describe the program that generates the instances (and hence defines the Kolmogorov ...
We can view of a language $A$ as an infinite binary string (the infinite binary string corresponding to the characteristic function of $A$) and then use the notion of Kolmogorov complexity for such strings.Google for: Kolmogorov complexity infinite words
It could be because "useful" is hard to define. Say we have a highly-structured, information-rich message $x$ which can be compressed at most by a factor of $\alpha$ to the message $y$. Intuitively, $x$ and $y$ contain the same amount of useful information; indeed, they contain the same amount of information according to the usual definition. Now imagine a ...
Surprisingly, the Kolmogorov complexity of some arbitrary number is known to be uncomputable. So the general form of your question, is an arbitrary number algorithmically compressible is undecidable (i.e. the problem of computing the algorithmic complexity of a sequence). This can be proven by reduction to the Halting problem.Problems that are known to be ... |
On Monday, Celestalon kicked off the official Alpha Theorycrafting season by posting a Theorycrafting Discussion thread on the forums. And he was kind enough to toss a meaty chunk of information our way about
Resolve, the replacement for Vengeance.
Resolve: Increases your healing and absorption done to yourself, based on Stamina and damage taken (before avoidance and mitigation) in the last 10 sec.
In today’s post, I want to go over the mathy details about how Resolve works, how it differs from Vengeance, and how it may (or may not) fix some of the problems we’ve discussed in previous blog posts.
Mathemagic
Celestalon broke the formula up into two components: one from stamina and one from damage taken. But for completeness, I’m going to bolt them together into one formula for resolve $R$:
$$ R =\frac{\rm Stamina}{250~\alpha} + 0.25\sum_i \frac{D_i}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t_i )}{10} \right )$$
where $D_i$ is an individual damage event that occurred $\Delta t_i$ seconds ago, and $\alpha$ is a level-dependent constant, with $\alpha(100)=261$. The sum is carried out over all damaging events that have happened in the last 10 seconds.
The first term in the equation is the stamina-based contribution, which is always active, even when out of combat. There’s a helpful buff in-game to alert you to this:
My premade character has 1294 character sheet stamina, which after dividing by 250 and $\alpha(90)=67$, gives me 0.07725, or about 7.725% Resolve. It’s not clear at this point whether the tooltip is misleadingly rounding down to 7% (i.e. using floor instead of round) or whether Resolve is only affected by the stamina from gear. The Alpha servers went down as I was attempting to test this, so we’ll have to revisit it later. We’ve already been told that this will update dynamically with stamina buffs, so having Power Word: Fortitude buffed on you mid-combat will raise your Resolve.
Once you’re in combat and taking damage, the second term makes a contribution:
I’ve left this term in roughly the form Celestalon gave, even though it can obviously be simplified considerably by combining all of the constants, because this form does a better job of illustrating the behavior of the mechanic. Let’s ignore the sum for now, and just consider an isolated damage event that does $D$ damage:
$$0.25\times\frac{D}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t )}{10} \right )$$
The 0.25 just moderates the amount of Resolve you get from damaging attacks. It’s a constant multiplicative factor that they will likely tweak to achieve the desired balance between baseline (stamina-based) Resolve and dynamic (damage-based) Resolve.
The factor of $D/{\rm MaxHealth}$ means that we’re normalizing the damage by our max health. So if we have 1000 health and take an attack that deals 1000 damage (remember, this is before mitigation), this term gives us a factor of 1. Avoided auto-attacks also count here, though instead of performing an actual damage roll the game just uses the mean value of the boss’s auto-attack damage. Again, nothing particularly complicated here, it just makes Resolve depend on the percentage of your health the attack would have removed rather than the raw damage amount. Also note that we’ve been told that dynamic health effects from temporary multipliers (e.g. Last Stand) aren’t included here, so we’re not punished for using temporary health-increasing cooldowns.
The term in parentheses is the most important part, though. In the instant the attack lands, $\Delta t=0$ and the term in parentheses evaluates to $2(10-0)/10 = 2.$ So that attack dealing 1000 damage to our 1000-health tank would give $0.25\times 1 \times 2 = 0.5,$ or 50% Resolve.
However, one second later, $\Delta t = 1$, so the term in parentheses is only $2(10-1)/10 = 1.8$, and the amount of resolve it grants is reduced to 45%. The amount of Resolve granted continues to linearly decrease as time passes, and by the time ten seconds have elapsed it’s reduced to zero. Each attack is treated independently, so to get our total Resolve from all damage taken we just have to add up the Resolve granted by every attack we’ve taken, hence the sum in my equation.
You may note that the time-average of the term in parentheses is 1, which is how we get the advertised “averages to ~Damage/MaxHealth” that Celestalon mentioned in the post. In that regard, he’s specifically referring to just the part within the sum, not the constant factor of 0.25 outside of it. So in total, your average Resolve contribution from damage is 25% of Damage/MaxHealth.
Comparing to Vengeance
Mathematically speaking, there’s a world of difference between Resolve and Vengeance. First and foremost is the part we already knew: Resolve doesn’t grant any offensive benefit. We’ve talked about that a lot before, though, so it’s not territory worth re-treading.
Even in the defensive component though, there are major differences. Vengeance’s difference equation, if solved analytically, gives solutions that are exponentials. In other words, provided you were continuously taking damage (such that it didn’t fall off entirely), Vengeance would decay and adjust to your new damage intake rather smoothly. It also meant that damage taken at the very beginning of an encounter was still contributing some amount of Vengeance at the very end, again, assuming there was no interruption. And since it was only recalculated on a damage event, you could play some tricks with it, like taking a giant attack that gave you millions of Vengeance and then riding that wave for 20 seconds while your co-tank takes the boss.
Resolve does away with all of that. It flat-out says “look, the only thing that matters is the last 10 seconds.” The calculation doesn’t rely on a difference equation, meaning that when recalculating, it doesn’t care what your Resolve was at any time previously. And it forces a recalculation at fixed intervals, not just when you take damage. As a result, it’s much harder to game than Vengeance was.
Celestalon’s post also outlines a few other significant differences:
No more ramp-up mechanism No taunt-transfer mechanism Resolve persists through shapeshifts Resolve only affects self-healing and self-absorbs
The lack of ramp-up and taunt-transfer mechanisms may at first seem like a problem. But in practice, I don’t think we’ll miss either of them. Both of these effects served offensive (i.e. threat) and defensive purposes, and it’s pretty clear that the offensive purposes are made irrelevant by definition here since Resolve won’t affect DPS/threat. The defensive purpose they served was to make sure you had
some Vengeance to counter the boss’s first few hits, since Vengeance had a relatively slow ramp-up time but the boss’s attacks did not.
However, Resolve ramps up a
lot faster than Vengeance does. Again, this is in part thanks to the fact that it isn’t governed by a difference equation. The other part is because it only cares about the last ten seconds.
To give you a visual representation of that, here’s a plot showing both Vengeance and Resolve for a player being attacked by a boss. The tank has 100 health and the boss swings for 30 raw damage every 1.5 seconds. Vengeance is shown in arbitrary units here since we’re not interested in the exact magnitude of the effect, just in its dynamic properties. I’ve also ignored the baseline (stamina-based) contribution to Resolve for the same reason.
As a final note, while the blog post says that Resolve is recalculated every second, it seemed like it was updating closer to every half-second when I fooled with it on alpha, so these plots use 0.5-second update intervals. Changing to 1-second intervals doesn’t significantly change the results (they just look a little more fragmented).
The plot very clearly shows the 50% ramp-up mechanism and slow decay-like behavior of Vengeance. Note that while the ramp-up mechanism gets you to 50% of Vengeance’s overall value at the first hit (at t=2.5 seconds), Resolve hits this mark as soon as the second hit lands (at 4.0 seconds) despite not having
any ramp-up mechanism.
Resolve also hits its steady-state value much more quickly than Vengeance does. By definition, Resolve gets there after about 10 seconds of combat (t=12.5 seconds). But with Vengeance, it takes upwards of 30-40 seconds to even approach the steady-state value thanks to the decay effect (again, a result of the difference equation used to calculate Vengeance). Since most fights involve tank swaps more frequently than this, it meant that you were consistently getting stronger the longer you tanked a boss. This in turn helped encourage the sort of “solo-tank things that should not be solo-tanked” behavior we saw in Mists.
This plot assumes a boss who does exactly 30 damage per swing, but in real encounters the boss’s damage varies. Both Vengeance and Resolve adapt to mimic that change in the tank’s damage intake, but as you could guess, Resolve adapts much more quickly. If we allow the boss to hit for a random amount between 20 and 40 damage:
You can certainly see the similar changes in both curves, but Resolve reacts quickly to each change while Vengeance changes rather slowly.
One thing you’ve probably noticed by now is that the Resolve plot looks very jagged (in physics, we might call this a “sawtooth wave”). This happens because of the linear decay built into the formula. It peaks in the instant you take the attack – or more accurately, in the instant that Resolve is recalculated after that attack. But then every time it’s recalculated it linearly decreases by a fixed percent. If the boss swings in 1.5-second intervals, then Resolve will zig-zag between its max value and 85% of its max value in the manner shown.
The more frequently the boss attacks, the smoother that zig-zag becomes; conversely, a boss with a long swing timer will cause a larger variation in Resolve. This is apparent if we adjust the boss’s swing timer in either direction:
It’s worth noting that every plot here has a new randomly-generated sequence of attacks, so don’t be surprised that the plots don’t have the same profile as the original. The key difference is the size of the zig-zag on the Resolve curve.
I’ve also run simulations where the boss’ base damage is 50 rather than 30, but apart from the y-axis having large numbers there’s no real difference:
Note that even a raw damage of 50% is pretty conservative for a boss – heroic bosses in Siege have frequently had raw damages that were larger than the player’s health. But it’s not clear if that will still be the case with the new tanking and healing paradigm that’s been unveiled for Warlords.
If we make the assumption that raw damage will be lower, then these rough estimates give us an idea of how large an effect Resolve will be. If we guess at a 5%-10% baseline value from stamina, these plots suggest that Resolve will end up being anywhere from a 50% to 200% modifier on our healing. In other words, it has the potential to double or triple our healing output with the current tuning numbers. Of course, it’s anyone’s guess as to whether those numbers are even remotely close to what they’ll end up being by the end of beta.
Is It Fixed Yet?
If you look back over our old blog posts, the vast majority of our criticisms of Vengeance had to do with its tie-in to damage output. Those have obviously been addressed, which leaves me worrying that I’ll have nothing to rant about for the next two or three years.
But regarding everything else, I think Resolve stands a fair chance of addressing our concerns. One of the major issues with Vengeance was the sheer magnitude of the effect – you could go from having 50k AP to 600k AP on certain bosses, which meant your abilities got up to 10x more effective. Even though that’s an extreme case, I regularly noted having over 300k AP during progression bosses, a factor of around 6x improvement. Resolve looks like it’ll tamp down on that some. Reasonable bosses are unlikely to grant a multiplier larger than 2x, which will be easier to balance around.
It hasn’t been mentioned specifically in Celestalon’s post, but I think it’s a reasonable guess that they will continue to disable Resolve gains from damage that could be avoided through better play (i.e. intentionally “standing in the bad”). If so, there will be little (if any) incentive to take excess damage to get more Resolve. Our sheer AP scaling on certain effects created situations where this was a net survivability gain with Vengeance, but the lower multiplier should make that impossible with Resolve.
While I still don’t think it needs to affect anything other than active mitigation abilities, the fact that it’s a multiplier affecting everything equally rather than a flat AP boost should make it easier to keep talents with different AP coefficients balanced (Eternal Flame and Sacred Shield, specifically). And we already know that Eternal Flame is losing its Bastion of Glory interaction, another change which will facilitate making both talents acceptable choices.
All in all, I think it’s a really good system, if slightly less transparent. It’s too soon to tell whether we’ll see any unexpected problems, of course, but the mechanic doesn’t have any glaring issues that stand out upon first examination (unlike Vengeance). I still have a few lingering concerns about steady-state threat stability between tanks (ironically, due to the
removal of Vengeance), but that is the sort of thing which will become apparent fairly quickly during beta testing, and at any rate shouldn’t reflect on the performance of Resolve. |
I must use each number 2,0,1,9 (only once) to come up with an answer of 76
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
How about this
$\frac{9}{.\overline{1}} - \frac{0!}{.2} = 76$
where
$.\overline{1} = .111111\ldots$
$$-\log_{\sqrt{9}!-2}(\log(\underbrace{\sqrt{\sqrt{...\sqrt{10}}}}_\text{158 square roots})$$ The "158" is not part of the equation. Normally, you'd write down all 158 square roots.
I was doing a bit of research into factorials, and found both hyperfactorials (denoted by an $H$) and alternating factorials (denoted by an $AF$). Hopefully this answer fulfills your need.
$AF(\sqrt{9} + 1) * H(2 + 0)$
First we can take the hyperfactorial of $H(2 + 0)$.
$AF(\sqrt{9} + 1) * 4$
We'll then solve for the other pair of brackets.
$AF(4) * 4$
Now we'll take the alternating factorial.
$19 * 4$
And then some basic multiplication to get:
$76$
(2+1)! + 0! = 7, concatenate with 9 flipped over = 76.
Or:
you used 19 in your guess, so I'm assuming concatenating the original numbers is allowed: (9+2)!!!!! + 10
(9^2)-(2^2)-(1^2)-(0^2)=76
If we take the square of all the numbers and apply subtraction then we get 76.
$(((\sqrt{9})!)!!-10)*2 = 76$
Here is the answer! Finally!!!!! And thanks to all who helped in the spirit of solving a puzzle. |
CMS Collaboration,; Canelli, Florencia; Kilminster, Benjamin; Aarestad, Thea; Brzhechko, Danyyl; Caminada, Lea; De Cosa, Annapaoloa; Del Burgo, Riccardo; Donato, Silvio; Galloni, Camilla; Hreus, Tomas; Leontsinis, Stefanos; Mikuni, Vinicius Massami; Neutelings, Izaak; Rauco, Giorgia; Robmann, Peter; Salerno, Daniel; Schweiger, Korbinian; Seitz, Claudia; Takahashi, Yuta; Wertz, Sebastien; Zucchetta, Alberto; et al, (2018).
Search for dark matter produced in association with a Higgs boson decaying to $\gamma\gamma$ or $\tau^+\tau^-$ at $\sqrt s$ = 13 TeV. Journal of High Energy Physics, 09:046. Abstract
A search for dark matter particles is performed by looking for events with large transverse momentum imbalance and a recoiling Higgs boson decaying to either a pair of photons or a pair of τ leptons. The search is based on proton-proton collision data at a center-of-mass energy of 13 TeV collected at the CERN LHC in 2016 and corresponding to an integrated luminosity of 35.9 fb−1. No significant excess over the expected standard model background is observed. Upper limits at 95% confidence level are presented for the product of the production cross section and branching fraction in the context of two benchmark simplified models. For the Z′-two-Higgs-doublet model (where Z′ is a new massive boson mediator) with an intermediate heavy pseudoscalar particle of mass mA = 300 GeV and mDM = 100 GeV, the Z′ masses from 550 GeV to 1265 GeV are excluded. For a baryonic Z′ model, with mDM = 1 GeV, Z′ masses up to 615 GeV are excluded. Results are also presented for the spin-independent cross section for the dark matter-nucleon interaction as a function of the mass of the dark matter particle. This is the first search for dark matter particles produced in association with a Higgs boson decaying to two τ leptons.
Abstract
A search for dark matter particles is performed by looking for events with large transverse momentum imbalance and a recoiling Higgs boson decaying to either a pair of photons or a pair of τ leptons. The search is based on proton-proton collision data at a center-of-mass energy of 13 TeV collected at the CERN LHC in 2016 and corresponding to an integrated luminosity of 35.9 fb−1. No significant excess over the expected standard model background is observed. Upper limits at 95% confidence level are presented for the product of the production cross section and branching fraction in the context of two benchmark simplified models. For the Z′-two-Higgs-doublet model (where Z′ is a new massive boson mediator) with an intermediate heavy pseudoscalar particle of mass mA = 300 GeV and mDM = 100 GeV, the Z′ masses from 550 GeV to 1265 GeV are excluded. For a baryonic Z′ model, with mDM = 1 GeV, Z′ masses up to 615 GeV are excluded. Results are also presented for the spin-independent cross section for the dark matter-nucleon interaction as a function of the mass of the dark matter particle. This is the first search for dark matter particles produced in association with a Higgs boson decaying to two τ leptons.
Additional indexing |
Question
Starting with $EMF_2 = -M \dfrac{\Delta I_1}{\Delta t}$, show that the units of inductance are $\dfrac{\textrm{V} \cdot \textrm{s}}{\textrm{A}} = \Omega \cdot \textrm{s}$
Final Answer
Please see the dimensional analysis in the solution video.
Video Transcript
This is College Physics Answers with Shaun Dychko. We are going to show that the unit of inductance are volt second per amp or ohm second. So, we have this formula that tells us the voltage is inductance times rate of change of current and then we can solve this for
Mand say that its emf induced times the amount of time divided by change in current and we do that by multiplying both sides by delta tover delta Iand then we can look at the units we are left with here. So, this emf has unit of voltage, this time has unit of second and this current has unit of amp and this inductance here, mutual inductance has units of henry, That’s what we are going to show the unit of henry is volts second per amp and so we showed that already because that’s the unit of each of these factors in this equation for inductance and then volts per amp has units of ohm because ohms law says Vequals I Rand we can re-arrange that for Rto say that it is Vdivided by Iso that’s voltage divided by amps and so volts over amp is ohm and so we have also shown, showing that unit of henry is same as units of ohm second. |
I am taking an introductory course in complexity theory, and as an exercise, we were given the following problem.
Consider the bin packing problem, with objects of positive (rational) weights $W = \{w_1, w_2, ..., w_n\}$ ($0 < w_i \leq 1 \forall i$) that should be placed in bins of capacity 1. Imagine that you are given a "magical function" $\phi$ that, given a set $W$ of weights, gives you the minimum number $k$ of bins needed to pack the corresponding objects, i.e., $\phi(W) = k$. Let $T(n)$ denote the running time of the implementation of $\phi(W)$ and construct an algorithm that returns an optimal packing and runs in $O(f(n) + g(n)T(n))$ time, where $f(n)$ and $g(n)$ are polynomials.
First, I tried to recursively make binary partitions of $W$ by using that $\phi(W_1) + \phi(W_2) = \phi(W)$ for any partition $W_1 \cup W_2 = W$ such that $W_1$ and $W_2$ separates objects that are in different bins in an optimal solution (that is, there is an optimal solution such that no weights in $W_1$ share bins with any weight in $W_2$). When we eventually get $\phi(W_j) = 1$, we know that $W_j$ corresponds to the objects in bin $j$ in an optimal solution. However, my attempts here only yield algorithms with $g(n)$ being exponential in $n$.
Second, I tried to construct the bins one by one, using that, for all bins $B_j$, $\phi(B_j) + \phi(W \setminus B_j) = k$. Again, I have yet to find an algorithm where $g(n)$ is polynomial.
It's rather frustrating, as this seems to be a simple problem. Any suggestions on where I should begin? |
With only union and concatenation, you can't describe any infinite language. The union and concatenation can only produce finitely many strings. With only union and the Kleene star, you can't describe a language such as $L = \{ab\}$, because there's no way to concatenate an expression generating only $a$ with an expression generating only $b$. With only ...
A language is regular, by definition, if it is accepted by some DFA. (This is at least one common definition.) Can you think of a DFA accepting the language?A well-known result (that is proved in many textbooks) states that the language of a regular expression is regular. Since $a^* b^*$ is a regular expression, its language must be regular (if you believe ...
$\{a^* b^*\}$ is a regular language, since it's generated by a regular expression.The key difference between $L_* = \{a^* b^*\}$ and $L_= = \{a^n b^n\}$ is that $L_=$ requires counting the $a$'s and $b$'s to check whether there's the same number of them, whereas $L_*$ doesn't require any counting. Counting requires unbounded memory as the number grows ...
If you consider now the powers of a language $W$ you have$W^xW^y=W^{x+y}$ If you want this to be consistent over $\mathbb N_0$,i.e. the non-negative integers, you have to define$W^0=\{\epsilon\}$. If you took it to be $\emptyset$ you would have$W^x=W^{x+0}=W^xW^0=W^x\emptyset=\emptyset$ including, among others,for $x=1$. Thus we would have $W^1=W=\...
A perhaps more interesting question is that of star height. The other answer mentions that if you can't use star, then you can only generate finite languages. What if you are not allowed to nest stars (so something like $(a^*b^*c)^*$ is not allowed)? What if you are only allowed to nest stars two levels deep? $d$ levels deep? It turns out that for every $d$ ...
Short answer: Yes. $10100$ is included.Long answer:Let's go through how this particular set is constructed. First, the expression$$10^n$$(in this context) denotes a single string consisting of a $1$ followed by $n$ $0$s (so the string depends on one variable $n$). Then, the set$$\{10^n \mid n \ge 1\}$$denotes the set of strings $\{10, 100, 1000, ...
As chi pointed out in the comment, since $\varepsilon\in L^*$ and $\varepsilon$ may not belong to the new grammar, so adding $S\rightarrow SS$ does not always generate $L^*$. It makes more sense to ask whether it generates $L^+$, so the following answer focuses on $L^+$.General FormConsider the following grammar:\begin{align}S&\rightarrow aSb\\S&...
$L^2 = LL = \{uv | u,v\in L\}$. In words, $L^2$ is a set of all strings that formed by concatenation of all strings from $L$. For instance, if $u=a^kb^k \in L$ and $v=a^tb^t \in L$ then $uv = a^kb^ka^tb^t$. That's why $L^2 =\{ a^{n_1}b^{n_1}a^{n_2}b^{n_2}\}$ for all $n_1, n_2 \geq 1$.Similarly $L^3, L^4,\dots$Example with a finite set: $L=\{ab, cd\}$. ...
A report by Aceto contains an axiomatization of the equational theory of regular expressions, called "classical" by Conway (Table 1 on page 5). One of the axioms is very similar to what you're after: $(a+b)^* = (a^*b)^*a^*$. Using these axioms, we can derive your identity as follows (using associativity (C3, C10) for free):$$\begin{align*}(a+b)^* &= (...
Let me solve a simpler question first. Suppose we'd like to knowIf $C$ is context-free, must $F(C)=\{w: ww \in C\}$ be context-free?The answer is no: $F(\{a^i b^i c^j a^j b^k c^k\})=\{a^i b^i c^i\}$.As for the original problem, consider the context-free language that contains words of the form $\{a^i b^i c^j d a^j b^k c^k d\}$ or those which don't ...
Yes, it is true. But the argument to support the conjecture is incorrect.The correct proof is as follows: $L_1 \cap L_2 \subseteq L_1 $ and therefore $(L_1 \cap L_2)^* \subseteq L_1^*$. (argument: if $A \subseteq B$ then $A^* \subseteq B^*$).Similarly $(L_1 \cap L_2)^* \subseteq L_2^*$.Therefore $(L_1 \cap L_2)^* \subseteq (L_1^* \cap L_2^*)$ (argument:...
Another interesting case arises by allowing complementation as a possible operation.Using union, concatenation and complement (but no star), one can obtain the language $A^*$ as the complement of the empty language. One can also obtain languages like $(ab)^*$ and $(a(ab)^*b)^*$ (not so easy to see if you don't know the trick), but there is no way to obtain $...
Your confusion seems to stem from the incorrect assumption that if a language $L$ has a certain property, then $L^*$ must also have that property. Consider this simpler example. Over the 1-symbol alphabet $\{a\}$ consider the property$L$ contains no words of length more than threeThen one language with this property is $L=\{a, aaa\}$. Then by the ...
$L^*$ always exists. ${}^*$ is an operation on languages, just like ${}^2$ is an operation on numbers – whenever you have a number $x$, there is a number $x^2$ and, similarly, whenever you have a language $L$, there is a language $L^*$. Specifically, $L^*$ is the language of all strings that can be made by concatenating zero ...
Clearly not.Let $A=\{a\}$ and $B=\{aa\}$.Now,$A\cap B = \emptyset$so$(A\cap B)^* = \{\epsilon\}$but$A^*\cap B^*=B^*=\{a^{2i} : i \in \mathbb{N}\}$(all strings consisting of an even number of $a$).
If $\varepsilon \in L$, then necessarily $\varepsilon \in L^+$ (and the converse as well). This is because $L$ itself is contained in $L^+$ and $L^+$ is defined as the union over the powers $L^i$ of $L$ for $i \in \mathbb{N}_+$. Note $L^+$ is not defined as $L^\ast \setminus \{ \varepsilon \}$; this is a common mistake.Hence, $\varepsilon \not\in L^+$ ...
Let $A$ be the language consisting of the following words:$$(u,v)|\Sigma^*|(v,w),$$where $u,v,w$ are numbers encoded in binary. This language is in $\mathsf{L}$.Given a directed graph $G$, we can encode it as a list of edges in the following way:$$(x_1,y_1)(x_1,y_1)|(x_2,y_2)(x_2,y_2)|\cdots|(x_m,y_m)(x_m,y_m)$$where $x_i,y_i$ are vertex numbers ...
The concatenation of zero words from $\emptyset$ is the empty word $\epsilon$, so $\epsilon \in \emptyset^*$. More generally, for a language $L$, the Kleene star $L^*$ consists of all concatenation of any number of words from $L$, any number including zero words.
You can always use the current state of the automaton to remember the last three characters you've seen. Now, you can implement two phases. In the first phase, you're happy if you're ever in the situation where the last three characters were $010$. In the second phase, you become unhappy if you ever see $010$ again.
Simply saying that $L$ has the alphabet $\{x, y\}$ and that no string in $L$ contains the substring $xx$ doesn't define a single language - there are infinite such languages. What you call expressing $L = \{y, xy, yx\}$ is actually defining such a language $L$ with three words that happens to have the property of no strings containing $xx$. $L^*$ doesn't ...
Unless parentheses say otherwise, the customary precedence is ${}^*$ (like exponentiation), then concatenation (like multiplication), then union (like addition).Your second example is the concatenation of starred expressions, so it does contain the empty string.
What you have written (i.e., $\{ \varepsilon, 01, 0011, 000111, \ldots \}$) is simply $A$ itself.(Assuming $A^\ast$ is the Kleene star operation) you cannot prove $A = A^\ast$ because it is not correct.The Kleene star $A^\ast$ is defined as the union of the powers $A^i$, where $A^0 = \{ \varepsilon \}$ and $A^{i+1} = A^i \cdot A = \{ w \cdot w' \mid w \...
I figured out the answer and hope this will help anyone with similar confusion in the future.Since a pushdown automaton is essentially a finite automaton with an extra component called the stack which is used for storage, if we are presented with a regular language like the one given above, we simply need to create a PDA that does not use the stack. This ...
I assume that in your definition $a \leq b$ iff $a + b = b$.First, note that if $a \leq b$ and $b \leq a$, then $a = a + b = b + a = b$.Therefore, in order to show that $a = b$, it is sufficient to show that $a \leq b$ and $b \leq a$.Now, you want to show that $(a + b)^\ast = (a + ab + b)^\ast$. As explained in the previous paragraph, it is sufficient ...
Your language could be simplified as follows, using $\emptyset^* =\{\epsilon\}$:$$\begin{align*}L(\emptyset\emptyset^*+\emptyset)&=L( \emptyset . \{\epsilon\} + \emptyset) \\&=L(\emptyset +\emptyset) & (\emptyset.\{\epsilon\}=\emptyset) \\&=L(\emptyset) & (\emptyset + \emptyset = \emptyset)\end{align*}$$So the language L ...
Here's a hint; your language is, more or less, the concatenation of three languages: the strings not containing 010, the language consisting of the single string 010, and the strings not containing 010.Can you make a DFA which accepts all and only those strings not containing 010, perhaps as the complement of something?When I said "more or less" above, I ...
Given a language $L$, let $L_0 = \{\epsilon\}$ and, for $i\geq 1$, let $L_i = \{w_1\circ \dots\circ w_i \mid w_j\in L \text{ for each } j\}$, where $\circ$ denotes concatenation. Then the Kleene closure of $L$ is the language $L^* = \bigcup_{i\geq 0} L_i$. |
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Render Latex formula to image Notice: This plugin is unmaintained and availablefor adoption. Description
This plugin allows you to render a mathematical formula as an image using Latex, dvips, and ImageMagick.
The images are named based on a SHA1 hash of the formula and parameters, so that images don't have to be regenerated every time they are viewed.
To create the image from the formula text, the following is performed:
Create .tex file in temporary directory. Run latex on .tex file to produce .dvi file. Run dvips on .dvi file to produce .eps file. Run convert on .eps file to produce .jpg file.
By default, the Latex code is rendered in math mode. This allows for quick math expressions. There is also a built-in macro called "mat" which is used for formatting matrices, like this:
[[formula(\mat{R'\\G'\\B'} = \mathbf{A} \mat{R\\G\\B} + \mathbf{x})]]
which will produce the following formula:
For more complicated usage, use the {{{ ... }}} format and the "nomode" command to disable the display-math mode:
{{{ #!formula #nomode A more complicated equation should be typeset in {\em displayed math\/} mode, like this: \[ z \left( 1 \ +\ \sqrt{\omega_{i+1} + \zeta -\frac{x+1}{\Theta +1} y + 1} \ \right) \ \ \ =\ \ \ 1 \] }}}
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Recent Changes |
My other answer addressed the question in the title. Here I will try to address the two queries at the end of the OP question. @ybeltukov was the first to point out the key issue, which is that
ContourPlot works only when it finds sample points where the values of the function are greater than the contour level and points where they are less than it.
ContourPlot, like other plot functions, starts with a 2D rectangular grid of sample points determined by
PlotPoints, which is then recursively subdivided up to
MaxRecursion times. Where or when a recursive subdivision occurs depends on the function and the contour levels. Some discussion of how the subdivision works can be found in the answers to Specific initial sample points for 3D plots.
I hope to show that the answer to the first question about why increasing
PlotPoints results in a worse graph is mainly due to the alignment of the sample points with a certain region, the extremely small region where the function is less than
0.001. The region is so small that misalignment is possible even when
PlotPoints is set well over
100.
Auxiliary functions
There are a couple of helper functions (
cpGrid,
cpShow) below that create
ContourPlots of a function that show how the plot domain is subdivided together with the sample points at which the values of the function are less than the contour level. (See "Code dump" below at the end.)
The OP's functions, which I will call
f1 and
f2:
ClearAll[f1, f2];
SetAttributes[f1, Listable];
SetAttributes[f2, Listable];
f1[x_, y_] := (Cos[x] + Cos[y]) Exp[I x y] Exp[x/2];
f2[a_, a0_, k_, K0_] :=
a^2 Sech[(a a0)/2]^2 (-2 I (1 + E^(2 I a0 k)) k +
a (-1 + E^(2 I a0 k)) Tanh[(a a0)/2]) +
2 k (I E^(I a0 k) ((a - k) (a + k) Cos[a0 k] + (a^2 + k^2) Cos[
a0 K0]) + a (-1 + E^(2 I a0 k)) k Tanh[(a a0)/2]);
ContourPlot sampling
The table of plots below show
PlotPoints settings of 3, 4, 5 points (rows) and
MaxRecursion settings of 0, 1, 2 (columns). The initial sample points lie on a rectangular grid, but the actual subdivision is made of triangles with a SW-NE bias. This is shown below by the gray
Mesh lines. The red points are sample points where the value of the function
Abs@f1 lies below the level
0.001. The other intersections are sample points where the value of the function is above the level. Where a red point is next to a gray intersection, one may observe an increase in the number of subdivisions.
{cp, samp} = cpGrid[Abs@f1@## &, {x, 0, 4 Pi}, {y, 0, 4 Pi}, 0.001, {3, 4, 5}];
cpShow[cp, samp, Abs@f1@## &, 0.001] // GraphicsGrid
There are several things to observe. The zeroes of
f1 lies on lines where $x \pm y$ is a multiple of $\pi$. For the domain $[0,4\pi]\times[0,4\pi]$, the sample points will happen to lie on these lines when
PlotPoints is of the form $4n+1$. So we see red points for 5 points with no subdivision, but not initially in the other plots. Below are some more examples showing the contour plots with the initial sample points (
MaxRecursion -> 0) that exhibit the $4n+1$ pattern. Compare 100 vs. 101 and 25 vs. 101.
GraphicsGrid@Table[
With[{tolerance = 0.001, pp = p + 4 j},
Show[ContourPlot[
Abs[f1[x, y]] == tolerance , {x, 0, 4 Pi}, {y, 0, 4 Pi},
PlotPoints -> pp, MaxRecursion -> 0, PlotRange -> All,
PlotLabel -> pp], PlotRange -> {{0, 4 Pi}, {0, 4 Pi}}]
],
{j, {1, 2, 20}}, {p, 20, 23}
]
Now a single subdivision of every triangle would transform a grid of $n$ by $n$ points into one with $2n-1$ plot points in each direction, which, if repeated, would eventually bring all grids into the form of $4n+1$ points on a side; however depending on the function and contour level, some triangles are subdivided twice and some only once. For example, there is a pair at approximately
{5, 8} in the plot for 3 points, 1 subdivision, whose shared hypotenuse was not divided even though the midpoint is a zero of the function. I do not know how
ContourPlot decides whether to subdivide, but the lack of a division here makes a gap in the contour. (One can check that the contour does not connect these points. A mesh line connecting two red points seems to be interpreted as being connected by a region in which the value of the function is less than
0.001.) In the next step where
MaxRecursion increases from 1 to 2, it is subdivided and the gap is closed. One observes a similar gap in the plot for 4 points, 2 subdivisions around the coordinates
{8.5, 5}. That gap disappears with another increase to
MaxRecursion -> 3. In fact, I think that whatever the number of initial
PlotPoints, the contour will be drawn completely with
MaxRecursion -> 3 for the symmetric domain $[0,4\pi]\times[0,4\pi]$, but it depends precisely because the special symmetry leads to alignment of the sample points with the zeroes of
f1. For an odd number of initial
PlotPoints, one only needs
MaxRecursion -> 2 to get the whole contour, and for a number of the form $4n+1$, no recursion is necessary.
The dependence on the symmetry can be seen by perturbing the domain slightly:
ContourPlot[
Abs[f1[x, y]] == 0.001, {x, 0, 4 \[Pi] + #}, {y, 0, 4 \[Pi] + #},
MaxRecursion -> 3] & /@ {0., 0.001} // GraphicsRow
When the domain is extended to
16 Pi, the number of plot points needs to be of the form $16n+1$. The following is perhaps the simplest way to get the complete contour (
MaxRecursion -> 1 is needed to overcome the SW-NE bias and connect the NW-SE contours):
ContourPlot[Abs@f1[x, y] == 0.001, {x, 0, 16 Pi}, {y, 0, 16 Pi},
PlotPoints -> 17, MaxRecursion -> 1]
Thus the alignment of the grid with the region in $xy$ plane where
Abs[f1[x, y]] < 0.001 is the key to the somewhat odd dependence on the number of
PlotPoints.
The thinness of the region in which sample points must land
A single contour level divides the plane into two regions, one over which the value of the function is greater than the level and one over which the value of the function is less, in addition to the level set itself. Normally, if the function is not locally constant, the level set will be a curve. In the OP's examples, we have another sort of edge case in which desired contour curve is the minimum of the function.
When one of the two regions is extremely small, it is unlikely that many sample points will fall into it except "by accident" so to speak. This is what is happening with both of the OP's functions.
To give an illustration we can see, let's use a sequence of values for the level, say,
1.,
0.1,
0.01. Despite the images, the true regions are connected, but the width of the region was too narrow to contain enough (or any) points to define a boundary.
GraphicsRow @ RegionPlot[
Abs[f1[x, y]] <= #, {x, 0, 4 Pi}, {y, 0, 4 Pi},
PlotPoints -> 40, MaxRecursion -> 3] & /@ {1, 0.1, 0.01}]
With[{a0 = 10, a = 1.4},
GraphicsRow @ RegionPlot[Abs[f2[a, a0, y, x]] <= #,
{x, -2 Pi/a0, 2 Pi/a0}, {y, 0, 2},
PlotPoints -> 40, MaxRecursion -> 3] & /@ {1, 0.1, 0.01}]]
For the second function, I think it proves to be not as well-behaved as at first glance. It gets quite steep in half of the domain, which accounts for the narrowing of the region as the
y coordinate increases. (Note: My plot turns out to have a somewhat greater
PlotRange than that of the OP.)
With[{a0 = 10, a = 1.4},
Plot3D[Abs[f2[a, a0, y, x]], {x, -2 \[Pi]/a0, 2 \[Pi]/a0}, {y, 0, 2}, PlotPoints -> 100]]
)
Summary
I think the behavior of
ContourPlot can be accounted in both cases by the extremely small area in the plot domain for which $f < 0.001$. In the case of the OP's first function, there is accidental and sporadic alignment of the zero set of $f1$ with the sample points, which accounts for why a plot with much fewer
PlotPoints can look better than one with more.
Code dump
For investigating contour plots.
cpGrid[f_, xdom_, ydom_, level_, pp0_List, mr0_: 2] :=
Module[{cp, samp},
cp = samp = Table[{}, {pp, Length@pp0}, {mr, 0, mr0}];
Do[{cp[[pp, 1 + mr]] = First@#,
samp[[pp, 1 + mr]] = Hold @@ Last@#} &@
Reap @ ContourPlot[
f[x, y],
{x, xdom[[-2]], xdom[[-1]]}, {y, ydom[[-2]], ydom[[-1]]},
Contours -> {level}, PlotPoints -> pp0[[pp]],
MaxRecursion -> mr, Mesh -> All, EvaluationMonitor :> Sow[{x, y}],
PlotLabel -> Row[{pp0[[pp]], " points, ", mr, " subdivisions"}]],
{pp, Length@pp0}, {mr, 0, mr0}];
{cp, samp}];
ClearAll[cpShow];
SetAttributes[cpShow, Listable];
cpShow[cp_, Hold[samp_], f_, level_] :=
Show[cp,
Graphics[{Red, PointSize[0.02], Point[Pick[samp, UnitStep[f @@@ samp - level], 0]]}]
] |
It took my probably longer than necessary, but now I understand why it sums up to 200%. An example illustrated it to me. Let me enlighten you with a nuclear power plant. But first, the answer:
Let us assume that $^{235}$U would only split into 1 specific isotope. So when it splits, 2 nuclei are created. $2=200\%$
So, the fission yield does not state the percentage a certain isotope represents of possible splits of Uranium, it rather counts
how many nuclei you would have at the end, if you split 100 Uranium nuclei!
This is a helpful information, e.g. if you liked to know the production rate of $^{131}$ I (iodine) of a nuclear power plant. Let us assume we have a powerful power plant with an energy production of 1 GW, and an efficiency of 20%. Let us further assume that the energy released in the fission of one $^{235}$U atom is 200 MeV. The fission yield of $^{131}$ I is given in the table of nuclides: $Y_{^{131}I}=2.885%$, as well as the half-life: $T_{1/2}=8.02$ d. Let us sum up everything we know in tabular form:
$P_{el}=1$ GW $\eta=0.2$ $\delta E=200$ MeV $Y_{^{131}I}=2.885\%$ $T_{1/2}=8.02$ d $\Rightarrow\lambda=1\cdot 10^{-6}$ s$^{-1}$
From the electrical power production $P_{el}$ and the efficiency $\eta$ we can determine the thermal power production $P_{th}=\frac{P_{el}}{\eta}=5$ GW.
We know that one decay of uranium yields 200 MeV, therefore the number of $^{235}$U has to change according to: $\frac{\mathrm{d}N_{^{235}U}}{\mathrm{d}t}=\frac{P_{th}}{\delta E}$.
With one decay of $^{235}$U, $Y_{^{131}I}=2.885\%$ are created (on average), which means the production of $^{131}$I per second is given by
$$P_{^{131}I} = Y_{^{131}I}\frac{\mathrm{d}N_{^{235}U}}{\mathrm{d}t}.$$
To sum up: The fission yield states how many daughter nuclei are produced under the decay of 100 $^{235}$U isotopes. As the mother splits into two different daughters, the total number of the daughters sum up to 200. This is then broken down to percentage values, called the fission yield. This can be used to calculate the production of daughter nuclides. |
The other answers to the effect that one needs big optics to see fine detail are indeed true for are true for conventional imaging optics that sense the electromagnetic
farfield or radiative field i.e. that whose Fourier component at frequency $\omega$ can be represented as a linear superposition of plane waves with real-valued wave-vectors $(k_x,\,k_y,\,k_z)$ with $k_x^2+k_y^2+k_z^2 = k^2 = \omega^2/c^2$. This is the kind of field which the Abbe diffraction limit applies to and limits "eyes" like our own comprising imaging optics and retinas, or even compound eyes like those of an ant.
However, this is not the whole electromagnetic field: very near to the objects that interact with it, the electromagnetic field includes nearfield or
evanenescent field components. These are generalised plane waves for which:
The component of the wavevector in some direction $k_\parallel$ is
greater than the wavenumber $k$ and can thus encode spatial variations potentially much smaller than a wavelength;
The component of the wavevector $k_\perp$ orthogonal to this direction must therefore be
imaginary, so that $k_\parallel^2 + k_\perp^2 = k^2$ can be fulfilled.
So such fields decay exponentially with distance from the disturbance to the electromagnetic field that begat them and thus cannot normally contribute to an image formed by an imaging system.
However, if you can bring your image sensors near enough to the disturbance, you can still register the detail encoded in the finer-than-wavelength evanescent components. This is the principle of the Scanning Nearfield Optical Microscope.
The near field optical microscope sensor can be extremely small indeed, so that a bacterium sized lifeform could register below-wavelength detail in the World around it with receptors built of a few molecules as long as the lifeform were near enough to the detail in question. Note that, when $k_\parallel > k$ that the fields decay like $exp(-\sqrt{k_\parallel^2-k^2} z)$ with rising distance $z$ from their sources. So there is a tradeoff between how much finer than a wavelength we can see with such a sensor, and how near to the source we need to be to see it. If we want to see features one tenth of the wavelength of seeable light, then $k\approx 12{\rm \mu m^{-1}}$ and $k_\parallel \approx 120{\rm \mu m^{-1}}$, so that the amplitude of the nearfield decays by a factor of $e$ for each hundredth of a wavelength distant from the source the detector is. Thus we lose about 10dB signal to noise ratio for every hundredth of a wavelength distance that separates the detector and source. So to sense such fine detail (50nm structures) from a micron away would need extremely strong light sources, so that the detectors would have a very clean signal.
Of course, the above is an extreme example, but if you're a bacterium sized lifeform directly sensing the field using a finely spaced array of molecular sensors, you may well be able to "see" below-wavelength features of the World in your immediate neighbourhood. Moreover, it is possible to conceive of a tiny creature "feeling" its neighbourhood using molecular atomic force microscopes.
So, yes, if you include all physics and heed the proviso that you must get up really close to the sensed objects, it would be possible for a bacterium sized lifeform to see below-wavelength detail in its immediate neighbourhood, maybe even individual atoms if we include atomic force sensing.
Of course, packing all the signal processing "brain" into the lifeform needed to understand this information might be another matter altogether. |
Event detail On Gaussian-width gradient complexity and mean-field behavior of interacting particle systems and random graphs
Seminar | November 1 | 3:10-4 p.m. | 1011 Evans Hall
Ronen Eldan, Weizmann Institute of Science
The motivating question for this talk is: What does a sparse Erd\"os-R\'enyi random graph, conditioned to have twice the number of triangles than the expected number, typically look like? Motivated by this question, In 2014, Chatterjee and Dembo introduced a framework for obtaining Large Deviation Principles (LDP) for nonlinear functions of Bernoulli random variables (this followed an earlier work of Chatterjee-Varadhan which used limit graph theory to answer this question in the dense regime). The aforementioned framework relies on a notion of "low complexity" functions on the discrete cube, defined in terms of the covering numbers of their gradient. The central lemma used in their proof provides a method of estimating the log-normalizing constant $\log \sum_{x \in \{-1,1\}^n} e^{f(x)}$ by a corresponding mean-fieldfunctional.
In this talk, we will introduce a new notion of complexity for measures on the discrete cube, namely the mean-width of the gradient of the log-density. We prove a general structure theorem for such measures which goes beyond the discrete cube. In particular, we show that a measure $\nu$ attaining low complexity (with no extra smoothness assumptions needed) are close to a product measure in the following sense: there exists a measure $\tilde \nu$ a small "tilt" of $\nu$ in the sense that their log-densities differ by a linear function with small slope, such that $\tilde \nu$ is close to a product measure in transportation distance. An easy corollary of our result is a strengthening of the framework of Chatterjee-Dembo, which in particular simplifies the derivation of LDPs for subgraph counts, and improves the attained bounds. We will demonstrate how our framework can be used to study the behavior of low-complexity measures beyond the approximation of the partition function. As an example application, we prove that exponential random graphs behave roughly like mixtures of stochastic block models.
510-000-0000 |
... The sphere is placed over a hole, of radius a, in the tank bottom. Develop a general expression for the range of specific gravities for which the sphere will float to the surface.
I've attached the diagram for the problem.
As an attempt at the solution, I've summed the forces on the sphere, or the buoyant force $F_b$ - the weight of the sphere. Buoyant force is equal to the weight of the displaced water by the sphere, $\frac{4}{3}\pi R^3\rho_{water}\times32\frac{ft}{s^2}$. The weight of the sphere is $\frac{4}{3}\pi R^3\rho_{sphere}\times32\frac{ft}{s^2}$. Based on force equation, in order for the ball to float the buoyant force would have to be greater than the weight of the sphere, or $F_b-W>0$. After some algebra you can say that the specific gravity of the sphere has to be $<1$, which seems too simple to be the actual answer.
I'm sure I'd have to take the hole that the sphere is resting atop into consideration, but I'm not sure how to go about that. My only ideas on how to include that would be to include the force from the ambient pressure on the bottom of the ball pushing upward in the force equation, but even if that effects the specific gravity at which the ball will float, once it started to float and that force was no longer there, the ball would simply sink again, unless $S.G.<1$. My other thought was that I would have to consider the volume of the sphere not submerged due to the hole and thus not contributing to the buoyant force, but I have no idea how to calculate what that volume would be.
Is my first solution correct, that $S.G.<1$ would cause the ball to float? That again feels too simple for the given parameters.
Any direction would be appreciated. |
When sound waves diffract through a single slit, do they produce an interference pattern which is mathematically identical to that of light waves in the corresponding experiment?
The answer is the two are very close, but there are theoretical differences. It is VERY hard to construct actual scenarios where linear acoustic diffraction differs substantially from the diffraction of light.
As noted already, linear acoustic waves are scalar pressure fields, whereas electromagnetism is a vector phenomenon,
i.e. there is a polarisation state light that does not apply to acoustic waves.
(
Likeness) The Cartesian components of the electromagnetic field vectors, as well as the Cartesian components of the potential four vector all fulfill the D'Alembert wave equation (Helmholtz equation in the time harmonic case). The propagation for linear acoustic waves is also the D'Alembert wave equation;
(
Difference) In co-ordinate systems other than Cartesian, the components of the electromagnetic field do notin general fulfill the D'Alembert wave equation.
(
Likeness) For many kinds of problem, 2. is not a big difference, either because light either has a predominant polarisation in a constant (non spatially varying) direction (so the only significant component is a Cartesian one), or because it is a classical probabalistic mixture of pure polarisation states with such predominant, non spatially varying direction polarisation. This idea can be generalised to any pure polarisation state: there are orthonormal (complex) transformations on the co-ordinate axes which keep the D'Alembert wave equation - see for example the Riemann-Silberstein vectors, whose Cartesian components fulfill the D'Alembert wave equation. Instead of xand ycomponents, they have left and right circular components.
(
Difference) The intensity of the sound field is the square of the scalar field, i.e.the square of a field fulfilling the D'Alembert wave equation. The intensity of an electromagnetic field is the inner product between the Poynting vector and the surface through which the energy flux is being defined. It can also be construed as the energy density $\frac{1}{2}\epsilon\,|\vec{E}^2| + \frac{1}{2}\mu\,|\vec{H}^2|$ times the speed of light in the medium in question. Thus it is a more general quadratic function of several different scalar fields, each fulfilling the D'Alembert wave equation.
(
Likeness) Again in many situations, there is one predominant direction of the electromagnetic field, thus one Cartesian component seems to hold sway and so we again get a square of a single scalar field fulfilling the D'Alembert wave equation as the approximate intensity. Thus for both fields the square of a scalar is the intensity- this holds exactly for acoustics, but only approximately for electromagnetism;
(
Difference) The main difference to my mind is the boundary conditions. For acoustics, the pressure and normal velocity component are continuous across interfaces. These are simple scalars. For the EM field, the polarisations bear critically on the behaviour at interfaces. The acoustic reflexion and transmission co-efficients bear strong likenesses to the Fresnel equations for light, but there is only onereflexion and onetransmission co-efficient. For light, these two co-efficients vary widely, depending on the incident polarisation, as described by the Fresnel equations.
(
Likeness) The Snell law, Eikonal equation and raytracing apply equally in both cases, being valid approximations under exactly analogous conditions.
Leaving the vector nature of light aside: a great deal of optical diffraction theory is scalar theory. Scalar optical diffraction theory is valid especially in the paraxial (fields comprising only a narrow spread of wavevector directions) case, and when one Cartesian component of a field vector is predominant.
There is no difference between scalar optical diffraction theory, and the theory of linear acoustic diffraction
When sound waves diffract through a single slit, do they produce an interference pattern which is mathematically identical to that of light waves in the corresponding experiment?
I think the answer is almost, but not quite. It's a little complicated.
First off, there's the question of whether you're talking about the far field (Fraunhofer diffraction) or the near field (Fresnel diffraction). In the Fraunhofer approximation, we can take all the waves reaching a given point to be plane waves arriving from the same direction. But in the near field, the waves being superposed can arrive from different directions, so depending on the polarization, you can get nontrivial vector addition of the fields. This causes an actual change in the shape of the diffraction pattern, not just a renormalization.
Huygens' principle doesn't say anything about polarization, so if you believed in Huygens' principle, then you would think that all results in diffraction would be independent of polarization. But Huygens' principle is an approximation. One example where it fails is in the case where the slit is small compared to the wavelength: Diffraction by small holes From Huygens' principle, you'd expect that 100% of the power impinging on the slit would emerge as a diffracted wave. But in reality the transmission in this limit is proportional to $(a/\lambda)^n$, where $a$ is the width of the slit and the exponent $n$ depends on the multipolarity of the wave. As explained in an answer to the linked question, we have $n=4$ for light waves, but $n=2$ for sound. This causes a difference in normalization in the two cases.
Finally, a physically realistic diffracting object will interact with the polarization of the wave. For example, a conducting metal slit will interact differently with a wave depending on the direction of the wave's polarization. These are basically the same phenomena you get in a polarizing filter. If the slit has a finite thickness, it can act like a waveguide, and waveguides are sensitive to polarization.
These are three independent, well known, noncontroversial reasons why the diffraction can and will be different for light than for sound. Having said all that, I think that under most circumstances the two diffraction patterns would look almost the same. For $\lambda \ll a$, I think the normalization would even be the same, in the sense that the unitless transmission factors would both be approximately 1. It would be an interesting project to try to detect some of the differences in the lab. I don't think it would be easy at all.
When sound waves diffract through a single slit, do they produce an interference pattern which is mathematically identical to that of light waves ...?
The answer is no. The diffraction pattern of sound and of light behind a slit is not similar and so the mathematical description is not complete. This is because the real patterns on an observation screen (or an imaginery line for sound waves) are never the same for sound and light. The fringes from the light diffraction are not moving patterns. More, the reflection of these light (from the observation screen) does not interact with the incomming waves. The diffracted water (or sound) waves don't stand still (picture), they are moving to the left and to the right if one look on them at the position there is the screen for light fringes. If one put a screen in the reflected waves interact with the waves from the slit and produce more complicated patterns.
We have to state that the mathematical model describes for water waves a instantaneous situation. The position of the spherical water waves behind a slit depends from the postion of the wave in front of the slit at some time. Opposite situation for light, the fringes from light stand still.
As pointed out in the answer from Ben Crowell, the slit can work as a waveguide. As pointed out from Floris, the edges could "electrically conducting at the frequency of interest for the electromagnetic case". In both cases the light interacts with the slits material. This do the light with every edge too, fringes appear behind every edge. As I pointed out here, "light is a lot of photons, and if you take monochromatic light, a lot of same energy photons and they interact with the electric field of the electrons of the edge. The last point one has to imagine is that the field they "built" together is quatitized. The fringes on the observers screen show how the field is quantized."
So a mathematical model for sound or water can predict instantaneous distances between maximum and minimum points of the sperical wave behind a slit. And a matematical model for light can predict the positions of maxima and minima for fringes on an observation screen. But physically that are total different phenomena.
If you can ignore the effects of sounds attenuation in air, I believe the answer is "yes". Note that when the slits are electrically conducting at the frequency of interest for the electromagnetic case, then certain effects relating to polarization come into play as explained in Ben Crowell's answer, and the sound and light cases become more and more different as the slit dimension decreases. I would argue that since there is no equivalent mechanism (conductivity along the edge of the slit) for sound, you cannot talk of an "equivalent experiment" I that case - and that the comparison must be made with a non-conductive slit. |
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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 ... |
Organic Compounds Containing Nitrogen Physical and Chemical Properties of Amines
Hoffmann Bromamide Degradation Reaction Best method for removing of carbon: When amide is treated with Bromine in the presence of strong base gives amine which has 1 carbon less than that of substrate that's why reaction is called degradation reaction
Note:- The above method is useful in various conversions Toluene — Aniline PHYSICAL PROPERTIES Aniline is soluble in dil. HCl because it can form aniline ions. Amines are in general colorless due to interaction with atm. O 2 they turns to brown colour. Basic nature of amines:- When amines are treated with an acid forms salts, this inferences the basic nature of
Basic strength can be understood with reference of base dissociation constant k
b or power of base dissociation constant P kb. According to kinetics:- \tt R-NH_{2} + H_{2}O \rightleftharpoons R - NH_{3}^{\oplus} + OH^{\ominus} \tt K_{equi} = \frac{\left\{R-NH_{3}^{\oplus}\right\}\left\{OH^{-}\right\}}{\left\{R-NH_{2}\right\}\left\{H_{2}O\right\}} \tt K_{equi} \left\{H_{2}O\right\} = \frac{\left\{R-NH_{3}^{\oplus}\right\}\left\{OH^{-}\right\}}{\left\{R-NH_{2}\right\}}
Higher the k
b, ↑ the basic strength P kb = −log k b lower the Pk b higher the basic nature.
In gaseous phase: -
i.e, NH
3 < 1° < 2° < 3°.
In case of aq. phase.
Basic strength influenced by following 3 factors:- Solvation, Inductive effect, Steric effect.
All those three effects makes 2° amine more basic than either 1° or 3°
In case of methylamines :- [Solvation dominates + I]
(CH 3) 2NH > CH 3NH 2 > (CH 3) 3 N > NH 3
In case of ethyl or any other aliphatic amines:-
+I effect dominates solvation.
All aliphatic amines are more basic than ammonia and aromatic amines.
Isomeric amines alicyclic amines are more basic than aliphatic Since +I effect more in case alicyclic than aliphatic. In case of aromatic amines the lone pair of nitrogen is always engaged with aryl group through +R effect as aryl amines are less basic than ammonia. AFFECT OF GROUP TOWARDS BASIC STRENGTH OF ARYL AMINES ORTHO EFFECT: Whether it is e − releasing or withdrawing group present at ortho to NH 4 except (ortho amino phenol) remaining all ortho substituted amines are less basic than aniline. e Θ releasing group present at para position increases the e Θ density on nitrogen and makes it more basic than aniline. If e Θ withdrawing group present at para position decreases the e Θ density on nitrogen makes it less basic than aniline. The relative basic strength also influenced by inductive effect of corresponding substituent.
In case of ortho amino phenone due to ability to form intramolecular hydrogen bond ortho amino phenone is basic then aniline.
Special Case:
Chemical properties of aniline:
Chemical properties of amides:
Properties of cyanides:
Properties of Nitro compounds:- Part1: View the Topic in this Video from 25:05 to 41:50 Part2: View the Topic in this Video from 46:00 to 58:28
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The reason you are getting the extra space is the following: fractions are placed so that the fraction line lies on the "math axis", which is a little above the baseline, at about the height of a minus sign, and the stretchy delimiters produced by
\left...\right are centered on the math axis as well, so they extend above and below the math axis by the same amount. That will be the maximum of the extent above the math axis and the extent below it of the enclosed material. In your case, the material above the axis is much larger, so the delimiters extend that far below as well.
One way to get around this is to use
\vcenter to recenter the equation vertically so that the fraction line is not on the math axis. You need to use
\vcenter{\hbox{$\displaystyle...$}} inside the
\left...\right. For example
\vec{B}(\vec{m},\vec{r}) \sim
\left(\vcenter{\hbox{$\displaystyle
\frac{3\left[\begin{pmatrix}0\\0\\m\end{pmatrix} \cdot
\begin{pmatrix}
\sin\theta\cos\varphi\\
\sin\theta\sin\varphi\\
\cos\theta
\end{pmatrix}\right]
\cdot
\begin{pmatrix}
\sin\theta\cos\varphi\\
\sin\theta\sin\varphi\\
\cos\theta
\end{pmatrix}
-
\begin{pmatrix}0\\0\\m\end{pmatrix}}{r^3}
$}}\right)
produces
Note, however, that some will find this distasteful since the fraction is not properly centered, so the other answers may be preferable. |
Tagged: ideal 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 526
A ring is called
local if it has a unique maximal ideal. (a) Prove that a ring $R$ with $1$ is local if and only if the set of non-unit elements of $R$ is an ideal of $R$.
Add to solve later
(b) Let $R$ be a ring with $1$ and suppose that $M$ is a maximal ideal of $R$. Prove that if every element of $1+M$ is a unit, then $R$ is a local ring. Problem 525
Let
\[R=\left\{\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix} \quad \middle | \quad a, b\in \Q \,\right\}.\] Then the usual matrix addition and multiplication make $R$ an ring.
Let
\[J=\left\{\, \begin{bmatrix} 0 & b\\ 0& 0 \end{bmatrix} \quad \middle | \quad b \in \Q \,\right\}\] be a subset of the ring $R$. (a) Prove that the subset $J$ is an ideal of the ring $R$.
Add to solve later
(b) Prove that the quotient ring $R/J$ is isomorphic to $\Q$. Problem 524
Let $R$ be the ring of all $2\times 2$ matrices with integer coefficients:
\[R=\left\{\, \begin{bmatrix} a & b\\ c& d \end{bmatrix} \quad \middle| \quad a, b, c, d\in \Z \,\right\}.\]
Let $S$ be the subset of $R$ given by
\[S=\left\{\, \begin{bmatrix} s & 0\\ 0& s \end{bmatrix} \quad \middle | \quad s\in \Z \,\right\}.\] (a) True or False: $S$ is a subring of $R$.
Add to solve later
(b) True or False: $S$ is an ideal of $R$. Problem 432 (a) Let $R$ be an integral domain and let $M$ be a finitely generated torsion $R$-module. Prove that the module $M$ has a nonzero annihilator. In other words, show that there is a nonzero element $r\in R$ such that $rm=0$ for all $m\in M$. Here $r$ does not depend on $m$.
Add to solve later
(b) Find an example of an integral domain $R$ and a torsion $R$-module $M$ whose annihilator is the zero ideal. Problem 431
Let $R$ be a commutative ring and let $I$ be a nilpotent ideal of $R$.
Let $M$ and $N$ be $R$-modules and let $\phi:M\to N$ be an $R$-module homomorphism.
Prove that if the induced homomorphism $\bar{\phi}: M/IM \to N/IN$ is surjective, then $\phi$ is surjective.Add to solve later
Problem 417
Let $R$ be a ring with $1$ and let $M$ be an $R$-module. Let $I$ be an ideal of $R$.
Let $M’$ be the subset of elements $a$ of $M$ that are annihilated by some power $I^k$ of the ideal $I$, where the power $k$ may depend on $a$. Prove that $M’$ is a submodule of $M$. |
I tried to derive a preconditioner for self-consistent iteration similar to section IX in arXiv:0804.2583. For simplicity, consider here only one orbital (one or two electrons) systems.
Suppose that the iteration starts with function $f_1$ and write $f_1 = \psi_0 + \Delta \psi_0$, where $\psi_0$ is the normalized exact solution of the Schrödinger equation. Suppose also that $H$ is real and Hermitian, $H \psi_0 = E \psi_0$, and $\psi_0$ and $\Delta \psi_0$ are real. Assume that $\Delta \psi_0$ is small.
Define $\Lambda := \left< f_1 \vert H f_1 \right>$ and $g := H f_1 - \Lambda f_1$. We have $$\Lambda = E + 2 \left< H \psi_0 \vert \Delta \psi_0 \right> + \left< \Delta \psi_0 \vert H \Delta \psi_0 \right>$$ and we approximate $$ \Lambda \approx E + 2 \left< H \psi_0 \vert \Delta \psi_0 \right> . $$ We have $$ g = E \psi_0 + H \Delta \psi_0 - \Lambda \psi_0 - \Lambda \Delta \psi_0 $$ from which it follows that $$ (H - \Lambda) \Delta \psi_0 = (\Lambda - E) \psi_0 + g \approx 2 \left< H \psi_0 \vert \Delta \psi_0 \right> \psi_0 + g \tag{1} $$ We have \begin{eqnarray} 2 \left< H \psi_0 \vert \Delta \psi_0 \right> \psi_0 & = & 2( f_1 - \Delta \psi_0) \left< H(f_1 - \Delta \psi_0) \vert \Delta \psi_0 \right> \nonumber \\ & = & 2( f_1 - \Delta \psi_0) \left( \left< H f_1 \vert \Delta \psi_0 \right> - \left< H \Delta \psi_0 \vert \Delta \psi_0 \right> \right) \nonumber \end{eqnarray} and we approximate $$ 2 \left< H \psi_0 \vert \Delta \psi_0 \right> \psi_0 \approx 2 \left< H f_1 \vert \Delta \psi_0 \right> f_1 $$ We define linear function $G$ by $$ G(x) := 2 \left< H f_1 \vert x \right> f_1 . $$ Now it follows from equation (1) that $$ (H - \Lambda - G) \Delta \psi_0 \approx g $$ Consequently we approximate $\Delta \psi_0$ by $$ \tilde{g} := (H - \Lambda - G)^{-1} g $$ and set $$ f_2 := \frac{f_1 - \tilde{g}}{\Vert f_1 - \tilde{g} \Vert} $$ for the next iteration. Unless we have an one electron system we also compute the new Hartree potential and the new Hamiltonian operator for the next iteration.
I tried this method for the hydrogen atom and the hydrogen molecule. For hydrogen atom I computed the approximate wavefunction with Arnoldi method and then tried to enhance the result with this method. The iteration converged well until both the gradient norm and quantity $\Vert H \psi - E \psi \Vert$ were about $4 \cdot 10^{-15}$ and then stopped converging. For hydrogen molecule I computed the initial orbitals with Arnoldi method setting Hartree potential to zero. The SC iteration converged well until the gradient norm was about $10^{-4}$ and $\Vert H \psi - E \psi \Vert \approx 10^{-6}$. After that the SC iteration stopped converging.
Q: Is the derivation of the preconditioner correct?
Q: Is there some reason why the method does not work when the iteration gets close to a self-consistent solution? |
Applications of the Integrals Area between Two Curves Area between the two parabolas y 2= 4ax and x 2= 4by is = \frac{16\ ab}{3} sq.units The area region bounded by y 2= 4ax and y = mx is = \frac{8\ a^{2}}{3 m^{3}} sq.units The area region bounded by y 2= 4ax and its latus rectum x = a is \frac{8\ a^{2}}{3} sq.units The area region bounded by y 2= 4ax and the line x = b is \frac{8\ a^{\frac{1}{2}}b^{\frac{3}{2}}}{3} sq.units The area region bounded by x 2= 4by and the line y = a is \frac{8\ b^{\frac{1}{2}}a^{\frac{3}{2}}}{3} sq.units The area region bounded by x 2= 4ay and y = mx is \frac{8}{3}a^{2}m^{3} sq.units The area region bounded by y 2= 4ax and the lines x = b, x = c is \frac{8}{3}a^{\frac{1}{2}}\left(c^{\frac{3}{2}}-b^{\frac{3}{2}}\right) sq.units. The area region bounded by y = ax 2+ bx + c and y = mx + k is \frac{\Delta^{\frac{3}{2}}}{6a^{2}}, where Δ is the discriminent of ax 2+ (b − m) x + (c − k) The area region bounded by x = ay 2+ by + c and x = my + k is \frac{\Delta^{\frac{3}{2}}}{6a^{2}}, where Δ is the discriminent of ay 2+ (b − m) y + (c − k) The area of the parallelogram formed by the lines a 1x + b 1y + c 1= 0, a 1x + b 1y + d 1= 0 & a 2x + b 2y + c 2= 0, a 2x + b 2y + d 2= 0 is \begin{vmatrix}\frac{(c_{1}-d_{1})(c_{2}-d_{2})}{a_{1}b_{2}-a_{2}b_{1}}\end{vmatrix} sq.units. The area of the rhombus formed by ax ± by ± c = 0 is \frac{2c^{2}}{|ab|} sq.units The area of the triangle by y = |ax + b| with y-axis (or) x = |ay + b| with x-axis is \frac{b^{2}}{a}. The area of the triangle formed by the tangent and normal at P(x 1, y 1) and x-axis is = \frac{y_{1}^{2}}{2}\begin{vmatrix}M+\frac{1}{M}\end{vmatrix} Where M is slope of tangent at P The area of the triangle formed by the tangent and normal at P(x 1, y 1) with y-axis is = \frac{x_{1}^{2}}{2}\begin{vmatrix}M+\frac{1}{M}\end{vmatrix} Where M is slope of tangent at P View the Topic in this video From 12:52 To 53:08
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1. The area of the region enclosed between two curves y = f(x), y = g(x) and the lines x = a, x = b is given by the formula,
Area =\int_{a}^{b} [f(x)-g(x)] \ dx, where, f(x) ≥ g(x) in [a, b]
2. If f(x) ≥ g(x) in [a, c] and f(x) ≤ g(x) in [c, b], a < c < b, then
Area = \int_{a}^{c} [f(x)-g(x)] \ dx+\int_{c}^{b} [g(x)-f(x)] \ dx. |
Taylor Series
$\frac {1} {\sqrt[]{2}} +\frac{x-\frac{\pi}{4}}{\sqrt[]{2}}+\frac{(x-\frac{\pi}{4})^2}{2{\sqrt[]{2}}}+\frac{(x-\frac{\pi}{4})^3}{6{\sqrt[]{2}}}+\frac{(x-\frac{\pi}{4})^4}{24{\sqrt[]{2}}}+\frac{(x-\frac{\pi}{4})^5}{120{\sqrt[]{2}}}+O(({x}-{\frac{\pi}{4})^6})$
Taylor series represent an approximation of a function, and also give a quantitative estimate on the error introduced by the approximation of the function. $O\left(({x}-{\frac{\pi}{4})^6}\right)$ symbolizes the asymptotic behavior associated with the error introduced by this series at the point $x=\frac{\pi}{4}$ for the function $\sin x$. Taylor series apply to real and complex variables, and lattice schemes use both real and complex functions in their construction.
$\alpha$-BDD
$\alpha$-BDD, or alpha bounded distance decoding, is a formulation of the LWE problem, such that given a lattice $\frak L$ and vector $y$, find a lattice point $x\in\frak L$ within distance $\alpha \cdot \lambda _1(\frak L)$. The paper on $\alpha$-BDD states:
"... a function $f(x)$ which approximates the distance of a target point $x$ from a lattice within a factor of $O\left(\sqrt{\frac{n}{\log n}}\right)$."
One Way Functions
Drawing from Micciancio speaking at the PQCrypto '14 Summer School, you can represent the noise of a lattice vector by looking at a specific point in that lattice. Drawing from what Micciancio said about the one-way function, there is a method of studying lattice cryptography using analytic geometry.
Micciancio covers the different forms of functions associated with the CVP:
Injective Surjective
The following mathematical expressions are taken from the video lecture of Miccancio:
Let $\beta \gg \mu:g_{\frak L}(\overrightarrow{x}) $ be nearly uniform. The vector $\overrightarrow{x} \in \beta$ denotes error for an arbitrary hard lattice $\frak L$, with input $\overrightarrow{x}$, and $\| \overrightarrow{x} \| \leq \beta$.
Then:
$\beta < \frac{\lambda_1}{2}:f(\frak L)$ is injective.
$\beta > \frac{\lambda_1}{2}:f(\frak L)$ is not injective.
$\beta \geq{\mu}:g_\frak L$ is surjective.
All three of these functions can be associated with how error is mapped to a specific point in the lattice.
For example, Micciancio shows in the video that the three inequalities are a OWF relying on the CVP. The error is what creates the "noise," which in turn makes the CVP computationally difficult. The inequalities are the bounds associated with the error, or another way of defining an $\alpha$ bound as discussed.
The Taylor series also evaluates an approximate function at a specific point. So in the function from the Taylor series example, for argument's sake, assume that $\sin x$ is the Gaussian distribution with basis vectors $b_1 \dots b_n \in \beta$ for a lattice $\frak L$ generated by the Taylor series.
Question
My question assumes that 1 is true, while 2 may or may not be true.
My understanding is that the approximate functions used to calculate the error in lattice cryptography helps established the security in that proof. So we can compare this to the boundary conditions to determine if $\sin x$ is injective or surjective, correct? Can't we just drop most of the terms from the Taylor series to derive $O(({x}-{\frac{\pi}{4})^6})$? This would become $O(({n}-{\frac{k}{4})^6})$ for some $n$ and constant $k$, right?
If both Big-O terms from the Taylor series and $\alpha$-BDD examples are error associated with their approximate functions, is it possible to use a Taylor series as a way to evaluate error associated with a point to determine if the scheme is an instance of LWE or SIVP? |
Let X be a normal random variable with mean $\mu$ and standard deviation $\sigma^2$. I am wondering how to calculate the third moment of a normal random variable without having a huge mess to integrate. Is there a quicker way to do this?
This is a general method to calculate any moment:
Let $X$ be standard normal. The moment generating function is:
$$M_X(t) = E(e^{tX}) = e^{t^2/2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-(x-t)^2/2} \:\:dx = e^{t^2/2}$$
since the integrand is the pdf of $N(t,1)$.
Specifically for the third moment you differentiate $M_X(t)$ three times
$M_X^{(3)}(t) = (t^3 + 3t)e^{t^2/2}$
and
$E[X^3] = M_X^{(3)}(0) = 0$
For a general normal variable $Y = \sigma X + \mu$ we have that
$$M_Y(t) = e^{\mu t} M_X(\sigma t) = e^{\mu t + \sigma^2 t^2 /2}$$
and you calculate the $n$th moment as a above, i.e. differentiating $n$ times and setting $t=0$: $$E[Y^n] = M_Y^{(n)}(0)$$.
$\mathbb{E}[(X-\mu)^3]=0$ since $X-\mu$ is normally distributed with mean zero, then expand out the cube.
If the distribution of a random variable $X$ is symmetric about $0$, meaning $\Pr(X>x)=\Pr(X<-x)$ for every $x>0$, then its third moment, if it exists at all, must be $0$, as must all of its odd-numbered moments.
If $\operatorname{E}\left[ \,|X^3|\, \right]<\infty$ then the third moment exists. Furthermore, Symmetry shows that the positive and negative parts of $\operatorname{E}\left[ X^3 \right]<0$ (without the absolute-value sign) cancel each other out. |
According to Einstein, light of frequency f consists of a photons each of energy hf. When a photon of light of frequency f is incident on a metal, the energy is completely transferred to a free electron in the metal. A part of the energy acquired by the electron is used to pull out the electron from the surface of the metal and the rest of its utilized in imparting K.E. to the emitted electron. Let W be the energy used up in librating the electrons from the surface of metal (work function) and \(\frac 12 mv^2 \) is K.E. of the photoelectron, then,
\begin{align*} \text {Energy of the photon} &= \text {Work function} + \text {K.E. of the electron} \\ \text {or} \: hf &= \phi + \frac 12 mv^2 \dots (i)\end{align*} This relation is known as the Einstein’s photoelectric equation. If f
0is the threshold frequency which just ejects an electron from the metal surface without any velocity, then. \begin{align*}\phi &= hf_0 \\ \therefore hf &= hf_0 + \frac 12 mv_{max}^2 \dots (ii) \\ \text {where}\: V_{max}\: \text {is the maximum velocity acquired by the electron} \\ \text {or,} \: \frac 12 mv_{max}^2 = h(f-f_0) \\ \end{align*}
In this experiment, alkali metal is used as emitters. Cylindrical blocks made by sodium, potassium or lithium are placed around a wheel W at the centre of the glass flask. To avoid tarnishing and the formation of oxide films on the metal surface, the metals were housed in a vacuum. Their surface was kept clean by a cutting knife K which could be moved and turned by the means of a magnet outside. When a beam of monochromatic light falls on the alkali metal, photoelectrons are emitted. The photoelectrons emitted reach the electrode C. the electrode C is usually kept negative with respect to the cylindrical blocks. Therefore, photo electrons will be repelled and only fast moving electrons will be able to reach the electrode C. Since the negative potential of electrode C is increased until the fastest moving photo electrons are repelled back and thud the current fall to zero. Therefore, the minimum potential and \(\frac 12 mV_{max}^2 \) is the K.E. of the fastest photoelectrons, then
\begin{align*} \frac 12 mV_{max}^2 &= eV_0 \dots (i) \\ \end{align*} where m, e, and v
maxbe the mass, charge and maximum velocity of the ejected electrons from Einstein’s photoelectric equation, we have \begin{align*} hf &= \phi + \frac 12 mV_{max}^2 \\ \text{or,} \: \frac 12 mV_{max}^2 &= hf – hf_0 \dots (ii) \\ \text {from equation} (i) \: \text {and} \: (ii), \text {we have,} \\ \text {or,} \: eV_0 &= hf –hf_0 \\ \text {or,} V_0 &= \frac he f - \frac he f_0 \dots (iii) \\ \end{align*}
Experimental graph between V
0 and f is in agreement with Einstein’s photoelectric equation. The slope of the straight line is
\begin{align*} m &= \tan \theta = \frac he \\ \therefore h &= e\tan \theta = 6.625 \times 10^{-34} Js \dots (iv) \end{align*}
Hence determining the slope of the curve V
0 and f. Planck’s constant h can be calculated by using equation (iv).
The photoelectric effect is used in photoelectric cells. The device used for converting light energy into electric energy is called the photoelectric cell. They are of three types:
It consists of two electrodes A and C called anode and cathode is a semi-cylindrical plate of metal coated with a photosensitive material of low work function such as cesium or silver oxide. The anode A is in a form of wire so that it does not obstruct the light falling on the cathode. A positive potential of about 100 volts is applied to the anode, the negative being connected through the galvanometer G.
When the light of frequency greater than the threshold frequency of cathode material is incident on the cathode C, photoelectrons are emitted from it. A small current flows through the cell and can be measured by the galvanometer.
It consists of a copper plate, on which a thin semiconductor layer of cuprous oxide is deposited. Over this, a very thin layer of silver or gold is deposited by the method of evaporation in a vacuum. When the light falls on the cell, it penetrates the film and ejects electrons from the interface between the gold film and the cuprous oxide layer. These electrons flow through the external low resistance R consulting a current, whose strength depends in the intensity of light.
The advantages of these cells are:
The cell works on the principle that the electrical resistance of the semiconductor, such as cadmium sulphide, decreases with the increase in the intensity of light falling on it. A film of selenium, deposited on one side of the iron plate, is also used. When the light of varying intensity is made to fall on the film, a current flows through the circuit, containing a galvanometer and a battery.
Applications of Photoelectric Cells
Reference
Manu Kumar Khatry, Manoj Kumar Thapa, Bhesha Raj Adhikari, Arjun Kumar Gautam, Parashu Ram Poudel.
Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.
S.K. Gautam, J.M. Pradhan.
A text Book of Physics. Kathmandu: Surya Publication, 2003.e
The minimum frequency, which can cause photoelectric emission, is called the threshold frequency.
According to Einstein, light of frequency f consists of a photons each of energy hf.
The minimum energy of photon required to just liberate an electron from the metal surface with zero velocity is known as work function W of that metal.
The device used for converting light energy into electric energy is called the photoelectric cell. They are of three types: Photo-emissive cells, Photo-voltaic cells and Photo-conductive cells. |
Inhomogeneous Neighbourhood Density Function
Computes spatially-weighted versions of the the local \(K\)-function or \(L\)-function.
Usage
localKinhom(X, lambda, ..., correction = "Ripley", verbose = TRUE, rvalue=NULL, sigma = NULL, varcov = NULL) localLinhom(X, lambda, ..., correction = "Ripley", verbose = TRUE, rvalue=NULL, sigma = NULL, varcov = NULL)
Arguments X
A point pattern (object of class
"ppp").
lambda
Optional. Values of the estimated intensity function. Either a vector giving the intensity values at the points of the pattern
X, a pixel image (object of class
"im") giving the intensity values at all locations, a fitted point process model (object of class
"ppm") or a
function(x,y)which can be evaluated to give the intensity value at any location.
…
Extra arguments. Ignored if
lambdais present. Passed to
density.pppif
lambdais omitted.
correction
String specifying the edge correction to be applied. Options are
"none",
"translate",
"Ripley",
"translation",
"isotropic"or
"best". Only one correction may be specified.
verbose
Logical flag indicating whether to print progress reports during the calculation.
rvalue
Optional. A
singlevalue of the distance argument \(r\) at which the function L or K should be computed. sigma, varcov
Optional arguments passed to
density.pppto control the kernel smoothing procedure for estimating
lambda, if
lambdais missing.
Details
Given a spatial point pattern
X, the inhomogeneous neighbourhood density function \(L_i(r)\) associated with the \(i\)th point in
X is computed by $$ L_i(r) = \sqrt{\frac 1 \pi \sum_j \frac{e_{ij}}{\lambda_j}} $$ where the sum is over all points \(j \neq i\) that lie within a distance \(r\) of the \(i\)th point, \(\lambda_j\) is the estimated intensity of the point pattern at the point \(j\), and \(e_{ij}\) is an edge correction term (as described in
Kest). The value of \(L_i(r)\) can also be interpreted as one of the summands that contributes to the global estimate of the inhomogeneous L function (see
Linhom).
By default, the function \(L_i(r)\) or \(K_i(r)\) is computed for a range of \(r\) values for each point \(i\). The results are stored as a function value table (object of class
"fv") with a column of the table containing the function estimates for each point of the pattern
X.
Alternatively, if the argument
rvalue is given, and it is a single number, then the function will only be computed for this value of \(r\), and the results will be returned as a numeric vector, with one entry of the vector for each point of the pattern
X.
Value
If
rvalue is given, the result is a numeric vector of length equal to the number of points in the point pattern.
the vector of values of the argument \(r\) at which the function \(K\) has been estimated
the theoretical value \(K(r) = \pi r^2\) or \(L(r)=r\) for a stationary Poisson process
See Also Aliases localKinhom localLinhom Examples
# NOT RUN { data(ponderosa) X <- ponderosa # compute all the local L functions L <- localLinhom(X) # plot all the local L functions against r plot(L, main="local L functions for ponderosa", legend=FALSE) # plot only the local L function for point number 7 plot(L, iso007 ~ r) # compute the values of L(r) for r = 12 metres L12 <- localL(X, rvalue=12)# }
Documentation reproduced from package spatstat, version 1.55-1, License: GPL (>= 2) |
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 $V$?
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 $\Fun(S, V)$.
Let $\mathrm{P}_3$ denote the set of polynomials of degree $3$ or less with real coefficients. Consider the ordered basis\[B = \left\{ 1+x , 1+x^2 , x – x^2 + 2x^3 , 1 – x – x^2 \right\}.\]Write the coordinate vector for the polynomial $f(x) = -3 + 2x^3$ in terms of the basis $B$.
Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$ matrices. Define the linear transformation $T : V \rightarrow W$ by\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a+b & 2d \\ 2b – d & -3c \\ 2b – c & -3a \end{bmatrix}.\]
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,\[T (f) (x) = x f(x).\]
Prove that $T$ is a linear transformation, and find its range and nullspace.
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
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 $W$.
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.
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)
Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\]be a subset in $C[-1, 1]$.
(a) Prove that $V$ is a subspace of $C[-1, 1]$.
(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.
(c) Prove that\[B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}\]is a basis for $V$. |
Classic thermodynamic stability theory states that heat capacities have to be positive [1,2]. For a system in thermal contact with a larger system, the second order virtual variation in entropy around an equilibrium state is \[ \delta^2 S = - \frac{C_V (\delta T)^2}{T^2} \lt 0 , \] with heat capacity at constant volume \[ C_V = \left( \frac{\partial E}{\partial T} \right)_V , \] and the negative sign required by stability implying that \( C_V \gt 0 \). Recall that the quantities used in thermodynamics are averages over microscopic realizations. For instance, the thermodynamic internal energy is given by \( E = \langle E^\mathrm{micr} \rangle \).
Statistical mechanics introduces a different proof of the positivity of heat capacities [3]. It starts from the expression for the energy average over a canonical ensemble \[ E = \langle E^\mathrm{micr} \rangle = \frac{\sum_i E_i^\mathrm{micr} \exp(-E_i^\mathrm{micr}/k_\mathrm{B}T)}{\sum_i \exp(-E_i^\mathrm{micr}/k_\mathrm{B}T)} , \] and gets \[ \frac{\diff E}{\diff T} = \frac{1}{k_\mathrm{B}T^2} \bigg\langle \big(E_i^\mathrm{micr} - \langle E^\mathrm{micr} \rangle \big)^2 \bigg\rangle , \] which is clearly a positive quantity. Notice that the canonical ensemble does not include the volume as variable, and the above total derivative is equivalent to the partial derivative that defines \( C_V \).
Part of the nanocluster community claims that small systems are not bound by the above macroscopic proofs, and that negative heat capacities exist at small scales. Michaelian and Santamaría-Holek show that these incorrect results (both experimental and theoretical) derive from two basic problems: (
i) the system is non-ergodic or ( ii) the model used to represent the system does not obey quantum laws [2].
On the other side of the size spectrum, very large systems, we find to astronomers claiming that the above macroscopic proofs are wrong and that heat capacity can be negative in gravitational systems [3]. Their starting point is the virial theorem that relates kinetic energy \( K = \langle K^\mathrm{micr} \rangle \) and potential energy \( \Phi = \langle \Phi^\mathrm{micr} \rangle \) for an isolated gravitational system with energy \( E = K + \Phi \) \[ 2 K + \Phi = 0 . \] Using \( K = \frac{3}{2} N k_\mathrm{B} T \) yields \[ \frac{\diff E}{\diff T} = - \frac{\diff K}{\diff T} = - \frac{3}{2} N k_\mathrm{B} . \] Their 'proof' concludes by associating \( C_V \) to this negative quantity. Astronomers and nuclear physicists not only pretend that negative heat capacities are theoretically admissible, but they also claim those capacities are routinely measured in gravitational systems and nuclear clusters [3], by plotting \( E \) vs \( T \) and then finding regions where energy decreases when temperature increases; i.e., finding regions where \( (\diff E /\diff T) \lt 0 \). A more accurate description of the astronomers' claim is
when heat is absorbed by a star, or star cluster,I bolded the relevant part that disproves their claims. The problem is that the quantity \( (\diff E/\diff T) \) that they are measuring is not \( C_V \) because volume is not held constant during differentiation. If we write a detailed model of the energy we find that the potential energy \( \Phi \) depends on volume through the interparticle distances \[ E(T,V) = K(T) + \Phi(V) = \frac{3}{2} N k_\mathrm{B} T + \Phi(V) . \] Now, using the definition (2), we find \[ C_V = \left( \frac{\partial E(T,V)}{\partial T} \right)_V = \frac{\diff K(T)}{\diff T} = \frac{3}{2} N k_\mathrm{B} \gt 0 . \] Therefore, what astronomers and others are really measuring is not the heat capacity \( C_V \) but the abstract quantity \[ \frac{\diff E}{\diff T} = C_V + \frac{\diff\Phi(V(T))}{\diff T} . \] The sign of this abstract quantity depends of the nature of the interactions. For systems satisfying the virial theorem this quantity is just it will expandand cool down. minusthe heat capacity \( (\diff E/\diff T) = - C_V \). Acknowledgement
I thank Prof. Karo Michaelian for further remarks.
References [1]Modern Thermodynamics 1998: John Wiley & Sons Ltd.; Chichester.Kondepudi, D. K.; Prigogine, I. [2]Critical analysis of negative heat capacity in nanoclusters 2007: EPL, 79, 43001.Michaelian K.; Santamaría-Holek I. [3]Negative Specific Heat in Astronomy, Physics and Chemistry 1998: arXiv:cond-mat/9812172 v1.Lynden-Bell, D. |
Structure of Atom Dual nature of matter and De-Broglie's relationship De - Broglie's wave theory Vs Bohr atomic model :
2\pi r = n\lambda
n = integral value 2\pi r = n\frac{h}{mv} mvr = \frac{nh}{2\pi} View the Topic in this Video from 42:00 to 51:00
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1. de-Broglie equation, \tt \lambda=\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mE_{k}}}=\frac{h}{\sqrt{2mqv}}
2.Planck's Equation, \tt E=h\nu=\frac{h.c}{\lambda}
3.Einstein's mass energy relationship, E = mc
2
4.
de Broglie wavelength associated with charged particles
(a) For electron \tt \lambda = \frac{12.27}{\sqrt{V}}Å
(b) For proton \tt \lambda = \frac{0.286}{\sqrt{V}}Å
(c) For α -particles \tt \lambda = \frac{0.101}{\sqrt{V}}Å
where V is the accelerating potential of these particles.
5.
de Broglie wavelength associated with uncharged particles (a) For neutrons \tt\lambda=\frac{h}{\sqrt{2Em}}=\frac{6.62\times10^{-34}}{\sqrt{2\times1.67\times 10^{-27}\times E}}=\frac{0.286}{\sqrt{E\left(eV\right)}}Å
(b) For gas molecules \tt\lambda=\frac{h}{m\times u_{rms}}=\frac{h}{\sqrt{3mkT}}
where k is the Boltzmann constant. |
Suppose that a polynomial $p(x,y)$ defined on $\mathbb{R}^2$ is identically zero on some open ball (in the Euclidean topology). How does one go about proving that this must be the zero polynomial?
WLOG suppose that the center of ball is the origin and write
$$ p(x,y)=\sum _{i,j=0}^ma_{i,j}x^iy^j $$
Plug in $x=y=0$. You find that $a_{0,0}=0$. Take the partial derivative with respect to $x$ and set $x=y=0$. You find that $a_{1,0}=0$. You should be able to finish it from here by continuining similarly. . .
To make notation simpler, let $(a,b)$ be the center of the open ball. Let $g(x,y)=f(a+x,b+y)$. Then the polynomial $g$ is identically $0$ in an open ball containing the origin. We show that $g(x,y)$ is identically $0$.
Consider any line through the origin. We will show that $g(x,y)=0$ at all points on that line. The lines are given by $y=kx$ where $k$ is a constant, and, easily forgotten, $x=0$.
Let $P(t)=g(t,kt)$ (for the line $x=0$, let $P(t)=g(0,t)$).
Then $P(t)$ is a polynomial, and is identically $0$ in an interval. In particular, $P(t)=0$ for infinitely many $t$. Thus $P(t)$ must be identically $0$ (a non-zero polynomial has only finitely many roots).
We conclude that $g$ is identically $0$ on every line through the origin, and hence everywhere.
Note that essentially the same argument works for polynomials in $n$ variables.
This follows purely algebraically by induction on degree using the fact that a polynomial has no more roots than its degree over a domain - see my prior post.
Write $p(x,y) = \sum_{i=0}^n q_i(x)y^i$, where $q_i(x)$ are polynomials in $x$.
Pick a point $(x_0,y_0)$ interior to $U$, where $U$ is your open ball. Then there exists a $r>0$ so that $B_r(x_0,y_0) \subset U$.
Pick any $x_1 \in (x_0-r,x_0+r)$, and chose some $\delta>0$ so that $\{x_1\} \times (y_0-\delta ,y_0+\delta) \subset U$.
Then $p(x_1,y)= \sum_{i=0}^n q_i(x_1)y^i$ is a polynomial in $y$ with constant coefficients $q_i(x_1)$ which is identically zero on on the interval $(y_0-\delta, y_0+\delta)$. Thus $p(x_1,y)$ is the zero polynomial.
Hence $q_i(x_1)=0$. But since $x_1$ was arbitrary in $(x_0-r,x_0+r)$, each $q_i$ has infinitelly many roots, thus each $q_i=0$.
A stronger, but still reasonably easy to prove, statement follows from the combinatorial Nullstellensatz. It's actually enough to require that $p$ is identically zero on a lattice with sufficiently many points. |
Answer
Since the speed was $23.0~mi/h$ at the moment the brakes were first applied, the police officer can not give a ticket for speeding in a 25 mi/h zone.
Work Step by Step
We can find the rate of deceleration: $F_N~\mu_k = ma$ $mg~\mu_k = ma$ $a = g~\mu_k$ $a = (9.80~m/s^2)(0.60)$ $a = 5.88~m/s^2$ We can find the initial velocity: $v_f^2 = v_0^2+2ax$ $v_0^2 = v_f^2-2ax$ $v_0 = \sqrt{v_f^2-2ax}$ $v_0 = \sqrt{0-(2)(-5.88~m/s^2)(9.0~m)}$ $v_0 = 10.29~m/s$ We can convert the speed to units of mi/h: $v_0 = 10.29~m/s\times \frac{1~mi}{1609~m} \times \frac{3600~s}{1~hr} = 23.0~mi/h$ Since the speed was $23.0~mi/h$ at the moment the brakes were first applied, the police officer can not give a ticket for speeding in a 25 mi/h zone. |
Here is pseudocode for the algorithm:
select(L,k){if (L has 10 or fewer elements){ sort L return the element in the kth position}partition L into subsets S[i] of five elements each (there will be n/5 subsets total).for (i = 1 to n/5) do x[i] = select(S[i],3)M = select({x[i]}, n/10)partition L into L1<M, L2=M, L3>Mif (k <= length(L1)) return select(L1,k)else if (k > length(L1)+length(L2)) return select(L3,k-length(L1)-length(L2))else return M}
Here is some analysis of the algorithm: http://www.ics.uci.edu/~eppstein/161/960130.html
The analysis suggests to use the recurrence relation $T(n) \leq \frac{12n}{5} + T(\frac{n}5) + T(\frac{7n}{10})$. Solving this we get linear work for a call. But aren't there log many recursive calls? So that would be $n\log n$.
Put another way, conceptually this algorithm seems like it could be described as "for each call, cut the search area kind of like binary search, but not guaranteed to cut the search area by as much as binary search, and add a linear time partition". Binary search runs in $O(\log(n))$, so adding a linear search per call would make it $O(n\log(n))$.
What am I not understanding about the linked analysis? |
Fix $t>0$ and consider the following Principal Value integral: $$ \mathscr{P}\int_{-\infty}^{\infty}d\omega \frac{e^{-i \omega t}}{\sqrt{\omega^2}} = - \log(t^2) $$
This is the function that Mathematica spits out (and this also matches a result that I'm finding in Lighthill's ``An Introduction to Fourier Analysis and Generalized Functions'', up to a constant).
I'm posting because when I check this calculation myself explicitly, I'm getting something strange. This is my method of computation: $$ \mathscr{P} \int_{-\infty}^{\infty} \frac{d\omega}{\sqrt{\omega^2}} e^{ - i \omega t } = \lim\limits_{\eta \to 0^{+}} \left\{ \int_{-\infty}^{-\eta} \frac{d\omega}{-\omega} e^{ - i \omega t } \ + \ \int_{\eta}^{\infty} \frac{d\omega}{\omega} e^{ - i \omega t } \right\} $$
After switching the variable $\omega \mapsto -\omega$ in the first integral, the above simplifies to: $$ \ldots = \lim\limits_{\eta \to 0^{+}} \left\{ 2 \int_{\eta}^{\infty} \frac{d\omega}{\omega} \cos\left( \omega t \right) \right\} = \lim\limits_{\eta \to 0^{+}} \left\{ - 2 \mathrm{Ci}\left( \eta t \right) \right\} $$
Where $\mathrm{Ci}$ is the cosine-integral function. The problem is that I cannot take the limit $\eta \to 0^{+}$. Upon an expansion about $\eta =0$, I find that the above looks like: $$ \ldots = \lim\limits_{\eta \to 0^{+}} \left\{ - 2 \gamma - \log(\eta^2) - \log(t^2) + \mathscr{O}(\eta^2) \right\} $$
So it seems to me like the integral has the value $- 2\gamma - \lim\limits_{\eta\to 0^+} \log(\eta^2) - \log(t^2)$. So I get the right functional form in $t$, but I have an extra two constant appearing - one of which is infinite!
The $\eta$ seems to be a 'regulator' for the normally divergent integral - how does one get rid of it? Am I using the wrong definition for the Cauchy Principal value in my computation? Mathematica and Lighthill seems to be throwing the $-2 \gamma - \log(\eta^2)$ away somehow.
P.S. $\gamma$ is the Euler-Mascheroni constant. |
On existence and nonexistence of the positive solutions of non-newtonian filtration equation
1.
Department of Mathematics, Hacettepe University, 06800 Beytepe - Ankara, Turkey
$ \rho (|x|) \frac{\partial u}{\partial t}- \sum_{i=1}^N D_i(u^{m-1}|D_i u|^{\lambda -1}D_i u)+g(u)+lu=f(x)$ (1)
or, after the change $v=u^{\sigma}$, $\sigma =\frac{m+\lambda -1}{\lambda }, $ of equation
$\rho (|x|) \frac{\partial v^{\frac{1}{ \sigma }}}{\partial t}-\sigma ^{-\lambda }\sum_{i=1} ^N D_i(|D_i v|^{\lambda -1}D_i v)+g(v^{\frac{1}{\sigma }}) +lv^{\frac{1}{ \sigma }}=f(x),$ (1')
in unbounded domain $R_+\times R^N,$ where the term $g(s)$ is supposed to satisfy just a lower polynomial growth condition and $g'(s) > -l_1$. The existence of the solution in $ L^{1+1/\sigma}(0, T; L^{1+1/\sigma}(R^N))\cap L^{\lambda +1}(0, T; W^{1,\lambda +1}(R^N))$ is proved. Also, under some condition on $g(s)$ and $u_0$ is shown a nonexistence of the solution.
Keywords:nonexistence., Nonlinear degenerate equation, existence, FSP(finite speed of propagation of perturbations). Mathematics Subject Classification:Primary: 35K15, 35K65; Secondary: 35B3. Citation:Emil Novruzov. On existence and nonexistence of the positive solutions of non-newtonian filtration equation. Communications on Pure & Applied Analysis, 2011, 10 (2) : 719-730. doi: 10.3934/cpaa.2011.10.719
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S. P. Degtyarev and A. F. Tedeev,
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[2]
D. Blanchard and G. A. Francfort,
[3]
S. P. Degtyarev and A. F. Tedeev,
[4] [5] [6]
H. A. Levine,
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J. L. Lions, "Quelques Methodes de Resolution des Problemes aux Limites Nonlineaires,",
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J. L. Lions and E. Magenes, "Non-Homogeneous Boundary Value Problems and Applications,",
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A. V. Martynenko, and A. F. Tedeev,
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G. Reyes and J. L. Vázquez,
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A. F. Tedeev,
[15] [16]
M. Tsutsumi,
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Z. Xiang, Ch. Mu and X. Hu,
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The shear stress is part of the pressure tensor. However, here, and many parts of the book, it will be treated as a separate issue. In solid mechanics, the shear stress is considered as the ratio of the force acting on area in the direction of the forces perpendicular to area. Different from solid, fluid cannot pull directly but through a solid surface. Consider liquid that undergoes a shear stress between a short distance of two plates as shown in Figure 1.3.
Figure 1.3: Schematics to describe the shear stress in fluid mechanics.
The upper plate velocity generally will be
\[U = \ff(A, F, h)\tag{2}\]
Where
Equations qref{intro:eq:Upropto} can be rearranged to be
\[U h \propto \dfrac{F}{A} \tag{4}\]
Shear stress was defined as
\[T_{xy} = \dfrac{F}{A}\tag{5}\]
The index
\[T_{xy} = \mu \dfrac{U}{h}\tag{6}\]
Where \mu is called the absolute viscosity or dynamic viscosity which will be discussed later in this chapter in a great length.
Fig. 1.4 The deformation of fluid due to shear stress as progression of time.
In steady state, the distance the upper plate moves after small amount of time, \delta t is
\[dll = U \delta t \tag{7}\]
From Figure 1.4 it can be noticed that for a small angle, \(\delta\beta \cong \sin\beta\), the regular approximation provides
\[dll = U \delta t = h\delta\beta\tag{8}\]
From equation qref{intro:eq:dUdx} it follows that
\[U=h\dfrac{\delta\beta}{\delta t}\tag{9}\]
Combining equation qref{intro:eq:dbdt} with equation qref{intro:eq:shearS} yields
\[T_{xy} = \mu\dfrac{\delta\beta}{\delta t}\tag{10}\]
If the velocity profile is linear between the plate (it will be shown later that it is consistent with derivations of velocity), then it can be written for small a angel that
\[\dfrac{\delta \beta}{\delta t} = \dfrac{dU}{dy}\tag{11}\]
Materials which obey equation qref{intro:eq:tau_dbdt} referred to as Newtonian fluid. For this kind of substance
\[T_{xy} = \mu\dfrac{dU}{dy}\tag{12}\]
Newtonian fluids are fluids which the ratio is constant. Many fluids fall into this category such as air, water etc. This approximation is appropriate for many other fluids but only within some ranges. Equation qref{intro:eq:dbdt} can be interpreted as momentum in the \x direction transferred into the \y direction. Thus, the viscosity is the resistance to the flow (flux) or the movement. The property of viscosity, which is exhibited by all fluids, is due to the existence of cohesion and interaction between fluid molecules. These cohesion and interactions hamper the flux in y–direction. Some referred to shear stress as viscous flux of x–momentum in the y–direction. The units of shear stress are the same as flux per time as following
\[\dfrac{F}{A}\left\dfrac{kg m}{sec^2}\dfrac{1}{m^2}\right = \dfrac{mU}{A}\left\dfrac{kg}{sec}\dfrac{m}{sec}\dfrac{1}{m^2}\right\tag\]
Thus, the notation of \(T_{xy}\) is easier to understand and visualize. In fact, this interpretation is more suitable to explain the molecular mechanism of the viscosity. The units of absolute viscosity are \([N sec / m^2]\). |
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 $V$?
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 $\Fun(S, V)$.
Let $\mathrm{P}_3$ denote the set of polynomials of degree $3$ or less with real coefficients. Consider the ordered basis\[B = \left\{ 1+x , 1+x^2 , x – x^2 + 2x^3 , 1 – x – x^2 \right\}.\]Write the coordinate vector for the polynomial $f(x) = -3 + 2x^3$ in terms of the basis $B$.
Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$ matrices. Define the linear transformation $T : V \rightarrow W$ by\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a+b & 2d \\ 2b – d & -3c \\ 2b – c & -3a \end{bmatrix}.\]
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,\[T (f) (x) = x f(x).\]
Prove that $T$ is a linear transformation, and find its range and nullspace.
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
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 $W$.
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.
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)
Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\]be a subset in $C[-1, 1]$.
(a) Prove that $V$ is a subspace of $C[-1, 1]$.
(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.
(c) Prove that\[B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}\]is a basis for $V$. |
Research
I study the distribution of algebraic numbers, mathematical statistical physics and roots/eigenvalues of random polynomials/matrices.
Projects in Progress
1
The distribution of values of the non-archimedean absolute Vandermonde determinant and the non-archimedean Selberg integral (with Jeff Vaaler). The Mellin transform of the distribution function of the non-archimedean absolute Vandermonde (on the ring of integers of a local field) is related to a non-archimedean analog of the Selberg/Mehta integral. A recursion for this integral allows us to find an analytic continuation to a rational function on a cylindrical Riemann surface. Information about the poles of this rational function allow us to draw conclusions about the range of values of the non-archimedean absolute Vandermonde.
2
Non-archimedean electrostatics. The study of charged particles in a non-archimedean local field whose interaction energy is proportional to the log of the distance between particles, at fixed coldness $\beta$. The microcanonical, canonical and grand canonical ensembles are considered, and the partition function is related to the non-archimedean Selberg integral considered in 1. Probabilities of cylinder sets are explicitly computable in both the canonical and grand canonical ensembles.
3
Adèlic electrostatics and global zeta functions (with Joe Webster). The non-archimedean Selberg integral/canonical partition function are examples of Igusa zeta functions, and as such local Euler factors in a global zeta function. This global zeta function (the exact definition of which is yet to be determined) is also the partition function for a canonical electrostatic ensemble defined on the adèles of a number field. The archimedean local factors relate to the ordinary Selberg integral, the Mehta integral, and the partition function for the complex asymmetric $\beta$ ensemble. The dream would be a functional equation for the global zeta function via Fourier analysis on the idèles, though any analytic continuation would tell us something about the distribution of energies in the adèlic ensemble.
4
Pair correlation in circular ensembles when $\beta$ is an even square integer (with Nate Wells and Elisha Hulbert). This can be expressed in terms of a form in a grading of an exterior algebra, the coefficients of which are products of Vandermonde determinants in integers. Hopefully an understanding of the asymptotics of these coefficients will lead to scaling limits for the pair correlation function for an infinite family of coldnesses via hyperpfaffian/Berezin integral techniques. This would partially generalize the Pfaffian point process arising in COE and CSE. There is a lot of work to do, but there is hope.
5
Martingales in the Weil height Banach space (with Nathan Hunter). Allcock and Vaaler produce a Banach space in which $\overline{\mathbb Q}^{\times}/\mathrm{Tor}$ embeds densely in a co-dimension 1 subspace, the (Banach space) norm of which extends the logarithmic Weil height. Field extensions of the maximal abelian extension of $\mathbb Q$ correspond to $\sigma$-algebras, and towers of fields to filtrations. Elements in the Banach space (including those from $\overline{\mathbb Q}^{\times}/\mathrm{Tor}$) represent random variables, and the set up is ready for someone to come along and use martingale techniques—including the optional stopping time theorem—to tell us something about algebraic numbers. Instruction
I have three current PhD students and one current departmental Honors student. I have supervised two completed PhDs and six completed honors theses. You can find a list of current and completed PhD and honors students on my CV.
My teaching load has been reduced for the last five years (or so) due to an FTE release for serving on the Executive Council of United Academics. As President of United Academics, and Immediate Past President of the University Senate I am not teaching in the 2018 academic year. In AY2019, I am scheduled to teach a two-quarter sequence on mathematical statistical physics.
I take my teaching seriously. I prepare detailed lecture notes for most courses (exceptions being introductory courses, where my notes are better characterized as well-organized outlines). When practical and appropriate I use active learning techniques, mostly through supervised group work. I am a tough, but fair grader.
Service
Service encompasses pretty much everything that an academic does outside of teaching and research. This includes advising, serving on university and departmental committees, reviewing papers, writing letters of recommendation, organizing seminars and conferences, serving on professional boards, etc. The impossibility of doing it all allows academics to decide what types of service they are going specialize based on their interests and abilities.
I have spent the last three years heavily engaged in university level service. I currently serve as the president of United Academics of the University of Oregon, and I am the immediate-past president of the University Senate. Before that I was the Vice President of the Senate and the chair of the Committee on Committees. All of these roles are difficult and require a large investment of thought and energy. The reward for this hard work is a good understanding of how the university works, who to go to when issues need resolution, and who can be safely ignored.
I know what academic initiatives are underway, being involved in several of them. I am spearheading, with the new Core Education Council, the reform of general education at UO. I am working on the New Faculty Success Program—an onboarding program for new faculty—with the Office of the Provost and United Academics. I am currently on the Faculty Salary Equity Committee and its Executive Committee. I have been a bit player in many other projects and initiatives including student evaluation reform, the re-envisioning of the undergraduate multicultural requirement, and the creation of an expedited tenure process to allow the institution alacrity when recruiting imminent scholars. This list is incomplete.
Next year, with high probability, I will be the chair of the bargaining committee for the next collective bargaining agreement between United Academics and the University of Oregon (this assumes I am elected UA president). I will also be working with the Core Ed Council to potentially redefine the BA/BS distinction, with a personal focus on ensuring quantitative/data/information literacy is distributed throughout our undergraduate curriculum. I will also be working to help pilot (and hopefully scale) the Core Ed “Runways” (themed, cohorted clusters of gen ed courses) with the aspirational goal of having 100% of traditional undergraduates in a high-support, high-engagement, uniquely-Oregon first-year experience within the next 3-5 years.
As important as the service I
am doing, is the service I am not doing. I do little to no departmental service (though part of this derives from the CAS dean’s interpretation of the CBA) and I avoid non-required departmental functions (for reasons). I do routinely serve on academic committees for graduate/honors students, etc. I decline most requests to referee papers/grants applications, and serve on no editorial boards. The national organizations for which I am an officer are not mathematical organizations, but rather organizations dedicated to shared governance. Diversity & Equity
The two principles which drive my professional work are
truth and fairness.
I remember after a particularly troubling departmental vote, a senior colleague attempted to assuage my anger at the department by explaining that “the world is not fair.” I hate this argument because it removes responsibility from those participating in such decisions, and places blame instead on a stochastic universe. And, while there
is stochasticity in the universe, we should be working toward ameliorating inequities caused by chance, and in instances where we have agency, making decisions which do not compound them.
I do not think the department does a very good job at recognizing nor ameliorating inequities. Indeed, there are individuals, policies and procedures that negatively impact diversity. See my recent post Women & Men in Mathematics for examples.
My work on diversity and equity issues has been primarily through the University Senate and United Academics. As Vice-president of the UO Senate, I sat on the committee which vetted the Diversity Action Plans of academic units. I also worked on, or presided over several motions put forth by the University Senate which address equity, diversity and inclusion. Obviously, the work of the Senate involves many people, and in many instances I played only a bit part, but nonetheless I am proud to have supported/negotiated/presided over the following motions which have addressed diversity and equity issues on campus:
Implementing A System for the Continuous Improvement and Evaluation of Teaching Proposed Changes to Multicultural Requirement Resolution denouncing White Supremacy & Hate Speech on Campus Proposed Change to Admissions Policies Requiring Disclosure of Criminal and Disciplinary Hearing A Resolution in support of LGBTQAI Student Rights Declaring UO a Sanctuary Campus Reaffirming our Shared Values of Respect for Diversity, Equity, and Inclusion Student Sexual and gender-Based Harassment and Violence Complaint and Response Policy
Besides my work with the Senate, I have also participated in diversity activities through my role(s) with United Academics of the University of Oregon. United Academics supports both a Faculty of Color and LGBTQ* Caucus which help identify barriers and propose solutions to problems affecting those communities on campus. United Academics bargained a tenure-track faculty equity study, and I am currently serving on a university committee identifying salary inequities based on protected class and proposing remedies for them.
I have attended in innumerable rallies supporting social justice, and marched in countless marches. I flew to Washington D.C. to attend the March for Science. I’ve participated in workshops and trainings on diversity provided by the American Federation of Teachers, and the American Association of University Professors.
I recognize that I am not perfect. I cannot represent all communities nor emulate the diversity of thought on campus. I have occasionally used out-moded words and am generally terrible at using preferred pronouns (though I try). I recognize my short-comings and continually work to address them.
There are different tactics for turning advocacy into action, and individuals may disagree on their appropriateness and if/when escalation is called for. My general outlook is to work within a system to address inequities until it becomes clear that change is impossible from within. In such instances, if the moral imperative for change is sufficient then I work for change from without. This is my current strategy when tackling departmental diversity issues; I work with administrative units, the Senate and the union to put forth/support policies which minimize bias, discrimination and caprice in departmental decisions. I ensure that appropriate administrators know when I feel the department has fallen down on our institutional commitment to diversity, and I report incidents of bias, discrimination and harassment to the appropriate institutional offices (subject to the policy on Student Directed Reporters).
Fairness is as important to me as truth, and I look forward to the day where I can focus more of my time uncovering the latter instead of continually battling for the former. |
berylium? really? okay then...toroidalet wrote:I Undertale hate it when people Emoji movie insert keywords so people will see their berylium page.
A forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for "non-academic" content.
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
When xq is in the middle of a different object's apgcode. "That's no ship!"
Airy Clave White It Nay
When you post something and someone else posts something unrelated and it goes to the next page.
Also when people say that things that haven't happened to them trigger them.
Also when people say that things that haven't happened to them trigger them.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
Huh. I've never seen a c/posts spaceship before.drc wrote:"The speed is actually" posts
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
It could be solved with a simple PM rather than an entire post.Gamedziner wrote:What's wrong with them?drc wrote:"The speed is actually" posts
An exception is if it's contained within a significantly large post.
I hate it when people post rule tables for non-totalistic rules. (Yes, I know some people are on mobile, but they can just generate them themselves. [citation needed])
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
OK this is a very niche one that I hadn't remembered until a few hours ago.
You know in some arcades they give you this string of cardboard tickets you can redeem for stuff, usually meant for kids. The tickets fold beautifully perfectly packed if you order them one right, one left - zigzagging. When people fold them randomly in any direction giving a clearly low density packing with loads of strain, I just think
You know in some arcades they give you this string of cardboard tickets you can redeem for stuff, usually meant for kids.
The tickets fold beautifully perfectly packed if you order them one right, one left - zigzagging.
When people fold them randomly in any direction giving a clearly low density packing with loads of strain, I just think
omg why on Earth would you do that?!Surely they'd have realised by now? It's not that crazy to realise? Surely there is a clear preference for having them well packed; nobody would prefer an unwieldy mess?!
Also when I'm typing anything and I finish writing it and it just goes to the next line or just goes to the next page. Especially when the punctuation mark at the end brings the last word down one line. This also applies to writing in a notebook: I finish writing something but the very last thing goes to a new page.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: ... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature.
ON A DIFFERENT NOTE.
When i want to rotate a hexagonal file but golly refuses because for some reason it calculates hexagonal patterns on a square grid and that really bugs me because if you want to show that something has six sides you don't show it with four and it makes more sense to have the grid be changed to hexagonal but I understand Von Neumann because no shape exists (that I know of) that has 4 corners and no edges but COME ON WHY?! WHY DO YOU REPRESENT HEXAGONS WITH SQUARES?!
In all seriousness this bothers me and must be fixed or I will SINGLEHANDEDLY eat a universe.
EDIT: possibly this one.
EDIT 2:
IT HAS BEGUN.
HAS
BEGUN.
Last edited by 83bismuth38 on September 19th, 2017, 8:25 pm, edited 1 time in total.
Actually, I don't remember who I was referencing, but I don't think it was you, and if it was, it wasn't personal.83bismuth38 wrote:... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature.
x₁=ηx
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
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
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: oh okay yeah of course sureA for awesome wrote:Actually, I don't remember who I was referencing, but I don't think it was you, and if it was, it wasn't personal.83bismuth38 wrote:... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature.
but really though, i wouldn't have cared.
When someone gives a presentation to a bunch of people and you
knowthat they're getting the facts wrong. Especially if this is during the Q&A section.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
When you watch a boring video in class but you understand it perfectly and then at the end your classmates dont get it so the teacher plays the borinh video again
Airy Clave White It Nay
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact:
when scientists decide to send a random guy into a black hole hovering directly above Earth for no reason at all.
hit; that random guy was me.
hit; that random guy was me.
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact:
When I see a "one-step" organic reaction that occurs in an exercise book for senior high school and simply takes place under "certain circumstance" like the one marked "?" here but fail to figure out how it works even if I have prepared for our provincial chemistry olympiadEDIT: In fact it's not that hard.Just do a Darzens reaction then hydrolysis and decarboxylate.
Current status: outside the continent of cellular automata. Specifically, not on the plain of life.
An awesome gun firing cool spaceships:
An awesome gun firing cool spaceships:
Code: Select all
x = 3, y = 5, rule = B2kn3-ekq4i/S23ijkqr4eikry2bo$2o$o$obo$b2o!
When there's a rule with a decently common puffer but it can't interact with itself
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
When that oscillator is just
When you're sooooooo close to a thing you consider amazing but miss... not sparky enough.
When you're sooooooo close to a thing you consider amazing but miss...
Airy Clave White It Nay
People posting tons of "new" discoveries that have been known for decades, showing that they've not observed standard netiquette by reading the forums a while before posting, nor done the most minimal research about whether things have been already known, despit repeated posts about where to find such resources (e.g. jslife, wiki, Life lexicon, etc.).
People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive"). Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades. Posts where the quoted text is substantially longer than added text. Especially "me too" posts. People whose signatures are longer than the actual text of their posts. People whose signatures include graphics or pattern files, especially ones that are just human-readable text. Improper grammar, spelling, and punctuation (although I've gotten used to that; long-term use of the internet has made me rather fluent in typo, both reading and writing). Imperfect English is not unreasonable from people for whom English is not a primary language, but from English speakers, it is a symptom of sloppiness that can also manifest in other areas.
People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive").
Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades.
Posts where the quoted text is substantially longer than added text. Especially "me too" posts.
People whose signatures are longer than the actual text of their posts.
People whose signatures include graphics or pattern files, especially ones that are just human-readable text.
Improper grammar, spelling, and punctuation (although I've gotten used to that; long-term use of the internet has made me rather fluent in typo, both reading and writing). Imperfect English is not unreasonable from people for whom English is not a primary language, but from English speakers, it is a symptom of sloppiness that can also manifest in other areas.
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X That's G U S T A V O right theremniemiec wrote:People posting tons of "new" discoveries that have been known for decades, showing that they've not observed standard netiquette by reading the forums a while before posting, nor done the most minimal research about whether things have been already known, despit repeated posts about where to find such resources (e.g. jslife, wiki, Life lexicon, etc.). People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive"). Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades.
Also, when you walk into a wall slowly and carefully but you hit your teeth on the wall and it hurts so bad.
Airy Clave White It Nay |
This overview does the following:
Outlines the theory for recommendation systems based on matrix factorization. Describes the weighted alternating least squares (WALS) algorithm used to perform the matrix factorization. Provides an overview for a set of tutorials that provide step-by-step guidance for implementing a recommendation system on GCP.
The recommendation system in the tutorial uses the weighted alternating leastsquares (WALS) algorithm. WALS is included in the
contrib.factorizationpackage of the TensorFlow code base, and is used to factorize a large matrixof user and item ratings.
Tutorials in this series
The tutorials that go with this overview include the following:
Apply to Data from Google Analytics (Part 3) shows you how to apply the recommendation system to data imported directly from Google Analytics 360 in order to perform recommendations for websites that use Analytics.
Deploy the Recommendation System (Part 4) shows you how to deploy a production system on GCP to make real-time recommendations for a website.
Background theory for recommendations
This article outlines the background theory for matrix factorization-based collaborative filtering as applied to recommendation systems. The following topics are covered in depth, with some links provided for further reading:
Collaborative filtering. What is collaborative filtering, and how can you use insights gained from it? Matrix factorization. What is a ratings matrix, what are latent factors, and how can you mathematically transform the ratings matrix to account for them? WALS method for matrix factorization. How do you perform matrix factorization using the WALS method? Collaborative filtering for recommendation systems
The collaborative filtering technique is a powerful method for generating user recommendations. Collaborative filtering relies only on observed user behavior to make recommendations—no profile data or content access is necessary.
The technique is based on the following observations:
Users who interact with items in a similar manner (for example, buying the same products or viewing the same articles) share one or more hidden preferences. Users with shared preferences are likely to respond in the same way to the same items.
Combining these basic observations allows a recommendation engine to function without needing to determine the precise nature of the shared user preferences. All that's required is that the preferences exist and are meaningful. The basic assumption is that similar user behavior reflects similar fundamental preferences, allowing a recommendation engine to make suggestions accordingly.
For example, suppose User 1 has viewed items A, B, C, D, E, and F. User 2 has viewed items A, B, D, E and F, but not C. Because both users viewed five of the same six items, it's likely that they share some basic preferences. User 1 liked item C, and it's probable that User 2 would also like item C if the user were aware of its existence. This is where the recommendation engine steps in: it informs User 2 about item C, piquing that user's interest.
Matrix factorization for collaborative filtering
The collaborative filtering problem can be solved using matrix factorization. Suppose you have a matrix consisting of user IDs and their interactions with your products. Each row corresponds to a unique user, and each column corresponds to an item. The item could be an product in a catalog, an article, or a video. Each entry in the matrix captures a user's rating or preference for a single item. The rating could be explicit, directly generated by user feedback, or it could be implicit, based on user purchases or time spent interacting with an article or video.
If a user has never rated an item or shown any implied interest in it, thematrix entry is zero. Figure 1 shows a representation of a
MovieLens ratingmatrix.
Ratings in the
MovieLens dataset range from 1 to 5. Empty rating entries havevalue 0, meaning that a given user hasn't rated the item.
Defining the matrix factorization method
A ratings matrix consists of a matrix
, where entries\(r_{ij}\) are ratings of user \(i\) for item \(j\). For manyinternet applications, these matrices are large, with millions of users andmillions of different items. They are alsosparse,meaning that each user has typically rated, viewed, or purchased only a smallnumber of items relative to the entire set. The vast majority of matrix entries,often greater than 99%, are zero. R
The matrix factorization method assumes that there is a set of attributes commonto all items, with items differing in the degree to which they express theseattributes. Furthermore, the matrix factorization method assumes that each userhas their own expression for each of these attributes, independent of the items.In this way, a user's item rating can be approximated by summing the user'sstrength for each attribute weighted by the degree to which the item expressesthis attribute. These attributes are sometimes called hidden or
latentfactors.
Intuitively, it's easy to see that these hypothetical latent factors actually exist. In the case of movies, it's clear that many users prefer certain genres, actors, or directors. These categories represent latent factors that, while obvious, are still quite useful. For example, genre preferences manifest themselves in the movies that users tend to like, and people with similar genre preferences presumably like similar movies. Much of the power of the matrix factorization approach to collaborative filtering derives from the fact that it's not necessary to know the number of genres or actors or other categories that might comprise the entirety of a given user's preferences. It's simply assumed there are an arbitrary number of them.
Transforming the matrix to represent latent factors
To translate the existence of latent factors into the matrix of ratings, youdo this: for a set of users
of size U uand items of size I i, you pick an arbitrary number kof latent factors and factorize the large matrix into two much smaller matrices R (the "row factor") and X (the "column factor"). Matrix Y has dimension X u× k, and has dimension Y k× i. This is shown in figure 2.
In linear algebra this is called alow-rank approximation.You can view this process as compressing the sparse information in
intothe much lower dimensional spaces R u× kand k× i. The matrix , obtained when the low-rank matrices R' and X are multiplied, represents an approximation of Y . R
Every user is represented by a vector in this
k-dimensional space, andeach row \( \mathbf{x}_{u}\) in corresponds to the strength ofthe user's preferences for these X kfactors. Similarly, every item represented by a vector in this k-dimensional space has a column \(\mathbf{y}_{i}\) in corresponding to the item's expression of the same Y kfactors.
To calculate the rating of user
u for item i, you take the dot productof the two vectors:
This product is a real number \(r_{ui}\) and represents an approximation,or in ML terms a
prediction, of user u's rating for item i.This is illustrated in figure 3.
You can define a loss function measuring the accuracy of the approximation in the following way:
This equation does the following: over all users and items, sum the squared difference between the approximate rating \(\mathbf{x}^T_{u} \cdot \mathbf{y}_{i}\) and the actual rating from the training set \(r_{ui}\).
Here, \(\lambda \) is a regularization constant, one of the model’s hyperparameters, which we discuss in Part 2 of the tutorial series.
The WALS method of matrix factorization
This section discusses two methods of matrix factorization.
Alternating least squares method
The alternating least squares method of matrix factorization is an iterativemethod for determining the optimal factors
and X that best approximate Y . In each iteration, one of the row or column factors is held fixed and the other is computed by minimizing the loss function with respect to the other factor. R
First, you begin with the row factors:
Set the column factors to constant values. Take the derivative of the loss function with respect to the row factors. Set the equation equal to zero.
The iteration proceeds by holding the solved-for row factors fixed and solving the analogous equation for the column factors. Alternating row and column factors, the iterative process is repeated until convergence, which typically occurs within a small (< 20) number of iterations even for very large matrices consisting of tens of millions of rows or columns.
Weighted alternating least squares (WALS) method
The weighted version of the algorithm introduces different weights for the zero, or unobserved, entries and the non-zero entries in the matrix:
In this equation:
\(w_{ui} = \omega_{0}\) for zero (unobserved) entries in the ratings matrix, and \(w_{ui} = \omega_{0} + f(c_{i})\) for observed entries, where \(c_{i} = \sum_{u,i}1 \text{ if } r_{ui} > 0\) the sum of the number of nonzero entries for column i
The weight is scaled by the sum of the nonzero entries in a row to normalize the weight for users who have rated a different number of items.
This type of weighting allows for a more flexible model of the user's preferences and produces better empirical results than the unweighted version. (For more details, see the paper Collaborative Filtering for Implicit Feedback Datasets.) The function \(f\) varies according to the dataset and whether ratings are explicit or implicit.
The
MovieLens dataset contains explicit ratings. In this case, abetter choice for \(f\) is one that weighs the observed entries linearly:
\(f = \omega_{k}c_{i}\) where \(\omega_{k}\) is the observed weight.
For implicit ratings related to dynamic events, where each rating corresponds to the number of times a video has been watched, an article read, or a web page viewed, the rating itself may have an exponential distribution due to user behavior. There will be many low-value ratings as users click on a page or video and navigate away quickly. There will be fewer large implicit ratings as users read an entire article, watch an entire video, or watch a given scheduled show multiple times. In this case, an appropriate \(f\) is one that weights the observed ratings to account for this distribution:
\(f = \left(\frac{1}{c_{i}}\right)^{e}\) where \(e\) is the feature weight exponent. WALS compared to other techniques
Many matrix factorization techniques are used for collaborative filtering, including SVD and Stochastic Gradient Descent. In some cases these techniques give better reduced-rank approximations than WALS. It's beyond the scope of this article to compare the strengths and weaknesses of different collaborative filtering methods, but it's worth noting the following advantages of WALS:
The weights used in WALS make it suitable for implicit ratings, butWALS can also be applied to explicit rating datasets such as
MovieLens.
WALS includes algorithmic optimizations that make it easy to incorporate weights and efficiently calculate row and column factor updates. For details, see the paper Collaborative Filtering for Implicit Feedback Datasets. These optimizations are built into the WALS TensorFlow code base. It's relatively straightforward to create a distributed version of the algorithm. Distributed versions can process very large matrices with potentially hundreds of millions of rows. Discussion of the distributed version of WALS is beyond the scope of this article. What's next
Start working through the tutorials in this series: |
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